(i) 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.
(ii) 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 and capacitance of the capacitor is gradually reduced?

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Solution

(i) Power dissipated in a circuit is given by
$P = V_{r m s} I_{r m s} c o s ϕ$
where $c o s ϕ = \frac{R}{Z}$
For a ideal capacitor, $R = 0$
Hence $c o s ϕ = 0$
$⟹ P = 0$

(ii) When AC source is connected, the reactance of capacitor is $X_{C} = \frac{1}{ω C}$ .
On reducing $C$ , the reactance increases, and lesser current flows through the bulb and brightness reduces.

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