In the picture A and B are equivalent bulbs C is a capacitor and V is the voltag

Question

In the picture A and B are equivalent bulbs C is a capacitor and V is the voltage. Describe the behavior of bulb A from just after the switch is closed until a long time later? Describe the behavior of bulb B from just after the switch is closed until a long time later? A long time after the switch is closed, how does the brightness of bulb B compare to that of bulb A? I am confused because technically A and B are in a loop with the voltage and are not directly dependent on the capacitor. Does that mean the won't shut off after a long period of time? Please help :) Describe the behavior of bulb B from just after the switch is closed until a long time later? A long time after the switch is closed, how does the brightness of bulb B compare to that of bulb A? I am confused because technically A and B are in a loop with the voltage and are not directly dependent on the capacitor. Does that mean the won't shut off after a long period of time? Please help :)

Explanation / Answer

Just after the switch is closed, the capacitor is like a common wire with zero resistance, so no current is through A, but a current is through B and C. As time goes by, the current through B and C decreases exponentially. The voltage on C increases, it equals to the voltage on A, so there is a current through A. The voltage on B decreases, so the current through B decreases. After a long time, no current through C, the current through A and B is the same. Therefore, from just after the switch is closed until a long time later, the bulb A starts dark and is getting brighter; the bulb B starts very bright and is getting dimmer until same brightness.-- (if A and B are same bulb)

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