JEE PYQ: Electromagnetic Induction Question 15
Question 15 - 2019 (08 Apr Shift 1)
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:
(1) $\frac{2}{\ln 2}$
(2) $\frac{1}{2}\ln 2$
(3) $2\ln 2$
(4) $\ln 2$
Show Answer
Answer: (3)
Solution
$i^2 R = Li\frac{di}{dt}$. For an LR circuit, $i = \frac{E}{R}(1 - e^{-Rt/L})$. Rate of energy dissipation in $R$ equals rate of energy storage in $L$ when $i = \frac{E}{2R}$, i.e., $e^{-Rt/L} = \frac{1}{2}$, giving $t = \frac{L}{R}\ln 2 = \frac{20}{10}\ln 2 = 2\ln 2$.
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