In the circuit shown below,the key K is close | vedclass

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(C) At $t = 0$,the inductor $L$ acts as an open circuit. The current flows only through resistor $R_2 = 10 \ \Omega$. Therefore,the current $I_0 = V /

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$1 \ A$ at $t = 0$ and $3 \ A$ at $t = \infty$

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(C) At $t = 0$,the inductor $L$ acts as an open circuit. The current flows only through resistor $R_2 = 10 \ \Omega$. Therefore,the current $I_0 = V / R_2 = 10 \ V / 10 \ \Omega = 1 \ A$.At $t = \infty$,the inductor $L$ acts as a short circuit. The circuit consists of two resistors $R_1 = 5 \ \Omega$ and $R_2 = 10 \ \Omega$ in parallel. The equivalent resistance $R_{eq} = (R_1 	imes R_2) / (R_1 + R_2) = (5 	imes 10) / (5 + 10) = 50 / 15 = 10 / 3 \ \Omega$.The current $I_{\infty} = V / R_{eq} = 10 / (10 / 3) = 3 \ A$.

In the circuit shown below,the key $K$ is closed at $t = 0$. The current through the battery is

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