JEE PYQ: Alternating Current Question 23
Question 23 - 2020 (04 Sep Shift 2)
A series $L$-$R$ circuit is connected to a battery of emf $V$. If the circuit is switched on at $t = 0$, then the time at which the energy stored in the inductor reaches $\left(\frac{1}{n}\right)$ times of its maximum value, is:
(a) $\frac{L}{R}\ln\left(\frac{\sqrt{n}}{\sqrt{n} - 1}\right)$ (b) $\frac{L}{R}\ln\left(\frac{\sqrt{n}+1}{\sqrt{n}-1}\right)$ (c) $\frac{L}{R}\ln\left(\frac{\sqrt{n}}{\sqrt{n}+1}\right)$ (d) $\frac{L}{R}\ln\left(\frac{\sqrt{n}-1}{\sqrt{n}}\right)$
Show Answer
Answer: (a)
Solution
$e^{-Rt/L} = \frac{\sqrt{n}-1}{\sqrt{n}}$, so $t = \frac{L}{R}\ln\left(\frac{\sqrt{n}}{\sqrt{n}-1}\right)$.
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