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MCQEasyJEE 2025Self & Mutual Inductance

JEE Physics 2025 Question with Solution

Circuit diagram showing a 12 V source in series with a 3 H inductor and a variable resistor R with sliding contact.

In the given circuit the sliding contact is pulled outwards such that electric current in the circuit changes at the rate of 8 A/s8 \, \text{A/s}. At an instant when RR is 12 Ω12 \, \Omega, the value of the current in the circuit will be _____ A.

  • A

    2 A2 \, \text{A}

  • B

    4 A4 \, \text{A}

  • C

    3 A3 \, \text{A}

  • D

    5 A5 \, \text{A}

Answer

Correct answer:C

Step-by-step solution

Standard Method

Given: A series RL circuit with supply voltage V=12 VV = 12 \, \text{V}, inductance L=3 HL = 3 \, \text{H}, resistance R=12 ΩR = 12 \, \Omega, and the current is changing at the rate dIdt=−8 A/s\frac{dI}{dt} = -8 \, \text{A/s} because the sliding contact is pulled outwards.

Find: The instantaneous current II.

For an RL circuit, the governing equation is

V=LdIdt+IRV = L\frac{dI}{dt} + IR

Substitute the given values:

12=3(−8)+12I12 = 3(-8) + 12I

Now simplify:

12=−24+12I12 = -24 + 12I 36=12I36 = 12I I=3 AI = 3 \, \text{A}

Therefore, the current in the circuit at that instant is 3 A3 \, \text{A}. The correct option is C.

Using induced emf idea

Given: L=3 HL = 3 \, \text{H} and dIdt=8 A/s\frac{dI}{dt} = 8 \, \text{A/s} in magnitude, with R=12 ΩR = 12 \, \Omega and source voltage 12 V12 \, \text{V}.

Find: The current II at that instant.

The induced emf across the inductor has magnitude

E=LdIdt=3×8=24 VE = L\frac{dI}{dt} = 3 \times 8 = 24 \, \text{V}

Since the current is decreasing, this induced emf opposes the source term in the circuit equation.

Using Kirchhoff's loop equation in sign-consistent form:

12=3(−8)+12I12 = 3(-8) + 12I

so

12=−24+12I12 = -24 + 12I 36=12I36 = 12I I=3 AI = 3 \, \text{A}

Hence, the instantaneous current is 3 A3 \, \text{A}.

Common mistakes

  • Using dIdt=+8 A/s\frac{dI}{dt} = +8 \, \text{A/s} instead of recognizing that the current is decreasing. This gives the wrong sign in the RL equation. Use the rate with the correct sign based on whether current is increasing or decreasing.

  • Applying only E=−LdIdtE = -L\frac{dI}{dt} and forgetting the resistor drop IRIR. In a series RL circuit, the loop equation must include both the inductor term and the resistor term.

  • Substituting R=12 ΩR = 12 \, \Omega and then dividing incorrectly by 1212 at the final step. After obtaining 36=12I36 = 12I, divide both sides by 1212 to get I=3 AI = 3 \, \text{A}.

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