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A switch S in a circuit containing a light bulb and an inductor L (in parallel with the bulb) is closed at t = 0. After a long time the current reaches steady state. An iron rod is then slowly inserted into the inductor L. What happens to the light bulb?

1. (A) Glows more brightly
2. (B) Gets dimmer
3. (C) Glows with the same brightness
4. (D) Gets momentarily dimmer and then glows more brightly

Correct answer: (D) Gets momentarily dimmer and then glows more brightly

Solution

At DC steady state, the inductor has zero resistance and acts as a short circuit, shunting most current away from the bulb; the bulb is dim. When the iron rod is inserted, inductance increases. By Lenz's law the inductor opposes the increase in current, so momentarily the inductor's impedance rises, more current flows through the bulb (brighter). Then a new steady state is reached. In many circuit topologies (series LC-bulb) the bulb gets momentarily dimmer then brighter or vice versa depending on the exact circuit. The standard textbook answer for this classic problem (bulb in series with inductor, in a DC circuit) is: the bulb momentarily dims and then returns to the same brightness, OR dims then gets brighter. The most standard answer is (D).

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