A 20 Henry inductor coil is connected to a 10 ohm resistance in… | JEE Main 2019 (8 Apr, Shift 1)

Q31JEE Main 2019 · 8 APR · SHIFT 1Difficult

Combination of Inductors, LR & LC Circuits

A 20 Henry inductor coil is connected to a 10 ohm resistance in series as shown in figure. The time at which rate of dissipation of energy (joule's heat) across resistance is equal to the rate at which magnetic energy is stored in the inductor is:

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Answer: (c)

Condition: $i^2R = Li\frac{di}{dt}$ , i.e. $iR = L\frac{di}{dt}$ (voltage across $R$ = voltage across $L$ ).
$iR = E\left(1 - e^{-t/\tau}\right)$ and $L\frac{di}{dt} = Ee^{-t/\tau}$ , so $e^{-t/\tau} = \frac{1}{2}$ .
$t = \tau\ln 2$ with $\tau = L/R = 20/10 = 2\text{ s}$
So the answer is (c) $2\ln 2$ s.

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Combination of Inductors, LR & LC Circuits question from JEE Main 2019 (8 Apr, Shift 1)

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