Alternating Current: Conceptual Doubts Cleared (1)

easy 3 min read

Question

In a purely capacitive AC circuit, the average power dissipated over one cycle is zero. Yet, the bulb in our home glows on AC and consumes real power. Where does this contradiction come from? Students mix up “purely capacitive” with “real circuits” — let’s clear it up.

Solution — Step by Step

In a pure capacitor, voltage and current are 90°90° out of phase. Instantaneous power P(t)=V(t)I(t)P(t) = V(t)I(t) oscillates between positive and negative values symmetrically. Over one full cycle:

Pavg=VrmsIrmscosϕ=VrmsIrmscos90°=0P_{\text{avg}} = V_{rms} \cdot I_{rms} \cdot \cos\phi = V_{rms} I_{rms} \cos 90° = 0

So a pure capacitor does not dissipate energy. It stores energy in its electric field during half a cycle and returns it to the source during the other half.

A bulb is essentially a resistor (its filament). For a pure resistor, voltage and current are in phase, ϕ=0°\phi = 0°, so cosϕ=1\cos\phi = 1 and Pavg=VrmsIrmsP_{\text{avg}} = V_{rms} I_{rms} — fully dissipated as heat and light.

A real AC home circuit is a mix of resistive (bulbs, heaters) and reactive (capacitors in fans, inductors in motors) loads. The reactive parts store and return energy without consuming it; the resistive parts consume it. The “power factor” cosϕ\cos\phi measures what fraction of the apparent power VrmsIrmsV_{rms}I_{rms} is actually consumed.

There’s no contradiction — pure capacitor zero average power is a textbook idealization. Real bulbs are resistors, not capacitors, so they consume.

Why This Works

The phase difference between VV and II is the key. In a resistor, VV and II rise and fall together, so power is always non-negative. In a capacitor, when VV is high, II is zero (capacitor fully charged), and when V=0V = 0, II is at maximum (capacitor discharging through the source). The average over a full cycle cancels.

This is why power companies hate inductive loads — they draw RMS current that doesn’t translate to real power, increasing transmission losses without delivering useful energy.

For JEE Main, remember the three special cases: pure RRcosϕ=1\cos\phi = 1, pure CC or LLcosϕ=0\cos\phi = 0, series LCRLCR at resonance → cosϕ=1\cos\phi = 1. Power factor questions almost always test these limits.

Alternative Method

Energy approach: in one cycle, the capacitor’s electric field energy goes from 0 → 12CVmax2\frac{1}{2}CV_{\max}^2 → 0 → 12CVmax2\frac{1}{2}CV_{\max}^2 → 0. Net energy delivered to the capacitor over one cycle = 0. Same conclusion as the phasor method.

Common Mistake

Students sometimes write P=Irms2RP = I_{rms}^2 R and then substitute capacitive reactance XCX_C for RR — that gives a non-zero “power”, but XCX_C is not a resistance, it’s a reactance. Reactive elements don’t dissipate. The power formula for AC is P=VrmsIrmscosϕP = V_{rms}I_{rms}\cos\phi, full stop.

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