An inductor (L = 0.03 H) and a resistor (R = | vedclass

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(C) When $K_1$ is closed for a long time,the inductor acts as a short circuit. The initial current $I_0$ through the inductor is:$I_0 = \frac{V}{R} =

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$0.67$

Vedclass · Answer

(C) When $K_1$ is closed for a long time,the inductor acts as a short circuit. The initial current $I_0$ through the inductor is:$I_0 = \frac{V}{R} = \frac{15 \ V}{0.15 	imes 10^3 \ \Omega} = \frac{15}{150} = 0.1 \ A = 100 \ mA$.When $K_1$ is opened and $K_2$ is closed,the battery is removed from the circuit,and the inductor discharges through the resistor. This is a decaying $LR$ circuit.The current at time $t$ is given by $I(t) = I_0 e^{-Rt/L}$.Given $R = 150 \ \Omega$,$L = 0.03 \ H$,and $t = 1 \ ms = 10^{-3} \ s$.The exponent is $-\frac{Rt}{L} = -\frac{150 	imes 10^{-3}}{0.03} = -\frac{0.15}{0.03} = -5$.Thus,$I(t) = 100 \ mA 	imes e^{-5}$.Using $e^5 \cong 150$,we have $e^{-5} = \frac{1}{e^5} \cong \frac{1}{150}$.$I(1 \ ms) = 100 	imes \frac{1}{150} = \frac{10}{15} = \frac{2}{3} \approx 0.67 \ mA$.

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