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A 2 V battery, two resistors of 10 ohm and 20 ohm, and an ideal 10 H inductor are connected as shown. The key K is closed (shorted) at t = 0. Find the current through the battery immediately after closing (t = 0) and a long time later (t -> infinity).

1. 1/15 A, 1/10 A
2. 1/10 A, 1/15 A
3. 2/15 A, 1/10 A
4. 1/15 A, 2/25 A

Correct answer: 1/15 A, 1/10 A

Solution

At t = 0 the inductor branch is open, so the battery current flows through 20 ohm and 10 ohm in series: I0 = 2/(20+10) = 1/15 A. At t -> infinity the inductor shorts the 10 ohm resistor, leaving only the 20 ohm: I_inf = 2/20 = 1/10 A.

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