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Question:

When an AC source is connected to an ideal capacitor, show that the average power supplied by the source over a complete cycle is zero. A bulb is connected in series with a variable capacitor and an AC source. What happens to the brightness of the bulb when the key is plugged in and capacitance of the capacitor is gradually reduced?

Solution:

(i) Power dissipated in a circuit is given by P = Vrms Irms cosφ where cosφ = R/Z. For an ideal capacitor, R = 0. Hence cosφ = 0 ⇒ P = 0.
(ii) When AC source is connected, the reactance of capacitor is XC = 1/ωC. On reducing C, the reactance increases, and lesser current flows through the bulb and brightness reduces.