A bulb and a capacitor are connected in series across an ac supply. A dielectric is then placed between the plates of the capacitor. The glow of the bulb :

Solution

Physics sir · 4 steps

A dielectric raises C, a bigger C means less reactance, so more current flows through the bulb.

1. Dielectric raises C. A dielectric of constant K between the plates makes the capacitance C' = KC. Since K > 1, the capacitance goes up.

   X_C = 1/(ω C), Z = √(R² + X_C²)

2. Reactance falls. The capacitive reactance X_C is inversely proportional to C, so it drops. The bulb's resistance R is unchanged, so the impedance Z of the series circuit decreases.
3. Current rises. With the same supply voltage, Iᵣₘₛ = Vᵣₘₛ/Z increases. The power in the bulb is Iᵣₘₛ² R, so it goes up too.
4. So. The bulb glows brighter: the glow increases, option (c).

The idea to keep. In AC, a capacitor opposes current through X_C = 1/ω C. Anything that increases C lowers that opposition, so more current flows in a series circuit.

Answer from the NTA final answer key. Solution by PrepWiser.