(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 the capacitance of the capacitor is gradually reduced?
Answer: (i) Average power =0 because the current leads the voltage by π/2. (ii) The brightness decreases.
Solution
(i) For an ideal capacitor the current leads the voltage by 2π, so the power factor is cos2π=0. The average power over a cycle is
Pavg=VrmsIrmscosϕ=VrmsIrmscos2π=0
Energy is alternately stored and returned by the capacitor, so no net power is consumed. (ii) The capacitive reactance is XC=ωC1. Reducing C increases XC, which increases the circuit impedance and lowers the current. With less current through it, the bulb grows dimmer.
CBSE Class 12 Board · 2016 · All India · Subjective