An e.m.f. of 15 V is applied in a circuit co | vedclass

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(B) The current in an $LR$ circuit at time $t$ is given by $I(t) = I_0(1 - e^{-Rt/L})$,where $I_0 = \frac{V}{R}$ is the steady-state current.Given $V

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

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(B) The current in an $LR$ circuit at time $t$ is given by $I(t) = I_0(1 - e^{-Rt/L})$,where $I_0 = \frac{V}{R}$ is the steady-state current.Given $V = 15 \ V$,$L = 5 \ H$,and $R = 10 \ \Omega$.Steady-state current $I_{\infty} = \frac{15}{10} = 1.5 \ A$.At $t = 1 \ s$,$I(1) = 1.5(1 - e^{-10(1)/5}) = 1.5(1 - e^{-2})$.The ratio of currents is $\frac{I_{\infty}}{I(1)} = \frac{1.5}{1.5(1 - e^{-2})} = \frac{1}{1 - e^{-2}} = \frac{1}{1 - 1/e^2} = \frac{e^2}{e^2 - 1}$.

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

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