A light bulb and a parallel-plate capacitor with air between the plates are connected in series to an ac source. What happens to the brightness of the bulb when a dielectric is inserted between the plates of the capacitor? Explain

A capacitor is a device consisting of two conductors in close proximity that are used to store electrical energy. These conductors are insulated from each other. When a capacitor is attached to an AC supply, the resistance produced by it is called capacitive reactance (X_C).

Resistance is measure of opposition to the flow of current in a closed electrical circuit. It is measured in Ohm (Ω).

Impedance is defined as the effective resistance of an electric circuit to the flow of current due to the combined effect of resistance (offered by resistor) and reactance (offered by capacitor and inductor).

Power dissipated in the bulb is given by

$P=I_{rms}{R}^2$

By inserting a dielectric material capacitance of the capacitor will increase, and thus more energy will be stored in the capacitor.

If Z is the impedance of the circuit, then current in the circuit is given by

$I_{rms}=V/Z$

As the capacitance increase, the impedance decreases and since the rms current is inversely proportional to impedance, the current consumed by bulb increases which means the consumed power will increase and thus bulb will be more bright.

Therefore, the brightness of the bulb increases.