A small signal voltage V(t) = V0 t is applied across an ideal capacitor C
- Current I(t) is in phase with voltage V(t).
- Current I(t) leads voltage V(t) by 180
- Current I(t) lags voltage V(t) by 90 .
- Over a full cycle the capacitor C does not consume any energy from the voltage source.
Answer
(4) Over a full cycle the capacitor C does not consume any energy from the voltage source.
Solution
For an ideal capacitor: I(t) = C dVdt = C V0 t = I0 ( t + 2 )
The current leads the voltage by 90 , so options (1), (2) and (3) are wrong.
Average power: P = Vrms Irms = Vrms Irms 90 = 0
Over a full cycle the capacitor does not consume any energy from the source (energy stored in one quarter cycle is returned in the next).
Answer: (4)