A capacitor is connected to a switch, a battery and a bulb as shown in the diagram - Leaving Cert Physics - Question d - 2012

The bulb lights when the switch is in position B because the capacitor discharges through the circuit. At this point, the stored energy in the capacitor is released, causing current to flow momentarily, which makes the bulb illuminate.

The bulb lights only briefly because the capacitor discharges quickly. As it releases its stored energy, the potential difference decreases rapidly, leading to a transient current that diminishes until the capacitor is discharged.

To calculate the charge (Q) on the capacitor, we use the formula:

$Q = C imes V$

Given that the capacitance (C) is 200 µF (which is 200 × 10^{-6} F) and the voltage (V) is 6 V, we have:

$Q = 200 imes 10^{-6} imes 6 = 1.2 imes 10^{-3} ext{ C}$

Thus, the charge on the capacitor is 1.2 mC.

Capacitors can be used to store energy for applications such as radio tuning, filtering signals, or providing a quick discharge current for electronic circuits.