What is the average power dissipation in an ideal capacitor in an AC circuit?
(A) $2C{V^2}$
(B) $\dfrac{1}{2}C{V^2}$
(C) Zero
(D) $C{V^2}$

Hint
To solve this problem we can use the formula to calculate average power in AC circuits, where we know that the phase angle between the voltage source and the current in a capacitive circuit is $90^\circ $ .
To solve this sum we just need to know the formula for average power,
$P = {V_{rms}}{I_{rms}}\cos \phi $
where ${V_{rms}}$= the rms voltage from the source
 ${I_{rms}}$= the rms current in the circuit
 $\phi $ = the phase angle between the voltage source and the current.

Complete step by step answer
We know that to calculate the average power in any AC circuit we use the formula$P = {V_{rms}}{I_{rms}}\cos \phi $
In the case of a purely capacitive circuit, that is, a circuit that consists only of an ideal capacitor, the phase angle between the voltage source and the current source is $90^\circ $ .
So by putting the phase angle in the formula,
$P = {V_{rms}}{I_{rms}}\cos 90^\circ $
Now the value of $\cos 90^\circ $ is zero.
So by putting that in the equation,
 $P = {V_{rms}}{I_{rms}} \times 0$
$ \Rightarrow P = 0$
Therefore the correct answer to this problem is option (C); Zero.

Additional Information
In a purely capacitive circuit, the voltage wave is $ - 90^\circ $ out of phase with the current wave. This means that graphically, the current wave leads and the voltage wave lags behind
In the case of a purely inductive circuit the situation is the opposite, that is, the voltage wave leads and the current wave lags.

Note
We should take a proper note that the phase angle in the capacitive circuit is $ - 90^\circ $ and that in an inductive circuit is $ + 90^\circ $. The capacitors have a wide range of uses in electrical circuits as they can store electrical energy when connected to its charging circuit.