JEE Advanced 2018 · Paper 1 · Physics Q4

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PhysicsMultiple correct+4 / -2Question 4 of 54

In the figure below, the switches $S_1$ and $S_2$ are closed simultaneously at $t = 0$ 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 $I$ in the middle wire reaches its maximum magnitude $I_{max}$ at time $t = \tau$ . Which of the following statements is (are) true?

Two loops sharing a middle wire (current I, downward). Left loop: battery V, switch S₁, resistor R and inductor L. Right loop: resistor R, inductor 2L, battery V and switch S₂. The two batteries have the same emf, with polarities as drawn.

I_{max} = \dfrac V{2R}

I_{max} = \dfrac V{4R}

\tau = \dfrac LR\ln2

\tau = \dfrac{2L}R\ln2

Answer and worked solution

Official answer: B, D

Two independent loops. The middle wire has no resistance, so each loop behaves as its own LR circuit with emf $V$ :
$i_1=\frac VR\left(1-e^{-Rt/L}\right),\qquad i_2=\frac VR\left(1-e^{-Rt/2L}\right).$
With the polarities shown, the two loop currents run opposite ways in the middle wire:
$I=i_1-i_2=\frac VR\left(e^{-Rt/2L}-e^{-Rt/L}\right).$

Maximum. Let $x=e^{-Rt/2L}$ . Then $I=\dfrac VR(x-x^2)$ , which is largest at $x=\tfrac12$ :
$I_{max}=\frac VR\cdot\frac14=\frac V{4R}.$

Time. $e^{-R\tau/2L}=\tfrac12\implies\tau=\dfrac{2L}R\ln2$ .

(B), (D)

Topics:LR circuits Kirchhoff's laws

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JEE Advanced 2018 · Paper 1, Physics question 4