Q 12-06-126JEE MainJEE Main 2020 (4 Sep, Shift 2)Medium
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 $\left(\dfrac1n\right)$ times of its maximum value, is:
Answer: (A) $\dfrac LR\ln\left(\dfrac{\sqrt n}{\sqrt n - 1}\right)$
$U \propto i^{2}$, so $U = \dfrac{U_{max}}{n}$ when $i = \dfrac{i_0}{\sqrt n}$.
$$i_0\left(1 - e^{-Rt/L}\right) = \frac{i_0}{\sqrt n} \Rightarrow e^{-Rt/L} = \frac{\sqrt n - 1}{\sqrt n} \Rightarrow t = \frac LR\ln\left(\frac{\sqrt n}{\sqrt n - 1}\right)$$
Solution by Sreeraj P, M.Sc Physics