Concept:In a pure capacitor, current leads voltage by \(\phi = 90^\circ\). Energy stored during charging is completely returned to the source during discharging.
Formula / Reaction:$$P_{\text{avg}} = V_{\text{rms}} I_{\text{rms}} \cos\phi = V_{\text{rms}} I_{\text{rms}} \cos(90^\circ) = 0$$
Solution:- During the first quarter-cycle, the capacitor charges and stores electrostatic energy \(U = \frac{1}{2} C v^2\).
- During the next quarter-cycle, the capacitor discharges and returns this energy back to the circuit.
- The net energy dissipated as heat over any full cycle is exactly zero.
Why other options are incorrect:- Opt_A: Represents true power dissipation in a pure resistor where \(\cos(0^\circ) = 1\).
- Opt_B: Equivalent to \(V_{\text{rms}} I_{\text{rms}}\) for a resistor.
- Opt_D: Represents reactive volt-amperes, not real power dissipation.
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