What is the average power consumed \[cycl{e^{ - 1}}\] in an ideal capacitor?

Hint: The definition of an ideal capacitor is a fully reactive device with no resistive effect. The perfect capacitor is unaffected by atmospheric conditions, and it has a high level of thermal stability. Because of the contact resistance or can in the arcs of commutation for inductance and internal loss for load capacitor), it consumes very little power. When the voltage or current increases during a half cycle of alternating current, the energy is used and then recovered when the voltage or current decreases.



 Complete step by step solution:
Both the current and voltage waveforms are positive in value between the angles of ${0^ \circ }$ and ${90^ \circ }$ in the positive half of the voltage waveform, resulting in positive power consumption. The capacitor current is negative between ${90^ \circ }$ and ${180^ \circ }$ , yet the supply voltage is still positive. As a result, the volt-ampere product produces a negative power since a negative multiplied by a positive equals a negative. The coil is returning stored electrical energy to the source with this negative power.
Then, throughout one entire cycle of the voltage waveform, we have two identical positive and negative power pulses with $0$ average value in the purely capacitive circuit. Then, exactly like in the purely inductive circuit, we have two identical positive and negative power pulses with $0$ average value over one full cycle of the voltage waveform.

Note: In a completely inductive or purely capacitive circuit with reactance, $(X)$ the current will trail or lag the voltage by exactly ${90^ \circ }$ (the phase angle), storing and returning power to the source. As a result, the average power computed for one whole periodic cycle will be $0$.