What is power factor? The definition
Power factor is the ratio of real power to apparent power in an AC circuit: the fraction of the volt-amperes a load draws that it turns into useful work. It runs from 0 to 1. A power factor of 1 means every volt-ampere does work; 0.75 means only three-quarters does, and the supply has to carry a third more current than the work requires.
Every AC load draws a current, and the supply has to deliver all of it, whatever the load does with it. Multiply the RMS voltage by the RMS current and you get the apparent power, S, in volt-amperes: the size of the electrical pipe the load needs. Only part of it becomes heat, light or shaft power. That part is the real power, P, in watts, and the ratio of the two is the power factor:
A heater draws current exactly in step with the voltage, so every volt-ampere is a watt and the power factor is 1. A motor also has to build a magnetic field, and the part of its current that does that lags the voltage by a quarter of a cycle. The energy goes into the field and comes back out a quarter-cycle later. On average it does no work, but it flows through every conductor, breaker and transformer between the source and the motor.
At power factor 0.75 the slices of returned energy look small, but the current that carries them is not: for every kilowatt delivered, the supply has to provide 1.33 kVA. That extra current, rather than the energy itself, is what a low power factor costs, in heat, in capacity and often on the bill.
Real, reactive and apparent power: the power triangle
Split the current into a part in phase with the voltage, which carries real power, and a part a quarter-cycle out of phase, which carries reactive power. The two are at right angles, so their powers add like the sides of a right triangle:
- Real power P, in watts (W, kW): turned into heat, light and motion.
- Reactive power Q, in volt-amperes reactive (var; kVAR on most equipment): exchanged with magnetic and electric fields every half-cycle.
- Apparent power S, in volt-amperes (VA, kVA): V × I, which wiring, switchgear and transformers have to be rated for.
For sine waves, φ is the phase angle between voltage and current, and the power factor is cos φ. A load at 0.75 has its current 41.4° behind the voltage, and for every kilowatt it also draws 0.882 kVAR. Generators, transformers and UPS units carry kVA ratings for the same reason: the heating in their windings follows the current, so a kVA rating holds at any power factor where a kW rating would not. That is also the answer to “what is the power factor of a transformer?”: a transformer has no power factor of its own worth quoting; it passes on the power factor of the load it feeds, and how a transformer works explains why its rating is in kVA.
Power factor formula
The power factor formula is PF = P ÷ S = kW ÷ kVA, and for sinusoidal waveforms PF = cos φ. Rearranged with the power equations, it gives the formulas used every day for single-phase and balanced three-phase loads:
A single-phase motor drawing 10 A at 120 V with a power factor of 0.8 takes 960 W of real power but 1,200 VA of apparent power; working back from 960 W at the same power factor returns 10 A. Convert amps to watts for a load you have measured, or convert watts to amps to find the current a planned load will draw. Leaving the power factor at 1 for a motor overstates its watts and understates its current.
Power factor formula for three-phase
For a balanced three-phase load, PF = P ÷ (√3 × VLL × I), with the line-to-line voltage and the line current. Measured 150 kW on a 480 V supply drawing 240.6 A gives 150 kW ÷ (1.732 × 480 × 240.6) = 0.75.
Power factor unit and symbol
Power factor has no unit. It is watts divided by volt-amperes, so the units cancel and it is a pure number between 0 and 1, often written as a percentage. It is labelled PF on meters and nameplates, cos φ on motor nameplates and in most of Europe and India (strictly, the displacement power factor), and λ (lambda) in the IEC’s International Electrotechnical Vocabulary. A power factor is also described as lagging or leading, which tells you which way the current is shifted; the number alone does not. Don’t confuse it with pF, the picofarad.
Power factor of common loads
These are the usual ranges, the same guidance the calculators give. A real load’s power factor varies with how hard it is working, so measure it, or take it from the nameplate, when the number matters.
| Load | Typical power factor | Why |
|---|---|---|
| Resistance heaters, kettles, incandescent lamps | 1 | Current in step with the voltage |
| Induction motor at full load | about 0.8–0.9 | Lagging; check the nameplate |
| Induction motor, lightly loaded | much lower | The magnetising current stays while the load falls |
| Electronics with power-factor correction | 0.9 or better | Computer supplies, quality LED drivers, drives with active front ends |
| Small electronics without correction | often 0.5–0.7 | Mostly distortion, not displacement |
Calculate kW, kVA and current at any power factor
The three-phase calculator takes the line-to-line voltage, the line current or the load in kW or kVA, and the power factor, and returns real, apparent and reactive power with the line current. Change only the power factor and watch the current and kVA move while the kW stays put. The full three-phase power calculator adds wye and delta phase values.
Lagging vs leading power factor
Lagging and leading say which way the current is shifted. In an inductive load (induction motors, transformers, magnetic ballasts, reactors) the current lags the voltage. In a capacitive load (capacitor banks, long lightly loaded cables, the input filters of some electronics) it leads. The old memory aid is “ELI the ICE man”: in an inductor, L, the voltage, E, comes before the current, I; in a capacitor, C, the current comes before the voltage.
At the same power factor, lagging and leading loads draw the same current and the same kVA. What differs is the sign of the reactive power: by convention an inductive load absorbs vars and a capacitive load supplies them, which is how a capacitor bank cancels a motor’s reactive power. Most installations run lagging. A leading power factor usually means too much correction: capacitors left on at light load push the voltage up, and a generator feeding a leading load has to cut its field current, so at strongly leading power factors its voltage regulator can lose control. That is why automatic banks switch capacitor steps in and out with the load.
Displacement and true power factor
cos φ is the whole story only when the current is a sine wave. Rectifier-input loads, such as variable-frequency drives, switch-mode power supplies, LED drivers and UPS units, draw current in pulses. The pulses contain harmonics: currents at multiples of the supply frequency, which carry no average power against a sinusoidal voltage but still add to the RMS current. The definitions in IEEE 1459 therefore separate two quantities:
- Displacement (fundamental) power factor, cos φ₁: the cosine of the angle between the fundamental voltage and current.
- True power factor, P ÷ S, using the full RMS current with all its harmonics.
With a sinusoidal voltage they are linked by the current’s total harmonic distortion, THD:
| Current THD | 5 % | 20 % | 50 % | 80 % | 100 % | 150 % |
|---|---|---|---|---|---|---|
| Distortion factor | 0.999 | 0.981 | 0.894 | 0.781 | 0.707 | 0.555 |
An ideal six-pulse rectifier with a large DC choke draws a stepped current whose harmonics are the 5th, 7th, 11th, 13th and so on, each about 1/n of the fundamental (Schneider Electric’s Cahier Technique 152). That current has a THD of 31.1 %, so even with a displacement power factor of 1 its true power factor is 0.955. Single-phase rectifiers that charge a capacitor draw much narrower peaks and far more distortion: a power supply with, say, cos φ₁ of 0.98 and 80 % THD has a true power factor of only 0.77.
The practical consequence: capacitors supply reactive power at the supply frequency, so they correct displacement and do nothing for distortion. A drive whose nameplate says cos φ 0.98 can still have a true power factor well below that. Distortion is reduced at its source instead, with line reactors, 12-pulse rectifiers (which cancel the 5th and 7th harmonics), active front ends, harmonic filters (IEEE 1531) or the power-factor-correction stage built into better electronic supplies. IEEE 519 sets the limits on harmonic current at the point where a customer connects to the utility.
Why low power factor costs money
The process needs kilowatts; the installation has to carry amperes. For a fixed real power the current is inversely proportional to the power factor, and resistive losses go with the square of the current. Take the 150 kW load on a 480 V three-phase supply:
| Power factor | Apparent power | Line current | I²R loss vs PF 1 | Copper conductor | Transformer |
|---|---|---|---|---|---|
| 1 | 150 kVA | 180.4 A | — | 4/0 AWG | 150 kVA |
| 0.95 | 157.9 kVA | 189.9 A | +11 % | 250 kcmil | 225 kVA |
| 0.85 | 176.5 kVA | 212.3 A | +38 % | 300 kcmil | 225 kVA |
| 0.75 | 200 kVA | 240.6 A | +78 % | 350 kcmil | 225 kVA |
| 0.65 | 230.8 kVA | 277.6 A | +137 % | 500 kcmil | 300 kVA |
Three costs follow from the same current:
- Losses. At 0.75 the I²R losses in every cable and transformer winding upstream of the load are 78 % higher than at unity, for the same useful output.
- Capacity. Sized at 125 % for a continuous load (215.2(A)(1) in the 2023 NEC, 215.4(A)(1) in 2026), the conductors grow from 4/0 AWG at unity to 350 kcmil at 0.75 and 500 kcmil at 0.65, and the transformer from 150 kVA to 300 kVA. Use the transformer calculator to check full-load amps against a rating.
- Voltage drop. More current means more drop on every run, which shows up as dim lighting, slow motor starts and hot equipment at the far end.
Utilities face the same losses on their own network, and many recover them through the tariff: the US Department of Energy notes that many charge penalties when a plant’s power factor falls below 0.90 or 0.95. Two common structures are demand billed in kVA instead of kW, where the 150 kW load at 0.75 is billed as 200 kVA, 33 % more; and kW demand scaled up below a threshold, for example by target ÷ actual, where a 0.90 target bills the same load as 180 kW. Some tariffs charge for kVARh directly. Read your own rate schedule for the rule that applies; homes are usually billed for kWh only.
Power factor correction
The fix is to supply the reactive power locally. A capacitor draws a leading current that cancels part of the motor’s lagging current, so less reactive power travels from the utility. The real power does not change. The capacitor kVAR needed to raise the power factor from PF₁ to PF₂ at constant real power is:
For the example load, φ₁ = 41.4° with tan φ₁ = 0.8819, and φ₂ = 18.2° with tan φ₂ = 0.3287. So Qc = 150 kW × (0.8819 − 0.3287) = 82.98 kVAR. After correction the apparent power falls from 200 kVA to 157.9 kVA, the line current from 240.6 A to 189.9 A, and the I²R losses upstream of the capacitors by 38 %. For quick sizing, multiply the load in kW by the factor in this table:
| Present PF | to 0.90 | to 0.95 | to 1.00 |
|---|---|---|---|
| 0.60 | 0.849 | 1.005 | 1.333 |
| 0.65 | 0.685 | 0.840 | 1.169 |
| 0.70 | 0.536 | 0.692 | 1.020 |
| 0.75 | 0.398 | 0.553 | 0.882 |
| 0.80 | 0.266 | 0.421 | 0.750 |
| 0.85 | 0.135 | 0.291 | 0.620 |
| 0.90 | — | 0.156 | 0.484 |
Choosing and installing capacitors
- Check light load, not just full load. A bank sized to correct the full load leaves the site leading when the load drops. Automatic banks with switched steps follow the load; fixed capacitors suit loads that run steadily.
- Mind the voltage rating. A capacitor’s kVAR goes with the square of the applied voltage, so a 600 V-rated unit on a 480 V system delivers only 64 % of its nameplate rating. For a delta-connected three-phase bank, Q = 3 × 2πf × C × V², so the 82.98 kVAR above needs about 318 µF per phase at 480 V and 60 Hz.
- Choose the location. Capacitors can go at individual motors, at the motor control center or at the service, and they relieve only the wiring upstream of where they connect. A capacitor at a motor switches with it and also unloads the feeder.
- Respect the motor. Capacitors switched with a motor are kept below the motor’s no-load magnetizing kVAR; with more, the motor can self-excite and generate overvoltage while it coasts down after being disconnected. When the capacitor is on the load side of the overload relay, the relay sees less current, and NEC Article 460 requires its setting to be based on the improved power factor. Article 460 also covers the conductors, disconnecting means and discharge of stored energy for capacitor installations.
Resonance, harmonics and detuned banks
A capacitor bank and the inductance of the supply transformer form a parallel resonant circuit. Its resonant frequency, as a multiple of the supply frequency, is approximately:
where Ssc is the short-circuit power at the bank and QC the bank’s rating. If nr lands on or near a harmonic that drives or rectifiers inject, the 5th, 7th, 11th or 13th for six-pulse equipment, that harmonic is amplified: capacitors overheat, fuses blow and the voltage distorts. Take a 1,000 kVA, 480 V transformer with 5.75 % impedance. With an infinite source its secondary fault current is 20.9 kA, a short-circuit power of 17.4 MVA:
| Bank | Resonant harmonic order | Risk |
|---|---|---|
| 100 kVAR | 13.19 | within 10 % of the 13th harmonic |
| 200 kVAR | 9.33 | clear of the 5th, 7th, 11th and 13th |
| 300 kVAR | 7.61 | within 10 % of the 7th harmonic |
| 400 kVAR | 6.59 | within 10 % of the 7th harmonic |
| 500 kVAR | 5.90 | clear of the 5th, 7th, 11th and 13th |
| 600 kVAR | 5.38 | within 10 % of the 5th harmonic |
More capacitance pulls the resonance down: here a 300 kVAR bank sits at 7.61 and a 600 kVAR bank at 5.38. The utility’s own impedance lowers Ssc and moves it lower still. Where drives are present, the usual answer is a detuned bank: a reactor in series with each capacitor step tunes the combination below the lowest significant harmonic, so the bank looks inductive to the harmonics and capacitive at the supply frequency. Schneider Electric gives 135–225 Hz on 50 Hz networks, harmonic orders 2.7–4.5, which corresponds to reactors of 4.9 %–13.7 % of the capacitor’s reactance. Where the harmonics themselves must come down, a tuned filter (IEEE 1531) is the next step, and the capacitor ratings themselves are set by IEEE 18.
Power factor in three-phase systems
In a balanced load every phase has the same power factor, and P = √3 × VLL × I × PF holds for the whole load. Balanced three-phase power has a neat property: each phase’s instantaneous power pulses, and dips below zero when the power factor is less than 1, but the three add up to a constant. The reactive power still circulates, so the line currents are just as high.
Unbalanced loads are harder, because the result depends on how the apparent power is totalled. Three single-phase loads of 30 kW each, at power factors 1.0, 0.8, 0.6, add up to 90 kW and 62.5 kVAR. Taking the vector sum gives 109.6 kVA and a power factor of 0.821; adding the three kVA figures gives 117.5 kVA and 0.766. Both are in use, and IEEE 1459 adds a third, effective apparent power, for unbalanced and distorted systems. When two meters disagree, check which definition each one applies.
The two-wattmeter method
On a three-wire system, two wattmeters measure the total power whatever the balance. With a balanced load they also give the power factor:
At 480 V, 100 A and power factor 0.85, the meters read 47.98 kW and 22.69 kW. Their sum, 70.67 kW, is the same as √3 × V × I × PF = 70.67 kW, and the formula recovers a power factor of 0.85. At a power factor of 0.5 the second meter reads zero, and below that it reads negative. On a motor circuit a negative wattmeter means a low power factor, not reverse power.
Measuring power factor
- Use an instrument that reports both values. A power analyzer or power-quality meter shows the true power factor (PF) and the displacement factor (DPF or cos φ). A large gap between them points to harmonics, which capacitors will not fix.
- Connect it the right way round. Current clamps go on with the arrow pointing to the load, and each clamp must be on the same phase as its voltage lead. A reversed clamp reads negative power; a swapped phase gives a power factor that makes no sense.
- Measure at real operating load. Lightly loaded motors run at far lower power factors than their full-load nameplate figures, so a reading taken at a quiet moment can mislead in either direction.
- Read the bill. With kWh and kVARh on the meter, the month’s average is PF = kWh ÷ √(kWh² + kVARh²): 120,000 kWh and 90,000 kVARh average 0.80.
To watch a phase shift and its correction, open the power factor circuit and the power factor correction circuit in the simulator. For the resistive side of the story, see how Ohm’s law relates volts, amps and ohms.