CR · AC power

AC Power and the Power Factor

In AC, not all of voltage times current turns into work. Whatever is out of phase merely sloshes back and forth without working. Learn to separate the real work from the idle part and read it as the power factor.

Power has a right triangle too

Slide the phase angle θ and watch the power triangle. The hypotenuse (apparent power S) keeps its length, but as θ grows the real power P shrinks and the reactive power Q grows. Find unity power factor, where everything becomes work — that is θ = 0.

Phase angle θθ = 50°

Power factor and split

PF = 0.64  ·  P = 6.4 kW  ·  Q = 7.7 kvar

Low power factor

Apparent, real and reactive — three powers

Multiplying the rms voltage by the rms current gives the apparent power S, measured in VA. The part where voltage and current move in step does the actual work — the real power P, measured in W. The part 90 degrees out of phase does no work and only shuttles between source and load — the reactive power Q, measured in var. For the same S, how it splits into P and Q depends on the phase.

The phase makes a right triangle

Real and reactive power are 90 degrees apart, so they meet at a right angle in the complex plane. The apparent power is therefore not their plain sum but the hypotenuse, S² = P² + Q². Real power is P = S cosθ and reactive power Q = S sinθ, where θ is exactly the phase angle of the load impedance. Scale the impedance triangle (R, X, |Z|) by the current squared and you get this very power triangle.

A low power factor needs more current for the same work

The power factor PF = cosθ = P/S is the fraction of apparent power that became real work. With a low power factor you must push a larger apparent power — a larger current — to deliver the same real power, and the line losses grow with it. So factories hang capacitors in parallel to cancel the inductors’ reactive power and pull the power factor close to one. This is power-factor correction.

ObserveS² = P² + Q²

A right angle, so combine as a hypotenuse.

ChooseP = S ?

Real power is the cosine component.

Fill inPF = ?

Power factor is real over apparent.

On your ownθ = 0 → PF = ?

Pure resistance has zero reactive power.

Back to the first screen

As you slid the phase angle, the hypotenuse S kept its length while the real power P shrank and the reactive power Q grew. At θ = 0 the triangle flattened horizontally, Q vanished, P equaled S, and the power factor became one. The phase of the impedance carried straight over to the phase of the power, and a single cosine told you how efficiently the circuit does work.

The apparent power S = V·I (VA) is the right-angle sum of real power P = S cosθ (W) and reactive power Q = S sinθ (var): S² = P² + Q², where θ is the phase angle of the load impedance. The power factor PF = cosθ = P/S is the fraction of apparent power that became real work; for pure resistance (θ=0) it is one and the reactive power Q is zero. A low power factor needs more current for the same work.