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

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(B) The current in an $R-L$ circuit at any time $t$ is given by $i(t) = i_0(1 - e^{-Rt/L})$,where $i_0 = E/R$ is the steady-state current.Given: $E =

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

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(B) The current in an $R-L$ circuit at any time $t$ is given by $i(t) = i_0(1 - e^{-Rt/L})$,where $i_0 = E/R$ is the steady-state current.Given: $E = 15 \ V$,$L = 5 \ H$,$R = 10 \ \Omega$.Steady-state current $(t = \infty)$: $i_{\infty} = E/R = 15/10 = 1.5 \ A$.At $t = 1 \ s$: $i_1 = i_0(1 - e^{-R(1)/L}) = 1.5(1 - e^{-10/5}) = 1.5(1 - e^{-2})$.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 $t = \infty$ and at $t = 1 \ s$ is:

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