An inductor (L = 100 , mH),a resistor (R = 10 | vedclass

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(A) When the circuit is connected for a long time,the inductor acts as a short circuit,and the steady-state current $I_0$ is given by:$I_0 = \frac{E}{

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$1/e \, A$

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(A) When the circuit is connected for a long time,the inductor acts as a short circuit,and the steady-state current $I_0$ is given by:$I_0 = \frac{E}{R} = \frac{100 \, V}{100 \, \Omega} = 1 \, A$When the battery is disconnected and points $A$ and $B$ are short-circuited,the circuit becomes a decaying $LR$ circuit. The current at time $t$ is given by:$I(t) = I_0 e^{-t/	au}$Here,the time constant $	au$ is:$	au = \frac{L}{R} = \frac{100 	imes 10^{-3} \, H}{100 \, \Omega} = 1 	imes 10^{-3} \, s = 1 \, ms$Given $t = 1 \, ms$,the current is:$I = 1 	imes e^{-1 \, ms / 1 \, ms} = 1 	imes e^{-1} = \frac{1}{e} \, A$

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