Power factor is crucial in industrial applications:
Example: A factory with 1000kW load at 0.7 PF requires 1429kVA. Improving to 0.95 PF reduces requirement to 1053kVA.
Electric motors are major contributors to poor power factor:
Correction Methods: Capacitor banks, synchronous condensers, or power electronic devices.
Power factor in residential applications:
Standards: Many countries require residential equipment to maintain PF > 0.9.
Financial advantages of power factor correction:
Payback Period: Typical payback for PF correction is 1-3 years.
Power Factor (PF) is a measure of how effectively electrical power is being used. It represents the ratio of real power (doing useful work) to apparent power (total power drawn from the source).
A 50HP motor draws 100A at 480V with a power factor of 0.85. Calculate the real and reactive power.
Solution:
Apparent Power: S = √3 × V × I = 1.732 × 480 × 100 = 83.1 kVA
Real Power: P = S × PF = 83.1 × 0.85 = 70.6 kW
Reactive Power: Q = S × sin(φ) = 83.1 × 0.527 = 43.8 kVAR
A 500kW load operates at 0.7 PF. Calculate the capacitor size needed to improve PF to 0.95.
Solution:
φ₁ = arccos(0.7) = 45.57°, φ₂ = arccos(0.95) = 18.19°
Qc = 500 × (tan(45.57°) - tan(18.19°)) = 500 × (1.02 - 0.33) = 345 kVAR
A factory pays $0.10/kVAR penalty for PF below 0.9. Current load: 1000kW at 0.75 PF.
Before Correction: S = 1000/0.75 = 1333 kVA, Q = 882 kVAR
Monthly Penalty: 882 × $0.10 × 730 hours = $64,386
After Correction to 0.95 PF: Q = 329 kVAR, No penalty
Power factor is the ratio of real power to apparent power, but there are two distinct ways it can fall below unity, and most explanations only cover one of them.
This is the classic case: an inductive load such as a motor makes current lag voltage, and the displacement power factor is simply cos φ, the cosine of that phase angle. It is corrected with capacitors, which draw a leading current that cancels the lagging one locally instead of letting it travel back to the supply.
Electronic loads — LED drivers, computer supplies, variable-speed drives — often draw current in short pulses rather than sinusoids. That current may be perfectly in phase with the voltage, giving cos φ = 1, yet the power factor is still poor because the current is not a sine wave at all. Harmonic current heats conductors without delivering real power.
True power factor combines both effects:
PFtrue = cosφ × 1/√(1 + THD2)
This distinction matters practically: fitting capacitors to a harmonic load does not fix it. Worse, the capacitors can resonate with the supply inductance at a harmonic frequency and amplify the problem rather than correcting it. Harmonic distortion needs filters or an active front end, not power-factor capacitors.
For a given real power, current is inversely proportional to power factor, and losses go as the square of current. A load at PF 0.6 draws 67% more current than the same load at unity and dissipates about 2.8 times the copper loss. Every cable, breaker, contactor and transformer in the path must be sized for that current even though no extra useful work is being done — which is why utilities levy power-factor penalties on industrial customers.
The ratio of real power in watts to apparent power in volt-amperes. It describes how much of the current drawn is actually doing useful work.
Displacement power factor is cos φ, caused by current lagging or leading voltage. Distortion power factor is caused by non-sinusoidal current, typical of electronic loads. True power factor combines both.
No. Capacitors correct phase displacement, not harmonic distortion. On a harmonic load they can resonate with supply inductance and make matters worse. Harmonics require filters or an active front end.
Current is inversely proportional to power factor, and losses go as current squared. A load at PF 0.6 draws 67% more current than at unity and causes about 2.8 times the copper loss, while delivering identical useful power.
Because the reactive and harmonic current still has to be carried by their cables, transformers and switchgear, consuming capacity and causing losses without producing billable real energy.
Above 0.95 is generally regarded as good, and many tariffs set their penalty threshold near there. Induction motors at full load typically reach 0.85–0.90, and considerably less when lightly loaded.
Written and reviewed by
Lead Electronic Engineer (R&D) · Power Electronics Specialist · M.E. Power Electronics
Founder of Power4All. Works in power-electronics R&D and writes the converter guides, calculators and simulators on this site.
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