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

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Question

(A) When the circuit is connected for a long time,the inductor acts as a short circuit (ideal inductor). The steady-state current $I_0$ is given by $I

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$\frac{1}{e} \, A$

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(A) When the circuit is connected for a long time,the inductor acts as a short circuit (ideal inductor). 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 an $LR$ decay circuit. The current at time $t$ is given by $I(t) = I_0 e^{-t/	au}$,where $	au$ is the time constant.The time constant $	au = \frac{L}{R} = \frac{100 	imes 10^{-3} \, H}{100 \, \Omega} = 10^{-3} \, s = 1 \, ms$.Given $t = 1 \, ms = 10^{-3} \, s$,the current is $I = 1 	imes e^{-(10^{-3})/(10^{-3})} = 1 	imes e^{-1} = \frac{1}{e} \, A$.

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