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

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(A) At $t = 0$,the inductor $L$ acts as an open circuit because it opposes any change in current. Therefore,no current flows through the branch contai

Vedclass · Accepted Answer

$\frac{V}{R_2}$ at $t = 0$ and $\frac{V(R_1 + R_2)}{R_1 R_2}$ at $t = \infty$

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(A) At $t = 0$,the inductor $L$ acts as an open circuit because it opposes any change in current. Therefore,no current flows through the branch containing $L$ and $R_1$.Thus,the current through the battery is $I = \frac{V}{R_2}$.At $t = \infty$,the inductor $L$ acts as a short circuit (ideal inductor with zero resistance). The resistors $R_1$ and $R_2$ are now in parallel.The effective resistance of the circuit is $R_{eff} = \frac{R_1 R_2}{R_1 + R_2}$.Therefore,the current through the battery is $I = \frac{V}{R_{eff}} = \frac{V(R_1 + R_2)}{R_1 R_2}$.

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

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