AC power
Reactive power and the power triangle
- where
- reactive power (var)
- apparent power (VA)
- active power (W)
Explain this equation
Finds the reactive power from known active and apparent power. Use it when sizing power-factor correction or checking how a supply is loaded.
S is the apparent power in VA, the total supplied. P is the active power in W, the working part. Q is the reactive power in var: the energy that surges back and forth each cycle between the supply and the load's magnetic or electric fields without being consumed.
For sine waves, P and Q act at right angles to each other, with S as the hypotenuse of a right-angled triangle. Pythagoras gives S² = P² + Q², rearranged here for Q.
S = 1000 VA and P = 800 W give Q = 600 var, the familiar 3-4-5 triangle. At PF 0.8 the reactive power is always 0.75 × P.
Watch out: square both before subtracting; S − P is not Q. The answer is a size only and does not say whether it is lagging (inductive) or leading (capacitive). Harmonics break the simple triangle.
The right-angled power triangle relates active, reactive and apparent power for sinusoidal conditions. Squaring matching units keeps the calculation consistent.
This equation gives a magnitude. State separately whether reactive power is inductive (lagging) or capacitive (leading); the square root does not identify its sign.
Enter your values to see the result.
Common mistake
Do not subtract S − P without first squaring both terms.
Limits
Sinusoidal loads only for S² = P² + Q². Harmonic distortion can add a distortion contribution; do not infer kvar from this triangle in that case.
Sources
The tools calculate the displayed equation. They do not automatically choose cable sizes, approve installations, supply current regulatory tables, certify equipment or replace inspection and testing.