A second identical bulb is now added to the circuit as shown

A second identical bulb is now added to the circuit as shown. The capacitor is discharged. i. The switch is now moved to position I. Describe the behavior of bulbs B and C from just after the switch is closed until a long time later. Explain. How does ihc initial brightness of bulb C compare to the initial brightness of bulb A in question i of pan a? Explain your reasoning. A long time after the switch is closed, is the absolute value of the potential difference across the battery? Explain.

Solution

1.

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.

b. As the switch is put ON to position 1, there is a charging up of capacitor. Thus a non zero current flows. As the non zero current flows, both bulbs glow up. Then as the capacitor gets fully charged, the bulb dims off. Then on running it position 2, the discharging of capacitor happens, and bulbs glow. Both fade out with time.

Brightness is less. Simple reason, current withdrawn for the same potential difference is know V/2R <considering both have identical resistance> . Then calculate power by I²R.

E = V(1-e^-t/RC), so as t reaches infinite, E=V, and its almost equal to the potential difference