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 $\frac{I_{\max }}{\sqrt{n}}$
$\frac{I_{\max }}{\sqrt{n}}=I_{\max }\left(1-e^{-Rt/L}\right)$
$e^{-Rt/L}=1-\frac{1}{\sqrt{n}}=\frac{\sqrt{n}-1}{\sqrt{n}}$
$-\frac{Rt}{L}=\ln \left(\frac{\sqrt{n}-1}{\sqrt{n}}\right)$
$t=\frac{L}{R}\ln \left(\frac{\sqrt{n}}{\sqrt{n}-1}\right)$