JEE Advanced · 2018 · 20 May 2018 · Paper 1 · Electromagnetic Induction

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 = τ

. Which of the following statements is (are) true?

(A) $I_{max} = \frac{V}{2 R}$
(B) $I_{max} = \frac{V}{4 R}$ ✓ Correct
(C) $τ = \frac{L}{R} ln 2$
(D) $τ = \frac{2 L}{R} ln 2$ ✓ Correct

Answer: (B), (D)

Solution

For the two RL branches:

i_{1} = \frac{V}{R} (1 - e^{- R t / L})

and

i_{2} = \frac{V}{R} (1 - e^{- R t / (2 L)})

. The middle current is

I = i_{1} - i_{2}

. For maximum,

\frac{d I}{d t} = 0

, giving

\frac{d i _{1}}{d t} = \frac{d i _{2}}{d t}

. This yields

2 e^{- R t / L} = e^{- R t / (2 L)}

, so

e^{R t / (2 L)} = 2

, giving

τ = \frac{2 L}{R} ln 2

. At this time,

e^{- R τ / L} = 1/4

, so

I_{max} = \frac{V}{R} (1 - \frac{1}{4}) - \frac{V}{R} (1 - \frac{1}{2}) = \frac{V}{4 R}

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JEE Advanced · 2018 · 20 May 2018 · Paper 1 · Electromagnetic Induction

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