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In the circuit shown in the figure, switch $S$ is closed at $t=0$. Then:

1. After a long time interval, the potential difference across the capacitor and inductor will be equal.
2. After a long time interval, the charge on the capacitor will be $EC$.
3. After a long time interval, the current in the inductor will be $E/R$.
4. After a long time interval, the current through the battery will be the same as the current through it initially.

Correct answer: After a long time interval, the current through the battery will be the same as the current through it initially.

Solution

After a long time, the capacitor branch carries no current and the inductor branch offers negligible opposition to steady current. The final current through the battery is determined by the resistive part of the circuit and is the same as the initial current in the given arrangement. Hence the correct statement is that the battery current remains unchanged from its initial value.

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