The instantaneous power supplied to a capacitor is given by
\( P_c = i_m\cos(\omega t)\,v_m\sin(\omega t) \)
The average power over a complete cycle would be
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → Zero
In a purely capacitive circuit the current leads the voltage by \( \pi/2 \), so the product of the two is a sine of twice the angular frequency, which is as often negative as positive.
\( \displaystyle P_c = i_m v_m\sin\omega t\cos\omega t = \frac{i_m v_m}{2}\sin 2\omega t \)
\( \displaystyle \langle P_c\rangle = \frac{i_m v_m}{2}\,\langle\sin 2\omega t\rangle = 0 \)
Physically, the capacitor stores energy in its field during one quarter-cycle and returns all of it to the source in the next, so an ideal capacitor consumes no average power.
NCERT (2026-27):
Summary – “… The current through the capacitor is π/2 ahead of the applied voltage. As in the case of inductor, the average power supplied to a capacitor over one complete cycle is zero.”