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An inductor L, a resistor R, and two identical bulbs B1 and B2 are connected to a battery through a switch S. B1 is in series with the inductor L (whose coil resistance is also R) and B2 is in series with the resistor R, the two branches being in parallel. When switch S is closed, which statement correctly describes what happens?

1. B2 lights up earlier than B1 and finally both bulbs shine equally bright.
2. B1 lights up earlier and finally both bulbs acquire equal brightness.
3. B2 lights up earlier and finally B1 shines brighter than B2.
4. B1 and B2 light up together with equal brightness all the time.

Correct answer: B2 lights up earlier than B1 and finally both bulbs shine equally bright.

Solution

The branch with the inductor (B1) opposes the sudden rise of current, so B1 brightens slowly. The branch with only R (B2) reaches its current immediately, so B2 lights up first. At steady state the inductor's coil resistance is R, making both branches identical in resistance, so both bulbs end up equally bright.

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