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In a circuit containing a battery of EMF V, a resistor R1 in series, and an inductor in parallel with resistor R2 (standard L-R configuration), the switch K is closed at t = 0. Find the current drawn from the battery just after closing (t = 0) and a long time later (t = infinity).

1. V*(R1 + R2)/(R1*R2) at t = 0 and V/R2 at t = infinity
2. V*R1*R2/sqrt(R1² + R2²) at t = 0 and V/R2 at t = infinity
3. V/R2 at t = 0 and V*(R1 + R2)/(R1*R2) at t = infinity
4. V/R2 at t = 0 and V*R1*R2/sqrt(R1² + R2²) at t = infinity

Correct answer: V/R2 at t = 0 and V*(R1 + R2)/(R1*R2) at t = infinity

Solution

At the instant of closing, the inductor opposes any sudden current and behaves as an open branch, so the battery current flows through the R2 path only: I(0) = V/R2. After a long time the inductor behaves as a short circuit, bypassing its parallel resistor so that R1 and R2 act in parallel across the source, giving total current I(infinity) = V/(R1||R2) = V*(R1 + R2)/(R1*R2).

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