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In the circuit shown, switches S1 and S2 are both closed at t = 0 and current begins to flow. The two batteries have equal emf magnitude V and the polarities shown. Mutual inductance between the inductors is negligible. The current I in the central wire attains its maximum magnitude I_max at t = tau. Which of the following is/are correct?

1. I_max = V/(2R)
2. I_max = V/(4R)
3. tau = (L/R) ln 2
4. tau = (2L/R) ln 2

Correct answer: I_max = V/(2R)

Solution

This is JEE Advanced 2021. Each inductor branch reaches a final current of V/R, but the central wire carries the difference of the contributions and reaches a maximum value I_max = V/(2R) at the instant tau = (L/R) ln 2. Both statement A (I_max = V/(2R)) and statement C (tau = (L/R) ln 2) are the accepted correct answers.

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