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

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(C) $1$. When key $K_1$ is closed for a long time,the inductor acts as a short circuit. The steady-state current $i_0$ in the circuit is given by $i_0

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$0.67$

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(C) $1$. When key $K_1$ is closed for a long time,the inductor acts as a short circuit. The steady-state current $i_0$ in the circuit is given by $i_0 = \frac{V}{R} = \frac{15}{0.15 	imes 10^3} = 0.1 \; A$.$2$. At $t = 0$,$K_1$ 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.$3$. The current at time $t$ is given by $i = i_0 e^{-\frac{Rt}{L}}$.$4$. Substituting the values: $i = 0.1 	imes e^{-\frac{0.15 	imes 10^3 	imes 10^{-3}}{0.03}} = 0.1 	imes e^{-\frac{0.15}{0.03}} = 0.1 	imes e^{-5}$.$5$. Given $e^5 \cong 150$,then $e^{-5} = \frac{1}{150}$.$6$. Therefore,$i = 0.1 	imes \frac{1}{150} = \frac{0.1}{150} \; A = \frac{0.1}{150} 	imes 1000 \; mA = \frac{100}{150} \; mA = \frac{2}{3} \; mA \approx 0.67 \; mA$.

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