Pure capacitor circuit, zero real power

A pure capacitor has capacitance C only (no resistance, no leakage).

Derivation of current

Applied voltage: v = Vm sin ωt. Charge on the plates: q = C v. Current is the rate of flow of charge:

i = dq/dt = C dv/dt = C Vm ω cos ωt

i = (Vm / X_C) sin(ωt + 90°) = Im sin(ωt + 90°)

where X_C = 1/(ωC) = 1/(2πfC) is the capacitive reactance (ohm) and Im = Vm/X_C. So the current leads the voltage by 90°.

Waveforms and phasor diagram

       i        v
     .--.     .--.
 ___/    \___/    \___ ωt
  i peaks a quarter cycle before v

        I (leads by 90°)
        ▲
        │
 ───────┴──────────► V (reference)

Power

p = v i = Vm sin ωt × Im cos ωt = (Vm Im / 2) sin 2ωt

Average power over one cycle:

P = (1/2π) ∫₀^{2π} (Vm Im/2) sin 2ωt d(ωt) = 0

Also P = V I cos 90° = 0.

Graphical proof

  p
   │  .-.      .-.
 ──┼─/───\────/───\── ωt
   │       \_/      \_/
 positive loops = negative loops

The power wave has twice the supply frequency. During the quarter-cycles when |v| rises, p is positive and the capacitor charges, storing energy ½CV². In the next quarter-cycles, p is negative and the capacitor returns the same energy to the source. The positive and negative areas are equal, so net energy used is zero. A pure capacitor consumes no real power; it only exchanges reactive power.