Verified Solution • 2025–26 Session moderate

QJEEAdv18Phdead: JEE Advanced 2018 Physics - Class 12 JEEAdv18Phdead Solution: In the figure below, the switches and are closed simultaneously …

In the figure below, the switches S1{S_1} and S2{S_2} are closed simultaneously at t=0t=0 and a current starts to flow in the circuit. Both the batteries have the same magnitude of the electromotive force (emf) and the polarities are as indicated in the figure. Ignore mutual inductance between the inductors. The current II in the middle wire reaches its maximum magnitude Imax⁡{I_{\max }} at time t=τt = \tau . Which of the following statements is (are) true?

JEE Advanced 2018 Paper 1 Offline Physics - Electromagnetic Induction Question 16 English

Multiple Choice Options

  • A Imax⁡=V2R{I_{\max }} = {V \over {2R}}
  • B Imax=V4R{I_{max}} = {V \over {4R}}
  • C τ=LRln⁡2\tau = {L \over R}\ln 2
  • D τ=2LRln⁡2\tau = {{2L} \over R}\ln 2

Step-by-Step Verified Solution

I1=VR(1−e−tRL){I_1} = {V \over R}\left( {1 - {e^{ - {{tR} \over L}}}} \right)

JEE Advanced 2018 Paper 1 Offline Physics - Electromagnetic Induction Question 16 English Explanation

I2=VR(1−e−tR2L){I_2} = {V \over R}\left( {1 - {e^{ - {{tR} \over {2L}}}}} \right)

From principle of superposition,

I=I1−I2⇒I=VRe−tR2L(1−e−tR2L)I = {I_1} - {I_2} \Rightarrow I = {V \over R}{e^{ - {{tR} \over {2L}}}}\left( {1 - {e^{ - {{tR} \over {2L}}}}} \right) ...... (i)

I is maximum when dIdt=0{{dI} \over {dt}} = 0 , which gives

e−tR2L=12{e^{ - {{tR} \over {2L}}}} = {1 \over 2} or t=2LRln⁡2t = {{2L} \over R}\ln 2

Substituting this time in Eq. (i), we get

Imax⁡=V4R{I_{\max }} = {V \over {4R}}

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