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Phasor relation in a pure capacitor

Short answer

In a pure capacitor, the current leads the voltage by 90° (π/2). The phasor diagram shows voltage I lagging, with magnitude V/X_C where X_C=1/(ωC). In complex form, V = V∠0°, I = (V/X_C)∠90°.

Describe and illustrate the phasor relationship that exist between the voltage that appears across the terminals of a pure capacitor and the current that flows through it in steady state when the capacitor is excited by a sinusoidal source.

Statement: In a pure capacitor the current leads the voltage by 90° (π/2). Equivalently, the voltage lags the current by 90°.

Description

Let the applied voltage be v = Vm sin ωt. The charge is q = C v, so the current is

i = dq/dt = C dv/dt = ωC Vm cos ωt = (Vm / X_C) sin(ωt + 90°)

where X_C = 1/(ωC) is the capacitive reactance in ohm. So Im = Vm / X_C and in rms terms I = V / X_C.

  • The current is maximum when the voltage passes through zero (rate of change of v is highest).
  • The current is zero when the voltage is at its peak.

Phasor diagram (V as reference)

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

In complex form: V = V∠0°, I = (V / X_C)∠90° = V / (−jX_C).

Waveforms

      v        i
        .--.  .--.
   ____/    \/    \____ ωt
        i leads v by 90°

The current wave reaches its peak a quarter-cycle before the voltage wave does.

Power

p = v i = (Vm Im / 2) sin 2ωt. Its average over a cycle is zero, so a pure capacitor consumes no real power. It takes energy in one quarter-cycle and returns it in the next.

✓ Answer checked by Claude Opus 5.5 (high effort)

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