Question:
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:
Correct Option: 1
Solution:
(1) Potential energy stored in the inductor
$U=\frac{1}{2} L I^{2}$
During growth of current,
$i=I_{\max }\left(1-e^{-R t / L}\right)$
For $U$ to be $\frac{U_{\max }}{n} ; i$ has to be $\frac{I_{\max }}{\sqrt{n}}$
$\therefore \quad \frac{I_{\max }}{\sqrt{n}}=I_{\max }\left(1-e^{-R t / L}\right)$
$\Rightarrow e^{-R t / L}=1-\frac{1}{\sqrt{n}}=\frac{\sqrt{n}-1}{\sqrt{n}}$
$\Rightarrow-\frac{R t}{L}=\ln \left(\frac{\sqrt{n}-1}{\sqrt{n}}\right)$
$\Rightarrow t=\frac{L}{R} \ln \left(\frac{\sqrt{n}}{\sqrt{n}-1}\right)$