In the circuit shown, the switch S1 is closed at time t = 0 and the… | JEE Main 2019 (11 Jan, Shift 1)

Q21JEE Main 2019 · 11 JAN · SHIFT 1Difficult

Inductors & DC Transients (Mixed)

In the circuit shown,

the switch S1 is closed at time t = 0 and the switch S2 is kept open. At some later time(t0), the switch S1 is opened and S2 is closed. The behavior of the current I as a function of time 't' is given by:

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Answer: (a)

S1 closed: the current grows as $I = I_0(1 - e^{-Rt/L})$ , a rising curve.
At $t_0$ , S1 opens and S2 closes: L and R form a closed loop, so the current cannot jump; it decays exponentially as $I(t_0)e^{-R(t-t_0)/L}$ .
Only graph (a) shows a smooth rise followed by an exponential decay (the others decay linearly or rise linearly).
So the answer is (a).

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