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Q.

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

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a

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

b

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

c

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

d

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

answer is D.

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Detailed Solution

$\begin{array}{l} i = i_{0} \{1 - e^{- Rt / L}\} \\ U = \frac{1}{2} {Li}^{2} \Rightarrow U = U_{0} {(1 - e^{- Rt / L})}^{2} \\ \Rightarrow {(1 - e^{- R t / L})}^{2} = \frac{1}{n} \Rightarrow 1 - e^{- Rt / L} = \frac{1}{\sqrt{n}} \\ \Rightarrow e^{Rt / L} = \frac{\sqrt{n}}{\sqrt{n} - 1} \Rightarrow t = \frac{L}{R} \ln (\frac{\sqrt{n}}{\sqrt{n} - 1}) \end{array}$

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