In the circuits (a) and (b) switches S1 and S2 are closed at t = 0… | JEE Main 2015 (10 Apr, Shift 1)

Q24JEE Main 2015 · 10 APR · SHIFT 1Medium

Inductors & DC Transients (Mixed)

In the circuits (a) and (b) switches S1 and S2 are closed at t = 0 and are kept closed for a long time. The variation of currents in the two circuits for t $\geq$ 0 are roughly shown by (figures are schematic and not drawn to scale):

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Answer: (a)

Circuit (a), R-C: the uncharged capacitor acts as a short at $t = 0$ , so $i = \frac{E}{R}$ at first and then decays as $\frac{E}{R}e^{-t/RC}$ to zero.
Circuit (b), L-R: the inductor stops a sudden current, so $i = 0$ at $t = 0$ and then grows as $\frac{E}{R}(1 - e^{-Rt/L})$ to $\frac{E}{R}$ .
So the answer is (a): curve (a) decays from $E/R$ , curve (b) rises to $E/R$ .

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Inductors & DC Transients (Mixed) question from JEE Main 2015 (10 Apr, Shift 1)

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