Let's say we have a normal circuit with a light bulb, with wires and a battery.
When one places a capacitor in this circuit, how is the light bulb able to light up, even when the capacitor prevents the flow of charge? Also, why does it dim and then go out eventually?
Then when the battery is removed from this circuit, how is the light bulb still able to light up? And what is happening when the light bulb dims and goes out in this situation as well?
Best Answer
First, note that the light bulb is essentially just a glorified resistor. As current flows through the filament, Joule heating causes the filament to get hot and emit light.
When one places a capacitor in a circuit containing a light bulb and a battery, the capacitor will initially charge up, and as this charging up is happening, there will be a nonzero current in the circuit, so the light bulb will light up. However, the capacitor will eventually be fully charged at which point the potential between its plates will match the voltage of the battery, and the current in the circuit will drop to zero. This is when the light bulb will dim and then fizzle out.
When the battery is removed from the circuit, there is nothing to maintain the potential difference between the plates, and the capacitor will discharge. As this happens, there will once again be a nonzero current flowing through the circuit, and the bulb will light up. However, the current will steadily decrease as the capacitor discharges and will eventually drop to zero at which point the bulb will go off.