Discuss power in an $AC$ circuit containing only a capacitor.

(N/A) The instantaneous power supplied to the capacitor is given by:
$p = v \cdot i$
Let the voltage across the capacitor be $v = V_m \sin(\omega t)$.
The current in a purely capacitive circuit leads the voltage by $\pi/2$,so $i = I_m \sin(\omega t + \pi/2) = I_m \cos(\omega t)$.
Thus,the instantaneous power is:
$p = (V_m \sin(\omega t)) \cdot (I_m \cos(\omega t))$
$p = V_m I_m \sin(\omega t) \cos(\omega t)$
Using the identity $\sin(2\theta) = 2 \sin \theta \cos \theta$,we get:
$p = \frac{V_m I_m}{2} \sin(2\omega t)$
The average power $P$ over a complete cycle is the average of the instantaneous power over one period $T = 2\pi/\omega$:
$P = \langle p \rangle = \frac{V_m I_m}{2} \langle \sin(2\omega t) \rangle$
Since the average value of $\sin(2\omega t)$ over a complete cycle is zero,the average power dissipated in a purely capacitive circuit is:
$P = 0$
This indicates that a capacitor does not consume any real power in an $AC$ circuit; it only stores and releases energy.