The circuit below contains an ideal battery (epsilon), three light bulbs, and an

Question

The circuit below contains an ideal battery (epsilon), three light bulbs, and an initially uncharged capacitor. At the instant the switch S is closed, which light bulb is brightest? After a long period of time, which is brightest? At the instant S closed, A & B brightest; after a long time A & B brightest At the instant S closed, A & B brightest; after a long time C brightest At the instant S closed, C brightest; after a long time C brightest At the instant S closed, C brightest; after a long time A & B brightest

Explanation / Answer

At the instant S is closed, C is brightest; after a long time, A and B are brightest.

EXPLANATION

As the capacitor charges it gains an emf across it. After a period of time the capacitor will have the E volts across it and that leaves zero volts across bulb C

A and B will each have E/2 volts across them.
at first when the switch is closed the capacitor has zero volts across it so that leaves E volts across bulb C and at that instant it will be bright.
It gradually diminishes in brightness but the other two remain with E volts at all times.so after some time, A and B will be brightest

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