The average power associated with an ideal capacitor of capacitance 5…

The correct answer is Option (1) → zero

In an ideal capacitor the current leads the voltage by \( \pi/2 \), so the instantaneous power \( p = iv \) is positive for a quarter of a cycle and negative for the next quarter. The capacitor takes energy from the source and returns it, so the average power over a complete cycle is zero.

    \( v = v_m\sin\omega t,\quad i = i_m\cos\omega t \)

    \( p = iv = i_mv_m\sin\omega t\cos\omega t \)

    \( \displaystyle p = \frac{i_mv_m}{2}\sin2\omega t \)

    \( \displaystyle \langle p\rangle = \frac{i_mv_m}{2}\langle\sin2\omega t\rangle = 0 \)

The values of the capacitance and of the emf play no part.

NCERT (2026-27):
AC Voltage Applied to a Capacitor – “… The instantaneous power supplied to the capacitor is

    \( \displaystyle p_c = iv = i_m\cos(\omega t)\,v_m\sin(\omega t) \)

    \( \displaystyle = i_mv_m\cos(\omega t)\sin(\omega t) \)

    \( \displaystyle = \frac{i_mv_m}{2}\sin(2\omega t) \qquad (7.19) \)

So, as in the case of an inductor, the average power

    \( \displaystyle P_C = \left\langle\frac{i_mv_m}{2}\sin(2\omega t)\right\rangle = \frac{i_mv_m}{2}\langle\sin(2\omega t)\rangle = 0 \)

since \( \langle\sin(2\omega t)\rangle = 0 \) over a complete cycle.”