Real (true) power in an ideal capacitor – Is the average power zero? Assume a sinusoidal steady-state source and an ideal lossless capacitor.

Introduction / Context:
Reactive elements exchange energy with the source but ideally do not dissipate it. Distinguishing between instantaneous power and average (true) power is vital for power factor correction and reactive power management.

Given Data / Assumptions:

• Ideal capacitor with zero loss (no ESR or dielectric loss).
• Sinusoidal voltage and current; current leads voltage by 90°.
• Average power is defined over a complete cycle.

Concept / Approach:

Instantaneous power p(t) = v(t) * i(t). For a capacitor, i(t) is 90° ahead of v(t), so their product is a sinusoid with equal positive and negative halves over a cycle. The net average (true) power is zero, although reactive power Q is nonzero.

Step-by-Step Solution:

Verification / Alternative check:

Phasor power: P = VI cos(φ). For a capacitor, φ = −90°, so cos(φ) = 0, yielding P = 0 while reactive power Q = ± VI sin(φ) is nonzero.

Why Other Options Are Wrong:

Frequency (50/60 Hz) does not change the ideal result. ESR values matter only in nonideal capacitors; any loss introduces a small positive real power, but the ideal model remains zero.

Common Pitfalls:

Confusing reactive power with real power; assuming “current flows” means “power consumed.” Reactive elements store and return energy without net dissipation in the ideal case.

Final Answer:

True.