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(D) Given: Inductance $L = 5 H$,resistance $R = 10 \Omega$,and emf $V = 15 V$.The time constant $	au_L$ for an $LR$ circuit is given by:$	au_L = \fr

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$\frac{e^2}{e^2-1}$

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(D) Given: Inductance $L = 5 H$,resistance $R = 10 \Omega$,and emf $V = 15 V$.The time constant $	au_L$ for an $LR$ circuit is given by:$	au_L = \frac{L}{R} = \frac{5}{10} = 0.5 s$The current $i(t)$ in the $LR$ circuit at any time $t$ is given by:$i(t) = i_0(1 - e^{-t/	au_L})$where $i_0 = \frac{V}{R}$ is the maximum current at $t = \infty$.We need to find the ratio of the current at $t = \infty$ $(i_0)$ to the current at $t = 1 s$ $(i(1))$:$\frac{i_0}{i(1)} = \frac{i_0}{i_0(1 - e^{-1/0.5})} = \frac{1}{1 - e^{-2}}$Simplifying the expression:$\frac{1}{1 - \frac{1}{e^2}} = \frac{1}{\frac{e^2 - 1}{e^2}} = \frac{e^2}{e^2 - 1}$Thus,the required ratio is $\frac{e^2}{e^2 - 1}$.

An emf of $15 V$ is applied to a circuit containing $5 H$ inductance and $10 \Omega$ resistance. The ratio of currents at time $t=\infty$ and $t=1 s$ after closing the switch is:

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