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(A) At $t=0$,the inductor acts as an open circuit because it opposes any change in current. The circuit consists of the $10 \ \Omega$ resistor and $20

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$\frac{1}{15} \ A, \frac{1}{10} \ A$

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(A) At $t=0$,the inductor acts as an open circuit because it opposes any change in current. The circuit consists of the $10 \ \Omega$ resistor and $20 \ \Omega$ resistor in series. The total resistance is $R_{eq} = 10 \ \Omega + 20 \ \Omega = 30 \ \Omega$. The initial current is $I_0 = \frac{E}{R_{eq}} = \frac{2 \ V}{30 \ \Omega} = \frac{1}{15} \ A$.At $t \rightarrow \infty$,the inductor acts as a short circuit (ideal conductor) because the current becomes steady. The $10 \ \Omega$ resistor is bypassed by the inductor. The circuit now only contains the $20 \ \Omega$ resistor in series with the battery. The final current is $I_{\infty} = \frac{E}{20 \ \Omega} = \frac{2 \ V}{20 \ \Omega} = \frac{1}{10} \ A$.

Two resistors of $10 \Omega$ and $20 \Omega$ and an ideal inductor of $10 \ H$ are connected to a $2 \ V$ battery as shown. The key $K$ is closed at time $t=0$. Find the initial $(t=0)$ and final $(t \rightarrow \infty)$ currents through the battery.

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