Question 125 of 190

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 $\left(\dfrac{1}{n}\right)$ times of its maximum value, is :

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

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Answer: (A) $\dfrac{L}{R}\ln \left(\dfrac{\sqrt{n}}{\sqrt{n}-1}\right)$

Potential energy stored in inductor is given by $U=\dfrac{1}{2}{L}^{2}$
$\cup \propto {l}^{2}$
$\dfrac{U}{{U}_{0}}={\left(\dfrac{1}{{I}_{0}}\right)}^{2}$
$\dfrac{1}{n}={\left(\dfrac{1}{{I}_{0}}\right)}^{2}$
$\dfrac{1}{{I}_{0}}=1-{e}^{-RT/L}=\dfrac{1}{\sqrt{n}}$
$t=\dfrac{L}{R}\ln \dfrac{\sqrt{n}}{\sqrt{n}-1}$