AC Power & Power Factor

Phasors give you the steady state: with v(t) = V₀ cos ωt across a load, the current is i(t) = I₀ cos(ωt − φ), where φ comes from the impedance. Power is not a phasor — it is the ordinary product p(t) = v(t)i(t), taken instant by instant. Use cos A cos B = ½[cos(A − B) + cos(A + B)] and it separates cleanly: p(t) = ½V₀I₀[cos φ + cos(2ωt − φ)]. A constant term, and a ripple at twice the drive frequency. The ripple is observable — a filament on a 50 Hz supply brightens and dims at 100 Hz, and a single-phase motor delivers a torque that pulses at 100 Hz too. Note the size of the ripple, ½V₀I₀, against the constant offset, ½V₀I₀ cos φ. Whenever φ ≠ 0 the ripple is the larger of the two, so p(t) goes negative over part of every cycle. During those intervals the load is not taking energy from the source; it is pushing energy back.

Average p(t) over one full period. The 2ω term completes two whole cycles in that time, so it averages to exactly zero and only the constant survives: P = ½V₀I₀ cos φ. Substitute V(rms) = V₀/√2 and I(rms) = I₀/√2 — the substitution that made rms worth defining in the first place — and the halves are absorbed: P = V(rms)I(rms) cos φ. The factor cos φ is the power factor. It is not an efficiency and not a fudge; it is the cosine of the phase angle between current and voltage, and for a series circuit the impedance triangle reads it off directly as cos φ = R/Z. Put numbers on it: R = 30 Ω, X(L) = 60 Ω and X(C) = 20 Ω give Z = √(30² + 40²) = 50 Ω, so cos φ = 0.60. On a 120 V rms supply, I(rms) = 2.40 A and P = 120 × 2.40 × 0.60 = 173 W. Forget the cos φ and you would claim 288 W — 67% too high.

Set φ = 90°, a pure inductor. Then cos φ = 0 and P = 0 — yet the current is not zero, and the instantaneous power is not zero either. It is p(t) = ½V₀I₀ sin 2ωt, a pure oscillation at 2ω: for a quarter cycle the source pours energy into the magnetic field, raising the stored ½Li² to its peak of ½LI₀², and over the next quarter it pulls every joule back out. A capacitor does the same with ½Cv², a quarter cycle out of step, which is why an inductor and a capacitor can trade with each other. Nothing is dissipated in either case. That is why the average power in any series RLC circuit can be written P = I(rms)²R, with no reactive term anywhere: dissipation lives in the resistance alone. Check it on the same circuit — (2.40 A)² × 30 Ω = 173 W, identical to V(rms)I(rms) cos φ, because V(rms) cos φ is just I(rms)R.

The three quantities are kept apart by their units rather than by their symbols. Apparent power S = V(rms)I(rms) is quoted in volt-amperes: it is what conductors, breakers and transformers must be sized for, because they respond to current and are indifferent to phase. Real power P = S cos φ is in watts — what the load converts and what the meter bills. Reactive power Q = S sin φ carries the unit var precisely to flag that it is not watts; it measures the amplitude of the borrow-and-return exchange. The three form a right triangle, S² = P² + Q², which is the impedance triangle multiplied through by I(rms)². For the circuit above: S = 288 V⋅A, P = 173 W, and Q = 288 × 0.80 = 230 var, confirmed by I(rms)²(X(L) − X(C)) = 5.76 × 40 = 230 var. Q is signed — positive for a lagging inductive load, negative for a leading capacitive one — and that sign is what makes cancellation possible.

An inductive load's Q is positive and a capacitor's is negative, so a capacitor wired across the load lets the two swap energy with each other instead of with the generator. In parallel, not in series, because the load's voltage must not change. Size it from the reactive powers. A factory draws P = 20.0 kW at 400 V rms and 50 Hz with cos φ₁ = 0.70 lagging, so it pulls I = 20000/(400 × 0.70) = 71.4 A and needs Q₁ = P tan φ₁ = 20.4 kvar. At the target cos φ₂ = 0.95 the supply need only provide Q₂ = P tan φ₂ = 6.57 kvar, so the capacitor must furnish ΔQ = 13.8 kvar. A capacitor's reactive power is V(rms)²ωC, so C = ΔQ/(ωV(rms)²) = 13830/(314 × 400²) = 275 μF. The current falls to 20000/(400 × 0.95) = 52.6 A. The real power is untouched — the machines do the same work — but the supply carries 26% less current.

Follow that current back out to the line. Delivering P at V(rms) demands I(rms) = P/(V(rms) cos φ), so the conductor heating is P(loss) = I(rms)²R(line) = R(line)P²/(V(rms) cos φ)², inverse in cos²φ. Put the factory on a feeder of 0.10 Ω round trip: 71.4 A wastes 510 W before correction and 52.6 A wastes 277 W after — 233 W saved by a component that consumes nothing itself. That 1/cos²φ, and the 1/V(rms)² sitting beside it, is why utilities meter reactive power for industrial customers and why transmission runs at high voltage. Two limits. Everything here assumes sinusoids: rectifiers and switching supplies draw distorted current whose harmonics carry no average power against a sinusoidal voltage, so the true power factor P/S drops below the displacement factor cos φ. And over-correcting swings the load capacitive, which raises the current again and can resonate with the supply inductance.