How to Calculate Power Factor
In an AC circuit, power factor (PF) is the ratio of real power (P, measured in watts or kW) — the power that actually does useful work, like turning a motor shaft or producing heat — to apparent power (S, measured in volt-amps or kVA) — the total power the source must supply to deliver that real power. PF = P ÷ S = cos(θ), where θ is the phase angle between the voltage and current waveforms. Power factor ranges from 0 to 1 (or 0-100%); a purely resistive load (a heater, an incandescent bulb) has PF = 1, while inductive loads like motors and transformers pull PF below 1.
The power triangle: real, reactive, and apparent power
- Real power (P): the working power, in watts (W) or kilowatts (kW), that performs mechanical work or is converted to heat and light.
- Reactive power (Q): in volt-amps-reactive (VAR) or kVAR, the power that oscillates back and forth between the source and the magnetic field of an inductive load (motors, transformers, ballasts) or the electric field of a capacitive load. It does no net work but still requires current-carrying capacity.
- Apparent power (S): in volt-amps (VA) or kVA, the vector sum of real and reactive power: S² = P² + Q², so Q = √(S² − P²). This is the total current × voltage the generator, transformer, and wiring must be sized to deliver, even though only the P portion does useful work.
- Phase angle (θ): the angle between the voltage and current waveforms in the power triangle, where PF = cos(θ) and θ = arccos(PF).
Single-phase vs. three-phase apparent power
- Single-phase circuits: apparent power is simply S = V × I, where V is the RMS voltage and I is the RMS current.
- Balanced three-phase circuits: using line-to-line voltage (V_LL) and line current (I_L), apparent power is S = √3 × V_LL × I_L ≈ 1.732 × V_LL × I_L. The √3 factor accounts for the 120° phase separation between the three phase conductors.
- Once you know S and the real power P (read from a wattmeter or nameplate), power factor follows directly: PF = P ÷ S.
Common mistakes and correcting a low power factor
- Real power can never exceed apparent power — if your calculated P is larger than V × I, double-check that the voltage and current are RMS values for the same load and phase configuration.
- Mixing per-phase and line values in a three-phase system produces a result off by a factor of √3 — be consistent about whether voltage/current are per-phase or line quantities.
- A low power factor (below ~0.90) is usually caused by inductive loads such as motors, transformers, and fluorescent/HID lighting ballasts running under light load.
- Utilities often add a surcharge when PF drops below 0.90-0.95, since they must deliver more current — and lose more energy to I²R heating — for the same billed kWh. Adding capacitor banks near the load supplies local reactive power and raises the measured PF toward unity without changing the real power consumed.
Frequently Asked Questions
What is power factor?
Power factor (PF) is the ratio of real power (P, in kW) — the power that actually does useful work — to apparent power (S, in kVA) — the total power the source must supply. PF = P ÷ S = cos(θ), where θ is the phase angle between the voltage and current waveforms. It ranges from 0 to 1 (or is expressed as 0-100%).
What is considered a good power factor?
A power factor of 0.95 or higher is generally considered good to excellent. Many electric utilities charge a penalty when a commercial or industrial customer's power factor drops below 0.90-0.95, because low PF forces the utility to deliver more current for the same amount of useful work.
How do I calculate apparent power from voltage and current?
For a single-phase circuit, apparent power is S = V × I (volts × amps = volt-amps). For a balanced three-phase circuit using line-to-line voltage and line current, S = √3 × V_LL × I_L. Once you know S and the real power P, power factor is simply PF = P ÷ S.
Why is a low power factor a problem?
A low power factor means a large share of the current flowing through the wiring is reactive current that does no useful work. To deliver the same real power (kW), the source, transformers, and conductors must be sized for a higher apparent power (kVA) and current, which increases I²R losses and can trigger utility surcharges.