AC power
Power factor from measured power
- where
- power factor
- active power (W)
- apparent power (VA)
Explain this equation
Finds a load's power factor from measured active and apparent power. Use it to judge whether correction is worthwhile, or to explain why a load draws more current than its wattage suggests.
P is the active power in watts, the power actually used. S is the apparent power in VA, the voltage times the current supplied. PF is their ratio, a number from 0 to 1.
PF tells you what fraction of the supplied volt-amperes does useful work. For pure sine waves it equals cos φ, where φ is the phase shift between voltage and current. With harmonics present, P/S also captures the distortion, so it gives the true power factor.
8 kW out of 10 kVA gives 0.8. Resistive heating is close to 1. A lightly loaded induction motor can fall to 0.5 or lower. On a large installation, anything much under 0.9 is worth a look.
Watch out: use matching units, kW with kVA or W with VA. P can never be larger than S.
Power factor compares active power with apparent power. An importing load with 8 kW and 10 kVA has a power factor of 0.8.
With harmonics present, true power factor includes distortion as well as phase displacement. A cos φ reading alone may not describe the complete load.
Enter your values to see the result.
Common mistake
Use matching prefixes: kW/kVA or W/VA, never W/kVA.
Limits
Use RMS voltage and current. The cos φ form assumes sinusoidal voltage and current; for distorted waveforms use measured true power factor, P/S.
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.