A series L-R circuit is connected to a battery of emf V. If the current is switched on at t = 0, then the time at which the energy stored in inductor reaches $(\frac{1}{n})$ times of it’s maximum value is

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$\frac{L}{R} \ln (\frac{\sqrt{n}}{\sqrt{n} + 1})$

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

$\frac{L}{R} \ln (\frac{\sqrt{n} + 1}{\sqrt{n} - 1})$

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

answer is C.

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Detailed Solution

Maximum energy stored in an inductor current $u_{\max} = \frac{1}{2} L i_{\max}^{2}$

For U to be $\frac{U_{\max}}{n};$ i has to be $\frac{I_{\max}}{\sqrt{n}}$

$i = i_{\max} (1 - e^{- \frac{R t}{L}})$

$\frac{I_{\max}}{\sqrt{n}} = i_{\max} (1 - e^{- \frac{R t}{L}})$

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

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