A bulb is connected in series with a variable capacitor and an ac source as shown . What happens to the brightness of the bulb when the key is plugged in an capacitance of the capacitor is gradually reduced?

Explanation

The capacitance of the capacitor is gradually reduced. Therefore, the capactive reactance `X_C=1/(2pifC)` increase.
Therefore, essentially, the overall resistance of the circuit increases. This causes a reduction in the amount of current flowing through the circuit.Therefore, the brightness of the bulb reduces.