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CR · AC power

AC Power and the Power Factor

In AC circuits, the product of voltage and current does not all turn into work. The part that is out of phase does no work and merely shuttles between source and load, and this unit shows how to separate the working share from the idling share and read the result as the power factor.

Power has a right triangle too

In the right triangle, the hypotenuse is the apparent power, the horizontal side the real power, and the vertical side the reactive power. Change the phase angle θ and the hypotenuse (apparent power S) keeps its length, but as θ grows the horizontal side shrinks and the vertical side grows. Find a power factor of 1, that is, θ = 0, where all the power turns into work.

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

The rms voltage times the rms current is the apparent power S, measured in VA. The part of it in which voltage and current move in step is the real power P that actually does work, measured in W. The part 90 degrees out of step, which does no work and only shuttles between source and load, is the reactive power Q, measured in var. Even for the same S, how P and Q divide it depends on the phase.

The phase makes a right triangle

Real power and reactive power are components 90 degrees apart, so they meet at a right angle in the complex plane. The apparent power is therefore not their simple sum but the hypotenuse, S² = P² + Q². Real power is P = S cosθ and reactive power is Q = S sinθ, where θ is exactly the phase angle of the load impedance. Multiplying each side of the impedance triangle (R, X, |Z|) by the square of the current gives 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 the apparent power that has actually become work. The apparent power in the plot is fixed at 10 kVA, so at the starting angle θ = 50 degrees the real power is 6.4 kW and the reactive power is 7.7 kvar. A low power factor means a larger apparent power, and therefore a larger current, is needed to deliver the same real power, and line losses grow accordingly. That is why factories connect capacitors in parallel to cancel the inductive reactive power and raise the power factor close to 1. This is called power factor correction.

ObserveS² = P² + Q²
The two components are at right angles, so they combine along the hypotenuse.
ChooseP = S ?
Real power is the cosine component.
Fill inPF = ?
The power factor is real power divided by apparent power.
On your ownθ = 0 → PF = ?
A pure resistance has 0 reactive power.

Back to the first screen

As you changed the phase angle, the hypotenuse of apparent power S kept its length while the real power shrank and the reactive power grew. At θ = 0 the triangle flattened horizontally, the reactive power vanished, the real power equaled the apparent power, and the power factor became 1. The phase of the impedance carried straight over into the phase of the power, and how productively the circuit works came down to a single cosine.

The apparent power S = V·I (VA) is the right-angle sum of the real power P = S cosθ (W) and the reactive power Q = S sinθ (var), so S² = P² + Q². Here θ is the phase angle of the load impedance. The power factor PF = cosθ = P/S is the fraction of the apparent power that becomes work; for a pure resistance (θ=0) it is 1 and the reactive power Q is 0. A low power factor means more current is needed for the same work.