The formula
S² = P² + Q² ; P = S × cos φ ; pf = P / S
- P
- real (active) power — does the work (watts)
- Q
- reactive power — stored and returned each cycle (VAr)
- S
- apparent power — what the cable and transformer must carry (VA)
- pf
- power factor, cos φ (0 to 1)
Worked example
A load draws 8 kW at a power factor of 0.8 lagging.
- S = P / pf = 8 / 0.8 = 10 kVA
- Q = √(S² − P²) = √(100 − 64)
- Q = √36
S = 10 kVA and Q = 6 kVAr — the supply carries 10 kVA to deliver 8 kW of useful work.
Where you'll use it
Sizing cables, transformers and generators, which are rated in kVA precisely because they must carry the apparent power. It is also the setup for every power-factor correction question.