A small signal voltage V(t) = V₀ sin ωt is applied across an ideal capacitor C:
Answer: (B) Over a full cycle the capacitor C does not consume any energy from the voltage source. Answer: (B) Over a full cycle the capacitor C does not consume any energy from the voltage source Solution: In a purely capacitive AC circuit the current leads the voltage by 90° (not lags, not 180°, not in phase).
- A.Current I(t), lags voltage V(t) by 90°
- B.Over a full cycle the capacitor C does not consume any energy from the voltage source✓
- C.Current I(t) is in phase with voltage V(t).
- D.Current I(t) leads voltage V(t) by 180°
Correct Answer
(B) Over a full cycle the capacitor C does not consume any energy from the voltage source
Solution & Explanation
Answer: (B) Over a full cycle the capacitor C does not consume any energy from the voltage source Solution: In a purely capacitive AC circuit the current leads the voltage by 90° (not lags, not 180°, not in phase). Since the phase difference is 90°, the average power = V_rms·I_rms·cos 90° = 0, so over a full cycle the ideal capacitor consumes no energy (wattless current).
