Pure inductive circuit, zero real power

A pure inductive circuit has inductance L only (R = 0).

Derivation of current

Applied voltage v = Vm sin ωt. The induced emf is e = −L di/dt and it balances the applied voltage:

v = L di/dt ⟹ di = (Vm/L) sin ωt dt

Integrating: i = −(Vm/ωL) cos ωt = (Vm/ωL) sin(ωt − 90°)

i = Im sin(ωt − 90°), where Im = Vm/(ωL) = Vm/X_L

X_L = ωL = 2πfL (ohm) is the inductive reactance. The current lags the voltage by 90°.

Waveforms and phasor diagram

       v        i (lags 90°)
        .--.     .--.
   ____/    \___/    \___  ωt
  i reaches peak a quarter cycle after v.

   ─────────────► V (reference)
   │
   │
   ▼
   I (lags V by 90°)

Power

Instantaneous power:

p = v i = Vm sin ωt × Im sin(ωt − 90°) = −Vm Im sin ωt cos ωt

p = −(Vm Im / 2) sin 2ωt

Average power over a cycle:

P = (1/2π) ∫₀^{2π} p d(ωt) = 0, since the mean of sin 2ωt is zero.

Also P = V I cos φ = V I cos 90° = 0.

Graphical proof

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

The power wave has twice the supply frequency. In the quarter-cycles when v and i have the same sign, p is positive and the inductor stores energy (½Li²) in its magnetic field. In the next quarter-cycles p is negative and it returns the same energy to the source. The positive and negative areas are equal, so the net energy consumed is zero. The inductor consumes no real power; it only exchanges reactive power Q = V I VAR.