A series L-R circuit is connected to a battery of emf V.
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:

  1. (1) $\frac{L}{R} \ln \left(\frac{\sqrt{n}}{\sqrt{n}-1}\right)$

  2. (2) $\frac{L}{R} \ln \left(\frac{\sqrt{n}+1}{\sqrt{n}-1}\right)$

  3. (3) $\frac{L}{R} \ln \left(\frac{\sqrt{n}}{\sqrt{n}+1}\right)$

  4. (4) $\frac{L}{R} \ln \left(\frac{\sqrt{n}-1}{\sqrt{n}}\right)$


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 $\frac{I_{\max }}{\sqrt{n}}$

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

$\Rightarrow e^{-R t / L}=1-\frac{1}{\sqrt{n}}=\frac{\sqrt{n}-1}{\sqrt{n}}$

$\Rightarrow-\frac{R t}{L}=\ln \left(\frac{\sqrt{n}-1}{\sqrt{n}}\right)$

$\Rightarrow t=\frac{L}{R} \ln \left(\frac{\sqrt{n}}{\sqrt{n}-1}\right)$

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