Calculation
λ = P / (U·I)
Power factor is the share of apparent power U·I converted into real power.
Busch 2006, Gleichung zum Leistungsfaktor
Power factor from real and apparent power
Power factor is the share of apparent power U·I converted into real power.
MINTSI
02
Result
Select a target and calculate.
Understand the inputs
- Power factor λ — Ratio of real to apparent power; equal to cos φ only for sinusoidal quantities without harmonics.
- Real power P — Measured average real power consumed by the load, not apparent power in VA.
- RMS voltage U — RMS voltage directly across the single-phase load under consideration.
- RMS current I — RMS total load current in the same operating state as P and U.
Example
P=1,800 W, U=230 V and I=10 A give S=2,300 VA and λ≈0.783.
Assumptions and limits
Single-phase steady operation with simultaneous measurements; displacement and distortion power factor are not separated.
Understand Power factor from real and apparent power
This calculator determines what share of apparent electrical power is converted into real power. It is intended for simultaneous measurements on a single-phase AC load.
What does this quantity describe?
Apparent power is S=U·I and λ=P/S ideally lies between 0 and 1. Only with sinusoidal voltage and current and no distortion may λ be read directly as cos φ.
Formula and variables
λ = P / (U·I)
S = U · Iλ = P/S = P/(U·I)
| Symbol / input | Meaning |
|---|---|
| Power factor λ | Ratio of real to apparent power; equal to cos φ only for sinusoidal quantities without harmonics. |
| Real power P | Measured average real power consumed by the load, not apparent power in VA. |
| RMS voltage U | RMS voltage directly across the single-phase load under consideration. |
| RMS current I | RMS total load current in the same operating state as P and U. |
Choose the inputs correctly
P comes from a real-power meter or reliable watt rating. U and I are RMS values measured directly at the same load, simultaneously and in the same operating state.
How to use the calculator
Measure real power, voltage and current after the load settles. Enter all three and interpret λ together with load type and possible harmonic content.
Worked example
1,800 W at 230 V and 10 A corresponds to 2,300 VA apparent power and λ≈0.783.
Understand the result and units
λ≈0.783 means conductors, switches and source carry 2.3 kVA although only 1.8 kW is converted. It does not identify whether phase shift or distortion is responsible.
P in W/kW, U in V and I in A; λ is dimensionless and displayed as a decimal. 0.9 means 90%.
Typical applications
Plausibility checks for motors, transformers, power supplies and other single-phase loads, and preparation for reactive-power or harmonic analysis.
Assumptions, limits and common mistakes
No three-phase evaluation and no separation of displacement factor cos φ from distortion factor. Measurement error, startup and changing loads are excluded.
Common mistake: Do not enter VA as W or combine nameplate values from different operating points. For switch-mode supplies do not automatically equate λ with cos φ.
Frequently asked questions
What is “Power factor from measurements” used for?
Plausibility checks for motors, transformers, power supplies and other single-phase loads, and preparation for reactive-power or harmonic analysis.
Where do the input values come from?
P comes from a real-power meter or reliable watt rating. U and I are RMS values measured directly at the same load, simultaneously and in the same operating state.
What does the result not cover?
No three-phase evaluation and no separation of displacement factor cos φ from distortion factor. Measurement error, startup and changing loads are excluded.