Power factor is the ratio of active power to apparent power in an AC circuit, equal to the cosine of the phase angle between sinusoidal voltage and current. It indicates how much of the current is doing useful work.
Learning objectives
- Define power factor and relate it to the phase angle
- Calculate active, reactive and apparent power
- Determine the reactive power needed for correction
- Explain why low power factor increases current and losses
- Measure power factor with a power analyser
Progressive explanation
Start with the plain-language idea, then move into the engineering detail. Each level stands on its own, so you can stop at the depth you need.
Beginner The core idea
Motors, transformers, ballasts and many electronic loads have a power factor below 1. The utility must supply the extra current even though the customer uses only the active power, so a low power factor increases losses and may be penalised.
Intermediate How it behaves in a circuit
Capacitor correction reduces the reactive component: Qc = P × (tan φ1 − tan φ2), where φ1 is the initial angle and φ2 the target angle. Correction raises the power factor and reduces the supply current without changing the active power of the load.
Advanced Limits, assumptions, engineering detail
Formulas and variables
Units of measurement
| Quantity | Symbol | Unit |
|---|---|---|
| Power factor | PF | dimensionless |
| Active power | P | watt (W) |
| Reactive power | Q | volt-ampere reactive (var) |
| Apparent power | S | volt-ampere (VA) |
Worked examples
Power triangle of a motor
Problem. A single-phase motor draws 10 A at 230 V with a power factor of 0.8 lagging. Find the apparent, active and reactive power.
Solution. S = 230 × 10 = 2300 VA. P = 2300 × 0.8 = 1840 W. Q = √(2300² − 1840²) = √(5 290 000 − 3 385 600) = √1 904 400.
Capacitor correction
Problem. The same 1840 W load has a power factor of 0.8. How much reactive power must be added to raise the power factor to 0.95?
Solution. φ1 = 36.87° so tan φ1 = 0.75. φ2 = 18.19° so tan φ2 = 0.329. Qc = 1840 × (0.75 − 0.329).
Current after correction
Problem. After correction to 0.95, what current does the 1840 W load draw from the 230 V supply?
Solution. S = P/PF = 1840/0.95 = 1937 VA. I = S/V = 1937/230.
Practical applications
- Industrial electricity billing and demand charges
- Motor and transformer loading
- Generator and UPS sizing
- Cable and switchgear sizing
- Capacitor bank design
- Energy-efficiency assessments
Automotive applications
Alternators, inverters and on-board chargers have their own power factor and distortion behaviour on the AC side. Electric-vehicle AC charging is specified with power-factor requirements so that the mains supply is not overloaded by reactive current.
Industrial applications
Factories with many induction motors often correct power factor with capacitor banks to avoid utility penalties and release transformer capacity. Where harmonics are present, detuned reactors or active filters are used instead of plain capacitors to avoid resonance.
Measurement procedure
- Use a power analyser that displays active, reactive and apparent power and the power factor together.
- Measure on the load side of the supply to distinguish load power factor from the corrected installation.
- Check whether the instrument reports displacement or true power factor when harmonics are present.
- Verify that voltage and current channels are connected with the correct polarity and phase.
Common mistakes
- Confusing power factor with efficiency; a motor can have a good efficiency and a poor power factor.
- Adding capacitors without checking for resonance with the supply and harmonics.
- Assuming that correcting power factor reduces the energy bill; it mainly reduces current and kVA demand.
- Ignoring distortion and trusting a displacement power-factor reading on a drive load.
Safety considerations
- Capacitor banks retain charge after isolation; discharge them and verify before touching terminals.
- Over-correction can raise the voltage and cause equipment problems; design correction for the actual load.
- Arc-flash risk increases with large reactive banks and high fault levels; follow the switching procedure.