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 :

• A. $\frac{L}{R} \ln \left(\frac{\sqrt{n}-1}{\sqrt{n}}\right)$
• B. $\frac{L}{R} \ln \left(\frac{\sqrt{n}}{\sqrt{n}+1}\right)$
• C. $\frac{L}{R} \ln \left(\frac{\sqrt{n}}{\sqrt{n}-1}\right)$
• D. $\frac{L}{R} \ln \left(\frac{\sqrt{n}+1}{\sqrt{n}-1}\right)$

Answer 1

Answer: (C)

$U _{\max }=\frac{1}{2} L I _{\text{ max }}^{2}$
$i = I _{\max }\left(1- e ^{- Rt / L }\right)$
For $U$ to be $\frac{U_{\max }}{n} $ ; i has to be $\frac{I_{\max }}{\sqrt{n}}$
$\frac{ I _{\max }}{\sqrt{ n }}= I _{\max }\left(1- e ^{- Rt / L }\right)$
$e ^{- Rt / L }=1-\frac{1}{\sqrt{ n }}=\frac{\sqrt{ n }-1}{\sqrt{ n }}$
$-\frac{ Rt }{ L }=\ln \left(\frac{\sqrt{ n }-1}{\sqrt{ n }}\right)$
$t =\frac{ L }{ R } \ln \left(\frac{\sqrt{ n }}{\sqrt{ n }-1}\right)$