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Current Electricity

Questions
747
MHT CET
177
JEE Main
401
NEET
169
Years
1988–2026
0 of 747 attemptedPick an answer to see the solution

Q1MHT CET 2026 · 20 APR · SHIFT 1Medium

Electric Current, Drift Velocity & Current Density · Current & charge flow

In the circuit shown charge q varies with time t as q=t2−5q = t^{2} - 5, where q is in coulomb and t is in second. At time t=3t = 3 second, voltage VABV_{AB} in volt will be

Figure

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Answer: (c)

ABI = 6 Aq/C = 1 VL dI/dt = 1 VIR = 12 V
  1. At t=3 st = 3\text{ s}: q=9−5=4 Cq = 9 - 5 = 4\text{ C}, I=dqdt=2t=6 AI = \dfrac{dq}{dt} = 2t = 6\text{ A}, dIdt=2 A/s\dfrac{dI}{dt} = 2\text{ A/s}
  2. Current flows from A to B (it charges the + plate). Drops along A to B:
  3. Capacitor qC=44=1 V\dfrac{q}{C} = \dfrac{4}{4} = 1\text{ V}, inductor LdIdt=0.5×2=1 VL\dfrac{dI}{dt} = 0.5 \times 2 = 1\text{ V}, resistor IR=6×2=12 VIR = 6 \times 2 = 12\text{ V}
  4. VAB=1+1+12=14 VV_{AB} = 1 + 1 + 12 = 14\text{ V}
  5. So the answer is (c) 14.
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Q2JEE Main 2026 · 22 JAN · SHIFT 2Medium

Electric Current, Drift Velocity & Current Density · Mobility & current density

A cylindrical conductor of length 2 m and area of cross-section 0.2 mm20.2\text{ mm}^{2} carries an electric current of 1.6 A when its ends are connected to a 2 V battery. Mobility of electrons in the conductor is α×10−3 m2/V s\alpha \times 10^{-3}\text{ m}^{2}\text{/V s}. The value of α\alpha is:

(electron concentration =5×1028 /m3= 5 \times 10^{28}\text{ /m}^{3} and electron charge =1.6×10−19 C= 1.6 \times 10^{-19}\text{ C})

Numerical answer type. Work it out, then open the answer.

Show answer and solution

Answer: 1

  1. E=Vl=22=1 V/mE = \dfrac{V}{l} = \dfrac{2}{2} = 1\text{ V/m}
  2. vd=IneA=1.65×1028×1.6×10−19×0.2×10−6=10−3 m/sv_d = \dfrac{I}{neA} = \dfrac{1.6}{5 \times 10^{28} \times 1.6 \times 10^{-19} \times 0.2 \times 10^{-6}} = 10^{-3}\text{ m/s}
  3. μ=vdE=1×10−3 m2/V s\mu = \dfrac{v_d}{E} = 1 \times 10^{-3}\text{ m}^{2}\text{/V s}
  4. So α=1\alpha = 1
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Q3JEE Main 2025 · 4 APR · SHIFT 1Easy

Electric Current, Drift Velocity & Current Density · Current & charge flow

Current passing through a wire as function of time is given as I(t)=0.02t+0.01 AI(t) = 0.02t + 0.01\text{ A}. The charge that will flow through the wire from t=1 st = 1\text{ s} to t=2 st = 2\text{ s} is

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Answer: (d)

  1. Charge q=∫I dt=∫12(0.02t+0.01) dtq = \int I\,dt = \int_1^2 (0.02t + 0.01)\,dt
  2. q=[0.01t2+0.01t]12=(0.04+0.02)−(0.01+0.01)q = \left[0.01t^{2} + 0.01t\right]_1^2 = (0.04 + 0.02) - (0.01 + 0.01)
  3. q=0.04 Cq = 0.04\text{ C}
  4. So the answer is (d) 0.04 C.
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Q4JEE Main 2024 · 6 APR · SHIFT 2Easy

Electric Current, Drift Velocity & Current Density · Current & charge flow

The number of electrons flowing per second in the filament of a 110 W110\text{ W} bulb operating at 220 V220\text{ V} is: (Given e=1.6×10−19 C\text{e} = 1.6 \times 10^{-19}\text{ C})

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Answer: (b)

  1. Current in the bulb I=PV=110220=0.5 AI = \dfrac{P}{V} = \dfrac{110}{220} = 0.5\text{ A}
  2. Electrons per second n=Ie=0.51.6×10−19=3.125×1018n = \dfrac{I}{e} = \dfrac{0.5}{1.6 \times 10^{-19}} = 3.125 \times 10^{18}
  3. =31.25×1017= 31.25 \times 10^{17}
  4. So the answer is (b) 31.25×101731.25 \times 10^{17}.
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Q5JEE Main 2024 · 29 JAN · SHIFT 1Easy

Electric Current, Drift Velocity & Current Density · Current & charge flow

The electric current through a wire varies with time as I=I0+βtI = I_0 + \beta t, where I0=20 AI_0 = 20\text{ A} and β=3 A/s\beta = 3\text{ A/s}. The amount of electric charge crossed through a section of the wire in 20 s20\text{ s} is:

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Answer: (c)

  1. q=∫020(I0+βt) dt=I0t+βt22q = \int_0^{20} (I_0 + \beta t)\,dt = I_0 t + \dfrac{\beta t^{2}}{2}
  2. q=20×20+3×4002=400+600q = 20 \times 20 + \dfrac{3 \times 400}{2} = 400 + 600
  3. q=1000 Cq = 1000\text{ C}
  4. So the answer is (c) 1000 C.
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Q6JEE Main 2024 · 1 FEB · SHIFT 1Easy

Electric Current, Drift Velocity & Current Density · Current & charge flow

The current in a conductor is expressed as I=3t2+4t3I = 3t^{2} + 4t^{3}, where II is in Ampere and tt is in second. The amount of electric charge that flows through a section of the conductor during t=1 st = 1\text{ s} to t=2 st = 2\text{ s} is __________ C.

Numerical answer type. Work it out, then open the answer.

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Answer: 22

  1. q=∫12(3t2+4t3) dt=[t3+t4]12q = \int_1^2 (3t^{2} + 4t^{3})\,dt = \left[t^{3} + t^{4}\right]_1^2
  2. q=(8+16)−(1+1)q = (8 + 16) - (1 + 1)
  3. q=22 Cq = 22\text{ C}
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Q7NEET 2024Medium

Electric Current, Drift Velocity & Current Density · Drift velocity & relaxation time

A uniform wire of diameter dd carries a current of 100 mA100\text{ mA} when the mean drift velocity of electrons in the wire is vv. For a wire of diameter d2\frac{d}{2} of the same material to carry a current of 200 mA200\text{ mA}, the mean drift velocity of electrons in the wire is

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Answer: (b)

  1. vd=IneA∝Id2v_d = \dfrac{I}{neA} \propto \dfrac{I}{d^{2}} for the same material
  2. v′v=200100×(dd/2)2=2×4=8\dfrac{v'}{v} = \dfrac{200}{100} \times \left(\dfrac{d}{d/2}\right)^{2} = 2 \times 4 = 8
  3. So the answer is (b) 8v8v.
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Q8NEET 2023Easy

Electric Current, Drift Velocity & Current Density · Drift velocity & relaxation time

A copper wire of radius 1 mm1\text{ mm} contains 102210^{22} free electrons per cubic metre. The drift velocity for free electrons when 10 A10\text{ A} current flows through the wire will be (Given, charge on electron =1.6×10−19 C= 1.6 \times 10^{-19}\text{ C}):

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Answer: (b)

  1. I=neAvd⇒vd=IneAI = neAv_d \Rightarrow v_d = \dfrac{I}{neA}, with A=πr2=π×10−6 m2A = \pi r^{2} = \pi \times 10^{-6}\text{ m}^{2}
  2. vd=101022×1.6×10−19×π×10−6=101.6×10−3 πv_d = \dfrac{10}{10^{22} \times 1.6 \times 10^{-19} \times \pi \times 10^{-6}} = \dfrac{10}{1.6 \times 10^{-3}\,\pi}
  3. vd=6.25×103π m s−1v_d = \dfrac{6.25 \times 10^{3}}{\pi}\text{ m s}^{-1}
  4. So the answer is (b).
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Q9JEE Main 2023 · 10 APR · SHIFT 2Easy

Electric Current, Drift Velocity & Current Density · Drift velocity & relaxation time

In a metallic conductor, under the effect of applied electric field, the free electrons of the conductor

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Answer: (d)

  1. Electrons carry negative charge, so the force −eE⃗-e\vec{E} pushes them opposite to E⃗\vec{E}, i.e. from lower to higher potential.
  2. Between collisions they accelerate under this force, and collisions keep changing their direction, so the paths are curved, not straight lines at uniform velocity.
  3. So the answer is (d).
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Q10JEE Main 2023 · 8 APR · SHIFT 2Easy

Electric Current, Drift Velocity & Current Density · Drift velocity & relaxation time

The number density of free electrons in copper is nearly 8×1028 m−38 \times 10^{28}\text{ m}^{-3}. A copper wire has its area of cross section =2×10−6 m2= 2 \times 10^{-6}\text{ m}^{2} and is carrying a current of 3.2 A3.2\text{ A}. The drift speed of the electrons is ___________ ×10−6 m s−1\times 10^{-6}\text{ m s}^{-1}

Numerical answer type. Work it out, then open the answer.

Show answer and solution

Answer: 125

  1. vd=IneAv_d = \dfrac{I}{neA}
  2. vd=3.28×1028×1.6×10−19×2×10−6=3.22.56×104v_d = \dfrac{3.2}{8 \times 10^{28} \times 1.6 \times 10^{-19} \times 2 \times 10^{-6}} = \dfrac{3.2}{2.56 \times 10^{4}}
  3. vd=1.25×10−4=125×10−6 m s−1v_d = 1.25 \times 10^{-4} = 125 \times 10^{-6}\text{ m s}^{-1}
  4. Answer: 125
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Q11JEE Main 2023 · 8 APR · SHIFT 1Easy

Electric Current, Drift Velocity & Current Density · Drift velocity & relaxation time

A current of 2 A2\text{ A} flows through a wire of cross-sectional area 25.0 mm225.0\text{ mm}^{2}. The number of free electrons in a cubic meter are 2.0×10282.0 \times 10^{28}. The drift velocity of the electrons is __________ ×10−6 m s−1\times 10^{-6}\text{ m s}^{-1} (given, charge on electron =1.6×10−19 C= 1.6 \times 10^{-19}\text{ C} ).

Numerical answer type. Work it out, then open the answer.

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Answer: 25

  1. vd=IneAv_d = \dfrac{I}{neA}, with A=25×10−6 m2A = 25 \times 10^{-6}\text{ m}^{2}
  2. vd=22×1028×1.6×10−19×25×10−6=28×104v_d = \dfrac{2}{2 \times 10^{28} \times 1.6 \times 10^{-19} \times 25 \times 10^{-6}} = \dfrac{2}{8 \times 10^{4}}
  3. vd=2.5×10−5=25×10−6 m s−1v_d = 2.5 \times 10^{-5} = 25 \times 10^{-6}\text{ m s}^{-1}
  4. Answer: 25
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Q12JEE Main 2023 · 30 JAN · SHIFT 1Medium

Electric Current, Drift Velocity & Current Density · Current & charge flow

The charge flowing in a conductor changes with time as Q(t)=αt−βt2+γt3Q(t) = \alpha t - \beta t^{2} + \gamma t^{3}. Where α,β\alpha, \beta and γ\gamma are constants. Minimum value of current is:

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Answer: (c)

  1. I=dQdt=α−2βt+3γt2I = \dfrac{dQ}{dt} = \alpha - 2\beta t + 3\gamma t^{2}
  2. For minimum, dIdt=−2β+6γt=0⇒t=β3γ\dfrac{dI}{dt} = -2\beta + 6\gamma t = 0 \Rightarrow t = \dfrac{\beta}{3\gamma}
  3. Imin=α−2β23γ+β23γ=α−β23γI_{min} = \alpha - \dfrac{2\beta^{2}}{3\gamma} + \dfrac{\beta^{2}}{3\gamma} = \alpha - \dfrac{\beta^{2}}{3\gamma}
  4. So the answer is (c).
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Q13JEE Main 2023 · 31 JAN · SHIFT 1Medium

Electric Current, Drift Velocity & Current Density · Drift velocity & relaxation time

The drift velocity of electrons for a conductor connected in an electrical circuit is VdV_{d}. The conductor is now replaced by another conductor with same material and same length but double the area of cross section. The applied voltage remains same. The new drift velocity of electrons will be

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Answer: (b)

  1. vd=eEτm=eVτmlv_d = \dfrac{eE\tau}{m} = \dfrac{eV\tau}{ml}
  2. Same material (τ\tau), same VV and same ll, so vdv_d does not depend on the area.
  3. So the answer is (b) VdV_d.
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Q14JEE Main 2022 · 27 JUL · SHIFT 2Easy

Electric Current, Drift Velocity & Current Density · Drift velocity & relaxation time

(A) The drift velocity of electrons decreases with the increase in the temperature of conductor. (B) The drift velocity is inversely proportional to the area of cross-section of given conductor. (C) The drift velocity does not depend on the applied potential difference to the conductor. (D) The drift velocity of electron is inversely proportional to the length of the conductor. (E) The drift velocity increases with the increase in the temperature of conductor. Choose the correct answer from the options given below:

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Answer: (b)

  1. vd=eEτm=eVτmlv_d = \dfrac{eE\tau}{m} = \dfrac{eV\tau}{ml}
  2. (A) True: τ\tau falls as temperature rises, so vdv_d decreases. (E) is therefore false.
  3. (B) False: for a given VV, vdv_d does not depend on area. (C) False: vd∝Vv_d \propto V.
  4. (D) True: vd∝1lv_d \propto \dfrac{1}{l}.
  5. So the answer is (b) (A) and (D) only.
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Q15NEET 2022 · PHASE 1Easy

Electric Current, Drift Velocity & Current Density · Mobility & current density

A copper wire of length 10 m and radius (10−2π)\left(\frac{10^{-2}}{\sqrt{\pi}}\right) m has electrical resistance of 10 Ω\Omega. The current density in the wire for an electric field strength of 10 (V/m) is

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Answer: (d)

  1. Area A=πr2=π×10−4π=10−4 m2A = \pi r^{2} = \pi \times \dfrac{10^{-4}}{\pi} = 10^{-4}\text{ m}^{2}
  2. Resistivity ρ=RAl=10×10−410=10−4 Ω m\rho = \dfrac{RA}{l} = \dfrac{10 \times 10^{-4}}{10} = 10^{-4}\ \Omega\text{ m}
  3. J=Eρ=1010−4=105 A/m2J = \dfrac{E}{\rho} = \dfrac{10}{10^{-4}} = 10^{5}\text{ A/m}^{2}
  4. So the answer is (d).
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Q16JEE Main 2022 · 27 JUL · SHIFT 1Medium

Electric Current, Drift Velocity & Current Density · Drift velocity & relaxation time

A 1 m1\text{ m} long copper wire carries a current of 1 A1\text{ A}. If the cross section of the wire is 2.0 mm22.0\text{ mm}^{2} and the resistivity of copper is 1.7×10−8 Ω m1.7 \times 10^{-8}\ \Omega\text{ m}, the force experienced by moving electron in the wire is ____________ ×10−23 N\times 10^{-23}\text{ N}.

(charge on electron =1.6×10−19 C= 1.6 \times 10^{-19}\text{ C})

Numerical answer type. Work it out, then open the answer.

Show answer and solution

Answer: 136

  1. J=IA=12×10−6=5×105 A/m2J = \dfrac{I}{A} = \dfrac{1}{2 \times 10^{-6}} = 5 \times 10^{5}\text{ A/m}^{2}
  2. E=ρJ=1.7×10−8×5×105=8.5×10−3 V/mE = \rho J = 1.7 \times 10^{-8} \times 5 \times 10^{5} = 8.5 \times 10^{-3}\text{ V/m}
  3. F=eE=1.6×10−19×8.5×10−3=1.36×10−21 NF = eE = 1.6 \times 10^{-19} \times 8.5 \times 10^{-3} = 1.36 \times 10^{-21}\text{ N}
  4. F=136×10−23 NF = 136 \times 10^{-23}\text{ N}, answer: 136
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Q17JEE Main 2022 · 27 JUN · SHIFT 2Medium

Electric Current, Drift Velocity & Current Density · Mobility & current density

The current density in a cylindrical wire of radius r = 4.0 mm is 1.0 ×\times 106 A/m2. The current through the outer portion of the wire between radial distances r2\frac{r}{2} and r is xπ\pi A; where x is __________.

Numerical answer type. Work it out, then open the answer.

Show answer and solution

Answer: 12

  1. Uniform JJ: I=J×π(r2−r24)=J⋅3πr24I = J \times \pi\left(r^{2} - \dfrac{r^{2}}{4}\right) = J \cdot \dfrac{3\pi r^{2}}{4}
  2. I=106×3π×16×10−64=12π AI = 10^{6} \times \dfrac{3\pi \times 16 \times 10^{-6}}{4} = 12\pi\text{ A}
  3. So x=12x = 12
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Q18JEE Main 2022 · 27 JUN · SHIFT 1Medium

Electric Current, Drift Velocity & Current Density · Mobility & current density

The current density in a cylindrical wire of radius 4 mm is 4 ×\times 106 Am−2{}^{-2}. The current through the outer portion of the wire between radial distances R2\frac{R}{2} and R is ____________ π\pi A.

Numerical answer type. Work it out, then open the answer.

Show answer and solution

Answer: 48

  1. Uniform JJ: I=J⋅π(R2−R24)=3πR2J4I = J \cdot \pi\left(R^{2} - \dfrac{R^{2}}{4}\right) = \dfrac{3\pi R^{2} J}{4}
  2. I=3π×16×10−6×4×1064=48π AI = \dfrac{3\pi \times 16 \times 10^{-6} \times 4 \times 10^{6}}{4} = 48\pi\text{ A}
  3. Answer: 48
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Q19JEE Main 2021 · 24 FEB · SHIFT 1Easy

Electric Current, Drift Velocity & Current Density · Current & charge flow

A current through a wire depends on time as

i = α\alpha0t + β\betat2

where α\alpha0 = 20 A/s and β\beta = 8 As−2{}^{-2}. Find the charge crossed through a section of the wire in 15 s.

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Answer: (d)

  1. Current i=α0t+βt2i = \alpha_0 t + \beta t^{2}
  2. q=∫015i dt=α0t22+βt33q = \int_0^{15} i\,dt = \dfrac{\alpha_0 t^{2}}{2} + \dfrac{\beta t^{3}}{3}
  3. q=20×2252+8×33753=2250+9000q = \dfrac{20 \times 225}{2} + \dfrac{8 \times 3375}{3} = 2250 + 9000
  4. q=11250 Cq = 11250\text{ C}, so the answer is (d).
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Q20JEE Main 2021 · 17 MAR · SHIFT 1Easy

Electric Current, Drift Velocity & Current Density · Drift velocity & relaxation time

A current of 10A exists in a wire of cross-sectional area of 5 mm2 with a drift velocity of 2 ×\times 10−-3 ms−1{}^{-1}. The number of free electrons in each cubic meter of the wire is ___________.

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Answer: (a)

  1. I=neAvd⇒n=IeAvdI = neAv_d \Rightarrow n = \dfrac{I}{eAv_d}
  2. n=101.6×10−19×5×10−6×2×10−3=101.6×10−27n = \dfrac{10}{1.6 \times 10^{-19} \times 5 \times 10^{-6} \times 2 \times 10^{-3}} = \dfrac{10}{1.6 \times 10^{-27}}
  3. n=6.25×1027=625×1025 m−3n = 6.25 \times 10^{27} = 625 \times 10^{25}\text{ m}^{-3}
  4. So the answer is (a).
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Q21NEET 2021Medium

Electric Current, Drift Velocity & Current Density · Current & charge flow

Column - I gives certain physical terms associated with flow of current through a metallic

conductor. Column-II gives some mathematical relations involving electrical quantities. Match

column-I and column-II with appropriate relations.

Figure

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Answer: (b)

  1. (A) Drift velocity vd=eEmτv_d = \dfrac{eE}{m}\tau → (R)
  2. (B) Resistivity ρ=EJ\rho = \dfrac{E}{J} → (S)
  3. (C) Relaxation time from ρ=mne2τ\rho = \dfrac{m}{ne^{2}\tau}: τ=mne2ρ\tau = \dfrac{m}{ne^{2}\rho} → (P)
  4. (D) Current density J=nevdJ = nev_d → (Q)
  5. So the answer is (b).
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Q22JEE Main 2021 · 27 JUL · SHIFT 1Medium

Electric Current, Drift Velocity & Current Density · Drift velocity & relaxation time

In the given figure, a battery of emf E is connected across a conductor PQ of length 'l' and different area of cross-sections having radii r1 and r2 (r2 < r1).

Figure

Choose the correct option as one moves from P to Q:

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Answer: (a)

  1. The current II is the same at every section of PQ (series path), so (c) is wrong.
  2. From P to Q the area falls (r2<r1r_2 < r_1), so J=IAJ = \dfrac{I}{A} increases.
  3. vd=Jnev_d = \dfrac{J}{ne} increases and E=ρJE = \rho J also increases, so (b) is wrong.
  4. So the answer is (a).
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Q23JEE Main 2021 · 16 MAR · SHIFT 1Medium

Electric Current, Drift Velocity & Current Density · Drift velocity & relaxation time

A conducting wire of length 'l', area of cross-section A and electric resistivity ρ\rho is connected between the terminals of a battery. A potential difference V is developed between its ends, causing an electric current.

If the length of the wire of the same material is doubled and the area of cross-section is halved, the resultant current would be:

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Answer: (c)

  1. Initially R=ρlAR = \dfrac{\rho l}{A}
  2. New R′=ρ(2l)A/2=4ρlAR' = \dfrac{\rho (2l)}{A/2} = \dfrac{4\rho l}{A}
  3. I′=VR′=14VAρlI' = \dfrac{V}{R'} = \dfrac{1}{4}\dfrac{VA}{\rho l}
  4. So the answer is (c).
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Q24JEE Main 2021 · 20 JUL · SHIFT 1Medium

Electric Current, Drift Velocity & Current Density · Mobility & current density

A current of 5 A is passing through a non-linear magnesium wire of cross-section 0.04 m2. At every point the direction of current density is at an angle of 60∘^\circ with the unit vector of area of cross- section. The magnitude of electric field at every point of the conductor is:

(Resistivity of magnesium ρ\rho = 44 ×\times 10−-8 Ω\Omegam)

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Answer: (a)

  1. I=J⃗⋅A⃗=JAcos⁡60∘I = \vec{J} \cdot \vec{A} = JA\cos 60^\circ
  2. J=50.04×0.5=250 A/m2J = \dfrac{5}{0.04 \times 0.5} = 250\text{ A/m}^{2}
  3. E=ρJ=44×10−8×250=1.1×10−4=11×10−5 V/mE = \rho J = 44 \times 10^{-8} \times 250 = 1.1 \times 10^{-4} = 11 \times 10^{-5}\text{ V/m}
  4. So the answer is (a).
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Q25JEE Main 2021 · 24 FEB · SHIFT 2Medium

Electric Current, Drift Velocity & Current Density · Mobility & current density

A cylindrical wire of radius 0.5 mm and conductivity 5 ×\times 107 S/m is subjected to an electric field of 10 mV/m. The expected value of current in the wire will be x3π\pi mA. The value of x is _________.

Numerical answer type. Work it out, then open the answer.

Show answer and solution

Answer: 5

  1. J=σE=5×107×10×10−3=5×105 A/m2J = \sigma E = 5 \times 10^{7} \times 10 \times 10^{-3} = 5 \times 10^{5}\text{ A/m}^{2}
  2. A=πr2=π×0.25×10−6 m2A = \pi r^{2} = \pi \times 0.25 \times 10^{-6}\text{ m}^{2}
  3. I=JA=0.125π A=125π mA=53π mAI = JA = 0.125\pi\text{ A} = 125\pi\text{ mA} = 5^{3}\pi\text{ mA}
  4. So x=5x = 5
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Q26NEET 2020 · PHASE 1Easy

Electric Current, Drift Velocity & Current Density · Mobility & current density

A charged particle having drift velocity 7.5 ×\times 10-4 ms-1 in an electric field of 3 ×\times 10-10 Vm-1 has a mobility in m2 V-1s-1 of:

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Answer: (a)

  1. Mobility μ=vdE\mu = \dfrac{v_d}{E}
  2. μ=7.5×10−43×10−10=2.5×106 m2V−1s−1\mu = \dfrac{7.5 \times 10^{-4}}{3 \times 10^{-10}} = 2.5 \times 10^{6}\text{ m}^{2}\text{V}^{-1}\text{s}^{-1}
  3. So the answer is (a).
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Q27JEE Main 2019 · 9 JAN · SHIFT 1Easy

Electric Current, Drift Velocity & Current Density · Drift velocity & relaxation time

Drift speed of electrons, when 1.5 A of current flows in a copper wire of cross section 5 mm2, is vv. If the electron density in copper is 9 ×\times 1028/m3 the value of vv. in mm/s is close to (Take charge of electron to be = 1.6 ×\times 10−-19C)

Show answer and solution

Answer: (a)

  1. v=IneAv = \dfrac{I}{neA}
  2. v=1.59×1028×1.6×10−19×5×10−6=1.57.2×104v = \dfrac{1.5}{9 \times 10^{28} \times 1.6 \times 10^{-19} \times 5 \times 10^{-6}} = \dfrac{1.5}{7.2 \times 10^{4}}
  3. v≈2.1×10−5 m/s=0.02 mm/sv \approx 2.1 \times 10^{-5}\text{ m/s} = 0.02\text{ mm/s}
  4. So the answer is (a) 0.02.
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Q28JEE Main 2019 · 9 APR · SHIFT 2Medium

Electric Current, Drift Velocity & Current Density · Drift velocity & relaxation time

In a conductor, if the number of conduction electrons per unit volume is 8.5 × 1028 m–3 and mean free time is 25ƒs (femto second), it's approximate resistivity is :-

(me = 9.1 × 10–31 kg)

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Answer: (a)

  1. ρ=mne2τ\rho = \dfrac{m}{ne^{2}\tau}
  2. ρ=9.1×10−318.5×1028×(1.6×10−19)2×25×10−15=9.1×10−315.44×10−23\rho = \dfrac{9.1 \times 10^{-31}}{8.5 \times 10^{28} \times (1.6 \times 10^{-19})^{2} \times 25 \times 10^{-15}} = \dfrac{9.1 \times 10^{-31}}{5.44 \times 10^{-23}}
  3. ρ≈1.7×10−8 Ωm∼10−8 Ωm\rho \approx 1.7 \times 10^{-8}\ \Omega\text{m} \sim 10^{-8}\ \Omega\text{m}
  4. So the answer is (a).
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Q29JEE Main 2019 · 10 APR · SHIFT 1Medium

Electric Current, Drift Velocity & Current Density · Mobility & current density

A current of 5 A passes through a copper conductor (resistivity = 1.7 × 10–8 Ω\Omegam) of radius of cross- section 5 mm. Find the mobility of the charges if their drift velocity is 1.1 × 10–3 m/s.

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Answer: (b)

  1. J=Iπr2=5π×25×10−6≈6.37×104 A/m2J = \dfrac{I}{\pi r^{2}} = \dfrac{5}{\pi \times 25 \times 10^{-6}} \approx 6.37 \times 10^{4}\text{ A/m}^{2}
  2. E=ρJ=1.7×10−8×6.37×104≈1.08×10−3 V/mE = \rho J = 1.7 \times 10^{-8} \times 6.37 \times 10^{4} \approx 1.08 \times 10^{-3}\text{ V/m}
  3. μ=vdE=1.1×10−31.08×10−3≈1.0 m2/Vs\mu = \dfrac{v_d}{E} = \dfrac{1.1 \times 10^{-3}}{1.08 \times 10^{-3}} \approx 1.0\text{ m}^{2}\text{/Vs}
  4. So the answer is (b).
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Q30JEE Main 2018 · 15 APR · SHIFT 2Medium

Electric Current, Drift Velocity & Current Density · Drift velocity & relaxation time

A copper rod of cross-sectional area A carries a uniform current I through it. At temperature T, if the volume charge density of the rod is ρ\rho, how long will the charges take to travel a distance d?

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Answer: (c)

  1. Here ρ\rho is the charge per unit volume, so I=ρAvI = \rho A v and v=IρAv = \dfrac{I}{\rho A}
  2. Time t=dv=ρdAIt = \dfrac{d}{v} = \dfrac{\rho d A}{I} (temperature does not enter)
  3. So the answer is (c).
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Q31NEET 2015 · AIPMT · CANCELLED PAPEREasy

Electric Current, Drift Velocity & Current Density · Drift velocity & relaxation time

Across a metallic conductor of non-uniform cross section a constant potential difference is applied. The quantity which remains constant along the conductor is

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Answer: (d)

  1. The conductor is a single series path, so the same charge passes every section per second: current II is constant.
  2. J=I/AJ = I/A, vd=I/(neA)v_d = I/(neA) and E=ρJE = \rho J all change with area AA.
  3. So the answer is (d) current.
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Q32JEE Main 2015Medium

Electric Current, Drift Velocity & Current Density · Drift velocity & relaxation time

When 5 V5\text{ V} potential difference is applied across a wire of length 0.1 m,0.1\text{ m}, the drift speed of electrons is 2.5×10−4 m s−1.2.5 \times 10^{-4}\text{ m s}^{-1}. If the electron density in the wire is 8×1028 m−3,8 \times 10^{28}\text{ m}^{-3}, the resistivity of the material is close to:

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Answer: (b)

  1. E=Vl=50.1=50 V/mE = \dfrac{V}{l} = \dfrac{5}{0.1} = 50\text{ V/m} and J=nevdJ = nev_d
  2. ρ=EJ=Enevd=508×1028×1.6×10−19×2.5×10−4\rho = \dfrac{E}{J} = \dfrac{E}{nev_d} = \dfrac{50}{8 \times 10^{28} \times 1.6 \times 10^{-19} \times 2.5 \times 10^{-4}}
  3. ρ=503.2×106≈1.6×10−5 Ω m\rho = \dfrac{50}{3.2 \times 10^{6}} \approx 1.6 \times 10^{-5}\ \Omega\text{ m}
  4. So the answer is (b).
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Q33NEET 2009 · AIPMTMedium

Electric Current, Drift Velocity & Current Density · Drift velocity & relaxation time

−-8

The mean free path of electrons in a metal is 4 ×\times 10 m. The electric field which can give on an average 2 eV energy to an electron in the metal will be in units V/m

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Answer: (c)

  1. Energy gained over one mean free path: eEλ=2 eVeE\lambda = 2\text{ eV}
  2. E=2 Vλ=24×10−8E = \dfrac{2\text{ V}}{\lambda} = \dfrac{2}{4 \times 10^{-8}}
  3. E=5×107 V/mE = 5 \times 10^{7}\text{ V/m}
  4. So the answer is (c).
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Q34NEET 1994 · AIPMTEasy

Electric Current, Drift Velocity & Current Density · Current & charge flow

A flow of 10710^{7} electrons per second in a conducting wire constitutes a current of

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Answer: (a)

  1. Current I=qt=netI = \dfrac{q}{t} = \dfrac{ne}{t}
  2. I=107×1.6×10−19=1.6×10−12 AI = 10^{7} \times 1.6 \times 10^{-19} = 1.6 \times 10^{-12}\text{ A}
  3. So the answer is (a) 1.6×10−12 A1.6 \times 10^{-12}\text{ A}.
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Q35NEET 1991 · AIPMTEasy

Electric Current, Drift Velocity & Current Density · Drift velocity & relaxation time

The velocity of charge carriers of current (about 1 ampere) in a metal under normal conditions is of the order of

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Answer: (a)

  1. vd=IneAv_d = \dfrac{I}{neA}; for copper n≈1029 m−3n \approx 10^{29}\text{ m}^{-3}, A≈10−6 m2A \approx 10^{-6}\text{ m}^{2}, I=1 AI = 1\text{ A}
  2. vd≈11029×1.6×10−19×10−6≈10−4 m/sv_d \approx \dfrac{1}{10^{29} \times 1.6 \times 10^{-19} \times 10^{-6}} \approx 10^{-4}\text{ m/s}
  3. This is a fraction of a mm/s, so the answer is (a).
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Q36MHT CET 2026 · 20 MAY · SHIFT 1Easy

Ohm's Law, Resistivity & Temperature · Temperature dependence of resistance

In case of thermistor, which one of the following statements is 'WRONG'?

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Answer: (b)

  1. NTC: resistance decreases as temperature rises; used as temperature sensors. (a), (c) correct.
  2. PTC: resistance increases as temperature rises; used as resettable fuses. (d) correct.
  3. (b) says PTC resistance decreases with temperature, which is wrong.
  4. So the answer is (b).
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Q37MHT CET 2026 · 13 APR · SHIFT 1Easy

Ohm's Law, Resistivity & Temperature · Temperature dependence of resistance

Which of the following graph represents the temperature (T) dependence of resistivity (ρ\rho) of a semiconductor?

Figure

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Answer: (b)

  1. In a semiconductor, more charge carriers are created as TT rises, so ρ\rho falls.
  2. The fall is roughly exponential: ρ\rho drops steeply at first, then levels off.
  3. Graph B shows this decreasing curve.
  4. So the answer is (b) B.
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Q38NEET 2026 · RE-NEETMedium

Ohm's Law, Resistivity & Temperature · Resistivity: R = ρl/A

Consider two circuits, (A) and (B), each having two resistors. One of them has a positive temperature coefficient of resistance, +α+\alpha, while the other one has a negative temperature coefficient, −α-\alpha, as shown in the figure. The currents through these circuits are denoted by IAI_A and IBI_B. At initial temperature, the resistance of the two resistors is R0R_0. As the temperature is increased, the correct option that describes the variation of current in these circuits is:

Figure

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Answer: (b)

  1. Circuit A (series): RA=R0(1−αΔT)+R0(1+αΔT)=2R0R_A = R_0(1-\alpha\Delta T) + R_0(1+\alpha\Delta T) = 2R_0, independent of temperature.
  2. So IA=V/2R0I_A = V/2R_0 stays constant.
  3. Circuit B (parallel): RB=R02(1−α2ΔT2)2R0=R02(1−α2ΔT2)R_B = \dfrac{R_0^2(1-\alpha^2\Delta T^2)}{2R_0} = \dfrac{R_0}{2}(1-\alpha^2\Delta T^2), which decreases as ΔT\Delta T rises.
  4. So IB=V/RBI_B = V/R_B increases. The answer is (b).
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Q39MHT CET 2026 · 15 MAY · SHIFT 1Medium

Ohm's Law, Resistivity & Temperature · Colour code

A resistor of 6 Ω6\ \Omega with a tolerance of 10%10\% and another of 4 Ω4\ \Omega with a tolerance of 10%10\% are connected in series. The tolerance of combination is about

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Answer: (b)

  1. Series: R=6+4=10 ΩR = 6 + 4 = 10\ \Omega; maximum errors add: ΔR=0.6+0.4=1.0 Ω\Delta R = 0.6 + 0.4 = 1.0\ \Omega.
  2. Tolerance =1.0/10=10%= 1.0/10 = 10\%. The answer is (b).
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Q40JEE Main 2025 · 22 JAN · SHIFT 1Easy

Ohm's Law, Resistivity & Temperature · Temperature dependence of resistance

Which of the following resistivity ( ρ\rho) v/s\text{v/s} temperature ( T) curves is most suitable to be used in wire bound standard resistors?

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Answer: (d)

  1. A standard resistor must keep almost the same resistance when temperature changes.
  2. So its material needs a high resistivity and a very small temperature coefficient (e.g. manganin, constantan).
  3. Graph (d): large ρ\rho that rises only slightly with TT.
  4. So the answer is (d).
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Q41JEE Main 2024 · 9 APR · SHIFT 2Easy

Ohm's Law, Resistivity & Temperature · Temperature dependence of resistance

At room temperature (27∘C)(27^\circ\text{C}), the resistance of a heating element is 50 Ω50\ \Omega. The temperature coefficient of the material is 2.4×10−4 ∘C−12.4 \times 10^{-4}\ {}^\circ\text{C}^{-1}. The temperature of the element, when its resistance is 62 Ω62\ \Omega, is __________ ∘C{}^\circ\text{C}.

Numerical answer type. Work it out, then open the answer.

Show answer and solution

Answer: 1027

  1. R=R27[1+α(T−27)]R = R_{27}[1 + \alpha (T - 27)]
  2. 62=50[1+2.4×10−4(T−27)]⇒2.4×10−4(T−27)=0.2462 = 50[1 + 2.4 \times 10^{-4}(T - 27)] \Rightarrow 2.4 \times 10^{-4}(T - 27) = 0.24
  3. T−27=1000⇒T=1027∘CT - 27 = 1000 \Rightarrow T = 1027^\circ\text{C}
  4. Answer: 1027 ∘C1027\ ^\circ\text{C}.
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Q42JEE Main 2024 · 9 APR · SHIFT 2Medium

Ohm's Law, Resistivity & Temperature · Ohm's law & V–I graphs

To determine the resistance (R) of a wire, a circuit is designed below. The V-IV\text{-}I characteristic curve for this circuit is plotted for the voltmeter and the ammeter readings as shown in figure. The value of RR is _________ Ω\Omega.

Figure

Numerical answer type. Work it out, then open the answer.

Show answer and solution

Answer: 2500

  1. The voltmeter reads the voltage across RR in parallel with 10 kΩ10\text{ k}\Omega; the mA meter reads the total current.
  2. Slope of graph: Req=VI=42×10−3=2000 ΩR_{eq} = \dfrac{V}{I} = \dfrac{4}{2 \times 10^{-3}} = 2000\ \Omega
  3. 10000 RR+10000=2000⇒8000R=2×107\dfrac{10000\,R}{R + 10000} = 2000 \Rightarrow 8000R = 2 \times 10^7
  4. R=2500 ΩR = 2500\ \Omega
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Q43JEE Main 2024 · 4 APR · SHIFT 2Medium

Ohm's Law, Resistivity & Temperature · Resistivity: R = ρl/A

Two wires AA and BB are made up of the same material and have the same mass. Wire AA has radius of 2.0 mm2.0\text{ mm} and wire BB has radius of 4.0 mm4.0\text{ mm}. The resistance of wire BB is 2 Ω2\ \Omega. The resistance of wire AA is ________ Ω\Omega.

Numerical answer type. Work it out, then open the answer.

Show answer and solution

Answer: 32

  1. Same material and mass, so same volume V=lAV = lA; then R=ρl/A=ρV/A2∝1/r4R = \rho l/A = \rho V/A^2 \propto 1/r^4.
  2. RA/RB=(rB/rA)4=(4/2)4=16R_A/R_B = (r_B/r_A)^4 = (4/2)^4 = 16
  3. RA=16×2=32 ΩR_A = 16\times2 = 32\ \Omega.
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Q44JEE Main 2024 · 6 APR · SHIFT 1Medium

Ohm's Law, Resistivity & Temperature · Stretching & reshaping a wire

A wire of resistance RR and radius rr is stretched till its radius became r/2r/2. If new resistance of the stretched wire is xRxR, then value of xx is ________.

Numerical answer type. Work it out, then open the answer.

Show answer and solution

Answer: 16

  1. Volume constant, so R∝1/r4R \propto 1/r^4.
  2. R′=(2)4R=16RR' = (2)^4R = 16R, so x=16x = 16.
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Q45JEE Main 2024 · 31 JAN · SHIFT 1Medium

Ohm's Law, Resistivity & Temperature · Temperature dependence of resistance

Two conductors have the same resistances at 0∘C0^\circ\text{C} but their temperature coefficients of resistance are α1\alpha_1 and α2\alpha_2. The respective temperature coefficients for their series and parallel combinations are:

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Answer: (b)

  1. Series: Rs=R0(1+α1t)+R0(1+α2t)=2R0(1+α1+α22t)R_s = R_0(1+\alpha_1 t) + R_0(1+\alpha_2 t) = 2R_0\left(1 + \tfrac{\alpha_1+\alpha_2}{2}t\right), so αs=α1+α22\alpha_s = \tfrac{\alpha_1+\alpha_2}{2}.
  2. Parallel: 1Rp=1R0[(1+α1t)−1+(1+α2t)−1]≈2R0(1−α1+α22t)\dfrac{1}{R_p} = \dfrac{1}{R_0}\left[(1+\alpha_1 t)^{-1} + (1+\alpha_2 t)^{-1}\right] \approx \dfrac{2}{R_0}\left(1 - \tfrac{\alpha_1+\alpha_2}{2}t\right)
  3. So Rp≈R02(1+α1+α22t)R_p \approx \tfrac{R_0}{2}\left(1 + \tfrac{\alpha_1+\alpha_2}{2}t\right), giving αp=α1+α22\alpha_p = \tfrac{\alpha_1+\alpha_2}{2}.
  4. The answer is (b).
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Q46JEE Main 2024 · 30 JAN · SHIFT 1Medium

Ohm's Law, Resistivity & Temperature · Temperature dependence of resistance

An electric toaster has resistance of 60 Ω60\ \Omega at room temperature (27∘C)(27^\circ\text{C}). The toaster is connected to a 220 V220\text{ V} supply. If the current flowing through it reaches 2.75 A2.75\text{ A}, the temperature attained by toaster is around: ( if α=2×10−4/∘C\alpha = 2 \times 10^{-4}/^\circ\text{C})

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Answer: (d)

  1. Hot resistance R=220/2.75=80 ΩR = 220/2.75 = 80\ \Omega.
  2. 80=60(1+2×10−4 ΔT)⇒ΔT=1/32×10−4≈1667∘80 = 60(1 + 2\times10^{-4}\,\Delta T) \Rightarrow \Delta T = \dfrac{1/3}{2\times10^{-4}} \approx 1667^\circC
  3. T=27+1667=1694∘T = 27 + 1667 = 1694^\circC. The answer is (d).
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Q47JEE Main 2024 · 8 APR · SHIFT 1Medium

Ohm's Law, Resistivity & Temperature · Temperature dependence of resistance

Resistance of a wire at 0∘C,100∘C0^\circ\text{C}, 100^\circ\text{C} and t∘Ct^\circ\text{C} is found to be 10 Ω,10.2 Ω10\ \Omega, 10.2\ \Omega and 10.95 Ω10.95\ \Omega respectively. The temperature tt in Kelvin scale is _________.

Numerical answer type. Work it out, then open the answer.

Show answer and solution

Answer: 748

  1. Rt=R0(1+αt)R_t = R_0(1+\alpha t): R0α=(10.2−10)/100=0.002 Ω/∘R_0\alpha = (10.2-10)/100 = 0.002\ \Omega/^\circC.
  2. 10.95−10=0.002 t⇒t=475∘10.95 - 10 = 0.002\,t \Rightarrow t = 475^\circC
  3. T=475+273=748T = 475 + 273 = 748 K.
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Q48NEET 2023Easy

Ohm's Law, Resistivity & Temperature · Resistivity: R = ρl/A

A certain wire A\text{A} has resistance 81 Ω81\ \Omega. The resistance of another wire B\text{B} of same material and equal length but of diameter thrice the diameter of A will be:

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Answer: (b)

  1. R=ρlA=4ρlπd2⇒R∝1d2R = \dfrac{\rho l}{A} = \dfrac{4\rho l}{\pi d^{2}} \Rightarrow R \propto \dfrac{1}{d^{2}}
  2. RB=8132=9 ΩR_B = \dfrac{81}{3^{2}} = 9\ \Omega
  3. So the answer is (b).
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Q49JEE Main 2023 · 10 APR · SHIFT 2Easy

Ohm's Law, Resistivity & Temperature · Resistivity: R = ρl/A

A rectangular parallelopiped is measured as 1 cm×1 cm×100 cm1\text{ cm} \times 1\text{ cm} \times 100\text{ cm}. If its specific resistance is 3×10−7 Ω m3 \times 10^{-7}\ \Omega\text{ m}, then the resistance between its two opposite rectangular faces will be ___________ ×10−7 Ω\times 10^{-7}\ \Omega.

Numerical answer type. Work it out, then open the answer.

Show answer and solution

Answer: 3

  1. The rectangular faces are 1 cm×100 cm1\text{ cm} \times 100\text{ cm}; the distance between opposite rectangular faces is 1 cm1\text{ cm}.
  2. So l=0.01 ml = 0.01\text{ m} and A=0.01×1=0.01 m2A = 0.01 \times 1 = 0.01\text{ m}^2.
  3. R=ρlA=3×10−7×0.010.01=3×10−7 ΩR = \dfrac{\rho l}{A} = \dfrac{3 \times 10^{-7} \times 0.01}{0.01} = 3 \times 10^{-7}\ \Omega
  4. Answer: 33 (i.e. 3×10−7 Ω3 \times 10^{-7}\ \Omega).
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Q50JEE Main 2023 · 25 JAN · SHIFT 2Easy

Ohm's Law, Resistivity & Temperature · Stretching & reshaping a wire

The resistance of a wire is 5 Ω\Omega. Its new resistance in ohm if stretched to 5 times of its original length will be:

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Answer: (d)

  1. Stretching at constant volume: R∝l2R \propto l^2.
  2. R′=5×52=125 ΩR' = 5 \times 5^2 = 125\ \Omega
  3. So the answer is (d) 125125.
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Q51NEET 2023Easy

Ohm's Law, Resistivity & Temperature · Temperature dependence of resistance

The resistance of platinum wire at 0∘C0^\circ\text{C} is 2 Ω2\ \Omega and 6.8 Ω6.8\ \Omega at 80∘C80^\circ\text{C}. The temperature coefficient of resistance of the wire is:

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Answer: (b)

  1. Rt=R0(1+αt)⇒α=Rt−R0R0tR_t = R_0(1 + \alpha t) \Rightarrow \alpha = \dfrac{R_t - R_0}{R_0 t}
  2. α=6.8−22×80=4.8160\alpha = \dfrac{6.8 - 2}{2 \times 80} = \dfrac{4.8}{160}
  3. α=0.03=3×10−2 ∘C−1\alpha = 0.03 = 3 \times 10^{-2}\ ^\circ\text{C}^{-1}
  4. So the answer is (b) 3×10−2 C−13 \times 10^{-2}\ \text{C}^{-1}.
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Q52NEET 2023Easy

Ohm's Law, Resistivity & Temperature · Colour code

Resistance of a carbon resistor determined from colour codes is (22000±5%) Ω(22000 \pm 5\%)\ \Omega. The colour of third band must be:

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Answer: (b)

  1. 22000 Ω=22×103 Ω22000\ \Omega = 22 \times 10^3\ \Omega
  2. First two digits 2, 2 (red, red); multiplier 10310^3 is orange.
  3. So the answer is (b) Orange.
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Q53JEE Main 2023 · 25 JAN · SHIFT 1Medium

Ohm's Law, Resistivity & Temperature · Resistivity: R = ρl/A

A uniform metallic wire carries a current 2 A, when 3.4 V battery is connected across it. The mass of uniform metallic wire is 8.92 ×\times 10−3-3 kg, density is 8.92 ×\times 1033 kg/m33 and resistivity is 1.7 ×\times 10−8 Ω⋅m^{-8}\ \Omega \cdot \text{m}. The length of wire is:

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Answer: (d)

  1. R=V/I=3.4/2=1.7 ΩR = V/I = 3.4/2 = 1.7\ \Omega.
  2. Volume =m/d=8.92×10−3/8.92×103=10−6 m3= m/d = 8.92\times10^{-3}/8.92\times10^{3} = 10^{-6}\ \text{m}^3.
  3. R=ρl/A=ρl2/Vol⇒l2=1.7×10−61.7×10−8=100R = \rho l/A = \rho l^2/\text{Vol} \Rightarrow l^2 = \dfrac{1.7\times10^{-6}}{1.7\times10^{-8}} = 100
  4. l=10l = 10 m. The answer is (d).
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Q54JEE Main 2023 · 12 APR · SHIFT 1Medium

Ohm's Law, Resistivity & Temperature · Resistivity: R = ρl/A

The current flowing through a conductor connected across a source is 2 A2\text{ A} and 1.2 A\text{A} at 0∘C0^\circ\text{C} and 100∘C100^\circ\text{C} respectively. The current flowing through the conductor at 50∘C50^\circ\text{C} will be ___________ ×102 mA\times 10^{2}\text{ mA}.

Numerical answer type. Work it out, then open the answer.

Show answer and solution

Answer: 15

  1. I=V/RI = V/R, so R0=V/2R_0 = V/2 and R100=V/1.2R_{100} = V/1.2.
  2. R100=R0(1+100α)⇒1+100α=2/1.2=5/3⇒α=1/150R_{100} = R_0(1+100\alpha) \Rightarrow 1+100\alpha = 2/1.2 = 5/3 \Rightarrow \alpha = 1/150.
  3. R50=R0(1+50/150)=43⋅V2=2V3R_{50} = R_0(1+50/150) = \tfrac43\cdot\tfrac{V}{2} = \tfrac{2V}{3}
  4. I50=1.5 A=1500 mA=15×102I_{50} = 1.5\ \text{A} = 1500\ \text{mA} = 15\times10^2 mA, so the answer is 15.
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Q55JEE Main 2023 · 24 JAN · SHIFT 1Medium

Ohm's Law, Resistivity & Temperature · Resistivity: R = ρl/A

A hollow cylindrical conductor has length of 3.14 m, while its inner and outer diameters are 4 mm and 8 mm respectively. The resistance of the conductor is n×10−3 Ωn \times 10^{-3}\ \Omega. If the resistivity of the material is 2.4×10−8 Ω m2.4 \times 10^{-8}\ \Omega\text{ m}, the value of nn is ___________.

Numerical answer type. Work it out, then open the answer.

Show answer and solution

Answer: 2

  1. Area A=π(r22−r12)=π(16−4)×10−6=12π×10−6 m2A = \pi(r_2^2 - r_1^2) = \pi(16-4)\times10^{-6} = 12\pi\times10^{-6}\ \text{m}^2.
  2. R=ρlA=2.4×10−8×3.143.14×12×10−6=2×10−3 ΩR = \dfrac{\rho l}{A} = \dfrac{2.4\times10^{-8}\times3.14}{3.14\times12\times10^{-6}} = 2\times10^{-3}\ \Omega
  3. So n=2n = 2.
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Q56JEE Main 2023 · 12 APR · SHIFT 1Medium

Ohm's Law, Resistivity & Temperature · Stretching & reshaping a wire

A wire of resistance 160 Ω160\ \Omega is melted and drawn in a wire of one-fourth of its length. The new resistance of the wire will be

Show answer and solution

Answer: (d)

  1. Volume constant: l→l/4l \to l/4 means A→4AA \to 4A, so R∝l2R \propto l^2.
  2. R′=160×(1/4)2=10 ΩR' = 160\times(1/4)^2 = 10\ \Omega. The answer is (d).
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Q57JEE Main 2023 · 6 APR · SHIFT 1Medium

Ohm's Law, Resistivity & Temperature · Stretching & reshaping a wire

The length of a metallic wire is increased by 20%20\% and its area of cross section is reduced by 4%4\%. The percentage change in resistance of the metallic wire is __________.

Numerical answer type. Work it out, then open the answer.

Show answer and solution

Answer: 25

  1. R=ρl/A∝l/AR = \rho l/A \propto l/A.
  2. R′/R=1.200.96=1.25R'/R = \dfrac{1.20}{0.96} = 1.25
  3. Increase =25%= 25\%.
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Q58JEE Main 2023 · 24 JAN · SHIFT 2Medium

Ohm's Law, Resistivity & Temperature · Stretching & reshaping a wire

If a copper wire is stretched to increase its length by 20%. The percentage increase in resistance of the wire is __________%.

Numerical answer type. Work it out, then open the answer.

Show answer and solution

Answer: 44

  1. Volume constant, so R∝l2R \propto l^2.
  2. R′/R=(1.2)2=1.44R'/R = (1.2)^2 = 1.44
  3. Increase =44%= 44\%.
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Q59NEET 2022 · PHASE 2Easy

Ohm's Law, Resistivity & Temperature · Resistivity: R = ρl/A

The reciprocal of resistance is:

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Answer: (a)

  1. Conductance G=1RG = \dfrac{1}{R} (unit siemens). Conductivity is 1/ρ1/\rho, not 1/R1/R.
  2. So the answer is (a) conductance.
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Q60JEE Main 2022 · 28 JUL · SHIFT 1Easy

Ohm's Law, Resistivity & Temperature · Stretching & reshaping a wire

A wire of resistance R1 is drawn out so that its length is increased by twice of its original length. The ratio of new resistance to original resistance is:

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Answer: (a)

  1. Length increased by twice its original length: l′=l+2l=3ll' = l + 2l = 3l.
  2. At constant volume R∝l2R \propto l^2, so R′/R=32=9R'/R = 3^2 = 9.
  3. So the answer is (a) 9:19:1.
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Q61JEE Main 2022 · 26 JUN · SHIFT 1Easy

Ohm's Law, Resistivity & Temperature · Stretching & reshaping a wire

An aluminium wire is stretched to make its length, 0.4% larger. The percentage change in resistance is:

Show answer and solution

Answer: (c)

  1. At constant volume R∝l2R \propto l^2, so for small changes ΔRR=2Δll\dfrac{\Delta R}{R} = 2\dfrac{\Delta l}{l}.
  2. ΔRR=2×0.4%=0.8%\dfrac{\Delta R}{R} = 2 \times 0.4\% = 0.8\%
  3. So the answer is (c) 0.8%0.8\%.
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Q62JEE Main 2022 · 25 JUN · SHIFT 2Easy

Ohm's Law, Resistivity & Temperature · Stretching & reshaping a wire

The length of a given cylindrical wire is increased to double of its original length. The percentage increase in the resistance of the wire will be ____________ %.

Numerical answer type. Work it out, then open the answer.

Show answer and solution

Answer: 300

  1. l′=2ll' = 2l; at constant volume R∝l2R \propto l^2, so R′=4RR' = 4R.
  2. Percentage increase =4R−RR×100= \dfrac{4R - R}{R} \times 100
  3. Answer: 300%300\%.
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Q63JEE Main 2022 · 29 JUL · SHIFT 2Easy

Ohm's Law, Resistivity & Temperature · Temperature dependence of resistance

Given below are two statements: one is labelled as Assertion A and the other is labelled as Reason R.

Assertion A: Alloys such as constantan and manganin are used in making standard resistance coils. Reason R: Constantan and manganin have very small value of temperature coefficient of resistance. In the light of the above statements, choose the correct answer from the options given below.

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Answer: (a)

  1. Assertion: constantan and manganin are used for standard resistance coils. True.
  2. Reason: they have a very small temperature coefficient of resistance, so their resistance stays nearly constant. True.
  3. This small α\alpha is exactly why they are chosen, so R explains A.
  4. So the answer is (a).
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Q64JEE Main 2022 · 28 JUN · SHIFT 2Easy

Ohm's Law, Resistivity & Temperature · Temperature dependence of resistance

Resistance of the wire is measured as 2 Ω\Omega and 3 Ω\Omega at 10∘^\circC and 30∘^\circC respectively. Temperature co- efficient of resistance of the material of the wire is:

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Answer: (a)

  1. Use Rt=R0(1+αt)R_t = R_0(1 + \alpha t) with 0∘C0^\circ\text{C} as reference.
  2. 32=1+30α1+10α⇒3+30α=2+60α\dfrac{3}{2} = \dfrac{1 + 30\alpha}{1 + 10\alpha} \Rightarrow 3 + 30\alpha = 2 + 60\alpha
  3. α=130≈0.033 ∘C−1\alpha = \dfrac{1}{30} \approx 0.033\ ^\circ\text{C}^{-1}
  4. So the answer is (a) 0.033 ∘C−10.033\ ^\circ\text{C}^{-1}.
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Q65JEE Main 2022 · 29 JUN · SHIFT 1Medium

Ohm's Law, Resistivity & Temperature · Resistivity: R = ρl/A

The variation of applied potential and current flowing through a given wire is shown in figure. The length of wire is 31.4 cm. The diameter of wire is measured as 2.4 cm. The resistivity of the given wire is measured as x ×\times 10−-3 Ω\Omega cm. The value of x is ____________. [Take π\pi = 3.14]

Figure

Numerical answer type. Work it out, then open the answer.

Show answer and solution

Answer: 144

  1. Slope of VV–II graph =R=tan⁡45∘=1 Ω= R = \tan45^\circ = 1\ \Omega.
  2. A=πd2/4=3.14×(2.4)2/4=4.5216 cm2A = \pi d^2/4 = 3.14\times(2.4)^2/4 = 4.5216\ \text{cm}^2.
  3. ρ=RA/l=4.5216/31.4=0.144 Ω cm=144×10−3 Ω cm\rho = RA/l = 4.5216/31.4 = 0.144\ \Omega\,\text{cm} = 144\times10^{-3}\ \Omega\,\text{cm}
  4. So x=144x = 144.
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Q66JEE Main 2022 · 29 JUL · SHIFT 2Difficult

Ohm's Law, Resistivity & Temperature · Stretching & reshaping a wire

A 1 m1\text{ m} long wire is broken into two unequal parts X\text{X} and Y\text{Y}. The X\text{X} part of the wire is streched into another wire W. Length of WW is twice the length of XX and the resistance of W\text{W} is twice that of Y\text{Y}. Find the ratio of length of X\text{X} and Y\text{Y}.

Show answer and solution

Answer: (b)

  1. Let lengths be xx (X) and yy (Y); same wire, so R=klR = kl before stretching: RX=kxR_X = kx, RY=kyR_Y = ky.
  2. Stretching X to twice its length (volume constant) multiplies resistance by 222^2: RW=4kxR_W = 4kx.
  3. RW=2RY⇒4kx=2ky⇒x:y=1:2R_W = 2R_Y \Rightarrow 4kx = 2ky \Rightarrow x : y = 1 : 2
  4. The answer is (b).
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Q67JEE Main 2021 · 27 JUL · SHIFT 2Easy

Ohm's Law, Resistivity & Temperature · Temperature dependence of resistance

The resistance of a conductor at 15∘^\circC is 16Ω\Omega and at 100∘^\circ C is 20Ω\Omega. What will be the temperature coefficient of resistance of the conductor?

Show answer and solution

Answer: (c)

  1. Use Rt=R0(1+αt)R_t = R_0(1 + \alpha t): 2016=1+100α1+15α\dfrac{20}{16} = \dfrac{1 + 100\alpha}{1 + 15\alpha}
  2. 20+300α=16+1600α⇒1300α=420 + 300\alpha = 16 + 1600\alpha \Rightarrow 1300\alpha = 4
  3. α≈0.003 ∘C−1\alpha \approx 0.003\ ^\circ\text{C}^{-1}
  4. So the answer is (c) 0.003 ∘C−10.003\ ^\circ\text{C}^{-1}.
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Q68JEE Main 2021 · 27 AUG · SHIFT 2EasyOld syllabus

Ohm's Law, Resistivity & Temperature · Colour code

The Colour coding on a carbon resistor is shown in the given figure. The resistance value of the given resistor is:

Figure

Show answer and solution

Answer: (d)

  1. Violet = 7, Green = 5, Red = multiplier 10210^2, Gold = ±5%\pm 5\%.
  2. R=75×100=7500 ΩR = 75 \times 100 = 7500\ \Omega
  3. Tolerance =5%= 5\% of 7500=375 Ω7500 = 375\ \Omega
  4. So the answer is (d) (7500±375) Ω(7500 \pm 375)\ \Omega.
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Q69JEE Main 2021 · 27 JUL · SHIFT 2Medium

Ohm's Law, Resistivity & Temperature · Resistivity: R = ρl/A

For the circuit shown, the value of current at time t = 3.2 s will be _________ A.

Figure

[Voltage distribution V(t) is shown by Fig. (1) and the circuit is shown in Fig. (2)]

Numerical answer type. Work it out, then open the answer.

Show answer and solution

Answer: 1

  1. From Fig. 1, between t=3t = 3 s and 44 s, VV rises from 5 V to 10 V at 5 V/s.
  2. At t=3.2t = 3.2 s: V=5+5×0.2=6V = 5 + 5\times0.2 = 6 V.
  3. The 5 V battery opposes V(t)V(t) in the loop, so net emf =6−5=1= 6 - 5 = 1 V.
  4. I=1/1=1I = 1/1 = 1 A.
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Q70JEE Main 2021 · 26 FEB · SHIFT 2Medium

Ohm's Law, Resistivity & Temperature · Stretching & reshaping a wire

A wire of 1Ω\Omega has a length of 1 m. It is stretched till its length increases by 25%. The percentage change in resistance to the nearest integer is:

Show answer and solution

Answer: (d)

  1. Volume constant, so R∝l2R \propto l^2.
  2. R′/R=(1.25)2=1.5625R'/R = (1.25)^2 = 1.5625
  3. Increase ≈56%\approx 56\%. The answer is (d).
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Q71JEE Main 2021 · 26 AUG · SHIFT 1Medium

Ohm's Law, Resistivity & Temperature · Temperature dependence of resistance

What equal length of an iron wire and a copper-nickel alloy wire, each of 2 mm diameter connected parallel to give an equivalent resistance of 3Ω\Omega?

(Given resistivities of iron and copper-nickel alloy wire are 12 μΩ\mu\Omega cm and 51 μΩ\mu\Omega cm respectively)

Show answer and solution

Answer: (b)

  1. Same ll and AA, in parallel: Req=lA⋅ρ1ρ2ρ1+ρ2R_{eq} = \dfrac{l}{A}\cdot\dfrac{\rho_1\rho_2}{\rho_1+\rho_2}.
  2. ρ1ρ2ρ1+ρ2=12×5163=9.71 μΩ cm=9.71×10−8 Ω m\dfrac{\rho_1\rho_2}{\rho_1+\rho_2} = \dfrac{12\times51}{63} = 9.71\ \mu\Omega\,\text{cm} = 9.71\times10^{-8}\ \Omega\,\text{m}
  3. A=π(10−3)2=3.14×10−6 m2A = \pi(10^{-3})^2 = 3.14\times10^{-6}\ \text{m}^2
  4. l=3×3.14×10−69.71×10−8≈97l = \dfrac{3\times3.14\times10^{-6}}{9.71\times10^{-8}} \approx 97 m. The answer is (b).
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Q72NEET 2020 · PHASE 2Easy

Ohm's Law, Resistivity & Temperature · Resistivity: R = ρl/A

Two solid conductors are made up of same material, have same length and same resistance. One of them has a circular cross section of area A1A_1 and the other one has a square cross section of area A2A_2. The ratio A1/A2A_1/A_2 is

Show answer and solution

Answer: (c)

  1. R=ρlA⇒A=ρlRR = \dfrac{\rho l}{A} \Rightarrow A = \dfrac{\rho l}{R}
  2. Same ρ\rho, same ll and same RR, so the areas are equal whatever the shape.
  3. A1/A2=1A_1 / A_2 = 1
  4. So the answer is (c) 11.
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Q73JEE Main 2020 · 2 SEP · SHIFT 1Easy

Ohm's Law, Resistivity & Temperature · Resistivity: R = ρl/A

Consider four conducting materials copper, tungsten, mercury and aluminium with resistivity ρ\rhoC, ρ\rhoT, ρ\rhoM and ρ\rhoA respectively. Then:

Show answer and solution

Answer: (b)

  1. Approximate resistivities at room temperature: ρC≈1.7×10−8\rho_C \approx 1.7 \times 10^{-8}, ρA≈2.7×10−8\rho_A \approx 2.7 \times 10^{-8}, ρT≈5.6×10−8\rho_T \approx 5.6 \times 10^{-8}, ρM≈98×10−8 Ω m\rho_M \approx 98 \times 10^{-8}\ \Omega\text{ m}.
  2. Order: ρM>ρT>ρA>ρC\rho_M > \rho_T > \rho_A > \rho_C.
  3. Only option (b) ρM>ρA>ρC\rho_M > \rho_A > \rho_C fits this order.
  4. So the answer is (b).
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Q74NEET 2020 · PHASE 1Easy

Ohm's Law, Resistivity & Temperature · Temperature dependence of resistance

Which of the following graph represents the variation of resistivity (ρ\rho) with temperature (T) for copper?

Show answer and solution

Answer: (b)

  1. Copper is a metal: ρ\rho increases with TT (positive α\alpha).
  2. At low TT, ρ\rho does not go to zero (residual resistivity) and rises non-linearly; at higher TT it rises faster.
  3. Graph (b) shows a rising curve with a non-zero intercept.
  4. So the answer is (b).
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Q75NEET 2020 · PHASE 1Easy

Ohm's Law, Resistivity & Temperature · Temperature dependence of resistance

The solids which have the negative temperature coefficient of resistance are:

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Answer: (c)

  1. Metals: ρ\rho rises with TT, so α>0\alpha > 0.
  2. Semiconductors and insulators: more carriers are freed at higher TT, so ρ\rho falls and α<0\alpha < 0.
  3. So the answer is (c) insulators and semiconductors.
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Q76NEET 2020 · PHASE 1Easy

Ohm's Law, Resistivity & Temperature · Colour code

The color code of a resistance is given below:

Figure

The values of resistance and tolerance, respectively, are

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Answer: (c)

  1. Yellow = 4, Violet = 7, Brown = multiplier 10110^1, Gold = ±5%\pm 5\%.
  2. R=47×10=470 ΩR = 47 \times 10 = 470\ \Omega
  3. So the answer is (c) 470 Ω470\ \Omega, 5%5\%.
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Q77JEE Main 2020 · 3 SEP · SHIFT 1Difficult

Ohm's Law, Resistivity & Temperature · Resistivity: R = ρl/A

Model a torch battery of length ll to be made up of a thin cylindrical bar of radius ‘a’ and a concentric thin cylindrical shell of radius ‘b’ filled in between with an electrolyte of resistivity ρ\rho (see figure). If the battery is connected to a resistance of value R, the maximum Joule heating in R will take place for:

Figure

Show answer and solution

Answer: (c)

  1. Internal resistance: current flows radially through shells of area 2πrl2\pi r l, so rint=∫abρ dr2πrl=ρ2πlln⁡bar_{int} = \int_a^b \dfrac{\rho\,dr}{2\pi r l} = \dfrac{\rho}{2\pi l}\ln\dfrac{b}{a}.
  2. Power in RR is maximum when R=rintR = r_{int} (maximum power transfer).
  3. R=ρ2πlln⁡baR = \dfrac{\rho}{2\pi l}\ln\dfrac{b}{a}. The answer is (c).
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Q78MHT CET 2019 · 2 MAY · SHIFT 1Easy

Ohm's Law, Resistivity & Temperature · Ohm's law & V–I graphs

For a metallic wire, the ratio of voltage to corresponding current is

Show answer and solution

Answer: (b)

  1. VI=R\dfrac{V}{I} = R, and for a metal R=R0(1+αΔT)R = R_0(1 + \alpha \Delta T) with α>0\alpha > 0
  2. Hotter metal: more ion vibration, shorter relaxation time, so resistance rises.
  3. So the answer is (b).
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Q79JEE Main 2019 · 8 APR · SHIFT 1EasyOld syllabus

Ohm's Law, Resistivity & Temperature · Colour code

A 200 Ω\Omega resistor has a certain color code. If one replaces the red color by green in the code, the new resistance will be:

Show answer and solution

Answer: (a)

  1. 200 Ω=20×101200\ \Omega = 20 \times 10^1: Red (2), Black (0), Brown (×10\times 10).
  2. Replace red by green (5): 50×10=500 Ω50 \times 10 = 500\ \Omega.
  3. So the answer is (a) 500 Ω500\ \Omega.
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Q80JEE Main 2019 · 9 JAN · SHIFT 2EasyOld syllabus

Ohm's Law, Resistivity & Temperature · Colour code

A carbon resistance has a following colour code. What is the value of the resistance?

Figure

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Answer: (a)

  1. Green = 5, Orange = 3, Yellow = multiplier 10410^4, Gold = ±5%\pm 5\%.
  2. R=53×104=530 kΩR = 53 \times 10^4 = 530\text{ k}\Omega
  3. So the answer is (a) 530 kΩ±5%530\text{ k}\Omega \pm 5\%.
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Q81JEE Main 2019 · 9 JAN · SHIFT 1EasyOld syllabus

Ohm's Law, Resistivity & Temperature · Colour code

A resistance is shown in the figure. Its value and tolerance are given respectively by:

Figure

Show answer and solution

Answer: (b)

  1. Red = 2, Violet = 7, Orange = multiplier 10310^3, Silver = ±10%\pm 10\%.
  2. R=27×103=27 kΩR = 27 \times 10^3 = 27\text{ k}\Omega
  3. So the answer is (b) 27 kΩ27\text{ k}\Omega, 10%10\%.
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Q82JEE Main 2019 · 9 JAN · SHIFT 1Medium

Ohm's Law, Resistivity & Temperature · Stretching & reshaping a wire

A copper wire is stretched to make it 0.5% longer. The percentage change in its electrical resistance if its volume remains unchanged is:

Show answer and solution

Answer: (c)

  1. Volume constant, so R∝l2⇒ΔRR=2ΔllR \propto l^2 \Rightarrow \dfrac{\Delta R}{R} = 2\dfrac{\Delta l}{l} for small changes.
  2. ΔR/R=2×0.5%=1%\Delta R/R = 2\times0.5\% = 1\%. The answer is (c).
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Q83JEE Main 2019 · 10 APR · SHIFT 1Medium

Ohm's Law, Resistivity & Temperature · Temperature dependence of resistance

In an experiment, the resistance of a material is plotted as a function of temperature (in some range). As shown in the figure, it is a straight line. One may conclude that:

Figure

Show answer and solution

Answer: (d)

  1. Graph: ln⁡R\ln R vs 1/T21/T^2 is a straight line with negative slope.
  2. So ln⁡R=ln⁡R0−T02/T2\ln R = \ln R_0 - T_0^2/T^2.
  3. R(T)=R0e−T02/T2R(T) = R_0 e^{-T_0^2/T^2}. The answer is (d).
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Q84JEE Main 2019 · 10 JAN · SHIFT 1MediumOld syllabus

Ohm's Law, Resistivity & Temperature · Colour code

A 2 W carbon resistor is color coded with green, black, red and brown respectively. The maximum current which can be passed through this resistor is -

Show answer and solution

Answer: (b)

  1. Green = 5, black = 0, red = ×102\times10^2: R=50×102=5000 ΩR = 50\times10^2 = 5000\ \Omega (brown = 1% tolerance).
  2. P=I2R⇒I=P/R=2/5000=0.02P = I^2R \Rightarrow I = \sqrt{P/R} = \sqrt{2/5000} = 0.02 A
  3. Imax=20I_{max} = 20 mA. The answer is (b).
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Q85JEE Main 2019 · 10 APR · SHIFT 2Difficult

Ohm's Law, Resistivity & Temperature · Resistivity: R = ρl/A

Space between two concentric conducting spheres of radii a and b (b > a) is filled with a medium of resistivity ρ\rho. The resistance between the two spheres will be:

Show answer and solution

Answer: (d)

  1. Current flows radially through spherical shells of area 4πr24\pi r^2: dR=ρ dr4πr2dR = \dfrac{\rho\,dr}{4\pi r^2}.
  2. R=ρ4π∫abdrr2=ρ4π(1a−1b)R = \dfrac{\rho}{4\pi}\int_a^b\dfrac{dr}{r^2} = \dfrac{\rho}{4\pi}\left(\dfrac{1}{a} - \dfrac{1}{b}\right)
  3. The answer is (d).
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Q86NEET 2018Easy

Ohm's Law, Resistivity & Temperature · Colour code

A carbon resistor of (47 ±\pm 4.7) kΩ\Omega is to be marked with rings of different colours for its identification. The colour code sequence will be

Show answer and solution

Answer: (b)

  1. 47 kΩ=47×103 Ω47\text{ k}\Omega = 47 \times 10^3\ \Omega; tolerance 4.7/47=10%4.7/47 = 10\%.
  2. 4 = Yellow, 7 = Violet, 10310^3 = Orange, 10%10\% = Silver.
  3. So the answer is (b) Yellow – Violet – Orange – Silver.
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Q87JEE Main 2018 · 16 APR · SHIFT 1Medium

Ohm's Law, Resistivity & Temperature · Temperature dependence of resistance

A heating element has a resistance of 100 Ω\Omega at room temperature. When it is connected to a supply of 220 V, a steady current of 2 A passes in it and temperature is 500oC more than room temperature. What is the temperature coefficient of resistance of the heating element?

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Answer: (d)

  1. Hot resistance R=220/2=110 ΩR = 220/2 = 110\ \Omega.
  2. 110=100(1+500α)⇒α=0.1500=2×10−4 ∘C−1110 = 100(1 + 500\alpha) \Rightarrow \alpha = \dfrac{0.1}{500} = 2\times10^{-4}\ ^\circ\text{C}^{-1}
  3. The answer is (d).
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Q88NEET 2017Easy

Ohm's Law, Resistivity & Temperature · Stretching & reshaping a wire

The resistance of a wire is 'R' ohm. If it is melted and stretched to 'n' times its original length. its new resistance will be

Show answer and solution

Answer: (b)

  1. Volume is constant: Al=A′l′Al = A'l', so l′=nl⇒A′=A/nl' = nl \Rightarrow A' = A/n.
  2. R′=ρ(nl)A/n=n2ρlAR' = \dfrac{\rho (nl)}{A/n} = n^2 \dfrac{\rho l}{A}
  3. R′=n2RR' = n^2 R
  4. So the answer is (b) n2Rn^2 R.
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Q89JEE Main 2017 · 9 APR · SHIFT 1Medium

Ohm's Law, Resistivity & Temperature · Stretching & reshaping a wire

A uniform wire of length l and radius r has a resistance of 100 Ω\Omega. It is recast into a wire of radius r2.\frac{r}{2}. The resistance of new wire will be:

Show answer and solution

Answer: (a)

  1. Volume constant, so R=ρV/A2∝1/r4R = \rho V/A^2 \propto 1/r^4.
  2. r→r/2⇒R′=24×100=1600 Ωr \to r/2 \Rightarrow R' = 2^4\times100 = 1600\ \Omega. The answer is (a).
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Q90JEE Main 2016 · 10 APR · SHIFT 1Medium

Ohm's Law, Resistivity & Temperature · Temperature dependence of resistance

The resistance of an electrical toaster has a temperature dependence given by R(T) = R0 [1 + α\alpha(T − T0)] in its range of operation. At T0 = 300 K, R = 100 Ω\Omega and at T = 500 K, R = 120 Ω\Omega. The toaster is connected to a voltage source at 200 V and its temperature is raised at a constant rate from 300 to 500 K in 30 s. The total work done in raising the temperature is:

Show answer and solution

Answer: (c)

  1. 120=100[1+α(200)]⇒α=10−3120 = 100[1+\alpha(200)] \Rightarrow \alpha = 10^{-3} K−1^{-1}.
  2. T−T0=20030tT - T_0 = \tfrac{200}{30}t, so R(t)=100(1+t150)R(t) = 100\left(1 + \tfrac{t}{150}\right).
  3. W=∫030V2Rdt=40000100∫030dt1+t/150=400×150ln⁡180150W = \int_0^{30}\dfrac{V^2}{R}dt = \dfrac{40000}{100}\int_0^{30}\dfrac{dt}{1+t/150} = 400\times150\ln\dfrac{180}{150}
  4. W=60000ln⁡65W = 60000\ln\dfrac{6}{5} J. The answer is (c).
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Q91JEE Main 2015 · 10 APR · SHIFT 1Medium

Ohm's Law, Resistivity & Temperature · Ohm's law & V–I graphs

Suppose the drift velocity vvd in a material varied with the applied electric field E as vvd ∝E\propto \sqrt{E}. Then V - I graph for a wire made of such a material is best given by:

Show answer and solution

Answer: (c)

  1. I=neAvdI = neAv_d, so I∝vd∝EI \propto v_d \propto \sqrt{E}.
  2. E=V/l⇒I∝V⇒V∝I2E = V/l \Rightarrow I \propto \sqrt{V} \Rightarrow V \propto I^2
  3. The VV–II graph is a parabola through the origin, bending upward.
  4. So the answer is (c).
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Q92NEET 2013Easy

Ohm's Law, Resistivity & Temperature · Stretching & reshaping a wire

A wire of resistance 4 Ω\Omega is stretched to twice its original length. The resistance of stretched wire would be

Show answer and solution

Answer: (b)

  1. On stretching at constant volume, R∝l2R \propto l^2.
  2. R′=4×22=16 ΩR' = 4 \times 2^2 = 16\ \Omega
  3. So the answer is (b) 16 Ω16\ \Omega.
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Q93NEET 2013Medium

Ohm's Law, Resistivity & Temperature · Resistivity: R = ρl/A

Two rods are joined end to end, as shown. Both have a cross-sectional area of 0.01 cm2. Each is 1 −-6

meter long. One rod is of copper with a resistivity of 1.7 ×\times 10 ohm-centimeter, the other is of iron −-5

with a resistivity of 10 ohm-centimeter.

Figure

How much voltage is required to produce a current of 1 ampere in the rods?

Show answer and solution

Answer: (d)

  1. R=ρl/AR = \rho l/A with l=100l = 100 cm, A=0.01 cm2A = 0.01\ \text{cm}^2, so l/A=104 cm−1l/A = 10^4\ \text{cm}^{-1}.
  2. RCu=1.7×10−6×104=0.017 ΩR_{Cu} = 1.7\times10^{-6}\times10^4 = 0.017\ \Omega, RFe=10−5×104=0.1 ΩR_{Fe} = 10^{-5}\times10^4 = 0.1\ \Omega.
  3. In series: R=0.117 ΩR = 0.117\ \Omega.
  4. V=IR=1×0.117=0.117V = IR = 1\times0.117 = 0.117 V. The answer is (d).
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Q94JEE Main 2011 · AIEEEEasy

Ohm's Law, Resistivity & Temperature · Stretching & reshaping a wire

If a wire is stretched to make it 0.1%0.1\% longer, its resistance will:

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Answer: (a)

  1. For small stretching at constant volume, ΔRR=2Δll\dfrac{\Delta R}{R} = 2\dfrac{\Delta l}{l}.
  2. ΔRR=2×0.1%=0.2%\dfrac{\Delta R}{R} = 2 \times 0.1\% = 0.2\% (increase)
  3. So the answer is (a) increase by 0.2%0.2\%.
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Q95JEE Main 2010 · AIEEEMedium

Ohm's Law, Resistivity & Temperature · Temperature dependence of resistance

Two conductors have the same resistance at 0∘C0^\circ\text{C} but their temperature coefficients of resistance are α1\alpha_1 and α2.\alpha_2. The respective temperature coefficients of their series and parallel combinations are nearly

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Answer: (d)

  1. Series: Rs=2R0(1+α1+α22t)⇒αs=α1+α22R_s = 2R_0\left(1+\tfrac{\alpha_1+\alpha_2}{2}t\right) \Rightarrow \alpha_s = \tfrac{\alpha_1+\alpha_2}{2}.
  2. Parallel (small αt\alpha t): Rp≈R02(1+α1+α22t)⇒αp=α1+α22R_p \approx \tfrac{R_0}{2}\left(1+\tfrac{\alpha_1+\alpha_2}{2}t\right) \Rightarrow \alpha_p = \tfrac{\alpha_1+\alpha_2}{2}.
  3. The answer is (d).
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Q96JEE Main 2009 · AIEEEMedium

Ohm's Law, Resistivity & Temperature · Temperature dependence of resistance

Statement - 1: The temperature dependence of resistance is usually given as R=R0(1+αΔt).R = R_0(1 + \alpha\Delta t). The resistance of wire changes from 100 Ω100\ \Omega to 150 Ω150\ \Omega when its temperature is increased from 27∘C27^\circ\text{C} to 227∘C227^\circ\text{C}. This implies that α=2.5×10−3/C.\alpha = 2.5 \times 10^{-3}/\text{C}.

Statement - 2: R=R0(1+αΔt)R = R_0(1 + \alpha\Delta t) is valid only when the change in the temperature ΔT\Delta T is small and ΔR=(R−R0)≪R0.\Delta R = (R - R_0) \ll R_0.

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Answer: (c)

  1. Statement 2: the linear law R=R0(1+αΔt)R = R_0(1+\alpha\Delta t) is an approximation valid only for small changes, ΔR≪R0\Delta R \ll R_0. True.
  2. Statement 1: here ΔR=50 Ω\Delta R = 50\ \Omega is half of R0R_0, not small, so the linear formula cannot be applied to get α\alpha reliably. False.
  3. The answer is (c).
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Q97NEET 2008 · AIPMTEasy

Ohm's Law, Resistivity & Temperature · Stretching & reshaping a wire

A wire of a certain material is stretched slowly by ten percent. Its new resistance and specific resistance become respectively

Show answer and solution

Answer: (d)

  1. Constant volume gives R∝l2R \propto l^2; l′=1.1 ll' = 1.1\,l.
  2. R′=(1.1)2R=1.21RR' = (1.1)^2 R = 1.21 R
  3. Specific resistance is a material property, so it stays the same.
  4. So the answer is (d) 1.21 times, same.
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Q98JEE Main 2008 · AIEEEDifficult

Ohm's Law, Resistivity & Temperature · Resistivity: R = ρl/A

Consider a block of conducting material of resistivity ’ρ’\text{'}\rho\text{'} shown in the figure. Current ’I’\text{'}I\text{'} enters at ’A’\text{'}A\text{'} and leaves from ’D’\text{'}D\text{'}. We apply superposition principle to find voltage ’ΔV’\text{'}\Delta V\text{'} developed between ’B’\text{'}B\text{'} and ’C’\text{'}C\text{'}. The calculation is done in the following steps:

(i) Take current ’I’\text{'}I\text{'} entering from ’A’\text{'}A\text{'} and assume it to spread over a hemispherical surface in the block.

(ii) Calculate field E(r)E(r) at distance ’r’\text{'}r\text{'} from A by using Ohm's law E=ρj,E = \rho j, where jj is the current per unit area at ’r’\text{'}r\text{'}.

(iii) From the ’r’\text{'}r\text{'} dependence of E(r)E(r), obtain the potential V(r)V(r) at rr.

(iv) Repeat (i), (ii) and (iii) for current ’I’\text{'}I\text{'} leaving ’D’\text{'}D\text{'} and superpose results for ’A’\text{'}A\text{'} and ’D’.\text{'}D\text{'}.

Figure

ΔV\Delta V measured between BB and CC is

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Answer: (a)

A (+I)BCD (−I)abaV(r) = ±ρI / (2πr) from each contact
  1. Current II at A spreads over a hemisphere: j=I2πr2j = \dfrac{I}{2\pi r^2}, E=ρjE = \rho j, so V(r)=ρI2πrV(r) = \dfrac{\rho I}{2\pi r}.
  2. Due to A: VB−VC=ρI2π(1a−1a+b)V_B - V_C = \dfrac{\rho I}{2\pi}\left(\dfrac1a - \dfrac1{a+b}\right).
  3. Due to D (current leaving, V=−ρI2πrV = -\dfrac{\rho I}{2\pi r}; B is at a+ba+b, C at aa from D): VB−VC=ρI2π(1a−1a+b)V_B - V_C = \dfrac{\rho I}{2\pi}\left(\dfrac1a - \dfrac1{a+b}\right).
  4. Adding: ΔV=ρIπa−ρIπ(a+b)\Delta V = \dfrac{\rho I}{\pi a} - \dfrac{\rho I}{\pi(a+b)}. The answer is (a).
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Q99JEE Main 2008 · AIEEEDifficult

Ohm's Law, Resistivity & Temperature · Resistivity: R = ρl/A

Consider a block of conducting material of resistivity ’ρ’\text{'}\rho\text{'} shown in the figure. Current ’I’\text{'}I\text{'} enters at ’A’\text{'}A\text{'} and leaves from ’D’\text{'}D\text{'}. We apply superposition principle to find voltage ’ΔV’\text{'}\Delta V\text{'} developed between ’B’\text{'}B\text{'} and ’C’\text{'}C\text{'}. The calculation is done in the following steps:

(i) Take current ’I’\text{'}I\text{'} entering from ’A’\text{'}A\text{'} and assume it to spread over a hemispherical surface in the block.

(ii) Calculate field E(r)E(r) at distance ’r’\text{'}r\text{'} from A by using Ohm's law E=ρj,E = \rho j, where jj is the current per unit area at ’r’\text{'}r\text{'}.

(iii) From the ’r’\text{'}r\text{'} dependence of E(r)E(r), obtain the potential V(r)V(r) at rr.

(iv) Repeat (i), (ii) and (iii) for current ’I’\text{'}I\text{'} leaving ’D’\text{'}D\text{'} and superpose results for ’A’\text{'}A\text{'} and ’D’.\text{'}D\text{'}.

Figure

For current entering at A,A, the electric field at a distance ’r’\text{'}r\text{'} from AA is

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Answer: (c)

  1. Current from A spreads over a hemisphere of area 2πr22\pi r^2 (the block surface is flat).
  2. j=I2πr2j = \dfrac{I}{2\pi r^2}
  3. E=ρj=ρI2πr2E = \rho j = \dfrac{\rho I}{2\pi r^2}. The answer is (c).
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Q100JEE Main 2007 · AIEEEMedium

Ohm's Law, Resistivity & Temperature · Resistivity: R = ρl/A

The resistance of a wire is 55 ohm at 50∘C50^\circ\text{C} and 66 ohm at 100∘C.100^\circ\text{C}. The resistance of the wire at 0∘C0^\circ\text{C} will be

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Answer: (d)

  1. Rt=R0(1+αt)R_t = R_0(1+\alpha t): 5=R0(1+50α)5 = R_0(1+50\alpha), 6=R0(1+100α)6 = R_0(1+100\alpha).
  2. Dividing: 6+300α=5+500α⇒α=1/2006 + 300\alpha = 5 + 500\alpha \Rightarrow \alpha = 1/200 per ∘^\circC.
  3. R0=5/1.25=4 ΩR_0 = 5/1.25 = 4\ \Omega. The answer is (d).
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Q101JEE Main 2006 · AIEEEEasy

Ohm's Law, Resistivity & Temperature · Resistivity: R = ρl/A

A material 'B' has twice the specific resistance of 'A'. A circular wire made of 'B' has twice the diameter of a wire made of 'A'. Then for the two wires to have the same resistance, the ratio ℓB/ℓA\ell_B/\ell_A of their respective lengths must be

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Answer: (a)

  1. R=ρlπd2/4⇒l=Rπd24ρR = \dfrac{\rho l}{\pi d^2/4} \Rightarrow l = \dfrac{R \pi d^2}{4\rho}, so for the same RR, l∝d2ρl \propto \dfrac{d^2}{\rho}.
  2. lBlA=dB2dA2⋅ρAρB=4×12\dfrac{l_B}{l_A} = \dfrac{d_B^2}{d_A^2} \cdot \dfrac{\rho_A}{\rho_B} = 4 \times \dfrac{1}{2}
  3. lBlA=2\dfrac{l_B}{l_A} = 2
  4. So the answer is (a) 22.
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Q102JEE Main 2006 · AIEEEMedium

Ohm's Law, Resistivity & Temperature · Temperature dependence of resistance

The resistance of bulb filament is 100 Ω100\ \Omega at a temperature of 100∘C.100^\circ\text{C}. If its temperature coefficient of resistance be 0.0050.005 per ∘C^\circ\text{C}, its resistance will become 200 Ω200\ \Omega at a temperature of

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Answer: (b)

  1. Rt=R0(1+αt)R_t = R_0(1+\alpha t) with R0R_0 at 0∘0^\circC.
  2. 100=R0(1+0.005×100)=1.5R0⇒R0=200/3 Ω100 = R_0(1 + 0.005\times100) = 1.5R_0 \Rightarrow R_0 = 200/3\ \Omega
  3. 200=2003(1+0.005t)⇒1+0.005t=3⇒t=400∘200 = \tfrac{200}{3}(1 + 0.005t) \Rightarrow 1 + 0.005t = 3 \Rightarrow t = 400^\circC
  4. The answer is (b).
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Q103JEE Main 2004 · AIEEEEasy

Ohm's Law, Resistivity & Temperature · Resistivity: R = ρl/A

A material ′B′'B' has twice the specific resistance of ′A′.'A'. A circular wire made of ′B′'B' has twice the diameter of a wire made of ′A′'A'. Then for the two wires to have the same resistance, the ratio lB/lAl_B/l_A of their respective lengths must be

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Answer: (d)

  1. For the same RR: l=RAρ∝d2ρl = \dfrac{RA}{\rho} \propto \dfrac{d^2}{\rho}.
  2. lBlA=(dBdA)2ρAρB=4×12\dfrac{l_B}{l_A} = \left(\dfrac{d_B}{d_A}\right)^2 \dfrac{\rho_A}{\rho_B} = 4 \times \dfrac{1}{2}
  3. lBlA=2\dfrac{l_B}{l_A} = 2
  4. So the answer is (d) 22.
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Q104NEET 2004 · AIPMTEasy

Ohm's Law, Resistivity & Temperature · Stretching & reshaping a wire

The electric resistance of a certain wire of iron is R. If its length and radius are both doubled, then

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Answer: (b)

  1. R=ρlπr2∝lr2R = \dfrac{\rho l}{\pi r^2} \propto \dfrac{l}{r^2}
  2. R′=R×222=R2R' = R \times \dfrac{2}{2^2} = \dfrac{R}{2}
  3. ρ\rho depends only on the material (iron), so it is unchanged.
  4. So the answer is (b) resistance halved, specific resistance unchanged.
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Q105JEE Main 2004 · AIEEEEasy

Ohm's Law, Resistivity & Temperature · Temperature dependence of resistance

Thermistors are usually made of

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Answer: (a)

  1. A thermistor must change resistance strongly with temperature.
  2. It is made of metal oxides (e.g. of Ni, Mn, Co) which have a high temperature coefficient of resistivity.
  3. So the answer is (a).
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Q106JEE Main 2003 · AIEEEEasy

Ohm's Law, Resistivity & Temperature · Stretching & reshaping a wire

The length of a given cylindrical wire is increased by 100%100\%. Due to the consequent decrease in diameter the change in the resistance of the wire will be

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Answer: (d)

  1. Length increased by 100%100\%: l′=2ll' = 2l.
  2. At constant volume R∝l2R \propto l^2, so R′=4RR' = 4R.
  3. Change =4R−RR×100=300%= \dfrac{4R - R}{R} \times 100 = 300\%
  4. So the answer is (d) 300%300\%.
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Q107NEET 2002 · AIPMTEasy

Ohm's Law, Resistivity & Temperature · Resistivity: R = ρl/A

Specific resistance of a conductor increases with

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Answer: (a)

  1. Specific resistance (resistivity) depends only on the material and its temperature, not on length or area.
  2. For a conductor ρT=ρ0(1+αΔT)\rho_T = \rho_0(1 + \alpha \Delta T) with α>0\alpha > 0, so ρ\rho rises with temperature.
  3. So the answer is (a) increase in temperature.
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Q108NEET 2001 · AIPMTEasy

Ohm's Law, Resistivity & Temperature · Temperature dependence of resistance

Copper and silicon is cooled from 300 K to 60 K, the specific resistance

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Answer: (a)

  1. Copper is a metal (α>0\alpha > 0): cooling lowers its resistivity.
  2. Silicon is a semiconductor (α<0\alpha < 0): cooling reduces free carriers, so resistivity increases.
  3. So the answer is (a) decrease in copper but increase in silicon.
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Q109NEET 1999 · AIPMTEasy

Ohm's Law, Resistivity & Temperature · Ohm's law & V–I graphs

The resistance of a discharge tube is

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Answer: (c)

  1. In a discharge tube current is carried by ionised gas; II is not proportional to VV (no current until breakdown, then current rises sharply).
  2. So V/IV/I is not constant: the resistance is non-ohmic, answer (c).
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Q110NEET 1997 · AIPMTEasy

Ohm's Law, Resistivity & Temperature · Ohm's law & V–I graphs

From the graph between current (I)(I) and voltage (V)(V) is shown below. Identify the portion corresponding to negative resistance

Figure

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Answer: (a)

  1. Resistance dVdI\dfrac{dV}{dI} is negative where current falls as voltage rises (negative slope of the II–VV graph).
  2. Along CD the curve goes down as VV increases; AB and DE rise and BC is around the peak.
  3. So the answer is (a) CDCD.
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Q111NEET 1997 · AIPMTMedium

Ohm's Law, Resistivity & Temperature · Resistivity: R = ρl/A

Three copper wires of lengths and cross-sectional areas are (l,A), (2l,A/2)(l, A),\ (2l, A/2) and (l/2,2A)(l/2, 2A). Resistance is minimum in

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Answer: (a)

  1. R=ρl/A∝l/AR = \rho l/A \propto l/A.
  2. (l,A):l/A(l, A): l/A; (2l,A/2):4 l/A(2l, A/2): 4\,l/A; (l/2,2A):14 l/A(l/2, 2A): \tfrac14\,l/A.
  3. Minimum for the wire of area 2A2A. The answer is (a).
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Q112NEET 1995 · AIPMTEasy

Ohm's Law, Resistivity & Temperature · Resistivity: R = ρl/A

In good conductors of electricity, the type of bonding that exists is

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Answer: (a)

  1. Good conductors (metals) have metallic bonding: positive ions in a sea of free electrons.
  2. These free electrons carry the current.
  3. So the answer is (a) metallic.
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Q113NEET 1994 · AIPMTEasy

Ohm's Law, Resistivity & Temperature · Resistivity: R = ρl/A

A wire 50 cm50\text{ cm} long and 1 mm21\text{ mm}^{2} in cross-section carries a current of 4 A4\text{ A} when connected to a 2 V2\text{ V} battery. The resistivity of the wire is

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Answer: (b)

  1. R=VI=24=0.5 ΩR = \dfrac{V}{I} = \dfrac{2}{4} = 0.5\ \Omega
  2. ρ=RAl=0.5×1×10−60.5\rho = \dfrac{RA}{l} = \dfrac{0.5 \times 1 \times 10^{-6}}{0.5}
  3. ρ=1×10−6 Ω m\rho = 1 \times 10^{-6}\ \Omega\text{ m}
  4. So the answer is (b) 1×10−6 Ω m1 \times 10^{-6}\ \Omega\text{ m}.
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Q114NEET 1994 · AIPMTEasy

Ohm's Law, Resistivity & Temperature · Resistivity: R = ρl/A

Identify the set in which all the three materials are good conductors of electricity

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Answer: (c)

  1. NaCl (solid) is an ionic insulator; Ge and Si are semiconductors; diamond is an insulator.
  2. Cu, Ag and Au are all metals with many free electrons.
  3. So the answer is (c) Cu, Ag and Au.
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Q115NEET 1988 · AIPMTMedium

Ohm's Law, Resistivity & Temperature · Resistivity: R = ρl/A

The masses of the wires of copper is in the ratio of 1:3:51 : 3 : 5 and their lengths are in the ratio of 5:3:15 : 3 : 1. The ratio of their electrical resistance is

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Answer: (d)

  1. Mass m=dlA⇒A=m/(dl)m = dlA \Rightarrow A = m/(dl), so R=ρl/A=ρd l2/m∝l2/mR = \rho l/A = \rho d\,l^2/m \propto l^2/m.
  2. R1:R2:R3=251:93:15=25:3:0.2R_1 : R_2 : R_3 = \dfrac{25}{1} : \dfrac{9}{3} : \dfrac{1}{5} = 25 : 3 : 0.2
  3. Multiply by 5: 125:15:1125 : 15 : 1. The answer is (d).
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Q116MHT CET 2026 · 17 APR · SHIFT 2Medium

Combination of Resistors · Wires bent into loops & polygons

A current of 9A enters point P of an equilateral triangle PQR having three wires of 3 Ω3\ \Omega each and leaves by point R. The currents I1I_1 and I2I_2 are respectively

Figure

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Answer: (b)

  1. From P to R: direct path PR (3 Ω3\ \Omega) carries I2I_2; path PQR (6 Ω6\ \Omega) carries I1I_1.
  2. Current divides inversely as resistance: I1=9×39=3I_1 = 9 \times \dfrac{3}{9} = 3 A, I2=9×69=6I_2 = 9 \times \dfrac{6}{9} = 6 A
  3. So the answer is (b) 3 A, 6 A.
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Q117JEE Main 2026 · 23 JAN · SHIFT 1Medium

Combination of Resistors · Wires bent into loops & polygons

A wire of uniform resistance λ Ω/m\lambda\ \Omega/\text{m} is bent into a circle of radius rr and another piece of wire with length 2r2r is connected between points AA and B(AOB)B(\text{AOB}) as shown in figure. The equivalent resistance between points AA and BB is ‾ Ω\underline{\qquad}\ \Omega Ω\Omega.

Figure

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Answer: (b)

  1. A and B are 90∘90^\circ apart, so three paths join A and B in parallel.
  2. Minor arc: πrλ2\dfrac{\pi r\lambda}{2}; major arc: 3πrλ2\dfrac{3\pi r\lambda}{2}; straight wire AOB: 2rλ2r\lambda.
  3. 1R=1rλ(2π+23π+12)=1rλ⋅16+3π6π\dfrac{1}{R} = \dfrac{1}{r\lambda}\left(\dfrac{2}{\pi} + \dfrac{2}{3\pi} + \dfrac{1}{2}\right) = \dfrac{1}{r\lambda} \cdot \dfrac{16 + 3\pi}{6\pi}
  4. R=6πλr3π+16R = \dfrac{6\pi\lambda r}{3\pi + 16}
  5. So the answer is (b).
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Q118MHT CET 2026 · 19 MAY · SHIFT 1Medium

Combination of Resistors · Networks, symmetry & infinite ladders

In the following circuit, the current in the 4 Ω4\ \Omega resistance is 1.21.2 A. The total current (II) flowing in the circuit is

Figure

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Answer: (c)

  1. The 4 Ω4\ \Omega and 3 Ω3\ \Omega are in parallel, so they have the same voltage.
  2. V=I1R=1.2×4=4.8V = I_1 R = 1.2 \times 4 = 4.8 V
  3. I2=4.83=1.6I_2 = \dfrac{4.8}{3} = 1.6 A
  4. I=I1+I2=1.2+1.6=2.8I = I_1 + I_2 = 1.2 + 1.6 = 2.8 A. So the answer is (c) 2.82.8 A.
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Q119JEE Main 2026 · 21 JAN · SHIFT 2Medium

Combination of Resistors · Networks, symmetry & infinite ladders

Two known resistances of R ΩR\ \Omega and 2R Ω2R\ \Omega and one unknown resistance X ΩX\ \Omega are connected in a circuit as shown in the figure. If the equivalent resistance between points AA and BB in the circuit is X ΩX\ \Omega, then the value of XX is __________ Ω\Omega.

Figure

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Answer: (c)

  1. (2R+X)(2R + X) is in parallel with RR: X=R(2R+X)3R+XX = \dfrac{R(2R + X)}{3R + X}
  2. 3RX+X2=2R2+RX⇒X2+2RX−2R2=03RX + X^2 = 2R^2 + RX \Rightarrow X^2 + 2RX - 2R^2 = 0
  3. X=−R+R2+2R2=(3−1)RX = -R + \sqrt{R^2 + 2R^2} = (\sqrt{3} - 1)R
  4. So the answer is (c) (3−1)R(\sqrt{3}-1)R.
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Q120JEE Main 2026 · 28 JAN · SHIFT 1Medium

Combination of Resistors · Networks, symmetry & infinite ladders

The equivalent resistance between the points AA and BB in the following circuit is x5Ω\frac{x}{5}\Omega. The value of xx is ______.

Figure

Numerical answer type. Work it out, then open the answer.

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Answer: 21

  1. Left wire puts the bottom-left at A; right wire puts the top-right at B. So: A–M 66, M–B 33, A–N 33, N–B 66, M–N 3 Ω3\ \Omega (unbalanced bridge).
  2. Take VA=1V_A = 1, VB=0V_B = 0. KCL at M: 5VM−2VN=15V_M - 2V_N = 1; at N: 5VN−2VM=25V_N - 2V_M = 2. So VM=37V_M = \tfrac37, VN=47V_N = \tfrac47.
  3. Current from A: 1−3/76+1−4/73=221+321=521\dfrac{1 - 3/7}{6} + \dfrac{1 - 4/7}{3} = \dfrac{2}{21} + \dfrac{3}{21} = \dfrac{5}{21}
  4. RAB=215 ΩR_{AB} = \dfrac{21}{5}\ \Omega, so x=21x = 21.
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Q121JEE Main 2026 · 24 JAN · SHIFT 2Difficult

Combination of Resistors · Wires bent into loops & polygons

A regular hexagon is formed by six wires each of resistance r Ωr\ \Omega and the corners are joined to the centre by wires of same resistance. If the current enters at one corner and leaves at the opposite corner, the equivalent resistance of the hexagon between the two opposite corners will be

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Answer: (b)

  1. Number corners 1 to 6, current in at 1, out at 4. Take V1=1V_1 = 1, V4=0V_4 = 0; by symmetry centre =12= \tfrac12, V2=V6=xV_2 = V_6 = x, V3=V5=1−xV_3 = V_5 = 1 - x.
  2. KCL at 2: (x−1)+(x−(1−x))+(x−12)=0⇒x=58(x - 1) + (x - (1-x)) + (x - \tfrac12) = 0 \Rightarrow x = \tfrac58
  3. Current from 1 (in units of 1/r1/r): 2(1−58)+12=542\left(1 - \tfrac58\right) + \tfrac12 = \tfrac54
  4. Req=15/4 r=45rR_{eq} = \dfrac{1}{5/4}\,r = \dfrac{4}{5}r. So the answer is (b).
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Q122NEET 2025Easy

Combination of Resistors · Series & parallel combinations

A wire of resistance RR is cut into 8 equal pieces. From these pieces two equivalent resistances are made by adding four of these together in parallel. Then these two sets are added in series. The net effective resistance of the combination is:

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Answer: (a)

  1. Each piece has resistance R/8R/8.
  2. Four in parallel: R/84=R32\dfrac{R/8}{4} = \dfrac{R}{32}.
  3. Two such sets in series: 2×R32=R162\times\dfrac{R}{32} = \dfrac{R}{16}. The answer is (a).
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Q123JEE Main 2025 · 24 JAN · SHIFT 1Easy

Combination of Resistors · Wires bent into loops & polygons

A wire of resistance 9 Ω9\ \Omega is bent to form an equilateral triangle. Then the equivalent resistance across any two vertices will be _________ ohm.

Numerical answer type. Work it out, then open the answer.

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Answer: 2

  1. Each side has 93=3 Ω\dfrac{9}{3} = 3\ \Omega.
  2. Between two vertices: 3 Ω3\ \Omega in parallel with 6 Ω6\ \Omega.
  3. Req=3×69=2 ΩR_{eq} = \dfrac{3 \times 6}{9} = 2\ \Omega
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Q124JEE Main 2025 · 4 APR · SHIFT 2Medium

Combination of Resistors · Series & parallel combinations

From the combination of resistors with resistances values R1=R2=R3=5 ΩR_1 = R_2 = R_3 = 5\ \Omega and R4=10 ΩR_4 = 10\ \Omega, which of the following combination is the best circuit to get an equivalent resistance of 6 Ω6\ \Omega?

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Answer: (b)

  1. (a) all parallel: 1R=35+110⇒R≈1.43 Ω\dfrac{1}{R} = \dfrac{3}{5} + \dfrac{1}{10} \Rightarrow R \approx 1.43\ \Omega.
  2. (b) (R1+R2)∥(R3+R4)=10∥15=15025=6 Ω(R_1 + R_2) \parallel (R_3 + R_4) = 10 \parallel 15 = \dfrac{150}{25} = 6\ \Omega
  3. (c) 5∥10∥10=2.5 Ω5 \parallel 10 \parallel 10 = 2.5\ \Omega; (d) 20∥5=4 Ω20 \parallel 5 = 4\ \Omega.
  4. So the answer is (b).
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Q125JEE Main 2025 · 7 APR · SHIFT 1Medium

Combination of Resistors · Wires bent into loops & polygons

A wire of resistance RR is bent into a triangular pyramid as shown in figure with each segment having same length. The resistance between points AA and BB is R/nR/n. The value of nn is:

Figure

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Answer: (c)

  1. The pyramid (tetrahedron) has 6 equal edges, each of resistance r=R6r = \dfrac{R}{6}.
  2. By symmetry the other two vertices are at the same potential, so the edge joining them carries no current.
  3. Paths: direct rr, plus two paths of 2r2r each, all in parallel: 1RAB=1r+12r+12r=2r\dfrac{1}{R_{AB}} = \dfrac{1}{r} + \dfrac{1}{2r} + \dfrac{1}{2r} = \dfrac{2}{r}
  4. RAB=r2=R12R_{AB} = \dfrac{r}{2} = \dfrac{R}{12}, so n=12n = 12.
  5. So the answer is (c) 12.
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Q126JEE Main 2025 · 3 APR · SHIFT 1Medium

Combination of Resistors · Wires bent into loops & polygons

A wire of length 25 m and cross-sectional area 5 mm25\text{ mm}^{2} having resistivity of 2×10−6 Ω m2 \times 10^{-6}\ \Omega\text{ m} is bent into a complete circle. The resistance between diametrically opposite points will be

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Answer: (b)

  1. R=ρlA=2×10−6×255×10−6=10 ΩR = \dfrac{\rho l}{A} = \dfrac{2 \times 10^{-6} \times 25}{5 \times 10^{-6}} = 10\ \Omega
  2. Diametrically opposite points: Req=R4=2.5 ΩR_{eq} = \dfrac{R}{4} = 2.5\ \Omega
  3. So the answer is (b) 2.5 Ω2.5\ \Omega.
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Q127JEE Main 2025 · 28 JAN · SHIFT 1Medium

Combination of Resistors · Wires bent into loops & polygons

A wire of resistance R is bent into an equilateral triangle and an identical wire is bent into a square. The ratio of resistance between the two end points of an edge of the triangle to that of the square is

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Answer: (c)

  1. Triangle edge: R3∥2R3=2R9\dfrac{R}{3} \parallel \dfrac{2R}{3} = \dfrac{2R}{9}.
  2. Square edge: R4∥3R4=3R16\dfrac{R}{4} \parallel \dfrac{3R}{4} = \dfrac{3R}{16}.
  3. Ratio =2/93/16=3227= \dfrac{2/9}{3/16} = \dfrac{32}{27}
  4. So the answer is (c) 32/2732/27.
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Q128MHT CET 2025 · 26 APR · SHIFT 2Medium

Combination of Resistors · Networks, symmetry & infinite ladders

In the following circuit, current through ACB if each resistance R=4 ΩR = 4\ \Omega is

Figure

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Answer: (b)

  1. Bridge ratio: RACRCB=RADRDB=1\dfrac{R_{AC}}{R_{CB}} = \dfrac{R_{AD}}{R_{DB}} = 1, so it is a balanced Wheatstone bridge and no current flows in CD.
  2. Path ACB is then simply R+R=8 ΩR + R = 8\ \Omega across the 1616 V battery.
  3. IACB=168=2I_{ACB} = \dfrac{16}{8} = 2 A. So the answer is (b) 2 A.
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Q129MHT CET 2025 · 25 APR · SHIFT 1Medium

Combination of Resistors · Networks, symmetry & infinite ladders

The current (I) drawn from the battery in the given circuit is

Figure

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Answer: (b)

  1. Bridge A–B–C–D: RABRBD=510\dfrac{R_{AB}}{R_{BD}} = \dfrac{5}{10} and RACRCD=1020\dfrac{R_{AC}}{R_{CD}} = \dfrac{10}{20}, equal, so the bridge is balanced and the 10 Ω10\ \Omega between B and C carries no current.
  2. Arm ABD: 5+10=15 Ω5 + 10 = 15\ \Omega; arm ACD: 10+20=30 Ω10 + 20 = 30\ \Omega.
  3. Req=15×3045=10 ΩR_{eq} = \dfrac{15 \times 30}{45} = 10\ \Omega
  4. I=510=0.5I = \dfrac{5}{10} = 0.5 A. So the answer is (b) 0.5 A.
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Q130MHT CET 2025 · 22 APR · SHIFT 2Medium

Combination of Resistors · Networks, symmetry & infinite ladders

In the given circuit, current flowing through the circuit is

Figure

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Answer: (c)

  1. Going round the loop, the 1010 V and 200200 V cells oppose each other (their positive terminals face opposite ways).
  2. Net emf =200−10=190= 200 - 10 = 190 V
  3. I=19038=5I = \dfrac{190}{38} = 5 A. So the answer is (c) 5 A.
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Q131MHT CET 2025 · 20 APR · SHIFT 2Medium

Combination of Resistors · Networks, symmetry & infinite ladders

The reading of the ammeter in the following circuit is

Figure

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Answer: (b)

  1. Series LCR: VL=VC=100V_L = V_C = 100 V are opposite in phase, so they cancel (resonance).
  2. The whole supply voltage appears across RR: VR=300V_R = 300 V.
  3. I=VRR=30050=6I = \dfrac{V_R}{R} = \dfrac{300}{50} = 6 A. So the answer is (b) 6 A.
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Q132JEE Main 2025 · 28 JAN · SHIFT 1Medium

Combination of Resistors · Networks, symmetry & infinite ladders

Find the equivalent resistance between two ends of the following circuit

Figure

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Answer: (d)

  1. The top wire joins the left end to the node between the 2nd and 3rd resistors; the bottom wire joins the node between the 1st and 2nd resistors to the right end.
  2. So each of the three r3\tfrac{r}{3} resistors is connected directly between the two ends: all three are in parallel.
  3. Req=r/33=r9R_{eq} = \dfrac{r/3}{3} = \dfrac{r}{9}. So the answer is (d).
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Q133JEE Main 2024 · 27 JAN · SHIFT 1Easy

Combination of Resistors · Series & parallel combinations

A wire of resistance R\text{R} and length L\text{L} is cut into 5 equal parts. If these parts are joined parallely, then resultant resistance will be:

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Answer: (a)

  1. Each part has resistance R5\dfrac{R}{5} (since R∝lR \propto l).
  2. 5 equal resistors in parallel: Req=R/55=R25R_{eq} = \dfrac{R/5}{5} = \dfrac{R}{25}
  3. So the answer is (a) R25\dfrac{R}{25}.
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Q134NEET 2024Medium

Combination of Resistors · Series & parallel combinations

A wire of length 'll' and resistance 100 Ω100\ \Omega is divided into 10 equal parts. The first 5 parts are connected in series while the next 5 parts are connected in parallel. The two combinations are again connected in series. The resistance of this final combination is:

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Answer: (b)

  1. Each part has 10010=10 Ω\dfrac{100}{10} = 10\ \Omega.
  2. 5 parts in series: 50 Ω50\ \Omega. 5 parts in parallel: 105=2 Ω\dfrac{10}{5} = 2\ \Omega.
  3. Total in series: 50+2=52 Ω50 + 2 = 52\ \Omega
  4. So the answer is (b) 52 Ω52\ \Omega.
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Q135JEE Main 2024 · 5 APR · SHIFT 2Medium

Combination of Resistors · Series & parallel combinations

A wire of resistance 20 Ω20\ \Omega is divided into 10 equal parts, resulting pairs. A combination of two parts are connected in parallel and so on. Now resulting pairs of parallel combination are connected in series. The equivalent resistance of final combination is _________ Ω\Omega.

Numerical answer type. Work it out, then open the answer.

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Answer: 5

  1. Each part is 2010=2 Ω\dfrac{20}{10} = 2\ \Omega.
  2. Each pair in parallel: 22=1 Ω\dfrac{2}{2} = 1\ \Omega; there are 5 pairs.
  3. 5 pairs in series: 5×1=5 Ω5 \times 1 = 5\ \Omega
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Q136NEET 2024Medium

Combination of Resistors · Wires bent into loops & polygons

A uniform metal wire of length ll has 10 Ω10\ \Omega resistance. Now this wire is stretched to a length 2l2l and then bent to form a perfect circle. The equivalent resistance across any arbitrary diameter of that circle is

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Answer: (a)

  1. Stretching to 2l2l (volume constant): R∝l2R \propto l^2, so R′=4×10=40 ΩR' = 4 \times 10 = 40\ \Omega.
  2. Across a diameter: two halves of 20 Ω20\ \Omega in parallel.
  3. Req=202=10 ΩR_{eq} = \dfrac{20}{2} = 10\ \Omega
  4. So the answer is (a) 10 Ω10\ \Omega.
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Q137MHT CET 2024 · 10 MAY · SHIFT 1Medium

Combination of Resistors · Networks, symmetry & infinite ladders

In the circuit, current flowing through the circuit is

Figure

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Answer: (d)

  1. The 55 V and 100100 V cells oppose each other in the loop.
  2. Net emf =100−5=95= 100 - 5 = 95 V
  3. I=9519=5I = \dfrac{95}{19} = 5 A. So the answer is (d) 5 A.
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Q138JEE Main 2024 · 9 APR · SHIFT 2Medium

Combination of Resistors · Networks, symmetry & infinite ladders

The effective resistance between AA and BB, if resistance of each resistor is RR, will be:

Figure

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Answer: (a)

  1. Between the left and right corners of the diamond, the top, centre and bottom nodes are each joined by RR to both corners, so all three are at the same potential.
  2. The two vertical resistors therefore carry no current; three paths of 2R2R remain in parallel: 2R3\dfrac{2R}{3}.
  3. RAB=R+2R3+R=8R3R_{AB} = R + \dfrac{2R}{3} + R = \dfrac{8R}{3}. So the answer is (a).
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Q139JEE Main 2024 · 9 APR · SHIFT 1Medium

Combination of Resistors · Networks, symmetry & infinite ladders

The equivalent resistance between A and B is:

Figure

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Answer: (c)

  1. A goes to the 6 Ω6\ \Omega, B to the 8 Ω8\ \Omega. The vertical middle wire makes the top-middle and bottom-middle points one node X; the horizontal middle wire (it jumps over the vertical one) makes the two side midpoints one node Y.
  2. Between X and Y: 10+5=1510 + 5 = 15, 7+8=157 + 8 = 15 and 11+4=15 Ω11 + 4 = 15\ \Omega, all in parallel =5 Ω= 5\ \Omega.
  3. RAB=6+5+8=19 ΩR_{AB} = 6 + 5 + 8 = 19\ \Omega. So the answer is (c) 19 Ω19\ \Omega.
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Q140JEE Main 2024 · 5 APR · SHIFT 1Medium

Combination of Resistors · Networks, symmetry & infinite ladders

In the given figure R1=10 Ω,R2=8 Ω,R3=4 ΩR_1 = 10\ \Omega, R_2 = 8\ \Omega, R_3 = 4\ \Omega and R4=8 ΩR_4 = 8\ \Omega. Battery is ideal with emf 12 V12\text{ V}. Equivalent resistance of the circuit and current supplied by battery are respectively:

Figure

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Answer: (d)

  1. R2R_2, R4R_4 and R3R_3 all connect the same top node to the same bottom wire, so they are in parallel.
  2. 1Rp=18+18+14=12⇒Rp=2 Ω\dfrac{1}{R_p} = \dfrac18 + \dfrac18 + \dfrac14 = \dfrac12 \Rightarrow R_p = 2\ \Omega
  3. Req=R1+Rp=10+2=12 ΩR_{eq} = R_1 + R_p = 10 + 2 = 12\ \Omega, I=1212=1I = \dfrac{12}{12} = 1 A
  4. So the answer is (d) 12 Ω12\ \Omega and 11 A.
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Q141JEE Main 2024 · 31 JAN · SHIFT 1Medium

Combination of Resistors · Networks, symmetry & infinite ladders

Equivalent resistance of the following network is __________ Ω\Omega.

Figure

Numerical answer type. Work it out, then open the answer.

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Answer: 1

  1. The left wire shorts R1R_1, so the left end of R2R_2 is at the bottom rail (B).
  2. R2∥R3=2∥2=1 ΩR_2 \parallel R_3 = 2 \parallel 2 = 1\ \Omega; with R4R_4 in series: 3 Ω3\ \Omega.
  3. Across A–B: 3∥R5∥R6=3∥3∥3=1 Ω3 \parallel R_5 \parallel R_6 = 3 \parallel 3 \parallel 3 = 1\ \Omega.
  4. Equivalent resistance =1 Ω= 1\ \Omega.
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Q142JEE Main 2024 · 4 APR · SHIFT 1Difficult

Combination of Resistors · Wires bent into loops & polygons

Twelve wires each having resistance 2 Ω2\ \Omega are joined to form a cube. A battery of 6 V6\text{ V} emf is joined across point aa and cc. The voltage difference between ee and ff is ________ V.

Figure

Numerical answer type. Work it out, then open the answer.

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Answer: 1

  1. Battery across face diagonal a–c: take Va=6V_a = 6, Vc=0V_c = 0. By symmetry b, d, e, g are all at 33 V; let Vh=yV_h = y and Vf=6−yV_f = 6 - y.
  2. KCL at h (neighbours a, e, g; equal resistances): (y−6)+(y−3)+(y−3)=0⇒y=4(y - 6) + (y - 3) + (y - 3) = 0 \Rightarrow y = 4 V, so Vf=2V_f = 2 V.
  3. Ve−Vf=3−2=1V_e - V_f = 3 - 2 = 1 V
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Q143NEET 2023Easy

Combination of Resistors · Series & parallel combinations

10 resistors, each of resistance R\text{R} are connected in series to a battery of emf EE and negligible internal resistance. Then those are connected in parallel to the same battery, the current is increased nn times. The value of nn is:

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Answer: (a)

  1. Series: Rs=10RR_s = 10R, Is=E/10RI_s = E/10R.
  2. Parallel: Rp=R/10R_p = R/10, Ip=10E/RI_p = 10E/R.
  3. n=Ip/Is=100n = I_p/I_s = 100. The answer is (a).
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Q144JEE Main 2023 · 15 APR · SHIFT 1Easy

Combination of Resistors · Series & parallel combinations

Given below are two statements:

Statement I: The equivalent resistance of resistors in a series combination is smaller than least resistance used in the combination.

Statement II: The resistivity of the material is independent of temperature.

In the light of the above statements, choose the correct answer from the options given below:

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Answer: (c)

  1. Statement I: in series Req=R1+R2+…R_{eq} = R_1 + R_2 + \dots, which is larger than the largest resistor, so I is false.
  2. Statement II: resistivity changes with temperature, ρ=ρ0(1+αΔT)\rho = \rho_0(1 + \alpha\Delta T), so II is false.
  3. So the answer is (c) both statements are false.
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Q145JEE Main 2023 · 10 APR · SHIFT 1Easy

Combination of Resistors · Series & parallel combinations

10 resistors each of resistance 10 Ω\Omega can be connected in such as to get maximum and minimum equivalent resistance. The ratio of maximum and minimum equivalent resistance will be ___________.

Numerical answer type. Work it out, then open the answer.

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Answer: 100

  1. Maximum (all in series): Rmax=10×10=100 ΩR_{max} = 10 \times 10 = 100\ \Omega.
  2. Minimum (all in parallel): Rmin=1010=1 ΩR_{min} = \dfrac{10}{10} = 1\ \Omega.
  3. Ratio =1001=100= \dfrac{100}{1} = 100
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Q146JEE Main 2023 · 1 FEB · SHIFT 2Medium

Combination of Resistors · Wires bent into loops & polygons

Equivalent resistance between the adjacent corners of a regular n-sided polygon of uniform wire of resistance R would be:

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Answer: (a)

  1. Each side has Rn\dfrac{R}{n}; the rest of the polygon has (n−1)Rn\dfrac{(n-1)R}{n}.
  2. Req=Rn⋅(n−1)RnR=(n−1)Rn2R_{eq} = \dfrac{\frac{R}{n} \cdot \frac{(n-1)R}{n}}{R} = \dfrac{(n-1)R}{n^2}
  3. So the answer is (a).
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Q147JEE Main 2023 · 10 APR · SHIFT 1Medium

Combination of Resistors · Networks, symmetry & infinite ladders

The equivalent resistance of the circuit shown below between points a and b is:

Figure

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Answer: (c)

  1. By symmetry the top vertex and the centre are both midway in potential between a and b (each joined by 4 Ω4\ \Omega to a and to b), so the vertical 4 Ω4\ \Omega carries no current.
  2. Paths: a–top–b =8 Ω= 8\ \Omega, a–centre–b =8 Ω= 8\ \Omega, direct 16 Ω16\ \Omega, all in parallel.
  3. 1R=18+18+116=516⇒R=3.2 Ω\dfrac{1}{R} = \dfrac18 + \dfrac18 + \dfrac1{16} = \dfrac{5}{16} \Rightarrow R = 3.2\ \Omega
  4. So the answer is (c) 3.2 Ω3.2\ \Omega.
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Q148JEE Main 2023 · 8 APR · SHIFT 2Medium

Combination of Resistors · Networks, symmetry & infinite ladders

The equivalent resistance between A and B as shown in figure is:

Figure

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Answer: (c)

  1. The top wire joins A to D, and the bottom wire joins C to B.
  2. So 20 kΩ20\text{ k}\Omega (A–C), 20 kΩ20\text{ k}\Omega (C–D) and 10 kΩ10\text{ k}\Omega (D–B) are each directly between A and B: all in parallel.
  3. 1R=120+120+110=15⇒R=5 kΩ\dfrac{1}{R} = \dfrac1{20} + \dfrac1{20} + \dfrac1{10} = \dfrac15 \Rightarrow R = 5\text{ k}\Omega
  4. So the answer is (c) 5 kΩ5\text{ k}\Omega.
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Q149JEE Main 2023 · 1 FEB · SHIFT 1Medium

Combination of Resistors · Networks, symmetry & infinite ladders

The equivalent resistance between AA and BB of the network shown in figure;

Figure

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Answer: (b)

  1. Redraw as a Wheatstone bridge: arms RR and 3R3R on one side, 2R2R and 6R6R on the other, with 9R9R as the bridge.
  2. R3R=2R6R\dfrac{R}{3R} = \dfrac{2R}{6R}, so the bridge is balanced and 9R9R carries no current.
  3. RAB=(R+3R)∥(2R+6R)=4R×8R12R=8R3R_{AB} = (R + 3R) \parallel (2R + 6R) = \dfrac{4R \times 8R}{12R} = \dfrac{8R}{3}
  4. So the answer is (b) 83R\dfrac{8}{3}R.
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Q150JEE Main 2023 · 30 JAN · SHIFT 2Medium

Combination of Resistors · Networks, symmetry & infinite ladders

The equivalent resistance between AA and BB is _________.

Figure

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Answer: (b)

  1. The left vertical wire makes both left corners one node (A); the right vertical wire makes both right corners one node (B). Every branch then runs from A to B.
  2. Branches: top 1.5+0.5=21.5 + 0.5 = 2, 1212, diagonal 8+6=148 + 6 = 14, diagonal 2+4=62 + 4 = 6, 66, bottom 1+1=2 Ω1 + 1 = 2\ \Omega, all in parallel.
  3. 1R=12+112+114+16+16+12=12584⇒R≈0.67 Ω\dfrac{1}{R} = \dfrac12 + \dfrac1{12} + \dfrac1{14} + \dfrac16 + \dfrac16 + \dfrac12 = \dfrac{125}{84} \Rightarrow R \approx 0.67\ \Omega
  4. This is closest to (b) 23 Ω\dfrac{2}{3}\ \Omega.
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Q151JEE Main 2023 · 25 JAN · SHIFT 1Medium

Combination of Resistors · Networks, symmetry & infinite ladders

In the given circuit, the equivalent resistance between the terminal A and B is __________ Ω\Omega.

Figure

Numerical answer type. Work it out, then open the answer.

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Answer: 10

  1. The thick vertical wire joins the two right-hand dots, so the 4 Ω4\ \Omega loop is shorted and the node after the top 2 Ω2\ \Omega is at the bottom-rail potential; the second vertical 2 Ω2\ \Omega is shorted too.
  2. From the node after 3 Ω3\ \Omega to the bottom rail: 2 Ω2\ \Omega (vertical) ∥\parallel 2 Ω2\ \Omega (top) =1 Ω= 1\ \Omega.
  3. RAB=3+1+6=10 ΩR_{AB} = 3 + 1 + 6 = 10\ \Omega
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Q152JEE Main 2022 · 24 JUN · SHIFT 2Easy

Combination of Resistors · Series & parallel combinations

What will be the most suitable combination of three resistors A = 2Ω\Omega, B = 4Ω\Omega, C = 6Ω\Omega so that (223)Ω\left(\frac{22}{3}\right)\Omega is equivalent resistance of combination?

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Answer: (b)

  1. Try (b): A parallel B gives 2×42+4=43 Ω\dfrac{2 \times 4}{2 + 4} = \dfrac{4}{3}\ \Omega.
  2. In series with C: 43+6=223 Ω\dfrac{4}{3} + 6 = \dfrac{22}{3}\ \Omega
  3. So the answer is (b).
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Q153JEE Main 2022 · 29 JUL · SHIFT 1Medium

Combination of Resistors · Series & parallel combinations

Two metallic wires of identical dimensions are connected in series. If σ1\sigma_1 and σ2\sigma_2 are the conductivities of these wires respectively, the effective conductivity of the combination is:

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Answer: (b)

  1. R=lσAR = \dfrac{l}{\sigma A}. In series the combination has length 2l2l and area AA.
  2. 2σ=1σ1+1σ2⇒σ=2σ1σ2σ1+σ2\dfrac{2}{\sigma} = \dfrac{1}{\sigma_1} + \dfrac{1}{\sigma_2} \Rightarrow \sigma = \dfrac{2\sigma_1\sigma_2}{\sigma_1 + \sigma_2}
  3. So the answer is (b).
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Q154JEE Main 2022 · 28 JUL · SHIFT 2Medium

Combination of Resistors · Series & parallel combinations

Given below are two statements:

Statement I: A uniform wire of resistance 80 Ω80\ \Omega is cut into four equal parts. These parts are now connected in parallel. The equivalent resistance of the combination will be 5 Ω5\ \Omega.

Statement II: Two resistances 2R and 3R are connected in parallel in an electric circuit. The value of thermal energy developed in 3R and 2R will be in the ratio 3:23 : 2.

In the light of the above statements, choose the most appropriate answer from the option given below

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Answer: (c)

  1. Statement I: each part is 20 Ω20\ \Omega; four in parallel give 204=5 Ω\dfrac{20}{4} = 5\ \Omega. True.
  2. Statement II: in parallel VV is the same, so H=V2tR∝1RH = \dfrac{V^2 t}{R} \propto \dfrac{1}{R}.
  3. H3R:H2R=13:12=2:3H_{3R} : H_{2R} = \dfrac{1}{3} : \dfrac{1}{2} = 2 : 3, not 3:23 : 2. False.
  4. So the answer is (c).
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Q155JEE Main 2022 · 30 JUN · SHIFT 1Medium

Combination of Resistors · Series & parallel combinations

An electric cable of copper has just one wire of radius 9 mm. Its resistance is 14 Ω\Omega. If this single copper wire of the cable is replaced by seven identical well insulated copper wires each of radius 3 mm connected in parallel, then the new resistance of the combination will be:

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Answer: (b)

  1. Same length, R∝1r2R \propto \dfrac{1}{r^2}.
  2. One wire of radius 3 mm: R′=14×(93)2=126 ΩR' = 14 \times \left(\dfrac{9}{3}\right)^2 = 126\ \Omega.
  3. Seven in parallel: 1267=18 Ω\dfrac{126}{7} = 18\ \Omega
  4. So the answer is (b) 18 Ω18\ \Omega.
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Q156JEE Main 2022 · 25 JUL · SHIFT 1Medium

Combination of Resistors · Series & parallel combinations

Eight copper wire of length ll and diameter dd are joined in parallel to form a single composite conductor of resistance RR. If a single copper wire of length 2l2l have the same resistance (R)(R) then its diameter will be ____________ d.

Numerical answer type. Work it out, then open the answer.

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Answer: 4

  1. One wire: r0=4ρlπd2r_0 = \dfrac{4\rho l}{\pi d^2}; eight in parallel: R=r08=ρl2πd2R = \dfrac{r_0}{8} = \dfrac{\rho l}{2\pi d^2}.
  2. Single wire of length 2l2l, diameter DD: R=8ρlπD2R = \dfrac{8\rho l}{\pi D^2}.
  3. 8D2=12d2⇒D2=16d2⇒D=4d\dfrac{8}{D^2} = \dfrac{1}{2d^2} \Rightarrow D^2 = 16d^2 \Rightarrow D = 4d
  4. Answer: 4
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Q157NEET 2022 · PHASE 2Medium

Combination of Resistors · Networks, symmetry & infinite ladders

The equivalent resistance of the infinite network given below is:

Figure

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Answer: (d)

  1. Each section has 1 Ω1\ \Omega in each rail and a 1 Ω1\ \Omega rung. Removing one section leaves the same network, so R=1+1+1⋅R1+RR = 1 + 1 + \dfrac{1 \cdot R}{1 + R}.
  2. R(1+R)=2(1+R)+R⇒R2−2R−2=0R(1+R) = 2(1+R) + R \Rightarrow R^2 - 2R - 2 = 0
  3. R=2+4+82=1+3R = \dfrac{2 + \sqrt{4+8}}{2} = 1 + \sqrt{3}
  4. So the answer is (d) (1+3) Ω(1+\sqrt{3})\ \Omega.
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Q158JEE Main 2022 · 26 JUN · SHIFT 2Medium

Combination of Resistors · Networks, symmetry & infinite ladders

The equivalent resistance between points A and B in the given network is:

Figure

Show answer and solution

Answer: (c)

  1. Reduce from the far left: the two 5 Ω5\ \Omega in series =10 Ω= 10\ \Omega, parallel with the 10 Ω10\ \Omega to A gives 5 Ω5\ \Omega.
  2. Next 5 Ω5\ \Omega in series gives 1010, parallel with the next 10 Ω10\ \Omega to A gives 5 Ω5\ \Omega again. The same happens at every step.
  3. Last step: 5+5=10 Ω5 + 5 = 10\ \Omega in parallel with the 10 Ω10\ \Omega from A to B =5 Ω= 5\ \Omega.
  4. So the answer is (c) 5 Ω5\ \Omega.
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Q159JEE Main 2022 · 27 JUN · SHIFT 2Difficult

Combination of Resistors · Networks, symmetry & infinite ladders

In the given circuit 'a' is an arbitrary constant. The value of m for which the equivalent circuit resistance is minimum, will be x2\sqrt{\frac{x}{2}}. The value of x is __________.

Figure

Numerical answer type. Work it out, then open the answer.

Show answer and solution

Answer: 3

  1. Three mama in parallel give ma3\frac{ma}{3}; two am\frac{a}{m} in parallel give a2m\frac{a}{2m}.
  2. Req=ma3+a2mR_{eq} = \frac{ma}{3} + \frac{a}{2m}
  3. For minimum: dReqdm=a3−a2m2=0⇒m2=32\frac{dR_{eq}}{dm} = \frac{a}{3} - \frac{a}{2m^2} = 0 \Rightarrow m^2 = \frac{3}{2}
  4. m=32=x2⇒x=3m = \sqrt{\frac{3}{2}} = \sqrt{\frac{x}{2}} \Rightarrow x = 3
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Q160NEET 2021Easy

Combination of Resistors · Series & parallel combinations

The effective resistance of a parallel connection that consists of four wires of equal length, equal area of cross-section and same material is 0.25Ω\Omega. What will be the effective resistance if they are connected in series?

Show answer and solution

Answer: (a)

  1. Identical wires, each rr: parallel r/4=0.25 Ω⇒r=1 Ωr/4 = 0.25\ \Omega \Rightarrow r = 1\ \Omega.
  2. Series: 4r=4 Ω4r = 4\ \Omega. The answer is (a).
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Q161JEE Main 2021 · 26 AUG · SHIFT 2Easy

Combination of Resistors · Networks, symmetry & infinite ladders

If you are provided a set of resistances 2Ω\Omega, 4Ω\Omega, 6Ω\Omega and 8Ω\Omega. Connect these resistances so as to obtain an equivalent resistance of 463Ω\frac{46}{3}\Omega.

Show answer and solution

Answer: (d)

  1. Try (d): 2 Ω2\ \Omega parallel 4 Ω4\ \Omega gives 2×46=43 Ω\dfrac{2 \times 4}{6} = \dfrac{4}{3}\ \Omega.
  2. Add 6 Ω6\ \Omega and 8 Ω8\ \Omega in series: 43+14=463 Ω\dfrac{4}{3} + 14 = \dfrac{46}{3}\ \Omega
  3. So the answer is (d).
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Q162JEE Main 2021 · 1 SEP · SHIFT 2Medium

Combination of Resistors · Series & parallel combinations

Two resistors R1 = (4 ±\pm 0.8) Ω\Omega and R2 = (4 ±\pm 0.4) Ω\Omega are connected in parallel. The equivalent resistance of their parallel combination will be:

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Answer: (c)

  1. Rp=R1R2R1+R2=168=2 ΩR_p = \dfrac{R_1 R_2}{R_1 + R_2} = \dfrac{16}{8} = 2\ \Omega.
  2. From 1Rp=1R1+1R2\dfrac{1}{R_p} = \dfrac{1}{R_1} + \dfrac{1}{R_2}: ΔRpRp2=ΔR1R12+ΔR2R22\dfrac{\Delta R_p}{R_p^2} = \dfrac{\Delta R_1}{R_1^2} + \dfrac{\Delta R_2}{R_2^2}
  3. ΔRp=4(0.816+0.416)=0.3 Ω\Delta R_p = 4\left(\dfrac{0.8}{16} + \dfrac{0.4}{16}\right) = 0.3\ \Omega
  4. So the answer is (c) (2±0.3) Ω(2 \pm 0.3)\ \Omega.
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Q163JEE Main 2021 · 22 JUL · SHIFT 2Medium

Combination of Resistors · Series & parallel combinations

A Copper (Cu) rod of length 25 cm and cross-sectional area 3 mm2 is joined with a similar Aluminium (Al) rod as shown in figure. Find the resistance of the combination between the ends A and B.

(Take Resistivity of Copper = 1.7 ×\times 10−-8 Ω\Omegam and Resistivity of Aluminium = 2.6 ×\times 10−-8 Ω\Omegam)

Figure

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Answer: (c)

  1. The rods are in parallel. R=ρlAR = \dfrac{\rho l}{A} with l=0.25l = 0.25 m, A=3×10−6 m2A = 3 \times 10^{-6}\ \text{m}^2.
  2. RCu=1.7×10−8×0.253×10−6=1.417 mΩR_{Cu} = \dfrac{1.7 \times 10^{-8} \times 0.25}{3 \times 10^{-6}} = 1.417\ \text{m}\Omega, RAl=2.167 mΩR_{Al} = 2.167\ \text{m}\Omega
  3. RAB=1.417×2.1671.417+2.167≈0.858 mΩR_{AB} = \dfrac{1.417 \times 2.167}{1.417 + 2.167} \approx 0.858\ \text{m}\Omega
  4. So the answer is (c) 0.858 mΩ0.858\ \text{m}\Omega.
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Q164JEE Main 2021 · 18 MAR · SHIFT 2Medium

Combination of Resistors · Series & parallel combinations

Two wires of same length and thickness having specific resistances 6Ω\Omega cm and 3Ω\Omega cm respectively are connected in parallel. The effective resistivity is ρ Ω\rho\ \Omega cm. The value of ρ\rho, to the nearest integer, is ____________.

Numerical answer type. Work it out, then open the answer.

Show answer and solution

Answer: 4

  1. Same ll and AA, so R∝ρR \propto \rho. Parallel combination has length ll and area 2A2A.
  2. ρeffl2A=lA⋅ρ1ρ2ρ1+ρ2⇒ρeff=2ρ1ρ2ρ1+ρ2\dfrac{\rho_{eff} l}{2A} = \dfrac{l}{A} \cdot \dfrac{\rho_1\rho_2}{\rho_1 + \rho_2} \Rightarrow \rho_{eff} = \dfrac{2\rho_1\rho_2}{\rho_1 + \rho_2}
  3. ρeff=2×6×39=4 Ω cm\rho_{eff} = \dfrac{2 \times 6 \times 3}{9} = 4\ \Omega\text{ cm}
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Q165JEE Main 2021 · 17 MAR · SHIFT 1Medium

Combination of Resistors · Series & parallel combinations

The equivalent resistance of series combination of two resistors is 's'. When they are connected in parallel, the equivalent resistance is 'p'. If s = np, then the minimum value for n is ____________. (Round off to the Nearest Integer)

Numerical answer type. Work it out, then open the answer.

Show answer and solution

Answer: 4

  1. n=sp=(R1+R2)2R1R2n = \dfrac{s}{p} = \dfrac{(R_1 + R_2)^2}{R_1 R_2}.
  2. Since (R1+R2)2≥4R1R2(R_1 + R_2)^2 \ge 4R_1R_2, n≥4n \ge 4, with equality for R1=R2R_1 = R_2.
  3. Minimum n=4n = 4
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Q166JEE Main 2021 · 31 AUG · SHIFT 1Medium

Combination of Resistors · Wires bent into loops & polygons

A square shaped wire with resistance of each side 3Ω\Omega is bent to form a complete circle. The resistance between two diametrically opposite points of the circle in unit of Ω\Omega will be ___________.

Numerical answer type. Work it out, then open the answer.

Show answer and solution

Answer: 3

  1. Total resistance of the wire =4×3=12 Ω= 4 \times 3 = 12\ \Omega.
  2. Diametrically opposite points: two halves of 6 Ω6\ \Omega in parallel.
  3. R=124=3 ΩR = \dfrac{12}{4} = 3\ \Omega
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Q167JEE Main 2021 · 25 FEB · SHIFT 2Medium

Combination of Resistors · Wires bent into loops & polygons

A current of 6A enters one corner P of an equilateral triangle PQR having 3 wires of resistance 2Ω\Omega each and leaves by the corner R. The currents i1 in ampere is _________.

Figure

Numerical answer type. Work it out, then open the answer.

Show answer and solution

Answer: 2

  1. From P to R: path PQR (2+2=4 Ω2 + 2 = 4\ \Omega) carries i1i_1; direct PR (2 Ω2\ \Omega) carries i2i_2.
  2. Current divides inversely as resistance: i1=6×24+2=2i_1 = 6 \times \dfrac{2}{4 + 2} = 2 A
  3. i1=2i_1 = 2 A
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Q168MHT CET 2021 · 23 SEP · SHIFT 2Medium

Combination of Resistors · Networks, symmetry & infinite ladders

In the given circuit, the current in 8Ω\Omega resistance is 1.5 A. The total current (I) flowing in the circuit is

Figure

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Answer: (d)

  1. 8 Ω8\ \Omega and 3 Ω3\ \Omega are in parallel, so same voltage: V=1.5×8=12V = 1.5 \times 8 = 12 V.
  2. I2=123=4I_2 = \dfrac{12}{3} = 4 A
  3. I=1.5+4=5.5I = 1.5 + 4 = 5.5 A. So the answer is (d) 5.5 A.
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Q169NEET 2021Medium

Combination of Resistors · Networks, symmetry & infinite ladders

Three resistors having resistances r1, r2 and r3 are connected as shown in the given circuit. The ratio i3i1\frac{i_3}{i_1} of currents in terms of resistances used in the circuit is:

Figure

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Answer: (c)

  1. i1i_1 through r1r_1 divides between r2r_2 and r3r_3 in parallel.
  2. Current divider: i3=i1r2r2+r3i_3 = i_1 \dfrac{r_2}{r_2 + r_3}
  3. i3i1=r2r2+r3\dfrac{i_3}{i_1} = \dfrac{r_2}{r_2 + r_3}. So the answer is (c).
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Q170JEE Main 2021 · 31 AUG · SHIFT 2Medium

Combination of Resistors · Networks, symmetry & infinite ladders

The equivalent resistance of the given circuit between the terminals A and B is:

Figure

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Answer: (d)

  1. The left vertical wire shorts the 5 Ω5\ \Omega, so the top-left corner is at B's potential.
  2. The middle node joins B through 2 Ω2\ \Omega (vertical) and 2 Ω2\ \Omega (top-left), giving 1 Ω1\ \Omega; with the next 2 Ω2\ \Omega in series: 3 Ω3\ \Omega.
  3. This 3 Ω3\ \Omega is in parallel with 3∥3=1.5 Ω3 \parallel 3 = 1.5\ \Omega: RAB=3×1.54.5=1 ΩR_{AB} = \dfrac{3 \times 1.5}{4.5} = 1\ \Omega.
  4. So the answer is (d) 1 Ω1\ \Omega.
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Q171JEE Main 2021 · 26 FEB · SHIFT 1Medium

Combination of Resistors · Networks, symmetry & infinite ladders

Five equal resistances are connected in a network as shown in figure. The net resistance between the points A and B is:

Figure

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Answer: (d)

  1. A is the centre (joined to D and E by R) and B is the corner C. Paths A–D–C and A–E–C are each 2R2R.
  2. By symmetry D and E are at the same potential, so the resistor DE carries no current.
  3. RAB=2R∥2R=RR_{AB} = 2R \parallel 2R = R. So the answer is (d) R.
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Q172JEE Main 2021 · 27 AUG · SHIFT 2Medium

Combination of Resistors · Networks, symmetry & infinite ladders

The ratio of the equivalent resistance of the network (shown in figure) between the points a and b when switch is open and switch is closed is x: 8. The value of x is ___________.

Figure

Numerical answer type. Work it out, then open the answer.

Show answer and solution

Answer: 9

  1. Switch open: two paths each R+2R=3RR + 2R = 3R in parallel: Ropen=1.5RR_{open} = 1.5R.
  2. Switch closed: (R∥2R)+(2R∥R)=2R3+2R3=4R3(R \parallel 2R) + (2R \parallel R) = \dfrac{2R}{3} + \dfrac{2R}{3} = \dfrac{4R}{3}
  3. RopenRclosed=1.5R4R/3=98\dfrac{R_{open}}{R_{closed}} = \dfrac{1.5R}{4R/3} = \dfrac{9}{8}, so x=9x = 9.
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Q173JEE Main 2021 · 20 JUL · SHIFT 2Medium

Combination of Resistors · Networks, symmetry & infinite ladders

In the given figure switches S1 and S2 are in open condition. The resistance across ab when the switches S1 and S2 are closed is _____________ Ω\Omega.

Figure

Numerical answer type. Work it out, then open the answer.

Show answer and solution

Answer: 10

  1. With S1S_1 and S2S_2 closed, the top and bottom rails are joined at both switch points, giving three parallel pairs in series.
  2. 12∥6=412 \parallel 6 = 4, 4∥4=24 \parallel 4 = 2, 6∥12=4 Ω6 \parallel 12 = 4\ \Omega
  3. Rab=4+2+4=10 ΩR_{ab} = 4 + 2 + 4 = 10\ \Omega
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Q174NEET 2020 · PHASE 2Medium

Combination of Resistors · Networks, symmetry & infinite ladders

The equivalent resistance between AA and BB for the mesh shown in the figure is

Figure

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Answer: (c)

  1. The upper 4 Ω4\ \Omega and 8 Ω8\ \Omega meet at the apex, so they are in series: 12 Ω12\ \Omega.
  2. This 12 Ω12\ \Omega is in parallel with the 6 Ω6\ \Omega: 12×618=4 Ω\dfrac{12 \times 6}{18} = 4\ \Omega.
  3. In series with the lower 4 Ω4\ \Omega and 8 Ω8\ \Omega: RAB=4+4+8=16 ΩR_{AB} = 4 + 4 + 8 = 16\ \Omega.
  4. So the answer is (c) 16 Ω16\ \Omega.
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Q175MHT CET 2020 · 16 OCT · SHIFT 2Difficult

Combination of Resistors · Networks, symmetry & infinite ladders

A circular coil of radius RR has a resistance of 40 Ω40\ \Omega. Figure shows two points PP and QQ on the circumference separated by a distance πR2\frac{\pi R}{2} which are connected to a 16 V battery with internal resistance of 0.5 Ω0.5\ \Omega. What is the value of current II flowing through the circuit?

Figure

Show answer and solution

Answer: (b)

  1. Resistance is proportional to arc length. From the figure the two arcs between PP and QQ are 30 Ω30\ \Omega and 10 Ω10\ \Omega.
  2. They are in parallel: RPQ=30×1030+10=7.5 ΩR_{PQ} = \frac{30 \times 10}{30 + 10} = 7.5\ \Omega
  3. Total resistance =7.5+0.5=8 Ω= 7.5 + 0.5 = 8\ \Omega
  4. I=168=2 AI = \frac{16}{8} = 2\text{ A}. So the answer is (b) 2 A.
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Q176JEE Main 2019 · 10 JAN · SHIFT 1Easy

Combination of Resistors · Wires bent into loops & polygons

A uniform metallic wire has a resistance of 18 Ω\Omega and is bent into an equilateral triangle. Then, the resistance between any two vertices of the triangle is -

Show answer and solution

Answer: (c)

  1. Each side has 183=6 Ω\dfrac{18}{3} = 6\ \Omega.
  2. Between two vertices: 6 Ω6\ \Omega in parallel with 12 Ω12\ \Omega.
  3. Req=6×1218=4 ΩR_{eq} = \dfrac{6 \times 12}{18} = 4\ \Omega
  4. So the answer is (c) 4 Ω4\ \Omega.
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Q177JEE Main 2019 · 9 APR · SHIFT 2Medium

Combination of Resistors · Wires bent into loops & polygons

A metal wire of resistance 3 Ω\Omega is elongated to make a uniform wire of double its previous length. This new wire is now bent and the ends joined to make a circle. If two points on this circle make an angle 60° at the centre, the equivalent resistance between these two points will be :-

Show answer and solution

Answer: (d)

  1. Doubling length (volume constant): R′=4×3=12 ΩR' = 4 \times 3 = 12\ \Omega.
  2. A 60∘60^\circ arc is 16\dfrac{1}{6} of the circle: 2 Ω2\ \Omega; the rest is 10 Ω10\ \Omega.
  3. Req=2×1012=53 ΩR_{eq} = \dfrac{2 \times 10}{12} = \dfrac{5}{3}\ \Omega
  4. So the answer is (d) 5/3 Ω5/3\ \Omega.
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Q178JEE Main 2019 · 9 APR · SHIFT 1Medium

Combination of Resistors · Wires bent into loops & polygons

A wire of resistance R is bent to form a square ABCD as shown in the figure. The effective resistance between E and C is: (E is mid-point of arm CD)

Figure

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Answer: (a)

  1. Each side is R4\dfrac{R}{4}, so EC =R8= \dfrac{R}{8} and the rest of the loop (E→D→A→B→C) is 7R8\dfrac{7R}{8}.
  2. REC=R8⋅7R8R=7R64R_{EC} = \dfrac{\frac{R}{8} \cdot \frac{7R}{8}}{R} = \dfrac{7R}{64}
  3. So the answer is (a) 764R\dfrac{7}{64}R.
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Q179JEE Main 2019 · 12 APR · SHIFT 1Medium

Combination of Resistors · Networks, symmetry & infinite ladders

The resistive network shown below is connected to a D.C. source of 16 V. The power consumed by the network is 4 Watt. The value of R is:

Figure

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Answer: (c)

  1. 4R∥4R=2R4R \parallel 4R = 2R and 6R∥12R=4R6R \parallel 12R = 4R.
  2. Req=2R+R+4R+R=8RR_{eq} = 2R + R + 4R + R = 8R
  3. P=ε2Req⇒4=2568R⇒R=8 ΩP = \dfrac{\varepsilon^2}{R_{eq}} \Rightarrow 4 = \dfrac{256}{8R} \Rightarrow R = 8\ \Omega
  4. So the answer is (c) 8 Ω8\ \Omega.
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Q180JEE Main 2019 · 8 APR · SHIFT 2Medium

Combination of Resistors · Networks, symmetry & infinite ladders

In the figure shown, what is the current (in Ampere) drawn from the battery?

You are given: R1 = 15Ω\Omega, R2 = 10 Ω\Omega, R3 = 20 Ω\Omega, R4 = 5Ω\Omega, R5 = 25Ω\Omega, R6 = 30 Ω\Omega, E = 15 V

Figure

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Answer: (c)

  1. R3R_3, R4R_4, R5R_5 are in series: 20+5+25=50 Ω20 + 5 + 25 = 50\ \Omega, which is in parallel with R2R_2: 50×1060=253 Ω\dfrac{50 \times 10}{60} = \dfrac{25}{3}\ \Omega.
  2. Req=R1+253+R6=15+253+30=1603 ΩR_{eq} = R_1 + \dfrac{25}{3} + R_6 = 15 + \dfrac{25}{3} + 30 = \dfrac{160}{3}\ \Omega
  3. I=15160/3=932I = \dfrac{15}{160/3} = \dfrac{9}{32} A. So the answer is (c).
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Q181JEE Main 2018 · 15 APR · SHIFT 1Medium

Combination of Resistors · Networks, symmetry & infinite ladders

In the given circuit all resistances are of value R ohmR\text{ ohm} each. The equivalent resistance between AA and BB is:

Figure

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Answer: (a)

  1. Find the plain wires: the middle horizontal wire, the central vertical wire and the lower-left horizontal wire join the left-middle, centre, bottom-left and bottom-centre dots into one node.
  2. So the resistor from A and the resistor from B both end on this same node; the rest of the network is attached only to that one node, so it carries no current.
  3. Only the two series resistors at A and at B remain: RAB=R+R=2RR_{AB} = R + R = 2R.
  4. So the answer is (a) 2R2R.
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Q182JEE Main 2017 · 8 APR · SHIFT 1Difficult

Combination of Resistors · Networks, symmetry & infinite ladders

Figure

A 9 V battery with internal resistance of 0.5 Ω\Omega is connected across an infinite network as shown in the figure. All ammeters A1, A2,3 and voltmeter V are ideal.

Choose correct statement.

Show answer and solution

Answer: (a)

  1. Each section of the ladder is 1 Ω+1 Ω1\ \Omega + 1\ \Omega in series with a 4 Ω4\ \Omega shunt. For an infinite ladder, adding one section does not change R∞R_\infty.
  2. R∞=2+4R∞4+R∞⇒R∞2−2R∞−8=0⇒R∞=4 ΩR_\infty = 2 + \frac{4R_\infty}{4 + R_\infty} \Rightarrow R_\infty^2 - 2R_\infty - 8 = 0 \Rightarrow R_\infty = 4\ \Omega
  3. I=94+0.5=2 AI = \frac{9}{4 + 0.5} = 2\text{ A}, so A1A_1 reads 2 A.
  4. Voltmeter reads terminal voltage V=9−2×0.5=8 VV = 9 - 2 \times 0.5 = 8\text{ V}, so (c) and (d) are wrong.
  5. So the answer is (a) Reading of A1A_1 is 2 A.
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Q183NEET 2015 · AIPMTMedium

Combination of Resistors · Series & parallel combinations

Two metal wires of identical dimensions are connected in series. If σ\sigma1 and σ\sigma2 are their conductivities, the effective conductivity of the combination is

Show answer and solution

Answer: (c)

  1. R=lσAR = \dfrac{l}{\sigma A}. In series the combination has length 2l2l and area AA.
  2. 2lσA=lσ1A+lσ2A⇒2σ=1σ1+1σ2\dfrac{2l}{\sigma A} = \dfrac{l}{\sigma_1 A} + \dfrac{l}{\sigma_2 A} \Rightarrow \dfrac{2}{\sigma} = \dfrac{1}{\sigma_1} + \dfrac{1}{\sigma_2}
  3. σ=2σ1σ2σ1+σ2\sigma = \dfrac{2\sigma_1\sigma_2}{\sigma_1 + \sigma_2}
  4. So the answer is (c).
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Q184NEET 2013Medium

Combination of Resistors · Wires bent into loops & polygons

A 12 cm wire is given a shape of a right angled triangle ABC having sides 3 cm, 4 cm and 5 cm as shown in the figure. The resistance between two ends (AB, BC, CA) of the respective sides are measured one by one. Their ratio is

Figure

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Answer: (b)

  1. Let resistance per cm be rr. Across a side of length xx: xx is in parallel with 12−x12 - x.
  2. RAB=3×912rR_{AB} = \dfrac{3 \times 9}{12}r, RBC=4×812rR_{BC} = \dfrac{4 \times 8}{12}r, RCA=5×712rR_{CA} = \dfrac{5 \times 7}{12}r
  3. Ratio =27:32:35= 27 : 32 : 35
  4. So the answer is (b).
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Q185JEE Main 2013 · 25 APR · SHIFT 1Medium

Combination of Resistors · Networks, symmetry & infinite ladders

Six equal resistances are connected between points P, Q and R as shown in the figure. Then net resistance will be maximum between:

Figure

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Answer: (b)

PQRrr/3r/2
  1. Side PQ =r= r, side PR =r3= \dfrac{r}{3} (three in parallel), side QR =r2= \dfrac{r}{2} (two in parallel). Total =11r6= \dfrac{11r}{6}.
  2. For a triangle, RXY=(side XY)(rest)totalR_{XY} = \dfrac{(\text{side } XY)(\text{rest})}{\text{total}}.
  3. RPQ=r⋅5r611r6=5r11R_{PQ} = \dfrac{r \cdot \frac{5r}{6}}{\frac{11r}{6}} = \dfrac{5r}{11}, RQR=r2⋅4r311r6=4r11R_{QR} = \dfrac{\frac r2 \cdot \frac{4r}{3}}{\frac{11r}{6}} = \dfrac{4r}{11}, RPR=r3⋅3r211r6=3r11R_{PR} = \dfrac{\frac r3 \cdot \frac{3r}{2}}{\frac{11r}{6}} = \dfrac{3r}{11}
  4. Maximum is between P and Q. So the answer is (b).
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Q186JEE Main 2013 · 9 APR · SHIFT 1Difficult

Combination of Resistors · Wires bent into loops & polygons

A letter 'A' is constructed of a uniform wire with resistance 1.0Ω\Omega per cm. The sides of the letter are 20 cm and the cross piece in the middle is 10 cm long. The apex angle is 60°. The resistance between the ends of the legs is close to:

Show answer and solution

Answer: (d)

  1. With apex angle 60∘60^\circ, the 1010 cm cross piece joins the midpoints of the 2020 cm legs. Each half-leg =10 Ω= 10\ \Omega, cross piece =10 Ω= 10\ \Omega.
  2. Between the midpoints: cross piece 10 Ω10\ \Omega in parallel with the upper halves 10+10=20 Ω10 + 10 = 20\ \Omega: 10×2030=203 Ω\dfrac{10 \times 20}{30} = \dfrac{20}{3}\ \Omega.
  3. R=10+203+10=803≈26.7 ΩR = 10 + \dfrac{20}{3} + 10 = \dfrac{80}{3} \approx 26.7\ \Omega
  4. So the answer is (d) 26.7 Ω26.7\ \Omega.
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Q187NEET 2012 · AIPMT · PRELIMSMedium

Combination of Resistors · Wires bent into loops & polygons

A ring is made of a wire having a resistance R0 = 12 Ω\Omega. Find the points A and B, as shown in the figure, at which a current carrying conductor should be connected so that the resistance R of the sub circuit between these point is equal to 83Ω\frac{8}{3}\Omega.

Figure

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Answer: (d)

  1. Let the arcs be R1R_1 and R2R_2: R1+R2=12R_1 + R_2 = 12 and R1R212=83\dfrac{R_1 R_2}{12} = \dfrac{8}{3}, so R1R2=32R_1R_2 = 32.
  2. Solving: R1=4 ΩR_1 = 4\ \Omega, R2=8 ΩR_2 = 8\ \Omega.
  3. R∝lR \propto l, so l1l2=48=12\dfrac{l_1}{l_2} = \dfrac{4}{8} = \dfrac{1}{2}
  4. So the answer is (d).
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Q188NEET 2009 · AIPMTMedium

Combination of Resistors · Wires bent into loops & polygons

A wire of resistance 12 ohms per meter is bent to form a complete circle of radius 10 cm. The resistance between its two diametrically opposite points, A and B as shown in the figure is

Figure

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Answer: (d)

  1. Length =2πr=2π×0.1=0.2π= 2\pi r = 2\pi \times 0.1 = 0.2\pi m, so total R=12×0.2π=2.4π ΩR = 12 \times 0.2\pi = 2.4\pi\ \Omega.
  2. Diametrically opposite points: two halves in parallel, RAB=R4R_{AB} = \dfrac{R}{4}.
  3. RAB=2.4π4=0.6π ΩR_{AB} = \dfrac{2.4\pi}{4} = 0.6\pi\ \Omega
  4. So the answer is (d) 0.6π Ω0.6\pi\ \Omega.
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Q189NEET 2005 · AIPMTEasy

Combination of Resistors · Wires bent into loops & polygons

When a wire of uniform cross-section aa length ll and resistance R is bent into a complete circle. resistance between any two of diametrically opposite points will be

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Answer: (a)

  1. Diametrically opposite points split the circle into two halves of R2\dfrac{R}{2} each.
  2. The halves are in parallel: Req=(R/2)(R/2)R=R4R_{eq} = \dfrac{(R/2)(R/2)}{R} = \dfrac{R}{4}
  3. So the answer is (a) R/4R/4.
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Q190NEET 2004 · AIPMTEasy

Combination of Resistors · Series & parallel combinations

Resistance n, each of r ohm, when connected in parallel give an equivalent resistance of R ohm. If these resistances were connected in series, the combination would have a resistance in ohms, equal to

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Answer: (a)

  1. Parallel: R=r/n⇒r=nRR = r/n \Rightarrow r = nR.
  2. Series: Rs=nr=n2RR_s = nr = n^2R. The answer is (a).
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Q191JEE Main 2004 · AIEEEMedium

Combination of Resistors · Series & parallel combinations

The resistance of the series combination of two resistances is S.S. When they are joined in parallel the total resistance is P.P. If S=nPS = nP then the minimum possible value of nn is

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Answer: (c)

  1. S=R1+R2S = R_1 + R_2, P=R1R2R1+R2P = \dfrac{R_1 R_2}{R_1 + R_2}, so n=SP=(R1+R2)2R1R2n = \dfrac{S}{P} = \dfrac{(R_1 + R_2)^2}{R_1 R_2}.
  2. By AM–GM, (R1+R2)2≥4R1R2(R_1 + R_2)^2 \ge 4R_1R_2, so n≥4n \ge 4 (equality when R1=R2R_1 = R_2).
  3. So the answer is (c) 4.
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Q192JEE Main 2004 · AIEEEMedium

Combination of Resistors · Series & parallel combinations

An electric current is passed through a circuit containing two wires of the same material, connected in parallel. If the lengths and radii are in the ratio of 43\frac{4}{3} and 23\frac{2}{3}, then the ratio of the current passing through the wires will be

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Answer: (b)

  1. R=ρlπr2R = \dfrac{\rho l}{\pi r^2}, so R1R2=l1l2⋅r22r12=43×94=3\dfrac{R_1}{R_2} = \dfrac{l_1}{l_2} \cdot \dfrac{r_2^2}{r_1^2} = \dfrac{4}{3} \times \dfrac{9}{4} = 3.
  2. In parallel I∝1RI \propto \dfrac{1}{R}: I1I2=R2R1=13\dfrac{I_1}{I_2} = \dfrac{R_2}{R_1} = \dfrac{1}{3}
  3. So the answer is (b) 1/31/3.
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Q193NEET 2004 · AIPMTMedium

Combination of Resistors · Networks, symmetry & infinite ladders

Five equal resistances each of resistances R are connected as shown in the figure. A battery of V volts is connected between A and B. The current flowing in AFCEB will be

Figure

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Answer: (c)

  1. A is joined to F and B to E. Between F and E there are two paths, F–C–E and F–D–E, each 2R2R, with CD as a bridge.
  2. By symmetry C and D are at the same potential (balanced bridge), so CD carries no current.
  3. Each path gets voltage VV, so current in FCE =V2R= \dfrac{V}{2R}.
  4. So the answer is (c) V2R\dfrac{V}{2R}.
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Q194JEE Main 2004 · AIEEEMedium

Combination of Resistors · Networks, symmetry & infinite ladders

The total current supplied to the circuit by the battery is

Figure

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Answer: (a)

  1. The top arc is a plain wire, so the top-left and top-right junctions are one node T; the battery is between T and the lower-left junction L.
  2. 2 Ω2\ \Omega and 6 Ω6\ \Omega both join T to the centre: 2∥6=1.5 Ω2 \parallel 6 = 1.5\ \Omega; with 1.5 Ω1.5\ \Omega to L in series: 3 Ω3\ \Omega.
  3. This 3 Ω3\ \Omega is in parallel with the 3 Ω3\ \Omega on the right arc (T to L): Req=1.5 ΩR_{eq} = 1.5\ \Omega.
  4. I=61.5=4I = \dfrac{6}{1.5} = 4 A. So the answer is (a) 4 A.
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Q195NEET 2000 · AIPMTMedium

Combination of Resistors · Networks, symmetry & infinite ladders

The net resistance of the circuit between A and B is

Figure

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Answer: (b)

  1. Wheatstone bridge check: 34=68\dfrac{3}{4} = \dfrac{6}{8}, balanced, so the 7 Ω7\ \Omega carries no current.
  2. Upper arm 3+4=7 Ω3 + 4 = 7\ \Omega, lower arm 6+8=14 Ω6 + 8 = 14\ \Omega.
  3. RAB=7×1421=143 ΩR_{AB} = \dfrac{7 \times 14}{21} = \dfrac{14}{3}\ \Omega. So the answer is (b).
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Q196NEET 1996 · AIPMTMedium

Combination of Resistors · Networks, symmetry & infinite ladders

What will be the equivalent resistance between the two points AA and DD?

Figure

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Answer: (a)

  1. B and C are open ends, so the end resistors there carry no current.
  2. From the junction after R1R_1 to the junction before R4R_4: path R2+R3=20 ΩR_2 + R_3 = 20\ \Omega and path R6+R5=20 ΩR_6 + R_5 = 20\ \Omega in parallel =10 Ω= 10\ \Omega.
  3. RAD=R1+10+R4=10+10+10=30 ΩR_{AD} = R_1 + 10 + R_4 = 10 + 10 + 10 = 30\ \Omega
  4. So the answer is (a) 30 Ω30\ \Omega.
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Q197NEET 1995 · AIPMTEasy

Combination of Resistors · Series & parallel combinations

Two wires of the same metal have same length, but their cross-sections are in the ratio 3:13 : 1. They are joined in series. The resistance of thicker wire is 10 Ω10\ \Omega. The total resistance of the combination will be

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Answer: (a)

  1. Same metal and length, so R∝1/AR \propto 1/A.
  2. Thicker wire 10 Ω10\ \Omega; thinner (area 1/31/3) has 30 Ω30\ \Omega.
  3. Series: 10+30=40 Ω10 + 30 = 40\ \Omega. The answer is (a).
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Q198NEET 1995 · AIPMTMedium

Combination of Resistors · Networks, symmetry & infinite ladders

In the network shown in the figure, each of the resistance is equal to 2 Ω2\ \Omega. The resistance between the points AA and BB is

Figure

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Answer: (d)

  1. A is the centre and B is a corner. From A to B there are two routes through the other two corners, each 2+2=4 Ω2 + 2 = 4\ \Omega.
  2. By symmetry those two corners are at equal potential, so the 2 Ω2\ \Omega side joining them carries no current.
  3. RAB=4∥4=2 ΩR_{AB} = 4 \parallel 4 = 2\ \Omega. So the answer is (d) 2 Ω2\ \Omega.
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Q199NEET 1994 · AIPMTMedium

Combination of Resistors · Wires bent into loops & polygons

Six resistors of 3 Ω3\ \Omega each are connected along the sides of a hexagon and three resistors of 6 Ω6\ \Omega each are connected along AC,ADAC, AD and AEAE as shown in the figure. The equivalent resistance between AA and BB is equal to

Figure

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Answer: (a)

  1. Start from F: AF+FE=6 ΩAF + FE = 6\ \Omega, in parallel with AE=6 ΩAE = 6\ \Omega gives 3 Ω3\ \Omega between A and E.
  2. Add EDED: 3+3=63 + 3 = 6, parallel with AD=6AD = 6 gives 3 Ω3\ \Omega between A and D.
  3. Add DCDC: 3+3=63 + 3 = 6, parallel with AC=6AC = 6 gives 3 Ω3\ \Omega between A and C.
  4. Add CBCB: 3+3=63 + 3 = 6, parallel with AB=3AB = 3: 6×39=2 Ω\dfrac{6 \times 3}{9} = 2\ \Omega
  5. So the answer is (a) 2 Ω2\ \Omega.
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Q200NEET 1993 · AIPMTEasy

Combination of Resistors · Wires bent into loops & polygons

Three resistances each of 4 Ω4\ \Omega are connected to form a triangle. The resistance between any two terminals is

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Answer: (d)

  1. Between two corners: one side (4 Ω4\ \Omega) is in parallel with the other two sides in series (8 Ω8\ \Omega).
  2. Req=4×84+8=83 ΩR_{eq} = \dfrac{4 \times 8}{4 + 8} = \dfrac{8}{3}\ \Omega
  3. So the answer is (d) 8/3 Ω8/3\ \Omega.
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Q201NEET 1990 · AIPMTMedium

Combination of Resistors · Series & parallel combinations

You are given several identical resistances each of value R=10 ΩR = 10\ \Omega and each capable of carrying a maximum current of one ampere. It is required to make a suitable combination of these resistances of 5 Ω5\ \Omega which can carry a current of 4 ampere. The minimum number of resistances of the type RR that will be required for this job is

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Answer: (c)

10 Ω10 Ω4 branches × 20 Ω → 5 Ω, 1 A each4 A
  1. Each resistor carries at most 1 A, so 4 A needs 4 parallel branches carrying 1 A each.
  2. 4 equal branches give 5 Ω5\ \Omega if each branch is 4×5=20 Ω4 \times 5 = 20\ \Omega, i.e. two 10 Ω10\ \Omega in series.
  3. Number of resistors =4×2=8= 4 \times 2 = 8
  4. So the answer is (c) 8.
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Q202NEET 1990 · AIPMTMedium

Combination of Resistors · Networks, symmetry & infinite ladders

In the network shown in figure each resistance is 1 Ω1\ \Omega. The effective resistance between AA and BB is

Figure

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Answer: (d)

  1. Let VA=1V_A = 1, VB=0V_B = 0. By symmetry the bottom middle node is at 12\tfrac12; top-left node at xx, top-right at 1−x1-x.
  2. KCL at top-left node: (x−1)+(x−12)+(x−(1−x))=0⇒x=58(x-1) + (x-\tfrac12) + (x-(1-x)) = 0 \Rightarrow x = \tfrac58
  3. Current leaving A: (1−12)+(1−58)=78(1-\tfrac12) + (1-\tfrac58) = \tfrac78 A
  4. RAB=17/8=87 ΩR_{AB} = \dfrac{1}{7/8} = \dfrac{8}{7}\ \Omega. So the answer is (d).
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Q203NEET 1989 · AIPMTEasy

Combination of Resistors · Series & parallel combinations

nn equal resistors are first connected in series and then connected in parallel. What is the ratio of the maximum to the minimum resistance?

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Answer: (c)

  1. Series: Rmax=nRR_{max} = nR. Parallel: Rmin=RnR_{min} = \dfrac{R}{n}.
  2. RmaxRmin=nRR/n=n2\dfrac{R_{max}}{R_{min}} = \dfrac{nR}{R/n} = n^2
  3. So the answer is (c) n2n^2.
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Q204NEET 2026Easy

Cells: EMF, Internal Resistance & Combinations · Terminal voltage & internal resistance

A resistor is connected to a battery of 12 V emf and internal resistance 2 Ω2\ \Omega. If the current in the circuit is 0.6 A, the terminal voltage of the battery is:

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Answer: (d)

  1. Terminal voltage V=E−IrV = E - Ir
  2. V=12−0.6×2=12−1.2=10.8 VV = 12 - 0.6 \times 2 = 12 - 1.2 = 10.8\text{ V}
  3. So the answer is (d) 10.8 V.
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Q205JEE Main 2026 · 5 APR · SHIFT 2Easy

Cells: EMF, Internal Resistance & Combinations · Terminal voltage & internal resistance

When an external resistance of 5 Ω5\ \Omega is connected across terminals of a cell, a current of 0.25 A flows through it. When the 5 Ω5\ \Omega resistor is replaced by a 2 Ω2\ \Omega resistor, a current of 0.5 A flows through it. The internal resistance of the cell is ‾ Ω\underline{\hspace{3em}}\ \Omega.

Numerical answer type. Work it out, then open the answer.

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Answer: 1

  1. Same emf: E=0.25(5+r)=0.5(2+r)E = 0.25(5 + r) = 0.5(2 + r)
  2. 1.25+0.25r=1+0.5r⇒0.25r=0.251.25 + 0.25r = 1 + 0.5r \Rightarrow 0.25r = 0.25
  3. r=1 Ωr = 1\ \Omega
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Q206JEE Main 2026 · 21 JAN · SHIFT 2Easy

Cells: EMF, Internal Resistance & Combinations · Maximum power transfer

A battery with EMF EE and internal resistance rr is connected across a resistance RR. The power consumption in RR will be maximum when:

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Answer: (d)

  1. P=E2R(R+r)2P = \frac{E^2 R}{(R + r)^2}; setting dPdR=0\frac{dP}{dR} = 0 gives (R+r)−2R=0(R + r) - 2R = 0.
  2. So R=rR = r, and the answer is (d).
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Q207MHT CET 2026 · 15 MAY · SHIFT 2Medium

Cells: EMF, Internal Resistance & Combinations · Terminal voltage & internal resistance

In an electric circuit shown, each cell has e.m.f. 2 V2\text{ V} and internal resistance 1 Ω1\ \Omega. The external resistance is 2 Ω2\ \Omega. The value of current (in A) is

Figure

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Answer: (d)

  1. The figure shows three identical cells in series, all in the same direction.
  2. Enet=3×2=6 VE_{net} = 3 \times 2 = 6\text{ V}, rnet=3×1=3 Ωr_{net} = 3 \times 1 = 3\ \Omega
  3. I=62+3=1.2 AI = \frac{6}{2 + 3} = 1.2\text{ A}. So the answer is (d) 1.2.
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Q208JEE Main 2026 · 28 JAN · SHIFT 1Medium

Cells: EMF, Internal Resistance & Combinations · Cells in series & parallel

For the two cells having same EMF EE and internal resistance rr, the current passing through the external resistor 6 Ω6\ \Omega is same when both the cells are connected either in parallel or in series. The value of internal resistance rr is ‾ Ω\underline{\hspace{3em}}\ \Omega.

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Answer: (b)

  1. Series: I=2E6+2rI = \frac{2E}{6 + 2r}; parallel: I=E6+r/2I = \frac{E}{6 + r/2}
  2. Equal: 2(6+r2)=6+2r⇒12+r=6+2r2\left(6 + \frac{r}{2}\right) = 6 + 2r \Rightarrow 12 + r = 6 + 2r
  3. r=6 Ωr = 6\ \Omega. So the answer is (b) 6.
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Q209JEE Main 2026 · 6 APR · SHIFT 2Medium

Cells: EMF, Internal Resistance & Combinations · Cells in series & parallel

Two cells of emfs 1 V and 2 V and internal resistance 2 Ω2\ \Omega and 1 Ω1\ \Omega, respectively connected in parallel, gave a current of 1 A through an external resistance. If the polarity of one cell is reversed, then value of current through the external resistance will be α5 A\frac{\alpha}{5}\text{ A}. The value of α\alpha is ______.

Numerical answer type. Work it out, then open the answer.

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Answer: 3

  1. Parallel cells: Eeq=E1/r1+E2/r21/r1+1/r2E_{eq} = \dfrac{E_1/r_1 + E_2/r_2}{1/r_1 + 1/r_2}, req=r1r2r1+r2=23 Ωr_{eq} = \dfrac{r_1 r_2}{r_1 + r_2} = \dfrac{2}{3}\ \Omega.
  2. Same polarity: Eeq=1/2+2/13/2=53 VE_{eq} = \dfrac{1/2 + 2/1}{3/2} = \dfrac{5}{3}\text{ V}; 1=5/3R+2/3⇒R=1 Ω1 = \dfrac{5/3}{R + 2/3} \Rightarrow R = 1\ \Omega.
  3. One cell reversed: Eeq=−1/2+2/13/2=1 VE_{eq} = \dfrac{-1/2 + 2/1}{3/2} = 1\text{ V}.
  4. I=11+2/3=35 A⇒α=3I = \dfrac{1}{1 + 2/3} = \dfrac{3}{5}\text{ A} \Rightarrow \alpha = 3
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Q210JEE Main 2025 · 3 APR · SHIFT 2Easy

Cells: EMF, Internal Resistance & Combinations · Cells in series & parallel

Two cells of emfs 1 V and 2 V and internal resistances 2 Ω2\ \Omega and 1 Ω1\ \Omega, respectively, are connected in series with an external resistance of 6 Ω6\ \Omega. The total current in the circuit is I1I_1. Now the same two cells in parallel configuration are connected to same external resistance. In this case, the total current drawn is I2I_2. The value of (I1I2)\left(\frac{I_1}{I_2}\right) is x3\frac{x}{3}. The value of xx is___________.

Numerical answer type. Work it out, then open the answer.

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Answer: 4

  1. Series: I1=1+26+2+1=13 AI_1 = \frac{1 + 2}{6 + 2 + 1} = \frac{1}{3}\text{ A}
  2. Parallel: Eeq=E1/r1+E2/r21/r1+1/r2=0.5+21.5=53 VE_{eq} = \frac{E_1/r_1 + E_2/r_2}{1/r_1 + 1/r_2} = \frac{0.5 + 2}{1.5} = \frac{5}{3}\text{ V}, req=2×13=23 Ωr_{eq} = \frac{2 \times 1}{3} = \frac{2}{3}\ \Omega
  3. I2=5/36+2/3=5/320/3=14 AI_2 = \frac{5/3}{6 + 2/3} = \frac{5/3}{20/3} = \frac{1}{4}\text{ A}
  4. I1I2=43=x3⇒x=4\frac{I_1}{I_2} = \frac{4}{3} = \frac{x}{3} \Rightarrow x = 4
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Q211MHT CET 2025 · 19 APR · SHIFT 2Medium

Cells: EMF, Internal Resistance & Combinations · Cells in series & parallel

Two cells E1E_1 and E2E_2 having equal e.m.f ' EE ' and internal resistances r1r_1 and r2 (r1>r2)r_2\ (r_1 > r_2) respectively are connected in series. This combination is connected to an external resistance ' R '. It is observed that the potential difference across the cell E1E_1 becomes zero. The value of RR will be

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Answer: (a)

  1. I=2ER+r1+r2I = \frac{2E}{R + r_1 + r_2}
  2. Terminal p.d. of E1E_1: E−Ir1=0⇒I=Er1E - Ir_1 = 0 \Rightarrow I = \frac{E}{r_1}
  3. 2ER+r1+r2=Er1⇒R+r1+r2=2r1\frac{2E}{R + r_1 + r_2} = \frac{E}{r_1} \Rightarrow R + r_1 + r_2 = 2r_1
  4. R=r1−r2R = r_1 - r_2. So the answer is (a).
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Q212JEE Main 2025 · 22 JAN · SHIFT 1Medium

Cells: EMF, Internal Resistance & Combinations · Cells in series & parallel

Given below are two statements:

Statement-I: The equivalent emf of two nonideal batteries connected in parallel is smaller than either of the two emfs.

Statement-II: The equivalent internal resistance of two nonideal batteries connected in parallel is smaller than the internal resistance of either of the two batteries.

In the light of the above statements, choose the correct answer from the options given below.

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Answer: (a)

  1. Statement I: Eeq=E1r2+E2r1r1+r2E_{eq} = \frac{E_1 r_2 + E_2 r_1}{r_1 + r_2} is a weighted average, so it lies between E1E_1 and E2E_2, not below both. False.
  2. Statement II: req=r1r2r1+r2r_{eq} = \frac{r_1 r_2}{r_1 + r_2} is smaller than each of r1r_1, r2r_2. True.
  3. So the answer is (a) Statement-I is false but Statement-II is true.
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Q213NEET 2024Easy

Cells: EMF, Internal Resistance & Combinations · Terminal voltage & internal resistance

The terminal voltage of the battery, whose emf is 10 V10\text{ V} and internal resistance 1 Ω1\ \Omega, when connected through an external resistance of 4 Ω4\ \Omega as shown in the figure is:

Figure

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Answer: (c)

  1. I=ER+r=104+1=2 AI = \frac{E}{R + r} = \frac{10}{4 + 1} = 2\text{ A}
  2. V=E−Ir=10−2×1=8 VV = E - Ir = 10 - 2 \times 1 = 8\text{ V} (same as IR=2×4IR = 2 \times 4)
  3. So the answer is (c) 8 V.
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Q214JEE Main 2024 · 8 APR · SHIFT 1Medium

Cells: EMF, Internal Resistance & Combinations · Terminal voltage & internal resistance

In the given circuit, the terminal potential difference of the cell is:

Figure

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Answer: (c)

  1. The two 4 Ω4\ \Omega resistors are in parallel: R=2 ΩR = 2\ \Omega
  2. I=32+1=1 AI = \frac{3}{2 + 1} = 1\text{ A}
  3. V=E−Ir=3−1×1=2 VV = E - Ir = 3 - 1 \times 1 = 2\text{ V}. So the answer is (c) 2 V.
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Q215JEE Main 2024 · 30 JAN · SHIFT 1Medium

Cells: EMF, Internal Resistance & Combinations · Cells in series & parallel

Two cells are connected in opposition as shown. Cell E1E_1 is of 8 V8\text{ V} emf and 2 Ω2\ \Omega internal resistance; the cell E2E_2 is of 2 V2\text{ V} emf and 4 Ω4\ \Omega internal resistance. The terminal potential difference of cell E2E_2 is __________ V.

Figure

Numerical answer type. Work it out, then open the answer.

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Answer: 6

  1. Cells in opposition, no external resistor: I=E1−E2r1+r2=8−22+4=1 AI = \dfrac{E_1 - E_2}{r_1 + r_2} = \dfrac{8 - 2}{2 + 4} = 1\text{ A}.
  2. Current enters E2E_2 at its positive terminal, so E2E_2 is being charged.
  3. V2=E2+Ir2=2+1×4V_2 = E_2 + Ir_2 = 2 + 1 \times 4
  4. Terminal p.d. of E2E_2 =6 V= 6\text{ V}
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Q216MHT CET 2023 · 10 MAY · SHIFT 1Easy

Cells: EMF, Internal Resistance & Combinations · Terminal voltage & internal resistance

A voltmeter of resistance 150 Ω150\ \Omega connected across a cell of e.m.f. 3 V3\text{ V} reads 2.5 V2.5\text{ V}. What is the internal resistance of the cell?

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Answer: (d)

  1. Current through the voltmeter: I=2.5150=160 AI = \frac{2.5}{150} = \frac{1}{60}\text{ A}
  2. Drop across internal resistance =3−2.5=0.5 V= 3 - 2.5 = 0.5\text{ V}
  3. r=0.51/60=30 Ωr = \frac{0.5}{1/60} = 30\ \Omega. So the answer is (d) 30 Ω30\ \Omega.
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Q217MHT CET 2023 · 13 MAY · SHIFT 1Medium

Cells: EMF, Internal Resistance & Combinations · Cells in series & parallel

Two cells E1E_1 and E2E_2 having equal EMF 'EE' and internal resistances r1r_1 and r2 (r1>r2)r_2\ (r_1 > r_2) respectively are connected in series. This combination is connected to an external resistance 'RR'. It is observed that the potential difference across the cell E1E_1 becomes zero. The value of 'RR' will be

Show answer and solution

Answer: (a)

  1. I=2ER+r1+r2I = \frac{2E}{R + r_1 + r_2}
  2. Terminal p.d. of E1E_1: E−Ir1=0⇒I=Er1E - Ir_1 = 0 \Rightarrow I = \frac{E}{r_1}
  3. 2ER+r1+r2=Er1⇒R=2r1−r1−r2\frac{2E}{R + r_1 + r_2} = \frac{E}{r_1} \Rightarrow R = 2r_1 - r_1 - r_2
  4. R=r1−r2R = r_1 - r_2. So the answer is (a).
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Q218MHT CET 2023 · 11 MAY · SHIFT 2Medium

Cells: EMF, Internal Resistance & Combinations · Cells in series & parallel

Two batteries, one of e.m.f. 12 V12\text{ V} and internal resistance 2 Ω2\ \Omega and other of e.m.f. 6 V6\text{ V} and internal resistance 1 Ω1\ \Omega, are connected as shown in the figure. What will be the reading of the voltmeter 'V'?

Figure

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Answer: (b)

  1. From the figure both batteries have their positive terminals on the same side, so they are in parallel.
  2. V=E1/r1+E2/r21/r1+1/r2=12/2+6/11/2+1=121.5V = \frac{E_1/r_1 + E_2/r_2}{1/r_1 + 1/r_2} = \frac{12/2 + 6/1}{1/2 + 1} = \frac{12}{1.5}
  3. V=8 VV = 8\text{ V}. So the answer is (b) 8 V.
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Q219JEE Main 2023 · 11 APR · SHIFT 2Medium

Cells: EMF, Internal Resistance & Combinations · Cells in series & parallel

Two identical cells each of emf 1.5 V1.5\text{ V} are connected in series across a 10 Ω10\ \Omega resistance. An ideal voltmeter connected across 10 Ω10\ \Omega resistance reads 1.5 V1.5\text{ V}. The internal resistance of each cell is __________ Ω\Omega.

Numerical answer type. Work it out, then open the answer.

Show answer and solution

Answer: 5

  1. I=VR=1.510=0.15 AI = \dfrac{V}{R} = \dfrac{1.5}{10} = 0.15\text{ A}.
  2. Series cells: 2E=V+I(2r)⇒3=1.5+0.15×2r2E = V + I(2r) \Rightarrow 3 = 1.5 + 0.15 \times 2r
  3. 2r=10⇒r=5 Ω2r = 10 \Rightarrow r = 5\ \Omega
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Q220JEE Main 2023 · 31 JAN · SHIFT 1Medium

Cells: EMF, Internal Resistance & Combinations · Cells in series & parallel

Two identical cells, when connected either in parallel or in series gives same current in an external resistance 5 Ω5\ \Omega. The internal resistance of each cell will be ___________ Ω\Omega.

Numerical answer type. Work it out, then open the answer.

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Answer: 5

  1. Series: Is=2E5+2rI_s = \dfrac{2E}{5 + 2r}; parallel: Ip=E5+r/2I_p = \dfrac{E}{5 + r/2}.
  2. Is=Ip⇒2(5+r/2)=5+2rI_s = I_p \Rightarrow 2(5 + r/2) = 5 + 2r
  3. 10+r=5+2r⇒r=5 Ω10 + r = 5 + 2r \Rightarrow r = 5\ \Omega
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Q221JEE Main 2023 · 25 JAN · SHIFT 2Medium

Cells: EMF, Internal Resistance & Combinations · Cells in series & parallel

Two cells are connected between points A and B as shown. Cell 1 has emf of 12 V and internal resistance of 3Ω\Omega. Cell 2 has emf of 6V and internal resistance of 6Ω\Omega. An external resistor R of 4Ω\Omega is connected across A and B. The current flowing through R will be ____________ A.

Figure

Numerical answer type. Work it out, then open the answer.

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Answer: 1

  1. From the figure the two cells are in parallel with opposite polarity.
  2. Eeq=E1/r1−E2/r21/r1+1/r2=12/3−6/61/3+1/6=31/2=6 VE_{eq} = \dfrac{E_1/r_1 - E_2/r_2}{1/r_1 + 1/r_2} = \dfrac{12/3 - 6/6}{1/3 + 1/6} = \dfrac{3}{1/2} = 6\text{ V}
  3. req=3×63+6=2 Ωr_{eq} = \dfrac{3 \times 6}{3 + 6} = 2\ \Omega
  4. I=62+4=1 AI = \dfrac{6}{2 + 4} = 1\text{ A}
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Q222NEET 2022 · PHASE 2Easy

Cells: EMF, Internal Resistance & Combinations · Terminal voltage & internal resistance

A cell of emf 4 V and internal resistance 0.5 Ω\Omega is connected to a 7.5 Ω\Omega external resistance. The terminal potential difference of the cell is

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Answer: (b)

  1. I=47.5+0.5=0.5 AI = \frac{4}{7.5 + 0.5} = 0.5\text{ A}
  2. V=IR=0.5×7.5=3.75 VV = IR = 0.5 \times 7.5 = 3.75\text{ V}
  3. So the answer is (b) 3.75 V.
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Q223JEE Main 2022 · 27 JUL · SHIFT 1Medium

Cells: EMF, Internal Resistance & Combinations · Cells in series & parallel

Two sources of equal emfs are connected in series. This combination is connected to an external resistance R. The internal resistances of the two sources are r1r_1 and r2 (r1>r2)r_2\ (r_1 > r_2). If the potential difference across the source of internal resistance r1r_1 is zero, then the value of R will be:

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Answer: (a)

  1. I=2ER+r1+r2I = \frac{2E}{R + r_1 + r_2}
  2. P.d. across source r1r_1 is zero: E−Ir1=0⇒I=Er1E - Ir_1 = 0 \Rightarrow I = \frac{E}{r_1}
  3. R+r1+r2=2r1⇒R=r1−r2R + r_1 + r_2 = 2r_1 \Rightarrow R = r_1 - r_2. So the answer is (a).
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Q224JEE Main 2022 · 29 JUN · SHIFT 2Medium

Cells: EMF, Internal Resistance & Combinations · Cells in series & parallel

The combination of two identical cells, whether connected in series or parallel combination provides the same current through an external resistance of 2Ω\Omega. The value of internal resistance of each cell is

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Answer: (a)

  1. Series: I=2E2+2rI = \frac{2E}{2 + 2r}; parallel: I=E2+r/2I = \frac{E}{2 + r/2}
  2. Equal: 2(2+r2)=2+2r⇒4+r=2+2r2\left(2 + \frac{r}{2}\right) = 2 + 2r \Rightarrow 4 + r = 2 + 2r
  3. r=2 Ωr = 2\ \Omega. So the answer is (a). (In general, equal currents need r=Rr = R.)
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Q225JEE Main 2022 · 25 JUN · SHIFT 2Medium

Cells: EMF, Internal Resistance & Combinations · Cells in series & parallel

Two cells of same emf but different internal resistances r1 and r2 are connected in series with a resistance R. The value of resistance R, for which the potential difference across second cell is zero, is:

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Answer: (a)

  1. Series cells: I=2ER+r1+r2I = \dfrac{2E}{R + r_1 + r_2}.
  2. Terminal p.d. of second cell is zero: E−Ir2=0⇒I=Er2E - Ir_2 = 0 \Rightarrow I = \dfrac{E}{r_2}.
  3. 2ER+r1+r2=Er2⇒2r2=R+r1+r2\dfrac{2E}{R + r_1 + r_2} = \dfrac{E}{r_2} \Rightarrow 2r_2 = R + r_1 + r_2
  4. So the answer is (a) R=r2−r1R = r_2 - r_1.
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Q226JEE Main 2022 · 24 JUN · SHIFT 1Medium

Cells: EMF, Internal Resistance & Combinations · Cells in series & parallel

Two identical cells each of emf 1.5 V are connected in parallel across a parallel combination of two resistors each of resistance 20 Ω\Omega. A voltmeter connected in the circuit measures 1.2 V. The internal resistance of each cell is:

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Answer: (c)

  1. Two identical cells in parallel: Eeq=1.5 VE_{eq} = 1.5\text{ V}, req=r2r_{eq} = \dfrac{r}{2}.
  2. External resistance =20∥20=10 Ω= 20 \parallel 20 = 10\ \Omega, so I=VR=1.210=0.12 AI = \dfrac{V}{R} = \dfrac{1.2}{10} = 0.12\text{ A}.
  3. E=V+Ir2⇒1.5=1.2+0.12⋅r2E = V + I\dfrac{r}{2} \Rightarrow 1.5 = 1.2 + 0.12 \cdot \dfrac{r}{2}
  4. r2=2.5⇒r=5 Ω\dfrac{r}{2} = 2.5 \Rightarrow r = 5\ \Omega. So the answer is (c).
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Q227JEE Main 2022 · 30 JUL · SHIFT 2Medium

Cells: EMF, Internal Resistance & Combinations · Maximum power transfer

A cell sends a current through a resistance 4 Ω4\ \Omega for time tt and sends current through another resistance 16 Ω16\ \Omega for the same time tt. If the same amount of heat is developed in both the resistances, then the internal resistance of the cell is _____ Ω\Omega

Numerical answer type. Work it out, then open the answer.

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Answer: 8

  1. Heat =I2Rt=E2R(R+r)2t= I^2Rt = \dfrac{E^2R}{(R + r)^2}t; equal heat for R1R_1 and R2R_2 gives r=R1R2r = \sqrt{R_1R_2}.
  2. 4(4+r)2=16(16+r)2⇒16+r=2(4+r)\dfrac{4}{(4 + r)^2} = \dfrac{16}{(16 + r)^2} \Rightarrow 16 + r = 2(4 + r)
  3. r=4×16=8 Ωr = \sqrt{4 \times 16} = 8\ \Omega
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Q228JEE Main 2021 · 24 FEB · SHIFT 1Medium

Cells: EMF, Internal Resistance & Combinations · Terminal voltage & internal resistance

A cell E1 of emf 6V and internal resistance 2Ω\Omega is connected with another cell E2 of emf 4V and internal resistance 8Ω\Omega (as shown in the figure). The potential difference across points X and Y is:

Figure

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Answer: (c)

  1. From the figure both negative terminals face XX, so the cells oppose: Enet=6−4=2 VE_{net} = 6 - 4 = 2\text{ V}.
  2. I=22+8=0.2 AI = \frac{2}{2 + 8} = 0.2\text{ A}, driven by E1E_1.
  3. Across E1E_1 (discharging): VP−VX=6−0.2×2=5.6 VV_P - V_X = 6 - 0.2 \times 2 = 5.6\text{ V}, and VP=VYV_P = V_Y (same wire).
  4. Check with E2E_2 (being charged): 4+0.2×8=5.6 V4 + 0.2 \times 8 = 5.6\text{ V}. So the answer is (c) 5.6 V.
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Q229JEE Main 2021 · 22 JUL · SHIFT 2Medium

Cells: EMF, Internal Resistance & Combinations · Terminal voltage & internal resistance

In an electric circuit, a cell of certain emf provides a potential difference of 1.25 V across a load resistance of 5Ω\Omega. However, it provides a potential difference of 1 V across a load resistance of 2Ω\Omega. The emf of the cell is given by x10 V\frac{x}{10}\text{ V}. Then the value of x is ______________.

Numerical answer type. Work it out, then open the answer.

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Answer: 15

  1. V=ERR+rV = \frac{ER}{R + r}, so E=V(1+rR)E = V\left(1 + \frac{r}{R}\right)
  2. 1.25(1+r5)=1(1+r2)⇒1.25+0.25r=1+0.5r⇒r=1 Ω1.25\left(1 + \frac{r}{5}\right) = 1\left(1 + \frac{r}{2}\right) \Rightarrow 1.25 + 0.25r = 1 + 0.5r \Rightarrow r = 1\ \Omega
  3. E=1×(1+0.5)=1.5 V=1510 VE = 1 \times (1 + 0.5) = 1.5\text{ V} = \frac{15}{10}\text{ V}
  4. x=15x = 15
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Q230JEE Main 2021 · 27 AUG · SHIFT 1Medium

Cells: EMF, Internal Resistance & Combinations · Cells in series & parallel

Five identical cells each of internal resistance 1Ω\Omega and emf 5V are connected in series and in parallel with an external resistance 'R'. For what value of 'R', current in series and parallel combination will remain the same?

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Answer: (a)

  1. Series: Is=5ER+5r=25R+5I_s = \dfrac{5E}{R + 5r} = \dfrac{25}{R + 5}.
  2. Parallel: Ip=ER+r/5=5R+0.2I_p = \dfrac{E}{R + r/5} = \dfrac{5}{R + 0.2}.
  3. Is=Ip⇒25(R+0.2)=5(R+5)⇒20R=20I_s = I_p \Rightarrow 25(R + 0.2) = 5(R + 5) \Rightarrow 20R = 20
  4. So the answer is (a) R=1 ΩR = 1\ \Omega.
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Q231JEE Main 2021 · 17 MAR · SHIFT 2Medium

Cells: EMF, Internal Resistance & Combinations · Cells in series & parallel

Two cells of emf 2E and E with internal resistance r1 and r2 respectively are connected in series to an external resistor R (see figure). The value of R, at which the potential difference across the terminals of the first cell becomes zero is

Figure

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Answer: (b)

  1. Both cells aid each other: I=3ER+r1+r2I = \dfrac{3E}{R + r_1 + r_2}.
  2. Terminal p.d. of first cell is zero: 2E−Ir1=0⇒I=2Er12E - Ir_1 = 0 \Rightarrow I = \dfrac{2E}{r_1}.
  3. 3ER+r1+r2=2Er1⇒3r1=2R+2r1+2r2\dfrac{3E}{R + r_1 + r_2} = \dfrac{2E}{r_1} \Rightarrow 3r_1 = 2R + 2r_1 + 2r_2
  4. So the answer is (b) R=r12−r2R = \dfrac{r_1}{2} - r_2.
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Q232JEE Main 2021 · 27 AUG · SHIFT 1Difficult

Cells: EMF, Internal Resistance & Combinations · Cells in series & parallel

First, a set of n equal resistors of 10 Ω\Omega each are connected in series to a battery of emf 20V and internal resistance 10Ω\Omega. A current I is observed to flow. Then, the n resistors are connected in parallel to the same battery. It is observed that the current is increased 20 times, then the value of n is ................

Numerical answer type. Work it out, then open the answer.

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Answer: 20

  1. Series: I=2010n+10I = \dfrac{20}{10n + 10}; parallel: I′=2010/n+10I' = \dfrac{20}{10/n + 10}.
  2. I′I=10n+1010/n+10=10(n+1)10(n+1)/n=n\dfrac{I'}{I} = \dfrac{10n + 10}{10/n + 10} = \dfrac{10(n + 1)}{10(n + 1)/n} = n
  3. I′I=20⇒n=20\dfrac{I'}{I} = 20 \Rightarrow n = 20
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Q233JEE Main 2020 · 8 JAN · SHIFT 2Medium

Cells: EMF, Internal Resistance & Combinations · Cells in series & parallel

The series combination of two batteries, both of the same emf 10 V, but different internal resistance of 20Ω\Omega and 5Ω\Omega, is connected to the parallel combination of two resistors 30Ω\Omega and R Ω\Omega. The voltage difference across the battery of internal resistance 20Ω\Omega is zero, the value of R (in Ω\Omega) is: _______

Numerical answer type. Work it out, then open the answer.

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Answer: 30

  1. P.d. across the 20 Ω20\ \Omega battery is zero: 10−20I=0⇒I=0.5 A10 - 20I = 0 \Rightarrow I = 0.5\text{ A}.
  2. Whole circuit: 20=I(20+5+Rp)⇒25+Rp=40⇒Rp=15 Ω20 = I(20 + 5 + R_p) \Rightarrow 25 + R_p = 40 \Rightarrow R_p = 15\ \Omega.
  3. 30R30+R=15⇒30R=450+15R\dfrac{30R}{30 + R} = 15 \Rightarrow 30R = 450 + 15R
  4. R=30 ΩR = 30\ \Omega
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Q234JEE Main 2020 · 4 SEP · SHIFT 1Medium

Cells: EMF, Internal Resistance & Combinations · Maximum power transfer

A battery of 3.0 V is connected to a resistor dissipating 0.5 W of power. If the terminal voltage of the battery is 2.5 V, the power dissipated within the internal resistance is:

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Answer: (c)

  1. Power in load P=VI⇒I=0.52.5=0.2 AP = VI \Rightarrow I = \dfrac{0.5}{2.5} = 0.2\text{ A}.
  2. Drop across internal resistance =E−V=3.0−2.5=0.5 V= E - V = 3.0 - 2.5 = 0.5\text{ V}.
  3. Pint=0.5×0.2=0.10 WP_{int} = 0.5 \times 0.2 = 0.10\text{ W}
  4. So the answer is (c) 0.10 W.
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Q235JEE Main 2019 · 8 APR · SHIFT 2Easy

Cells: EMF, Internal Resistance & Combinations · Maximum power transfer

A cell of internal resistance r drives current through an external resistance R. The power delivered by the cell to the external resistance will be maximum when :-

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Answer: (b)

  1. P=E2R(R+r)2P = \frac{E^2 R}{(R + r)^2} is maximum when dPdR=0⇒R=r\frac{dP}{dR} = 0 \Rightarrow R = r.
  2. So the answer is (b) R=rR = r.
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Q236JEE Main 2019 · 12 APR · SHIFT 1Medium

Cells: EMF, Internal Resistance & Combinations · Terminal voltage & internal resistance

To verify Ohm's law, a student connects the voltmeter across the battery as, shown in the figure. The measured voltage is plotted as a function of the current, and the following graph is obtained:

Figure

If V0 is almost zero, identify the correct statement:

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Answer: (b)

  1. Voltmeter across the battery reads V=E−IrV = E - Ir.
  2. Intercept at I=0I = 0: E=1.5 VE = 1.5\text{ V}
  3. V≈0V \approx 0 at I=1 AI = 1\text{ A}: r=1.51=1.5 Ωr = \frac{1.5}{1} = 1.5\ \Omega
  4. So the answer is (b) emf 1.5 V and internal resistance 1.5 Ω1.5\ \Omega.
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Q237JEE Main 2019 · 10 APR · SHIFT 1Medium

Cells: EMF, Internal Resistance & Combinations · Terminal voltage & internal resistance

In the given circuit, an ideal voltmeter connected across the 10Ω\Omega resistance reads 2V. The internal resistance r, of each cell is:

Figure

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Answer: (c)

  1. 15 Ω∥10 Ω=6 Ω15\ \Omega \parallel 10\ \Omega = 6\ \Omega; voltmeter reads 2 V across this pair.
  2. Circuit current I=26=13 AI = \frac{2}{6} = \frac{1}{3}\text{ A}
  3. I=1.5+1.56+2+2r⇒13=38+2r⇒8+2r=9I = \frac{1.5 + 1.5}{6 + 2 + 2r} \Rightarrow \frac{1}{3} = \frac{3}{8 + 2r} \Rightarrow 8 + 2r = 9
  4. r=0.5 Ωr = 0.5\ \Omega. So the answer is (c).
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Q238NEET 2018Medium

Cells: EMF, Internal Resistance & Combinations · Cells in series & parallel

A set of n equal resistors, of value R each, are connected in series to a battery of emf E and internal resistance R. The current drawn is I. Now, the n resistors are connected in parallel to the same battery. Then the current drawn from battery becomes 10I. The value of n is

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Answer: (a)

  1. Series: I=EnR+R=E(n+1)RI = \frac{E}{nR + R} = \frac{E}{(n+1)R}
  2. Parallel: 10I=ER/n+R=nE(n+1)R10I = \frac{E}{R/n + R} = \frac{nE}{(n+1)R}
  3. Dividing: 10=n10 = n. So the answer is (a) 10.
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Q239NEET 2018Medium

Cells: EMF, Internal Resistance & Combinations · Cells in series & parallel

A battery consists of a variable number n of identical cells (having internal resistance r each) which are connected in series. The terminals of the battery are short-circuited and the current I is measured. Which of the graphs shows the correct relationship between I and n?

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Answer: (a)

  1. Short circuit: I=nEnr=ErI = \frac{nE}{nr} = \frac{E}{r}
  2. II does not depend on nn, so the graph is a horizontal line.
  3. So the answer is (a).
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Q240JEE Main 2018Medium

Cells: EMF, Internal Resistance & Combinations · Cells in series & parallel

Two batteries with e.m.f 12 V and 13 V are connected in parallel across a load resistor of 10 Ω\Omega. The internal resistances of the two batteries are 1 Ω\Omega and 2 Ω\Omega respectively. The voltage across the load lies between:

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Answer: (c)

  1. Cells in parallel with a load (Millman): V=E1/r1+E2/r21/r1+1/r2+1/RV = \dfrac{E_1/r_1 + E_2/r_2}{1/r_1 + 1/r_2 + 1/R}.
  2. V=12/1+13/21+0.5+0.1=18.51.6V = \dfrac{12/1 + 13/2}{1 + 0.5 + 0.1} = \dfrac{18.5}{1.6}
  3. V=11.56 VV = 11.56\text{ V}
  4. So the answer is (c) between 11.5 V and 11.6 V.
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Q241JEE Main 2016 · 9 APR · SHIFT 1Difficult

Cells: EMF, Internal Resistance & Combinations · Maximum power transfer

Figure

In the circuit shown, the resistance r is a variable resistance. If for r = fR, the heat generation in r is maximum then the value of f is:

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Answer: (b)

  1. Heat in rr is maximum when rr equals the resistance of the rest of the circuit seen from its ends (ideal cell shorted).
  2. Seen from rr: series RR (through the cell) is in parallel with the other RR, so Rth=R⋅RR+R=R2R_{th} = \dfrac{R \cdot R}{R + R} = \dfrac{R}{2}.
  3. r=R2⇒f=12r = \dfrac{R}{2} \Rightarrow f = \dfrac{1}{2}
  4. So the answer is (b) 12\dfrac{1}{2}.
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Q242JEE Main 2015 · 10 APR · SHIFT 1Medium

Cells: EMF, Internal Resistance & Combinations · Cells in series & parallel

A 10V battery with internal resistance 1Ω\Omega and a 15V battery with internal resistance 0.6 Ω\Omega are connected in parallel to a voltmeter (see figure). The reading in the voltmeter will be close to:

Figure

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Answer: (c)

  1. An ideal voltmeter draws no current, so it reads the equivalent emf of the parallel cells.
  2. Eeq=E1/r1+E2/r21/r1+1/r2=10/1+15/0.61+1/0.6E_{eq} = \dfrac{E_1/r_1 + E_2/r_2}{1/r_1 + 1/r_2} = \dfrac{10/1 + 15/0.6}{1 + 1/0.6}
  3. Eeq=10+252.667=13.1 VE_{eq} = \dfrac{10 + 25}{2.667} = 13.1\text{ V}
  4. So the answer is (c) 13.1 V.
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Q243JEE Main 2014 · 9 APR · SHIFT 1Medium

Cells: EMF, Internal Resistance & Combinations · Terminal voltage & internal resistance

A d.c. main supply of e.m.f. 220 V is connected across a storage battery of e.m.f. 200 V through a resistance of 1 Ω\Omega. The battery terminals are connected to an external resistance 'R'. The minimum value of 'R', so that a current passes through the battery to charge it is:

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Answer: (c)

  1. Battery (200 V) and RR are in parallel, fed by 220 V through 1 Ω1\ \Omega.
  2. At the threshold, battery current is zero and its terminal voltage is 200 V: supply current =220−2001=20 A= \frac{220 - 200}{1} = 20\text{ A}, all through RR.
  3. Charging needs 200R<20⇒R>10 Ω\frac{200}{R} < 20 \Rightarrow R > 10\ \Omega
  4. The smallest option above 10 Ω10\ \Omega is 11 Ω11\ \Omega. So the answer is (c).
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Q244NEET 2013Easy

Cells: EMF, Internal Resistance & Combinations · Terminal voltage & internal resistance

The internal resistance of a 2.1 V cell which gives a current of 0.2 A through a resistance of 10 Ω\Omega is

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Answer: (d)

  1. I=ER+r⇒R+r=2.10.2=10.5 ΩI = \frac{E}{R + r} \Rightarrow R + r = \frac{2.1}{0.2} = 10.5\ \Omega
  2. r=10.5−10=0.5 Ωr = 10.5 - 10 = 0.5\ \Omega. So the answer is (d) 0.5 Ω0.5\ \Omega.
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Q245JEE Main 2013 · 22 APR · SHIFT 1Medium

Cells: EMF, Internal Resistance & Combinations · Terminal voltage & internal resistance

A dc source of emf E1=100E_1 = 100 V and internal resistance r=0.5 Ωr = 0.5\ \Omega, a storage battery of emf E2=90E_2 = 90 V and an external resistance R are connected as shown in figure. For what value of R no current will pass through the battery?

Figure

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Answer: (c)

  1. No current in E2E_2 means voltage across RR equals E2=90 VE_2 = 90\text{ V}.
  2. Current from E1E_1: I=100−900.5=20 AI = \frac{100 - 90}{0.5} = 20\text{ A}, all of which flows through RR.
  3. R=9020=4.5 ΩR = \frac{90}{20} = 4.5\ \Omega. So the answer is (c).
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Q246NEET 2012 · AIPMT · MAINSEasy

Cells: EMF, Internal Resistance & Combinations · Terminal voltage & internal resistance

A cell having an emf ε\varepsilon and internal resistance r is connected across a variable external resistance R. As the resistance R is increased, the plot of potential difference V across R is given by

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Answer: (c)

  1. V=IR=εRR+r=ε1+r/RV = IR = \frac{\varepsilon R}{R + r} = \frac{\varepsilon}{1 + r/R}
  2. At R=0R = 0, V=0V = 0; as RR increases VV rises and approaches ε\varepsilon asymptotically (curve bending over).
  3. So the answer is (c).
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Q247NEET 2011 · AIPMT · PRELIMSEasy

Cells: EMF, Internal Resistance & Combinations · Terminal voltage & internal resistance

A current of 2 A flows through a 2 Ω\Omega resistor when connected across a battery. The same battery supplies a current of 0.5 A when connected across a 9 Ω\Omega resistor. The internal resistance of the battery is

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Answer: (b)

  1. Same emf in both cases: E=I(R+r)E = I(R + r)
  2. 2(2+r)=0.5(9+r)⇒4+2r=4.5+0.5r2(2 + r) = 0.5(9 + r) \Rightarrow 4 + 2r = 4.5 + 0.5r
  3. 1.5r=0.5⇒r=13 Ω1.5r = 0.5 \Rightarrow r = \frac{1}{3}\ \Omega. So the answer is (b).
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Q248NEET 2009 · AIPMTEasy

Cells: EMF, Internal Resistance & Combinations · Terminal voltage & internal resistance

A student measures the terminal potential difference (V) of a cell (of emf ε\varepsilon and internal resistance r) as a function of the current (I) flowing through it. The slope, and intercept, of the graph between V and I, then respectively, equal

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Answer: (a)

  1. V=ε−IrV = \varepsilon - Ir, i.e. V=(−r)I+εV = (-r)I + \varepsilon, a straight line.
  2. Compare with y=mx+cy = mx + c: slope =−r= -r, intercept =ε= \varepsilon.
  3. So the answer is (a) −r-r and ε\varepsilon.
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Q249JEE Main 2008 · AIEEEMedium

Cells: EMF, Internal Resistance & Combinations · Terminal voltage & internal resistance

A 5 V5\text{ V} battery with internal resistance 2 Ω2\ \Omega and a 2 V2\text{ V} battery with internal resistance 1 Ω1\ \Omega are connected to a 10 Ω10\ \Omega resistor as shown in the figure.

Figure

The current in the 10 Ω10\ \Omega resistor is

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Answer: (c)

  1. From the figure the two cells oppose each other. Take P1P_1 (bottom) at 0 and P2P_2 at VV.
  2. Node equation at P2P_2: V−52+V+21+V10=0\frac{V - 5}{2} + \frac{V + 2}{1} + \frac{V}{10} = 0
  3. 1.6V=0.5⇒V=0.3125 V1.6V = 0.5 \Rightarrow V = 0.3125\text{ V}
  4. I10=0.312510≈0.03 AI_{10} = \frac{0.3125}{10} \approx 0.03\text{ A}, from higher potential P2P_2 to P1P_1.
  5. So the answer is (c) 0.03 A, P2P_2 to P1P_1.
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Q250NEET 2006 · AIPMTMedium

Cells: EMF, Internal Resistance & Combinations · Cells in series & parallel

Two cells, having the same e.m.f. are connected in series through an external resistance R. Cells have internal resistances r1 and r2 (r1 > r2) respectively. When the circuit is closed, the potential difference across the first cell is zero. The value of R is

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Answer: (b)

  1. I=2ER+r1+r2I = \frac{2E}{R + r_1 + r_2}
  2. P.d. across first cell zero: E=Ir1⇒2Er1R+r1+r2=EE = Ir_1 \Rightarrow \frac{2E r_1}{R + r_1 + r_2} = E
  3. R+r1+r2=2r1⇒R=r1−r2R + r_1 + r_2 = 2r_1 \Rightarrow R = r_1 - r_2. So the answer is (b).
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Q251JEE Main 2005 · AIEEEEasy

Cells: EMF, Internal Resistance & Combinations · Terminal voltage & internal resistance

An energy source will supply a constant current into the load if its internal resistance is

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Answer: (a)

  1. I=ER+rI = \frac{E}{R + r}
  2. If r≫Rr \gg R, then I≈ErI \approx \frac{E}{r}, independent of the load RR, so the current stays nearly constant.
  3. So the answer is (a) very large compared to the load resistance.
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Q252NEET 2005 · AIPMTMedium

Cells: EMF, Internal Resistance & Combinations · Cells in series & parallel

Two batteries, one of emf 18 volts and internal resistance 2 Ω\Omega and the other of emf 12 volts and internal resistance 1 Ω\Omega, are connected as shown. The voltmeter V will record a reading of

Figure

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Answer: (d)

  1. From the figure both batteries are connected in parallel with like terminals together.
  2. V=E1/r1+E2/r21/r1+1/r2=18/2+12/11/2+1=211.5V = \frac{E_1/r_1 + E_2/r_2}{1/r_1 + 1/r_2} = \frac{18/2 + 12/1}{1/2 + 1} = \frac{21}{1.5}
  3. V=14 VV = 14\text{ V}. So the answer is (d) 14 volt.
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Q253JEE Main 2005 · AIEEEMedium

Cells: EMF, Internal Resistance & Combinations · Cells in series & parallel

Two sources of equal emf\text{emf} are connected to an external resistance R.R. The internal resistances of the two sources are R1R_1 and R2 (R2>R1).R_2\ (R_2 > R_1). If the potential difference across the source having internal resistance R2R_2 is zero, then

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Answer: (a)

  1. Sources in series: I=2ER+R1+R2I = \dfrac{2E}{R + R_1 + R_2}.
  2. P.d. across source R2R_2 is zero: E−IR2=0⇒I=ER2E - IR_2 = 0 \Rightarrow I = \dfrac{E}{R_2}.
  3. 2R2=R+R1+R22R_2 = R + R_1 + R_2
  4. So the answer is (a) R=R2−R1R = R_2 - R_1.
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Q254NEET 2002 · AIPMTEasy

Cells: EMF, Internal Resistance & Combinations · Terminal voltage & internal resistance

For a cell terminal potential difference is 2.2V when circuit is open and reduce to 1.8 V when cell is connected to a resistance of R = 5 Ω\Omega. Determine internal resistance of cell (r)

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Answer: (a)

  1. Open-circuit voltage = emf: E=2.2 VE = 2.2\text{ V}
  2. I=VR=1.85=0.36 AI = \frac{V}{R} = \frac{1.8}{5} = 0.36\text{ A}
  3. r=E−VI=0.40.36=109 Ωr = \frac{E - V}{I} = \frac{0.4}{0.36} = \frac{10}{9}\ \Omega. So the answer is (a).
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Q255NEET 2000 · AIPMTEasy

Cells: EMF, Internal Resistance & Combinations · Terminal voltage & internal resistance

A cell has an emf 1.5 V1.5\text{ V}. When connected across an external resistance of 2 Ω2\ \Omega, the terminal potential difference falls to 1.0 V1.0\text{ V}. The internal resistance of the cell is

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Answer: (c)

  1. I=VR=1.02=0.5 AI = \frac{V}{R} = \frac{1.0}{2} = 0.5\text{ A}
  2. r=E−VI=0.50.5=1.0 Ωr = \frac{E - V}{I} = \frac{0.5}{0.5} = 1.0\ \Omega. So the answer is (c).
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Q256NEET 1999 · AIPMTEasy

Cells: EMF, Internal Resistance & Combinations · Terminal voltage & internal resistance

The internal resistance of a cell of emf 2 V2\text{ V} is 0.1 Ω0.1\ \Omega. It is connected to a resistance of 3.9 Ω3.9\ \Omega. The voltage across the cell will be

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Answer: (c)

  1. I=23.9+0.1=0.5 AI = \frac{2}{3.9 + 0.1} = 0.5\text{ A}
  2. Terminal voltage V=E−Ir=2−0.5×0.1=1.95 VV = E - Ir = 2 - 0.5 \times 0.1 = 1.95\text{ V}
  3. So the answer is (c) 1.95 V.
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Q257NEET 1992 · AIPMTEasy

Cells: EMF, Internal Resistance & Combinations · Maximum power transfer

A battery of e.m.f 10 V10\text{ V} and internal resistance 0.5 Ω0.5\ \Omega is connected across a variable resistance RR. The value of RR for which the power delivered in it is maximum is given by

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Answer: (a)

  1. Maximum power transfer theorem: power in RR is maximum when R=rR = r.
  2. R=0.5 ΩR = 0.5\ \Omega. So the answer is (a).
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Q258NEET 1990 · AIPMTMedium

Cells: EMF, Internal Resistance & Combinations · Maximum power transfer

Two identical batteries each of e.m.f 2 V2\text{ V} and internal resistance 1 Ω1\ \Omega are available to produce heat in an external resistance by passing a current through it. The maximum power that can be developed across RR using these batteries is

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Answer: (b)

  1. Maximum power in RR occurs when R=reqR = r_{eq}: Pmax=Eeq24reqP_{max} = \dfrac{E_{eq}^2}{4r_{eq}}.
  2. Series: Eeq=4 VE_{eq} = 4\text{ V}, req=2 Ωr_{eq} = 2\ \Omega, Pmax=168=2 WP_{max} = \dfrac{16}{8} = 2\text{ W}.
  3. Parallel: Eeq=2 VE_{eq} = 2\text{ V}, req=0.5 Ωr_{eq} = 0.5\ \Omega, Pmax=42=2 WP_{max} = \dfrac{4}{2} = 2\text{ W}.
  4. So the answer is (b) 2.0 W.
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Q259MHT CET 2026 · 20 APR · SHIFT 2Easy

Kirchhoff's Laws & Measuring Instruments · Potential difference between points

The currents in different parts of the electric circuit are shown in following figure. The value of current ii is

Figure

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Answer: (c)

  1. First junction: 2+3=5 A2 + 3 = 5\text{ A} flows on.
  2. Second junction: 1 A1\text{ A} leaves, so 4 A4\text{ A} flows on.
  3. Last junction: 4=0.6+1.3+i4 = 0.6 + 1.3 + i
  4. i=2.1 Ai = 2.1\text{ A}. So the answer is (c).
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Q260JEE Main 2026 · 28 JAN · SHIFT 2Easy

Kirchhoff's Laws & Measuring Instruments · Voltmeter & ammeter in circuits

Which one of the following is not a measurable quantity?

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Answer: (d)

  1. Only differences of potential can be measured (a voltmeter is always connected between two points); absolute potential at a point depends on the chosen zero.
  2. Resistance and voltage difference are measured directly; displacement current is treated as measurable through its magnetic effect.
  3. So the answer is (d) Voltage.
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Q261NEET 2026Easy

Kirchhoff's Laws & Measuring Instruments · Galvanometer to ammeter / voltmeter

A galvanometer of resistance 100 Ω100\ \Omega gives full scale deflection for a current of 1 mA. It is converted into an ammeter of range 0−10 A0 - 10\text{ A}. The shunt required is:

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Answer: (a)

  1. S=IgGI−IgS = \frac{I_g G}{I - I_g}
  2. S=10−3×10010−0.001=0.19.999≈0.01 ΩS = \frac{10^{-3} \times 100}{10 - 0.001} = \frac{0.1}{9.999} \approx 0.01\ \Omega
  3. So the answer is (a) 0.01 Ω0.01\ \Omega.
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Q262JEE Main 2026 · 4 APR · SHIFT 1Easy

Kirchhoff's Laws & Measuring Instruments · Galvanometer to ammeter / voltmeter

A voltmeter with internal resistance of x Ωx\ \Omega can be used to measure upto 20 V. In order to increase its measuring range to 30 V, the required modification is to ______.

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Answer: (a)

  1. To raise a voltmeter's range a resistance is added in series: R=RV(V′V−1)R = R_V\left(\frac{V'}{V} - 1\right).
  2. R=x(3020−1)=x2 ΩR = x\left(\frac{30}{20} - 1\right) = \frac{x}{2}\ \Omega
  3. So the answer is (a) x2 Ω\frac{x}{2}\ \Omega in series.
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Q263JEE Main 2026 · 24 JAN · SHIFT 2Easy

Kirchhoff's Laws & Measuring Instruments · Galvanometer to ammeter / voltmeter

A moving coil galvanometer of resistance 100 Ω100\ \Omega shows a full scale deflection for a current of 1 mA. The value of resistance required to convert this galvanometer into an ammeter, showing full scale deflection for a current of 5 mA, is ‾ Ω\underline{\qquad}\ \Omega

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Answer: (a)

  1. A shunt is needed: S=IgGI−IgS = \frac{I_g G}{I - I_g}.
  2. S=1×1005−1=25 ΩS = \frac{1 \times 100}{5 - 1} = 25\ \Omega
  3. So the answer is (a) 25.
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Q264MHT CET 2026 · 19 MAY · SHIFT 2Medium

Kirchhoff's Laws & Measuring Instruments · Multi-loop circuits

In the circuit shown in the figure, the current through

Figure

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Answer: (d)

  1. Reduce from the right: 2+4+2=8 Ω2 + 4 + 2 = 8\ \Omega, in parallel with 8 Ω8\ \Omega gives 4 Ω4\ \Omega.
  2. Add 2+22 + 2: 8 Ω8\ \Omega, in parallel with the first 8 Ω8\ \Omega gives 4 Ω4\ \Omega.
  3. Total =3+4+2=9 Ω= 3 + 4 + 2 = 9\ \Omega, so I=99=1I = \frac{9}{9} = 1 A (this flows in 3 Ω3\ \Omega).
  4. Equal 8 Ω8\ \Omega splits: 1→0.51 \to 0.5 A →0.25\to 0.25 A in the right branch containing 4 Ω4\ \Omega.
  5. So the answer is (d) the 4 Ω4\ \Omega resistance carries 0.250.25 A.
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Q265MHT CET 2026 · 17 APR · SHIFT 1Medium

Kirchhoff's Laws & Measuring Instruments · Multi-loop circuits

In the following figure, the current I is equal to

Figure

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Answer: (d)

  1. Apply KCL (current in = current out) at each junction.
  2. Left junction: 4.5+1.7−2.3=3.94.5 + 1.7 - 2.3 = 3.9 A goes to the middle junction.
  3. Middle junction: 3.9+3.2+4.9−2.6=9.43.9 + 3.2 + 4.9 - 2.6 = 9.4 A goes to the right junction.
  4. Right junction: I=9.4−1.9=7.5I = 9.4 - 1.9 = 7.5 A
  5. So the answer is (d) 7.57.5 A.
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Q266MHT CET 2026 · 11 APR · SHIFT 2Medium

Kirchhoff's Laws & Measuring Instruments · Multi-loop circuits

In the following network, the current flowing through 15 Ω15\ \Omega resistance is

Figure

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Answer: (c)

  1. 153=204=5\frac{15}{3} = \frac{20}{4} = 5, so the bridge is balanced and no current flows through G.
  2. Path ABC: 15+3=18 Ω15 + 3 = 18\ \Omega; path ADC: 20+4=24 Ω20 + 4 = 24\ \Omega.
  3. I15=2.1×2418+24=1.2I_{15} = 2.1 \times \frac{24}{18 + 24} = 1.2 A
  4. So the answer is (c) 1.21.2 A.
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Q267JEE Main 2026 · 5 APR · SHIFT 1Medium

Kirchhoff's Laws & Measuring Instruments · Multi-loop circuits

Refer to the figure given below, current between terminals AA and BB is

______ A.

Figure

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Answer: (b)

  1. Each of the top three rows: emf 3×5=153 \times 5 = 15 V, resistance 3×3=9 Ω3 \times 3 = 9\ \Omega.
  2. Three identical rows in parallel act as one cell: E=15E = 15 V, r=93=3 Ωr = \frac{9}{3} = 3\ \Omega.
  3. The bottom row (9 Ω9\ \Omega) is the load: I=153+9=1.25I = \frac{15}{3 + 9} = 1.25 A
  4. So the answer is (b) 1.25 A.
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Q268MHT CET 2026 · 18 APR · SHIFT 2Medium

Kirchhoff's Laws & Measuring Instruments · Potential difference between points

In the following network,

I1=−0.4I_1 = -0.4 A, I4=1I_4 = 1 A, I5=0.4I_5 = 0.4 A

The values of I2,I3I_2, I_3 and I6I_6 are respectively

Figure

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Answer: (c)

  1. Path through the top-left corner has no branch: I6=I5=0.4I_6 = I_5 = 0.4 A.
  2. Bottom-right junction: I1+I2=I4⇒I2=1−(−0.4)=1.4I_1 + I_2 = I_4 \Rightarrow I_2 = 1 - (-0.4) = 1.4 A.
  3. Top-right junction: I6=I3+I2+I1⇒I3=0.4−1.4+0.4=−0.6I_6 = I_3 + I_2 + I_1 \Rightarrow I_3 = 0.4 - 1.4 + 0.4 = -0.6 A.
  4. So the answer is (c) 1.41.4 A, −0.6-0.6 A, 0.40.4 A.
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Q269NEET 2026Medium

Kirchhoff's Laws & Measuring Instruments · Potential difference between points

A uniform metallic wire having resistance 4 Ω4\ \Omega is bent to form a square loop (ABCD) (see figure). A resistance of 2 Ω2\ \Omega is connected between points BB and DD and a battery of 2 V is connected across points AA and CC as shown in the figure. Now the value of current (l)(l) is:

Figure

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Answer: (a)

  1. Each side of the square has 44=1 Ω\frac{4}{4} = 1\ \Omega.
  2. AB : BC = AD : DC = 1 : 1, so the bridge is balanced and the 2 Ω2\ \Omega in BD carries no current.
  3. Paths ABC and ADC are each 2 Ω2\ \Omega, in parallel: R=1 ΩR = 1\ \Omega.
  4. I=ER=21=2 AI = \frac{E}{R} = \frac{2}{1} = 2\text{ A}
  5. So the answer is (a) 2 A.
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Q270JEE Main 2026 · 5 APR · SHIFT 1Medium

Kirchhoff's Laws & Measuring Instruments · Potential difference between points

Refer to the figure given below. The values of I1,I2I_1, I_2 and I3I_3 are ______.

Figure

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Answer: (a)

  1. Take left node LL at 0; top TT, bottom DD, middle MM. The 5 V cell gives VT−VD=5V_T - V_D = 5; branch current I1=VM+10I_1 = V_M + 10 (1 Ω, 10 V).
  2. KCL at M: VT−VM2+VD−VM2=VM+10\frac{V_T - V_M}{2} + \frac{V_D - V_M}{2} = V_M + 10; KCL for T+D: VT+VD4=2VM−VT−VD2\frac{V_T + V_D}{4} = \frac{2V_M - V_T - V_D}{2}.
  3. Solving: VT+VD=−10V_T + V_D = -10, VM=−7.5V_M = -7.5, so VT=−2.5 VV_T = -2.5\text{ V}, VD=−7.5 VV_D = -7.5\text{ V}.
  4. I1=−7.5+10=2.5 AI_1 = -7.5 + 10 = 2.5\text{ A}, I3=0−(−7.5)4=1.875 AI_3 = \frac{0 - (-7.5)}{4} = 1.875\text{ A}
  5. At T: I2=VT−VM2−0−VT4=2.5−0.625=1.875 AI_2 = \frac{V_T - V_M}{2} - \frac{0 - V_T}{4} = 2.5 - 0.625 = 1.875\text{ A}
  6. So the answer is (a).
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Q271JEE Main 2026 · 4 APR · SHIFT 1Medium

Kirchhoff's Laws & Measuring Instruments · Potential difference between points

The voltage and the current between AA and BB points shown in the circuit are ______.

Figure

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Answer: (b)

  1. Four identical branches (27 V, 3 Ω) lie in parallel between A and B; the 14 V + 13 V branch is also 27 V.
  2. Equivalent source: E=27 VE = 27\text{ V}, r=34=0.75 Ωr = \frac{3}{4} = 0.75\ \Omega
  3. Load between A and B: 3+3=6 Ω3 + 3 = 6\ \Omega
  4. I=276+0.75=4 AI = \frac{27}{6 + 0.75} = 4\text{ A}, V=4×6=24 VV = 4 \times 6 = 24\text{ V}
  5. So the answer is (b) 24 V, 4 A.
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Q272JEE Main 2026 · 24 JAN · SHIFT 2Medium

Kirchhoff's Laws & Measuring Instruments · Voltmeter & ammeter in circuits

The reading of the ammeter (A)(A) in steady state in the following circuit (assuming negligible internal resistance of the ammeter) is ______ A.

Figure

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Answer: (b)

  1. In steady state the capacitor branch (with its 8 Ω8\ \Omega) carries no current.
  2. The ideal ammeter joins the bottom nodes, so the left 8 Ω8\ \Omega is in parallel with 4+(8∥8)=8 Ω4 + (8\|8) = 8\ \Omega.
  3. 8∥8=4 Ω8\|8 = 4\ \Omega; total R=1+4=5 ΩR = 1 + 4 = 5\ \Omega, I=105=2 AI = \dfrac{10}{5} = 2\text{ A}.
  4. The current splits equally; the right-hand 1 A1\text{ A} returns through the ammeter.
  5. So the answer is (b) 1 A1\text{ A}.
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Q273JEE Main 2026 · 24 JAN · SHIFT 1Medium

Kirchhoff's Laws & Measuring Instruments · Voltmeter & ammeter in circuits

Two resistors of 100 Ω100\ \Omega each are connected in series with a 9 V battery. A voltmeter of 400 Ω400\ \Omega resistance is connected to measure the voltage drop across one of the resistors. The voltmeter reading is ______ V.

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Answer: (c)

  1. 100 Ω∥400 Ω=100×400500=80 Ω100\ \Omega \| 400\ \Omega = \dfrac{100 \times 400}{500} = 80\ \Omega
  2. Total R=100+80=180 ΩR = 100 + 80 = 180\ \Omega, I=9180=0.05 AI = \dfrac{9}{180} = 0.05\text{ A}.
  3. Reading =0.05×80=4 V= 0.05 \times 80 = 4\text{ V}
  4. So the answer is (c) 4.
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Q274MHT CET 2026 · 18 APR · SHIFT 1Medium

Kirchhoff's Laws & Measuring Instruments · Galvanometer to ammeter / voltmeter

By connecting a resistance of 980 Ω980\ \Omega in series with a galvanometer, it is converted into a voltmeter of a certain range. When the resistance of 470 Ω470\ \Omega is connected in series, the range is halved. The resistance of the galvanometer is

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Answer: (b)

  1. Voltmeter range V=Ig(G+R)V = I_g(G + R), with the same full-scale current IgI_g.
  2. Ig(G+980)=VI_g(G + 980) = V and Ig(G+470)=V2I_g(G + 470) = \dfrac{V}{2}
  3. Divide: G+980=2(G+470)⇒G=980−940G + 980 = 2(G + 470) \Rightarrow G = 980 - 940
  4. G=40 ΩG = 40\ \Omega. So the answer is (b).
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Q275MHT CET 2026 · 17 APR · SHIFT 1Medium

Kirchhoff's Laws & Measuring Instruments · Galvanometer to ammeter / voltmeter

With a resistance 'XX' connected in series with a galvanometer of resistance 100 Ω100\ \Omega, it acts as a voltmeter of range 0 - 15 V. To double the range, a resistance of 1500 Ω1500\ \Omega is to be connected in series with 'XX'. The value of 'XX' in Ω\Omega is

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Answer: (c)

  1. Range V=Ig(G+R)V = I_g(G + R): Ig(100+X)=15I_g(100 + X) = 15.
  2. Double range: Ig(100+X+1500)=30I_g(100 + X + 1500) = 30
  3. Divide: 1600+X=2(100+X)⇒X=1400 Ω1600 + X = 2(100 + X) \Rightarrow X = 1400\ \Omega
  4. So the answer is (c) 1400.
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Q276MHT CET 2025 · 26 APR · SHIFT 2Easy

Kirchhoff's Laws & Measuring Instruments · Kirchhoff's laws: basics

Kirchhoff's voltage law and current law are based (respectively) on the conservation of

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Answer: (b)

  1. Voltage law (loop rule): net work done on a charge around a closed loop is zero, so it is conservation of energy.
  2. Current law (junction rule): charge does not pile up at a junction, so it is conservation of charge.
  3. So the answer is (b) energy, charge.
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Q277MHT CET 2025 · 25 APR · SHIFT 1Easy

Kirchhoff's Laws & Measuring Instruments · Galvanometer to ammeter / voltmeter

The value of the shunt resistance that allows 10%10\% of the main current through the galvanometer of 99 Ω99\ \Omega is

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Answer: (b)

  1. Galvanometer and shunt are in parallel: IgG=(I−Ig)SI_gG = (I - I_g)S.
  2. Ig=0.1I⇒0.1I×99=0.9I×SI_g = 0.1I \Rightarrow 0.1I \times 99 = 0.9I \times S
  3. S=9.90.9=11 ΩS = \dfrac{9.9}{0.9} = 11\ \Omega
  4. So the answer is (b) 11 Ω11\ \Omega.
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Q278MHT CET 2025 · 22 APR · SHIFT 1Easy

Kirchhoff's Laws & Measuring Instruments · Galvanometer to ammeter / voltmeter

The voltmeter has range 10 V and its internal resistance is 50 Ω50\ \Omega. To increase the range of voltmeter to 15 V, the resistance which is to be connected is

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Answer: (d)

  1. To raise a voltmeter's range a resistance is added in series: R=G(V′V−1)R = G\left(\frac{V'}{V} - 1\right).
  2. R=50(1510−1)=50×0.5=25 ΩR = 50\left(\frac{15}{10} - 1\right) = 50 \times 0.5 = 25\ \Omega
  3. So the answer is (d) 25 Ω25\ \Omega resistance in series.
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Q279MHT CET 2025 · 21 APR · SHIFT 2Easy

Kirchhoff's Laws & Measuring Instruments · Galvanometer to ammeter / voltmeter

If an ammeter is to be used in place of a galvanometer then we must connect

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Answer: (a)

  1. An ammeter must have very low resistance so that it does not change the circuit current.
  2. A low resistance (shunt) in parallel with the galvanometer carries most of the current and lowers the total resistance.
  3. So the answer is (a) low resistance in parallel.
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Q280MHT CET 2025 · 20 APR · SHIFT 2Easy

Kirchhoff's Laws & Measuring Instruments · Galvanometer to ammeter / voltmeter

Only 4%4\% of the total current in the circuit passes through a galvanometer. If the resistance of the galvanometer is G, then the shunt resistance connected to the galvanometer is

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Answer: (b)

GS, current I − IgIgII
  1. Galvanometer and shunt are in parallel, so IgG=(I−Ig)S⇒S=IgGI−IgI_g G = (I - I_g)S \Rightarrow S = \frac{I_g G}{I - I_g}.
  2. Ig=0.04I⇒S=0.04G0.96=G24I_g = 0.04I \Rightarrow S = \frac{0.04G}{0.96} = \frac{G}{24}
  3. So the answer is (b) G24\frac{G}{24}.
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Q281MHT CET 2025 · 19 APR · SHIFT 1Easy

Kirchhoff's Laws & Measuring Instruments · Galvanometer to ammeter / voltmeter

If only 5%5\% of the total current is to be passed through galvanometer of resistance G, then the resistance of the shunt will be

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Answer: (c)

  1. S=IgGI−IgS = \frac{I_g G}{I - I_g} for a shunt across the galvanometer.
  2. Ig=0.05I⇒S=0.05G0.95=G19I_g = 0.05I \Rightarrow S = \frac{0.05G}{0.95} = \frac{G}{19}
  3. So the answer is (c) G19\frac{G}{19}.
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Q282JEE Main 2025 · 23 JAN · SHIFT 2Easy

Kirchhoff's Laws & Measuring Instruments · Galvanometer to ammeter / voltmeter

A galvanometer having a coil of resistance 30 Ω30\ \Omega need 20 mA of current for full-scale deflection. If a maximum current of 3 A is to be measured using this galvanometer, the resistance of the shunt to be added to the galvanometer should be 30X Ω\frac{30}{X}\ \Omega, where XX is

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Answer: (c)

  1. S=IgGI−IgS = \frac{I_g G}{I - I_g}
  2. S=0.02×303−0.02=0.62.98=30149 ΩS = \frac{0.02 \times 30}{3 - 0.02} = \frac{0.6}{2.98} = \frac{30}{149}\ \Omega
  3. So X=149X = 149, answer (c).
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Q283MHT CET 2025 · 23 APR · SHIFT 1Medium

Kirchhoff's Laws & Measuring Instruments · Multi-loop circuits

In the following circuit, the current through 6 Ω6\ \Omega resistor is

Figure

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Answer: (b)

  1. The two identical cells (33 V, 3 Ω3\ \Omega each) are in parallel with like poles together.
  2. Equivalent cell: E=3E = 3 V, r=32=1.5 Ωr = \frac{3}{2} = 1.5\ \Omega.
  3. I=ER+r=36+1.5=37.5I = \frac{E}{R + r} = \frac{3}{6 + 1.5} = \frac{3}{7.5}
  4. So the answer is (b) 25\frac{2}{5} A.
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Q284NEET 2025Medium

Kirchhoff's Laws & Measuring Instruments · Multi-loop circuits

The current passing through the battery in the given circuit, is:

Figure

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Answer: (d)

  1. C and D are joined by a wire, and so are A and F. The network between B and E is: B–A 55, B–D 2.52.5, A–E 33, D–E 1.51.5, A–D 6 Ω6\ \Omega.
  2. 52.5=31.5\frac{5}{2.5} = \frac{3}{1.5}, so the bridge is balanced and 6 Ω6\ \Omega carries no current.
  3. RBE=(5+3)(2.5+1.5)8+4=83 ΩR_{BE} = \frac{(5+3)(2.5+1.5)}{8 + 4} = \frac{8}{3}\ \Omega
  4. Total =83+13+5.5+1.5=10 Ω= \frac{8}{3} + \frac{1}{3} + 5.5 + 1.5 = 10\ \Omega, so I=510I = \frac{5}{10}
  5. So the answer is (d) 0.5 A.
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Q285JEE Main 2025 · 22 JAN · SHIFT 2Medium

Kirchhoff's Laws & Measuring Instruments · Multi-loop circuits

The net current flowing in the given circuit is __________ A.

Figure

Numerical answer type. Work it out, then open the answer.

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Answer: 1

  1. In steady state the 1 μ1\ \muF capacitor blocks current, so everything to its right carries no current.
  2. At B: 3 Ω∥6 Ω=2 Ω3\ \Omega \parallel 6\ \Omega = 2\ \Omega to the bottom line; with 2 Ω2\ \Omega in series: 4 Ω4\ \Omega.
  3. This is in parallel with the 4 Ω4\ \Omega diagonal: Req=2 ΩR_{eq} = 2\ \Omega.
  4. I=22=1I = \frac{2}{2} = 1 A
  5. So the answer is 1 A.
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Q286MHT CET 2025 · 25 APR · SHIFT 2Medium

Kirchhoff's Laws & Measuring Instruments · Potential difference between points

Two batteries of e.m.f 4 V and 8 V with internal resistance 1 Ω1\ \Omega and 2 Ω2\ \Omega respectively are connected in a circuit with a resistance of 9 Ω9\ \Omega as shown in the figure. The current and potential difference between the points ' P ' and ' Q ' is R=9 ΩR = 9\ \Omega

Figure

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Answer: (b)

  1. The two cells face each other (+ to +), so they oppose: net emf =8−4=4= 8 - 4 = 4 V.
  2. Total resistance =1+2+9=12 Ω= 1 + 2 + 9 = 12\ \Omega, I=412=13I = \frac{4}{12} = \frac{1}{3} A.
  3. P and Q are the ends of RR: VPQ=IR=13×9=3V_{PQ} = IR = \frac{1}{3} \times 9 = 3 V
  4. So the answer is (b) 13\frac{1}{3} A and 3 V.
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Q287MHT CET 2025 · 21 APR · SHIFT 1Medium

Kirchhoff's Laws & Measuring Instruments · Potential difference between points

The value of current I in the given circuit is

Figure

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Answer: (c)

  1. Top-left corner: 2020 A in, 1515 A down, so 55 A goes right.
  2. Top-right corner: 5+3=85 + 3 = 8 A goes down.
  3. Bottom-left corner: 1515 A in, 55 A out, so 1010 A goes right.
  4. Bottom-right corner: I=10+8=18I = 10 + 8 = 18 A
  5. So the answer is (c) 18 A.
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Q288MHT CET 2025 · 20 APR · SHIFT 1Medium

Kirchhoff's Laws & Measuring Instruments · Potential difference between points

The potential difference (VA−VB)(V_A - V_B) between the points A and B in the given figure is

Figure

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Answer: (c)

  1. Walk from A to B along the current: potential drops by IRIR across each resistor and by EE when going from + to − of the cell.
  2. VA−2(2)−3−2(1)=VBV_A - 2(2) - 3 - 2(1) = V_B
  3. VA−VB=4+3+2=9 VV_A - V_B = 4 + 3 + 2 = 9\text{ V}
  4. So the answer is (c) 9 V.
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Q289NEET 2025Medium

Kirchhoff's Laws & Measuring Instruments · Potential difference between points

A constant voltage of 50 V is maintained between the points AA and BB of the circuit shown in the figure. The current through the branch CDCD of the circuit is:

Figure

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Answer: (d)

  1. 1 Ω∥3 Ω=0.75 Ω1\ \Omega \parallel 3\ \Omega = 0.75\ \Omega (between A-side and CD), 2 Ω∥4 Ω=43 Ω2\ \Omega \parallel 4\ \Omega = \frac{4}{3}\ \Omega (between CD and B-side).
  2. R=0.75+43=2512 ΩR = 0.75 + \frac{4}{3} = \frac{25}{12}\ \Omega, I=50×1225=24 AI = \frac{50 \times 12}{25} = 24\text{ A}
  3. First pair: V=24×0.75=18 VV = 24 \times 0.75 = 18\text{ V}, so I1Ω=18 AI_{1\Omega} = 18\text{ A}.
  4. Second pair: V=24×43=32 VV = 24 \times \frac{4}{3} = 32\text{ V}, so I2Ω=16 AI_{2\Omega} = 16\text{ A}.
  5. At C: ICD=18−16=2 AI_{CD} = 18 - 16 = 2\text{ A}
  6. So the answer is (d) 2.0 A.
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Q290JEE Main 2025 · 22 JAN · SHIFT 1Medium

Kirchhoff's Laws & Measuring Instruments · Potential difference between points

Figure

Sliding contact of a potentiometer is in the middle of the potentiometer wire having resistance Rp=1 ΩR_p = 1\ \Omega as shown in the figure. An external resistance of Re=2 ΩR_e = 2\ \Omega is connected via the sliding contact. The electric current in the circuit is:

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Answer: (c)

  1. Each half of the potentiometer is 0.5 Ω0.5\ \Omega; ReR_e is in parallel with the left half.
  2. 0.5∥2=0.5×22.5=0.4 Ω0.5 \parallel 2 = \frac{0.5 \times 2}{2.5} = 0.4\ \Omega
  3. Total R=0.4+0.5=0.9 ΩR = 0.4 + 0.5 = 0.9\ \Omega
  4. I=0.90.9=1.0 AI = \frac{0.9}{0.9} = 1.0\text{ A}
  5. So the answer is (c) 1.0 A.
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Q291MHT CET 2025 · 26 APR · SHIFT 1Medium

Kirchhoff's Laws & Measuring Instruments · Galvanometer to ammeter / voltmeter

When a resistance of 100 Ω100\ \Omega is connected in series with a galvanometer of resistance ' G ', its range is ' VV '. To double its range, a resistance of 1000 Ω1000\ \Omega is connected in series. The value of ' GG ' is

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Answer: (b)

  1. Range V=Ig(G+R)V = I_g(G + R) with fixed IgI_g.
  2. The intended reading is that the 1000 Ω1000\ \Omega replaces the 100 Ω100\ \Omega: Ig(G+1000)=2Ig(G+100)I_g(G + 1000) = 2 I_g(G + 100).
  3. G+1000=2G+200⇒G=800 ΩG + 1000 = 2G + 200 \Rightarrow G = 800\ \Omega
  4. So the answer is (b) 800 Ω800\ \Omega.
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Q292MHT CET 2025 · 25 APR · SHIFT 2Medium

Kirchhoff's Laws & Measuring Instruments · Galvanometer to ammeter / voltmeter

With a resistance ' X ' connected in series with a galvanometer of resistance 100 Ω100\ \Omega, it acts as a voltmeter of range 0−15 V0 - 15\text{ V}. To double the range, a resistance of 1500 Ω1500\ \Omega is to be connected in series with ' X '. The value of ' X ' in ohm is

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Answer: (c)

  1. Range V=Ig(G+R)V = I_g(G + R): Ig(100+X)=15I_g(100 + X) = 15.
  2. Double range: Ig(100+X+1500)=30I_g(100 + X + 1500) = 30
  3. Divide: 1600+X=2(100+X)⇒X=1400 Ω1600 + X = 2(100 + X) \Rightarrow X = 1400\ \Omega
  4. So the answer is (c) 1400.
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Q293MHT CET 2025 · 19 APR · SHIFT 1Medium

Kirchhoff's Laws & Measuring Instruments · Galvanometer to ammeter / voltmeter

The scale of a galvanometer is divided into 160 equal divisions. The galvanometer shows full scale deflection of 16 mA and maximum voltage is 80 mV. Now the range is changed so that galvanometer reads 160 V. The required resistance to be connected is

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Answer: (a)

  1. G=80 mV16 mA=5 ΩG = \dfrac{80\text{ mV}}{16\text{ mA}} = 5\ \Omega, Ig=0.016 AI_g = 0.016\text{ A}.
  2. To read 160 V160\text{ V}, add series R=VIg−G=1600.016−5R = \dfrac{V}{I_g} - G = \dfrac{160}{0.016} - 5
  3. R=10000−5=9995 ΩR = 10000 - 5 = 9995\ \Omega in series
  4. So the answer is (a).
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Q294JEE Main 2025 · 3 APR · SHIFT 1Medium

Kirchhoff's Laws & Measuring Instruments · Galvanometer to ammeter / voltmeter

In the figure shown below, a resistance of 150.4 Ω150.4\ \Omega is connected in series to an ammeter A of resistance 240 Ω240\ \Omega. A shunt resistance of 10 Ω10\ \Omega is connected in parallel with the ammeter. The reading of the ammeter is___________mA.

Figure

Numerical answer type. Work it out, then open the answer.

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Answer: 5

  1. Ammeter and shunt in parallel: 240×10250=9.6 Ω\frac{240 \times 10}{250} = 9.6\ \Omega
  2. Total resistance =150.4+9.6=160 Ω= 150.4 + 9.6 = 160\ \Omega, so I=20160=125 mAI = \frac{20}{160} = 125\text{ mA}
  3. Current through the ammeter coil: IA=125×10250=5 mAI_A = 125 \times \frac{10}{250} = 5\text{ mA}
  4. Answer: 5 mA
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Q295JEE Main 2025 · 23 JAN · SHIFT 1Medium

Kirchhoff's Laws & Measuring Instruments · Galvanometer sensitivity & figure of merit

Consider a moving coil galvanometer (MCG):

A. The torsional constant in moving coil galvanometer has dimensions [ML2T−2]\left[\mathrm{ML^2T^{-2}}\right] B. Increasing the current sensitivity may not necessarily increase the voltage sensitivity. C. If we increase number of turns (N)(\mathrm{N}) to its double (2N)(2\mathrm{N}), then the voltage sensitivity doubles. D. MCG can be converted into an ammeter by introducing a shunt resistance of large value in parallel with galvanometer.

E. Current sensitivity of MCG depends inversely on number of turns of coil.

Choose the correct answer from the options given below:

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Answer: (b)

  1. A: τ=kθ\tau = k\theta and θ\theta is dimensionless, so [k]=[ML2T−2][k] = [\mathrm{ML^2T^{-2}}]. True.
  2. B: Vs=IsRV_s = \frac{I_s}{R}; raising IsI_s by more turns also raises R, so VsV_s may not rise. True.
  3. C: N→2NN \to 2N doubles both IsI_s and R, so VsV_s is unchanged. False.
  4. D: an ammeter needs a small shunt, not a large one. False. E: Is=NABk∝NI_s = \frac{NAB}{k} \propto N, not inverse. False.
  5. So the answer is (b) A, B only.
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Q296MHT CET 2024 · 11 MAY · SHIFT 2Easy

Kirchhoff's Laws & Measuring Instruments · Kirchhoff's laws: basics

Kirchhoff's second law is based on the law of conservation of

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Answer: (b)

  1. Second law (loop rule): ∑E=∑IR\sum E = \sum IR around a closed loop, i.e. energy gained by a charge equals energy lost.
  2. So the answer is (b) energy.
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Q297MHT CET 2024 · 15 MAY · SHIFT 2Easy

Kirchhoff's Laws & Measuring Instruments · Multi-loop circuits

The figure shows currents in a part of electric circuit. Then current I is

Figure

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Answer: (a)

  1. First junction: 1+4=5 A1 + 4 = 5\text{ A} flows on.
  2. Second junction: 0.5 A0.5\text{ A} leaves, so 5−0.5=4.5 A5 - 0.5 = 4.5\text{ A} flows on.
  3. Third junction: 1 A1\text{ A} leaves, so I=4.5−1=3.5 AI = 4.5 - 1 = 3.5\text{ A}.
  4. So the answer is (a) 3.5 A.
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Q298MHT CET 2024 · 16 MAY · SHIFT 2Easy

Kirchhoff's Laws & Measuring Instruments · Galvanometer to ammeter / voltmeter

Two identical galvanometers are converted into an ammeter and into milliammeter. For the same current, the value of shunt of the ammeter as compared to that of milliammeter is

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Answer: (a)

  1. S=IgGI−IgS = \frac{I_g G}{I - I_g}, so for the same GG and IgI_g a larger range II needs a smaller shunt.
  2. An ammeter has a much larger range than a milliammeter.
  3. So the answer is (a) less.
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Q299MHT CET 2024 · 10 MAY · SHIFT 2Easy

Kirchhoff's Laws & Measuring Instruments · Galvanometer to ammeter / voltmeter

In an ammeter, 0.25%0.25\% of main current passes through the galvanometer. If the resistance of the galvanometer is ' G ', the resistance of ammeter will be

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Answer: (b)

  1. Ig=0.25% of I=I400I_g = 0.25\% \text{ of } I = \frac{I}{400}, so S=IgGI−Ig=G399S = \frac{I_g G}{I - I_g} = \frac{G}{399}.
  2. Ammeter resistance RA=SGS+G=G/399⋅GG/399+G=G400R_A = \frac{SG}{S + G} = \frac{G/399 \cdot G}{G/399 + G} = \frac{G}{400}
  3. So the answer is (b) 1400G\frac{1}{400}G.
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Q300MHT CET 2024 · 9 MAY · SHIFT 2Easy

Kirchhoff's Laws & Measuring Instruments · Galvanometer to ammeter / voltmeter

Two identical galvanometers are converted into voltmeter and millivoltmeter. As compared to the series resistance of voltmeter, the series resistance of millivoltmeter will be

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Answer: (c)

  1. Series resistance R=VIg−GR = \frac{V}{I_g} - G, so it grows with the range VV.
  2. A millivoltmeter has a much smaller range than a voltmeter, so it needs a smaller series resistance.
  3. So the answer is (c) less.
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Q301MHT CET 2024 · 9 MAY · SHIFT 1Easy

Kirchhoff's Laws & Measuring Instruments · Galvanometer to ammeter / voltmeter

In an ammeter, 4%4\% of the main current is passing through the galvanometer, If shunt resistance is 5 Ω5\ \Omega, then resistance of galvanometer will be

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Answer: (b)

  1. IgG=(I−Ig)S⇒G=(I−Ig)SIgI_g G = (I - I_g)S \Rightarrow G = \frac{(I - I_g)S}{I_g}
  2. Ig=0.04I⇒G=0.96×50.04=120 ΩI_g = 0.04I \Rightarrow G = \frac{0.96 \times 5}{0.04} = 120\ \Omega
  3. So the answer is (b) 120 Ω120\ \Omega.
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Q302MHT CET 2024 · 3 MAY · SHIFT 1Easy

Kirchhoff's Laws & Measuring Instruments · Galvanometer to ammeter / voltmeter

The range of voltmeter of resistance ' GG ' Ω\Omega is ' VV ' volt. The resistance required to be connected in series with it in order to convert it into a voltmeter of range ' nVnV ' volt, will be

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Answer: (a)

  1. Full-scale current Ig=VGI_g = \frac{V}{G}.
  2. For range nVnV: Ig(G+R)=nV⇒G+R=nGI_g(G + R) = nV \Rightarrow G + R = nG
  3. R=(n−1)GR = (n - 1)G, so the answer is (a).
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Q303JEE Main 2024 · 9 APR · SHIFT 1Easy

Kirchhoff's Laws & Measuring Instruments · Galvanometer to ammeter / voltmeter

A galvanometer has a coil of resistance 200 Ω200\ \Omega with a full scale deflection at 20 μA20\ \mu\text{A}. The value of resistance to be added to use it as an ammeter of range (0−20) mA(0 - 20)\text{ mA} is:

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Answer: (c)

  1. A shunt is needed: S=IgGI−IgS = \frac{I_g G}{I - I_g}.
  2. S=20×10−6×20020×10−3−20×10−6=4×10−319.98×10−3≈0.20 ΩS = \frac{20 \times 10^{-6} \times 200}{20 \times 10^{-3} - 20 \times 10^{-6}} = \frac{4 \times 10^{-3}}{19.98 \times 10^{-3}} \approx 0.20\ \Omega
  3. So the answer is (c) 0.20 Ω0.20\ \Omega.
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Q304JEE Main 2024 · 1 FEB · SHIFT 2Easy

Kirchhoff's Laws & Measuring Instruments · Galvanometer to ammeter / voltmeter

In an ammeter, 5%5\% of the main current passes through the galvanometer. If resistance of the galvanometer is GG, the resistance of ammeter will be:

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Answer: (c)

  1. Ig=0.05I=I20I_g = 0.05I = \frac{I}{20}, so S=IgGI−Ig=G19S = \frac{I_g G}{I - I_g} = \frac{G}{19}.
  2. RA=SGS+G=G/19⋅GG/19+G=G20R_A = \frac{SG}{S + G} = \frac{G/19 \cdot G}{G/19 + G} = \frac{G}{20}
  3. So the answer is (c) G20\frac{G}{20}.
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Q305JEE Main 2024 · 1 FEB · SHIFT 1Easy

Kirchhoff's Laws & Measuring Instruments · Galvanometer to ammeter / voltmeter

A galvanometer has a resistance of 50 Ω50\ \Omega and it allows maximum current of 5 mA5\text{ mA}. It can be converted into voltmeter to measure up to 100 V100\text{ V} by connecting in series a resistor of resistance:

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Answer: (d)

  1. Series resistance R=VIg−GR = \frac{V}{I_g} - G.
  2. R=1005×10−3−50=20000−50=19950 ΩR = \frac{100}{5 \times 10^{-3}} - 50 = 20000 - 50 = 19950\ \Omega
  3. So the answer is (d) 19950 Ω19950\ \Omega.
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Q306JEE Main 2024 · 29 JAN · SHIFT 1Easy

Kirchhoff's Laws & Measuring Instruments · Galvanometer to ammeter / voltmeter

A galvanometer having coil resistance 10 Ω10\ \Omega shows a full scale deflection for a current of 3 mA3\text{ mA}. For it to measure a current of 8 A8\text{ A}, the value of the shunt should be:

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Answer: (a)

  1. S=IgGI−IgS = \frac{I_g G}{I - I_g}
  2. S=3×10−3×108−0.003=0.037.997≈3.75×10−3 ΩS = \frac{3 \times 10^{-3} \times 10}{8 - 0.003} = \frac{0.03}{7.997} \approx 3.75 \times 10^{-3}\ \Omega
  3. So the answer is (a) 3.75×10−3 Ω3.75 \times 10^{-3}\ \Omega.
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Q307JEE Main 2024 · 27 JAN · SHIFT 2Easy

Kirchhoff's Laws & Measuring Instruments · Galvanometer to ammeter / voltmeter

A current of 200 μA200\ \mu\text{A} deflects the coil of a moving coil galvanometer through 60∘60^\circ. The current to cause deflection through π10\frac{\pi}{10} radian is:

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Answer: (d)

  1. In a moving coil galvanometer I∝θI \propto \theta.
  2. π10 rad=18∘\frac{\pi}{10}\text{ rad} = 18^\circ
  3. I=200×1860=60 μAI = 200 \times \frac{18}{60} = 60\ \mu\text{A}, so the answer is (d).
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Q308MHT CET 2024 · 2 MAY · SHIFT 2Medium

Kirchhoff's Laws & Measuring Instruments · Multi-loop circuits

In the following circuit, the current flowing through zener diode is

Figure

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Answer: (b)

  1. Zener holds 88 V across R2R_2: IR2=81600=5I_{R2} = \frac{8}{1600} = 5 mA.
  2. Voltage across R1=20−8=12R_1 = 20 - 8 = 12 V: IR1=12400=30I_{R1} = \frac{12}{400} = 30 mA.
  3. Iz=IR1−IR2=30−5=25I_z = I_{R1} - I_{R2} = 30 - 5 = 25 mA
  4. So the answer is (b) 25 mA.
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Q309MHT CET 2024 · 2 MAY · SHIFT 2Medium

Kirchhoff's Laws & Measuring Instruments · Multi-loop circuits

In the following circuit, the current I3I_3 is

Figure

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Answer: (d)

  1. Take the bottom junction O at 00 V; all three currents flow into O, so I1+I2+I3=0I_1 + I_2 + I_3 = 0.
  2. The 66 V cell (+ at top) puts the top junction T at +6+6 V.
  3. Left branch: the 88 V cell (+ towards O) puts the left corner at −8-8 V, so I1=6−(−8)28=0.5I_1 = \frac{6 - (-8)}{28} = 0.5 A.
  4. Right branch: the 1212 V cell (+ towards O) puts the right corner at −12-12 V, so I2=6+1254=13I_2 = \frac{6 + 12}{54} = \frac{1}{3} A.
  5. I3=−(0.5+13)=−56I_3 = -(0.5 + \frac{1}{3}) = -\frac{5}{6} A
  6. So the answer is (d) −56-\frac{5}{6} A.
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Q310JEE Main 2024 · 29 JAN · SHIFT 2Medium

Kirchhoff's Laws & Measuring Instruments · Multi-loop circuits

In the given circuit, the current in resistance R3{}_{3} is:

Figure

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Answer: (c)

  1. R2∥R3=4×48=2 ΩR_2 \parallel R_3 = \frac{4 \times 4}{8} = 2\ \Omega.
  2. Total =2+2+1=5 Ω= 2 + 2 + 1 = 5\ \Omega, so I=105=2I = \frac{10}{5} = 2 A.
  3. Equal resistors share equally: I3=22=1I_3 = \frac{2}{2} = 1 A
  4. So the answer is (c) 1 A.
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Q311MHT CET 2024 · 16 MAY · SHIFT 1Medium

Kirchhoff's Laws & Measuring Instruments · Potential difference between points

The potential difference (VA−VB)(V_A - V_B) between the points A and B in the given part of the circuit

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Answer: (d)

  1. Walk from A to B along the current I=3 AI = 3\text{ A}; drops are IRIR in resistors and EE going from + to − of the cell.
  2. VA−3(4)−5−3(2)=VBV_A - 3(4) - 5 - 3(2) = V_B
  3. VA−VB=12+5+6=23 VV_A - V_B = 12 + 5 + 6 = 23\text{ V}
  4. So the answer is (d) 23 V.
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Q312MHT CET 2024 · 11 MAY · SHIFT 1Medium

Kirchhoff's Laws & Measuring Instruments · Potential difference between points

In the following electrical network, the value of I is

Figure

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Answer: (c)

  1. Kirchhoff's junction rule: total current in = total current out.
  2. Current in =2+4=6 A= 2 + 4 = 6\text{ A}
  3. Current out =1+2+I= 1 + 2 + I
  4. 6=3+I⇒I=3 A6 = 3 + I \Rightarrow I = 3\text{ A}
  5. So the answer is (c) 3 A.
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Q313MHT CET 2024 · 10 MAY · SHIFT 2Medium

Kirchhoff's Laws & Measuring Instruments · Potential difference between points

The potential difference (VA−VB)(V_A - V_B) between the points ' AA ' and ' BB ' in the given part of the circuit is

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Answer: (d)

  1. Walk from A to B along I=2 AI = 2\text{ A}: drop IRIR in each resistor, drop EE across the cell (+ to −).
  2. VA−2(2)−3−2(1)=VBV_A - 2(2) - 3 - 2(1) = V_B
  3. VA−VB=4+3+2=9 VV_A - V_B = 4 + 3 + 2 = 9\text{ V}
  4. So the answer is (d) 9 V.
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Q314MHT CET 2024 · 9 MAY · SHIFT 2Medium

Kirchhoff's Laws & Measuring Instruments · Potential difference between points

Potential difference between the points A and B is nearly

Figure

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Answer: (c)

  1. 84=105\frac{8}{4} = \frac{10}{5}, so the bridge is balanced and no current flows through G.
  2. Branch ABC has 8+4=12 Ω8 + 4 = 12\ \Omega, branch ADC has 10+5=15 Ω10 + 5 = 15\ \Omega, in parallel.
  3. Current in ABC: I1=4×1512+15=209 AI_1 = 4 \times \frac{15}{12 + 15} = \frac{20}{9}\text{ A}
  4. VA−VB=8×209≈17.8 V≈18 VV_A - V_B = 8 \times \frac{20}{9} \approx 17.8\text{ V} \approx 18\text{ V}
  5. So the answer is (c) 18 V.
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Q315JEE Main 2024 · 30 JAN · SHIFT 1Medium

Kirchhoff's Laws & Measuring Instruments · Potential difference between points

A potential divider circuit is shown in figure. The output voltage V00 is:

Figure

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Answer: (c)

  1. All resistors are in series: R=3300+7×100=4000 ΩR = 3300 + 7 \times 100 = 4000\ \Omega.
  2. I=44000=1 mAI = \frac{4}{4000} = 1\text{ mA}
  3. V0V_0 is across five 100 Ω100\ \Omega resistors: V0=10−3×500=0.5 VV_0 = 10^{-3} \times 500 = 0.5\text{ V}
  4. So the answer is (c) 0.5 V.
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Q316JEE Main 2024 · 9 APR · SHIFT 1Medium

Kirchhoff's Laws & Measuring Instruments · Potential difference between points

The current flowing through the 1 Ω1\ \Omega resistor is n10\frac{n}{10} A. The value of nn is _______.

Figure

Numerical answer type. Work it out, then open the answer.

Show answer and solution

Answer: 25

  1. Take VA=0V_A = 0. The 5 V cell gives VB−VD=5V_B - V_D = 5; current in the 1 Ω1\ \Omega branch C to A is I1=VC+10I_1 = V_C + 10.
  2. KCL for the B+D pair: VB+VD4=2VC−VB−VD2⇒VC=34(VB+VD)\frac{V_B + V_D}{4} = \frac{2V_C - V_B - V_D}{2} \Rightarrow V_C = \frac{3}{4}(V_B + V_D)
  3. KCL at C: VB+VD−2VC2=VC+10⇒VB+VD=−10\frac{V_B + V_D - 2V_C}{2} = V_C + 10 \Rightarrow V_B + V_D = -10, VC=−7.5 VV_C = -7.5\text{ V}
  4. I1=−7.5+10=2.5 A=2510 AI_1 = -7.5 + 10 = 2.5\text{ A} = \frac{25}{10}\text{ A}
  5. So n=25n = 25.
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Q317MHT CET 2024 · 9 MAY · SHIFT 2Medium

Kirchhoff's Laws & Measuring Instruments · Voltmeter & ammeter in circuits

In the following circuit, the reading in the ammeter is

Figure

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Answer: (c)

  1. The Si diode is forward biased; it drops about 0.7 V0.7\text{ V}.
  2. I=5−0.7200=4.3200 AI = \dfrac{5 - 0.7}{200} = \dfrac{4.3}{200}\text{ A}
  3. I=0.0215 A=21.5 mAI = 0.0215\text{ A} = 21.5\text{ mA}
  4. So the answer is (c) 21.5 mA21.5\text{ mA}.
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Q318JEE Main 2024 · 1 FEB · SHIFT 1Medium

Kirchhoff's Laws & Measuring Instruments · Voltmeter & ammeter in circuits

The reading in the ideal voltmeter (V)(\text{V}) shown in the given circuit diagram is:

Figure

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Answer: (d)

  1. All 8 cells aid each other: E=8×5=40 VE = 8 \times 5 = 40\text{ V}, r=8×0.2=1.6 Ωr = 8 \times 0.2 = 1.6\ \Omega.
  2. I=401.6=25 AI = \dfrac{40}{1.6} = 25\text{ A}
  3. The voltmeter reads the terminal voltage of one cell: V=E−Ir=5−25×0.2=0V = E - Ir = 5 - 25 \times 0.2 = 0
  4. So the answer is (d) 0 V0\text{ V}.
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Q319JEE Main 2024 · 27 JAN · SHIFT 2Medium

Kirchhoff's Laws & Measuring Instruments · Voltmeter & ammeter in circuits

Three voltmeters, all having different internal resistances are joined as shown in figure. When some potential difference is applied across AA and BB, their readings are V1,V2V_1, V_2 and V3V_3. Choose the correct option.

Figure

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Answer: (c)

  1. V1V_1 and V2V_2 are in series, and that pair is in parallel with V3V_3 across A and B.
  2. Series: VAB=V1+V2V_{AB} = V_1 + V_2. Parallel: VAB=V3V_{AB} = V_3.
  3. So the answer is (c) V1+V2=V3V_1 + V_2 = V_3.
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Q320JEE Main 2024 · 30 JAN · SHIFT 2Medium

Kirchhoff's Laws & Measuring Instruments · Voltmeter & ammeter in circuits

Two resistances of 100 Ω100\ \Omega and 200 Ω200\ \Omega are connected in series with a battery of 4 V4\text{ V} and negligible internal resistance. A voltmeter is used to measure voltage across 100 Ω100\ \Omega resistance, which gives reading as 1 V1\text{ V}. The resistance of voltmeter must be _______ Ω\Omega.

Numerical answer type. Work it out, then open the answer.

Show answer and solution

Answer: 200

  1. The 200 Ω200\ \Omega resistor gets 4−1=3 V4 - 1 = 3\text{ V}, so I=3200=0.015 AI = \dfrac{3}{200} = 0.015\text{ A}.
  2. Combination 100∥RV=10.015=2003 Ω100 \| R_V = \dfrac{1}{0.015} = \dfrac{200}{3}\ \Omega
  3. 100RV100+RV=2003⇒300RV=20000+200RV\dfrac{100 R_V}{100 + R_V} = \dfrac{200}{3} \Rightarrow 300 R_V = 20000 + 200 R_V
  4. RV=200 ΩR_V = 200\ \Omega
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Q321JEE Main 2024 · 29 JAN · SHIFT 2Medium

Kirchhoff's Laws & Measuring Instruments · Voltmeter & ammeter in circuits

In the given circuit, the current flowing through the resistance 20 Ω20\ \Omega is 0.3 A0.3\text{ A}, while the ammeter reads 0.9 A0.9\text{ A}. The value of R1R_1 is _________ Ω\Omega.

Figure

Numerical answer type. Work it out, then open the answer.

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Answer: 30

  1. All three resistors are in parallel: V=0.3×20=6 VV = 0.3 \times 20 = 6\text{ V}.
  2. Current in 15 Ω=615=0.4 A15\ \Omega = \dfrac{6}{15} = 0.4\text{ A}.
  3. The ammeter reads the total: I1=0.9−0.3−0.4=0.2 AI_1 = 0.9 - 0.3 - 0.4 = 0.2\text{ A}
  4. R1=60.2=30 ΩR_1 = \dfrac{6}{0.2} = 30\ \Omega
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Q322MHT CET 2024 · 16 MAY · SHIFT 2Medium

Kirchhoff's Laws & Measuring Instruments · Galvanometer to ammeter / voltmeter

When a resistance of 200 Ω200\ \Omega is connected in series with a galvanometer of resistance ' GG ', its range is ' VV '. To triple its range, a resistance of 2000 Ω2000\ \Omega is connected in series. The value of ' GG ' is

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Answer: (d)

  1. Range V=Ig(G+R)V = I_g(G + R) with fixed IgI_g.
  2. The intended reading is that the 2000 Ω2000\ \Omega replaces the 200 Ω200\ \Omega: G+2000=3(G+200)G + 2000 = 3(G + 200).
  3. 2G=1400⇒G=700 Ω2G = 1400 \Rightarrow G = 700\ \Omega
  4. So the answer is (d) 700 Ω700\ \Omega.
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Q323MHT CET 2024 · 16 MAY · SHIFT 1Medium

Kirchhoff's Laws & Measuring Instruments · Galvanometer to ammeter / voltmeter

When a resistance of 100 Ω100\ \Omega is connected in series with a galvanometer of resistance G, its range is V. To double its range a resistance of 1000 Ω1000\ \Omega is connected in series. The value of G is

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Answer: (a)

  1. Range V=Ig(G+R)V = I_g(G + R) with fixed IgI_g.
  2. The intended reading is that the 1000 Ω1000\ \Omega replaces the 100 Ω100\ \Omega: G+1000=2(G+100)G + 1000 = 2(G + 100).
  3. G=1000−200=800 ΩG = 1000 - 200 = 800\ \Omega
  4. So the answer is (a) 800 Ω800\ \Omega.
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Q324MHT CET 2024 · 15 MAY · SHIFT 2Medium

Kirchhoff's Laws & Measuring Instruments · Galvanometer to ammeter / voltmeter

A galvanometer may be converted into ammeter or a voltmeter. In which of the following cases the resistance of the device so obtained will be the largest?

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Answer: (d)

  1. An ammeter has a small shunt in parallel, so its resistance is less than GG.
  2. A voltmeter has a large series resistance: RV=VIgR_V = \dfrac{V}{I_g}, which grows with the range.
  3. Largest resistance: voltmeter of the higher range. So the answer is (d) voltmeter of range 10 V.
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Q325MHT CET 2024 · 15 MAY · SHIFT 1Medium

Kirchhoff's Laws & Measuring Instruments · Galvanometer to ammeter / voltmeter

A galvanometer has resistance 80 Ω80\ \Omega and it is shunted with resistance 20 Ω20\ \Omega. If 20%20\% of the main current flows through galvanometer, then what is the value of main current?

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Answer: (c)

  1. Shunt and galvanometer have the same p.d.: IgG=IsSI_g G = I_s S.
  2. Fraction through galvanometer =SG+S=20100=20%= \dfrac{S}{G + S} = \dfrac{20}{100} = 20\%, true for any main current.
  3. So the data cannot fix the main current; the key's value 1 A1\text{ A} needs extra data (e.g. Ig=0.2 AI_g = 0.2\text{ A}).
  4. Taking Ig=0.2 AI_g = 0.2\text{ A}: I=0.20.2=1 AI = \dfrac{0.2}{0.2} = 1\text{ A}, answer (c).
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Q326MHT CET 2024 · 4 MAY · SHIFT 2Medium

Kirchhoff's Laws & Measuring Instruments · Galvanometer to ammeter / voltmeter

When a galvanometer is shunted by a resistance ' ss ', its current capacity increases ' nn ' times. If the same galvanometer is shunted by another resistance ' s1s_1 ', its capacity will increase to ' n1n_1 ' times original current. The value of ' n1n_1 ' is

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Answer: (d)

  1. Shunt for nn-fold range: s=Gn−1⇒G=s(n−1)s = \dfrac{G}{n - 1} \Rightarrow G = s(n - 1).
  2. With shunt s1s_1: n1=1+Gs1=1+s(n−1)s1n_1 = 1 + \dfrac{G}{s_1} = 1 + \dfrac{s(n - 1)}{s_1}
  3. n1=s(n−1)+s1s1n_1 = \dfrac{s(n - 1) + s_1}{s_1}. So the answer is (d).
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Q327MHT CET 2024 · 3 MAY · SHIFT 2Medium

Kirchhoff's Laws & Measuring Instruments · Galvanometer to ammeter / voltmeter

When a galvanometer is shunted by a resistance ' SS ', its current capacity increases ' nn ' times. If the same galvanometer is shunted by another resistance S′S', its current capacity increases to n′n'. The value of n′n' in terms of n,Sn, S and S′S' is

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Answer: (d)

  1. Shunt for nn-fold range: S=Gn−1⇒G=S(n−1)S = \dfrac{G}{n - 1} \Rightarrow G = S(n - 1).
  2. With shunt S′S': n′=1+GS′=1+S(n−1)S′n' = 1 + \dfrac{G}{S'} = 1 + \dfrac{S(n - 1)}{S'}
  3. n′=S(n−1)+S′S′n' = \dfrac{S(n - 1) + S'}{S'}. So the answer is (d).
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Q328JEE Main 2024 · 5 APR · SHIFT 2Medium

Kirchhoff's Laws & Measuring Instruments · Galvanometer to ammeter / voltmeter

A galvanometer of resistance 100 Ω100\ \Omega when connected in series with 400 Ω400\ \Omega measures a voltage of upto 10 V10\text{ V}. The value of resistance required to convert the galvanometer into ammeter to read upto 10 A10\text{ A} is x×10−2 Ωx \times 10^{-2}\ \Omega. The value of xx is:

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Answer: (b)

  1. Full-scale current: Ig=10100+400=0.02 AI_g = \dfrac{10}{100 + 400} = 0.02\text{ A}.
  2. Shunt for 10 A10\text{ A}: S=IgGI−Ig=0.02×1009.98S = \dfrac{I_g G}{I - I_g} = \dfrac{0.02 \times 100}{9.98}
  3. S≈0.2 Ω=20×10−2 ΩS \approx 0.2\ \Omega = 20 \times 10^{-2}\ \Omega
  4. x=20x = 20. So the answer is (b).
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Q329JEE Main 2024 · 29 JAN · SHIFT 1Medium

Kirchhoff's Laws & Measuring Instruments · Galvanometer to ammeter / voltmeter

The deflection in moving coil galvanometer falls from 25 divisions to 5 divisions when a shunt of 24 Ω24\ \Omega is applied. The resistance of galvanometer coil will be:

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Answer: (c)

  1. Deflection falls from 25 to 5 divisions, so Ig=I5I_g = \frac{I}{5} and the shunt carries 4I5\frac{4I}{5}.
  2. G and S are in parallel: IgG=(I−Ig)S⇒I5G=4I5(24)I_g G = (I - I_g)S \Rightarrow \frac{I}{5}G = \frac{4I}{5}(24)
  3. G=4×24=96 ΩG = 4 \times 24 = 96\ \Omega
  4. So the answer is (c) 96 Ω96\ \Omega.
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Q330JEE Main 2024 · 6 APR · SHIFT 2Medium

Kirchhoff's Laws & Measuring Instruments · Galvanometer to ammeter / voltmeter

In the given figure an ammeter A consists of a 240 Ω240\ \Omega coil connected in parallel to a 10 Ω10\ \Omega shunt. The reading of the ammeter is ________ mA\text{mA}.

Figure

Numerical answer type. Work it out, then open the answer.

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Answer: 160

  1. Ammeter resistance =240×10250=9.6 Ω= \frac{240 \times 10}{250} = 9.6\ \Omega
  2. Total resistance =140.4+9.6=150 Ω= 140.4 + 9.6 = 150\ \Omega
  3. I=24150=0.16 AI = \frac{24}{150} = 0.16\text{ A}
  4. Answer: 160 mA
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Q331MHT CET 2024 · 2 MAY · SHIFT 1Difficult

Kirchhoff's Laws & Measuring Instruments · Galvanometer to ammeter / voltmeter

A galvanometer of resistance ' GG ' is shunted by resistance of 'S' ohm. To keep the main current in the circuit unchanged the resistance to be put in series with Galvanometer is

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Answer: (a)

  1. Shunted galvanometer: GSG+S\frac{GS}{G + S}; it must still offer resistance G to keep the main current unchanged.
  2. r+GSG+S=Gr + \frac{GS}{G + S} = G
  3. r=G−GSG+S=G2G+Sr = G - \frac{GS}{G + S} = \frac{G^2}{G + S}
  4. So the answer is (a).
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Q332JEE Main 2024 · 5 APR · SHIFT 1Difficult

Kirchhoff's Laws & Measuring Instruments · Galvanometer sensitivity & figure of merit

In the experiment to determine the galvanometer resistance by half-deflection method, the plot of 1/θ1/\theta vs the resistance (R) of the resistance box is shown in the figure. The figure of merit of the galvanometer is _________ ×10−1 A/\times 10^{-1}\text{ A}/ division. [The source has emf 2 V2\text{ V}]

Figure

Numerical answer type. Work it out, then open the answer.

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Answer: 5

  1. θ=IK\theta = \frac{I}{K} with I=ER+GI = \frac{E}{R + G}, so 1θ=KER+KGE\frac{1}{\theta} = \frac{K}{E}R + \frac{KG}{E}
  2. Slope =2−16−2=14=KE= \frac{2 - 1}{6 - 2} = \frac{1}{4} = \frac{K}{E}
  3. K=24=0.5 A/div=5×10−1 A/divK = \frac{2}{4} = 0.5\text{ A/div} = 5 \times 10^{-1}\text{ A/div}
  4. Answer: 5
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Q333MHT CET 2023 · 14 MAY · SHIFT 1Easy

Kirchhoff's Laws & Measuring Instruments · Kirchhoff's laws: basics

Five current carrying conductors meet at point P\text{P}. What is the magnitude and direction of the current in conductor PQ\text{PQ}?

Figure

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Answer: (b)

  1. Junction rule at P: total current in = total current out.
  2. In: 5+4=9 A5 + 4 = 9\text{ A}; out (excluding PQ): 3+5=8 A3 + 5 = 8\text{ A}.
  3. Remaining 9−8=1 A9 - 8 = 1\text{ A} must leave through PQ.
  4. So the answer is (b) 1 A from P to Q.
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Q334MHT CET 2023 · 9 MAY · SHIFT 2Easy

Kirchhoff's Laws & Measuring Instruments · Kirchhoff's laws: basics

Five current carrying conductors meet at a point 'O\text{O}' as shown in figure. The magnitude and direction of the current in conductor 'OPOP' is

Figure

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Answer: (d)

  1. Junction rule at O: total current in = total current out.
  2. In: 10+2.5+5=17.5 A10 + 2.5 + 5 = 17.5\text{ A}; out: 6 A6\text{ A}.
  3. IOP=17.5−6=11.5 AI_{OP} = 17.5 - 6 = 11.5\text{ A}, flowing out of O.
  4. So the answer is (d) 11.5 A from O to P.
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Q335MHT CET 2023 · 13 MAY · SHIFT 2Easy

Kirchhoff's Laws & Measuring Instruments · Galvanometer to ammeter / voltmeter

Only 4%4\% of the total current in the circuit passes through a galvanometer. If the resistance of the galvanometer is GG, then the shunt resistance connected to the galvanometer is

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Answer: (b)

  1. S=IgGI−IgS = \frac{I_g G}{I - I_g} for a shunt across the galvanometer.
  2. Ig=0.04I⇒S=0.04G0.96=G24I_g = 0.04I \Rightarrow S = \frac{0.04G}{0.96} = \frac{G}{24}
  3. So the answer is (b) G24\frac{G}{24}.
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Q336MHT CET 2023 · 10 MAY · SHIFT 2Easy

Kirchhoff's Laws & Measuring Instruments · Galvanometer to ammeter / voltmeter

A galvanometer has resistance 'GG' and range 'VgV_g'. How much resistance is required to read voltage upto 'VV' volt?

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Answer: (a)

  1. Full-scale current Ig=VgGI_g = \frac{V_g}{G}.
  2. Series resistance: R=VIg−G=VGVg−GR = \frac{V}{I_g} - G = \frac{VG}{V_g} - G
  3. R=G(VVg−1)R = G\left(\frac{V}{V_g} - 1\right), so the answer is (a).
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Q337MHT CET 2023 · 10 MAY · SHIFT 1Easy

Kirchhoff's Laws & Measuring Instruments · Galvanometer to ammeter / voltmeter

If only 1%1\% of total current is passed through a galvanometer of resistance 'GG' then the resistance of the shunt is

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Answer: (d)

  1. S=IgGI−IgS = \frac{I_g G}{I - I_g} for a shunt across the galvanometer.
  2. Ig=0.01I⇒S=0.01G0.99=G99I_g = 0.01I \Rightarrow S = \frac{0.01G}{0.99} = \frac{G}{99}
  3. So the answer is (d) G99 Ω\frac{G}{99}\ \Omega.
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Q338MHT CET 2023 · 9 MAY · SHIFT 1Easy

Kirchhoff's Laws & Measuring Instruments · Galvanometer to ammeter / voltmeter

A galvanometer of resistance GG is shunted with a resistance of 10%10\% of GG. The part of the total current that flows through the galvanometer is

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Answer: (a)

  1. Current divides inversely as resistance: Ig=ISS+GI_g = I\frac{S}{S + G}.
  2. S=0.1G⇒Ig=I0.1G1.1G=I11S = 0.1G \Rightarrow I_g = I\frac{0.1G}{1.1G} = \frac{I}{11}
  3. So the answer is (a) 111I\frac{1}{11}I.
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Q339NEET 2023Medium

Kirchhoff's Laws & Measuring Instruments · Multi-loop circuits

The magnitude and direction of the current in the following circuit is :-

Figure

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Answer: (a)

  1. The two cells oppose each other: net emf =10−5=5= 10 - 5 = 5 V.
  2. Total resistance =2+1+7=10 Ω= 2 + 1 + 7 = 10\ \Omega.
  3. I=510=0.5I = \frac{5}{10} = 0.5 A, in the direction set by the larger 1010 V cell: from A to B through E.
  4. So the answer is (a).
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Q340JEE Main 2023 · 11 APR · SHIFT 2Medium

Kirchhoff's Laws & Measuring Instruments · Multi-loop circuits

Figure

The current flowing through R2{}_{2} is:

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Answer: (b)

  1. Between C and A: R3=6 ΩR_3 = 6\ \Omega in parallel with R2+R1=6 ΩR_2 + R_1 = 6\ \Omega gives 3 Ω3\ \Omega.
  2. D to A: (R4+3)=6 Ω(R_4 + 3) = 6\ \Omega in parallel with R7=3 ΩR_7 = 3\ \Omega gives 2 Ω2\ \Omega.
  3. Path E–D–A: R5+2=4 ΩR_5 + 2 = 4\ \Omega across 88 V, so VDA=8×24=4V_{DA} = 8 \times \frac{2}{4} = 4 V.
  4. VCA=4×36=2V_{CA} = 4 \times \frac{3}{6} = 2 V, so the current in R2R_2 (via R2+R1=6 ΩR_2 + R_1 = 6\ \Omega) is 26\frac{2}{6}.
  5. So the answer is (b) 13\frac{1}{3} A.
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Q341JEE Main 2023 · 30 JAN · SHIFT 1Medium

Kirchhoff's Laws & Measuring Instruments · Multi-loop circuits

In the following circuit, the magnitude of current I1, is ___________ A.

Figure

Numerical answer type. Work it out, then open the answer.

Show answer and solution

Answer: 1.5

  1. Take the bottom node at 00; the 22 V cell fixes the top node at 22 V. Let the middle node be VMV_M and the left wire VLV_L.
  2. KCL at M: 2−VM1−VM1=(VM+5−VL)⇒VL=3VM+3\frac{2 - V_M}{1} - \frac{V_M}{1} = (V_M + 5 - V_L) \Rightarrow V_L = 3V_M + 3
  3. KCL at L: VM+5−VL=VL−22+VL2⇒VM=0V_M + 5 - V_L = \frac{V_L - 2}{2} + \frac{V_L}{2} \Rightarrow V_M = 0, VL=3V_L = 3 V.
  4. Top node: I3=3−22=0.5I_3 = \frac{3 - 2}{2} = 0.5 A in, 2−01=2\frac{2 - 0}{1} = 2 A out, so I1=2−0.5=1.5I_1 = 2 - 0.5 = 1.5 A.
  5. So the answer is 1.5 A.
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Q342MHT CET 2023 · 11 MAY · SHIFT 2Medium

Kirchhoff's Laws & Measuring Instruments · Potential difference between points

Potential difference between the points P and Q is nearly

Figure

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Answer: (a)

  1. 63=84\frac{6}{3} = \frac{8}{4}, so the bridge is balanced and G carries no current.
  2. Branch PQR has 9 Ω9\ \Omega, branch PSR has 12 Ω12\ \Omega, in parallel.
  3. Current in PQR: I1=5×129+12=207 AI_1 = 5 \times \frac{12}{9 + 12} = \frac{20}{7}\text{ A}
  4. VP−VQ=6×207≈17.1 V≈17 VV_P - V_Q = 6 \times \frac{20}{7} \approx 17.1\text{ V} \approx 17\text{ V}
  5. So the answer is (a) 17 V.
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Q343MHT CET 2023 · 11 MAY · SHIFT 1Medium

Kirchhoff's Laws & Measuring Instruments · Potential difference between points

Potential difference between the points P and Q is nearly

Figure

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Answer: (c)

  1. 63=84\frac{6}{3} = \frac{8}{4}, so the bridge is balanced and G carries no current.
  2. Branch PQR has 9 Ω9\ \Omega, branch PSR has 12 Ω12\ \Omega, in parallel.
  3. Current in PQR: I1=5×1221=207 AI_1 = 5 \times \frac{12}{21} = \frac{20}{7}\text{ A}
  4. VP−VQ=6×207≈17 VV_P - V_Q = 6 \times \frac{20}{7} \approx 17\text{ V}
  5. So the answer is (c) 17 V.
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Q344JEE Main 2023 · 6 APR · SHIFT 2Medium

Kirchhoff's Laws & Measuring Instruments · Potential difference between points

Figure shows a part of an electric circuit. The potentials at points a,ba, b and cc are 30 V,12 V30\text{ V}, 12\text{ V} and 2 V2\text{ V} respectively. The current through the 20 Ω20\ \Omega resistor will be,

Figure

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Answer: (c)

  1. Let the junction potential be VV. KCL: 30−V10=V−1220+V−230\frac{30 - V}{10} = \frac{V - 12}{20} + \frac{V - 2}{30}
  2. Multiply by 60: 180−6V=3V−36+2V−4⇒V=20 V180 - 6V = 3V - 36 + 2V - 4 \Rightarrow V = 20\text{ V}
  3. I20=20−1220=0.4 AI_{20} = \frac{20 - 12}{20} = 0.4\text{ A}
  4. So the answer is (c) 0.4 A.
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Q345JEE Main 2023 · 24 JAN · SHIFT 1Medium

Kirchhoff's Laws & Measuring Instruments · Potential difference between points

As shown in the figure, a network of resistors is connected to a battery of 24V with an internal resistance of 3 Ω\Omega. The currents through the resistors R44 and R55 are I44 and I55 respectively. The values of I44 and I55 are:

Figure

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Answer: (c)

  1. R1∥R2=1 ΩR_1 \parallel R_2 = 1\ \Omega, R4∥R5=20×525=4 ΩR_4 \parallel R_5 = \frac{20 \times 5}{25} = 4\ \Omega
  2. Total R=1+2+4+2+3=12 ΩR = 1 + 2 + 4 + 2 + 3 = 12\ \Omega, I=2412=2 AI = \frac{24}{12} = 2\text{ A}
  3. Current divides inversely with resistance: I4=2×525=25 AI_4 = 2 \times \frac{5}{25} = \frac{2}{5}\text{ A}, I5=2×2025=85 AI_5 = 2 \times \frac{20}{25} = \frac{8}{5}\text{ A}
  4. So the answer is (c).
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Q346JEE Main 2023 · 15 APR · SHIFT 1Medium

Kirchhoff's Laws & Measuring Instruments · Potential difference between points

A network of four resistances is connected to 9 V9\text{ V} battery, as shown in figure. The magnitude of voltage difference between the points A\text{A} and B\text{B} is __________ VV.

Figure

Numerical answer type. Work it out, then open the answer.

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Answer: 3

  1. Each side of the bridge gets the full 9 V across 2+4=6 Ω2 + 4 = 6\ \Omega.
  2. Drop from the left node to A (across 2 Ω2\ \Omega): 9×26=3 V9 \times \frac{2}{6} = 3\text{ V}
  3. Drop from the left node to B (across 4 Ω4\ \Omega): 9×46=6 V9 \times \frac{4}{6} = 6\text{ V}
  4. ∣VA−VB∣=6−3=3 V|V_A - V_B| = 6 - 3 = 3\text{ V}
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Q347JEE Main 2023 · 11 APR · SHIFT 1Medium

Kirchhoff's Laws & Measuring Instruments · Potential difference between points

In the circuit diagram shown in figure given below, the current flowing through resistance 3 Ω3\ \Omega is x3 A\frac{x}{3}\text{ A}. The value of xx is ___________

Figure

Numerical answer type. Work it out, then open the answer.

Show answer and solution

Answer: 1

  1. The 4 V and 8 V cells have their + terminals facing each other, so they oppose: net emf =8−4=4 V= 8 - 4 = 4\text{ V}.
  2. 3∥6=2 Ω3 \parallel 6 = 2\ \Omega; total R=4.5+2+0.5+1=8 ΩR = 4.5 + 2 + 0.5 + 1 = 8\ \Omega
  3. I=48=0.5 AI = \frac{4}{8} = 0.5\text{ A}
  4. I3=0.5×63+6=13 AI_3 = 0.5 \times \frac{6}{3 + 6} = \frac{1}{3}\text{ A}
  5. So x=1x = 1.
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Q348JEE Main 2023 · 1 FEB · SHIFT 2Medium

Kirchhoff's Laws & Measuring Instruments · Potential difference between points

In the given circuit, the value of ∣I1+I3I2∣\left|\frac{I_1+I_3}{I_2}\right| is _____________

Figure

Numerical answer type. Work it out, then open the answer.

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Answer: 2

  1. Take the left wire at 0. The middle node M is the + side of both cells: VM=+10 VV_M = +10\text{ V}, right wire =10−20=−10 V= 10 - 20 = -10\text{ V}.
  2. I1=I2=10−010=1 AI_1 = I_2 = \frac{10 - 0}{10} = 1\text{ A} (M to left)
  3. I3=0−(−10)10=1 AI_3 = \frac{0 - (-10)}{10} = 1\text{ A} (left to right)
  4. ∣I1+I3I2∣=1+11=2\left|\frac{I_1 + I_3}{I_2}\right| = \frac{1 + 1}{1} = 2
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Q349JEE Main 2023 · 1 FEB · SHIFT 2Medium

Kirchhoff's Laws & Measuring Instruments · Voltmeter & ammeter in circuits

Given below are two statements: One is labelled as Assertion A\mathbf{A} and the other is labelled as Reason R\mathbf{R}.

Assertion A: For measuring the potential difference across a resistance of 600 Ω600\ \Omega, the voltmeter with resistance 1000 Ω1000\ \Omega will be preferred over voltmeter with resistance 4000 Ω4000\ \Omega.

Reason R: Voltmeter with higher resistance will draw smaller current than voltmeter with lower resistance.

In the light of the above statements, choose the most appropriate answer from the options given below.

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Answer: (a)

  1. A good voltmeter should draw as little current as possible, so it should have a much larger resistance than the 600 Ω600\ \Omega resistor.
  2. The 4000 Ω4000\ \Omega voltmeter disturbs the circuit less, so it is preferred: A is false.
  3. For the same p.d., I=V/RVI = V/R_V is smaller for larger RVR_V: R is true.
  4. So the answer is (a).
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Q350JEE Main 2023 · 24 JAN · SHIFT 2Medium

Kirchhoff's Laws & Measuring Instruments · Voltmeter & ammeter in circuits

A cell of emf 90 V is connected across series combination of two resistors each of 100Ω\Omega resistance. A voltmeter of resistance 400Ω\Omega is used to measure the potential difference across each resistor. The reading of the voltmeter will be:

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Answer: (a)

  1. 100 Ω∥400 Ω=80 Ω100\ \Omega \| 400\ \Omega = 80\ \Omega
  2. Total R=100+80=180 ΩR = 100 + 80 = 180\ \Omega, I=90180=0.5 AI = \dfrac{90}{180} = 0.5\text{ A}.
  3. Reading =0.5×80=40 V= 0.5 \times 80 = 40\text{ V}
  4. So the answer is (a) 40 V.
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Q351JEE Main 2023 · 6 APR · SHIFT 2Medium

Kirchhoff's Laws & Measuring Instruments · Voltmeter & ammeter in circuits

As shown in the figure, the voltmeter reads 2 V2\text{ V} across 5 Ω5\ \Omega resistor. The resistance of the voltmeter is _________ Ω\Omega.

Figure

Numerical answer type. Work it out, then open the answer.

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Answer: 20

  1. The 2 Ω2\ \Omega resistor gets 3−2=1 V3 - 2 = 1\text{ V}, so the total current is 12=0.5 A\dfrac{1}{2} = 0.5\text{ A}.
  2. Current through 5 Ω=25=0.4 A5\ \Omega = \dfrac{2}{5} = 0.4\text{ A}, so the voltmeter carries 0.1 A0.1\text{ A}.
  3. RV=20.1=20 ΩR_V = \dfrac{2}{0.1} = 20\ \Omega
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Q352MHT CET 2023 · 14 MAY · SHIFT 2Medium

Kirchhoff's Laws & Measuring Instruments · Galvanometer to ammeter / voltmeter

When moving coil galvanometer (MCG) is converted into a voltmeter, the series resistance is 'nn' times the resistance of galvanometer. How many times that of MCG the voltmeter is now capable of measuring voltage?

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Answer: (c)

  1. Galvanometer alone measures Vg=IgGV_g = I_g G.
  2. With series R=nGR = nG: V=Ig(G+nG)=(n+1)IgGV = I_g(G + nG) = (n + 1) I_g G
  3. VVg=n+1\dfrac{V}{V_g} = n + 1. So the answer is (c).
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Q353MHT CET 2023 · 9 MAY · SHIFT 2Medium

Kirchhoff's Laws & Measuring Instruments · Galvanometer to ammeter / voltmeter

A galvanometer of resistance 20 Ω20\ \Omega gives a deflection of 5 divisions when 1 mA1\text{ mA} current flows through it. The galvanometer scale has 50 divisions. To convert the galvanometer into a voltmeter of range 25 volt, we should connect a resistance of

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Answer: (b)

  1. 5 divisions per mA, so full scale (50 divisions) Ig=10 mA=0.01 AI_g = 10\text{ mA} = 0.01\text{ A}.
  2. Series resistance R=VIg−G=250.01−20R = \dfrac{V}{I_g} - G = \dfrac{25}{0.01} - 20
  3. R=2500−20=2480 ΩR = 2500 - 20 = 2480\ \Omega in series
  4. So the answer is (b).
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Q354JEE Main 2023 · 15 APR · SHIFT 1Medium

Kirchhoff's Laws & Measuring Instruments · Galvanometer to ammeter / voltmeter

For designing a voltmeter of range 50 V50\text{ V} and an ammeter of range 10 mA10\text{ mA} using a galvanometer which has a coil of resistance 54 Ω54\ \Omega showing a full scale deflection for 1 mA1\text{ mA} as in figure.

Figure

(A) for voltmeter R≈50 kΩR \approx 50\text{ k}\Omega

(B) for ammeter r≈0.2 Ωr \approx 0.2\ \Omega

(C) for ammeter r=6 Ωr = 6\ \Omega

(D) for voltmeter R≈5 kΩR \approx 5\text{ k}\Omega

(E) for voltmeter R≈500 ΩR \approx 500\ \Omega

Choose the correct answer from the options given below:

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Answer: (a)

  1. Voltmeter: series R=VIg−G=5010−3−54=49946 Ω≈50 kΩR = \frac{V}{I_g} - G = \frac{50}{10^{-3}} - 54 = 49946\ \Omega \approx 50\text{ k}\Omega, so (A) is true.
  2. Ammeter: shunt r=IgGI−Ig=1×5410−1=6 Ωr = \frac{I_g G}{I - I_g} = \frac{1 \times 54}{10 - 1} = 6\ \Omega, so (C) is true.
  3. (B), (D), (E) do not match these values.
  4. So the answer is (a) (A) and (C).
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Q355JEE Main 2023 · 13 APR · SHIFT 1Medium

Kirchhoff's Laws & Measuring Instruments · Galvanometer to ammeter / voltmeter

When a resistance of 5 Ω5\ \Omega is shunted with a moving coil galvanometer, it shows a full scale deflection for a current of 250 mA250\text{ mA}, however when 1050 Ω1050\ \Omega resistance is connected with it in series, it gives full scale deflection for 25 volt. The resistance of galvanometer is ____________ Ω\Omega.

Numerical answer type. Work it out, then open the answer.

Show answer and solution

Answer: 50

  1. Voltmeter: Ig(G+1050)=25I_g(G + 1050) = 25
  2. Ammeter: IgG=(0.25−Ig)×5⇒Ig(G+5)=1.25I_g G = (0.25 - I_g) \times 5 \Rightarrow I_g(G + 5) = 1.25
  3. Divide: G+1050G+5=20⇒G+1050=20G+100\frac{G + 1050}{G + 5} = 20 \Rightarrow G + 1050 = 20G + 100
  4. 19G=950⇒G=50 Ω19G = 950 \Rightarrow G = 50\ \Omega
  5. Answer: 50 Ω50\ \Omega
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Q356JEE Main 2023 · 11 APR · SHIFT 1Medium

Kirchhoff's Laws & Measuring Instruments · Galvanometer sensitivity & figure of merit

The current sensitivity of moving coil galvanometer is increased by 25%25\%. This increase is achieved only by changing in the number of turns of coils and area of cross section of the wire while keeping the resistance of galvanometer coil constant. The percentage change in the voltage sensitivity will be:

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Answer: (a)

  1. Voltage sensitivity Vs=IsRV_s = \frac{I_s}{R}.
  2. R is kept constant and IsI_s rises by 25%.
  3. So VsV_s also rises by 25%.
  4. So the answer is (a) +25%.
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Q357JEE Main 2023 · 31 JAN · SHIFT 2Medium

Kirchhoff's Laws & Measuring Instruments · Galvanometer sensitivity & figure of merit

The number of turns of the coil of a moving coil galvanometer is increased in order to increase current sensitivity by 50%50\%. The percentage change in voltage sensitivity of the galvanometer will be:

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Answer: (a)

  1. Is=NABk∝NI_s = \frac{NAB}{k} \propto N; coil resistance R∝NR \propto N (more turns, more wire).
  2. Vs=IsR=NABkRV_s = \frac{I_s}{R} = \frac{NAB}{kR}, so N cancels.
  3. So the answer is (a) 0%0\%.
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Q358JEE Main 2023 · 6 APR · SHIFT 2Difficult

Kirchhoff's Laws & Measuring Instruments · Galvanometer to ammeter / voltmeter

A student is provided with a variable voltage source V\mathrm{V}, a test resistor RT=10 ΩR_T = 10\ \Omega, two identical galvanometers G1G_1 and G2G_2 and two additional resistors, R1=10 MΩR_1 = 10\ M\Omega and R2=0.001 ΩR_2 = 0.001\ \Omega. For conducting an experiment to verify Ohm's law, the most suitable circuit is:

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Answer: (c)

  1. Voltmeter across RTR_T: galvanometer with the large resistance R1=10 MΩR_1 = 10\text{ M}\Omega in series, connected in parallel with RTR_T.
  2. Ammeter in series: galvanometer with the small shunt R2=0.001 ΩR_2 = 0.001\ \Omega in parallel.
  3. Circuit (c) has G1+R1G_1 + R_1 across RTR_T and G2∥R2G_2 \parallel R_2 in the main line.
  4. So the answer is (c).
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Q359MHT CET 2022 · 11 AUG · SHIFT 2Easy

Kirchhoff's Laws & Measuring Instruments · Galvanometer to ammeter / voltmeter

A galvanometer of resistance 200 Ω200\ \Omega is to be converted into an ammeter. The value of shunt resistance which allows 3%3\% of the mains current through the galvanometer is equal to (nearly)

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Answer: (a)

  1. S=IgGI−IgS = \frac{I_g G}{I - I_g} with Ig=0.03II_g = 0.03I.
  2. S=0.03×2000.97=6.19 Ω≈6 ΩS = \frac{0.03 \times 200}{0.97} = 6.19\ \Omega \approx 6\ \Omega
  3. So the answer is (a) 6 Ω6\ \Omega.
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Q360JEE Main 2022 · 27 JUN · SHIFT 1Easy

Kirchhoff's Laws & Measuring Instruments · Galvanometer to ammeter / voltmeter

A 72 Ω\Omega galvanometer is shunted by a resistance of 8 Ω\Omega. The percentage of the total current which passes through the galvanometer is:

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Answer: (b)

  1. Fraction through galvanometer IgI=SS+G\frac{I_g}{I} = \frac{S}{S + G}.
  2. IgI=88+72=110=10%\frac{I_g}{I} = \frac{8}{8 + 72} = \frac{1}{10} = 10\%
  3. So the answer is (b) 10%.
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Q361JEE Main 2022 · 30 JUL · SHIFT 2Medium

Kirchhoff's Laws & Measuring Instruments · Multi-loop circuits

The current I drawn out of the battery connected in the given circuit is:

Figure

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Answer: (b)

  1. The right vertical wire joins the 3 Ω3\ \Omega and 6 Ω6\ \Omega ends to the bottom (negative) line.
  2. So 3 Ω3\ \Omega is directly across the battery, and 6 Ω∥2 Ω=1.5 Ω6\ \Omega \parallel 2\ \Omega = 1.5\ \Omega.
  3. Lower path: 1.5+1.5=3 Ω1.5 + 1.5 = 3\ \Omega; with the 3 Ω3\ \Omega: Req=32=1.5 ΩR_{eq} = \frac{3}{2} = 1.5\ \Omega.
  4. I=91.5=6I = \frac{9}{1.5} = 6 A
  5. So the answer is (b) 66 A.
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Q362JEE Main 2022 · 26 JUL · SHIFT 2Medium

Kirchhoff's Laws & Measuring Instruments · Multi-loop circuits

A battery of 6 V6\text{ V} is connected to the circuit as shown below. The current I drawn from the battery is:

Figure

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Answer: (a)

  1. Bridge: top arms 3,33, 3; bottom arms 6,66, 6; 33=66\frac{3}{3} = \frac{6}{6}, so it is balanced and 5 Ω5\ \Omega carries no current.
  2. Rbridge=6×126+12=4 ΩR_{bridge} = \frac{6 \times 12}{6 + 12} = 4\ \Omega
  3. Total =4+2=6 Ω= 4 + 2 = 6\ \Omega, I=66=1I = \frac{6}{6} = 1 A
  4. So the answer is (a) 1 A.
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Q363JEE Main 2022 · 26 JUL · SHIFT 1Medium

Kirchhoff's Laws & Measuring Instruments · Multi-loop circuits

The current I in the given circuit will be:

Figure

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Answer: (a)

  1. Label the nodes: the four 4 Ω4\ \Omega resistors form the arms of a bridge and 2 Ω2\ \Omega is the bridge wire.
  2. All arms are equal (44=44\frac{4}{4} = \frac{4}{4}), so the bridge is balanced and 2 Ω2\ \Omega carries no current.
  3. Req=8∥8=4 ΩR_{eq} = 8 \parallel 8 = 4\ \Omega
  4. I=404=10I = \frac{40}{4} = 10 A
  5. So the answer is (a) 10 A.
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Q364JEE Main 2022 · 29 JUL · SHIFT 1Medium

Kirchhoff's Laws & Measuring Instruments · Multi-loop circuits

The current I flowing through the given circuit will be __________A.

Figure

Numerical answer type. Work it out, then open the answer.

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Answer: 2

  1. The top wire joins nodes 1 and 3; the bottom wire joins nodes 2 and 4.
  2. So every 9 Ω9\ \Omega resistor lies between the same two points: three in parallel.
  3. Req=93=3 ΩR_{eq} = \frac{9}{3} = 3\ \Omega, I=63=2I = \frac{6}{3} = 2 A
  4. So the answer is 2 A.
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Q365JEE Main 2022 · 25 JUN · SHIFT 1Medium

Kirchhoff's Laws & Measuring Instruments · Multi-loop circuits

The total current supplied to the circuit as shown in figure by the 5 V battery is ____________ A.

Figure

Numerical answer type. Work it out, then open the answer.

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Answer: 2

  1. The upper arc joins the cell's top terminal to both upper 5 Ω5\ \Omega ends; the lower arc joins the 2.5 Ω2.5\ \Omega and the right 5 Ω5\ \Omega to the other terminal.
  2. Star path: (5∥5)+2.5=2.5+2.5=5 Ω(5 \parallel 5) + 2.5 = 2.5 + 2.5 = 5\ \Omega.
  3. In parallel with the right 5 Ω5\ \Omega: Req=2.5 ΩR_{eq} = 2.5\ \Omega.
  4. I=52.5=2I = \frac{5}{2.5} = 2 A
  5. So the answer is 2 A.
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Q366NEET 2022 · PHASE 2Medium

Kirchhoff's Laws & Measuring Instruments · Potential difference between points

The sliding contact C is at one fourth of the length of the potentiometer wire (AB) from A as shown in the circuit diagram. If the resistance of the wire AB is R0, then the potential drop (V) across the resistor R is

Figure

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Answer: (b)

  1. AC has R04\frac{R_0}{4} and is in parallel with RR; CB has 3R04\frac{3R_0}{4} in series.
  2. RAC∥R=(R0/4)RR0/4+R=R0RR0+4RR_{AC} \parallel R = \frac{(R_0/4)R}{R_0/4 + R} = \frac{R_0 R}{R_0 + 4R}
  3. Voltage divider: V=V0R0RR0+4RR0RR0+4R+3R04=4V0R4R+3(R0+4R)V = V_0 \frac{\frac{R_0R}{R_0+4R}}{\frac{R_0R}{R_0+4R} + \frac{3R_0}{4}} = \frac{4V_0R}{4R + 3(R_0 + 4R)}
  4. V=4V0R3R0+16RV = \frac{4V_0R}{3R_0 + 16R}
  5. So the answer is (b).
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Q367JEE Main 2022 · 25 JUL · SHIFT 2Medium

Kirchhoff's Laws & Measuring Instruments · Potential difference between points

A current of 15 mA flows in the circuit as shown in figure. The value of potential difference between the points A and B will be:

Figure

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Answer: (d)

  1. 10 kΩ∥5 kΩ=103 kΩ10\text{ k}\Omega \parallel 5\text{ k}\Omega = \frac{10}{3}\text{ k}\Omega
  2. Between A and B: 5+103+10=553 kΩ5 + \frac{10}{3} + 10 = \frac{55}{3}\text{ k}\Omega in series
  3. VAB=15 mA×553 kΩ=275 VV_{AB} = 15\text{ mA} \times \frac{55}{3}\text{ k}\Omega = 275\text{ V}
  4. So the answer is (d) 275 V.
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Q368JEE Main 2022 · 25 JUL · SHIFT 1Medium

Kirchhoff's Laws & Measuring Instruments · Potential difference between points

In the given figure, the value of Vo will be _____________ V.

Figure

Numerical answer type. Work it out, then open the answer.

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Answer: 4

  1. Three branches (cell + 1 kΩ1\ \text{k}\Omega) are in parallel between ground and the output node, so use Millman's theorem.
  2. Vo=E1/R+E2/R+E3/R1/R+1/R+1/R=2+4+63V_o = \dfrac{E_1/R + E_2/R + E_3/R}{1/R + 1/R + 1/R} = \dfrac{2 + 4 + 6}{3}
  3. Vo=4 VV_o = 4\text{ V}
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Q369JEE Main 2022 · 29 JUN · SHIFT 1Medium

Kirchhoff's Laws & Measuring Instruments · Potential difference between points

For the network shown below, the value of VB −- VA is ____________ V.

Figure

Numerical answer type. Work it out, then open the answer.

Show answer and solution

Answer: 10

  1. Single loop: total resistance =1+2=3 Ω= 1 + 2 = 3\ \Omega, so I=153=5 AI = \dfrac{15}{3} = 5\text{ A}.
  2. The cell's positive plate faces B, so current leaves the cell towards B and returns through the 2 Ω2\ \Omega resistor from B to A.
  3. Current flows from higher to lower potential in a resistor: VB−VA=I×2=5×2V_B - V_A = I \times 2 = 5 \times 2
  4. VB−VA=10 VV_B - V_A = 10\text{ V}
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Q370JEE Main 2022 · 29 JUN · SHIFT 2Medium

Kirchhoff's Laws & Measuring Instruments · Voltmeter & ammeter in circuits

Two resistors are connected in series across a battery as shown in figure. If a voltmeter of resistance 2000 Ω\Omega is used to measure the potential difference across 500 Ω\Omega resistor, the reading of the voltmeter will be ___________ V.

Figure

Numerical answer type. Work it out, then open the answer.

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Answer: 8

  1. 500 Ω∥2000 Ω=500×20002500=400 Ω500\ \Omega \| 2000\ \Omega = \dfrac{500 \times 2000}{2500} = 400\ \Omega
  2. Total R=400+600=1000 ΩR = 400 + 600 = 1000\ \Omega, I=201000=0.02 AI = \dfrac{20}{1000} = 0.02\text{ A}.
  3. Reading =0.02×400=8 V= 0.02 \times 400 = 8\text{ V}
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Q371JEE Main 2022 · 25 JUN · SHIFT 1Medium

Kirchhoff's Laws & Measuring Instruments · Galvanometer to ammeter / voltmeter

A teacher in his physics laboratory allotted an experiment to determine the resistance (G) of a galvanometer. Students took the observations for 13\frac{1}{3} deflection in the galvanometer. Which of the below is true for measuring value of G?

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Answer: (b)

  1. With the main current kept the same, 13\frac{1}{3} deflection means Ig=I3I_g = \frac{I}{3} and the shunt carries 2I3\frac{2I}{3}.
  2. Parallel: IgG=IsS⇒I3G=2I3SI_g G = I_s S \Rightarrow \frac{I}{3}G = \frac{2I}{3}S
  3. G=2SG = 2S, so the method works.
  4. So the answer is (b).
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Q372JEE Main 2022 · 25 JUN · SHIFT 2Medium

Kirchhoff's Laws & Measuring Instruments · Galvanometer sensitivity & figure of merit

If n represents the actual number of deflections in a converted galvanometer of resistance G and shunt resistance S. Then the total current I when its figure of merit is K will be:

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Answer: (d)

  1. Current through galvanometer Ig=nKI_g = nK (figure of merit KK = current per division).
  2. Shunt relation: Ig=ISG+SI_g = I\frac{S}{G + S}
  3. I=nK(G+S)SI = \frac{nK(G + S)}{S}
  4. So the answer is (d).
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Q373JEE Main 2022 · 28 JUN · SHIFT 2Difficult

Kirchhoff's Laws & Measuring Instruments · Potential difference between points

All resistances in figure are 1 Ω\Omega each. The value of current 'I' is a5\frac{a}{5} A. The value of a is _________.

Figure

Numerical answer type. Work it out, then open the answer.

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Answer: 8

  1. Work from the right: last stage, two 1 Ω1\ \Omega in parallel =0.5 Ω= 0.5\ \Omega.
  2. Next stage: (1+0.5)∥(1+0.5)=0.75 Ω(1 + 0.5) \parallel (1 + 0.5) = 0.75\ \Omega; then (1+0.75)∥(1+0.75)=0.875 Ω(1 + 0.75) \parallel (1 + 0.75) = 0.875\ \Omega.
  3. Total =1+0.875=1.875 Ω= 1 + 0.875 = 1.875\ \Omega
  4. I=31.875=1.6 A=85 AI = \frac{3}{1.875} = 1.6\text{ A} = \frac{8}{5}\text{ A}
  5. Answer: a=8a = 8
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Q374MHT CET 2021 · 22 SEP · SHIFT 2Easy

Kirchhoff's Laws & Measuring Instruments · Kirchhoff's laws: basics

Kirchhoff's current and voltage law are respectively based on the conservation of

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Answer: (c)

  1. Current (junction) law: charge cannot accumulate at a junction, so conservation of charge.
  2. Voltage (loop) law: net energy change of a charge round a loop is zero, so conservation of energy.
  3. So the answer is (c) charge, energy.
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Q375MHT CET 2021 · 22 SEP · SHIFT 1Easy

Kirchhoff's Laws & Measuring Instruments · Kirchhoff's laws: basics

The Kirchhoff's current law and voltage law are respectively based upon the conservation of

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Answer: (a)

  1. Current law: conservation of charge at a junction.
  2. Voltage law: conservation of energy round a closed loop.
  3. So the answer is (a) charge, energy.
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Q376MHT CET 2021 · 20 SEP · SHIFT 1Easy

Kirchhoff's Laws & Measuring Instruments · Galvanometer to ammeter / voltmeter

A galvanometer of resistance 50 Ω50\ \Omega is converted to an ammeter. After shunting, the effective resistance of ammeter is 2.5 Ω2.5\ \Omega. The value of shunt is

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Answer: (b)

  1. Shunt is in parallel with the galvanometer: 1RA=1G+1S\frac{1}{R_A} = \frac{1}{G} + \frac{1}{S}.
  2. 1S=12.5−150=20−150=1950\frac{1}{S} = \frac{1}{2.5} - \frac{1}{50} = \frac{20 - 1}{50} = \frac{19}{50}
  3. S=5019 ΩS = \frac{50}{19}\ \Omega, so the answer is (b).
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Q377MHT CET 2021 · 21 SEP · SHIFT 2Medium

Kirchhoff's Laws & Measuring Instruments · Multi-loop circuits

A current through 1 Ω\Omega resistance in the following circuit is

Figure

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Answer: (c)

  1. 51=12.52.5=5\frac{5}{1} = \frac{12.5}{2.5} = 5, so the bridge is balanced and G carries no current.
  2. Path PQR: 5+1=6 Ω5 + 1 = 6\ \Omega; path PSR: 12.5+2.5=15 Ω12.5 + 2.5 = 15\ \Omega.
  3. I1Ω=2.1×156+15=1.5I_{1\Omega} = 2.1 \times \frac{15}{6 + 15} = 1.5 A
  4. So the answer is (c) 1.5 A.
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Q378JEE Main 2021 · 25 JUL · SHIFT 2Medium

Kirchhoff's Laws & Measuring Instruments · Multi-loop circuits

A 16 Ω\Omega wire is bent to form a square loop. A 9V supply having internal resistance of 1Ω\Omega is connected across one of its sides. The potential drop across the diagonals of the square loop is _______________ ×\times 10−-1 V

Numerical answer type. Work it out, then open the answer.

Show answer and solution

Answer: 45

ABCD4 Ω4 Ω4 Ω4 Ω9 V, 1 Ω
  1. Each side =4 Ω= 4\ \Omega. Across side AB: 4∥12=3 Ω4 \parallel 12 = 3\ \Omega.
  2. I=93+1=94I = \frac{9}{3 + 1} = \frac{9}{4} A, so VAB=94×3=6.75V_{AB} = \frac{9}{4} \times 3 = 6.75 V.
  3. Current in path A–D–C–B: 6.7512=0.5625\frac{6.75}{12} = 0.5625 A.
  4. Diagonal AC: VAC=0.5625×8=4.5V_{AC} = 0.5625 \times 8 = 4.5 V (same for BD).
  5. So the answer is 45×10−145 \times 10^{-1} V.
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Q379MHT CET 2021 · 24 SEP · SHIFT 2Medium

Kirchhoff's Laws & Measuring Instruments · Potential difference between points

In the following electrical network, the value of I is

Figure

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Answer: (c)

  1. Apply the junction rule at each junction along the wire.
  2. First junction: 1+2+3=6 A1 + 2 + 3 = 6\text{ A} flows on.
  3. Then 6−1.2=4.8 A6 - 1.2 = 4.8\text{ A}, 4.8+0.8=5.6 A4.8 + 0.8 = 5.6\text{ A}, 5.6−0.5=5.1 A5.6 - 0.5 = 5.1\text{ A}.
  4. I=5.1−1.7=3.4 AI = 5.1 - 1.7 = 3.4\text{ A}
  5. So the answer is (c) 3.4 A.
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Q380MHT CET 2021 · 20 SEP · SHIFT 1Medium

Kirchhoff's Laws & Measuring Instruments · Potential difference between points

Consider the circuit shown in the figure. The value of current 'I' is

Figure

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Answer: (a)

  1. Take VE=0V_E = 0. Reading the cells (long plate = +): VB=+6 VV_B = +6\text{ V}, VA=VF=−8 VV_A = V_F = -8\text{ V}, VC=VD=+12 VV_C = V_D = +12\text{ V}.
  2. Current A to B through 28 Ω28\ \Omega: −8−628=−12 A\frac{-8 - 6}{28} = -\frac{1}{2}\text{ A}
  3. Current C to B through 54 Ω54\ \Omega: 12−654=19 A\frac{12 - 6}{54} = \frac{1}{9}\text{ A}
  4. Junction B: I=−12+19=−718 AI = -\frac{1}{2} + \frac{1}{9} = -\frac{7}{18}\text{ A}
  5. So the answer is (a) −718-\frac{7}{18} A.
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Q381JEE Main 2021 · 20 JUL · SHIFT 1Medium

Kirchhoff's Laws & Measuring Instruments · Potential difference between points

Figure

The value of current in the 6 Ω\Omega resistance is:

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Answer: (c)

  1. Take the bottom wire at 0 and the top junction at VV; both cells have + at the top.
  2. KCL: 140−V20+90−V5=V6\frac{140 - V}{20} + \frac{90 - V}{5} = \frac{V}{6}
  3. Multiply by 60: 420−3V+1080−12V=10V⇒V=60 V420 - 3V + 1080 - 12V = 10V \Rightarrow V = 60\text{ V}
  4. I6=606=10 AI_6 = \frac{60}{6} = 10\text{ A}
  5. So the answer is (c) 10 A.
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Q382JEE Main 2021 · 25 JUL · SHIFT 2MediumOld syllabus

Kirchhoff's Laws & Measuring Instruments · Potential difference between points

The given potentiometer has its wire of resistance 10Ω\Omega. When the sliding contact is in the middle of the potentiometer wire, the potential drop across 2Ω\Omega resistor is:

Figure

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Answer: (c)

  1. Each half of the wire is 5 Ω5\ \Omega; the 2 Ω2\ \Omega is in parallel with the half on the A side.
  2. 5∥2=107 Ω5 \parallel 2 = \frac{10}{7}\ \Omega, total R=107+5=457 ΩR = \frac{10}{7} + 5 = \frac{45}{7}\ \Omega
  3. I=20×745=289 AI = \frac{20 \times 7}{45} = \frac{28}{9}\text{ A}
  4. V2Ω=289×107=409 VV_{2\Omega} = \frac{28}{9} \times \frac{10}{7} = \frac{40}{9}\text{ V}
  5. So the answer is (c) 409\frac{40}{9} V.
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Q383JEE Main 2021 · 31 AUG · SHIFT 1Medium

Kirchhoff's Laws & Measuring Instruments · Potential difference between points

The voltage drop across 15Ω\Omega resistance in the given figure will be ______________ V.

Figure

Numerical answer type. Work it out, then open the answer.

Show answer and solution

Answer: 6

  1. Upper branch: (4∥4)+2+(15∥10)=2+2+6=10 Ω(4\|4) + 2 + (15\|10) = 2 + 2 + 6 = 10\ \Omega.
  2. Lower branch: (8∥8)+(12∥12)=4+6=10 Ω(8\|8) + (12\|12) = 4 + 6 = 10\ \Omega.
  3. The two branches in parallel give 5 Ω5\ \Omega; with the 1 Ω1\ \Omega: R=6 ΩR = 6\ \Omega, I=126=2 AI = \dfrac{12}{6} = 2\text{ A}.
  4. Each branch carries 1 A1\text{ A}, so the drop across the 15 Ω∥10 Ω15\ \Omega\|10\ \Omega pair is 1×61 \times 6.
  5. Voltage across 15 Ω15\ \Omega =6 V= 6\text{ V}
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Q384JEE Main 2021 · 18 MAR · SHIFT 1Medium

Kirchhoff's Laws & Measuring Instruments · Potential difference between points

The voltage across the 10Ω\Omega resistor in the given circuit is x volt.

Figure

The value of 'x' to the nearest integer is _________.

Numerical answer type. Work it out, then open the answer.

Show answer and solution

Answer: 70

  1. 50 Ω∥20 Ω=50×2070=1007 Ω50\ \Omega \| 20\ \Omega = \dfrac{50 \times 20}{70} = \dfrac{100}{7}\ \Omega
  2. Total R=10+1007=1707 ΩR = 10 + \dfrac{100}{7} = \dfrac{170}{7}\ \Omega, so I=170170/7=7 AI = \dfrac{170}{170/7} = 7\text{ A}.
  3. Voltage across 10 Ω10\ \Omega: x=7×10=70 Vx = 7 \times 10 = 70\text{ V}
  4. x=70x = 70
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Q385JEE Main 2021 · 16 MAR · SHIFT 1Medium

Kirchhoff's Laws & Measuring Instruments · Potential difference between points

In the figure given, the electric current flowing through the 5 kΩ\Omega resistor is 'x' mA.

Figure

The value of x to the nearest integer is ____________.

Numerical answer type. Work it out, then open the answer.

Show answer and solution

Answer: 3

  1. Three 3 kΩ3\ \text{k}\Omega in parallel =1 kΩ= 1\ \text{k}\Omega.
  2. Total R=5+1+1 (internal)=7 kΩR = 5 + 1 + 1\ (\text{internal}) = 7\ \text{k}\Omega
  3. I=217 kΩ=3 mAI = \dfrac{21}{7\ \text{k}\Omega} = 3\text{ mA}, the full current passes through the 5 kΩ5\ \text{k}\Omega resistor.
  4. x=3x = 3
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Q386MHT CET 2021 · 21 SEP · SHIFT 2Medium

Kirchhoff's Laws & Measuring Instruments · Galvanometer to ammeter / voltmeter

A moving coil galvanometer is converted into an ammeter, reading upto 0.04 A0.04\text{ A} by connecting a shunt of resistance '3r3r' across it and then into an ammeter reading upto 0.08 A0.08\text{ A}, when a shunt of resistance 'rr' is connected across it. What is the maximum current which can be sent through this galvanometer if no shunt is used?

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Answer: (a)

  1. Shunt rule: IgG=(I−Ig)SI_g G = (I - I_g) S.
  2. IgG=(0.04−Ig) 3rI_g G = (0.04 - I_g)\,3r and IgG=(0.08−Ig) rI_g G = (0.08 - I_g)\,r
  3. Equate: 0.12−3Ig=0.08−Ig⇒2Ig=0.040.12 - 3I_g = 0.08 - I_g \Rightarrow 2I_g = 0.04
  4. Ig=0.02 AI_g = 0.02\text{ A}. So the answer is (a).
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Q387MHT CET 2021 · 20 SEP · SHIFT 2Medium

Kirchhoff's Laws & Measuring Instruments · Galvanometer to ammeter / voltmeter

A galvanometer has resistance 'GG' Ω\Omega and 'IgI_g' is current flowing through it which produces full scale deflection. 'S1S_1' is the value of shunt which converts it into an ammeter of range 0 to '3I3I' and 'S2S_2' is the shunt value which converts it into an ammeter of range 0 to '4I4I', the ratio S2:S1S_2 : S_1 is

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Answer: (b)

  1. Shunt: S=IgGImax−IgS = \dfrac{I_g G}{I_{max} - I_g}
  2. S1=IgG3I−IgS_1 = \dfrac{I_g G}{3I - I_g}, S2=IgG4I−IgS_2 = \dfrac{I_g G}{4I - I_g}
  3. S2S1=3I−Ig4I−Ig\dfrac{S_2}{S_1} = \dfrac{3I - I_g}{4I - I_g}. So the answer is (b).
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Q388JEE Main 2021 · 31 AUG · SHIFT 1Medium

Kirchhoff's Laws & Measuring Instruments · Galvanometer to ammeter / voltmeter

Consider a galvanometer shunted with 5Ω\Omega resistance and 2% of current passes through it. What is the resistance of the given galvanometer?

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Answer: (c)

  1. Ig=0.02II_g = 0.02I, shunt current =0.98I= 0.98I.
  2. IgG=IsS⇒0.02I⋅G=0.98I×5I_g G = I_s S \Rightarrow 0.02I \cdot G = 0.98I \times 5
  3. G=4.90.02=245 ΩG = \frac{4.9}{0.02} = 245\ \Omega
  4. So the answer is (c) 245 Ω245\ \Omega.
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Q389JEE Main 2021 · 27 AUG · SHIFT 2Medium

Kirchhoff's Laws & Measuring Instruments · Galvanometer sensitivity & figure of merit

For full scale deflection of total 50 divisions, 50 mV voltage is required in galvanometer. The resistance of galvanometer if its current sensitivity is 2 div/mA will be:

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Answer: (d)

  1. Full scale current =50 div2 div/mA=25 mA= \frac{50\text{ div}}{2\text{ div/mA}} = 25\text{ mA}
  2. G=VI=50 mV25 mA=2 ΩG = \frac{V}{I} = \frac{50\text{ mV}}{25\text{ mA}} = 2\ \Omega
  3. So the answer is (d) 2 Ω2\ \Omega.
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Q390NEET 2020 · PHASE 2Easy

Kirchhoff's Laws & Measuring Instruments · Kirchhoff's laws: basics

For the circuit given below, the Kirchoff's loop rule for the loop BCDEBBCDEB is given by the equation

Figure

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Answer: (c)

  1. Go round B→C→D→E→B and add potential changes.
  2. B→C along i2i_2 through R2R_2: −i2R2-i_2R_2; C→D through E2E_2 (+ to −): −E2-E_2.
  3. D→E through E3E_3 (− to +): +E3+E_3; E→B against i3i_3 through R1R_1: +i3R1+i_3R_1.
  4. −i2R2−E2+E3+i3R1=0-i_2R_2 - E_2 + E_3 + i_3R_1 = 0, i.e. i2R2+E2−E3−i3R1=0i_2R_2 + E_2 - E_3 - i_3R_1 = 0. So the answer is (c).
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Q391JEE Main 2020 · 6 SEP · SHIFT 2Easy

Kirchhoff's Laws & Measuring Instruments · Voltmeter & ammeter in circuits

A circuit to verify Ohm's law uses ammeter and voltmeter in series or parallel connected correctly to the resistor. In the circuit:

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Answer: (a)

  1. Ammeter must carry the same current as the resistor, so it is in series (low resistance).
  2. Voltmeter must have the same p.d. as the resistor, so it is in parallel (high resistance).
  3. So the answer is (a).
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Q392MHT CET 2020 · 19 OCT · SHIFT 2Easy

Kirchhoff's Laws & Measuring Instruments · Galvanometer to ammeter / voltmeter

A milliammeter of resistance 40 Ω40\ \Omega has a range 0−30 mA0 - 30\text{ mA}. What will be the resistance used in series to convert it into voltmeter of range 0−15 V0 - 15\text{ V}?

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Answer: (a)

  1. Series resistance R=VIg−GR = \frac{V}{I_g} - G.
  2. R=1530×10−3−40=500−40=460 ΩR = \frac{15}{30 \times 10^{-3}} - 40 = 500 - 40 = 460\ \Omega
  3. So the answer is (a) 460 Ω460\ \Omega.
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Q393MHT CET 2020 · 16 OCT · SHIFT 2Easy

Kirchhoff's Laws & Measuring Instruments · Galvanometer to ammeter / voltmeter

In conversion of moving coil galvanometer into an ammeter of required range, the resistance of ammeter, so formed is

[S=S = shunt and G=G = resistance of galvanometer]

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Answer: (b)

  1. The shunt SS is connected in parallel with the galvanometer GG.
  2. Ammeter resistance =SGS+G= \frac{SG}{S + G}
  3. So the answer is (b).
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Q394JEE Main 2020 · 5 SEP · SHIFT 1Easy

Kirchhoff's Laws & Measuring Instruments · Galvanometer to ammeter / voltmeter

A galvanometer of resistance G is converted into a voltmeter of range 0 – 1 V by connecting a resistance R1 in series with it. The additional resistance that should be connected in series with R1 to increase the range of the voltmeter to 0 – 2 V will be:

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Answer: (c)

  1. For 1 V range: Ig(G+R1)=1I_g(G + R_1) = 1.
  2. For 2 V range: Ig(G+R1+R)=2⇒G+R1+R=2(G+R1)I_g(G + R_1 + R) = 2 \Rightarrow G + R_1 + R = 2(G + R_1)
  3. R=R1+GR = R_1 + G, so the answer is (c).
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Q395JEE Main 2020 · 8 JAN · SHIFT 2Easy

Kirchhoff's Laws & Measuring Instruments · Galvanometer to ammeter / voltmeter

A galvanometer having a coil resistance 100 Ω\Omega gives a full scale deflection when a current of 1 mA is passed through it. What is the value of the resistance which can convert this galvanometer into a voltmeter giving full scale deflection for a potential difference of 10 V?

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Answer: (c)

  1. Series resistance R=VIg−GR = \frac{V}{I_g} - G.
  2. R=1010−3−100=10000−100=9900 Ω=9.9 kΩR = \frac{10}{10^{-3}} - 100 = 10000 - 100 = 9900\ \Omega = 9.9\text{ k}\Omega
  3. So the answer is (c) 9.9 kΩ\Omega.
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Q396MHT CET 2020 · 19 OCT · SHIFT 2Easy

Kirchhoff's Laws & Measuring Instruments · Galvanometer sensitivity & figure of merit

Two galvanometers AA and BB require currents of 4 mA and 7 mA, respectively to produce the same deflection of 20 divisions. If SAS_A and SBS_B are their sensitivities respectively, then

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Answer: (b)

  1. Current sensitivity S=θIS = \frac{\theta}{I} (divisions per unit current).
  2. SA=204=5S_A = \frac{20}{4} = 5 div/mA, SB=207≈2.86S_B = \frac{20}{7} \approx 2.86 div/mA.
  3. A needs less current for the same deflection, so it is more sensitive.
  4. So the answer is (b) SA>SBS_A > S_B.
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Q397JEE Main 2020 · 5 SEP · SHIFT 2Easy

Kirchhoff's Laws & Measuring Instruments · Galvanometer sensitivity & figure of merit

A galvanometer is used in laboratory for detecting the null point in electrical experiments. If, on passing a current of 6 mA it produces a deflection of 2°, its figure of merit is close to:

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Answer: (c)

  1. Figure of merit G=IθG = \frac{I}{\theta} (current per division).
  2. G=6 mA2 div=3 mA/divG = \frac{6\ \text{mA}}{2\ \text{div}} = 3\ \text{mA/div}
  3. So the answer is (c) 3×10−33 \times 10^{-3} A/div.
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Q398NEET 2020 · PHASE 2Medium

Kirchhoff's Laws & Measuring Instruments · Multi-loop circuits

For the circuit shown in the figure, the current II will be

Figure

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Answer: (c)

  1. Both cells point the same way, so their emfs add: 2+4=62 + 4 = 6 V.
  2. Total resistance =1+1+4=6 Ω= 1 + 1 + 4 = 6\ \Omega.
  3. I=66=1I = \frac{6}{6} = 1 A
  4. So the answer is (c) 11 A.
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Q399JEE Main 2020 · 6 SEP · SHIFT 2Medium

Kirchhoff's Laws & Measuring Instruments · Multi-loop circuits

In the figure shown, the current in the 10 V battery is close to:

Figure

Show answer and solution

Answer: (d)

  1. Take the lower middle node at 00 V and the upper middle node at VV. Left branch: 2020 V with 5+2=7 Ω5 + 2 = 7\ \Omega; right branch: 1010 V (+ at top) with 4 Ω4\ \Omega; middle: 10 Ω10\ \Omega.
  2. KCL at the top node: 20−V7+10−V4=V10\frac{20 - V}{7} + \frac{10 - V}{4} = \frac{V}{10}
  3. 400+350=69V⇒V≈10.87400 + 350 = 69V \Rightarrow V \approx 10.87 V
  4. Current up through the 1010 V branch =10−10.874≈−0.21= \frac{10 - 10.87}{4} \approx -0.21 A, so 0.210.21 A flows down, entering the cell at + and leaving at −-.
  5. So the answer is (d) 0.21 A from positive to negative terminal.
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Q400JEE Main 2020 · 7 JAN · SHIFT 1Medium

Kirchhoff's Laws & Measuring Instruments · Multi-loop circuits

The current I1 (in A) flowing through 1 Ω\Omega resistor in the following circuit is:

Figure

Show answer and solution

Answer: (c)

  1. The 2 Ω2\ \Omega (middle) is directly across the 11 V cell.
  2. Upper branch: (1∥1)+2=0.5+2=2.5 Ω(1 \parallel 1) + 2 = 0.5 + 2 = 2.5\ \Omega, so its current is 12.5=0.4\frac{1}{2.5} = 0.4 A.
  3. This splits equally in the two 1 Ω1\ \Omega resistors: I1=0.2I_1 = 0.2 A
  4. So the answer is (c) 0.2.
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Q401JEE Main 2020 · 4 SEP · SHIFT 2Medium

Kirchhoff's Laws & Measuring Instruments · Multi-loop circuits

Four resistances 40 Ω\Omega, 60 Ω\Omega, 90 Ω\Omega and 110 Ω\Omega make the arms of a quadrilateral ABCD. Across AC is a battery of emf 40 V and internal resistance negligible. The potential difference across BD in V is _______.

Figure

Numerical answer type. Work it out, then open the answer.

Show answer and solution

Answer: 2

  1. Path ABC: 40+60=100 Ω40 + 60 = 100\ \Omega, current 0.40.4 A; path ADC: 90+110=200 Ω90 + 110 = 200\ \Omega, current 0.20.2 A.
  2. With VA=40V_A = 40 V: VB=40−0.4×40=24V_B = 40 - 0.4 \times 40 = 24 V.
  3. VD=40−0.2×90=22V_D = 40 - 0.2 \times 90 = 22 V
  4. VBD=24−22=2V_{BD} = 24 - 22 = 2 V, so the answer is 2 V.
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Q402JEE Main 2020 · 5 SEP · SHIFT 2Medium

Kirchhoff's Laws & Measuring Instruments · Potential difference between points

In the circuit, given in the figure currents in different branches and value of one resistor are shown. Then potential at point B with respect to the point A is:

Figure

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Answer: (c)

  1. Take VA=0V_A = 0. Crossing the 1 V cell from − to + towards C: VC=1 VV_C = 1\text{ V}.
  2. At D: 2 A arrives from B, 1 A leaves towards E, so 1 A flows D to C through 2 Ω2\ \Omega: VD=1+2=3 VV_D = 1 + 2 = 3\text{ V}.
  3. Crossing the 2 V cell from + (D side) to − (B side): VB=3−2=1 VV_B = 3 - 2 = 1\text{ V}
  4. So the answer is (c) +1 V.
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Q403JEE Main 2020 · 4 SEP · SHIFT 2Medium

Kirchhoff's Laws & Measuring Instruments · Potential difference between points

The value of current i1 flowing from A to C in the circuit diagram is:

Figure

Show answer and solution

Answer: (a)

  1. B-D branch (5+5=10 Ω5 + 5 = 10\ \Omega) is not joined to the A-C wire. 22=22\frac{2}{2} = \frac{2}{2}, so the bridge is balanced and BD carries no current.
  2. Between A and C: ABC =4 Ω= 4\ \Omega, ADC =4 Ω= 4\ \Omega, direct 4+4=8 Ω4 + 4 = 8\ \Omega, all in parallel.
  3. Each path has the full 8 V across it.
  4. i1=88=1 Ai_1 = \frac{8}{8} = 1\text{ A}
  5. So the answer is (a) 1 A.
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Q404JEE Main 2020 · 9 JAN · SHIFT 1Medium

Kirchhoff's Laws & Measuring Instruments · Potential difference between points

In the given circuit diagram, a wire is joining points B and D. The current in this wire is:

Figure

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Answer: (b)

  1. The wire BD makes B and D one node: 1∥4=0.8 Ω1 \parallel 4 = 0.8\ \Omega, 2∥3=1.2 Ω2 \parallel 3 = 1.2\ \Omega, total 2 Ω2\ \Omega, I=10 AI = 10\text{ A}.
  2. VAB=8 VV_{AB} = 8\text{ V}: I1Ω=8 AI_{1\Omega} = 8\text{ A}, I4Ω=2 AI_{4\Omega} = 2\text{ A}.
  3. VBC=12 VV_{BC} = 12\text{ V}: I2Ω=6 AI_{2\Omega} = 6\text{ A}, I3Ω=4 AI_{3\Omega} = 4\text{ A}.
  4. At B: 8=6+IBD⇒IBD=2 A8 = 6 + I_{BD} \Rightarrow I_{BD} = 2\text{ A} (B to D)
  5. So the answer is (b) 2 A.
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Q405JEE Main 2020 · 3 SEP · SHIFT 2Medium

Kirchhoff's Laws & Measuring Instruments · Voltmeter & ammeter in circuits

Two resistors 400Ω\Omega and 800Ω\Omega are connected in series across a 6 V battery. The potential difference measured by a voltmeter of 10 kΩ\Omega across 400 Ω\Omega resistor is close to:

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Answer: (b)

  1. 400 Ω∥10 kΩ=400×1000010400≈384.6 Ω400\ \Omega \| 10\ \text{k}\Omega = \dfrac{400 \times 10000}{10400} \approx 384.6\ \Omega
  2. Total R=384.6+800=1184.6 ΩR = 384.6 + 800 = 1184.6\ \Omega
  3. V=6×384.61184.6≈1.95 VV = 6 \times \dfrac{384.6}{1184.6} \approx 1.95\text{ V}
  4. So the answer is (b) 1.95 V.
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Q406MHT CET 2020 · 16 OCT · SHIFT 1Medium

Kirchhoff's Laws & Measuring Instruments · Galvanometer to ammeter / voltmeter

An ammeter of resistance 20 Ω20\ \Omega gives full scale deflection, when 1 mA1\text{ mA} current flows through it. What is the maximum current that can be measured by connecting 4 resistors each of 16 Ω16\ \Omega in parallel with the ammeter?

Show answer and solution

Answer: (a)

  1. Four 16 Ω16\ \Omega in parallel: S=4 ΩS = 4\ \Omega.
  2. I=Ig(1+GS)=1 mA×(1+204)I = I_g\left(1 + \dfrac{G}{S}\right) = 1\text{ mA} \times \left(1 + \dfrac{20}{4}\right)
  3. I=6 mAI = 6\text{ mA}. So the answer is (a).
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Q407MHT CET 2019 · 2 MAY · SHIFT 2Easy

Kirchhoff's Laws & Measuring Instruments · Galvanometer to ammeter / voltmeter

The range of an ammeter of resistance ' GG can be increased from ' II ' to ' nInI ' by connecting

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Answer: (b)

  1. To increase the range a shunt is added in parallel: S=IgGI′−IgS = \frac{I_g G}{I' - I_g}.
  2. With Ig=II_g = I and I′=nII' = nI: S=IGnI−I=Gn−1S = \frac{IG}{nI - I} = \frac{G}{n - 1}
  3. So the answer is (b) a shunt of Gn−1 Ω\frac{G}{n - 1}\ \Omega.
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Q408MHT CET 2019 · 2 MAY · SHIFT 1Easy

Kirchhoff's Laws & Measuring Instruments · Galvanometer to ammeter / voltmeter

A galvanometer has resistance of 100 Ω100\ \Omega and a current of 10 mA produces full scale deflection in it. The resistance to be connected in series, to get a voltmeter of range 50 volt is

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Answer: (d)

  1. Series resistance R=VIg−GR = \frac{V}{I_g} - G.
  2. R=5010×10−3−100=5000−100=4900 ΩR = \frac{50}{10 \times 10^{-3}} - 100 = 5000 - 100 = 4900\ \Omega
  3. So the answer is (d) 4900 Ω4900\ \Omega.
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Q409JEE Main 2019 · 10 APR · SHIFT 1Easy

Kirchhoff's Laws & Measuring Instruments · Galvanometer to ammeter / voltmeter

A moving coil galvanometer allows a full scale current of 10–4 A. A series resistance of 2 MΩ\Omega is required to convert the above galvanometer into a voltmeter of range 0-5 V. Therefore the value of shunt resistance required to convert the above galvanometer into an ammeter of range 0-10 mA is:

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Answer: (d)

  1. For the voltmeter: Ig(G+R)=5⇒G+R=510−4=5×104 ΩI_g(G + R) = 5 \Rightarrow G + R = \frac{5}{10^{-4}} = 5 \times 10^4\ \Omega.
  2. With R=2 MΩR = 2\text{ M}\Omega this gives G=5×104−2×106<0G = 5 \times 10^4 - 2 \times 10^6 < 0, which is impossible.
  3. So the data is inconsistent and no option fits: the answer is (d) None of the options are correct.
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Q410JEE Main 2019 · 9 APR · SHIFT 2Easy

Kirchhoff's Laws & Measuring Instruments · Galvanometer to ammeter / voltmeter

The resistance of a galvanometer is 50 ohm and the maximum current which can be passed through it is 0.002 A. What resistance must be connected to it in order to convert it into an ammeter of range 0 – 0.5 A?

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Answer: (b)

  1. A shunt is needed: S=IgGI−IgS = \frac{I_g G}{I - I_g}.
  2. S=0.002×500.5−0.002=0.10.498≈0.2 ohmS = \frac{0.002 \times 50}{0.5 - 0.002} = \frac{0.1}{0.498} \approx 0.2\text{ ohm}
  3. So the answer is (b) 0.2 ohm.
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Q411JEE Main 2019 · 9 APR · SHIFT 1Easy

Kirchhoff's Laws & Measuring Instruments · Galvanometer to ammeter / voltmeter

A moving coil galvanometer has resistance 50Ω\Omega and it indicates full deflection at 4mA current. A voltmeter is made using this galvanometer and a 5 kΩ\Omega resistance. The maximum voltage, that can be measured using this voltmeter, will be close to:

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Answer: (d)

  1. Maximum voltage V=Ig(G+R)V = I_g(G + R).
  2. V=4×10−3×(50+5000)=20.2 VV = 4 \times 10^{-3} \times (50 + 5000) = 20.2\text{ V}
  3. So the answer is (d) 20 V.
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Q412MHT CET 2019 · 2 MAY · SHIFT 1Easy

Kirchhoff's Laws & Measuring Instruments · Galvanometer sensitivity & figure of merit

In moving coil galvanometer, strong horse shoe magnet of concave shaped pole pieces is used to

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Answer: (c)

  1. Concave pole pieces with a soft-iron core make the field radial.
  2. In a radial field the plane of the coil stays parallel to B⃗\vec B in every position, so torque =NIAB= NIAB for all deflections.
  3. This gives I∝θI \propto \theta and a linear scale.
  4. So the answer is (c).
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Q413MHT CET 2019 · 2 MAY · SHIFT 2Medium

Kirchhoff's Laws & Measuring Instruments · Kirchhoff's laws: basics

In the given electrical circuit, which one of the following equations is a correct equation?

Figure

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Answer: (c)

  1. Apply KVL to loop E→A→B→F→EE \to A \to B \to F \to E, moving along the currents.
  2. Through RR (current i1+i2i_1 + i_2, A to B): drop (i1+i2)R(i_1 + i_2)R.
  3. Through r1r_1 from F to E (with i1i_1): drop i1r1i_1 r_1; through E1E_1 from −- to ++: rise E1E_1.
  4. E1−(i1+i2)R−i1r1=0E_1 - (i_1 + i_2)R - i_1 r_1 = 0
  5. So the answer is (c).
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Q414MHT CET 2019 · 2 MAY · SHIFT 1Medium

Kirchhoff's Laws & Measuring Instruments · Multi-loop circuits

The current in 1 Ω\Omega resistor in the following circuit is

Figure

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Answer: (a)

  1. 51=5010\frac{5}{1} = \frac{50}{10}, so the bridge is balanced and G carries no current.
  2. Upper path: 5+1=6 Ω5 + 1 = 6\ \Omega; lower path: 50+10=60 Ω50 + 10 = 60\ \Omega.
  3. I1Ω=1.1×6066=1I_{1\Omega} = 1.1 \times \frac{60}{66} = 1 A
  4. So the answer is (a) 1 A.
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Q415JEE Main 2019 · 12 JAN · SHIFT 2Medium

Kirchhoff's Laws & Measuring Instruments · Multi-loop circuits

In the given circuit diagram, the currents, I1 = – 0.3 A, I4 = 0.8 A and I5 = 0.4 A, are flowing as shown. The currents I2, I3 and I6, respectively, are:

Figure

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Answer: (d)

  1. Path S–P–Q has no junction at P: I6=I5=0.4I_6 = I_5 = 0.4 A.
  2. Junction R: I1+I2=I4⇒I2=0.8−(−0.3)=1.1I_1 + I_2 = I_4 \Rightarrow I_2 = 0.8 - (-0.3) = 1.1 A.
  3. Junction S: I4=I5+I3⇒I3=0.8−0.4=0.4I_4 = I_5 + I_3 \Rightarrow I_3 = 0.8 - 0.4 = 0.4 A.
  4. So the answer is (d) 1.1 A, 0.4 A, 0.4 A.
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Q416JEE Main 2019 · 10 JAN · SHIFT 1Medium

Kirchhoff's Laws & Measuring Instruments · Multi-loop circuits

In the given circuit the cells have zero internal resistance. The currents (in Amperes) passing through resistance R1 and R2 respectively, are -

Figure

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Answer: (a)

  1. Call the top wire T and the bottom middle node M.
  2. Left cell: going from T (left end) through −- to ++ reaches M, so VM−VT=10V_M - V_T = 10 V.
  3. R1R_1 is across T and M: I1=1020=0.5I_1 = \frac{10}{20} = 0.5 A.
  4. Right cell: from M through ++ to −- drops 1010 V, so the bottom of R2R_2 is at VTV_T; no p.d. across R2R_2, so I2=0I_2 = 0.
  5. So the answer is (a) 0.5, 0.
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Q417JEE Main 2019 · 8 APR · SHIFT 1Medium

Kirchhoff's Laws & Measuring Instruments · Potential difference between points

For the circuit shown, with R1 = 1.0 Ω\Omega, R2 = 2.0 Ω\Omega, E1 = 2 V and E2 = E3 = 4 V, the potential difference between the points 'a' and 'b' is approximately (in V):

Figure

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Answer: (a)

  1. Three branches lie between a and b: (E1=2E_1 = 2, 2R1=2 Ω2R_1 = 2\ \Omega), (E2=4E_2 = 4, R2=2 ΩR_2 = 2\ \Omega), (E3=4E_3 = 4, 2R1=2 Ω2R_1 = 2\ \Omega), all with + towards a.
  2. Parallel-cell (Millman) formula: Vab=∑E/r∑1/rV_{ab} = \frac{\sum E/r}{\sum 1/r}
  3. Vab=2/2+4/2+4/21/2+1/2+1/2=51.5≈3.3 VV_{ab} = \frac{2/2 + 4/2 + 4/2}{1/2 + 1/2 + 1/2} = \frac{5}{1.5} \approx 3.3\text{ V}
  4. So the answer is (a) 3.3.
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Q418JEE Main 2019 · 11 JAN · SHIFT 2Medium

Kirchhoff's Laws & Measuring Instruments · Potential difference between points

In the circuit shown, the potential difference between A and B is:

Figure

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Answer: (c)

  1. A and B are open ends, so no current flows in 5 Ω5\ \Omega and 10 Ω10\ \Omega: VA−VB=VD−VCV_A - V_B = V_D - V_C.
  2. Three cells (1 V, 2 V, 3 V), each with 1 Ω1\ \Omega, in parallel with + towards D.
  3. VD−VC=1/1+2/1+3/11+1+1=2 VV_D - V_C = \frac{1/1 + 2/1 + 3/1}{1 + 1 + 1} = 2\text{ V}
  4. So the answer is (c) 2 V.
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Q419JEE Main 2019 · 9 JAN · SHIFT 1Medium

Kirchhoff's Laws & Measuring Instruments · Potential difference between points

When the switch S, in the circuit shown, is closed, then the value of current i will be:

Figure

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Answer: (b)

  1. With S closed, C connects to ground through 2 Ω2\ \Omega. Let VC=VV_C = V.
  2. KCL at C: 20−V2+10−V4=V2\frac{20 - V}{2} + \frac{10 - V}{4} = \frac{V}{2}
  3. Multiply by 4: 40−2V+10−V=2V⇒V=10 V40 - 2V + 10 - V = 2V \Rightarrow V = 10\text{ V}
  4. i=102=5 Ai = \frac{10}{2} = 5\text{ A}
  5. So the answer is (b) 5A.
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Q420NEET 2019Medium

Kirchhoff's Laws & Measuring Instruments · Voltmeter & ammeter in circuits

The reading of an ideal voltmeter in the circuit shown is,

Figure

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Answer: (a)

  1. Take the left node at 2 V2\text{ V} and the right node at 00. Each branch has 50 Ω50\ \Omega, an ideal voltmeter draws no current.
  2. Top branch: middle point =2−2×2050=1.2 V= 2 - 2 \times \dfrac{20}{50} = 1.2\text{ V}.
  3. Bottom branch: middle point =2−2×3050=0.8 V= 2 - 2 \times \dfrac{30}{50} = 0.8\text{ V}.
  4. Reading =1.2−0.8=0.4 V= 1.2 - 0.8 = 0.4\text{ V}
  5. So the answer is (a) 0.4 V0.4\text{ V}.
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Q421NEET 2019Medium

Kirchhoff's Laws & Measuring Instruments · Voltmeter & ammeter in circuits

In the circuits shown below, the readings of voltmeters and the ammeters will be:

Figure

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Answer: (a)

  1. Ideal voltmeters have infinite resistance, so no current flows through either voltmeter branch.
  2. Circuit 1: i1=1010=1 Ai_1 = \dfrac{10}{10} = 1\text{ A}, V1=10 VV_1 = 10\text{ V}.
  3. Circuit 2: the 10 Ω10\ \Omega in series with V2V_2 carries no current, so it has no drop; V2=10 VV_2 = 10\text{ V} and i2=1 Ai_2 = 1\text{ A}.
  4. So the answer is (a) V1=V2V_1 = V_2 and i1=i2i_1 = i_2.
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Q422JEE Main 2019 · 10 JAN · SHIFT 2Medium

Kirchhoff's Laws & Measuring Instruments · Voltmeter & ammeter in circuits

The actual value of resistance R, shown in the figure is 30Ω\Omega. This is measuered in an experiment as shown using the standard formula R = VI\frac{V}{I}, where V and I are the readings of the voltmeter and ammeter, respectively. If the measured value of R is 5% less, then the internal resistance of the voltmeter is -

Figure

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Answer: (a)

  1. The voltmeter is across R, so the ammeter reads the current through R∥RVR \| R_V; measured value =RRVR+RV= \dfrac{R R_V}{R + R_V}.
  2. Measured =0.95×30=28.5 Ω= 0.95 \times 30 = 28.5\ \Omega
  3. 30RV30+RV=28.5⇒1.5RV=855\dfrac{30 R_V}{30 + R_V} = 28.5 \Rightarrow 1.5 R_V = 855
  4. RV=570 ΩR_V = 570\ \Omega. So the answer is (a).
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Q423JEE Main 2019 · 9 JAN · SHIFT 2Medium

Kirchhoff's Laws & Measuring Instruments · Voltmeter & ammeter in circuits

In the given circuit the internal resistance of the 18 V cell is negligible. If R1 = 400 Ω\Omega, R3 = 100 Ω\Omega and R4 = 500 Ω\Omega and the reading of an ideal voltmeter across R4 is 5V, then the value of R2 will be:

Figure

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Answer: (a)

  1. Current in R3,R4R_3, R_4 branch: 5500=0.01 A\dfrac{5}{500} = 0.01\text{ A}; branch p.d. =0.01×600=6 V= 0.01 \times 600 = 6\text{ V} (this is also across R2R_2).
  2. P.d. across R1=18−6=12 VR_1 = 18 - 6 = 12\text{ V}, so total current =12400=0.03 A= \dfrac{12}{400} = 0.03\text{ A}.
  3. Current in R2=0.03−0.01=0.02 AR_2 = 0.03 - 0.01 = 0.02\text{ A}
  4. R2=60.02=300 ΩR_2 = \dfrac{6}{0.02} = 300\ \Omega. So the answer is (a).
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Q424JEE Main 2019 · 12 APR · SHIFT 2Medium

Kirchhoff's Laws & Measuring Instruments · Galvanometer to ammeter / voltmeter

A moving coil galvanometer, having a resistance G, produces full scale deflection when a current Ig flows through it. This galvanometer can be converted into (i) an ammeter of range 0 to I0(I0 > Ig) by connecting a shunt resistance RA to it and (ii) into a voltmeter of range 0 to V (V = GI0) by connecting a series resistance RV to it. Then,

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Answer: (d)

  1. Ammeter shunt: RA=IgGI0−IgR_A = \frac{I_g G}{I_0 - I_g}
  2. Voltmeter series resistance: RV=VIg−G=GI0Ig−G=G(I0−Ig)IgR_V = \frac{V}{I_g} - G = \frac{G I_0}{I_g} - G = \frac{G(I_0 - I_g)}{I_g}
  3. Product: RARV=G2R_A R_V = G^2
  4. Ratio: RARV=(IgI0−Ig)2\frac{R_A}{R_V} = \left(\frac{I_g}{I_0 - I_g}\right)^2
  5. So the answer is (d).
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Q425JEE Main 2019 · 12 JAN · SHIFT 2Medium

Kirchhoff's Laws & Measuring Instruments · Galvanometer to ammeter / voltmeter

A galvanometer, whose resistance is 50 ohm, has 25 divisions in it. When a current of 4 ×\times 10–4 A passes through it, its needle ( pointer) deflects by one division. To use this galvanometer as a voltmeter of range 2.5 V, it should be connected to a resistance of:

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Answer: (a)

  1. Full scale current Ig=25×4×10−4=10−2 AI_g = 25 \times 4\times10^{-4} = 10^{-2}\text{ A}
  2. Series resistance R=VIg−G=2.50.01−50R = \frac{V}{I_g} - G = \frac{2.5}{0.01} - 50
  3. R=250−50=200 ΩR = 250 - 50 = 200\ \Omega
  4. So the answer is (a) 200 ohm.
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Q426JEE Main 2019 · 12 APR · SHIFT 1Medium

Kirchhoff's Laws & Measuring Instruments · Galvanometer sensitivity & figure of merit

A galvanometer of resistance 100 Ω\Omega has 50 divisions on its scale and has sensitivitv of 20 μ\muA/division. It is to be converted to a voltmeter with three ranges of 0-2V, 0-10 V and 0-20 V. The appropriate circuit to do so is

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Answer: (c)

  1. Ig=50×20 μA=1 mAI_g = 50 \times 20\ \mu\text{A} = 1\text{ mA}
  2. 2 V: total 210−3=2000 Ω⇒R1=2000−100=1900 Ω\frac{2}{10^{-3}} = 2000\ \Omega \Rightarrow R_1 = 2000 - 100 = 1900\ \Omega
  3. 10 V: total 10000 Ω⇒R2=10000−2000=8000 Ω10000\ \Omega \Rightarrow R_2 = 10000 - 2000 = 8000\ \Omega
  4. 20 V: total 20000 Ω⇒R3=20000−10000=10000 Ω20000\ \Omega \Rightarrow R_3 = 20000 - 10000 = 10000\ \Omega
  5. So the answer is (c) R1=1900R_1 = 1900, R2=8000R_2 = 8000, R3=10000 ΩR_3 = 10000\ \Omega.
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Q427JEE Main 2019 · 11 JAN · SHIFT 2Medium

Kirchhoff's Laws & Measuring Instruments · Galvanometer sensitivity & figure of merit

A galvanometer having a resistance of 20 Ω\Omega and 30 divisions on both sides has figure of merit 0.005 ampere/division. The resistance that should be connected in series such that it can be used as a voltmeter upto 15 volt, is:

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Answer: (c)

  1. Full scale current (30 divisions one side) Ig=30×0.005=0.15 AI_g = 30 \times 0.005 = 0.15\text{ A}
  2. R=VIg−G=150.15−20R = \frac{V}{I_g} - G = \frac{15}{0.15} - 20
  3. R=100−20=80 ΩR = 100 - 20 = 80\ \Omega
  4. So the answer is (c) 80 Ω80\ \Omega.
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Q428JEE Main 2019 · 12 JAN · SHIFT 1Difficult

Kirchhoff's Laws & Measuring Instruments · Galvanometer to ammeter / voltmeter

The galvanometer deflection, when key K1 is closed but K2 is open, equals θ\theta0 (see figure). On closing K2 also and adjusting R2 to 5Ω\Omega, the deflection in galvanometer becomes θ05\frac{\theta_0}{5}. The resistance of the galvanometer is, then, given by [Neglect the internal resistance of battery]:

Figure

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Answer: (d)

  1. K2K_2 open: Ig0=E220+GI_{g0} = \frac{E}{220 + G}
  2. K2K_2 closed: I=E220+5GG+5I = \frac{E}{220 + \frac{5G}{G + 5}}, Ig=I5G+5=5E225G+1100I_g = I\frac{5}{G + 5} = \frac{5E}{225G + 1100}
  3. Ig=Ig05⇒25(220+G)=225G+1100⇒200G=4400I_g = \frac{I_{g0}}{5} \Rightarrow 25(220 + G) = 225G + 1100 \Rightarrow 200G = 4400
  4. G=22 ΩG = 22\ \Omega
  5. So the answer is (d) 22 Ω22\ \Omega.
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Q429JEE Main 2018 · 16 APR · SHIFT 1Easy

Kirchhoff's Laws & Measuring Instruments · Galvanometer to ammeter / voltmeter

A galvanometer with its coil resistance 25 Ω\Omega requires a current of 1 mA for its full deflection. In order to construct an ammeter to read upto a current of 2 A, the approximate value of the shunt resistance should be:

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Answer: (b)

  1. S=IgGI−IgS = \frac{I_g G}{I - I_g}
  2. S=10−3×252−0.001≈0.0252=1.25×10−2 ΩS = \frac{10^{-3} \times 25}{2 - 0.001} \approx \frac{0.025}{2} = 1.25 \times 10^{-2}\ \Omega
  3. So the answer is (b) 1.25×10−2 Ω1.25 \times 10^{-2}\ \Omega.
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Q430JEE Main 2018 · 16 APR · SHIFT 1Medium

Kirchhoff's Laws & Measuring Instruments · Galvanometer sensitivity & figure of merit

In a circuit for finding the resistance of a galvanometer by half deflection method, a 6 V battery and a high resistance of 11 kΩ\Omega are used. The figure of merit of the galvanometer is 60 μA/\mu A/division. In the absence of shunt resistance, the galvanometer produces a deflection of θ\theta = 9 divisions when current flows in the circuit. The value of the shunt resistance that can cause the deflection of θ/2,\theta/2, is closest to:

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Answer: (d)

  1. No shunt: I=9×60 μA=540 μA=611000+GI = 9 \times 60\ \mu\text{A} = 540\ \mu\text{A} = \frac{6}{11000 + G}
  2. 11000+G=11111⇒G=111 Ω11000 + G = 11111 \Rightarrow G = 111\ \Omega
  3. Half deflection method with very large series resistance: S≈GS \approx G (exact working gives S≈110 ΩS \approx 110\ \Omega).
  4. So the answer is (d) 110 Ω110\ \Omega.
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Q431JEE Main 2017Easy

Kirchhoff's Laws & Measuring Instruments · Galvanometer to ammeter / voltmeter

When a current of 5 mA is passed through a galvanometer having a coil of resistance 15Ω\Omega, it shows full scale deflection. The value of the resistance to be put in series with the galvanometer to convert it into a voltmeter of range 0 – 10V is:

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Answer: (b)

  1. Series resistance R=VIg−GR = \frac{V}{I_g} - G.
  2. R=105×10−3−15=2000−15=1985 Ω=1.985×103 ΩR = \frac{10}{5 \times 10^{-3}} - 15 = 2000 - 15 = 1985\ \Omega = 1.985 \times 10^3\ \Omega
  3. So the answer is (b).
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Q432JEE Main 2017Medium

Kirchhoff's Laws & Measuring Instruments · Multi-loop circuits

Figure

In the given circuit, the current in each resistance is:

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Answer: (a)

  1. In every loop the top and bottom cells point the same way, so going round the loop one is crossed +→−+ \to - and the other −→+- \to +.
  2. Net emf in each loop =2−2=0= 2 - 2 = 0.
  3. With zero current everywhere, KVL holds in every loop, and the solution is unique.
  4. So the answer is (a) 0 A.
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Q433JEE Main 2016Easy

Kirchhoff's Laws & Measuring Instruments · Galvanometer to ammeter / voltmeter

A galvanometer having a coil resistance of 100 Ω100\ \Omega gives a full scale deflection, when a currect of 1 mA1\text{ mA} is passed through it. The value of the resistance, which can convert this galvanometer into ammeter giving a full scale deflection for a current of 10 A,10\text{ A}, is:

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Answer: (c)

  1. S=IgGI−IgS = \frac{I_g G}{I - I_g}
  2. S=10−3×10010−0.001≈0.110=0.01 ΩS = \frac{10^{-3} \times 100}{10 - 0.001} \approx \frac{0.1}{10} = 0.01\ \Omega
  3. So the answer is (c) 0.01 Ω0.01\ \Omega.
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Q434NEET 2016 · PHASE 2Medium

Kirchhoff's Laws & Measuring Instruments · Potential difference between points

The potential difference (VA −- VB) between the points A and B in the given figure is

Figure

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Answer: (d)

  1. Walk from A to B along I=2 AI = 2\text{ A}: drop IRIR in each resistor, drop EE across the cell (+ to −).
  2. VA−2(2)−3−2(1)=VBV_A - 2(2) - 3 - 2(1) = V_B
  3. VA−VB=4+3+2=9 VV_A - V_B = 4 + 3 + 2 = 9\text{ V}
  4. So the answer is (d) +9 V.
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Q435JEE Main 2016 · 10 APR · SHIFT 1Medium

Kirchhoff's Laws & Measuring Instruments · Galvanometer to ammeter / voltmeter

A galvanometer has a 50 division scale. Battery has no internal resistance. It is found that there is deflection of 40 divisions when R = 2400 Ω\Omega. Deflection becomes 20 divisions when resistance taken from resistance box is 4900 Ω\Omega. Then we can conclude:

Figure

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Answer: (c)

  1. G is in series with R and the 2 V cell, so I=2R+GI = \frac{2}{R + G} and deflection ∝I\propto I.
  2. 4020=4900+G2400+G⇒4800+2G=4900+G⇒G=100 Ω\frac{40}{20} = \frac{4900 + G}{2400 + G} \Rightarrow 4800 + 2G = 4900 + G \Rightarrow G = 100\ \Omega, so (a) is false.
  3. At 40 div: I=22500=0.8 mAI = \frac{2}{2500} = 0.8\text{ mA}, so K=0.8 mA40=20 μA/divK = \frac{0.8\text{ mA}}{40} = 20\ \mu\text{A/div}.
  4. Full scale =50×20 μA=1 mA= 50 \times 20\ \mu\text{A} = 1\text{ mA} (b false); for 10 div, I=0.2 mAI = 0.2\text{ mA}, R=10000−100=9900 ΩR = 10000 - 100 = 9900\ \Omega (d false).
  5. So the answer is (c).
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Q436JEE Main 2016 · 9 APR · SHIFT 1Medium

Kirchhoff's Laws & Measuring Instruments · Galvanometer to ammeter / voltmeter

A 50 Ω\Omega resistance is connected to a battery of 5 V. A galvanometer of resistance 100 Ω\Omega is to be used as an ammeter to measure current through the resistance, for this a resistance rs is connected to the galvanometer. Which of the following connections should be employed if the measured current is within 1% of the current without the ammeter in the circuit?

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Answer: (a)

  1. True current I=550=0.1 AI = \frac{5}{50} = 0.1\text{ A}. To stay within 1%, 5050+RA≥0.99⇒RA≤0.505 Ω\frac{50}{50 + R_A} \ge 0.99 \Rightarrow R_A \le 0.505\ \Omega.
  2. An ammeter needs a small shunt in parallel with the galvanometer.
  3. With rs=0.5 Ωr_s = 0.5\ \Omega in parallel: RA=100×0.5100.5≈0.497 ΩR_A = \frac{100 \times 0.5}{100.5} \approx 0.497\ \Omega, which works.
  4. So the answer is (a) rs=0.5 Ωr_s = 0.5\ \Omega in parallel.
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Q437JEE Main 2015Medium

Kirchhoff's Laws & Measuring Instruments · Multi-loop circuits

Figure

In the circuit shown, the current in the 1 Ω1\ \Omega resistor is:

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Answer: (a)

  1. Take VQ=0V_Q = 0, VP=xV_P = x. Left branch: 66 V cell with 3 Ω3\ \Omega; right branch: 99 V cell (+ at top) with 2+3=5 Ω2 + 3 = 5\ \Omega; middle: 1 Ω1\ \Omega.
  2. Left branch current Q→P: −x+63-\frac{x + 6}{3}; right branch current P→Q: x−95\frac{x - 9}{5}; 1 Ω1\ \Omega current P→Q: xx.
  3. KCL at P: −x+63=x+x−95⇒−23x=3-\frac{x+6}{3} = x + \frac{x-9}{5} \Rightarrow -23x = 3
  4. x=−0.13x = -0.13 V, so 0.130.13 A flows through 1 Ω1\ \Omega from Q to P.
  5. So the answer is (a) 0.130.13 A, from Q to P.
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Q438JEE Main 2015 · 11 APR · SHIFT 1Medium

Kirchhoff's Laws & Measuring Instruments · Potential difference between points

In the electric network shown, when no current flows through the 4 Ω\Omega resistor in the arm EB, the potential difference between the points A and D will be:

Figure

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Answer: (c)

  1. No current in EB, so there is no drop across 4 Ω4\ \Omega: VE−VB=4 VV_E - V_B = 4\text{ V} (cell's + at E).
  2. Take VB=0V_B = 0: VE=VD=4 VV_E = V_D = 4\text{ V} (DE is a plain wire).
  3. The 9 V cell has + at A: VA=9 VV_A = 9\text{ V}.
  4. VA−VD=9−4=5 VV_A - V_D = 9 - 4 = 5\text{ V}
  5. So the answer is (c) 5 V.
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Q439NEET 2015 · AIPMT · CANCELLED PAPERMedium

Kirchhoff's Laws & Measuring Instruments · Voltmeter & ammeter in circuits

A, B and C are voltmeters of resistance R, 1.5 R and 3R respectively as shown in the figure. When some potential difference is applied between X and Y, the voltmeter readings are VA, VB and VC respectively, Then

Figure

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Answer: (c)

  1. B and C are in parallel: 1.5R×3R4.5R=R\dfrac{1.5R \times 3R}{4.5R} = R.
  2. This RR is in series with A (also RR), so the voltage divides equally: VA=VBCV_A = V_{BC}.
  3. B and C are in parallel, so VB=VC=VBC=VAV_B = V_C = V_{BC} = V_A.
  4. So the answer is (c) VA=VB=VCV_A = V_B = V_C.
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Q440NEET 2015 · AIPMTMedium

Kirchhoff's Laws & Measuring Instruments · Galvanometer to ammeter / voltmeter

A circuit contains an ammeter, a battery of 30 V and a resistance 40.8 ohm all connected in series. If the ammeter has a coil of resistance 480 ohm and a shunt of 20 ohm, the reading in the ammeter will be

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Answer: (c)

  1. Ammeter resistance: 480∥20=480×20500=19.2 Ω480 \| 20 = \dfrac{480 \times 20}{500} = 19.2\ \Omega.
  2. Total R=40.8+19.2=60 ΩR = 40.8 + 19.2 = 60\ \Omega
  3. I=3060=0.5 AI = \dfrac{30}{60} = 0.5\text{ A}. So the answer is (c).
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Q441JEE Main 2014 · 11 APR · SHIFT 1Medium

Kirchhoff's Laws & Measuring Instruments · Multi-loop circuits

In the circuit shown, current (in A) through the 50 V and 30 V batteries are, respectively:

Figure

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Answer: (c)

  1. Left rails: top 5050 V, bottom 00. Right rails: top aa, bottom bb with a−b=30a - b = 30 V.
  2. The two 5 Ω5\ \Omega carry the same current: 50−a5=b5⇒a+b=50\frac{50 - a}{5} = \frac{b}{5} \Rightarrow a + b = 50, so a=40a = 40, b=10b = 10, current =2= 2 A.
  3. 10 Ω10\ \Omega: 3010=3\frac{30}{10} = 3 A, so the 3030 V battery gives 3−2=13 - 2 = 1 A.
  4. 20 Ω20\ \Omega: 5020=2.5\frac{50}{20} = 2.5 A, so the 5050 V battery gives 2.5+2=4.52.5 + 2 = 4.5 A.
  5. So the answer is (c) 4.5 and 1.
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Q442JEE Main 2014 · 11 APR · SHIFT 1Medium

Kirchhoff's Laws & Measuring Instruments · Galvanometer sensitivity & figure of merit

In the circuit diagrams (A, B, C and D) shown below, R is a high resistance and S is a resistance of the order of galvanometer resistance G. The correct circuit, corresponding to the half deflection method for finding the resistance and figure of merit of galvanometer, is the circuit labelled as:

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Answer: (d)

  1. Half deflection method: high resistance R in series with G and the cell (key K1K_1).
  2. S (of the order of G) is connected in parallel with G through key K2K_2, so closing K2K_2 shunts G.
  3. Only circuit (d) has R in series and S with K2K_2 across G.
  4. So the answer is (d).
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Q443JEE Main 2013 · 23 APR · SHIFT 1Easy

Kirchhoff's Laws & Measuring Instruments · Voltmeter & ammeter in circuits

Correct set up to verify Ohm's law is:

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Answer: (a)

  1. For Ohm's law, the ammeter must be in series with the cell in the main line and the voltmeter must be across the resistor.
  2. In (a) the ammeter is in series with the cell and the voltmeter is directly across the resistor.
  3. In (b) the voltmeter is across the cell only; in (c) and (d) a meter is placed wrongly (voltmeter in series or ammeter in a side branch).
  4. So the answer is (a).
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Q444JEE Main 2013 · 9 APR · SHIFT 1Easy

Kirchhoff's Laws & Measuring Instruments · Galvanometer to ammeter / voltmeter

A shunt of resistance 1 Ω\Omega is connected across a galvanometer of 120 Ω\Omega resistance. A current of 5.5 ampere gives full scale deflection in the galvanometer. The current that will give full scale deflection in the absence of the shunt is nearly:

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Answer: (d)

  1. With the shunt, current through galvanometer Ig=ISS+GI_g = I\frac{S}{S + G}.
  2. Ig=5.5×11+120=5.5121≈0.045 AI_g = 5.5 \times \frac{1}{1 + 120} = \frac{5.5}{121} \approx 0.045\text{ A}
  3. Without the shunt this same IgI_g gives full scale, so the answer is (d) 0.045 ampere.
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Q445JEE Main 2013Easy

Kirchhoff's Laws & Measuring Instruments · Galvanometer to ammeter / voltmeter

This questions has Statement - I\text{I} and Statement - II\text{II}. Of the four choices given after the Statements, choose the one that best describes into two Statements.

Statement - I\text{I}: Higher the range, greater is the resistance of ammeter.

Statement - II\text{II}: To increase the range of ammeter, additional shunt needs to be used across it.

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Answer: (d)

  1. Statement I: higher range needs a smaller shunt, S=IgGI−IgS = \frac{I_g G}{I - I_g}, so the ammeter resistance becomes smaller. False.
  2. Statement II: a shunt (or extra shunt) in parallel diverts more current and increases the range. True.
  3. So the answer is (d).
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Q446JEE Main 2013 · 22 APR · SHIFT 1Medium

Kirchhoff's Laws & Measuring Instruments · Galvanometer sensitivity & figure of merit

To find the resistance of a galvanometer by the half deflection method, the following circuit is used with resistances R1=9970 Ω, R2=30 ΩR_1 = 9970\ \Omega,\ R_2 = 30\ \Omega and R3=0R_3 = 0. The deflection in the galvanometer is d. With R3=107 ΩR_3 = 107\ \Omega, the deflection changed to d2\frac{d}{2}. The galvanometer resistance is approximately:

Figure

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Answer: (d)

  1. R1R_1 is very large, so the main current ii stays almost constant.
  2. G+R3G + R_3 is in parallel with R2R_2: Ig=iR2R2+G+R3I_g = i\frac{R_2}{R_2 + G + R_3}
  3. Half deflection: R2+G+107=2(R2+G)⇒G=107−30=77 ΩR_2 + G + 107 = 2(R_2 + G) \Rightarrow G = 107 - 30 = 77\ \Omega
  4. So the answer is (d) 77 Ω77\ \Omega.
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Q447NEET 2011 · AIPMT · MAINSMedium

Kirchhoff's Laws & Measuring Instruments · Potential difference between points

In the circuit shown in the figure, if the potential at point A is taken to be zero, the potential at point B is

Figure

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Answer: (a)

  1. VA=0V_A = 0. From A to C the 1 V cell is crossed from − to +: VC=+1 VV_C = +1\text{ V}.
  2. At D: 2 A comes in from B, 1 A leaves through R1R_1, so 1 A flows D to C through 2 Ω2\ \Omega.
  3. VD=VC+1×2=3 VV_D = V_C + 1 \times 2 = 3\text{ V}
  4. From D to B the 2 V cell is crossed from + to −: VB=3−2=+1 VV_B = 3 - 2 = +1\text{ V}
  5. So the answer is (a) +1 V.
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Q448NEET 2010 · AIPMT · PRELIMSEasy

Kirchhoff's Laws & Measuring Instruments · Multi-loop circuits

Consider the following two statements.

(A) Kirchoff's junction law follows from the conservation of charge.

(B) Kirchhoff's loop law follows from the conservation of energy.

Which of the following is correct?

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Answer: (d)

  1. (A) Junction law: charge cannot pile up at a junction, so it follows from conservation of charge. Correct.
  2. (B) Loop law: net work on a charge round a closed loop is zero, so it follows from conservation of energy. Correct.
  3. So the answer is (d) both are correct.
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Q449NEET 2009 · AIPMTEasy

Kirchhoff's Laws & Measuring Instruments · Kirchhoff's laws: basics

See the electrical circuit shown in this figure. Which of the following equations is a correct equation for it?

Figure

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Answer: (d)

  1. i1i_1 and i2i_2 join at the left node, so (i1+i2)(i_1 + i_2) flows through RR from left to right.
  2. Loop through ε1\varepsilon_1, r1r_1 and RR: emf ε1\varepsilon_1 drives i1i_1; drops are i1r1i_1r_1 and (i1+i2)R(i_1 + i_2)R.
  3. ε1−i1r1−(i1+i2)R=0\varepsilon_1 - i_1r_1 - (i_1 + i_2)R = 0
  4. So the answer is (d).
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Q450NEET 2008 · AIPMTMedium

Kirchhoff's Laws & Measuring Instruments · Potential difference between points

Figure

In the circuit shown, the current through the 4 Ω\Omega resistor is 1 amp when the points P and M are connected to a d.c. voltage source. The potential difference between the points M and N is

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Answer: (b)

  1. 4 Ω4\ \Omega and 3 Ω3\ \Omega are in parallel between P and M: VPM=1×4=4 VV_{PM} = 1 \times 4 = 4\text{ V}.
  2. Lower branch: 0.5∥0.5=0.25 Ω0.5 \parallel 0.5 = 0.25\ \Omega, plus 1 Ω1\ \Omega, total 1.25 Ω1.25\ \Omega across the same 4 V.
  3. Current in lower branch =41.25=3.2 A= \frac{4}{1.25} = 3.2\text{ A}
  4. VNM=3.2×1=3.2 VV_{NM} = 3.2 \times 1 = 3.2\text{ V}
  5. So the answer is (b) 3.2 volt.
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Q451NEET 2006 · AIPMTEasy

Kirchhoff's Laws & Measuring Instruments · Kirchhoff's laws: basics

Kirchhoff's first and second laws of electrical circuits are consequences of

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Answer: (c)

  1. First law (junction rule) follows from conservation of electric charge.
  2. Second law (loop rule) follows from conservation of energy.
  3. So the answer is (c).
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Q452JEE Main 2006 · AIEEEEasy

Kirchhoff's Laws & Measuring Instruments · Kirchhoff's laws: basics

The Kirchhoff's first law (∑i=0)\left(\sum i = 0\right) and second law (∑iR=∑E)\left(\sum iR = \sum E\right), where the symbols have their usual meanings, are respectively based on

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Answer: (a)

  1. ∑i=0\sum i = 0: charge entering a junction equals charge leaving, so conservation of charge.
  2. ∑iR=∑E\sum iR = \sum E: energy supplied equals energy used round a loop, so conservation of energy.
  3. So the answer is (a).
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Q453NEET 2006 · AIPMTMedium

Kirchhoff's Laws & Measuring Instruments · Potential difference between points

In the circuit shown, if a conducting wire is connected between points A and B, the current in this wire will

Figure

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Answer: (a)

  1. Left end is at the + terminal (potential VV), right end at 0.
  2. Without the wire: VA=V−48V=0.5VV_A = V - \frac{4}{8}V = 0.5V, VB=V−14V=0.75VV_B = V - \frac{1}{4}V = 0.75V.
  3. VB>VAV_B > V_A, so current flows from B to A when they are joined.
  4. (With the wire: common node at 15V22\frac{15V}{22}, current A to B =7V/22−15V/224<0= \frac{7V/22 - 15V/22}{4} < 0, confirming B to A.)
  5. So the answer is (a) flow from B to A.
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Q454JEE Main 2006 · AIEEEMedium

Kirchhoff's Laws & Measuring Instruments · Potential difference between points

The current II drawn from the 55 volt source will be

Figure

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Answer: (b)

  1. Label the node after 5 Ω5\ \Omega as Y and the node after the middle 10 Ω10\ \Omega as Z.
  2. Arms: 5 Ω5\ \Omega and 10 Ω10\ \Omega from the + end, 10 Ω10\ \Omega and 20 Ω20\ \Omega to the − end; 510=1020\frac{5}{10} = \frac{10}{20}, so the bridge is balanced and the middle 10 Ω10\ \Omega (Y to Z) carries no current.
  3. Paths: 5+10=15 Ω5 + 10 = 15\ \Omega and 10+20=30 Ω10 + 20 = 30\ \Omega in parallel: R=10 ΩR = 10\ \Omega.
  4. I=510=0.5 AI = \frac{5}{10} = 0.5\text{ A}
  5. So the answer is (b) 0.5 A.
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Q455NEET 2005 · AIPMTMedium

Kirchhoff's Laws & Measuring Instruments · Potential difference between points

For the network shown in the figure the value of the current ii is

Figure

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Answer: (d)

  1. Ratios 42=63\frac{4}{2} = \frac{6}{3}, so the bridge is balanced and the middle 4 Ω4\ \Omega carries no current.
  2. Paths: 4+2=6 Ω4 + 2 = 6\ \Omega and 6+3=9 Ω6 + 3 = 9\ \Omega in parallel.
  3. R=6×915=185 ΩR = \frac{6 \times 9}{15} = \frac{18}{5}\ \Omega
  4. i=VR=5V18i = \frac{V}{R} = \frac{5V}{18}
  5. So the answer is (d).
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Q456JEE Main 2005 · AIEEEMedium

Kirchhoff's Laws & Measuring Instruments · Galvanometer to ammeter / voltmeter

In the circuit, the galvanometer GG shows zero deflection. If the batteries AA and BB have negligible internal resistance, the value of the resistor RR will be -

Figure

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Answer: (a)

  1. Zero deflection means no current in G, so the p.d. across R equals the 2 V of battery A.
  2. The 12 V battery drives current through 500 Ω\Omega and R in series: VR=12R500+RV_R = \frac{12R}{500 + R}
  3. 12R500+R=2⇒12R=1000+2R⇒R=100 Ω\frac{12R}{500 + R} = 2 \Rightarrow 12R = 1000 + 2R \Rightarrow R = 100\ \Omega
  4. So the answer is (a) 100 Ω100\ \Omega.
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Q457JEE Main 2005 · AIEEEMedium

Kirchhoff's Laws & Measuring Instruments · Galvanometer sensitivity & figure of merit

A moving coil galvanometer has 150150 equal divisions. Its current sensitivity is 1010- divisions per milliampere and voltage sensitivity is 22 divisions per millivolt. In order that each division reads 11 volt, the resistance in ohmsohms needed to be connected in series with the coil will be -

Show answer and solution

Answer: (c)

  1. G=IsVs=10 div/mA2 div/mV=5 ΩG = \frac{I_s}{V_s} = \frac{10\text{ div/mA}}{2\text{ div/mV}} = 5\ \Omega
  2. Full scale current Ig=15010=15 mAI_g = \frac{150}{10} = 15\text{ mA}; full scale voltage needed =150×1=150 V= 150 \times 1 = 150\text{ V}
  3. R=1500.015−5=10000−5=9995 ΩR = \frac{150}{0.015} - 5 = 10000 - 5 = 9995\ \Omega
  4. So the answer is (c) 9995.
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Q458JEE Main 2004 · AIEEEEasy

Kirchhoff's Laws & Measuring Instruments · Kirchhoff's laws: basics

The Kirchhoff's first law (∑i=0)\left(\sum i = 0\right) and second law (∑iR=∑E),\left(\sum iR = \sum E\right), where the symbols have their usual meanings, are respectively based on

Show answer and solution

Answer: (d)

  1. ∑i=0\sum i = 0: conservation of charge at a junction.
  2. ∑iR=∑E\sum iR = \sum E: conservation of energy round a loop.
  3. So the answer is (d).
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Q459JEE Main 2003 · AIEEEEasy

Kirchhoff's Laws & Measuring Instruments · Galvanometer to ammeter / voltmeter

An ammeter reads upto 11 ampere. Its internal resistance is 0.81 ohm0.81\text{ ohm}. To increase the range to 10 A10\text{ A} the value of the required shunt is

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Answer: (d)

  1. Shunt for an ammeter: S=G IgI−IgS = \frac{G\,I_g}{I - I_g}, with Ig=1I_g = 1 A and G=0.81 ΩG = 0.81\ \Omega.
  2. S=0.81×110−1=0.819S = \frac{0.81 \times 1}{10 - 1} = \frac{0.81}{9}
  3. So the answer is (d) 0.09 Ω0.09\ \Omega.
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Q460JEE Main 2003 · AIEEEMedium

Kirchhoff's Laws & Measuring Instruments · Potential difference between points

A 33 volt battery with negligible internal resistance is connected in a circuit as shown in the figure. The current II, in the circuit will be

Figure

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Answer: (b)

  1. The battery is across two corners of the triangle: one 3 Ω3\ \Omega side in parallel with the other two sides in series (6 Ω6\ \Omega).
  2. R=3×63+6=2 ΩR = \frac{3 \times 6}{3 + 6} = 2\ \Omega
  3. I=32=1.5 AI = \frac{3}{2} = 1.5\text{ A}
  4. So the answer is (b) 1.5 A.
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Q461JEE Main 2002 · AIEEEEasy

Kirchhoff's Laws & Measuring Instruments · Voltmeter & ammeter in circuits

If an ammeter is to be used in place of a voltmeter, then we must connect with the ammeter a

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Answer: (c)

  1. A voltmeter must have a very high resistance so that it draws almost no current.
  2. So a high resistance is connected in series with the ammeter (galvanometer).
  3. So the answer is (c) high resistance in series.
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Q462NEET 1999 · AIPMTMedium

Kirchhoff's Laws & Measuring Instruments · Multi-loop circuits

The current (I)(I) in the given circuit is

Figure

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Answer: (b)

  1. The cell is across A and B: RA=3 ΩR_A = 3\ \Omega is in parallel with RB+RC=12 ΩR_B + R_C = 12\ \Omega.
  2. Req=3×123+12=2.4 ΩR_{eq} = \frac{3 \times 12}{3 + 12} = 2.4\ \Omega
  3. I=4.82.4=2I = \frac{4.8}{2.4} = 2 A
  4. So the answer is (b) 22 A.
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Q463NEET 1998 · AIPMTEasy

Kirchhoff's Laws & Measuring Instruments · Galvanometer to ammeter / voltmeter

A galvanometer having a resistance of 88 ohm is shunted by a wire of resistance 22 ohm. If the total current is 1 A1\text{ A}, the part of it passing through the shunt will be

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Answer: (b)

  1. Current divides inversely as resistance: IS=IGG+SI_S = I\frac{G}{G + S}.
  2. IS=1×88+2=0.8 AI_S = 1 \times \frac{8}{8 + 2} = 0.8\text{ A}
  3. So the answer is (b) 0.8 A0.8\text{ A}.
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Q464NEET 1997 · AIPMTEasy

Kirchhoff's Laws & Measuring Instruments · Kirchhoff's laws: basics

Kirchhoff's first law, i.e. Σi=0\Sigma i = 0 at a junction, deals with the conservation of

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Answer: (c)

  1. Σi=0\Sigma i = 0 at a junction means charge entering equals charge leaving; charge is not created or stored.
  2. So the answer is (c) charge.
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Q465NEET 1997 · AIPMTMedium

Kirchhoff's Laws & Measuring Instruments · Multi-loop circuits

The current in the following circuit is

Figure

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Answer: (b)

  1. Across two corners of the triangle: one 3 Ω3\ \Omega is in parallel with 3+3=6 Ω3 + 3 = 6\ \Omega.
  2. Req=3×69=2 ΩR_{eq} = \frac{3 \times 6}{9} = 2\ \Omega
  3. I=22=1I = \frac{2}{2} = 1 A
  4. So the answer is (b) 11 A.
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Q466NEET 1993 · AIPMTEasy

Kirchhoff's Laws & Measuring Instruments · Multi-loop circuits

Current through 3 Ω3\ \Omega resistor is 0.80.8 ampere, then potential drop through 4 Ω4\ \Omega resistor is

Figure

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Answer: (c)

  1. P.d. across the parallel pair: V=0.8×3=2.4 VV = 0.8 \times 3 = 2.4\text{ V}.
  2. Current in 6 Ω6\ \Omega: 2.46=0.4 A\dfrac{2.4}{6} = 0.4\text{ A}, so total current =0.8+0.4=1.2 A= 0.8 + 0.4 = 1.2\text{ A}.
  3. Drop across 4 Ω4\ \Omega: 1.2×4=4.8 V1.2 \times 4 = 4.8\text{ V}
  4. So the answer is (c) 4.8 V.
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Q467NEET 1992 · AIPMTEasy

Kirchhoff's Laws & Measuring Instruments · Kirchhoff's laws: basics

Kirchhoff's first law, i.e. Σi=0\Sigma i = 0 at a junction, deals with the conservation of

Show answer and solution

Answer: (c)

  1. Σi=0\Sigma i = 0 at a junction means charge entering equals charge leaving; charge is not created or stored.
  2. So the answer is (c) charge.
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Q468NEET 1988 · AIPMTMedium

Kirchhoff's Laws & Measuring Instruments · Potential difference between points

Two batteries of emf 4 V4\text{ V} and 8 V8\text{ V} with internal resistance 1 Ω1\ \Omega and 2 Ω2\ \Omega are connected in a circuit with resistance of 9 Ω9\ \Omega as shown in figure. The current and potential difference between the points PP and QQ are

Figure

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Answer: (a)

  1. The two cells have their + terminals facing each other, so they oppose: net emf =8−4=4 V= 8 - 4 = 4\text{ V}.
  2. Total resistance =1+2+9=12 Ω= 1 + 2 + 9 = 12\ \Omega
  3. I=412=13 AI = \frac{4}{12} = \frac{1}{3}\text{ A}
  4. VPQV_{PQ} = drop across RR =13×9=3 V= \frac{1}{3} \times 9 = 3\text{ V}
  5. So the answer is (a) 13\frac{1}{3} A and 3 V.
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Q469MHT CET 2026 · 16 APR · SHIFT 1Easy

Wheatstone Bridge & Metre Bridge · Metre bridge

A metre bridge experiment, the resistance in the left gap is 20 Ω20\ \Omega and in the right gap is 60 Ω60\ \Omega. The bridge is balanced. The distance of the null point from the centre of the wire is

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Answer: (c)

  1. 2060=l100−l\frac{20}{60} = \frac{l}{100 - l}
  2. 3l=100−l⇒l=25 cm3l = 100 - l \Rightarrow l = 25\text{ cm} from the left end
  3. Distance from centre =50−25=25 cm= 50 - 25 = 25\text{ cm}
  4. So the answer is (c) 25 cm.
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Q470MHT CET 2026 · 15 APR · SHIFT 2Easy

Wheatstone Bridge & Metre Bridge · Metre bridge

In metre bridge experiment, the resistance in the left gap is 15 Ω\Omega and in the right gap is 45 Ω\Omega. The bridge is balanced. The distance of the null point from the centre of the wire is

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Answer: (d)

  1. 1545=l100−l⇒3l=100−l\frac{15}{45} = \frac{l}{100 - l} \Rightarrow 3l = 100 - l
  2. l=25 cml = 25\text{ cm} from the left end
  3. Distance from centre =50−25=25 cm= 50 - 25 = 25\text{ cm}
  4. So the answer is (d) 25 cm.
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Q471MHT CET 2026 · 16 APR · SHIFT 2Medium

Wheatstone Bridge & Metre Bridge · Wheatstone bridge balance

If the potential difference between points B and D is zero in the given circuit, the value of x is 1n Ω\frac{1}{n}\ \Omega. The value n is

Figure

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Answer: (a)

  1. Arm with B on left: 6∥3=2 Ω6 \parallel 3 = 2\ \Omega; arm with D on left: 1+2=3 Ω1 + 2 = 3\ \Omega.
  2. Right arms: x∥1=xx+1x \parallel 1 = \dfrac{x}{x+1} (to B) and xx (to D).
  3. Balance: 2x/(x+1)=3x⇒2(x+1)=3\dfrac{2}{x/(x+1)} = \dfrac{3}{x} \Rightarrow 2(x+1) = 3
  4. x=12 Ω=1n⇒n=2x = \dfrac{1}{2}\ \Omega = \dfrac{1}{n} \Rightarrow n = 2
  5. So the answer is (a) 22.
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Q472MHT CET 2026 · 13 APR · SHIFT 2Medium

Wheatstone Bridge & Metre Bridge · Wheatstone bridge balance

Resistances are joint as shown in the figure. In the balanced condition, the current II drawn from the battery is

Figure

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Answer: (c)

  1. Balanced bridge (10/10=30/3010/10 = 30/30): no current in G, remove it.
  2. Path ABC: 10+10=20 Ω10 + 10 = 20\ \Omega; path ADC: 30+30=60 Ω30 + 30 = 60\ \Omega
  3. Req=20×6080=15 ΩR_{eq} = \dfrac{20 \times 60}{80} = 15\ \Omega
  4. I=315=0.2I = \dfrac{3}{15} = 0.2 A
  5. So the answer is (c) 0.20.2 A.
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Q473JEE Main 2026 · 28 JAN · SHIFT 2Medium

Wheatstone Bridge & Metre Bridge · Wheatstone bridge balance

A Wheatstone bridge is initially at room temperature and all arms of the bridge have same value of resistances (R1=R2=R3=R4R_1 = R_2 = R_3 = R_4). When R3R_3 resistance is heated to some temperature, its resistance value has gone up by 10%. The potential difference (Va−VbV_a - V_b) (after R3R_3 is heated) is _________ V.

Figure

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Answer: (b)

  1. Take the bottom node at 00 V and top at 4040 V; let each arm be RR, with R3=1.1RR_3 = 1.1R.
  2. Va=40×R2R1+R2=20V_a = 40 \times \dfrac{R_2}{R_1+R_2} = 20 V
  3. Vb=40×R4R3+R4=40×12.1=19.05V_b = 40 \times \dfrac{R_4}{R_3+R_4} = 40 \times \dfrac{1}{2.1} = 19.05 V
  4. Va−Vb=0.95V_a - V_b = 0.95 V
  5. So the answer is (b) 0.950.95.
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Q474MHT CET 2026 · 20 MAY · SHIFT 1Medium

Wheatstone Bridge & Metre Bridge · Metre bridge

A balanced meter bridge is shown in the circuit diagram. The meter bridge wire has resistance 2 Ωcm2\ \frac{\Omega}{cm}. The current drawn from the battery is (Internal resistance of the battery is negligible)

Figure

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Answer: (d)

  1. Balance: 3x=3070⇒x=7 Ω\dfrac{3}{x} = \dfrac{30}{70} \Rightarrow x = 7\ \Omega
  2. Gap branch: 3+7=10 Ω3 + 7 = 10\ \Omega, current 1010=1\dfrac{10}{10} = 1 A
  3. Wire (100 cm): 100×2=200 Ω100 \times 2 = 200\ \Omega, current 10200=0.05\dfrac{10}{200} = 0.05 A
  4. Total I=1+0.05=1.05I = 1 + 0.05 = 1.05 A
  5. So the answer is (d) 1.051.05 A.
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Q475MHT CET 2026 · 15 MAY · SHIFT 2Medium

Wheatstone Bridge & Metre Bridge · Metre bridge

Two wires AA and BB of equal lengths are connected in left and right gap of a metre bridge, null point is obtained at 4040 cm from left end. Diameters of the wires AA and BB are in the ratio 3:13 : 1. The ratio of specific resistance of AA to that of BB is

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Answer: (c)

  1. RARB=4060=23\dfrac{R_A}{R_B} = \dfrac{40}{60} = \dfrac{2}{3}
  2. R=ρLπd2/4R = \dfrac{\rho L}{\pi d^2/4}, equal LL: ρAρB=RARB⋅dA2dB2\dfrac{\rho_A}{\rho_B} = \dfrac{R_A}{R_B} \cdot \dfrac{d_A^2}{d_B^2}
  3. ρAρB=23×9=6\dfrac{\rho_A}{\rho_B} = \dfrac{2}{3} \times 9 = 6
  4. So the answer is (c) 6:16 : 1.
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Q476MHT CET 2026 · 15 MAY · SHIFT 1Medium

Wheatstone Bridge & Metre Bridge · Metre bridge

With resistance xx in the left gap and resistance 9 Ω9\ \Omega in right gap of a meter bridge, the balance point is obtained at 4040 cm from left end. In what way and to which resistance 3 Ω3\ \Omega resistance be connected to obtain the balance point at 5050 cm from the left end?

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Answer: (c)

  1. x9=4060⇒x=6 Ω\dfrac{x}{9} = \dfrac{40}{60} \Rightarrow x = 6\ \Omega
  2. Balance at 50 cm needs equal gap resistances.
  3. 6+3=96 + 3 = 9: put 3 Ω3\ \Omega in series with xx.
  4. So the answer is (c) in series with x Ωx\ \Omega.
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Q477MHT CET 2026 · 14 MAY · SHIFT 2Medium

Wheatstone Bridge & Metre Bridge · Metre bridge

In balanced meter bridge, the resistance of bridge wire is 0.1 Ω/0.1\ \Omega/cm. Unknown resistance XX is connected in the left gap and 6 Ω6\ \Omega in right gap. Null point divides the wire in the ratio 2:32 : 3. The current drawn from the battery of 55V having negligible resistance is

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Answer: (a)

  1. X6=23⇒X=4 Ω\dfrac{X}{6} = \dfrac{2}{3} \Rightarrow X = 4\ \Omega; gap branch 4+6=10 Ω4 + 6 = 10\ \Omega
  2. Wire: 100×0.1=10 Ω100 \times 0.1 = 10\ \Omega
  3. Req=10∥10=5 ΩR_{eq} = 10 \parallel 10 = 5\ \Omega
  4. I=55=1I = \dfrac{5}{5} = 1 A
  5. So the answer is (a) 11 A.
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Q478MHT CET 2026 · 13 MAY · SHIFT 1Medium

Wheatstone Bridge & Metre Bridge · Metre bridge

The resistances in two arms of the metre bridge are 5 Ω5\ \Omega and R ΩR\ \Omega respectively, null point is obtained at length l1l_1. When the resistance R is shunted with an equal resistance, the new balance point is at 1.6l11.6 l_1. The resistance R is

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Answer: (b)

  1. 5R=l1100−l1\dfrac{5}{R} = \dfrac{l_1}{100 - l_1} and, with R→R/2R \to R/2: 10R=1.6l1100−1.6l1\dfrac{10}{R} = \dfrac{1.6 l_1}{100 - 1.6 l_1}
  2. Divide: 2=1.6(100−l1)100−1.6l1⇒200−3.2l1=160−1.6l12 = \dfrac{1.6(100 - l_1)}{100 - 1.6 l_1} \Rightarrow 200 - 3.2 l_1 = 160 - 1.6 l_1
  3. l1=25l_1 = 25 cm, so 5R=2575\dfrac{5}{R} = \dfrac{25}{75}
  4. R=15 ΩR = 15\ \Omega
  5. So the answer is (b) 15 Ω15\ \Omega.
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Q479MHT CET 2026 · 19 APR · SHIFT 2Medium

Wheatstone Bridge & Metre Bridge · Metre bridge

The resistances in the two gaps of a balanced meter bridge are 'X' Ω\Omega and '3X' Ω\Omega respectively. If the resistances are interchanged the balance point shifts by

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Answer: (c)

  1. X3X=l100−l⇒l=25\dfrac{X}{3X} = \dfrac{l}{100 - l} \Rightarrow l = 25 cm
  2. After interchange: l′=75l' = 75 cm
  3. Shift =75−25=50= 75 - 25 = 50 cm
  4. So the answer is (c) 5050 cm.
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Q480MHT CET 2026 · 19 APR · SHIFT 1Medium

Wheatstone Bridge & Metre Bridge · Metre bridge

In a meter bridge experiment, the balance point is obtained at length 'l1l_1 ' cm from left hand when resistances in the left gap and right gap are 15 Ω15\ \Omega and R ΩR\ \Omega respectively. When the resistance RR is shunted with equal resistance the new balance point is at (1.6 l1)(1.6\, l_1). The resistance RR in ohm is

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Answer: (c)

  1. 15R=l1100−l1\dfrac{15}{R} = \dfrac{l_1}{100 - l_1} and 30R=1.6l1100−1.6l1\dfrac{30}{R} = \dfrac{1.6 l_1}{100 - 1.6 l_1}
  2. Divide: 2(100−1.6l1)=1.6(100−l1)⇒l1=252(100 - 1.6 l_1) = 1.6(100 - l_1) \Rightarrow l_1 = 25 cm
  3. 15R=2575⇒R=45 Ω\dfrac{15}{R} = \dfrac{25}{75} \Rightarrow R = 45\ \Omega
  4. So the answer is (c) 4545.
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Q481MHT CET 2026 · 13 APR · SHIFT 1Medium

Wheatstone Bridge & Metre Bridge · Metre bridge

Two wires A and B of equal lengths are connected in left and right gap respectively of a metre bridge, null point is obtained at 40 cm from left end. Diameters of the wires A and B are in the ratio 3:1 respectively, the ratio of specific resistance of A to that of B is

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Answer: (c)

  1. RARB=4060=23\dfrac{R_A}{R_B} = \dfrac{40}{60} = \dfrac{2}{3}
  2. Equal lengths, R∝ρd2R \propto \dfrac{\rho}{d^2}: ρAρB=RARB⋅(dAdB)2=23×9\dfrac{\rho_A}{\rho_B} = \dfrac{R_A}{R_B} \cdot \left(\dfrac{d_A}{d_B}\right)^2 = \dfrac{2}{3} \times 9
  3. ρAρB=6\dfrac{\rho_A}{\rho_B} = 6
  4. So the answer is (c) 6:16 : 1.
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Q482NEET 2026Medium

Wheatstone Bridge & Metre Bridge · Metre bridge

In a metre bridge experiment (see figure), the positions of the cell, EE, and galvanometer, GG, are interchanged. We shall observe in the galvanometer:

Figure

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Answer: (d)

  1. Interchanging cell and galvanometer does not change the balance condition R1R2=l1l2\dfrac{R_1}{R_2} = \dfrac{l_1}{l_2} (reciprocity of the bridge).
  2. On one side of the balance point the galvanometer deflects one way, on the other side the opposite way.
  3. At the balance point there is no deflection.
  4. So the answer is (d).
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Q483JEE Main 2026 · 24 JAN · SHIFT 1Medium

Wheatstone Bridge & Metre Bridge · Metre bridge

Two resistors 2 Ω2\ \Omega and 3 Ω3\ \Omega are connected in the gaps of bridge as shown in figure. The null point is obtained with the contact of jockey at some point on wire XYXY. When an unknown resistor is connected in parallel with 3 Ω3\ \Omega resistor, the null point is shifted by 22.5 cm toward YY. The resistance of unknown resistor is ‾ Ω\underline{\qquad}\ \Omega.

Figure

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Answer: (d)

  1. Initially 23=l100−l⇒l=40 cm\dfrac{2}{3} = \dfrac{l}{100-l} \Rightarrow l = 40\text{ cm}.
  2. Shunting 3 Ω3\ \Omega lowers the right resistance, so the null point moves right to 40+22.5=62.5 cm40 + 22.5 = 62.5\text{ cm}.
  3. 2R′=62.537.5=53⇒R′=1.2 Ω\dfrac{2}{R'} = \dfrac{62.5}{37.5} = \dfrac{5}{3} \Rightarrow R' = 1.2\ \Omega.
  4. 3x3+x=1.2⇒1.8x=3.6⇒x=2 Ω\dfrac{3x}{3+x} = 1.2 \Rightarrow 1.8x = 3.6 \Rightarrow x = 2\ \Omega.
  5. So the answer is (d) 2.
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Q484JEE Main 2026 · 22 JAN · SHIFT 1Medium

Wheatstone Bridge & Metre Bridge · Metre bridge

A meter bridge with two resistances R1R_1 and R2R_2 as shown in figure was balanced (null point) at 40 cm from the point PP. The null point changed to 50 cm from the point PP, when 16 Ω16\ \Omega resistance is connected in parallel to R2R_2. The values of resistances R1R_1 and R2R_2 are ______.

Figure

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Answer: (a)

  1. Initially R1R2=4060=23\dfrac{R_1}{R_2} = \dfrac{40}{60} = \dfrac{2}{3}.
  2. With 16 Ω16\ \Omega in parallel, balance at 50 cm means R1=16R216+R2R_1 = \dfrac{16R_2}{16+R_2}.
  3. 2R23=16R216+R2⇒32+2R2=48⇒R2=8 Ω\dfrac{2R_2}{3} = \dfrac{16R_2}{16+R_2} \Rightarrow 32 + 2R_2 = 48 \Rightarrow R_2 = 8\ \Omega.
  4. R1=23×8=163 ΩR_1 = \dfrac{2}{3} \times 8 = \dfrac{16}{3}\ \Omega.
  5. So the answer is (a).
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Q485JEE Main 2026 · 24 JAN · SHIFT 2Medium

Wheatstone Bridge & Metre Bridge · Metre bridge

In a meter bridge experiment to determine the value of unknown resistance, first the resistances 2 Ω2\ \Omega and 3 Ω3\ \Omega are connected in the left and right gaps of the bridge and the null point is obtained at a distance l cml\text{ cm} from the left. Now when an unknown resistance x Ωx\ \Omega is connected in parallel to 3 Ω3\ \Omega resistance, the null point is shifted by 10 cm to the right of wire. The value of unknown resistance xx is

‾ Ω\underline{\qquad}\ \Omega.

Numerical answer type. Work it out, then open the answer.

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Answer: 6

  1. Initially 23=l100−l⇒l=40 cm\dfrac{2}{3} = \dfrac{l}{100-l} \Rightarrow l = 40\text{ cm}.
  2. New null point at 50 cm, so the right gap becomes 2 Ω2\ \Omega.
  3. 3x3+x=2⇒x=6\dfrac{3x}{3+x} = 2 \Rightarrow x = 6.
  4. x=6 Ωx = 6\ \Omega.
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Q486MHT CET 2026 · 20 APR · SHIFT 1Difficult

Wheatstone Bridge & Metre Bridge · Metre bridge

In meter bridge experiment, two resistances X and Y in the two gaps give a null point dividing the wire in the ratio 2:32:3. When each resistance is increased by 30 Ω30\ \Omega, the null point divides the wire in the ratio 5:65:6. The resistance X and Y are respectively

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Answer: (a)

  1. XY=23\dfrac{X}{Y} = \dfrac{2}{3}, so X=2kX = 2k, Y=3kY = 3k.
  2. 2k+303k+30=56⇒12k+180=15k+150⇒k=10\dfrac{2k+30}{3k+30} = \dfrac{5}{6} \Rightarrow 12k + 180 = 15k + 150 \Rightarrow k = 10.
  3. X=20 ΩX = 20\ \Omega, Y=30 ΩY = 30\ \Omega. So the answer is (a).
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Q487MHT CET 2026 · 11 APR · SHIFT 1Difficult

Wheatstone Bridge & Metre Bridge · Metre bridge

In a metre bridge experiment the balance point is obtained if the gaps are closed by 2 Ω2\ \Omega and 3 Ω3\ \Omega. A shunt of X ΩX\ \Omega is added to 3 Ω3\ \Omega resistor to shift the balancing point by 22.522.5 cm. The value of XX is

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Answer: (b)

  1. Initially 23=l100−l⇒l=40 cm\dfrac{2}{3} = \dfrac{l}{100-l} \Rightarrow l = 40\text{ cm}.
  2. Shunting 3 Ω3\ \Omega lowers the right resistance, so the null point moves right to 40+22.5=62.5 cm40 + 22.5 = 62.5\text{ cm}.
  3. 2R′=62.537.5=53⇒R′=1.2 Ω\dfrac{2}{R'} = \dfrac{62.5}{37.5} = \dfrac{5}{3} \Rightarrow R' = 1.2\ \Omega.
  4. 3x3+x=1.2⇒1.8x=3.6⇒x=2 Ω\dfrac{3x}{3+x} = 1.2 \Rightarrow 1.8x = 3.6 \Rightarrow x = 2\ \Omega.
  5. So the answer is (b) 2.
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Q488MHT CET 2025 · 23 APR · SHIFT 1Medium

Wheatstone Bridge & Metre Bridge · Wheatstone bridge balance

The equivalent resistance of the following circuit when no current flows in the resistance of 5 Ω5\ \Omega is nearly

Figure

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Answer: (b)

  1. 1015=2030\dfrac{10}{15} = \dfrac{20}{30}, so the bridge is balanced and 5 Ω5\ \Omega carries no current.
  2. Path ABC: 10+15=25 Ω10 + 15 = 25\ \Omega; path ADC: 20+30=50 Ω20 + 30 = 50\ \Omega
  3. Req=25×5075=16.7≈17 ΩR_{eq} = \dfrac{25 \times 50}{75} = 16.7 \approx 17\ \Omega
  4. So the answer is (b) 17 Ω17\ \Omega.
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Q489MHT CET 2025 · 22 APR · SHIFT 1Medium

Wheatstone Bridge & Metre Bridge · Wheatstone bridge balance

In the given electrical network, the value of resistance ' RR ' when the current in the galvanometer will be zero, is

Figure

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Answer: (b)

  1. Zero current in G means the 66 V cell is balanced by the p.d. across RR: VR=6V_R = 6 V.
  2. Then the 1010 V cell drives current only through 4 Ω4\ \Omega and RR: I=104+RI = \dfrac{10}{4+R}
  3. 10R4+R=6⇒10R=24+6R⇒R=6 Ω\dfrac{10R}{4+R} = 6 \Rightarrow 10R = 24 + 6R \Rightarrow R = 6\ \Omega
  4. So the answer is (b) 6 Ω6\ \Omega.
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Q490MHT CET 2025 · 21 APR · SHIFT 2Medium

Wheatstone Bridge & Metre Bridge · Wheatstone bridge balance

In the circuit shown in the figure, P≠RP \neq R. The reading of the galvanometer remains the same with switch ' SS ' open or closed. Then

Figure

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Answer: (c)

  1. Reading unchanged with S open or closed means no current flows through S, so VB=VDV_B = V_D.
  2. At B (no current into S): IP=IQI_P = I_Q.
  3. At D (no current from S): current coming in through R goes out through G, so IR=IGI_R = I_G.
  4. So the answer is (c) IR=IGI_R = I_G.
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Q491JEE Main 2025 · 28 JAN · SHIFT 2Medium

Wheatstone Bridge & Metre Bridge · Wheatstone bridge balance

The value of current I in the electrical circuit as given below, when potential at A is equal to the potential at B, will be _______ A.

Figure

Numerical answer type. Work it out, then open the answer.

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Answer: 2

  1. VA=VBV_A = V_B: balanced bridge, 1020=R40⇒R=20 Ω\dfrac{10}{20} = \dfrac{R}{40} \Rightarrow R = 20\ \Omega
  2. No current in 30 Ω30\ \Omega. Top path 10+20=30 Ω10+20 = 30\ \Omega, bottom path 20+40=60 Ω20+40 = 60\ \Omega
  3. Req=30×6090=20 ΩR_{eq} = \dfrac{30 \times 60}{90} = 20\ \Omega
  4. I=4020=2I = \dfrac{40}{20} = 2 A
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Q492MHT CET 2025 · 5 MAY · SHIFT 2Difficult

Wheatstone Bridge & Metre Bridge · Metre bridge

In metre-bridge experiment the balance point is obtained if the gaps are closed by 2 Ω2\ \Omega and 3 Ω3\ \Omega. A shunt of x Ωx\ \Omega is added to 3 Ω3\ \Omega resistor to shift the balance point by 22.5 cm. The value of x is

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Answer: (b)

  1. Initially 23=l100−l⇒l=40 cm\dfrac{2}{3} = \dfrac{l}{100-l} \Rightarrow l = 40\text{ cm}.
  2. Shunting 3 Ω3\ \Omega lowers the right resistance, so the null point moves right to 40+22.5=62.5 cm40 + 22.5 = 62.5\text{ cm}.
  3. 2R′=62.537.5=53⇒R′=1.2 Ω\dfrac{2}{R'} = \dfrac{62.5}{37.5} = \dfrac{5}{3} \Rightarrow R' = 1.2\ \Omega.
  4. 3x3+x=1.2⇒1.8x=3.6⇒x=2 Ω\dfrac{3x}{3+x} = 1.2 \Rightarrow 1.8x = 3.6 \Rightarrow x = 2\ \Omega.
  5. So the answer is (b) 2.
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Q493MHT CET 2024 · 4 MAY · SHIFT 2Easy

Wheatstone Bridge & Metre Bridge · Wheatstone bridge balance

In a Wheatstone's bridge, the resistances in four arms are as shown in the figure. The balancing condition of the bridge is

Figure

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Answer: (d)

  1. S1S_1 and S2S_2 are in parallel: S=S1S2S1+S2S = \frac{S_1 S_2}{S_1 + S_2}
  2. Balance: PQ=RS\frac{P}{Q} = \frac{R}{S}
  3. PQ=R(S1+S2)S1S2\frac{P}{Q} = \frac{R(S_1 + S_2)}{S_1 S_2}
  4. So the answer is (d).
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Q494NEET 2024Easy

Wheatstone Bridge & Metre Bridge · Wheatstone bridge balance

Choose the correct circuit which can achieve the bridge balance.

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Answer: (a)

  1. A bridge is balanced when no current flows through G, i.e. the ratio of arms is equal on both sides.
  2. A diode passes current only when forward biased; reverse biased it acts as an open arm.
  3. Check the diode direction in each circuit against the cell polarity and see which arm values then satisfy the ratio condition.
  4. So the answer is (a).
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Q495JEE Main 2024 · 27 JAN · SHIFT 2Easy

Wheatstone Bridge & Metre Bridge · Wheatstone bridge balance

Wheatstone bridge principle is used to measure the specific resistance (S1)(S_1) of given wire, having length LL, radius rr. If XX is the resistance of wire, then specific resistance is; S1=X(πr2L)S_1 = X\left(\frac{\pi r^2}{L}\right). If the length of the wire gets doubled then the value of specific resistance will be:

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Answer: (d)

  1. Specific resistance (resistivity) is a property of the material, not of the wire's size.
  2. Doubling L doubles X, so S1=Xπr2LS_1 = X\frac{\pi r^2}{L} stays the same.
  3. So the answer is (d) S1S_1.
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Q496MHT CET 2024 · 2 MAY · SHIFT 1Easy

Wheatstone Bridge & Metre Bridge · Metre bridge

The resistances in the left and right gap of a metre bridge are 40 Ω40\ \Omega and 60 Ω60\ \Omega respectively. When the bridge is balanced, the distance of the null point from the centre of the wire towards left is

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Answer: (b)

  1. 4060=l100−l⇒3l=200−2l\frac{40}{60} = \frac{l}{100 - l} \Rightarrow 3l = 200 - 2l
  2. l=40 cml = 40\text{ cm} from the left end
  3. Distance from centre =50−40=10 cm= 50 - 40 = 10\text{ cm} towards left
  4. So the answer is (b) 10 cm.
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Q497MHT CET 2024 · 11 MAY · SHIFT 1Medium

Wheatstone Bridge & Metre Bridge · Wheatstone bridge balance

The current drawn from the battery in the given network is (Internal resistance of the battery is negligible)

Figure

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Answer: (b)

6 Ω4 Ω6 Ω4 Ω10 Ω (I = 0)XYLR
  1. Redraw: the two 6 Ω6\ \Omega go from the left terminal to nodes X and Y; the two 4 Ω4\ \Omega go from X and Y to the right terminal; 10 Ω10\ \Omega joins X and Y.
  2. 64=64\dfrac{6}{4} = \dfrac{6}{4}, so the bridge is balanced and 10 Ω10\ \Omega carries no current.
  3. Req=(6+4)∥(6+4)=5 ΩR_{eq} = (6+4) \parallel (6+4) = 5\ \Omega
  4. I=125=2.4I = \dfrac{12}{5} = 2.4 A
  5. So the answer is (b) 2.42.4 A.
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Q498JEE Main 2024 · 6 APR · SHIFT 1Medium

Wheatstone Bridge & Metre Bridge · Wheatstone bridge balance

The value of unknown resistance (x)(x) for which the potential difference between BB and DD will be zero in the arrangement shown, is:

Figure

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Answer: (c)

  1. AB: 24∥(12+12)=12 Ω24 \parallel (12+12) = 12\ \Omega; BC: 1∥1=0.5 Ω1 \parallel 1 = 0.5\ \Omega
  2. AD: x+(12∥12)=x+6x + (12 \parallel 12) = x + 6; DC: 0.5 Ω0.5\ \Omega
  3. Balance: 120.5=x+60.5⇒x+6=12\dfrac{12}{0.5} = \dfrac{x+6}{0.5} \Rightarrow x + 6 = 12
  4. x=6 Ωx = 6\ \Omega
  5. So the answer is (c) 6 Ω6\ \Omega.
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Q499MHT CET 2024 · 9 MAY · SHIFT 1Medium

Wheatstone Bridge & Metre Bridge · Metre bridge

When the two known resistances ' RR ' and ' SS ' are connected in the left and right gaps of a meter bridge respectively, the null point is found at a distance ' l1l_1 ' from the zero end of a meter bridge wire. An unknown resistance ' X ' is now connected in parallel with ' S ' and null point is found at a distance ' l2l_2 ' from zero end of meter bridge wire. The unknown resistance ' X ' is

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Answer: (a)

  1. RS=l1100−l1\dfrac{R}{S} = \dfrac{l_1}{100 - l_1} and R(S+X)SX=l2100−l2\dfrac{R(S+X)}{SX} = \dfrac{l_2}{100 - l_2}
  2. Divide: S+XX=l2(100−l1)l1(100−l2)\dfrac{S+X}{X} = \dfrac{l_2(100 - l_1)}{l_1(100 - l_2)}
  3. SX=l2(100−l1)−l1(100−l2)l1(100−l2)=100(l2−l1)l1(100−l2)\dfrac{S}{X} = \dfrac{l_2(100-l_1) - l_1(100-l_2)}{l_1(100-l_2)} = \dfrac{100(l_2 - l_1)}{l_1(100 - l_2)}
  4. X=Sl1(100−l2)100(l2−l1)X = \dfrac{S l_1 (100 - l_2)}{100(l_2 - l_1)}
  5. So the answer is (a).
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Q500MHT CET 2024 · 3 MAY · SHIFT 2Medium

Wheatstone Bridge & Metre Bridge · Metre bridge

Resistances in the left gap and right gap of a meter bridge are 10 Ω10\ \Omega and 30 Ω30\ \Omega respectively. If the resistances in the two gaps are interchanged, the balance point will shift to right by

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Answer: (c)

  1. 1030=l100−l⇒l=25\dfrac{10}{30} = \dfrac{l}{100 - l} \Rightarrow l = 25 cm
  2. After interchange: l′=75l' = 75 cm
  3. Shift to right =75−25=50= 75 - 25 = 50 cm
  4. So the answer is (c) 5050 cm.
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Q501NEET 2024Medium

Wheatstone Bridge & Metre Bridge · Metre bridge

The given circuit shows a uniform straight wire ABAB of 40 cm40\text{ cm} length fixed at both ends. In order to get zero reading in the galvanometer GG, the free end of JJ is to be placed from BB at:

Figure

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Answer: (d)

  1. Balance: 812=AJJB\dfrac{8}{12} = \dfrac{AJ}{JB} with AJ+JB=40 cmAJ + JB = 40\text{ cm}.
  2. AJ=25×40=16 cmAJ = \dfrac{2}{5} \times 40 = 16\text{ cm}, so JB=24 cmJB = 24\text{ cm}.
  3. So the answer is (d) 24 cm.
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Q502JEE Main 2024 · 1 FEB · SHIFT 2Medium

Wheatstone Bridge & Metre Bridge · Metre bridge

In a metre-bridge when a resistance in the left gap is 2 Ω2\ \Omega and unknown resistance in the right gap, the balance length is found to be 40 cm40\text{ cm}. On shunting the unknown resistance with 2 Ω2\ \Omega, the balance length changes by:

Show answer and solution

Answer: (b)

  1. 2X=4060⇒X=3 Ω\dfrac{2}{X} = \dfrac{40}{60} \Rightarrow X = 3\ \Omega.
  2. Shunted: X′=3×23+2=1.2 ΩX' = \dfrac{3 \times 2}{3+2} = 1.2\ \Omega.
  3. 21.2=l100−l⇒l=62.5 cm\dfrac{2}{1.2} = \dfrac{l}{100-l} \Rightarrow l = 62.5\text{ cm}.
  4. Change =62.5−40=22.5 cm= 62.5 - 40 = 22.5\text{ cm}. So the answer is (b).
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Q503JEE Main 2024 · 27 JAN · SHIFT 1Medium

Wheatstone Bridge & Metre Bridge · Metre bridge

A wire of length 10 cm10\text{ cm} and radius 7×10−4 m\sqrt{7} \times 10^{-4}\text{ m} is connected across the right gap of a meter bridge. When a resistance of 4.5 Ω4.5\ \Omega is connected on the left gap by using a resistance box, the balance length is found to be at 60 cm60\text{ cm} from the left end. If the resistivity of the wire is R×10−7 Ω m\text{R} \times 10^{-7}\ \Omega\text{ m}, then value of R\text{R} is:

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Answer: (c)

  1. 4.5Rw=6040⇒Rw=3 Ω\dfrac{4.5}{R_w} = \dfrac{60}{40} \Rightarrow R_w = 3\ \Omega.
  2. ρ=Rwπr2L=3×227×7×10−80.1\rho = \dfrac{R_w \pi r^2}{L} = \dfrac{3 \times \frac{22}{7} \times 7 \times 10^{-8}}{0.1}.
  3. ρ=66×10−7 Ω m\rho = 66 \times 10^{-7}\ \Omega\text{ m}.
  4. So the answer is (c) 66.
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Q504MHT CET 2024 · 4 MAY · SHIFT 1Difficult

Wheatstone Bridge & Metre Bridge · Metre bridge

In a meter bridge experiment, the balance point is obtained if the gaps are closed by 2 Ω2\ \Omega and 3 Ω3\ \Omega. A shunt of X ΩX\ \Omega is added to 3 Ω3\ \Omega resistor to shift the null point by 22.5 cm. The value of ' xx ' is

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Answer: (b)

  1. Initially 23=l100−l⇒l=40 cm\dfrac{2}{3} = \dfrac{l}{100-l} \Rightarrow l = 40\text{ cm}.
  2. Shunting 3 Ω3\ \Omega lowers the right resistance, so the null point moves right to 40+22.5=62.5 cm40 + 22.5 = 62.5\text{ cm}.
  3. 2R′=62.537.5=53⇒R′=1.2 Ω\dfrac{2}{R'} = \dfrac{62.5}{37.5} = \dfrac{5}{3} \Rightarrow R' = 1.2\ \Omega.
  4. 3x3+x=1.2⇒1.8x=3.6⇒x=2 Ω\dfrac{3x}{3+x} = 1.2 \Rightarrow 1.8x = 3.6 \Rightarrow x = 2\ \Omega.
  5. So the answer is (b) 2 Ω2\ \Omega.
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Q505JEE Main 2024 · 31 JAN · SHIFT 2Difficult

Wheatstone Bridge & Metre Bridge · Metre bridge

The resistance per centimeter of a meter bridge wire is rr, with X ΩX\ \Omega resistance in left gap. Balancing length from left end is at 40 cm40\text{ cm} with 25 Ω25\ \Omega resistance in right gap. Now the wire is replaced by another wire of 2r2r resistance per centimeter. The new balancing length for same settings will be at

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Answer: (c)

  1. Balance: X25=λlλ(100−l)\dfrac{X}{25} = \dfrac{\lambda l}{\lambda(100-l)}; the resistance per cm cancels.
  2. Doubling the resistance per cm does not change the ratio.
  3. So the answer is (c) 40 cm.
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Q506NEET 2023Easy

Wheatstone Bridge & Metre Bridge · Wheatstone bridge balance

If the galvanometer GG does not show any deflection in the circuit shown, the value of RR is given by:

Figure

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Answer: (b)

  1. No current through G, so the p.d. across R equals the 2 V cell.
  2. The 10 V cell sends current through 400 Ω\Omega and R in series: 10R400+R=2\frac{10R}{400 + R} = 2
  3. 10R=800+2R⇒R=100 Ω10R = 800 + 2R \Rightarrow R = 100\ \Omega
  4. So the answer is (b) 100 Ω100\ \Omega.
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Q507JEE Main 2023 · 30 JAN · SHIFT 2Medium

Wheatstone Bridge & Metre Bridge · Wheatstone bridge balance

If the potential difference between B\mathrm{B} and D\mathrm{D} is zero, the value of xx is 1n Ω\frac{1}{n}\ \Omega. The value of nn is __________.

Figure

Numerical answer type. Work it out, then open the answer.

Show answer and solution

Answer: 2

  1. Left arms: 6∥3=2 Ω6 \parallel 3 = 2\ \Omega (to B), 1+2=3 Ω1 + 2 = 3\ \Omega (to D).
  2. Right arms: x∥1=xx+1x \parallel 1 = \dfrac{x}{x+1} (to B), xx (to D).
  3. Balance: 2(x+1)x=3x⇒2x+2=3\dfrac{2(x+1)}{x} = \dfrac{3}{x} \Rightarrow 2x + 2 = 3
  4. x=12 Ωx = \dfrac{1}{2}\ \Omega, so n=2n = 2
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Q508MHT CET 2023 · 14 MAY · SHIFT 1Medium

Wheatstone Bridge & Metre Bridge · Metre bridge

A wire of length 3 m3\text{ m} connected in the left gap of a meter-bridge balances 8 Ω8\ \Omega resistance in the right gap at a point, which divides the bridge wire in the ratio 3:23 : 2. The length of the wire corresponding to resistance of 1 Ω1\ \Omega is

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Answer: (d)

  1. Rw8=32⇒Rw=12 Ω\dfrac{R_w}{8} = \dfrac{3}{2} \Rightarrow R_w = 12\ \Omega for 33 m of wire
  2. Length per ohm =312=0.25= \dfrac{3}{12} = 0.25 m
  3. So the answer is (d) 0.25 m0.25\text{ m}.
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Q509MHT CET 2023 · 13 MAY · SHIFT 1Medium

Wheatstone Bridge & Metre Bridge · Metre bridge

In a given meter bridge, the current flowing through 40 Ω40\ \Omega resistor is

Figure

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Answer: (d)

  1. Check balance: 510=2040\dfrac{5}{10} = \dfrac{20}{40}, so the bridge is balanced and Ig=0I_g = 0.
  2. The current I2I_2 entering the 20 Ω20\ \Omega arm continues fully into 40 Ω40\ \Omega.
  3. So the answer is (d) I2I_2.
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Q510MHT CET 2023 · 12 MAY · SHIFT 2Medium

Wheatstone Bridge & Metre Bridge · Metre bridge

Two resistance X\mathrm{X} and Y\mathrm{Y} are connected in the two gaps of a meterbridge and the null points is obtained at 20 cm20\text{ cm} from zero end. When the resistance of 20 Ω20\ \Omega is connected in series with the smaller of the two resistance X\mathrm{X} and Y\mathrm{Y}, the null point shifts to 40 cm40\text{ cm} from left end. The value of smaller resistance in ohm is

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Answer: (c)

  1. XY=2080⇒Y=4X\dfrac{X}{Y} = \dfrac{20}{80} \Rightarrow Y = 4X, so XX is smaller.
  2. With 20 Ω20\ \Omega in series with XX: X+204X=4060\dfrac{X + 20}{4X} = \dfrac{40}{60}
  3. 3X+60=8X⇒X=12 Ω3X + 60 = 8X \Rightarrow X = 12\ \Omega
  4. So the answer is (c) 12.
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Q511MHT CET 2023 · 10 MAY · SHIFT 2Medium

Wheatstone Bridge & Metre Bridge · Metre bridge

In meter bridge experiment, null point was obtained at a distance 'll' from left end. The values of resistances in the left and right gap are doubled and then interchanged. The new position of the null point is

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Answer: (a)

  1. Initially R1R2=l100−l\dfrac{R_1}{R_2} = \dfrac{l}{100 - l}
  2. New: left 2R22R_2, right 2R12R_1: l′100−l′=R2R1=100−ll\dfrac{l'}{100 - l'} = \dfrac{R_2}{R_1} = \dfrac{100 - l}{l}
  3. l′=100−ll' = 100 - l
  4. So the answer is (a) (100−l)(100 - l).
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Q512MHT CET 2023 · 9 MAY · SHIFT 1Medium

Wheatstone Bridge & Metre Bridge · Metre bridge

In a meter bridge experiment null point is obtained at l cml\text{ cm} from the left end. If the meter bridge wire is replaced by a wire of same material but twice the area of across-section, then the null point is obtained at a distance

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Answer: (b)

  1. Balance uses the ratio l100−l\dfrac{l}{100 - l} of resistances of two parts of a uniform wire.
  2. Changing the cross-section changes both parts by the same factor, so the ratio and the null point stay the same.
  3. So the answer is (b) l cml\text{ cm} from left end.
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Q513JEE Main 2023 · 29 JAN · SHIFT 2Medium

Wheatstone Bridge & Metre Bridge · Metre bridge

When two resistance R1R_1 and R2R_2 connected in series and introduced into the left gap of a meter bridge and a resistance of 10 Ω\Omega is introduced into the right gap, a null point is found at 60 cm from left side. When R1R_1 and R2R_2 are connected in parallel and introduced into the left gap, a resistance of 3 Ω\Omega is introduced into the right gap to get null point at 40 cm from left end. The product of R1R2R_1 R_2 is ____________Ω2\Omega^{2}

Numerical answer type. Work it out, then open the answer.

Show answer and solution

Answer: 30

  1. Series: R1+R210=6040⇒R1+R2=15 Ω\dfrac{R_1+R_2}{10} = \dfrac{60}{40} \Rightarrow R_1 + R_2 = 15\ \Omega.
  2. Parallel: Rp3=4060⇒Rp=2 Ω\dfrac{R_p}{3} = \dfrac{40}{60} \Rightarrow R_p = 2\ \Omega.
  3. Rp=R1R2R1+R2⇒R1R2=2×15R_p = \dfrac{R_1R_2}{R_1+R_2} \Rightarrow R_1R_2 = 2 \times 15.
  4. R1R2=30 Ω2R_1R_2 = 30\ \Omega^2.
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Q514JEE Main 2023 · 29 JAN · SHIFT 1Difficult

Wheatstone Bridge & Metre Bridge · Metre bridge

In a metre bridge experiment the balance point is obtained if the gaps are closed by 2Ω\Omega and 3Ω\Omega. A shunt of X Ω\Omega is added to 3Ω\Omega resistor to shift the balancing point by 22.5 cm. The value of X is ___________.

Numerical answer type. Work it out, then open the answer.

Show answer and solution

Answer: 2

  1. Initially 23=l100−l⇒l=40 cm\dfrac{2}{3} = \dfrac{l}{100-l} \Rightarrow l = 40\text{ cm}.
  2. Shunting 3 Ω3\ \Omega lowers the right resistance, so the null point moves right to 40+22.5=62.5 cm40 + 22.5 = 62.5\text{ cm}.
  3. 2R′=62.537.5=53⇒R′=1.2 Ω\dfrac{2}{R'} = \dfrac{62.5}{37.5} = \dfrac{5}{3} \Rightarrow R' = 1.2\ \Omega.
  4. 3x3+x=1.2⇒1.8x=3.6⇒x=2 Ω\dfrac{3x}{3+x} = 1.2 \Rightarrow 1.8x = 3.6 \Rightarrow x = 2\ \Omega.
  5. So X=2X = 2.
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Q515MHT CET 2022 · 11 AUG · SHIFT 2Medium

Wheatstone Bridge & Metre Bridge · Wheatstone bridge balance

In the following network, the current through galvanometer will

Figure

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Answer: (d)

  1. Take VP=VV_P = V (positive terminal) and VR=0V_R = 0.
  2. Branch PQR: 44 and 4 Ω4\ \Omega, so VQ=48V=0.5VV_Q = \dfrac{4}{8}V = 0.5V
  3. Branch PSR: 11 and 3 Ω3\ \Omega, so VS=34V=0.75VV_S = \dfrac{3}{4}V = 0.75V
  4. VS>VQV_S > V_Q, so current flows through G from S to Q.
  5. So the answer is (d) from S to Q.
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Q516NEET 2022 · PHASE 1Medium

Wheatstone Bridge & Metre Bridge · Wheatstone bridge balance

A wheatstone bridge is used to determine the value of unknown resistance X by adjusting the variable resistance Y as shown in the figure. For the most precise measurement of X, the resistance P and Q

Figure

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Answer: (b)

  1. The bridge is most sensitive when all four arm resistances are of the same order.
  2. With P≈QP \approx Q, balance gives X≈YX \approx Y directly, and small P,QP, Q allow larger galvanometer current near balance, so a small imbalance is easily detected.
  3. So the answer is (b) approximately equal and small.
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Q517JEE Main 2022 · 28 JUN · SHIFT 1Medium

Wheatstone Bridge & Metre Bridge · Wheatstone bridge balance

Current measured by the ammeter (A) in the reported circuit when no current flows through 10 Ω\Omega resistance, will be ________________ A.

Figure

Numerical answer type. Work it out, then open the answer.

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Answer: 10

  1. No current in 10 Ω10\ \Omega: R4=36⇒R=2 Ω\dfrac{R}{4} = \dfrac{3}{6} \Rightarrow R = 2\ \Omega
  2. Top path 2+4=6 Ω2 + 4 = 6\ \Omega, bottom path 3+6=9 Ω3 + 6 = 9\ \Omega
  3. Req=6×915=3.6 ΩR_{eq} = \dfrac{6 \times 9}{15} = 3.6\ \Omega
  4. I=363.6=10I = \dfrac{36}{3.6} = 10 A
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Q518JEE Main 2022 · 27 JUL · SHIFT 2Medium

Wheatstone Bridge & Metre Bridge · Metre bridge

In the given figure of meter bridge experiment, the balancing length AC corresponding to null deflection of the galvanometer is 40 cm40\text{ cm}. The balancing length, if the radius of the wire ABAB is doubled, will be ______________ cm\text{cm}.

Figure

Numerical answer type. Work it out, then open the answer.

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Answer: 40

  1. Balance: R1R2=λlλ(100−l)\dfrac{R_1}{R_2} = \dfrac{\lambda l}{\lambda(100-l)}; the resistance per cm λ\lambda cancels.
  2. Changing the wire radius changes λ\lambda only, not the ratio.
  3. Balancing length stays 40 cm40\text{ cm}.
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Q519JEE Main 2022 · 27 JUL · SHIFT 1Medium

Wheatstone Bridge & Metre Bridge · Metre bridge

In a meter bridge experiment, for measuring unknown resistance 'S', the null point is obtained at a distance 30 cm30\text{ cm} from the left side as shown at point D. If R is 5.6 kΩ5.6\text{ k}\Omega, then the value of unknown resistance 'S' will be __________ Ω\Omega.

Figure

Numerical answer type. Work it out, then open the answer.

Show answer and solution

Answer: 2400

  1. SR=3070\dfrac{S}{R} = \dfrac{30}{70}.
  2. S=5600×37=2400 ΩS = 5600 \times \dfrac{3}{7} = 2400\ \Omega.
  3. S=2400 ΩS = 2400\ \Omega.
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Q520JEE Main 2022 · 26 JUL · SHIFT 1Medium

Wheatstone Bridge & Metre Bridge · Metre bridge

Resistances are connected in a meter bridge circuit as shown in the figure. The balancing length l1l_1 is 40 cm40\text{ cm}. Now an unknown resistance xx is connected in series with PP and new balancing length is found to be 80 cm80\text{ cm} measured from the same end. Then the value of xx will be ____________ Ω\Omega.

Figure

Numerical answer type. Work it out, then open the answer.

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Answer: 20

  1. PQ=4060\dfrac{P}{Q} = \dfrac{40}{60} with P=4 Ω⇒Q=6 ΩP = 4\ \Omega \Rightarrow Q = 6\ \Omega.
  2. 4+x6=8020=4⇒4+x=24\dfrac{4+x}{6} = \dfrac{80}{20} = 4 \Rightarrow 4 + x = 24.
  3. x=20 Ωx = 20\ \Omega.
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Q521JEE Main 2022 · 28 JUN · SHIFT 1Medium

Wheatstone Bridge & Metre Bridge · Metre bridge

A meter bridge setup is shown in the figure. It is used to determine an unknown resistance R using a given resistor of 15 Ω\Omega. The galvanometer (G) shows null deflection when tapping key is at 43 cm mark from end A. If the end correction for end A is 2 cm, then the determined value of R will be ____________ Ω\Omega.

Figure

Numerical answer type. Work it out, then open the answer.

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Answer: 19

  1. End correction at A adds to the left length: lA=43+2=45 cml_A = 43 + 2 = 45\text{ cm}; right length =100−43=57 cm= 100 - 43 = 57\text{ cm}.
  2. 15R=4557\dfrac{15}{R} = \dfrac{45}{57}.
  3. R=15×5745=19 ΩR = 15 \times \dfrac{57}{45} = 19\ \Omega.
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Q522MHT CET 2021 · 24 SEP · SHIFT 2Medium

Wheatstone Bridge & Metre Bridge · Wheatstone bridge balance

The current drawn from the battery in the given network is (Internal resistance of the battery is negligible)

Figure

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Answer: (a)

  1. Redraw: two 3 Ω3\ \Omega from the left terminal to nodes X and Y; two 2 Ω2\ \Omega from X and Y to the right terminal; 5 Ω5\ \Omega joins X and Y.
  2. 32=32\dfrac{3}{2} = \dfrac{3}{2}, so the bridge is balanced and 5 Ω5\ \Omega carries no current.
  3. Req=(3+2)∥(3+2)=2.5 ΩR_{eq} = (3+2) \parallel (3+2) = 2.5\ \Omega
  4. I=62.5=2.4I = \dfrac{6}{2.5} = 2.4 A
  5. So the answer is (a) 2.42.4 A.
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Q523MHT CET 2021 · 22 SEP · SHIFT 1Medium

Wheatstone Bridge & Metre Bridge · Wheatstone bridge balance

In a wheatstone's bridge, three resistances P,Q\mathrm{P, Q} and R\mathrm{R} are connected in the three arms and the fourth arm is formed by two resistances S1S_1 and S2S_2 connected in parallel. The condition for the bridge to be balanced is

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Answer: (c)

  1. Fourth arm: S=S1S2S1+S2S = \dfrac{S_1 S_2}{S_1 + S_2}
  2. Balance: PQ=RS=R(S1+S2)S1S2\dfrac{P}{Q} = \dfrac{R}{S} = \dfrac{R(S_1+S_2)}{S_1 S_2}
  3. So the answer is (c).
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Q524MHT CET 2021 · 23 SEP · SHIFT 1Medium

Wheatstone Bridge & Metre Bridge · Metre bridge

A balanced bridge is shown in the circuit diagram. The metre bridge wire has resistance 1 Ω cm−11\ \Omega\,\text{cm}^{-1}. The current drawn from the battery is (Internal resistance of battery is negligible)

Figure

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Answer: (b)

  1. Balance: 4Y=4060⇒Y=6 Ω\dfrac{4}{Y} = \dfrac{40}{60} \Rightarrow Y = 6\ \Omega; gap branch 4+6=10 Ω4 + 6 = 10\ \Omega
  2. Wire: 100×1=100 Ω100 \times 1 = 100\ \Omega
  3. Req=10×100110=9.09 ΩR_{eq} = \dfrac{10 \times 100}{110} = 9.09\ \Omega
  4. I=69.09=0.66I = \dfrac{6}{9.09} = 0.66 A
  5. So the answer is (b) 0.660.66 A.
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Q525MHT CET 2021 · 21 SEP · SHIFT 1Medium

Wheatstone Bridge & Metre Bridge · Metre bridge

In a meter bridge experiment, the balance point is obtained at length ℓ1 cm\ell_1\text{ cm} from left end when resistances in the left gap and right gap are 5 Ω5\ \Omega and R ΩR\ \Omega respectively. When the resistance RR is shunted with equal resistance, the new balance point is at 1.6ℓ11.6\ell_1. The resitance RR in ohm is

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Answer: (b)

  1. Metre bridge balance: 5R=l1100−l1\dfrac{5}{R} = \dfrac{l_1}{100-l_1}.
  2. Shunting RR with an equal resistance makes it R2\dfrac{R}{2}, so 10R=1.6l1100−1.6l1\dfrac{10}{R} = \dfrac{1.6l_1}{100-1.6l_1}.
  3. Dividing: 2=1.6(100−l1)100−1.6l1⇒200−3.2l1=160−1.6l1⇒l1=25 cm2 = \dfrac{1.6(100-l_1)}{100-1.6l_1} \Rightarrow 200-3.2l_1 = 160-1.6l_1 \Rightarrow l_1 = 25\text{ cm}.
  4. 5R=2575⇒R=15 Ω\dfrac{5}{R} = \dfrac{25}{75} \Rightarrow R = 15\ \Omega.
  5. So the answer is (b) 15.
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Q526MHT CET 2021 · 21 SEP · SHIFT 1Medium

Wheatstone Bridge & Metre Bridge · Metre bridge

Two wires 'A\mathrm{A}' and 'B\mathrm{B}' of equal length are connected in left and right gap respectively of meter bridge, null point is obtained at 40 cm40\text{ cm}, from the left end. Diameters of the wires 'AA' and 'BB' are in the ratio 3:13 : 1 respectively, the ratio of specific resistance of 'A' to that of 'B' is

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Answer: (a)

  1. Balance: RARB=4060=23\dfrac{R_A}{R_B} = \dfrac{40}{60} = \dfrac{2}{3}.
  2. R=ρLπd2/4R = \dfrac{\rho L}{\pi d^2/4} with equal LL, so ρAρB=RARB(dAdB)2\dfrac{\rho_A}{\rho_B} = \dfrac{R_A}{R_B}\left(\dfrac{d_A}{d_B}\right)^2.
  3. ρAρB=23×9=6\dfrac{\rho_A}{\rho_B} = \dfrac{2}{3} \times 9 = 6.
  4. So the answer is (a) 6:16 : 1.
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Q527JEE Main 2021 · 18 MAR · SHIFT 2Medium

Wheatstone Bridge & Metre Bridge · Metre bridge

Consider a 72 cm long wire AB as shown in the figure. The galvanometer jockey is placed at P on AB at a distance x cm from A. The galvanometer shows zero deflection.

Figure

The value of x, to the nearest integer, is ___________.

Numerical answer type. Work it out, then open the answer.

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Answer: 48

  1. Balance: 126=x72−x\dfrac{12}{6} = \dfrac{x}{72-x}.
  2. x=2(72−x)⇒3x=144x = 2(72 - x) \Rightarrow 3x = 144.
  3. x=48 cmx = 48\text{ cm}.
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Q528JEE Main 2021 · 17 MAR · SHIFT 2Difficult

Wheatstone Bridge & Metre Bridge · Wheatstone bridge balance

The four arms of a Wheatstone bridge have resistances as shown in the figure. A galvanometer of 15Ω\Omega resistance is connected across BD. Calculate the current through the galvanometer when a potential difference of 10V is maintained across AC.

Figure

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Answer: (b)

  1. Let II = cell current, I1I_1 in AB, IgI_g in BD. Loop ABDA: 100I1+15Ig−60(I−I1)=0100I_1 + 15I_g - 60(I - I_1) = 0.
  2. Loop BCDB: 10(I1−Ig)−15Ig−5(I−I1+Ig)=0⇒I=3I1−6Ig10(I_1 - I_g) - 15I_g - 5(I - I_1 + I_g) = 0 \Rightarrow I = 3I_1 - 6I_g.
  3. Substituting: −20I1+375Ig=0⇒I1=18.75Ig-20I_1 + 375I_g = 0 \Rightarrow I_1 = 18.75I_g, I=50.25IgI = 50.25I_g.
  4. Loop ADC with cell: 60(I−I1)+5(I−I1+Ig)=10⇒2052.5 Ig=1060(I - I_1) + 5(I - I_1 + I_g) = 10 \Rightarrow 2052.5\,I_g = 10.
  5. Ig=4.87×10−3 AI_g = 4.87 \times 10^{-3}\text{ A}. So the answer is (b) 4.87 mA.
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Q529JEE Main 2020 · 8 JAN · SHIFT 1Medium

Wheatstone Bridge & Metre Bridge · Wheatstone bridge balance

Four resistances of 15Ω\Omega, 12Ω\Omega, 4Ω\Omega and 10Ω\Omega are connected respectively in cyclic order to form Wheatstone's network. The resistance that is to be connected in parallel with the resistance of 10Ω\Omega to balance the network is _____Ω\Omega.

Numerical answer type. Work it out, then open the answer.

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Answer: 10

  1. Cyclic order: arms 15,12,4,1015, 12, 4, 10; balance needs products of opposite arms equal: 15×4=12×S′15 \times 4 = 12 \times S'
  2. S′=5 ΩS' = 5\ \Omega is needed in place of 10 Ω10\ \Omega.
  3. 10y10+y=5⇒y=10\dfrac{10y}{10+y} = 5 \Rightarrow y = 10
  4. Resistance in parallel =10 Ω= 10\ \Omega
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Q530MHT CET 2020 · 16 OCT · SHIFT 1Medium

Wheatstone Bridge & Metre Bridge · Metre bridge

Two wires AA and BB of equal lengths are connected in left and right gap of a meter bridge, null point is obtained at 40 cm40\text{ cm} from left end. Diameters of the wire AA and BB are in that ratio 3:13 : 1. The ratio of specific resistance of AA to the of BB is

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Answer: (c)

  1. Balance: RARB=4060=23\dfrac{R_A}{R_B} = \dfrac{40}{60} = \dfrac{2}{3}.
  2. R=ρLπd2/4R = \dfrac{\rho L}{\pi d^2/4} with equal LL, so ρAρB=RARB(dAdB)2\dfrac{\rho_A}{\rho_B} = \dfrac{R_A}{R_B}\left(\dfrac{d_A}{d_B}\right)^2.
  3. ρAρB=23×9=6\dfrac{\rho_A}{\rho_B} = \dfrac{2}{3} \times 9 = 6.
  4. So the answer is (c) 6:16 : 1.
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Q531NEET 2020 · PHASE 1Medium

Wheatstone Bridge & Metre Bridge · Metre bridge

A resistance wire connected in the left gap of a metre bridge balances a 10 Ω\Omega resistance in the right gap at a point which divides the bridge wire in the ratio 3:2. If the length of the resistance wire is 1.5 m, then the length of 1 Ω\Omega of the resistance wire is:

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Answer: (a)

  1. Balance: R10=32⇒R=15 Ω\dfrac{R}{10} = \dfrac{3}{2} \Rightarrow R = 15\ \Omega.
  2. 15 Ω15\ \Omega of wire is 1.5 m long, so 1 Ω1\ \Omega corresponds to 1.515=0.1 m\dfrac{1.5}{15} = 0.1\text{ m}.
  3. So the answer is (a) 1.0×10−1 m1.0 \times 10^{-1}\text{ m}.
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Q532JEE Main 2020 · 9 JAN · SHIFT 2Medium

Wheatstone Bridge & Metre Bridge · Metre bridge

In a meter bridge experiment S is a standard resistance. R is a resistance wire. It is found that balancing length is ll = 25 cm. If R is replaced by a wire of half length and half diameter that of R of same material, then the balancing distance l′l' (in cm) will now be________.

Figure

Numerical answer type. Work it out, then open the answer.

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Answer: 40

  1. RS=2575=13\dfrac{R}{S} = \dfrac{25}{75} = \dfrac{1}{3}.
  2. R∝Ld2R \propto \dfrac{L}{d^2}: half length and half diameter give R′=1/21/4R=2RR' = \dfrac{1/2}{1/4}R = 2R.
  3. R′S=23=l′100−l′⇒l′=40 cm\dfrac{R'}{S} = \dfrac{2}{3} = \dfrac{l'}{100-l'} \Rightarrow l' = 40\text{ cm}.
  4. l′=40 cml' = 40\text{ cm}.
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Q533JEE Main 2019 · 11 JAN · SHIFT 1Medium

Wheatstone Bridge & Metre Bridge · Wheatstone bridge balance

In a Wheatstone bridge(see fig.), Resistances P and Q are approximately equal. When R = 400 Ω\Omega, the bridge is balanced. On interchanging P and Q, the value of R, for balance, is 405 Ω\Omega. The value of X is close to:

Figure

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Answer: (a)

  1. First balance: PQ=400X\dfrac{P}{Q} = \dfrac{400}{X} (R and X opposite P, Q).
  2. After interchange: QP=405X\dfrac{Q}{P} = \dfrac{405}{X}
  3. Multiply: 1=400×405X2⇒X=400×4051 = \dfrac{400 \times 405}{X^2} \Rightarrow X = \sqrt{400 \times 405}
  4. X≈402.5 ΩX \approx 402.5\ \Omega
  5. So the answer is (a) 402.5402.5 ohm.
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Q534JEE Main 2019 · 10 JAN · SHIFT 2MediumOld syllabus

Wheatstone Bridge & Metre Bridge · Wheatstone bridge balance

The Wheatstone bridge shown in figure, here, gets balanced when the carbon resistor used as R1 has the colour code (Orange, Red, Brown). The resistors R2 and R4 are 80Ω\Omega and 40Ω\Omega, respectively. Assuming that the colour code for the carbon resistors gives their accurate values, the colour code for the carbon resistor, used as R3, would be -

Figure

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Answer: (a)

  1. Orange, Red, Brown =32×101=320 Ω=R1= 32 \times 10^1 = 320\ \Omega = R_1
  2. Balance: R1R2=R3R4⇒R3=320×4080=160 Ω\dfrac{R_1}{R_2} = \dfrac{R_3}{R_4} \Rightarrow R_3 = \dfrac{320 \times 40}{80} = 160\ \Omega
  3. 160=16×101160 = 16 \times 10^1: Brown (1), Blue (6), Brown (×10\times 10)
  4. So the answer is (a) Brown, Blue, Brown.
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Q535MHT CET 2019 · 2 MAY · SHIFT 2Medium

Wheatstone Bridge & Metre Bridge · Metre bridge

With a resistance of ' XX ' in the left gap and resistance of 9 Ω9\ \Omega in the right gap of a meter bridge, the balance point is obtained at 40 cm from the left end. In what way and to which resistance 3 Ω3\ \Omega resistance be connected to obtain the balance at 50 cm from the left end?

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Answer: (c)

  1. X9=4060⇒X=6 Ω\dfrac{X}{9} = \dfrac{40}{60} \Rightarrow X = 6\ \Omega.
  2. Balance at 50 cm needs equal resistances in both gaps.
  3. 6+3=9 Ω6 + 3 = 9\ \Omega, so 3 Ω3\ \Omega must be added in series with XX.
  4. So the answer is (c) in series with X ΩX\ \Omega.
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Q536NEET 2019Medium

Wheatstone Bridge & Metre Bridge · Metre bridge

The metre bridge shown is in balance position with PQ=l1l2\frac{P}{Q} = \frac{l_1}{l_2}. If we now interchange the positions of galvanometer and cell, will the bridge work? If yes, what will be balance condition?

Figure

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Answer: (a)

  1. A Wheatstone bridge stays balanced when the cell and galvanometer are interchanged (reciprocity).
  2. The null point stays at the same position, so the condition is unchanged.
  3. So the answer is (a) yes, PQ=l1l2\dfrac{P}{Q} = \dfrac{l_1}{l_2}.
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Q537JEE Main 2019 · 10 APR · SHIFT 1Medium

Wheatstone Bridge & Metre Bridge · Metre bridge

In a meter bridge experiment, the circuit diagram and the corresponding observation table are shown in figure

Figure

Figure

Which of the readings is inconsistent?

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Answer: (a)

  1. X=R 100−llX = R\,\dfrac{100-l}{l} for each reading.
  2. 1: 1000×4060=667 Ω1000 \times \frac{40}{60} = 667\ \Omega; 2: 100×8713=669 Ω100 \times \frac{87}{13} = 669\ \Omega; 3: 10×98.51.5=657 Ω10 \times \frac{98.5}{1.5} = 657\ \Omega.
  3. 4: 1×991=99 Ω1 \times \frac{99}{1} = 99\ \Omega, far from the others.
  4. So the answer is (a) reading 4.
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Q538JEE Main 2019 · 11 JAN · SHIFT 2Medium

Wheatstone Bridge & Metre Bridge · Metre bridge

In the experimental set up of metre bridge shown in the figure, the null point is obtained at a distance of 40 cm from A. If a 10 Ω\Omega resistor is connected in series with R1, the null point shifts by 10 cm. The resistance that should be connected in parallel with (R1 + 10) Ω\Omega such that the null point shifts back to its initial position is:

Figure

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Answer: (d)

  1. R1R2=4060⇒R1=23R2\dfrac{R_1}{R_2} = \dfrac{40}{60} \Rightarrow R_1 = \dfrac{2}{3}R_2.
  2. Adding 10 Ω10\ \Omega to R1R_1 moves the null point to 50 cm: R1+10=R2R_1 + 10 = R_2.
  3. 23R2+10=R2⇒R2=30 Ω, R1=20 Ω\dfrac{2}{3}R_2 + 10 = R_2 \Rightarrow R_2 = 30\ \Omega,\ R_1 = 20\ \Omega.
  4. Need 30∥R=2030 \parallel R = 20: 30R30+R=20⇒R=60 Ω\dfrac{30R}{30+R} = 20 \Rightarrow R = 60\ \Omega.
  5. So the answer is (d) 60 Ω60\ \Omega.
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Q539MHT CET 2019 · 3 MAY · SHIFT 1Difficult

Wheatstone Bridge & Metre Bridge · Wheatstone bridge balance

In the network shown cell E has internal resistance rr and the galvanometer shows zero deflection. If the cell is replaced by a new cell of emf 2E2E and internal resistance 3r3r keeping everything else identical, then

Figure

Show answer and solution

Answer: (b)

  1. Balance condition R1R2=R3R4\dfrac{R_1}{R_2} = \dfrac{R_3}{R_4} depends only on the four arm resistances.
  2. It does not depend on the emf or internal resistance of the cell.
  3. So the bridge stays balanced. The answer is (b) zero deflection.
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Q540JEE Main 2019 · 12 JAN · SHIFT 1Difficult

Wheatstone Bridge & Metre Bridge · Metre bridge

In a meter bridge, the wire of length 1 m has a non-uniform cross-section such that, the variation dRdℓ\frac{dR}{d\ell} of its resistance R with length ℓ\ell is dRdℓ∝1ℓ\frac{dR}{d\ell} \propto \frac{1}{\sqrt{\ell}}. Two equal resistances are connected as shown in the figure. The galvanometer has zero deflection when the jockey is at point P. What is the length AP?

Figure

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Answer: (b)

  1. R(ℓ)∝∫0ℓdℓℓ∝ℓR(\ell) \propto \int_0^{\ell} \dfrac{d\ell}{\sqrt{\ell}} \propto \sqrt{\ell}.
  2. Equal resistances in the gaps, so wire resistance AP = PB: ℓ=1−ℓ\sqrt{\ell} = \sqrt{1} - \sqrt{\ell}.
  3. ℓ=12⇒ℓ=0.25 m\sqrt{\ell} = \dfrac{1}{2} \Rightarrow \ell = 0.25\text{ m}.
  4. So the answer is (b) 0.25 m.
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Q541JEE Main 2018Medium

Wheatstone Bridge & Metre Bridge · Metre bridge

On interchanging the resistances, the balance point of a meter bridge shifts to the left by 10 cm. The resistance of their series combination is 1 kΩ\Omega. How much was the resistance on the left slot before interchanging the resistances?

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Answer: (d)

  1. Before: R1R2=l100−l\dfrac{R_1}{R_2} = \dfrac{l}{100-l}. After interchange, balance at l−10l-10: R2R1=l−10110−l\dfrac{R_2}{R_1} = \dfrac{l-10}{110-l}.
  2. So l100−l=110−ll−10⇒l2−10l=11000−210l+l2⇒l=55 cm\dfrac{l}{100-l} = \dfrac{110-l}{l-10} \Rightarrow l^2 - 10l = 11000 - 210l + l^2 \Rightarrow l = 55\text{ cm}.
  3. R1=1000×55100=550 ΩR_1 = 1000 \times \dfrac{55}{100} = 550\ \Omega.
  4. So the answer is (d) 550 Ω550\ \Omega.
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Q542JEE Main 2018 · 15 APR · SHIFT 1Medium

Wheatstone Bridge & Metre Bridge · Metre bridge

Figure

In a meter bridge, as shown in the figure, it is given that resistance Y=12.5 ΩY = 12.5\ \Omega and that the balance is obtained at a distance 39.5 cm39.5\text{ cm} from end AA (by Jockey J). After interchanging the resistances XX and YY, a new balance point is found at a distance l2l_2 from end A.A. What are the value of XX and l2l_2?

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Answer: (a)

  1. XY=39.560.5⇒X=12.5×39.560.5=8.16 Ω\dfrac{X}{Y} = \dfrac{39.5}{60.5} \Rightarrow X = 12.5 \times \dfrac{39.5}{60.5} = 8.16\ \Omega.
  2. After interchanging, the balance length from A becomes the old right length: l2=60.5 cml_2 = 60.5\text{ cm}.
  3. So the answer is (a) 8.16 Ω8.16\ \Omega and 60.5 cm60.5\text{ cm}.
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Q543JEE Main 2017Easy

Wheatstone Bridge & Metre Bridge · Wheatstone bridge balance

Which of the following statements is false?

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Answer: (c)

  1. (a) Kirchhoff's loop law comes from energy conservation. True.
  2. (b) The bridge is most sensitive when all four resistances are of the same order. True.
  3. (c) Interchanging cell and galvanometer keeps the bridge balanced (PQ=RS\frac{P}{Q} = \frac{R}{S} is unchanged). False.
  4. (d) A rheostat can be used as a potential divider. True.
  5. So the false statement is (c).
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Q544JEE Main 2017 · 9 APR · SHIFT 1Medium

Wheatstone Bridge & Metre Bridge · Metre bridge

In a meter bridge experiment resistances are connected as shown in the figure. Initially resistance P = 4 Ω\Omega and the neutral point N is at 60 cm from A. Now an unknown resistance R is connected in series to P and the new position of the neutral point is at 80 cm from A. The value of unknown resistance R is:

Figure

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Answer: (d)

  1. PQ=6040⇒Q=83 Ω\dfrac{P}{Q} = \dfrac{60}{40} \Rightarrow Q = \dfrac{8}{3}\ \Omega.
  2. 4+RQ=8020=4⇒4+R=323\dfrac{4+R}{Q} = \dfrac{80}{20} = 4 \Rightarrow 4 + R = \dfrac{32}{3}.
  3. R=203 ΩR = \dfrac{20}{3}\ \Omega. So the answer is (d).
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Q545NEET 2014 · AIPMTMedium

Wheatstone Bridge & Metre Bridge · Metre bridge

The resistances in the two arms of the meter bridge are 5 Ω\Omega and RΩ\Omega respectively. When the resistance R is shunted with an equal resistance, the new balance point is at 1.6ll1. The resistance R is

Figure

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Answer: (b)

  1. Metre bridge balance: 5R=l1100−l1\dfrac{5}{R} = \dfrac{l_1}{100-l_1}.
  2. Shunting RR with an equal resistance makes it R2\dfrac{R}{2}, so 10R=1.6l1100−1.6l1\dfrac{10}{R} = \dfrac{1.6l_1}{100-1.6l_1}.
  3. Dividing: 2=1.6(100−l1)100−1.6l1⇒200−3.2l1=160−1.6l1⇒l1=25 cm2 = \dfrac{1.6(100-l_1)}{100-1.6l_1} \Rightarrow 200-3.2l_1 = 160-1.6l_1 \Rightarrow l_1 = 25\text{ cm}.
  4. 5R=2575⇒R=15 Ω\dfrac{5}{R} = \dfrac{25}{75} \Rightarrow R = 15\ \Omega.
  5. So the answer is (b) 15 Ω15\ \Omega.
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Q546JEE Main 2013 · 9 APR · SHIFT 1Easy

Wheatstone Bridge & Metre Bridge · Metre bridge

In a metre bridge experiment null point is obtained at 40 cm from one end of the wire when resistance X is balanced against another resistance Y. If X < Y, then the new position of the null point from the same end, if one decides to balance a resistance of 3X against Y, will be close to:

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Answer: (c)

  1. First balance: XY=4060=23\frac{X}{Y} = \frac{40}{60} = \frac{2}{3}
  2. New: 3XY=2=l100−l\frac{3X}{Y} = 2 = \frac{l}{100 - l}
  3. l=200−2l⇒l=66.7 cm≈67 cml = 200 - 2l \Rightarrow l = 66.7\text{ cm} \approx 67\text{ cm}
  4. So the answer is (c) 67 cm.
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Q547NEET 2013Medium

Wheatstone Bridge & Metre Bridge · Wheatstone bridge balance

The resistances of the four arms P, Q, R and S in a Wheatstone's bridge are 10 ohm, 30 ohm, 30 ohm and 90 ohm, respectively. The e.m.f. and internal resistance of the cell are 7 volt and 5 ohm respectively. If the galvanometer resistance is 50 ohm, the current drawn from the cell will be

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Answer: (d)

  1. PQ=1030=13=RS=3090\dfrac{P}{Q} = \dfrac{10}{30} = \dfrac{1}{3} = \dfrac{R}{S} = \dfrac{30}{90}: balanced, so no current through the galvanometer.
  2. (P+Q)∥(R+S)=40∥120=30 Ω(P+Q) \parallel (R+S) = 40 \parallel 120 = 30\ \Omega
  3. I=730+5=0.2I = \dfrac{7}{30 + 5} = 0.2 A
  4. So the answer is (d) 0.20.2 A.
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Q548NEET 2012 · AIPMT · PRELIMSMedium

Wheatstone Bridge & Metre Bridge · Wheatstone bridge balance

In the circuit shown the cells A and B have negligible resistances. For VA = 12 V, R1 = 500 Ω\Omega and R = 100 Ω\Omega the galvanometer (G) shows no deflection. The value of VB is

Figure

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Answer: (b)

  1. No current in G, so VAV_A drives current only through R1R_1 and RR, and p.d. across RR equals VBV_B.
  2. VB=VARR1+R=12×100600V_B = V_A \dfrac{R}{R_1 + R} = 12 \times \dfrac{100}{600}
  3. VB=2V_B = 2 V
  4. So the answer is (b) 22 V.
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Q549JEE Main 2008 · AIEEEEasy

Wheatstone Bridge & Metre Bridge · Metre bridge

Shown in the figure below is a meter-bridge set up with null deflection in the galvanometer.

Figure

The value of the unknown resister RR is

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Answer: (b)

  1. Metre bridge balance: 55R=l100−l\dfrac{55}{R} = \dfrac{l}{100-l}
  2. l=20l = 20 cm, so 55R=2080=14\dfrac{55}{R} = \dfrac{20}{80} = \dfrac{1}{4}
  3. R=4×55=220 ΩR = 4 \times 55 = 220\ \Omega
  4. So the answer is (b) 220 Ω220\ \Omega.
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Q550NEET 2007 · AIPMTMedium

Wheatstone Bridge & Metre Bridge · Wheatstone bridge balance

Three resistances, P, Q, R each of 2Ω\Omega and an unknown resistance S from the four arms of a Wheatstone bridge circuit. When a resistance of 6 Ω\Omega is connected in parallel to S the bridge gets balanced. What is the value of S?

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Answer: (a)

  1. P=Q=R=2 ΩP = Q = R = 2\ \Omega, so balance needs the fourth arm =2 Ω= 2\ \Omega.
  2. 6SS+6=2⇒6S=2S+12\dfrac{6S}{S+6} = 2 \Rightarrow 6S = 2S + 12
  3. S=3 ΩS = 3\ \Omega
  4. So the answer is (a) 3 Ω3\ \Omega.
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Q551JEE Main 2006 · AIEEEMedium

Wheatstone Bridge & Metre Bridge · Wheatstone bridge balance

In a Wheatstone's bridge, three resistance P,QP, Q and RR connected in the three arms and the fourth arm is formed by two resistances S1S_1 and S2S_2 connected in parallel. The condition for the bridge to be balanced will be

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Answer: (b)

  1. Fourth arm: S=S1S2S1+S2S = \dfrac{S_1 S_2}{S_1 + S_2}
  2. Balance: PQ=RS=R(S1+S2)S1S2\dfrac{P}{Q} = \dfrac{R}{S} = \dfrac{R(S_1+S_2)}{S_1 S_2}
  3. So the answer is (b).
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Q552JEE Main 2004 · AIEEEEasy

Wheatstone Bridge & Metre Bridge · Metre bridge

In a meter bridge experiment null point is obtained at 20 cm20\text{ cm}, from one end of the wire when resistance XX is balanced against another resistance Y.Y. If X<YX < Y, then where will be the new position of the null point from the same end, if one decides to balance a resistance of 4X4X against YY

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Answer: (c)

  1. Initially XY=2080=14\dfrac{X}{Y} = \dfrac{20}{80} = \dfrac{1}{4}
  2. Now 4XY=1=l100−l\dfrac{4X}{Y} = 1 = \dfrac{l}{100-l}
  3. l=100−l⇒l=50l = 100 - l \Rightarrow l = 50 cm
  4. So the answer is (c) 50 cm50\text{ cm}.
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Q553NEET 2003 · AIPMTEasy

Wheatstone Bridge & Metre Bridge · Wheatstone bridge balance

In a Wheatstone's bridge all the four arms have equal resistance R. If the resistance of the galvanometer arm is also R, the equivalent resistance of the combination as seen by the battery is

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Answer: (c)

  1. All four arms equal, so the bridge is balanced and no current flows in the galvanometer arm.
  2. Remove that arm: two branches of 2R2R each in parallel.
  3. Req=2R⋅2R4R=RR_{eq} = \frac{2R \cdot 2R}{4R} = R
  4. So the answer is (c) R.
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Q554NEET 2001 · AIPMTEasy

Wheatstone Bridge & Metre Bridge · Wheatstone bridge balance

The resistance of each arm of the Wheatstone's bridge is 10 ohm. A resistance of 10 ohm is connected in series with a galvanometer then the equivalent resistance across the battery will be

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Answer: (a)

  1. All arms are 10 Ω10\ \Omega, so the bridge is balanced; the galvanometer branch (with its series 10 Ω10\ \Omega) carries no current.
  2. Remaining: two 20 Ω20\ \Omega branches in parallel.
  3. Req=10 ΩR_{eq} = 10\ \Omega
  4. So the answer is (a) 10 ohm.
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Q555NEET 1999 · AIPMTEasy

Wheatstone Bridge & Metre Bridge · Metre bridge

In metre-bridge the balancing length from left is found to be 20 cm20\text{ cm} when standard resistance of 1 Ω1\ \Omega is in right gap. The value of unknown resistance is

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Answer: (a)

  1. X1=l100−l=2080\frac{X}{1} = \frac{l}{100 - l} = \frac{20}{80}
  2. X=0.25 ΩX = 0.25\ \Omega
  3. So the answer is (a) 0.25 Ω0.25\ \Omega.
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Q556MHT CET 2026 · 19 MAY · SHIFT 1Easy

Potentiometer · Potential gradient

The potential gradient along the length of 1 metre wire of uniform cross section is 66 V/m. Let 'BB' and 'CC' be the two points at 4040 cm and 6060 cm marks respectively from one end. The potential difference between points B and C is

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Answer: (a)

  1. VBC=k×BCV_{BC} = k \times BC with BC=60−40=20 cm=0.2 mBC = 60 - 40 = 20\text{ cm} = 0.2\text{ m}.
  2. VBC=6×0.2=1.2 VV_{BC} = 6 \times 0.2 = 1.2\text{ V}.
  3. So the answer is (a) 1.2 V.
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Q557MHT CET 2026 · 13 MAY · SHIFT 2Easy

Potentiometer · Potential gradient

A current of 0.010.01 A passes through the potentiometer wire of resistivity 106 Ω10^{6}\ \Omega cm and area of cross- section 10−210^{-2} cm. The potential gradient is (in V/m)

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Answer: (d)

  1. Potential gradient k=IRL=IρAk = \dfrac{IR}{L} = \dfrac{I\rho}{A} (area taken as 10−2 cm210^{-2}\text{ cm}^2).
  2. ρA=106 Ω cm10−2 cm2=108 Ω/cm=1010 Ω/m\dfrac{\rho}{A} = \dfrac{10^{6}\ \Omega\text{ cm}}{10^{-2}\text{ cm}^2} = 10^{8}\ \Omega/\text{cm} = 10^{10}\ \Omega/\text{m}.
  3. k=0.01×1010=108 V/mk = 0.01 \times 10^{10} = 10^{8}\text{ V/m}.
  4. So the answer is (d) 10810^{8}.
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Q558MHT CET 2026 · 16 APR · SHIFT 1Easy

Potentiometer · Potential gradient

A potentiometer wire is 4m long and potential difference of 3V is maintained between the ends. The e.m.f. of the cell which balances against a length of 100 cm of the potentiometer wire is

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Answer: (a)

  1. Potential gradient k=VL=34=0.75 V/mk = \dfrac{V}{L} = \dfrac{3}{4} = 0.75\text{ V/m}.
  2. E=k l=34×1E = k\,l = \dfrac{3}{4} \times 1.
  3. E=0.75 VE = 0.75\text{ V}. So the answer is (a) 0.750.75 V.
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Q559MHT CET 2026 · 15 APR · SHIFT 2Easy

Potentiometer · Potential gradient

A potentiometer wire is 4m long and potential difference of 2V is maintained between the ends. The e.m.f. of the cell which balances against a length of 80 cm of the potentiometer wire is

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Answer: (c)

  1. Potential gradient k=VL=24=0.5 V/mk = \dfrac{V}{L} = \dfrac{2}{4} = 0.5\text{ V/m}.
  2. E=k l=24×0.8E = k\,l = \dfrac{2}{4} \times 0.8.
  3. E=0.40 VE = 0.40\text{ V}. So the answer is (c) 0.400.40 V.
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Q560MHT CET 2026 · 13 APR · SHIFT 1Easy

Potentiometer · Potential gradient

A potentiometer wire is 4 m long and potential difference of 3V is maintained between the ends. The e.m.f. of the cell which balances against a length of 100 cm of the poteniometer wire is

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Answer: (c)

  1. Potential gradient k=VL=34=0.75 V/mk = \dfrac{V}{L} = \dfrac{3}{4} = 0.75\text{ V/m}.
  2. E=k l=34×1E = k\,l = \dfrac{3}{4} \times 1.
  3. E=0.75 VE = 0.75\text{ V}. So the answer is (c) 0.750.75 V.
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Q561MHT CET 2026 · 19 MAY · SHIFT 2Easy

Potentiometer · Comparison of emfs

In a potentiometer circuit when two cells of emf 1.61.6 V and 1.31.3 V are connected to assist each other, the balancing length is 290290 cm. What will be the balancing length in cm when these two cells are connected in opposition?

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Answer: (d)

  1. E1+E2E1−E2=l1l2\frac{E_1 + E_2}{E_1 - E_2} = \frac{l_1}{l_2}.
  2. 2.90.3=290l2\frac{2.9}{0.3} = \frac{290}{l_2}
  3. l2=30 cml_2 = 30\ \text{cm}.
  4. So the answer is (d) 30.
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Q562MHT CET 2026 · 19 APR · SHIFT 1Easy

Potentiometer · Comparison of emfs

In a potentiometer circuit when two cells of e.m.f. 1.51.5 volt and 1.21.2 volt are connected to assist each other, balancing length is 270270 cm. What will be the balancing length in cm when these two cells are connected in opposition?

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Answer: (d)

  1. E1+E2E1−E2=l1l2\frac{E_1 + E_2}{E_1 - E_2} = \frac{l_1}{l_2}.
  2. 2.70.3=270l2\frac{2.7}{0.3} = \frac{270}{l_2}
  3. l2=30 cml_2 = 30\ \text{cm}.
  4. So the answer is (d) 30.
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Q563MHT CET 2026 · 13 APR · SHIFT 2Easy

Potentiometer · Comparison of emfs

In an experiment to find emf of a cell using potentiometer, the length of null point for a cell of emf 1.51.5 V is found to be 6060 cm. If this cell is replaced by another cell of emf EE, the length of null point increases by 4040 cm. The value of EE is x/10x/10 V.

The value of xx is

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Answer: (b)

  1. E∝lE \propto l, new length =60+40=100 cm= 60 + 40 = 100\ \text{cm}.
  2. E=1.5×10060=2.5 V=2510 VE = 1.5 \times \frac{100}{60} = 2.5\ \text{V} = \frac{25}{10}\ \text{V}.
  3. So the answer is (b) x=25x = 25.
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Q564MHT CET 2026 · 13 MAY · SHIFT 1Easy

Potentiometer · Sensitivity & working of potentiometer

In a potentiometer, the null point is obtained on 7th7^{th} wire. What should we do to change the null point on 9th9^{th} wire by keeping all other quantities same?

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Answer: (b)

  1. Balancing length l=Ekl = \frac{E}{k}; to move the null point further (7th to 9th wire) kk must decrease.
  2. k=I×(resistance per length)k = I \times (\text{resistance per length}), so the current in the main circuit must decrease.
  3. Increasing the series (rheostat) resistance in the main circuit lowers the current.
  4. So the answer is (b) increase resistance in main circuit.
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Q565MHT CET 2026 · 15 MAY · SHIFT 1Medium

Potentiometer · Potential gradient

Potentiometer wire of length 11m is connected in series with 490 Ω490\ \Omega resistance and 22 V battery. If 0.20.2 mV/cm is the potential gradient then the resistance of the potentiometer wire is

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Answer: (a)

  1. k=0.2 mV/cm=0.02 V/mk = 0.2\ \text{mV/cm} = 0.02\ \text{V/m}, so p.d. across 1 m wire =0.02 V= 0.02\ \text{V}.
  2. 2r490+r=0.02⇒1.98r=9.8\frac{2r}{490 + r} = 0.02 \Rightarrow 1.98r = 9.8
  3. r≈4.95 Ω≈4.9 Ωr \approx 4.95\ \Omega \approx 4.9\ \Omega.
  4. So the answer is (a) 4.9 Ω4.9\ \Omega.
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Q566JEE Main 2026 · 21 JAN · SHIFT 2Medium

Potentiometer · Potential gradient

The total length of potentiometer wire AB is 50 cm in the arrangement as shown in the figure. If P is the point where the galvanometer shows zero reading then the length AP is ________ cm.

Figure

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Answer: (b)

  1. Zero galvanometer reading means a balanced Wheatstone bridge: 64=APPB\frac{6}{4} = \frac{AP}{PB}.
  2. AP=50×610=30 cmAP = 50 \times \frac{6}{10} = 30\ \text{cm}.
  3. So the answer is (b) 30.
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Q567MHT CET 2026 · 12 MAY · SHIFT 1Medium

Potentiometer · Comparison of emfs

Two cells of e.m.f.s approximately 55 V and 1010 V are to be compared using a potentiometer (sum and difference method) of length 400400 cm, then,

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Answer: (c)

  1. In the sum method the wire must balance E1+E2≈15 VE_1 + E_2 \approx 15\ \text{V}.
  2. So the p.d. across the whole wire, and hence the driving battery emf, must exceed 15 V.
  3. So the answer is (c).
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Q568MHT CET 2026 · 20 APR · SHIFT 1Medium

Potentiometer · Comparison of emfs

Two cells of e.m.f. E1E_1 and E2E_2 (E1>E2E_1 > E_2) are connected as shown in figure.

Figure

When a potentiometer is connected between points A and B the balancing length of potentiometer wire is 412412 cm. When same potentiometer wire is connected between points A and C the balancing length is 103103 cm. The ratio E1:E2E_1 : E_2 is

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Answer: (c)

  1. From the figure, E2E_2 is reversed relative to E1E_1, so between A and C the cells oppose.
  2. A–B: E1∝412E_1 \propto 412; A–C: E1−E2∝103E_1 - E_2 \propto 103.
  3. E1E1−E2=4⇒3E1=4E2\frac{E_1}{E_1 - E_2} = 4 \Rightarrow 3E_1 = 4E_2.
  4. So the answer is (c) 4:34:3.
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Q569MHT CET 2026 · 18 APR · SHIFT 2Medium

Potentiometer · Comparison of emfs

When cell of E.M.F. 'E1E_1' is connected to potentiometer wire the balancing length is 'l1l_1 '. Another cell of E.M.F. 'E2E_2' (E1>E2E_1 > E_2) is connected along with E1E_1 so as two cells oppose each other, the balancing length is 'l2l_2'. The ratio E1:E2E_1 : E_2 is

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Answer: (b)

  1. E1=kl1E_1 = k l_1 and E1−E2=kl2E_1 - E_2 = k l_2.
  2. So E2=k(l1−l2)E_2 = k(l_1 - l_2).
  3. E1:E2=l1:(l1−l2)E_1 : E_2 = l_1 : (l_1 - l_2).
  4. So the answer is (b).
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Q570JEE Main 2026 · 23 JAN · SHIFT 2Medium

Potentiometer · Comparison of emfs

To compare EMF of two cells using potentiometer the balancing lengths obtained are 200 cm and 150 cm. The least count of scale is 1 cm. The percentage error in the ratio of EMFs is ______

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Answer: (c)

  1. E1E2=l1l2\dfrac{E_1}{E_2} = \dfrac{l_1}{l_2}, so the fractional errors add: Δrr=Δl1l1+Δl2l2\dfrac{\Delta r}{r} = \dfrac{\Delta l_1}{l_1}+\dfrac{\Delta l_2}{l_2}.
  2. With Δl=1\Delta l = 1 cm: 1200+1150=0.005+0.00667=0.01167\dfrac{1}{200}+\dfrac{1}{150} = 0.005+0.00667 = 0.01167
  3. Percentage error =1.17%= 1.17\%
  4. So the answer is (c) 1.17.
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Q571MHT CET 2026 · 12 MAY · SHIFT 2Medium

Potentiometer · Internal resistance by potentiometer

To determine the internal resistance of a cell by using a potentiometer, the null point is obtained at 11m when cell is shunted by 3 Ω3\ \Omega resistance and at 1.51.5m length when cell is shunted by 6 Ω6\ \Omega resistance. The internal resistance of the cell is

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Answer: (c)

  1. Shunted cell: l=l0RR+rl = l_0\dfrac{R}{R+r}.
  2. 1=l033+r1 = l_0\dfrac{3}{3+r} and 1.5=l066+r1.5 = l_0\dfrac{6}{6+r}
  3. Dividing: 1.5=2(3+r)6+r⇒9+1.5r=6+2r1.5 = \dfrac{2(3+r)}{6+r} \Rightarrow 9+1.5r = 6+2r
  4. r=6 Ωr = 6\ \Omega
  5. So the answer is (c) 6 Ω6\ \Omega.
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Q572MHT CET 2026 · 20 APR · SHIFT 2Medium

Potentiometer · Internal resistance by potentiometer

Figure shows a potentiometer wire AB having resistance of 5 Ω5\ \Omega and length 10 m. An e.m.f. is 0.40.4 V of battery, the balancing length AP is (internal resistance is negligible)

Figure

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Answer: (a)

  1. From the figure, driver: 5 V cell, 45 Ω45\ \Omega in series with wire of 5 Ω5\ \Omega.
  2. I=545+5=0.1I = \dfrac{5}{45+5} = 0.1 A, so VAB=0.1×5=0.5V_{AB} = 0.1\times 5 = 0.5 V.
  3. k=0.510=0.05k = \dfrac{0.5}{10} = 0.05 V/m
  4. AP=0.40.05=8AP = \dfrac{0.4}{0.05} = 8 m
  5. So the answer is (a) 8 m.
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Q573MHT CET 2026 · 18 APR · SHIFT 1Medium

Potentiometer · Internal resistance by potentiometer

A potentiometer wire of length 44 m and resistance 5 Ω5\ \Omega is connected in series with a resistance of 992 Ω992\ \Omega and a cell of e.m.f. 44 V with internal resistance 3 Ω3\ \Omega. The length of 0.750.75 m on potentiometer wire balances the e.m.f. of

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Answer: (c)

  1. I=45+992+3=4×10−3I = \dfrac{4}{5+992+3} = 4\times10^{-3} A
  2. Vwire=4×10−3×5=0.02V_{wire} = 4\times10^{-3}\times 5 = 0.02 V, so k=0.024=5×10−3k = \dfrac{0.02}{4} = 5\times10^{-3} V/m.
  3. E=kl=5×10−3×0.75=3.75×10−3E = k l = 5\times10^{-3}\times 0.75 = 3.75\times10^{-3} V
  4. So the answer is (c) 3.75 mV.
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Q574JEE Main 2026 · 28 JAN · SHIFT 1Medium

Potentiometer · Internal resistance by potentiometer

In the potentiometer, when the cell in the secondary circuit is shunted with 4 Ω4\ \Omega resistance, the balance is obtained at the length 120 cm of wire. Now when the same cell is shunted with 12 Ω12\ \Omega resistance, the balance is shifted to a length of 180 cm. The internal resistance of cell is ‾ Ω\underline{\qquad}\ \Omega

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Answer: (a)

  1. l=l0RR+rl = l_0\dfrac{R}{R+r}: 120=l044+r120 = l_0\dfrac{4}{4+r}, 180=l01212+r180 = l_0\dfrac{12}{12+r}.
  2. Dividing: 1.5=3(4+r)12+r⇒18+1.5r=12+3r1.5 = \dfrac{3(4+r)}{12+r} \Rightarrow 18+1.5r = 12+3r
  3. r=4 Ωr = 4\ \Omega
  4. So the answer is (a) 4.
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Q575MHT CET 2026 · 14 MAY · SHIFT 1Medium

Potentiometer · Sensitivity & working of potentiometer

A potentiometer wire has length LL. For a given cell of e.m.f. EE, the balancing length is L3\frac{L}{3} from the positive end of the wire. If the length of potentiometer wire is increased by 50%50\% then for the same cell the balance point is obtained at length (from positive end)

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Answer: (a)

  1. Same driver, so VV across the wire is fixed and k=VLk = \dfrac{V}{L}.
  2. E=VL⋅L3=V3E = \dfrac{V}{L}\cdot\dfrac{L}{3} = \dfrac{V}{3}
  3. New length 1.5L1.5L: V1.5L x=V3⇒x=L2\dfrac{V}{1.5L}\,x = \dfrac{V}{3} \Rightarrow x = \dfrac{L}{2}
  4. So the answer is (a) L2\dfrac{L}{2}.
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Q576MHT CET 2026 · 11 APR · SHIFT 2Medium

Potentiometer · Sensitivity & working of potentiometer

In potentiometer experiment, null point is obtained at a particular point for a cell on potentiometer wire 'x' cm long. If length of potentiometer wire is increased by few cm without changing the cell, the balancing length will (Driving source is not changed)

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Answer: (b)

  1. Same driver, so the same p.d. is spread over a longer wire: k=VLk = \dfrac{V}{L} decreases.
  2. Balancing length l=Ekl = \dfrac{E}{k} increases.
  3. So the answer is (b) increase.
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Q577MHT CET 2025 · 5 MAY · SHIFT 2Easy

Potentiometer · Potential gradient

A 2.5 V battery is connected to a potentiometer wire. A cell of e.m.f. 1.08 V is balanced by the voltage drop across 2.16 m of wire. The length of the potentiometer wire is

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Answer: (c)

  1. Potential gradient k=El=1.082.16=0.5 V/mk = \frac{E}{l} = \frac{1.08}{2.16} = 0.5\ \text{V/m}.
  2. The whole wire carries the battery p.d.: L=Vk=2.50.5=5 mL = \frac{V}{k} = \frac{2.5}{0.5} = 5\ \text{m}.
  3. So the answer is (c) 5 m.
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Q578MHT CET 2025 · 26 APR · SHIFT 1Easy

Potentiometer · Potential gradient

The potentiometer wire is 5 m long and potential difference of 4 V is maintained between the ends. The e.m.f. of the cell which balances against a length of 200 cm of the potentiometer wire is

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Answer: (d)

  1. k=VL=45=0.8 V/mk = \frac{V}{L} = \frac{4}{5} = 0.8\ \text{V/m}.
  2. E=kl=0.8×2=1.6 VE = k l = 0.8 \times 2 = 1.6\ \text{V}.
  3. So the answer is (d) 1.6 V.
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Q579MHT CET 2025 · 20 APR · SHIFT 2Easy

Potentiometer · Sensitivity & working of potentiometer

If the length of the potentiometer wire is increased by keeping constant potential difference across the wire, then

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Answer: (c)

  1. k=VLk = \frac{V}{L}; with VV fixed, a longer wire gives smaller kk.
  2. l=Ekl = \frac{E}{k}, so the balancing length increases.
  3. So the answer is (c) the null point is obtained at larger distance.
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Q580MHT CET 2025 · 23 APR · SHIFT 2Medium

Potentiometer · Comparison of emfs

Two cells of e.m.f.s E1E_1 and E2 (E1>E2)E_2\ (E_1 > E_2) are connected as shown in figure.

Figure

When the potentiometer is connected between A and B, the balancing length of the potentiometer wire is 3.60 m. On connecting the potentiometer between A and C, the balancing length is 0.90 m. The ratio E1/E2E_1/E_2 is

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Answer:

  1. A–C gives a smaller length, so the cells oppose there.
  2. A–B: E1∝3.60E_1 \propto 3.60; A–C: E1−E2∝0.90E_1 - E_2 \propto 0.90.
  3. E2∝3.60−0.90=2.70E_2 \propto 3.60 - 0.90 = 2.70, so E1E2=3.602.70=43\frac{E_1}{E_2} = \frac{3.60}{2.70} = \frac{4}{3}.
  4. So the answer is (b) 4:34:3.
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Q581MHT CET 2025 · 22 APR · SHIFT 2Medium

Potentiometer · Comparison of emfs

When cell of e.m.f. ' E1E_1 ' is connected to potentiometer wire, the balancing length is ' l1l_1 '. Another cell of e.m.f. ' E2E_2 ' (E1>E2)(E_1 > E_2) is connected so that two cells oppose each other, the balancing length is ' l2l_2 '. The ratio E1:E2E_1 : E_2 is

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Answer: (b)

  1. E1=kl1E_1 = k l_1 and E1−E2=kl2E_1 - E_2 = k l_2.
  2. So E2=k(l1−l2)E_2 = k(l_1 - l_2).
  3. E1E2=l1l1−l2\frac{E_1}{E_2} = \frac{l_1}{l_1 - l_2}.
  4. So the answer is (b).
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Q582MHT CET 2025 · 20 APR · SHIFT 1Medium

Potentiometer · Internal resistance by potentiometer

A null point is obtained at 200 cm on potentiometer wire when cell in secondary circuit is shunted by 5 Ω5\ \Omega. When a resistance of 15 Ω15\ \Omega is used for shunting, null point moves to 300 cm. The internal resistance of the cell is

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Answer: (c)

  1. l=l0RR+rl = l_0\dfrac{R}{R+r}: 200=l055+r200 = l_0\dfrac{5}{5+r}, 300=l01515+r300 = l_0\dfrac{15}{15+r}.
  2. Dividing: 1.5=3(5+r)15+r⇒22.5+1.5r=15+3r1.5 = \dfrac{3(5+r)}{15+r} \Rightarrow 22.5+1.5r = 15+3r
  3. r=5 Ωr = 5\ \Omega
  4. So the answer is (c) 5 Ω5\ \Omega.
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Q583MHT CET 2025 · 19 APR · SHIFT 2Medium

Potentiometer · Internal resistance by potentiometer

To determine the internal resistance of a cell with potentiometer, when the cell is shunted by a resistance of 5 Ω5\ \Omega the balancing length is 250 cm. When the cell is shunted by 20 Ω20\ \Omega, the balancing length of potentiometer wire is 400 cm. The internal resistance, of the cell is

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Answer: (c)

  1. l=l0RR+rl = l_0\dfrac{R}{R+r}: 250=l055+r250 = l_0\dfrac{5}{5+r}, 400=l02020+r400 = l_0\dfrac{20}{20+r}.
  2. Dividing: 1.6=4(5+r)20+r⇒32+1.6r=20+4r1.6 = \dfrac{4(5+r)}{20+r} \Rightarrow 32+1.6r = 20+4r
  3. 2.4r=12⇒r=5 Ω2.4r = 12 \Rightarrow r = 5\ \Omega
  4. So the answer is (c) 5 Ω5\ \Omega.
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Q584MHT CET 2025 · 23 APR · SHIFT 2Medium

Potentiometer · Sensitivity & working of potentiometer

The length of a potentiometer wire is ' LL '. A cell of e.m.f. ' EE ' is balanced at a length L4\frac{L}{4} from the positive end of the wire. If the length of the original wire is increased by L3\frac{\mathrm{L}}{3}, then using the same cell null point is obtained at

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Answer: (b)

  1. k=VLk = \dfrac{V}{L}, so E=VL⋅L4=V4E = \dfrac{V}{L}\cdot\dfrac{L}{4} = \dfrac{V}{4}.
  2. New length 4L3\dfrac{4L}{3}: 3V4L x=V4\dfrac{3V}{4L}\,x = \dfrac{V}{4}
  3. x=L3x = \dfrac{L}{3}
  4. So the answer is (b) L3\dfrac{L}{3}.
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Q585MHT CET 2025 · 21 APR · SHIFT 1Medium

Potentiometer · Sensitivity & working of potentiometer

The length of a potentiometer wire is ' LL '. A cell of e.m.f. EE is balanced at a length L5\frac{L}{5} from the positive end of the wire. If the length of the wire is increased by L2\frac{L}{2}, the same cell will give balance point at distance ' xx '. The value of ' xx ' is

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Answer: (c)

  1. k=VLk = \dfrac{V}{L}, so E=VL⋅L5=V5E = \dfrac{V}{L}\cdot\dfrac{L}{5} = \dfrac{V}{5}.
  2. New length 3L2\dfrac{3L}{2}: 2V3L x=V5\dfrac{2V}{3L}\,x = \dfrac{V}{5}
  3. x=3L10x = \dfrac{3L}{10}
  4. So the answer is (c) 3L10\dfrac{3L}{10}.
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Q586MHT CET 2024 · 11 MAY · SHIFT 2Easy

Potentiometer · Potential gradient

A battery of 6 V is connected to the ends of uniform wire 3 m long and of resistance 100 Ω100\ \Omega. The difference of potential between two points 50 cm apart on the wire is

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Answer: (a)

  1. k=VL=63=2 V/mk = \frac{V}{L} = \frac{6}{3} = 2\ \text{V/m} (resistance value is not needed).
  2. V=kl=2×0.5=1 VV = k l = 2 \times 0.5 = 1\ \text{V}.
  3. So the answer is (a) 1 V.
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Q587MHT CET 2024 · 4 MAY · SHIFT 1Medium

Potentiometer · Potential gradient

A potentiometer wire of length 1 m is connected in series with 495 Ω495\ \Omega resistance and 2 V battery. If 0.2 mV/cm0.2\text{ mV/cm} is the potential gradient, then the resistance of the potentiometer wire is

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Answer: (d)

  1. k=0.2 mV/cm=0.02 V/mk = 0.2\ \text{mV/cm} = 0.02\ \text{V/m}, so p.d. across 1 m wire =0.02 V= 0.02\ \text{V}.
  2. 2r495+r=0.02⇒1.98r=9.9\frac{2r}{495 + r} = 0.02 \Rightarrow 1.98r = 9.9
  3. r=5 Ωr = 5\ \Omega.
  4. So the answer is (d) 5 Ω5\ \Omega.
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Q588MHT CET 2024 · 10 MAY · SHIFT 1Medium

Potentiometer · Comparison of emfs

In potentiometer experiment, cells of e.m.f. E1E_1 and E2E_2 are connected in series (E1>E2)(E_1 > E_2) the balancing length is 80 cm of the wire. If the polarity of E2E_2 is reversed, the balancing length becomes 20 cm. The ratio E1/E2E_1/E_2 is

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Answer: (d)

  1. E1+E2E1−E2=8020=4\frac{E_1 + E_2}{E_1 - E_2} = \frac{80}{20} = 4.
  2. E1+E2=4E1−4E2⇒3E1=5E2E_1 + E_2 = 4E_1 - 4E_2 \Rightarrow 3E_1 = 5E_2.
  3. So the answer is (d) 5:35:3.
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Q589MHT CET 2024 · 2 MAY · SHIFT 2Medium

Potentiometer · Comparison of emfs

When cell of E.M.F. ' E1E_1 ' is connected to potentiometer wire, the balancing length is l1l_1. Another cell of E.M.F. ' E2E_2 ' (E1>E2)(E_1 > E_2) is connected so that two cells oppose each other, then the balancing length is l2l_2. The ratio E1:E2E_1 : E_2 is

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Answer: (b)

  1. E1=kl1E_1 = k l_1 and E1−E2=kl2E_1 - E_2 = k l_2.
  2. So E2=k(l1−l2)E_2 = k(l_1 - l_2).
  3. E1E2=l1l1−l2\frac{E_1}{E_2} = \frac{l_1}{l_1 - l_2}.
  4. So the answer is (b).
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Q590MHT CET 2024 · 15 MAY · SHIFT 1Medium

Potentiometer · Internal resistance by potentiometer

A cell balances against a length of 150 cm on a potentiometer wire when it is shunted by a resistance of 5 Ω5\ \Omega. But when it is shunted by a resistance of 10 Ω10\ \Omega, then balancing length increases by 25 cm. The balancing length when the cell is in an open circuit is

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Answer: (b)

  1. l=l0RR+rl = l_0\dfrac{R}{R+r}: 150=l055+r150 = l_0\dfrac{5}{5+r}, 175=l01010+r175 = l_0\dfrac{10}{10+r}.
  2. Dividing: 76=2(5+r)10+r⇒70+7r=60+12r⇒r=2 Ω\dfrac{7}{6} = \dfrac{2(5+r)}{10+r} \Rightarrow 70+7r = 60+12r \Rightarrow r = 2\ \Omega
  3. l0=150×5+25=210l_0 = 150\times\dfrac{5+2}{5} = 210 cm
  4. So the answer is (b) 210 cm.
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Q591JEE Main 2024 · 4 APR · SHIFT 1Medium

Potentiometer · Internal resistance by potentiometer

To measure the internal resistance of a battery, potentiometer is used. For R=10 ΩR = 10\ \Omega, the balance point is observed at l=500 cml = 500\text{ cm} and for R=1 Ω\mathrm{R} = 1\ \Omega the balance point is observed at l=400 cml = 400\text{ cm}. The internal resistance of the battery is approximately:

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Answer: (b)

  1. l=l0RR+rl = l_0\dfrac{R}{R+r}: 500=l01010+r500 = l_0\dfrac{10}{10+r}, 400=l011+r400 = l_0\dfrac{1}{1+r}.
  2. Dividing: 54=10(1+r)10+r⇒50+5r=40+40r\dfrac{5}{4} = \dfrac{10(1+r)}{10+r} \Rightarrow 50+5r = 40+40r
  3. r=1035≈0.29 Ωr = \dfrac{10}{35} \approx 0.29\ \Omega
  4. So the answer is (b) 0.3 Ω0.3\ \Omega.
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Q592MHT CET 2023 · 12 MAY · SHIFT 1Easy

Potentiometer · Potential gradient

Resistance of a potentiometer wire is 2 Ω/m2\ \Omega/\text{m}. A cell of e.m.f. 1.5 V1.5\text{ V} balances at 300 cm300\text{ cm}. The current through the wire is

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Answer: (c)

  1. k=El=1.53=0.5 V/mk = \frac{E}{l} = \frac{1.5}{3} = 0.5\ \text{V/m}.
  2. Also k=I×(resistance per metre)k = I \times (\text{resistance per metre}), so I=0.52=0.25 AI = \frac{0.5}{2} = 0.25\ \text{A}.
  3. So the answer is (c) 250 mA.
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Q593JEE Main 2023 · 1 FEB · SHIFT 1EasyOld syllabus

Potentiometer · Comparison of emfs

In an experiment to find emf of a cell using potentiometer, the length of null point for a cell of emf 1.5 V1.5\text{ V} is found to be 60 cm60\text{ cm}. If this cell is replaced by another cell of emf E, the length-of null point increases by 40 cm40\text{ cm}. The value of EE is x10 V\frac{x}{10}\text{ V}. The value of xx is ____________.

Numerical answer type. Work it out, then open the answer.

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Answer: 25

  1. E∝lE \propto l, new length =60+40=100 cm= 60 + 40 = 100\ \text{cm}.
  2. E=1.5×10060=2.5 V=2510 VE = 1.5 \times \frac{100}{60} = 2.5\ \text{V} = \frac{25}{10}\ \text{V}.
  3. So x=25x = 25.
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Q594JEE Main 2023 · 29 JAN · SHIFT 2EasyOld syllabus

Potentiometer · Sensitivity & working of potentiometer

With the help of potentiometer, we can determine the value of emf of a given cell. The sensitivity of the potentiometer is

(A) directly proportional to the length of the potentiometer wire

(B) directly proportional to the potential gradient of the wire

(C) inversely proportional to the potential gradient of the wire

(D) inversely proportional to the length of the potentiometer wire

Choose the correct option for the above statements:

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Answer: (c)

  1. Sensitivity is the ability to show a small p.d. as a large length: higher for smaller kk.
  2. So sensitivity ∝1k\propto \frac{1}{k} (C true, B false).
  3. k=VLk = \frac{V}{L}, so sensitivity ∝L\propto L (A true, D false).
  4. So the answer is (c) A and C only.
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Q595MHT CET 2023 · 14 MAY · SHIFT 2Medium

Potentiometer · Comparison of emfs

In potentiometer experiment, the balancing length is 8 m8\text{ m} when two cells E1E_1 and E2E_2 are joined in series. When two cells are connected in opposition the balancing length is 4 m4\text{ m}. The ratio of the e.m.f. of the two cells (E1E2)\left(\frac{E_1}{E_2}\right) is

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Answer: (d)

  1. E1+E2E1−E2=84=2\frac{E_1 + E_2}{E_1 - E_2} = \frac{8}{4} = 2.
  2. E1+E2=2E1−2E2⇒E1=3E2E_1 + E_2 = 2E_1 - 2E_2 \Rightarrow E_1 = 3E_2.
  3. So the answer is (d) 3:13:1.
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Q596MHT CET 2023 · 11 MAY · SHIFT 1Medium

Potentiometer · Comparison of emfs

In potentiometer experiments, two cells of e. m. f. 'E1E_1' and 'E2E_2' are connected in series (E1>E2)(E_1 > E_2), the balancing length is 64 cm64\text{ cm} of the wire. If the polarity of E2E_2 is reversed, the balancing length becomes 32 cm32\text{ cm}. The ratio E1/E2\mathrm{E}_1/\mathrm{E}_2 is

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Answer: (d)

  1. Balancing length ∝\propto emf: E1+E2E1−E2=l1l2\dfrac{E_1+E_2}{E_1-E_2} = \dfrac{l_1}{l_2}.
  2. E1+E2E1−E2=6432=2⇒E1+E2=2E1−2E2\dfrac{E_1+E_2}{E_1-E_2} = \dfrac{64}{32} = 2 \Rightarrow E_1+E_2 = 2E_1-2E_2
  3. E1=3E2E_1 = 3E_2
  4. So the answer is (d) 3:13:1.
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Q597MHT CET 2023 · 13 MAY · SHIFT 2Medium

Potentiometer · Internal resistance by potentiometer

A 10 m10\text{ m} long wire of resistance 20 Ω20\ \Omega is connected in series with a battery of emf 3 V3\text{ V} (negligible internal resistance) and a resistance of 10 Ω10\ \Omega. The potential gradient along the wire is

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Answer: (c)

  1. I=320+10=0.1I = \dfrac{3}{20+10} = 0.1 A
  2. Vwire=0.1×20=2V_{wire} = 0.1\times 20 = 2 V
  3. k=210=0.2k = \dfrac{2}{10} = 0.2 V/m
  4. So the answer is (c) 0.2 V/m.
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Q598MHT CET 2023 · 12 MAY · SHIFT 2Medium

Potentiometer · Internal resistance by potentiometer

A potentiometer wire of length 4 m4\text{ m} and resistance 5 Ω5\ \Omega is connected in series with a resistance of 992 Ω992\ \Omega and a cell of e.m.f. 4 V4\text{ V} with internal resistance 3 Ω3\ \Omega. The length of 0.75 m0.75\text{ m} on potentiometer wire balances the e.m.f. of

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Answer: (b)

  1. I=45+992+3=4I = \dfrac{4}{5+992+3} = 4 mA
  2. k=4×10−3×54=5×10−3k = \dfrac{4\times10^{-3}\times 5}{4} = 5\times10^{-3} V/m
  3. E=5×10−3×0.75=3.75E = 5\times10^{-3}\times 0.75 = 3.75 mV
  4. So the answer is (b) 3.75 mV.
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Q599NEET 2023Medium

Potentiometer · Internal resistance by potentiometer

The emf of a cell having internal resistance 1 Ω1\ \Omega is balanced against a length of 330 cm330\text{ cm} on a potentiometer wire. When an external resistance of 2 Ω2\ \Omega is connected across the cell, the balancing length will be:

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Answer: (a)

  1. l=l0RR+rl = l_0\dfrac{R}{R+r}
  2. l=330×22+1=220l = 330\times\dfrac{2}{2+1} = 220 cm
  3. So the answer is (a) 220 cm.
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Q600JEE Main 2023 · 29 JAN · SHIFT 2MediumOld syllabus

Potentiometer · Internal resistance by potentiometer

A null point is found at 200 cm in potentiometer when cell in secondary circuit is shunted by 5Ω\Omega. When a resistance of 15Ω\Omega is used for shunting, null point moves to 300 cm. The internal resistance of the cell is ___________Ω\Omega.

Numerical answer type. Work it out, then open the answer.

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Answer: 5

  1. l=l0RR+rl = l_0\dfrac{R}{R+r}: 200=l055+r200 = l_0\dfrac{5}{5+r}, 300=l01515+r300 = l_0\dfrac{15}{15+r}.
  2. Dividing: 1.5=3(5+r)15+r⇒22.5+1.5r=15+3r1.5 = \dfrac{3(5+r)}{15+r} \Rightarrow 22.5+1.5r = 15+3r
  3. r=5 Ωr = 5\ \Omega
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Q601MHT CET 2023 · 12 MAY · SHIFT 1Difficult

Potentiometer · Internal resistance by potentiometer

A potentiometer wire has length of 5 m5\text{ m} and resistance of 16 Ω16\ \Omega. The driving cell has an e.m.f. of 5 V5\text{ V} and an internal resistance of 4 Ω4\ \Omega. When the two cells of e.m.f.s 1.3 V1.3\text{ V} and 1.1 V1.1\text{ V} are connected so as to assist each other and then oppose each other, the balancing lengths are respectively

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Answer: (a)

  1. Wire current I=516+4=0.25 AI = \frac{5}{16+4} = 0.25\ \text{A}; p.d. across wire =0.25×16=4 V= 0.25 \times 16 = 4\ \text{V}.
  2. Potential gradient k=45=0.8 V/mk = \frac{4}{5} = 0.8\ \text{V/m}.
  3. Assisting: l1=1.3+1.10.8=3 ml_1 = \frac{1.3+1.1}{0.8} = 3\ \text{m}.
  4. Opposing: l2=1.3−1.10.8=0.25 ml_2 = \frac{1.3-1.1}{0.8} = 0.25\ \text{m}.
  5. So the answer is (a) 3 m, 0.25 m.
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Q602JEE Main 2022 · 25 JUL · SHIFT 2EasyOld syllabus

Potentiometer · Comparison of emfs

In a potentiometer arrangement, a cell of emf 1.20 V gives a balance point at 36 cm length of wire. This cell is now replaced by another cell of emf 1.80 V. The difference in balancing length of potentiometer wire in above conditions will be ___________ cm.

Numerical answer type. Work it out, then open the answer.

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Answer: 18

  1. l2=l1E2E1=36×1.801.20=54 cml_2 = l_1 \frac{E_2}{E_1} = 36 \times \frac{1.80}{1.20} = 54\ \text{cm}.
  2. Difference =54−36=18 cm= 54 - 36 = 18\ \text{cm}.
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Q603JEE Main 2022 · 24 JUN · SHIFT 1EasyOld syllabus

Potentiometer · Comparison of emfs

In a potentiometer arrangement, a cell gives a balancing point at 75 cm length of wire. This cell is now replaced by another cell of unknown emf. If the ratio of the emf's of two cells respectively is 3: 2, the difference in the balancing length of the potentiometer wire in above two cases will be ___________ cm.

Numerical answer type. Work it out, then open the answer.

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Answer: 25

  1. E1E2=l1l2=32\frac{E_1}{E_2} = \frac{l_1}{l_2} = \frac{3}{2}.
  2. l2=75×23=50 cml_2 = 75 \times \frac{2}{3} = 50\ \text{cm}.
  3. Difference =75−50=25 cm= 75 - 50 = 25\ \text{cm}.
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Q604JEE Main 2022 · 28 JUL · SHIFT 1MediumOld syllabus

Potentiometer · Potential gradient

As shown in the figure, a potentiometer wire of resistance 20 Ω20\ \Omega and length 300 cm300\text{ cm} is connected with resistance box (R.B.) and a standard cell of emf 4 V4\text{ V}. For a resistance 'RR' of resistance box introduced into the circuit, the null point for a cell of 20 mV20\text{ mV} is found to be 60 cm60\text{ cm}. The value of 'RR' is ___________ Ω.\Omega.

Figure

Numerical answer type. Work it out, then open the answer.

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Answer: 780

  1. k=20 mV60 cmk = \frac{20\ \text{mV}}{60\ \text{cm}}, so p.d. across 300 cm =20×5=100 mV=0.1 V= 20 \times 5 = 100\ \text{mV} = 0.1\ \text{V}.
  2. I=0.120=5×10−3 AI = \frac{0.1}{20} = 5 \times 10^{-3}\ \text{A}.
  3. Total resistance =40.005=800 Ω= \frac{4}{0.005} = 800\ \Omega, so R=800−20=780 ΩR = 800 - 20 = 780\ \Omega.
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Q605JEE Main 2022 · 26 JUL · SHIFT 2MediumOld syllabus

Potentiometer · Potential gradient

A potentiometer wire of length 300 cm300\text{ cm} is connected in series with a resistance 780 Ω\Omega and a standard cell of emf 4 V4\text{ V}. A constant current flows through potentiometer wire. The length of the null point for cell of emf 20 mV20\text{ mV} is found to be 60 cm60\text{ cm}. The resistance of the potentiometer wire is ____________ Ω\Omega.

Numerical answer type. Work it out, then open the answer.

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Answer: 20

  1. P.d. across 300 cm =20 mV×30060=0.1 V= 20\ \text{mV} \times \frac{300}{60} = 0.1\ \text{V}.
  2. 4r780+r=0.1⇒3.9r=78\frac{4r}{780 + r} = 0.1 \Rightarrow 3.9r = 78
  3. r=20 Ωr = 20\ \Omega.
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Q606JEE Main 2022 · 24 JUN · SHIFT 2MediumOld syllabus

Potentiometer · Potential gradient

A potentiometer wire of length 10 m and resistance 20 Ω\Omega is connected in series with a 25 V battery and an external resistance 30 Ω\Omega. A cell of emf E in secondary circuit is balanced by 250 cm long potentiometer wire. The value of E (in volt) is x10\frac{x}{10}. The value of x is __________.

Numerical answer type. Work it out, then open the answer.

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Answer: 25

  1. I=2520+30=0.5 AI = \frac{25}{20 + 30} = 0.5\ \text{A}, p.d. across wire =0.5×20=10 V= 0.5 \times 20 = 10\ \text{V}.
  2. k=1010=1 V/mk = \frac{10}{10} = 1\ \text{V/m}.
  3. E=1×2.5=2.5 V=2510 VE = 1 \times 2.5 = 2.5\ \text{V} = \frac{25}{10}\ \text{V}, so x=25x = 25.
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Q607JEE Main 2022 · 30 JUN · SHIFT 1MediumOld syllabus

Potentiometer · Internal resistance by potentiometer

The circuit diagram of potentiometer used to measure the internal resistance of a cell (E) is shown in figure. The key 'K' is kept closed so as to send constant current through potentiometer wire. When key 'K1' is kept open the null point is found to be at 120 cm on the potentiometer wire. When the key 'K1' is closed the null point is shifted at 80 cm at the potentiometer wire. The internal resistance of the given cell is _____________ Ω\Omega.

Figure

Numerical answer type. Work it out, then open the answer.

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Answer: 2

  1. K1K_1 open: open-circuit length l0=120l_0 = 120 cm. K1K_1 closed: cell shunted by R=4 ΩR = 4\ \Omega, l=80l = 80 cm.
  2. r=Rl0−ll=4×120−8080r = R\dfrac{l_0-l}{l} = 4\times\dfrac{120-80}{80}
  3. r=2 Ωr = 2\ \Omega
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Q608JEE Main 2022 · 27 JUN · SHIFT 1MediumOld syllabus

Potentiometer · Internal resistance by potentiometer

A cell, shunted by a 8 Ω\Omega resistance, is balanced across a potentiometer wire of length 3 m. The balancing length is 2 m when the cell is shunted by 4 Ω\Omega resistance. The value of internal resistance of the cell will be ____________ Ω\Omega.

Numerical answer type. Work it out, then open the answer.

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Answer: 8

  1. l=l0RR+rl = l_0\dfrac{R}{R+r}: 3=l088+r3 = l_0\dfrac{8}{8+r}, 2=l044+r2 = l_0\dfrac{4}{4+r}.
  2. Dividing: 32=2(4+r)8+r⇒24+3r=16+4r\dfrac{3}{2} = \dfrac{2(4+r)}{8+r} \Rightarrow 24+3r = 16+4r
  3. r=8 Ωr = 8\ \Omega
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Q609NEET 2021Easy

Potentiometer · Comparison of emfs

In a potentiometer circuit a cell of EMF 1.5 V gives balance point at 36 cm length of wire. If another cell of EMF 2.5 V replaces the first cell, then at what length of the wire, the balance point occurs?

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Answer: (b)

  1. E2E1=l2l1\frac{E_2}{E_1} = \frac{l_2}{l_1}.
  2. l2=36×2.51.5=60 cml_2 = 36 \times \frac{2.5}{1.5} = 60\ \text{cm}.
  3. So the answer is (b) 60 cm.
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Q610MHT CET 2021 · 24 SEP · SHIFT 1Medium

Potentiometer · Comparison of emfs

In a potentiometer experiment, when three cells A,B\mathrm{A}, \mathrm{B} and C\mathrm{C} are connected in series, the balancing length is found to be 420 cm420\text{ cm}. If cells A\mathrm{A} and B\mathrm{B} are connected in series the balancing length is 220 cm220\text{ cm} and for cells B\mathrm{B} and C\mathrm{C} connected in series the balancing length is 320 cm320\text{ cm}. The emf of cells A,B\mathrm{A}, \mathrm{B} and C\mathrm{C} are respectively in the ratio of

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Answer: (c)

  1. Emf ∝\propto balancing length, so take A+B+C=420A+B+C = 420, A+B=220A+B = 220, B+C=320B+C = 320 (in units of kk).
  2. C=420−220=200C = 420-220 = 200, A=420−320=100A = 420-320 = 100, B=220−100=120B = 220-100 = 120.
  3. A:B:C=100:120:200=1:1.2:2A:B:C = 100:120:200 = 1:1.2:2
  4. So the answer is (c) 1:1.2:21:1.2:2.
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Q611MHT CET 2021 · 23 SEP · SHIFT 1Medium

Potentiometer · Comparison of emfs

In potentiometer experiment, cells of e.m.f. 'E1E_1 ' and 'E2E_2' are connected in series (E1>E2)(E_1 > E_2) the balancing length is 64 cm64\text{ cm} of the wire. If the polarity of E2E_2 is reversed, the balancing length becomes 32 cm32\text{ cm}. The ratio E1E2\frac{E_1}{E_2} is

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Answer: (c)

  1. Series aiding: E1+E2=k(64)E_1+E_2 = k(64); series opposing: E1−E2=k(32)E_1-E_2 = k(32).
  2. E1+E2E1−E2=2⇒E1=3E2\dfrac{E_1+E_2}{E_1-E_2} = 2 \Rightarrow E_1 = 3E_2
  3. So the answer is (c) 3:13:1.
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Q612JEE Main 2021 · 25 FEB · SHIFT 1MediumOld syllabus

Potentiometer · Comparison of emfs

In the given circuit of potentiometer, the potential difference E across AB (10 m length) is larger than E1 and E2 as well. For key K1 (closed), the jockey is adjusted to touch the wire at point J1 so that there is no deflection in the galvanometer. Now the first battery (E1) is replaced by second battery (E2) for working by making K1 open and K2 closed. The galvanometer gives then null deflection at J2. The value of E1E2\frac{E_1}{E_2} is ab\frac{a}{b}, where a = _________.

Figure

Numerical answer type. Work it out, then open the answer.

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Answer: 1

  1. The wire is 10 rows of 1 m each, folded; odd rows run left to right from A, even rows run right to left.
  2. J1J_1 is on row 4 (runs right to left), 20 cm from its left end: AJ1=3 m+80 cm=380AJ_1 = 3\text{ m} + 80\text{ cm} = 380 cm.
  3. J2J_2 is on row 8 (runs right to left), 60 cm from its right end: AJ2=7 m+60 cm=760AJ_2 = 7\text{ m} + 60\text{ cm} = 760 cm.
  4. E1E2=AJ1AJ2=380760=12\dfrac{E_1}{E_2} = \dfrac{AJ_1}{AJ_2} = \dfrac{380}{760} = \dfrac{1}{2}
  5. So a=1a = 1.
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Q613JEE Main 2021 · 25 JUL · SHIFT 2MediumOld syllabus

Potentiometer · Comparison of emfs

In the given potentiometer circuit arrangement, the balancing length AC is measured to be 250 cm. When the galvanometer connection is shifted from point (1) to point (2) in the given diagram, the balancing length becomes 400 cm. The ratio of the emf of two cells, ε1ε2\frac{\varepsilon_1}{\varepsilon_2} is:

Figure

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Answer: (a)

  1. With G at point (1), only ε1\varepsilon_1 is in the loop: ε1=k(250)\varepsilon_1 = k(250).
  2. With G at point (2), both cells aid each other: ε1+ε2=k(400)\varepsilon_1+\varepsilon_2 = k(400).
  3. ε2=k(150)\varepsilon_2 = k(150)
  4. ε1ε2=250150=53\dfrac{\varepsilon_1}{\varepsilon_2} = \dfrac{250}{150} = \dfrac{5}{3}
  5. So the answer is (a) 53\dfrac{5}{3}.
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Q614MHT CET 2021 · 23 SEP · SHIFT 2Medium

Potentiometer · Internal resistance by potentiometer

To determine the internal resistance of a cell by using a potentiometer, the null point is at 1 m1\text{ m} when shunted by 3 Ω3\ \Omega resistance and at a length 1.5 m1.5\text{ m}, when cell is shunted by 6 Ω6\ \Omega resistance The internal resistance of the cell is

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Answer: (d)

  1. l=l0RR+rl = l_0\dfrac{R}{R+r}: 1=l033+r1 = l_0\dfrac{3}{3+r}, 1.5=l066+r1.5 = l_0\dfrac{6}{6+r}.
  2. Dividing: 1.5=2(3+r)6+r⇒9+1.5r=6+2r1.5 = \dfrac{2(3+r)}{6+r} \Rightarrow 9+1.5r = 6+2r
  3. r=6 Ωr = 6\ \Omega
  4. So the answer is (d) 6 Ω6\ \Omega.
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Q615MHT CET 2021 · 22 SEP · SHIFT 2Medium

Potentiometer · Internal resistance by potentiometer

In a potentiometer experiment, the balancing length for a cell is 240 cm240\text{ cm}. On shunting the cell with a resistance of 2 Ω2\ \Omega, the balancing length becomes half the initial balancing length. The internal resistance of the cell is

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Answer: (d)

  1. r=R(l0l−1)r = R\left(\dfrac{l_0}{l}-1\right)
  2. l=2402=120l = \dfrac{240}{2} = 120 cm, so r=2(240120−1)r = 2\left(\dfrac{240}{120}-1\right)
  3. r=2 Ωr = 2\ \Omega
  4. So the answer is (d) 2 Ω2\ \Omega.
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Q616JEE Main 2021 · 25 JUL · SHIFT 1MediumOld syllabus

Potentiometer · Internal resistance by potentiometer

In the given figure, there is a circuit of potentiometer of length AB = 10 m. The resistance per unit length is 0.1 Ω\Omega per cm. Across AB, a battery of emf E and internal resistance 'r' is connected. The maximum value of emf measured by this potentiometer is:

Figure

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Answer: (a)

  1. Wire resistance: 1000 cm×0.1 Ω/cm=100 Ω1000\text{ cm}\times0.1\ \Omega/\text{cm} = 100\ \Omega.
  2. From the figure, driver is 6 V with 20 Ω20\ \Omega: I=6100+20=0.05I = \dfrac{6}{100+20} = 0.05 A.
  3. Maximum measurable emf =VAB=0.05×100=5= V_{AB} = 0.05\times100 = 5 V
  4. So the answer is (a) 5 V.
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Q617MHT CET 2021 · 20 SEP · SHIFT 2Medium

Potentiometer · Sensitivity & working of potentiometer

In potentiometer experiment, null point is obtained at a particular point for a cell on potentiometer wire 'x\mathrm{x}' cm long. If length of potentiometer wire is increased by few centimeter without changing the cell, the balancing length will [Driving source is not changed]

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Answer: (b)

  1. Same driver, so a longer wire gives a smaller gradient k=VLk = \dfrac{V}{L}.
  2. Balancing length l=Ekl = \dfrac{E}{k} increases.
  3. So the answer is (b) increase.
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Q618MHT CET 2020 · 19 OCT · SHIFT 2Easy

Potentiometer · Potential gradient

A 10 m long wire of resistance 20 Ω20\ \Omega is connected in series with a battery of emf 3 V and a resistance of 10 Ω10\ \Omega. The potential gradient along the wire in V/m\text{V/m} is

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Answer: (d)

  1. I=320+10=0.1 AI = \frac{3}{20 + 10} = 0.1\ \text{A}.
  2. P.d. across wire =0.1×20=2 V= 0.1 \times 20 = 2\ \text{V}.
  3. k=210=0.2 V/mk = \frac{2}{10} = 0.2\ \text{V/m}.
  4. So the answer is (d) 0.20.
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Q619MHT CET 2020 · 16 OCT · SHIFT 1Easy

Potentiometer · Potential gradient

A potentiometer wire is 4 m4\text{ m} long and potential difference of 3 V3\text{ V} is maintained between the ends. The emf of the cell, which balances against a length of 100 cm100\text{ cm} of the potentiometer wire is

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Answer: (d)

  1. k=34=0.75 V/mk = \frac{3}{4} = 0.75\ \text{V/m}.
  2. E=kl=0.75×1=0.75 VE = k l = 0.75 \times 1 = 0.75\ \text{V}.
  3. So the answer is (d) 0.75 V.
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Q620MHT CET 2020 · 16 OCT · SHIFT 2Easy

Potentiometer · Sensitivity & working of potentiometer

Which of the following instruments is not a direct reading instrument?

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Answer: (b)

  1. A potentiometer uses a null (comparison) method: the emf is found from a balancing length and a calculation, not read off a scale.
  2. Balance, ammeter and voltmeter give direct readings.
  3. So the answer is (b) Potentiometer.
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Q621MHT CET 2020 · 19 OCT · SHIFT 2Medium

Potentiometer · Comparison of emfs

Two cells having unknown emfs E1E_1 and E2 (E1>E2)E_2\ (E_1 > E_2) are connected in potentiometer circuit, so as to assist each other. The null point obtained is at 490 cm from the higher potential end. When cell E2E_2 is connected, so as to oppose cell E1E_1, the null point is obtained at 90 cm from the same end. The ratio of the emfs of two cells (E1E2)\left(\frac{E_1}{E_2}\right) is

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Answer: (c)

  1. E1+E2E1−E2=49090=499\dfrac{E_1+E_2}{E_1-E_2} = \dfrac{490}{90} = \dfrac{49}{9}
  2. 9E1+9E2=49E1−49E2⇒40E1=58E29E_1+9E_2 = 49E_1-49E_2 \Rightarrow 40E_1 = 58E_2
  3. E1E2=5840=1.45\dfrac{E_1}{E_2} = \dfrac{58}{40} = 1.45
  4. So the answer is (c) 1.45.
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Q622JEE Main 2020 · 2 SEP · SHIFT 2MediumOld syllabus

Potentiometer · Comparison of emfs

A potentiometer wire PQ of 1 m length is connected to a standard cell E1. Another cell E2 of emf 1.02 V is connected with a resistance ‘r’ and switch S (as shown in figure). With switch S open, the null position is obtained at a distance of 49 cm from Q. The potential gradient in the potentiometer wire is:

Figure

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Answer: (c)

  1. At null point no current flows through E2E_2 and rr, so the full emf E2E_2 is balanced.
  2. Null point is 49 cm from Q, so PJ=100−49=51PJ = 100-49 = 51 cm from P (where E2E_2 is connected).
  3. k=E2PJ=1.0251=0.02k = \dfrac{E_2}{PJ} = \dfrac{1.02}{51} = 0.02 V/cm
  4. So the answer is (c) 0.02 V/cm.
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Q623JEE Main 2020 · 7 JAN · SHIFT 2MediumOld syllabus

Potentiometer · Internal resistance by potentiometer

The balancing length for a cell is 560 cm in a potentiometer experiment. When an external resistance of 10 Ω\Omega is connected in parallel to the cell, the balancing length changes by 60 cm. If the internal resistance of the cell is N10 Ω\frac{N}{10}\ \Omega, where N is an integer then value of N is .............

Numerical answer type. Work it out, then open the answer.

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Answer: 12

  1. Shunting reduces the length: l=560−60=500l = 560-60 = 500 cm.
  2. r=Rl0−ll=10×60500=1.2 Ωr = R\dfrac{l_0-l}{l} = 10\times\dfrac{60}{500} = 1.2\ \Omega
  3. N10=1.2⇒N=12\dfrac{N}{10} = 1.2 \Rightarrow N = 12
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Q624JEE Main 2020 · 8 JAN · SHIFT 1MediumOld syllabus

Potentiometer · Internal resistance by potentiometer

The length of a potentiometer wire is 1200 cm and it carries a current of 60 mA. For a cell of emf 5V and internal resistance of 20Ω\Omega, the null point on it is found to be a 1000 cm. The resistance of whole wire is:

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Answer: (d)

  1. At null point no current flows in the cell, so its emf 5 V is balanced by 1000 cm.
  2. k=51000=5×10−3k = \dfrac{5}{1000} = 5\times10^{-3} V/cm, so Vwire=5×10−3×1200=6V_{wire} = 5\times10^{-3}\times1200 = 6 V.
  3. R=VI=60.06=100 ΩR = \dfrac{V}{I} = \dfrac{6}{0.06} = 100\ \Omega
  4. So the answer is (d) 100 Ω100\ \Omega.
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Q625JEE Main 2019 · 12 JAN · SHIFT 1MediumOld syllabus

Potentiometer · Potential gradient

An ideal battery of 4 V and resistance R are connected in series in the primary circuit of a potentiometer of length 1 m and resistance 5 Ω\Omega. The value of R, to give a potential difference of 5 mV across 10 cm of potentiometer wire, is:

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Answer: (d)

  1. k=5 mV10 cm=0.05 V/mk = \frac{5\ \text{mV}}{10\ \text{cm}} = 0.05\ \text{V/m}, so p.d. across 1 m wire =0.05 V= 0.05\ \text{V}.
  2. I=0.055=0.01 AI = \frac{0.05}{5} = 0.01\ \text{A}.
  3. Total resistance =40.01=400 Ω= \frac{4}{0.01} = 400\ \Omega, so R=400−5=395 ΩR = 400 - 5 = 395\ \Omega.
  4. So the answer is (d) 395 Ω395\ \Omega.
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Q626MHT CET 2019 · 3 MAY · SHIFT 1Medium

Potentiometer · Sensitivity & working of potentiometer

A potentiometer wire has length LL For given cell of emf EE, the balancing length is L3\frac{L}{3} from the positive end of the wire. If the length of potentiometer wire is increased by 50%50\%, then for the same cell, the balance point is obtained at length

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Answer: (a)

  1. k=VLk = \dfrac{V}{L}, so E=VL⋅L3=V3E = \dfrac{V}{L}\cdot\dfrac{L}{3} = \dfrac{V}{3}.
  2. New length 1.5L1.5L: V1.5L x=V3⇒x=L2\dfrac{V}{1.5L}\,x = \dfrac{V}{3} \Rightarrow x = \dfrac{L}{2}
  3. So the answer is (a) L2\dfrac{L}{2} from positive end.
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Q627JEE Main 2019 · 11 JAN · SHIFT 1DifficultOld syllabus

Potentiometer · Comparison of emfs

The resistance of the meter bridge AB in given figure is 4 Ω\Omega. With a cell of emf ε\varepsilon = 0.5 V and rheostat resistance Rh = 2 Ω\Omega the null point is obtained at some point J. When the cell is replaced by another one of emf ε=ε\varepsilon = \varepsilon2 the same null point J is found for Rh = 6 Ω\Omega. The emf ε\varepsilon2 is:

Figure

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Answer: (a)

  1. Primary current I=6RAB+RhI = \frac{6}{R_{AB} + R_h}; at balance ε=I⋅RAJ\varepsilon = I \cdot R_{AJ} (same J both times).
  2. Case 1: I1=64+2=1 AI_1 = \frac{6}{4+2} = 1\ \text{A}, so RAJ=0.51=0.5 ΩR_{AJ} = \frac{0.5}{1} = 0.5\ \Omega.
  3. Case 2: I2=64+6=0.6 AI_2 = \frac{6}{4+6} = 0.6\ \text{A}.
  4. ε2=I2RAJ=0.6×0.5=0.3 V\varepsilon_2 = I_2 R_{AJ} = 0.6 \times 0.5 = 0.3\ \text{V}.
  5. So the answer is (a) 0.3 V.
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Q628JEE Main 2019 · 10 JAN · SHIFT 1DifficultOld syllabus

Potentiometer · Internal resistance by potentiometer

A potentiometer wire AB having length L and resistance 12 r is joined to a cell D of emf ε\varepsilon and internal resistance r. A cell C having emf ε\varepsilon/2 and internal resistance 3r is connected. The length AJ at which the galvanometer as shown in figure shows no deflection is –

Figure

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Answer: (b)

  1. Current in wire: I=ε12r+r=ε13rI = \frac{\varepsilon}{12r + r} = \frac{\varepsilon}{13r}.
  2. Potential gradient k=I⋅12rL=12ε13Lk = \frac{I \cdot 12r}{L} = \frac{12\varepsilon}{13L}.
  3. At null point no current flows through C, so its internal resistance does not matter: ε2=k⋅AJ\frac{\varepsilon}{2} = k \cdot AJ.
  4. AJ=ε/212ε/13L=1324LAJ = \frac{\varepsilon/2}{12\varepsilon/13L} = \frac{13}{24}L.
  5. So the answer is (b) 1324L\frac{13}{24}L.
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Q629JEE Main 2019 · 8 APR · SHIFT 2DifficultOld syllabus

Potentiometer · Sensitivity & working of potentiometer

In the circuit shown, a four-wire potentiometer is made of a 400 cm long wire, which extends between A and B. The resistance per unit length of the potentiometer wire is r = 0.01 Ω\Omega/cm. If an ideal voltmeter is connected as shown with jockey J at 50 cm from end A, the expected reading of the voltmeter will be :-

Figure

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Answer: (a)

  1. Cells in series: emf =3 V= 3\ \text{V}, internal resistance =1 Ω= 1\ \Omega.
  2. Wire resistance =400×0.01=4 Ω= 400 \times 0.01 = 4\ \Omega; total R=4+1+1=6 ΩR = 4 + 1 + 1 = 6\ \Omega.
  3. I=36=0.5 AI = \frac{3}{6} = 0.5\ \text{A}.
  4. Resistance of AJ =50×0.01=0.5 Ω= 50 \times 0.01 = 0.5\ \Omega, so V=0.5×0.5=0.25 VV = 0.5 \times 0.5 = 0.25\ \text{V}.
  5. So the answer is (a) 0.25 V.
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Q630JEE Main 2018MediumOld syllabus

Potentiometer · Internal resistance by potentiometer

In a potentiometer experiment, it is found that no current passes through the galvanometer when the terminals of the cell are connected across 52 cm of the potentiometer wire. If the cell is shunted by a resistance of 5 Ω\Omega, a balance is found when the cell is connected across 40 cm of the wire. Find the internal resistance of the cell.

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Answer: (c)

  1. r=Rl0−llr = R\dfrac{l_0-l}{l}
  2. r=5×52−4040=1.5 Ωr = 5\times\dfrac{52-40}{40} = 1.5\ \Omega
  3. So the answer is (c) 1.5 Ω1.5\ \Omega.
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Q631NEET 2017Easy

Potentiometer · Sensitivity & working of potentiometer

A potentiometer is an accurate and versatile device to make electrical measurements of EMF because the method involves

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Answer: (b)

  1. At balance no current flows through the galvanometer, so no current is drawn from the cell being measured.
  2. Hence there is no internal-resistance drop and the true emf is measured.
  3. So the answer is (b).
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Q632JEE Main 2017 · 8 APR · SHIFT 1MediumOld syllabus

Potentiometer · Sensitivity & working of potentiometer

A potentiometer PQ is set up to compare two resistances as shown in the figure. The ammeter A in the circuit reads 1.0 A when two way key K3 is open. The balance point is at a length ℓ\ell1 cm from P when two way key K3 is plugged in between 2 and 1, while the balance point is at a length ℓ\ell2 cm from P when key K3 is plugged in between 3 and 1. The ratio of two resistances R1R2\frac{R_1}{R_2}, is found to be:

Figure

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Answer: (d)

  1. Same current II flows through R1R_1 and R2R_2.
  2. Key between 2 and 1: G measures p.d. across R1R_1: IR1=kl1IR_1 = kl_1.
  3. Key between 3 and 1: G measures p.d. across R1+R2R_1+R_2: I(R1+R2)=kl2I(R_1+R_2) = kl_2, so IR2=k(l2−l1)IR_2 = k(l_2-l_1).
  4. R1R2=l1l2−l1\dfrac{R_1}{R_2} = \dfrac{l_1}{l_2-l_1}
  5. So the answer is (d) l1l2−l1\dfrac{l_1}{l_2-l_1}.
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Q633NEET 2016 · PHASE 1Medium

Potentiometer · Comparison of emfs

A potentiometer wire is 100 cm long and a constant potential difference is maintained across it. Two cells are connected in series first to support one another and then in opposite direction. The balance points are obtained at 50 cm and 10 cm from the positive end of the wire in the two cases. The ratio of emf's is

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Answer: (b)

  1. E1+E2E1−E2=5010=5\dfrac{E_1+E_2}{E_1-E_2} = \dfrac{50}{10} = 5
  2. E1+E2=5E1−5E2⇒6E2=4E1E_1+E_2 = 5E_1-5E_2 \Rightarrow 6E_2 = 4E_1
  3. E1E2=32\dfrac{E_1}{E_2} = \dfrac{3}{2}
  4. So the answer is (b) 3:23:2.
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Q634NEET 2015 · AIPMTMedium

Potentiometer · Potential gradient

A potentiometer wire of length L and a resistance r are connected in series with a battery of e.m.f. E0 and a resistance r1. An unknown e.m.f. E is balanced at a length ll of the potentiometer wire. The e.m.f. E will be given by

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Answer: (d)

  1. Current I=E0r+r1I = \frac{E_0}{r + r_1}, p.d. across wire =E0rr+r1= \frac{E_0 r}{r + r_1}.
  2. k=E0r(r+r1)Lk = \frac{E_0 r}{(r + r_1)L}
  3. E=kl=E0rr+r1⋅lLE = k l = \frac{E_0 r}{r + r_1} \cdot \frac{l}{L}.
  4. So the answer is (d).
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Q635NEET 2015 · AIPMT · CANCELLED PAPERMedium

Potentiometer · Potential gradient

A potentiometer wire has length 4 m and resistance 8 Ω\Omega. The resistance that must be connected in series with the wire and an accumulator of e.m.f. 2 V, so as to get a potential gradient 1 mV per cm on the wire is

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Answer: (c)

  1. k=1 mV/cm=0.1 V/mk = 1\ \text{mV/cm} = 0.1\ \text{V/m}, p.d. across wire =0.1×4=0.4 V= 0.1 \times 4 = 0.4\ \text{V}.
  2. I=0.48=0.05 AI = \frac{0.4}{8} = 0.05\ \text{A}.
  3. Total resistance =20.05=40 Ω= \frac{2}{0.05} = 40\ \Omega, so series R=40−8=32 ΩR = 40 - 8 = 32\ \Omega.
  4. So the answer is (c) 32 Ω32\ \Omega.
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Q636NEET 2014 · AIPMTMedium

Potentiometer · Internal resistance by potentiometer

A potentiometer circuit has been set up for finding the internal resistance of a given cell. The main battery, used across the potentiometer wire, has an emf of 2.0 V and a negligible internal resistance. The potentiometer wire itself is 4 m long. When the resistance R, connected across the given cell, has values of (i) infinity, (ii) 9.5 Ω\Omega

the balancing lengths on the potentiometer wire are found to be 3 m and 2.85 m, respectively. The value of internal resistance of the cell is

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Answer: (c)

  1. R=∞R = \infty gives the open-circuit length l0=3l_0 = 3 m.
  2. r=Rl0−ll=9.5×3−2.852.85r = R\dfrac{l_0-l}{l} = 9.5\times\dfrac{3-2.85}{2.85}
  3. r=9.5×0.152.85=0.5 Ωr = \dfrac{9.5\times0.15}{2.85} = 0.5\ \Omega
  4. So the answer is (c) 0.5 Ω0.5\ \Omega.
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Q637JEE Main 2014 · 12 APR · SHIFT 1MediumOld syllabus

Potentiometer · Sensitivity & working of potentiometer

In the experiment of calibration of voltmeter, a standard cell of e.m.f. 1.1 volt is balanced against 440 cm of potentiometer wire. The potential difference across the ends of resistance is found to balance against 220 cm of the wire. The corresponding reading of voltmeter is 0.5 volt. The error in the reading of voltmeter will be:

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Answer: (d)

  1. Potential gradient from the standard cell: k=1.1440=0.0025 V/cmk = \frac{1.1}{440} = 0.0025\ \text{V/cm}.
  2. True p.d. =k×220=0.0025×220=0.55 V= k \times 220 = 0.0025 \times 220 = 0.55\ \text{V}.
  3. Error == voltmeter reading −- true value =0.5−0.55=−0.05 V= 0.5 - 0.55 = -0.05\ \text{V}.
  4. So the answer is (d) −0.05-0.05 V.
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Q638NEET 2010 · AIPMT · PRELIMSMedium

Potentiometer · Potential gradient

A potentiometer circuit is set up as shown. The potential gradient, across the potentiometer wire, is k volt/cm and the ammeter, present in the circuit, reads 1.0 A when two way key is switched off. The balance points, when the key between the terminals (i) 1 and 2 (ii) 1 and 3, is plugged in, are found to be at length ll1 cm and ll2 cm respectively. The magnitudes, of the resistors R and X, in ohms, are then, equal, respectively, to

Figure

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Answer: (b)

  1. Current through R and X is I=1.0 AI = 1.0\ \text{A}.
  2. Key 1–2: galvanometer balances p.d. across R: IR=kl1⇒R=kl1IR = k l_1 \Rightarrow R = k l_1.
  3. Key 1–3: it balances p.d. across R + X: I(R+X)=kl2⇒X=k(l2−l1)I(R + X) = k l_2 \Rightarrow X = k(l_2 - l_1).
  4. So the answer is (b) kl1k l_1 and k(l2−l1)k(l_2 - l_1).
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Q639NEET 2008 · AIPMTMedium

Potentiometer · Internal resistance by potentiometer

A cell can be balanced against 110 cm and 100 cm of potentiometer wire, respectively with and without being short circuited through a resistance of 10 Ω\Omega. Its internal resistance is

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Answer: (c)

  1. Open-circuit length is the larger one: l0=110l_0 = 110 cm; shunted by 10 Ω10\ \Omega: l=100l = 100 cm.
  2. r=Rl0−ll=10×110−100100r = R\dfrac{l_0-l}{l} = 10\times\dfrac{110-100}{100}
  3. r=1 Ωr = 1\ \Omega
  4. So the answer is (c) 1.0 ohm.
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Q640JEE Main 2005 · AIEEEMediumOld syllabus

Potentiometer · Internal resistance by potentiometer

In a potentiometer experiment the balancing with a cell is at length 240 cm.240\text{ cm}. On shunting the cell with a resistance of 2 Ω,2\ \Omega, the balancing length becomes 120 cm120\text{ cm}. The internal resistance of the cell is

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Answer: (c)

  1. r=R(l0l−1)r = R\left(\dfrac{l_0}{l}-1\right)
  2. r=2(240120−1)=2 Ωr = 2\left(\dfrac{240}{120}-1\right) = 2\ \Omega
  3. So the answer is (c) 2 Ω2\ \Omega.
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Q641NEET 2004 · AIPMTEasy

Potentiometer · Potential gradient

A 6 volt battery is connected to the terminals of a three metre long wire of uniform thickness and resistance of 100 ohm. The difference of potential between two points on the wire separated by a distance of 50 cm will be

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Answer: (c)

  1. k=63=2 V/mk = \frac{6}{3} = 2\ \text{V/m}.
  2. V=2×0.5=1 VV = 2 \times 0.5 = 1\ \text{V}.
  3. So the answer is (c) 1 volt.
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Q642JEE Main 2003 · AIEEEMediumOld syllabus

Potentiometer · Internal resistance by potentiometer

The length of a wire of a potentiometer is 100 cm100\text{ cm}, and the e.m.f.\text{e.m.f.} of its standard cell is EE volt. It is employed to measure the e.m.f.\text{e.m.f.} of a battery whose internal resistance is 0.5 Ω.0.5\ \Omega. If the balance point is obtained at l=30 cml = 30\text{ cm} from the positive end, the e.m.f.\text{e.m.f.} of the battery is

where ii is the current in the potentiometer wire.

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Answer: (d)

  1. Taking EE as the emf across the whole 100 cm wire: k=E100k = \dfrac{E}{100} per cm.
  2. At balance no current flows through the battery, so its internal resistance does not matter.
  3. ε=k×30=30E100\varepsilon = k\times 30 = \dfrac{30E}{100}
  4. So the answer is 30E100\dfrac{30E}{100}, which is option (d) in the original paper.
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Q643NEET 2001 · AIPMTEasy

Potentiometer · Potential gradient

If specific resistance of a potentiometer wire is 10-7Ω\Omega m and current flow through it is 0.1 amp., cross-sectional area of wire is 10−-6 m2 then potential gradient will be

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Answer: (a)

  1. k=VL=IRL=IρAk = \frac{V}{L} = \frac{IR}{L} = \frac{I\rho}{A}.
  2. k=0.1×10−710−6=10−2 V/mk = \frac{0.1 \times 10^{-7}}{10^{-6}} = 10^{-2}\ \text{V/m}.
  3. So the answer is (a) 10−210^{-2} volt/m.
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Q644NEET 2000 · AIPMTEasy

Potentiometer · Sensitivity & working of potentiometer

The potentiometer is best for measuring voltage, as

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Answer: (a)

  1. At balance the sensitive galvanometer shows null deflection, so no current is drawn from the source.
  2. The potentiometer therefore measures the true p.d. (emf) accurately.
  3. So the answer is (a).
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Q645NEET 1999 · AIPMTEasy

Potentiometer · Potential gradient

A potentiometer consists of a wire of length 4 m4\text{ m} and resistance 10 Ω10\ \Omega. It is connected to a cell of emf 2 V2\text{ V}. The potential gradient of wire is

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Answer: (a)

  1. Ideal cell, so the full 2 V is across the wire.
  2. k=24=0.5 V/mk = \frac{2}{4} = 0.5\ \text{V/m}.
  3. So the answer is (a) 0.5 V/m.
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Q646NEET 2026Easy

Electrical Power & Heating Effect · Bulbs & heaters in combination

A room heater is rated 400 W,220 V400\text{ W}, 220\text{ V}. If the supply voltage drops to 200 V, what will be the power consumed (approximately)?

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Answer: (c)

  1. Resistance from rating: R=V2P=2202400=121 ΩR = \frac{V^2}{P} = \frac{220^2}{400} = 121\ \Omega.
  2. At 200 V: P′=2002121=40000121≈331 WP' = \frac{200^2}{121} = \frac{40000}{121} \approx 331\ \text{W}.
  3. So the answer is (c) 331 W.
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Q647JEE Main 2026 · 22 JAN · SHIFT 2Medium

Electrical Power & Heating Effect · Power: P = VI = I²R = V²/R

An electric power line having total resistance of 2 Ω2\ \Omega, delivers 1 kW of power at 250 V. The percentage efficiency of transmission line is ______.

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Answer: (a)

  1. Line current: I=P/V=1000/250=4 AI = P/V = 1000/250 = 4\text{ A}.
  2. Line loss: I2R=16×2=32 WI^2 R = 16 \times 2 = 32\text{ W}.
  3. η=10001000+32×100=96.9%\eta = \dfrac{1000}{1000 + 32} \times 100 = 96.9\%.
  4. So the answer is (a) 96.9.
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Q648JEE Main 2026 · 6 APR · SHIFT 1Medium

Electrical Power & Heating Effect · Heat & electrical energy

Refer to the circuit diagram given below. The heat generated across the 6 Ω6\ \Omega resistance in 100 second is α100 J\frac{\alpha}{100}\text{ J}. The value of α\alpha is ______. (Nearest integer)

Figure

Numerical answer type. Work it out, then open the answer.

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Answer: 3477

  1. Take the right node as 0 V. The left end of 3 Ω3\ \Omega is at 2 V and the 3 V3\text{ V} cell puts the end of 6 Ω6\ \Omega at 3 V.
  2. KCL at the junction XX: V−23+V4+V−36=0⇒9V=14, V=149 V\dfrac{V - 2}{3} + \dfrac{V}{4} + \dfrac{V - 3}{6} = 0 \Rightarrow 9V = 14,\ V = \dfrac{14}{9}\text{ V}.
  3. Current in 6 Ω6\ \Omega: 3−14/96=1354 A\dfrac{3 - 14/9}{6} = \dfrac{13}{54}\text{ A}.
  4. H=I2Rt=(1354)2×6×100≈34.77 J=3477100 JH = I^2 R t = \left(\dfrac{13}{54}\right)^2 \times 6 \times 100 \approx 34.77\text{ J} = \dfrac{3477}{100}\text{ J}.
  5. So α=3477\alpha = 3477.
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Q649JEE Main 2026 · 21 JAN · SHIFT 1Medium

Electrical Power & Heating Effect · Heat & electrical energy

The heat generated in 1 minute between points AA and BB in the given circuit, when a battery of 9 V with internal resistance of 1 Ω1\ \Omega is connected across these points is ______ J.

Figure

Numerical answer type. Work it out, then open the answer.

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Answer: 1080

  1. 12=24\dfrac{1}{2} = \dfrac{2}{4}, so the bridge is balanced and the middle 1 Ω1\ \Omega carries no current.
  2. RAB=(1+2)∥(2+4)=3∥6=2 ΩR_{AB} = (1 + 2) \parallel (2 + 4) = 3 \parallel 6 = 2\ \Omega.
  3. I=92+1=3 AI = \dfrac{9}{2 + 1} = 3\text{ A}.
  4. H=I2RABt=9×2×60=1080 JH = I^2 R_{AB} t = 9 \times 2 \times 60 = 1080\text{ J}.
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Q650JEE Main 2026 · 2 APR · SHIFT 2Medium

Electrical Power & Heating Effect · Bulbs & heaters in combination

Two resistors of 200 Ω\Omega and 400 Ω\Omega are connected in series with a battery of 100 V. A bulb rated at 200 V, 100 W is connected across the 400 Ω\Omega resistance. The potential drop across the bulb is ________ V.

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Answer: (b)

  1. Bulb: Rb=2002100=400 ΩR_b = \dfrac{200^2}{100} = 400\ \Omega.
  2. 400∥400=200 Ω400 \parallel 400 = 200\ \Omega; total 200+200=400 Ω200 + 200 = 400\ \Omega.
  3. I=100/400=0.25 AI = 100/400 = 0.25\text{ A}, so Vbulb=0.25×200=50 VV_{bulb} = 0.25 \times 200 = 50\text{ V}.
  4. So the answer is (b) 50.
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Q651JEE Main 2025 · 3 APR · SHIFT 2Easy

Electrical Power & Heating Effect · Power: P = VI = I²R = V²/R

A motor operating on 100 V draws a current of 1 A. If the efficiency of the motor is 91.6%91.6\%, then the loss of power in units of cal/s\text{cal/s} is

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Answer: (b)

  1. Input power =VI=100×1=100 W= VI = 100 \times 1 = 100\ \text{W}.
  2. Loss =(100−91.6)%×100=8.4 W= (100 - 91.6)\% \times 100 = 8.4\ \text{W}.
  3. In cal/s: 8.44.2=2 cal/s\frac{8.4}{4.2} = 2\ \text{cal/s}.
  4. So the answer is (b) 2.
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Q652JEE Main 2025 · 2 APR · SHIFT 1Medium

Electrical Power & Heating Effect · Heat & electrical energy

The battery of a mobile phone is rated as 4.2 V, 5800 mAh4.2\text{ V},\ 5800\text{ mAh}. How much energy is stored in it when fully charged?

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Answer: (a)

  1. Charge q=5800 mAh=5.8×3600=20880 Cq = 5800\text{ mAh} = 5.8 \times 3600 = 20880\text{ C}.
  2. E=Vq=4.2×20880≈87696 JE = Vq = 4.2 \times 20880 \approx 87696\text{ J}.
  3. So the answer is (a) 87.7 kJ.
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Q653JEE Main 2025 · 4 APR · SHIFT 2Medium

Electrical Power & Heating Effect · Bulbs & heaters in combination

There are ' nn ' number of identical electric bulbs, each is designed to draw a power pp independently from the mains supply. They are now joined in series across the mains supply. The total power drawn by the combination is:

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Answer: (d)

  1. Each bulb: R=V2/pR = V^2/p.
  2. In series: Rtotal=nRR_{total} = nR, so P=V2nR=pnP = \dfrac{V^2}{nR} = \dfrac{p}{n}.
  3. So the answer is (d) pn\dfrac{p}{n}.
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Q654MHT CET 2024 · 2 MAY · SHIFT 2Easy

Electrical Power & Heating Effect · Bulbs & heaters in combination

The current flowing in a coil is 3 A and the power consumed is 108 W. If the a.c. source is of 120 V,50 Hz120\text{ V}, 50\text{ Hz}, the resistance in the circuit is

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Answer: (c)

  1. Average power is consumed only in the resistance: P=I2RP = I^2 R.
  2. R=PI2=1089=12 ΩR = \frac{P}{I^2} = \frac{108}{9} = 12\ \Omega.
  3. So the answer is (c) 12 Ω12\ \Omega.
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Q655JEE Main 2024 · 4 APR · SHIFT 2Easy

Electrical Power & Heating Effect · Bulbs & heaters in combination

An electric bulb rated 50 W−200 V50\text{ W} - 200\text{ V} is connected across a 100 V100\text{ V} supply. The power dissipation of the bulb is:

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Answer: (c)

  1. Fixed resistance: P∝V2P \propto V^2.
  2. P′=50×(100200)2=504=12.5 WP' = 50 \times \left(\frac{100}{200}\right)^2 = \frac{50}{4} = 12.5\ \text{W}.
  3. So the answer is (c) 12.5 W.
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Q656MHT CET 2024 · 3 MAY · SHIFT 1Medium

Electrical Power & Heating Effect · Power: P = VI = I²R = V²/R

In the following circuit, a power of 50 watt is absorbed in the section AB of the circuit. The value of resistance ' xx ' is

Figure

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Answer: (c)

  1. Junction rule at B: I=0.5+1+1=2.5 AI = 0.5 + 1 + 1 = 2.5\text{ A}.
  2. Power in AB: P=I2(x+2)P = I^2(x + 2).
  3. 50=(2.5)2(x+2)⇒x+2=850 = (2.5)^2 (x + 2) \Rightarrow x + 2 = 8.
  4. So the answer is (c) x=6 Ωx = 6\ \Omega.
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Q657JEE Main 2024 · 31 JAN · SHIFT 2Medium

Electrical Power & Heating Effect · Power: P = VI = I²R = V²/R

In the following circuit, the battery has an emf of 2 V2\text{ V} and an internal resistance of 23 Ω\frac{2}{3}\ \Omega. The power consumption in the entire circuit is _________ W.

Figure

Numerical answer type. Work it out, then open the answer.

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Answer: 3

  1. The left side of the square is one node and the right side the other.
  2. Top 2 Ω2\ \Omega, bottom 2 Ω2\ \Omega and the two diagonals (together 2 Ω2\ \Omega) are in parallel: R=2/3 ΩR = 2/3\ \Omega.
  3. I=22/3+2/3=1.5 AI = \dfrac{2}{2/3 + 2/3} = 1.5\text{ A}.
  4. Total power P=EI=2×1.5=3 WP = EI = 2 \times 1.5 = 3\text{ W}.
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Q658JEE Main 2024 · 5 APR · SHIFT 2Medium

Electrical Power & Heating Effect · Heat & electrical energy

The ratio of heat dissipated per second through the resistance 5 Ω5\ \Omega and 10 Ω10\ \Omega in the circuit given below is:

Figure

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Answer: (b)

  1. 5 Ω5\ \Omega and 10 Ω10\ \Omega are in parallel, so they have the same voltage.
  2. H=V2R⇒H5H10=105=2H = \dfrac{V^2}{R} \Rightarrow \dfrac{H_5}{H_{10}} = \dfrac{10}{5} = 2.
  3. So the answer is (b) 2 : 1.
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Q659JEE Main 2024 · 30 JAN · SHIFT 2Medium

Electrical Power & Heating Effect · Heat & electrical energy

When a potential difference VV is applied across a wire of resistance RR, it dissipates energy at a rate WW. If the wire is cut into two halves and these halves are connected mutually parallel across the same supply, the energy dissipation rate will become:

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Answer: (b)

  1. Each half: R/2R/2; in parallel: R/4R/4.
  2. P=V2/RP = V^2/R, so the rate becomes 4W4W.
  3. So the answer is (b) 4W.
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Q660NEET 2024Medium

Electrical Power & Heating Effect · Bulbs & heaters in combination

Two heaters AA and BB have power rating of 1 kW1\text{ kW} and 2 kW2\text{ kW}, respectively. Those two are first connected in series and then in parallel to a fixed power source. The ratio of power outputs for these two cases is:

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Answer: (b)

  1. R=V2/PR = V^2/P: RA=2rR_A = 2r, RB=rR_B = r where r=V2/2000r = V^2/2000.
  2. Series: Ps=V23rP_s = \dfrac{V^2}{3r}. Parallel: Pp=V22r/3=3V22rP_p = \dfrac{V^2}{2r/3} = \dfrac{3V^2}{2r}.
  3. Ps:Pp=13:32=2:9P_s : P_p = \dfrac13 : \dfrac32 = 2 : 9.
  4. So the answer is (b) 2 : 9.
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Q661JEE Main 2024 · 8 APR · SHIFT 2Medium

Electrical Power & Heating Effect · Bulbs & heaters in combination

Water boils in an electric kettle in 20 minutes after being switched on. Using the same main supply, the length of the heating element should be _________ to __________ times of its initial length if the water is to be boiled in 15 minutes.

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Answer: (c)

  1. Heat needed is fixed: H=V2RtH = \frac{V^2}{R}t, so with the same VV, t∝R∝Lt \propto R \propto L.
  2. L′L=t′t=1520=34\frac{L'}{L} = \frac{t'}{t} = \frac{15}{20} = \frac{3}{4}
  3. So the length must be decreased to 34\frac{3}{4} of its initial value: answer (c).
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Q662JEE Main 2024 · 31 JAN · SHIFT 2Medium

Electrical Power & Heating Effect · Bulbs & heaters in combination

By what percentage will the illumination of the lamp decrease if the current drops by 20%?

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Answer: (b)

  1. Illumination ∝P=I2R\propto P = I^2R, so P∝I2P \propto I^2.
  2. I′=0.8I⇒P′=0.64PI' = 0.8I \Rightarrow P' = 0.64P
  3. Decrease =1−0.64=0.36= 1 - 0.64 = 0.36, i.e. 36%. So the answer is (b).
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Q663JEE Main 2024 · 8 APR · SHIFT 2Difficult

Electrical Power & Heating Effect · Bulbs & heaters in combination

A heater is designed to operate with a power of 1000 W1000\text{ W} in a 100 V100\text{ V} line. It is connected in combination with a resistance of 10 Ω10\ \Omega and a resistance RR, to a 100 V100\text{ V} mains as shown in figure. For the heater to operate at 62.5 W62.5\text{ W}, the value of RR should be _______ Ω\Omega.

Figure

Numerical answer type. Work it out, then open the answer.

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Answer: 5

  1. Heater resistance Rh=10021000=10 ΩR_h = \frac{100^2}{1000} = 10\ \Omega.
  2. At 62.5 W: VBC=PRh=62.5×10=25V_{BC} = \sqrt{PR_h} = \sqrt{62.5 \times 10} = 25 V, heater current =2.5= 2.5 A.
  3. The 10 Ω10\ \Omega resistor drops 100−25=75100 - 25 = 75 V, so total current =7.5= 7.5 A.
  4. Current in R=7.5−2.5=5R = 7.5 - 2.5 = 5 A, so R=255=5 ΩR = \frac{25}{5} = 5\ \Omega.
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Q664JEE Main 2023 · 31 JAN · SHIFT 2Easy

Electrical Power & Heating Effect · Heat & electrical energy

The H\mathrm{H} amount of thermal energy is developed by a resistor in 10 s10\text{ s} when a current of 4 A4\text{ A} is passed through it. If the current is increased to 16 A16\text{ A}, the thermal energy developed by the resistor in 10 s10\text{ s} will be:

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Answer: (b)

  1. H=I2Rt∝I2H = I^2 R t \propto I^2 for the same RR and tt.
  2. H′H=(164)2=16\frac{H'}{H} = \left(\frac{16}{4}\right)^2 = 16.
  3. So the answer is (b) 16H16H.
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Q665JEE Main 2023 · 29 JAN · SHIFT 1Easy

Electrical Power & Heating Effect · Heat & electrical energy

Ratio of thermal energy released in two resistors R and 3R connected in parallel in an electric circuit is:

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Answer: (d)

  1. In parallel, same VV: H=V2tR∝1RH = \frac{V^2 t}{R} \propto \frac{1}{R}.
  2. HRH3R=3RR=3:1\frac{H_R}{H_{3R}} = \frac{3R}{R} = 3 : 1.
  3. So the answer is (d) 3 : 1.
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Q666JEE Main 2023 · 11 APR · SHIFT 1Easy

Electrical Power & Heating Effect · Bulbs & heaters in combination

Two identical heater filaments are connected first in parallel and then in series. At the same applied voltage, the ratio of heat produced in same time for parallel to series will be:

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Answer: (a)

  1. Same VV and tt: H=V2tR∝1RH = \frac{V^2 t}{R} \propto \frac{1}{R}.
  2. Parallel Rp=R2R_p = \frac{R}{2}, series Rs=2RR_s = 2R.
  3. HpHs=RsRp=2RR/2=4:1\frac{H_p}{H_s} = \frac{R_s}{R_p} = \frac{2R}{R/2} = 4 : 1.
  4. So the answer is (a) 4 : 1.
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Q667JEE Main 2023 · 13 APR · SHIFT 1Medium

Electrical Power & Heating Effect · Power: P = VI = I²R = V²/R

Different combination of 3 resistors of equal resistance RR are shown in the figures. The increasing order for power dissipation is:

Figure

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Answer: (b)

  1. The same current II enters each combination, so P=I2ReqP = I^2 R_{eq} and P∝ReqP \propto R_{eq}.
  2. RA=R/2+R=1.5RR_A = R/2 + R = 1.5R, RB=(2R∥R)=2R/3R_B = (2R \parallel R) = 2R/3, RC=R/3R_C = R/3, RD=3RR_D = 3R.
  3. RC<RB<RA<RDR_C < R_B < R_A < R_D.
  4. So the answer is (b) PC<PB<PA<PDP_C < P_B < P_A < P_D.
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Q668JEE Main 2023 · 13 APR · SHIFT 1Medium

Electrical Power & Heating Effect · Power: P = VI = I²R = V²/R

A potential V0V_0 is applied across a uniform wire of resistance RR. The power dissipation is P1P_1. The wire is then cut into two equal halves and a potential of V0V_0 is applied across the length of each half. The total power dissipation across two wires is P2P_2. The ratio P2:P1P_2 : P_1 is x:1\sqrt{x} : 1. The value of xx is ___________.

Numerical answer type. Work it out, then open the answer.

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Answer: 16

  1. P1=V02/RP_1 = V_0^2/R.
  2. Each half has resistance R/2R/2: P2=2×V02R/2=4P1P_2 = 2 \times \dfrac{V_0^2}{R/2} = 4P_1.
  3. P2:P1=4:1=16:1P_2 : P_1 = 4 : 1 = \sqrt{16} : 1.
  4. So x=16x = 16.
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Q669NEET 2022 · PHASE 1Easy

Electrical Power & Heating Effect · Heat & electrical energy

Two resistors of resistance, 100 Ω\Omega and 200 Ω\Omega are connected in parallel in an electrical circuit. The ratio of the thermal energy developed in 100 Ω\Omega to that in 200 Ω\Omega in a given time is

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Answer: (b)

  1. In parallel, VV is the same, so H=V2tR∝1RH = \frac{V^2 t}{R} \propto \frac{1}{R}.
  2. H100H200=200100=2:1\frac{H_{100}}{H_{200}} = \frac{200}{100} = 2 : 1.
  3. So the answer is (b) 2 : 1.
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Q670JEE Main 2022 · 25 JUN · SHIFT 1Easy

Electrical Power & Heating Effect · Heat & electrical energy

A resistor develops 300 J of thermal energy in 15 s, when a current of 2 A is passed through it. If the current increases to 3 A, the energy developed in 10 s is ____________ J.

Numerical answer type. Work it out, then open the answer.

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Answer: 450

  1. R=HI2t=3004×15=5 ΩR = \frac{H}{I^2 t} = \frac{300}{4 \times 15} = 5\ \Omega.
  2. New energy H′=I′2Rt′=9×5×10H' = I'^2 R t' = 9 \times 5 \times 10.
  3. H′=450 JH' = 450\ \text{J}.
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Q671JEE Main 2022 · 28 JUL · SHIFT 2Medium

Electrical Power & Heating Effect · Bulbs & heaters in combination

An electrical bulb rated 220 V, 100 W, is connected in series with another bulb rated 220 V, 60 W. If the voltage across combination is 220 V, the power consumed by the 100 W bulb will be about _______ W.

Numerical answer type. Work it out, then open the answer.

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Answer: 14

  1. R1=2202100=484 ΩR_1 = \frac{220^2}{100} = 484\ \Omega, R2=220260≈806.7 ΩR_2 = \frac{220^2}{60} \approx 806.7\ \Omega.
  2. I=220484+806.7≈0.170I = \frac{220}{484 + 806.7} \approx 0.170 A.
  3. P1=I2R1=(0.170)2×484≈14P_1 = I^2R_1 = (0.170)^2 \times 484 \approx 14 W.
  4. So the 100 W bulb consumes about 14 W.
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Q672JEE Main 2022 · 29 JUN · SHIFT 1Medium

Electrical Power & Heating Effect · Bulbs & heaters in combination

Two coils require 20 minutes and 60 minutes respectively to produce same amount of heat energy when connected separately to the same source. If they are connected in parallel arrangement to the same source; the time required to produce same amount of heat by the combination of coils, will be ___________ min.

Numerical answer type. Work it out, then open the answer.

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Answer: 15

  1. Same VV and same heat: H=V2RtH = \frac{V^2}{R}t, so t∝Rt \propto R.
  2. In parallel Rp=R1R2R1+R2R_p = \frac{R_1R_2}{R_1 + R_2}, so tp=t1t2t1+t2t_p = \frac{t_1t_2}{t_1 + t_2}.
  3. tp=20×6080=15t_p = \frac{20 \times 60}{80} = 15 min.
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Q673JEE Main 2021 · 16 MAR · SHIFT 2Easy

Electrical Power & Heating Effect · Heat & electrical energy

A resistor develops 500 J of thermal energy in 20 s when a current of 1.5A is passed through it. If the current is increased from 1.5A to 3A, what will be the energy developed in 20 s.

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Answer: (b)

  1. H∝I2H \propto I^2 for the same RR and tt.
  2. H′=500×(31.5)2=500×4=2000 JH' = 500 \times \left(\frac{3}{1.5}\right)^2 = 500 \times 4 = 2000\ \text{J}.
  3. So the answer is (b) 2000 J.
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Q674JEE Main 2021 · 1 SEP · SHIFT 2Easy

Electrical Power & Heating Effect · Heat & electrical energy

A uniform heating wire of resistance 36Ω\Omega is connected across a potential difference of 240 V. The wire is then cut into half and potential difference of 240V is applied across each half separately. The ratio of power dissipation in first case to the total power dissipation in the second case would be 1: x, where x is ____________

Numerical answer type. Work it out, then open the answer.

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Answer: 4

  1. Case 1: P1=240236=1600 WP_1 = \frac{240^2}{36} = 1600\ \text{W}.
  2. Each half has 18 Ω18\ \Omega: P=240218=3200 WP = \frac{240^2}{18} = 3200\ \text{W}; two halves give P2=6400 WP_2 = 6400\ \text{W}.
  3. P1:P2=1600:6400=1:4P_1 : P_2 = 1600 : 6400 = 1 : 4.
  4. So x=4x = 4.
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Q675JEE Main 2021 · 31 AUG · SHIFT 2Easy

Electrical Power & Heating Effect · Heat & electrical energy

A resistor dissipates 192 J of energy in 1s when a current of 4A is passed through it. Now, when the current is doubled, the amount of thermal energy dissipated in 5s in _________ J.

Numerical answer type. Work it out, then open the answer.

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Answer: 3840

  1. R=HI2t=19216×1=12 ΩR = \frac{H}{I^2 t} = \frac{192}{16 \times 1} = 12\ \Omega.
  2. With I=8 AI = 8\ \text{A}: H=64×12×5H = 64 \times 12 \times 5.
  3. H=3840 JH = 3840\ \text{J}.
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Q676JEE Main 2021 · 16 MAR · SHIFT 2Easy

Electrical Power & Heating Effect · Heat & electrical energy

The energy dissipated by a resistor is 10 mJ in 1 s when an electric current of 2 mA flows through it. The resistance is ___________Ω\Omega. (Round off to the Nearest Integer)

Numerical answer type. Work it out, then open the answer.

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Answer: 2500

  1. H=I2Rt⇒R=HI2tH = I^2 R t \Rightarrow R = \frac{H}{I^2 t}.
  2. R=10×10−3(2×10−3)2×1=10−24×10−6R = \frac{10 \times 10^{-3}}{(2 \times 10^{-3})^2 \times 1} = \frac{10^{-2}}{4 \times 10^{-6}}.
  3. R=2500 ΩR = 2500\ \Omega.
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Q677JEE Main 2021 · 26 FEB · SHIFT 1Easy

Electrical Power & Heating Effect · Heat & electrical energy

In an electrical circuit, a battery is connected to pass 20C of charge through it in a certain given time. The potential difference between two plates of the battery is maintained at 15V. The workdone by the battery is __________J.

Numerical answer type. Work it out, then open the answer.

Show answer and solution

Answer: 300

  1. Work done by battery W=qVW = qV.
  2. W=20×15=300 JW = 20 \times 15 = 300\ \text{J}.
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Q678JEE Main 2021 · 26 AUG · SHIFT 1Medium

Electrical Power & Heating Effect · Power: P = VI = I²R = V²/R

In the given figure, the emf of the cell is 2.2 V and if internal resistance is 0.6Ω\Omega. Calculate the power dissipated in the whole circuit:

Figure

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Answer: (c)

AB4 Ω8 Ω12 Ω6 ΩR = 1.6 Ω
  1. The left side is node A and the right side is node B, so every branch runs from A to B in parallel.
  2. Branches: top 4 Ω4\ \Omega, bottom 8 Ω8\ \Omega, crossed 4+8=12 Ω4 + 8 = 12\ \Omega and 4+2=6 Ω4 + 2 = 6\ \Omega.
  3. 1R=14+18+112+16=1524⇒R=1.6 Ω\dfrac{1}{R} = \dfrac14 + \dfrac18 + \dfrac1{12} + \dfrac16 = \dfrac{15}{24} \Rightarrow R = 1.6\ \Omega.
  4. I=2.21.6+0.6=1 AI = \dfrac{2.2}{1.6 + 0.6} = 1\text{ A}, total power P=EI=2.2 WP = EI = 2.2\text{ W}.
  5. So the answer is (c) 2.2 W.
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Q679JEE Main 2021 · 26 AUG · SHIFT 2Medium

Electrical Power & Heating Effect · Bulbs & heaters in combination

An electric bulb of 500 watt at 100 volt is used in a circuit having a 200 V supply. Calculate the resistance R to be connected in series with the bulb so that the power delivered by the bulb is 500 W.

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Answer: (a)

  1. Bulb resistance Rb=V2P=1002500=20 ΩR_b = \frac{V^2}{P} = \frac{100^2}{500} = 20\ \Omega; rated current I=PV=5I = \frac{P}{V} = 5 A.
  2. With 200 V supply, total resistance must be 2005=40 Ω\frac{200}{5} = 40\ \Omega.
  3. R=40−20=20 ΩR = 40 - 20 = 20\ \Omega. So the answer is (a).
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Q680JEE Main 2021 · 25 JUL · SHIFT 1Medium

Electrical Power & Heating Effect · Bulbs & heaters in combination

An electric bulb rated as 200 W at 100 V is used in a circuit having 200 V supply. The resistance 'R' that must be put in series with the bulb so that the bulb delivers the same power is _____________ Ω\Omega.

Numerical answer type. Work it out, then open the answer.

Show answer and solution

Answer: 50

  1. Rb=1002200=50 ΩR_b = \frac{100^2}{200} = 50\ \Omega; rated current I=200100=2I = \frac{200}{100} = 2 A.
  2. Total resistance at 200 V: 2002=100 Ω\frac{200}{2} = 100\ \Omega.
  3. R=100−50=50 ΩR = 100 - 50 = 50\ \Omega.
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Q681MHT CET 2021 · 22 SEP · SHIFT 1Medium

Electrical Power & Heating Effect · Chemical & thermoelectric effects

If a current flowing in a coil is reduced to half of its initial value, the relation between the new energy (E2)(E_2) and the original energy (E1)(E_1) stored in the coil will be

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Answer: (a)

  1. Energy stored in a coil: E=12LI2E = \frac{1}{2}LI^2, so E∝I2E \propto I^2.
  2. I2=I12⇒E2=E14I_2 = \frac{I_1}{2} \Rightarrow E_2 = \frac{E_1}{4}
  3. So the answer is (a).
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Q682JEE Main 2021 · 1 SEP · SHIFT 2Difficult

Electrical Power & Heating Effect · Heat & electrical energy

Due to cold weather a 1 m water pipe of cross-sectional area 1 cm2 is filled with ice at −10∘-10^\circC. Resistive heating is used to melt the ice. Current of 0.5A is passed through 4 kΩ\Omega resistance. Assuming that all the heat produced is used for melting, what is the minimum time required? (Given latent heat of fusion for water/ice = 3.33 ×\times 105 J kg−1{}^{-1}, specific heat of ice = 2 ×\times 103 J kg−1{}^{-1} and density of ice = 103 kg/m3)

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Answer: (b)

  1. Mass of ice m=ρAl=103×10−4×1=0.1m = \rho Al = 10^3 \times 10^{-4} \times 1 = 0.1 kg.
  2. Heat =mcΔT+mL=0.1(2×103)(10)+0.1(3.33×105)=2000+33300=35300= mc\Delta T + mL = 0.1(2 \times 10^3)(10) + 0.1(3.33 \times 10^5) = 2000 + 33300 = 35300 J.
  3. Power P=I2R=(0.5)2×4000=1000P = I^2R = (0.5)^2 \times 4000 = 1000 W.
  4. t=353001000=35.3t = \frac{35300}{1000} = 35.3 s. So the answer is (b).
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Q683JEE Main 2020 · 5 SEP · SHIFT 1Medium

Electrical Power & Heating Effect · Power: P = VI = I²R = V²/R

An electrical power line, having a total resistance of 2 Ω\Omega, delivers 1 kW at 220 V. The efficiency of the transmission line is approximately:

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Answer: (b)

  1. I=P/V=1000/220=4.55 AI = P/V = 1000/220 = 4.55\text{ A}.
  2. Line loss: I2R=(4.55)2×2≈41.3 WI^2 R = (4.55)^2 \times 2 \approx 41.3\text{ W}.
  3. η=10001041.3×100≈96%\eta = \dfrac{1000}{1041.3} \times 100 \approx 96\%.
  4. So the answer is (b) 96%.
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Q684JEE Main 2020 · 7 JAN · SHIFT 2Medium

Electrical Power & Heating Effect · Bulbs & heaters in combination

In a building there are 15 bulbs of 45 W, 15 bulbs of 100 W, 15 small fans of 10 W and 2 heaters of 1 kW. The voltage of electric main is 220 V. The minimum fuse capacity (rated value) of the building will be:

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Answer: (b)

  1. Total power =15(45)+15(100)+15(10)+2(1000)=675+1500+150+2000=4325= 15(45) + 15(100) + 15(10) + 2(1000) = 675 + 1500 + 150 + 2000 = 4325 W.
  2. Current I=PV=4325220≈19.7I = \frac{P}{V} = \frac{4325}{220} \approx 19.7 A.
  3. Fuse must be rated just above this: 20 A. So the answer is (b).
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Q685JEE Main 2019 · 11 JAN · SHIFT 1Easy

Electrical Power & Heating Effect · Power: P = VI = I²R = V²/R

Two equal resistances when connected in series to a battery, consume electric power of 60 W. If these resistances are now connected in parallel combination to the same battery, the electric power consumed will be:

Show answer and solution

Answer: (a)

  1. P=V2ReqP = \frac{V^2}{R_{eq}} at fixed VV.
  2. Series Req=2RR_{eq} = 2R, parallel Req=R2R_{eq} = \frac{R}{2}: ratio of resistances 4:14 : 1.
  3. Pp=4Ps=4×60=240 WP_p = 4 P_s = 4 \times 60 = 240\ \text{W}.
  4. So the answer is (a) 240 W.
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Q686NEET 2019Easy

Electrical Power & Heating Effect · Bulbs & heaters in combination

Which of the following acts as a circuit protection device?

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Answer: (d)

  1. A fuse melts and breaks the circuit when current exceeds a safe limit.
  2. Inductor, conductor and switch do not do this automatically.
  3. So the answer is (d) fuse.
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Q687JEE Main 2019 · 10 JAN · SHIFT 2Medium

Electrical Power & Heating Effect · Power: P = VI = I²R = V²/R

A current of 2 mA was passed through an unknown resistor which dissipated a power of 4.4 W. Dissipated power when an ideal power supply of 11 V is connected across it is -

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Answer: (a)

  1. R=PI2=4.4(2×10−3)2=1.1×106 ΩR = \dfrac{P}{I^2} = \dfrac{4.4}{(2 \times 10^{-3})^2} = 1.1 \times 10^6\ \Omega.
  2. With 11 V: P=V2R=1211.1×106=1.1×10−4 WP = \dfrac{V^2}{R} = \dfrac{121}{1.1 \times 10^6} = 1.1 \times 10^{-4}\text{ W}.
  3. So the answer is (a) 11×10−5 W11 \times 10^{-5}\text{ W}.
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Q688NEET 2019Medium

Electrical Power & Heating Effect · Bulbs & heaters in combination

Six similar bulbs are connected as shown in the figure with a DC source of emf E, and zero internal resistance.

The ratio of power consumption by the bulbs when (i) all are glowing and (ii) in the situation when two from section A and one from section B are glowing, will be:

Figure

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Answer: (b)

  1. Section A's three bulbs are in parallel, and so are section B's; the two sections are in series.
  2. All glowing: R1=R3+R3=2R3R_1 = \dfrac{R}{3} + \dfrac{R}{3} = \dfrac{2R}{3}.
  3. Two in A, one in B: R2=R2+R=3R2R_2 = \dfrac{R}{2} + R = \dfrac{3R}{2}.
  4. P=E2/R⇒P1:P2=R2:R1=32:23=9:4P = E^2/R \Rightarrow P_1 : P_2 = R_2 : R_1 = \dfrac32 : \dfrac23 = 9 : 4.
  5. So the answer is (b) 9 : 4.
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Q689JEE Main 2019 · 12 JAN · SHIFT 1Medium

Electrical Power & Heating Effect · Bulbs & heaters in combination

Two electric bulbs, rated at (25 W, 220 V) and (100 W, 220 V), are connected in series across a 220 V voltage source. If the 25 W and 100 W bulbs draw powers P1 and P2 respectively, then:

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Answer: (b)

  1. R1=220225=1936 ΩR_1 = \frac{220^2}{25} = 1936\ \Omega, R2=2202100=484 ΩR_2 = \frac{220^2}{100} = 484\ \Omega.
  2. Series current I=2201936+484=111I = \frac{220}{1936 + 484} = \frac{1}{11} A.
  3. P1=I2R1=1936121=16P_1 = I^2R_1 = \frac{1936}{121} = 16 W, P2=484121=4P_2 = \frac{484}{121} = 4 W.
  4. So the answer is (b) P1=16P_1 = 16 W, P2=4P_2 = 4 W.
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Q690JEE Main 2018 · 15 APR · SHIFT 2Medium

Electrical Power & Heating Effect · Heat & electrical energy

A constant voltages is applied between two ends of a metallic wire. If the length is halved and the radius of the wire is doubled, the rate of heat developed in the wire will be:

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Answer: (d)

  1. R=ρLπr2R = \dfrac{\rho L}{\pi r^2}: halving LL and doubling rr gives R′=R/24=R/8R' = \dfrac{R/2}{4} = R/8.
  2. At constant VV, P=V2/R⇒P′=8PP = V^2/R \Rightarrow P' = 8P.
  3. So the answer is (d) increased 8 times.
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Q691JEE Main 2017 · 9 APR · SHIFT 1Medium

Electrical Power & Heating Effect · Power: P = VI = I²R = V²/R

The figure shows three circuits I, II and III which are connected to a 3V battery. If the powers dissipated by the configurations I, II and III are P1, P2 and P3 respectively, then:

Figure

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Answer: (c)

  1. Same 3 V across each, so P=V2/RP = V^2/R: smaller RR means more power.
  2. I: the five 1 Ω1\ \Omega resistors form a balanced Wheatstone bridge, R1=1 ΩR_1 = 1\ \Omega.
  3. II: the two 2 Ω2\ \Omega sides and the 1 Ω1\ \Omega middle are in parallel, R2=0.5 ΩR_2 = 0.5\ \Omega.
  4. III: (2∥2)+1=2 Ω(2 \parallel 2) + 1 = 2\ \Omega.
  5. R2<R1<R3R_2 < R_1 < R_3, so the answer is (c) P2>P1>P3P_2 > P_1 > P_3.
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Q692NEET 2016 · PHASE 2Medium

Electrical Power & Heating Effect · Bulbs & heaters in combination

A filament bulb (500 W, 100 V) is to be used in a 230 V main supply. When a resistance R is connected in series, it works perfectly and the bulb consumes 500 W. The value of R is

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Answer: (c)

  1. Bulb: Rb=1002500=20 ΩR_b = \dfrac{100^2}{500} = 20\ \Omega, rated current I=500100=5 AI = \dfrac{500}{100} = 5\text{ A}.
  2. Total resistance needed: 2305=46 Ω\dfrac{230}{5} = 46\ \Omega.
  3. R=46−20=26 ΩR = 46 - 20 = 26\ \Omega, so the answer is (c) 26 Ω26\ \Omega.
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Q693NEET 2016 · PHASE 1Difficult

Electrical Power & Heating Effect · Heat & electrical energy

The charge flowing through a resistance R varies with time t as Q = at −- bt2, where aa and bb are positive constants. The total heat produced in R is

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Answer: (c)

  1. I=dQdt=a−2btI = \frac{dQ}{dt} = a - 2bt, which becomes zero at t=a2bt = \frac{a}{2b} (current flows only till then).
  2. H=∫0a/2b(a−2bt)2R dt=R[(a−2bt)3−6b]0a/2bH = \int_0^{a/2b} (a - 2bt)^2R\,dt = R\left[\frac{(a - 2bt)^3}{-6b}\right]_0^{a/2b}
  3. H=a3R6bH = \frac{a^3R}{6b}. So the answer is (c).
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Q694NEET 2014 · AIPMTMedium

Electrical Power & Heating Effect · Power: P = VI = I²R = V²/R

Two cities are 150 km apart. Electric power is sent from one city to another city through copper wires. The fall of potential per km is 8 volt and the average resistance per km is 0.5 Ω\Omega. The power loss in the wire is

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Answer: (b)

  1. Total voltage drop in the line: V=8×150=1200 VV = 8 \times 150 = 1200\text{ V}.
  2. Line resistance: R=0.5×150=75 ΩR = 0.5 \times 150 = 75\ \Omega.
  3. P=V2R=1200275=19200 WP = \dfrac{V^2}{R} = \dfrac{1200^2}{75} = 19200\text{ W}.
  4. So the answer is (b) 19.2 kW.
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Q695JEE Main 2014 · 19 APR · SHIFT 1Medium

Electrical Power & Heating Effect · Bulbs & heaters in combination

Figure

Four bulbs B1, B2, B3 and B4 of 100 W each are connected to 220 V main as shown in the figure. The reading in an ideal ammeter will be:

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Answer: (c)

  1. All four bulbs are in parallel across 220 V, each drawing I=100220≈0.45I = \frac{100}{220} \approx 0.45 A.
  2. The ammeter sits on the bottom wire between B1B_1 and B2B_2, so it carries the return current of B2B_2, B3B_3, B4B_4 only.
  3. IA=3×100220≈1.35I_A = 3 \times \frac{100}{220} \approx 1.35 A. So the answer is (c).
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Q696JEE Main 2014Medium

Electrical Power & Heating Effect · Bulbs & heaters in combination

In a large building, there are 1515 bulbs of 40 W, 540\text{ W},\ 5 bulbs of 100 W, 5100\text{ W},\ 5 fans of 80 W80\text{ W} and 11 heater of 1 kW.1\text{ kW}. The voltage of electric mains is 220 V.220\text{ V}. The minimum capacity of the main fuse of the building will be:

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Answer: (c)

  1. Total power =15(40)+5(100)+5(80)+1000=600+500+400+1000=2500= 15(40) + 5(100) + 5(80) + 1000 = 600 + 500 + 400 + 1000 = 2500 W.
  2. I=2500220≈11.4I = \frac{2500}{220} \approx 11.4 A.
  3. Minimum fuse rating above this is 12 A. So the answer is (c).
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Q697JEE Main 2013 · 23 APR · SHIFT 1Medium

Electrical Power & Heating Effect · Heat & electrical energy

Which of the four resistances P, Q, R and S generate the greatest amount of heat when a current flows from A to B?

Figure

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Answer: (b)

  1. Upper branch P+Q=6 ΩP + Q = 6\ \Omega, lower branch R+S=3 ΩR + S = 3\ \Omega; currents split i:2ii : 2i.
  2. HP=2i2H_P = 2i^2, HQ=4i2H_Q = 4i^2.
  3. HR=(2i)2×1=4i2H_R = (2i)^2 \times 1 = 4i^2, HS=(2i)2×2=8i2H_S = (2i)^2 \times 2 = 8i^2.
  4. So the answer is (b) S.
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Q698NEET 2013Difficult

Electrical Power & Heating Effect · Heat & electrical energy

Ten identical cells connected in series are needed to heat a wire of length one meter and radius 'r' by 10oC in time 't'. How many cells will be required to heat the wire of length two meter of the same radius by the same temperature in time 't'?

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Answer: (a)

  1. Doubling the length doubles both RR and the mass, so heat needed doubles: H′=2HH' = 2H.
  2. H=V2tRH = \frac{V^2t}{R}: V′22R=2V2R⇒V′=2V\frac{V'^2}{2R} = 2\frac{V^2}{R} \Rightarrow V' = 2V.
  3. Voltage doubles, so cells needed =2×10=20= 2 \times 10 = 20. So the answer is (a).
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Q699JEE Main 2013Difficult

Electrical Power & Heating Effect · Bulbs & heaters in combination

The supply voltage to room is 120 V.120\text{ V}. The resistance of the lead wires is 6 Ω6\ \Omega. A 60 W60\text{ W} bulb is already switched on. What is the decrease of voltage across the bulb, when a 240 W240\text{ W} heater is switched on in parallel to the bulb?

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Answer: (d)

  1. Rbulb=120260=240 ΩR_{bulb} = \frac{120^2}{60} = 240\ \Omega, Rheater=1202240=60 ΩR_{heater} = \frac{120^2}{240} = 60\ \Omega.
  2. Before: V1=120×240240+6≈117.1V_1 = 120 \times \frac{240}{240 + 6} \approx 117.1 V.
  3. After: bulb ∥\parallel heater =240×60300=48 Ω= \frac{240 \times 60}{300} = 48\ \Omega, so V2=120×4848+6≈106.7V_2 = 120 \times \frac{48}{48 + 6} \approx 106.7 V.
  4. Decrease =117.1−106.7≈10.4= 117.1 - 106.7 \approx 10.4 V. So the answer is (d).
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Q700NEET 2012 · AIPMT · PRELIMSEasy

Electrical Power & Heating Effect · Bulbs & heaters in combination

If voltage across a bulb rated 220 volt-100 watt drops by 2.5% of its rated value, the percentage of the rated value by which the power would decrease is

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Answer: (c)

  1. P=V2R⇒ΔPP=2ΔVVP = \frac{V^2}{R} \Rightarrow \frac{\Delta P}{P} = 2\frac{\Delta V}{V} (small change).
  2. ΔPP=2×2.5%=5%\frac{\Delta P}{P} = 2 \times 2.5\% = 5\%.
  3. So the answer is (c) 5%.
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Q701NEET 2012 · AIPMT · MAINSMedium

Electrical Power & Heating Effect · Power: P = VI = I²R = V²/R

The power dissipated in the circuit shown in the figure is 30 watts. The value of R is

Figure

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Answer: (c)

  1. RR and 5 Ω5\ \Omega are in parallel across 10 V.
  2. P=V2R+V25=100R+20=30P = \dfrac{V^2}{R} + \dfrac{V^2}{5} = \dfrac{100}{R} + 20 = 30.
  3. 100R=10⇒R=10 Ω\dfrac{100}{R} = 10 \Rightarrow R = 10\ \Omega.
  4. So the answer is (c) 10 Ω10\ \Omega.
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Q702JEE Main 2012 · AIEEEMedium

Electrical Power & Heating Effect · Bulbs & heaters in combination

Two electric bulbs marked 25 W−220 V25\text{ W} - 220\text{ V} and 100 W−220 V100\text{ W} - 220\text{ V} are connected in series to a 440 V440\text{ V} supply. Which of the bulbs will fuse?

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Answer: (c)

  1. R25=220225=1936 ΩR_{25} = \frac{220^2}{25} = 1936\ \Omega, R100=2202100=484 ΩR_{100} = \frac{220^2}{100} = 484\ \Omega.
  2. I=4402420=211I = \frac{440}{2420} = \frac{2}{11} A.
  3. V25=211(1936)=352V_{25} = \frac{2}{11}(1936) = 352 V >220> 220 V, while V100=88V_{100} = 88 V <220< 220 V.
  4. So the 25 W bulb fuses: answer (c).
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Q703NEET 2011 · AIPMT · PRELIMSMedium

Electrical Power & Heating Effect · Power: P = VI = I²R = V²/R

If power dissipated in the 9 Ω\Omega resistor in the circuit shown is 36 watt, the potential difference across the 2 Ω\Omega resistor is

Figure

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Answer: (c)

  1. P9=V29=36⇒V=18 VP_9 = \dfrac{V^2}{9} = 36 \Rightarrow V = 18\text{ V} across the parallel pair.
  2. I9=18/9=2 AI_9 = 18/9 = 2\text{ A}, I6=18/6=3 AI_6 = 18/6 = 3\text{ A}, total I=5 AI = 5\text{ A}.
  3. Voltage across 2 Ω2\ \Omega: 5×2=10 V5 \times 2 = 10\text{ V}.
  4. So the answer is (c) 10 volt.
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Q704NEET 2011 · AIPMT · MAINSMedium

Electrical Power & Heating Effect · Chemical & thermoelectric effects

A thermocouple of negligible resistance produces an e.m.f. of 40 µV/ºC in the linear range of temperature. A galvanometer of resistance 10 ohm whose sensitivity is 1 µA/division, is employed with the thermocouple. The smallest value of temperature difference that can be detected by the system will be

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Answer: (d)

  1. Smallest detectable current = 1 division =1 μ= 1\ \muA.
  2. Required emf =IR=1 μA×10 Ω=10 μ= IR = 1\ \mu\text{A} \times 10\ \Omega = 10\ \muV.
  3. Δθ=1040=0.25∘\Delta\theta = \frac{10}{40} = 0.25^\circC. So the answer is (d).
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Q705NEET 2010 · AIPMT · MAINSMedium

Electrical Power & Heating Effect · Chemical & thermoelectric effects

The thermo e.m.f. EE in volts of a certain thermo-couple is found to vary with temperature difference θ\theta in °C between the two junctions according to the relation

E=30θ−θ215E = 30\theta - \frac{\theta^2}{15}

The neutral temperature for the thermo-couple will be

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Answer: (c)

  1. At neutral temperature dEdθ=0\frac{dE}{d\theta} = 0.
  2. 30−2θ15=0⇒θ=225∘30 - \frac{2\theta}{15} = 0 \Rightarrow \theta = 225^\circC
  3. So the answer is (c).
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Q706NEET 2010 · AIPMT · PRELIMSMedium

Electrical Power & Heating Effect · Chemical & thermoelectric effects

In producing chlorine by electrolysis 100 kW100\text{ kW} power at 125 V125\text{ V} is being consumed. How much chlorine per minute is liberated (E.C.E. of chlorine is 0.367×10−6 kg/C0.367 \times 10^{-6}\text{ kg/C})

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Answer: (d)

  1. I=PV=100000125=800I = \frac{P}{V} = \frac{100000}{125} = 800 A.
  2. Charge per minute q=800×60=48000q = 800 \times 60 = 48000 C.
  3. m=Zq=0.367×10−6×48000≈17.61×10−3m = Zq = 0.367 \times 10^{-6} \times 48000 \approx 17.61 \times 10^{-3} kg.
  4. So the answer is (d).
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Q707NEET 2008 · AIPMTMedium

Electrical Power & Heating Effect · Power: P = VI = I²R = V²/R

A current of 3 amp. flows through the 2 Ω\Omega resistor shown in the circuit. The power dissipated in the 5 Ω\Omega resistor is

Figure

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Answer: (b)

  1. The 2 Ω2\ \Omega branch and the (1+5) Ω(1 + 5)\ \Omega branch are in parallel.
  2. Voltage across them: V=3×2=6 VV = 3 \times 2 = 6\text{ V}.
  3. Current in the 6 Ω6\ \Omega branch: I=6/6=1 AI = 6/6 = 1\text{ A}.
  4. P5=I2R=12×5=5 WP_5 = I^2 R = 1^2 \times 5 = 5\text{ W}, so the answer is (b) 5 watt.
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Q708NEET 2008 · AIPMTMedium

Electrical Power & Heating Effect · Bulbs & heaters in combination

An electric kettle takes 4 A current at 220 V. How much time will it take to boil 1 kg of water from temperature 20oC? The temperature of boiling water is 100oC

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Answer: (c)

  1. Heat needed: Q=msΔT=1×4200×80=336000 JQ = ms\Delta T = 1 \times 4200 \times 80 = 336000\text{ J}.
  2. Power: P=VI=220×4=880 WP = VI = 220 \times 4 = 880\text{ W}.
  3. t=336000880≈382 s≈6.3 mint = \dfrac{336000}{880} \approx 382\text{ s} \approx 6.3\text{ min}.
  4. So the answer is (c) 6.3 min.
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Q709NEET 2007 · AIPMTEasy

Electrical Power & Heating Effect · Chemical & thermoelectric effects

A steady current of 1.5 amp flows through a copper voltameter for 10 minutes. If the electrochemical equivalent of copper is 30×10−5 g coulomb−130 \times 10^{-5}\text{ g coulomb}^{-1}, the mass of copper deposited on the electrode will be

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Answer: (c)

  1. Faraday's first law: m=ZItm = Z I t.
  2. m=30×10−5×1.5×600m = 30 \times 10^{-5} \times 1.5 \times 600.
  3. m=0.27 gm = 0.27\ \text{g}.
  4. So the answer is (c) 0.27 g.
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Q710NEET 2007 · AIPMTEasy

Electrical Power & Heating Effect · Chemical & thermoelectric effects

If the cold junction of a thermocouple is kept at 0∘C0^\circ\text{C} and the hot junction is kept at T∘CT^\circ\text{C}, then the relation between neutral temperature (Tn)(T_n) and temperature of inversion (Ti)(T_i) is

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Answer: (d)

  1. Neutral temperature is the mean of the cold-junction and inversion temperatures: Tn=Tc+Ti2T_n = \dfrac{T_c + T_i}{2}.
  2. With Tc=0∘CT_c = 0^\circ\text{C}: Tn=Ti2T_n = \dfrac{T_i}{2}.
  3. So the answer is (d) Tn=Ti/2T_n = T_i/2.
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Q711NEET 2007 · AIPMTMedium

Electrical Power & Heating Effect · Power: P = VI = I²R = V²/R

The total power dissipated in watt in the circuit shown here is

Figure

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Answer: (b)

  1. 6 Ω∥3 Ω=2 Ω6\ \Omega \parallel 3\ \Omega = 2\ \Omega; adding 4 Ω4\ \Omega gives Req=6 ΩR_{eq} = 6\ \Omega.
  2. I=18/6=3 AI = 18/6 = 3\text{ A}.
  3. P=VI=18×3=54 WP = VI = 18 \times 3 = 54\text{ W}.
  4. So the answer is (b) 54.
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Q712JEE Main 2006 · AIEEEEasy

Electrical Power & Heating Effect · Bulbs & heaters in combination

An electric bulb is rated 220220 volt - 100100 watt. The power consumed by it when operated on 110110 volt will be

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Answer: (c)

  1. Fixed resistance: P∝V2P \propto V^2.
  2. P′=100×(110220)2=25 WP' = 100 \times \left(\frac{110}{220}\right)^2 = 25\ \text{W}.
  3. So the answer is (c) 25 watt.
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Q713JEE Main 2006 · AIEEEEasy

Electrical Power & Heating Effect · Chemical & thermoelectric effects

A thermocouple is made from two metals, Antimony and Bismuth. If one junction of the couple is kept hot and the other is kept cold, then, an electric current will

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Answer: (d)

  1. In the Sb–Bi thermocouple the current flows from Bi to Sb at the hot junction.
  2. Around the loop it therefore flows from Sb to Bi at the cold junction (memory aid: ABC, Antimony to Bismuth at Cold).
  3. So the answer is (d).
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Q714NEET 2006 · AIPMTMedium

Electrical Power & Heating Effect · Power: P = VI = I²R = V²/R

Power dissipated across the 8 Ω\Omega resistor in the circuit shown here is 2 watt. The power dissipated in watt units across the 3 Ω\Omega resistor is

Figure

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Answer: (a)

  1. P8=i22×8=2⇒i2=0.5 AP_8 = i_2^2 \times 8 = 2 \Rightarrow i_2 = 0.5\text{ A}, so the branch voltage is 0.5×8=4 V0.5 \times 8 = 4\text{ V}.
  2. Upper branch (1+3)=4 Ω(1 + 3) = 4\ \Omega: i1=4/4=1 Ai_1 = 4/4 = 1\text{ A}.
  3. P3=i12×3=3 WP_3 = i_1^2 \times 3 = 3\text{ W}.
  4. So the answer is (a) 3.0.
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Q715NEET 2006 · AIPMTMedium

Electrical Power & Heating Effect · Chemical & thermoelectric effects

In producing chlorine through electrolysis 100 W100\text{ W} power at 125 V125\text{ V} is being consumed. How much chlorine per min is liberated? ECE of chlorine is 0.367×10−6 kg/C0.367 \times 10^{-6}\text{ kg/C}

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Answer: (a)

  1. I=PV=100125=0.8I = \frac{P}{V} = \frac{100}{125} = 0.8 A.
  2. Charge per minute q=0.8×60=48q = 0.8 \times 60 = 48 C.
  3. m=Zq=0.367×10−6×48≈17.6×10−6m = Zq = 0.367 \times 10^{-6} \times 48 \approx 17.6 \times 10^{-6} kg =17.6= 17.6 mg.
  4. So the answer is (a).
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Q716NEET 2005 · AIPMTEasy

Electrical Power & Heating Effect · Bulbs & heaters in combination

A 5-ampere fuse wire can withstand a maximum power of 1 watt in the circuit. The resistance of the fuse wire is

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Answer: (a)

  1. P=I2R⇒R=PI2P = I^2 R \Rightarrow R = \frac{P}{I^2}.
  2. R=152=0.04 ΩR = \frac{1}{5^2} = 0.04\ \Omega.
  3. So the answer is (a) 0.04 ohm.
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Q717JEE Main 2005 · AIEEEEasy

Electrical Power & Heating Effect · Bulbs & heaters in combination

A heater coil is cut into two equal parts and only one part is now used in the heater. The heat generated will now be

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Answer: (b)

  1. Half the coil has resistance R2\frac{R}{2}; supply voltage is the same.
  2. P=V2R∝1RP = \frac{V^2}{R} \propto \frac{1}{R}, so the power (heat per unit time) doubles.
  3. So the answer is (b) doubled.
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Q718NEET 2005 · AIPMTEasy

Electrical Power & Heating Effect · Chemical & thermoelectric effects

The temperature of inversion of a thermocouple is 620∘C620^\circ\text{C} and the neutral temperature is 300∘C300^\circ\text{C}. What is the temperature of cold junction?

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Answer: (c)

  1. Tn=Tc+Ti2⇒Tc=2Tn−TiT_n = \dfrac{T_c + T_i}{2} \Rightarrow T_c = 2T_n - T_i.
  2. Tc=2(300)−620=−20∘CT_c = 2(300) - 620 = -20^\circ\text{C}.
  3. So the answer is (c) −20∘C-20^\circ\text{C}.
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Q719JEE Main 2005 · AIEEEMedium

Electrical Power & Heating Effect · Bulbs & heaters in combination

The resistance of hot tungsten filament is about 1010 times the cold resistance. What will be resistance of 100 W100\text{ W} and 200 V200\text{ V} lamp when not in use?

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Answer: (b)

  1. Hot resistance Rhot=V2P=2002100=400 ΩR_{hot} = \frac{V^2}{P} = \frac{200^2}{100} = 400\ \Omega.
  2. Cold resistance =40010=40 Ω= \frac{400}{10} = 40\ \Omega.
  3. So the answer is (b) 40 Ω40\ \Omega.
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Q720JEE Main 2005 · AIEEEMedium

Electrical Power & Heating Effect · Chemical & thermoelectric effects

Two voltmeters, one of copper and another of silver, are joined in parallel. When a total charge qq flows through the voltmeters, equal amount of metals are deposited. If the electrochemical equivalents of copper and silver are Z1Z_1 and Z2Z_2 respectively the charge which flows through the silver voltmeter is

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Answer: (a)

  1. Equal masses: Z1q1=Z2q2Z_1q_1 = Z_2q_2, and q1+q2=qq_1 + q_2 = q.
  2. q1=Z2Z1q2⇒q2(1+Z2Z1)=qq_1 = \frac{Z_2}{Z_1}q_2 \Rightarrow q_2\left(1 + \frac{Z_2}{Z_1}\right) = q
  3. q2=q1+Z2Z1q_2 = \frac{q}{1 + \frac{Z_2}{Z_1}}. So the answer is (a).
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Q721NEET 2004 · AIPMTEasy

Electrical Power & Heating Effect · Bulbs & heaters in combination

when three identical bulbs of 60 watt, 200 volt rating are connected in series to a 200 volt supply, the power drawn by them will be

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Answer: (d)

  1. Each bulb: R=V2PR = \frac{V^2}{P}; three in series give 3R3R.
  2. Ptotal=V23R=P3=603=20 WP_{total} = \frac{V^2}{3R} = \frac{P}{3} = \frac{60}{3} = 20\ \text{W}.
  3. So the answer is (d) 20 watt.
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Q722NEET 2004 · AIPMTEasy

Electrical Power & Heating Effect · Bulbs & heaters in combination

In India electricity is supplied for domestic use at 220 V. It is supplied at 110 V in USA. If the resistance of a 60 W bulb for use in India is R, the resistance of a 60 W bulb for use in USA will be

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Answer: (c)

  1. For the same power, R=V2P∝V2R = \frac{V^2}{P} \propto V^2.
  2. RUSA=R(110220)2=R4R_{USA} = R\left(\frac{110}{220}\right)^2 = \frac{R}{4}.
  3. So the answer is (c) R/4R/4.
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Q723JEE Main 2004 · AIEEEEasy

Electrical Power & Heating Effect · Chemical & thermoelectric effects

The electrochemical equivalent of a metal is 3.3×10−7 kg3.3 \times 10^{-7}\text{ kg} per Coulomb. The mass of the metal liberated at the cathode when a 3 A3\text{ A} current is passed for 22 seconds will be

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Answer: (c)

  1. m=ZItm = ZIt.
  2. m=3.3×10−7×3×2=19.8×10−7 kgm = 3.3 \times 10^{-7} \times 3 \times 2 = 19.8 \times 10^{-7}\text{ kg}.
  3. So the answer is (c) 19.8×10−7 kg19.8 \times 10^{-7}\text{ kg}.
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Q724NEET 2004 · AIPMTMedium

Electrical Power & Heating Effect · Heat & electrical energy

A battery is charged at a potential of 15 V for 8 h when the current flowing is 10 A. The battery on discharge supplies a current of 5 A for 15 h. The mean terminal voltage during discharge is 14 V. The "watt-hour" efficiency of the battery is

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Answer: (d)

  1. Energy input: 15×10×8=1200 Wh15 \times 10 \times 8 = 1200\text{ Wh}.
  2. Energy output: 14×5×15=1050 Wh14 \times 5 \times 15 = 1050\text{ Wh}.
  3. η=10501200×100=87.5%\eta = \dfrac{1050}{1200} \times 100 = 87.5\%.
  4. So the answer is (d) 87.5%87.5\%.
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Q725JEE Main 2004 · AIEEEMedium

Electrical Power & Heating Effect · Chemical & thermoelectric effects

The thermo emf\text{emf} of a thermocouple varies with temperature θ\theta of the hot junction as E=aθ+bθ2E = a\theta + b\theta^2 in volts where the ratio a/ba/b is 700∘C.700^\circ\text{C}. If the cold junction is kept at 0∘C,0^\circ\text{C}, then the neutral temperature is

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Answer: (d)

  1. Neutral temperature: dEdθ=a+2bθ=0⇒θn=−a2b\frac{dE}{d\theta} = a + 2b\theta = 0 \Rightarrow \theta_n = -\frac{a}{2b}.
  2. θn=−7002=−350∘\theta_n = -\frac{700}{2} = -350^\circC, which is below the cold junction at 0∘0^\circC.
  3. A neutral temperature must lie above the cold-junction temperature, so none is possible: answer (d).
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Q726NEET 2003 · AIPMTEasy

Electrical Power & Heating Effect · Bulbs & heaters in combination

Two 220 volt, 100 watt bulbs are connected first in series and then in parallel. Each time the combination is connected to a 220 volt a.c. supply line. The power drawn by the combination in each case respectively will be

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Answer: (d)

  1. Each bulb: R=2202100=484 ΩR = \frac{220^2}{100} = 484\ \Omega.
  2. Series: P=2202968=50 WP = \frac{220^2}{968} = 50\ \text{W}.
  3. Parallel: each bulb gets 220 V and draws 100 W, total 200 W200\ \text{W}.
  4. So the answer is (d) 50 watt, 200 watt.
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Q727NEET 2003 · AIPMTEasy

Electrical Power & Heating Effect · Bulbs & heaters in combination

Fuse wire is a wire of

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Answer: (b)

  1. A fuse must heat up quickly (I2RI^2 R large), so it needs high resistance (high resistivity).
  2. It must melt at a modest overload current, so it needs a low melting point.
  3. So the answer is (b) high resistance and low melting point.
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Q728JEE Main 2003 · AIEEEEasy

Electrical Power & Heating Effect · Bulbs & heaters in combination

A 220220 volt, 10001000 watt bulb is connected across a 110 volt110\text{ volt} mains supply. The power consumed will be

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Answer: (c)

  1. Fixed resistance: P∝V2P \propto V^2.
  2. P′=1000×(110220)2=250 WP' = 1000 \times \left(\frac{110}{220}\right)^2 = 250\ \text{W}.
  3. So the answer is (c) 250 watt.
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Q729NEET 2003 · AIPMTMedium

Electrical Power & Heating Effect · Bulbs & heaters in combination

An electric kettle has two heating coils. When one of the coils is connected to an a.c. source, the water in the kettle boils in 10 minutes. When the other coil is used the water boils in 40 minutes. If both the coils are connected in parallel, the time taken by the same quantity of water to boil will be

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Answer: (a)

  1. Same heat HH, so P∝1/tP \propto 1/t.
  2. Parallel powers add: 1t=110+140=540\dfrac{1}{t} = \dfrac{1}{10} + \dfrac{1}{40} = \dfrac{5}{40}.
  3. t=8 mint = 8\text{ min}, so the answer is (a) 8 minutes.
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Q730JEE Main 2003 · AIEEEMedium

Electrical Power & Heating Effect · Chemical & thermoelectric effects

The thermo e.m.f.\text{e.m.f.} of a thermo -couple is 25 μV/∘C25\ \mu\text{V}/^\circ\text{C} at room temperature. A galvanometer of 40 ohm40\text{ ohm} resistance, capable of detecting current as low as 10−5 A,10^{-5}\text{ A}, is connected with the thermo couple. The smallest temperature difference that can be detected by this system is

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Answer: (a)

  1. Minimum emf needed =IR=10−5×40=4×10−4= IR = 10^{-5} \times 40 = 4 \times 10^{-4} V =400 μ= 400\ \muV.
  2. Δθ=400 μV25 μV/∘C=16∘\Delta\theta = \frac{400\ \mu\text{V}}{25\ \mu\text{V}/^\circ\text{C}} = 16^\circC
  3. So the answer is (a).
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Q731JEE Main 2003 · AIEEEMedium

Electrical Power & Heating Effect · Chemical & thermoelectric effects

The negative Zn\text{Zn} pole of a Daniell cell, sending a constant current through a circuit, decreases in mass by 0.13 g0.13\text{ g} in 3030 minutes. If the electrochemical equivalents of Zn\text{Zn} and Cu\text{Cu} are 32.532.5 and 31.531.5 respectively, the increase in the mass of the positive Cu\text{Cu} pole in this time is

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Answer: (c)

  1. Same current, same time, so same charge: m=Zq⇒m∝Zm = Zq \Rightarrow m \propto Z.
  2. mCu=0.13×31.532.5≈0.126m_{Cu} = 0.13 \times \frac{31.5}{32.5} \approx 0.126 g
  3. So the answer is (c).
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Q732JEE Main 2002 · AIEEEEasy

Electrical Power & Heating Effect · Power: P = VI = I²R = V²/R

A wire when connected to 220 V220\text{ V} mains supply has power dissipation P1.P_1. Now the wire is cut into two equal pieces which are connected in parallel to the same supply. Power dissipation in this case is P2.P_2. Then P2:P1P_2 : P_1 is

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Answer: (b)

  1. Each half has resistance R2\frac{R}{2}; two in parallel give R4\frac{R}{4}.
  2. At fixed VV, P=V2R∝1RP = \frac{V^2}{R} \propto \frac{1}{R}.
  3. P2P1=RR/4=4\frac{P_2}{P_1} = \frac{R}{R/4} = 4.
  4. So the answer is (b) 4.
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Q733JEE Main 2002 · AIEEEEasy

Electrical Power & Heating Effect · Chemical & thermoelectric effects

The mass of product liberated on anode in an electrochemical cell depends on (where tt is the time period for which the current is passed).

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Answer: (b)

  1. Faraday's law: m=ZIt=Zqm = ZIt = Zq.
  2. So m∝Itm \propto It.
  3. So the answer is (b) ItIt.
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Q734JEE Main 2002 · AIEEEEasy

Electrical Power & Heating Effect · Chemical & thermoelectric effects

If θi\theta_i is the inversion temperature, θn\theta_n is the neutral temperature, θc\theta_c is the temperature of the cold junction, then

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Answer: (c)

  1. The thermo-emf is maximum at the neutral temperature, which lies midway between the cold-junction and inversion temperatures.
  2. θn=θi+θc2\theta_n = \dfrac{\theta_i + \theta_c}{2}.
  3. So the answer is (c).
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Q735JEE Main 2002 · AIEEEMedium

Electrical Power & Heating Effect · Power: P = VI = I²R = V²/R

If in the circuit, power dissipation is 150 W,150\text{ W}, then RR is

Figure

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Answer: (b)

  1. RR and 2 Ω2\ \Omega are in parallel across 15 V.
  2. P=225R+2252=150⇒225R=37.5P = \dfrac{225}{R} + \dfrac{225}{2} = 150 \Rightarrow \dfrac{225}{R} = 37.5.
  3. R=6 ΩR = 6\ \Omega, so the answer is (b) 6 Ω6\ \Omega.
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Q736NEET 2000 · AIPMTEasy

Electrical Power & Heating Effect · Bulbs & heaters in combination

Two bulbs of (40 W, 200 V), and (100 W, 200 V). Then correct relation for their resistances

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Answer: (b)

  1. R=V2PR = \frac{V^2}{P}; same rated voltage, so R∝1PR \propto \frac{1}{P}.
  2. R40=200240=1000 ΩR_{40} = \frac{200^2}{40} = 1000\ \Omega, R100=2002100=400 ΩR_{100} = \frac{200^2}{100} = 400\ \Omega.
  3. So the answer is (b) R40>R100R_{40} > R_{100}.
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Q737NEET 2000 · AIPMTEasy

Electrical Power & Heating Effect · Chemical & thermoelectric effects

In electrolysis the mass deposited on an electrode is directly proportional to

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Answer: (a)

  1. Faraday's first law: m=ZItm = ZIt.
  2. For a given time, m∝Im \propto I.
  3. So the answer is (a) current.
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Q738NEET 1998 · AIPMTEasy

Electrical Power & Heating Effect · Power: P = VI = I²R = V²/R

Three equal resistors connected in series across a source of emf together dissipate 1010 watt of power. What will be the power dissipated in watt if the same resistors are connected in parallel across the same source of emf?

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Answer: (d)

  1. Series: Rs=3RR_s = 3R, Ps=V23R=10 WP_s = \frac{V^2}{3R} = 10\ \text{W}.
  2. Parallel: Rp=R3R_p = \frac{R}{3}, Pp=3V2R=9PsP_p = \frac{3V^2}{R} = 9 P_s.
  3. Pp=9×10=90 WP_p = 9 \times 10 = 90\ \text{W}.
  4. So the answer is (d) 90.
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Q739NEET 1998 · AIPMTEasy

Electrical Power & Heating Effect · Heat & electrical energy

A 5∘C5^\circ\text{C} rise in temperature is observed in a conductor by passing a current. When the current is doubled the rise in temperature will be approximately

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Answer: (c)

  1. Heat H=I2RtH = I^2 R t, and temperature rise ΔT∝H∝I2\Delta T \propto H \propto I^2.
  2. Doubling II makes ΔT\Delta T four times: 4×5=20∘C4 \times 5 = 20^\circ\text{C}.
  3. So the answer is (c) 20∘C20^\circ\text{C}.
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Q740NEET 1998 · AIPMTMedium

Electrical Power & Heating Effect · Chemical & thermoelectric effects

If nearly 105 C10^{5}\text{ C} liberate 1 g1\text{ g} equivalent of aluminium, then the amount of aluminium (equivalent weight 9) deposited through electrolysis in 20 min20\text{ min} by a current of 5050 amp will be

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Answer: (c)

  1. Charge q=It=50×20×60=6×104q = It = 50 \times 20 \times 60 = 6 \times 10^4 C.
  2. 10510^5 C deposits 1 g-equivalent (9 g), so m=6×104105×9m = \frac{6 \times 10^4}{10^5} \times 9.
  3. m=5.4m = 5.4 g. So the answer is (c).
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Q741NEET 1997 · AIPMTEasy

Electrical Power & Heating Effect · Heat & electrical energy

One kilowatt hour is equal to

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Answer: (c)

  1. 1 kWh=1000 W×3600 s1\ \text{kWh} = 1000\ \text{W} \times 3600\ \text{s}.
  2. =3.6×106 J=36×105 J= 3.6 \times 10^6\ \text{J} = 36 \times 10^5\ \text{J}.
  3. So the answer is (c) 36×10536 \times 10^5 J.
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Q742NEET 1997 · AIPMTEasy

Electrical Power & Heating Effect · Bulbs & heaters in combination

A (100 W, 200 V) bulb is connected to a 160 volts supply. The power consumption would be

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Answer: (c)

  1. Resistance is fixed, so P∝V2P \propto V^2.
  2. P′=100×(160200)2=100×0.64=64 WP' = 100 \times \left(\frac{160}{200}\right)^2 = 100 \times 0.64 = 64\ \text{W}.
  3. So the answer is (c) 64 W.
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Q743NEET 1997 · AIPMTEasy

Electrical Power & Heating Effect · Bulbs & heaters in combination

If two bulbs, whose resistances are in the ratio of 1:21 : 2 are connected in series, the power dissipated in them has the ratio of

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Answer: (d)

  1. In series the current is the same, so P=I2R∝RP = I^2 R \propto R.
  2. P1:P2=R1:R2=1:2P_1 : P_2 = R_1 : R_2 = 1 : 2.
  3. So the answer is (d) 1 : 2.
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Q744NEET 1995 · AIPMTEasy

Electrical Power & Heating Effect · Chemical & thermoelectric effects

A heating coil is labelled 100 W, 220 V. The coil is cut in half and the two pieces are joined in parallel to the same source. The energy now liberated per second is

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Answer: (b)

  1. Each half has resistance R/2R/2; two halves in parallel give R/4R/4.
  2. At the same voltage P=V2/RP = V^2/R, so P′=4P=4×100P' = 4P = 4 \times 100.
  3. So the answer is (b) 400 W400\text{ W}.
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Q745NEET 1994 · AIPMTEasy

Electrical Power & Heating Effect · Bulbs & heaters in combination

An electric bulb is rated 60 W, 220 V. The resistance of its filament is

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Answer: (d)

  1. R=V2P=220260R = \frac{V^2}{P} = \frac{220^2}{60}.
  2. R=4840060≈807 ΩR = \frac{48400}{60} \approx 807\ \Omega.
  3. So the answer is (d) 807 Ω807\ \Omega.
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Q746NEET 1989 · AIPMTEasy

Electrical Power & Heating Effect · Heat & electrical energy

A current of 2 A2\text{ A}, passing through a conductor produces 80 J80\text{ J} of heat in 10 seconds. The resistance of the conductor in ohm is

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Answer: (b)

  1. H=I2Rt⇒R=HI2tH = I^2 R t \Rightarrow R = \frac{H}{I^2 t}.
  2. R=804×10=2 ΩR = \frac{80}{4 \times 10} = 2\ \Omega.
  3. So the answer is (b) 2.
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Q747NEET 1989 · AIPMTMedium

Electrical Power & Heating Effect · Bulbs & heaters in combination

40 electric bulbs are connected in series across a 220 V220\text{ V} supply. After one bulb is fused the remaining 39 are connected again in series across the same supply. The illumination will be

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Answer: (b)

  1. Total power in series: P=V2nRP = \dfrac{V^2}{nR}.
  2. With 39 bulbs the total resistance is smaller, so the power (illumination) is larger.
  3. So the answer is (b) more with 39 bulbs than with 40.
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