Case 1
- Real power P (W)
- 800W
- Apparent power S (VA)
- 1000VA
0.800
Open with these valuesPower Factor Calculator
0.800
Result: 0.800The power factor is simply real power divided by apparent power: 800 W out of 1000 VA is 0.8, so 80 % of the supplied power does useful work. It is a ratio, not an angle — the phase angle is its arc cosine, 36.9° in that example. A value close to 1 is ideal.
Worked examples
Case 2
- Real power P (W)
- 1500W
- Apparent power S (VA)
- 2000VA
0.750
Open with these valuesCase 3
- Real power P (W)
- 230W
- Apparent power S (VA)
- 400VA
0.575
Open with these valuesHow it's calculated
PF = P ÷ S
- StepEnter the real power in watts — what a wattmeter reads.
- StepEnter the apparent power in volt-amperes, the voltage times the current.
- ResultRead the power factor between 0 and 1; its arc cosine is the phase angle.
Reference table
| P, S | Phase angle | Power factor |
|---|---|---|
| 230, 400 | 54.9° | 0.575 |
| 600, 1000 | 53.1° | 0.600 |
| 1500, 2000 | 41.4° | 0.750 |
| 800, 1000 | 36.9° | 0.800 |
| 1000, 1000 | 0° | 1.000 |
Questions
How do I calculate the power factor?
Divide the real power by the apparent power. For 800 W of real power and 1000 VA of apparent power that is 800 ÷ 1000 = 0.8, meaning 80 % of the supplied power does useful work.
What is the difference between real and apparent power?
Real power in watts is the power that does useful work, which is what a wattmeter reads. Apparent power in volt-amperes is the voltage times the current actually drawn. When current and voltage are out of phase the apparent power is larger, and the ratio between them is the power factor.
What is a good power factor?
A power factor close to 1 is ideal, because nearly all the supplied power then does useful work. Purely resistive loads such as heaters sit at 1. Values below about 0.9 are often penalised for industrial customers, since the extra current raises losses and demands larger cables.
How do I get the reactive power from this?
Reactive power is the third side of the power triangle, Q = √(S² − P²), and it flows back and forth without doing net work. For 1000 VA and 800 W that is √(1000² − 800²) = 600 VAR.
Why must apparent power be at least the real power?
Apparent power contains the real power — it is the hypotenuse of the power triangle, with real power as one leg. A hypotenuse can never be shorter than a leg, so S is always at least P. Entering less returns a factor above 1, which has no physical meaning.
Sources and last check
Information, not professional advice.