# 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: Option: 1 Option: 2 Option: 3 Option: 4

\begin{aligned} &\frac{1}{2} \mathrm{Li}^{2}=\frac{1}{\mathrm{n}} \frac{1}{2} \mathrm{Li}_{0}^{2}\\ &\Rightarrow \quad \mathbf{i}=\frac{\mathbf{i}_{0}}{\sqrt{\mathbf{n}}}\\ &\text { During growth of current, }\\ &\mathbf{i}=\mathbf{i}_{0}\left(1-\mathbf{e}^{-t / \tau}\right) \quad\left(\tau=\frac{\mathbf{L}}{\mathbf{R}}\right)\\ &e^{t / \tau}=\frac{\sqrt{n}}{\sqrt{n}-1}\\ &t=\frac{L}{R} \ln \left(\frac{\sqrt{n}}{\sqrt{n}-1}\right) \end{aligned}

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