Who This Guide Is For
Use this guide when you need to understand why AC current can be high even when kW is modest, and how power factor affects transformers, generators, conductors, and correction capacitors.
Formula Summary
PF = kW ÷ kVA
kVA = kW ÷ PF
A 100 kVA load using 80 kW has 0.80 power factor. The remaining apparent-power burden is associated with reactive power and, for distorted waveforms, harmonic content.
Worked Example
A 100 kW load at 0.80 PF requires 125 kVA.
kVA = 100 ÷ 0.80 = 125 kVA
Improving PF to 0.95 reduces apparent power to about 105.3 kVA for the same 100 kW load.
Why Low PF Increases Current
At fixed real power and voltage, current rises as PF falls. Higher current increases I²R losses, voltage drop, and capacity used in conductors, transformers, generators, and switchgear.
Displacement and True Power Factor
Displacement PF describes phase shift between fundamental voltage and current. True PF also includes waveform distortion. Capacitors can correct inductive displacement PF but do not automatically remove harmonic current.
Capacitor Correction
Capacitors supply leading reactive power that offsets inductive reactive power from motors and transformers. Correction size is based on load kW and the tangent difference between initial and target phase angles.
Harmonics and resonance must be evaluated before applying capacitor banks, especially on systems with drives or rectifiers.
Common Mistakes
- Assuming low PF means the load uses less real energy.
- Confusing displacement PF with true PF on distorted waveforms.
- Correcting PF without checking harmonics or resonance.
- Comparing kW loads directly against kVA equipment ratings.
Reference Notes and Assumptions
These relationships use standard AC apparent-power and power-factor definitions. Utility billing, metering, and harmonic standards may define measurement windows and correction requirements differently.