How it works
Inductive loads such as motors and transformers draw reactive power that does no useful work but loads the wiring and the utility. Power factor is the ratio of real power to apparent power. A low power factor means more current for the same real power, and many utilities charge a penalty for it.
Capacitors supply the reactive power locally. The reactive power you need to remove is the difference between the load's reactive power now and at the target power factor. Reactive power equals real power times the tangent of the power-factor angle.
Worked example
A 100 kW three-phase, 480 V load at 0.80 power factor, corrected to 0.95:
- tan θ₁ = tan(arccos 0.8) = 0.75; tan θ₂ = tan(arccos 0.95) = 0.3287
- kVAR = 100 kW × (0.75 − 0.3287) = 42.1 kVAR
- Apparent power: 100 ÷ 0.8 = 125 kVA before; 100 ÷ 0.95 = 105.3 kVA after
- Line current: 150.4 A before, 126.6 A after
| Input | Value |
|---|---|
| Real power | 100 kW |
| Present power factor | 0.8 |
| Target power factor | 0.95 |
| Line voltage (line-to-line for three-phase) | 480 V |
| Frequency | 60 Hz |
| System | Three-phase |
| Result | Value |
|---|---|
| Capacitor kVAR needed | 42.1 kVAR |
| Capacitance (single-phase, or per phase of a delta bank) | 161.7 µF |
| Apparent power before | 125 kVA |
| Apparent power after | 105.3 kVA |
| Line current before | 150.4 A |
| Line current after | 126.6 A |
Assumptions and limits
- Sinusoidal voltage and current. Harmonics from drives and electronics change the result and can cause capacitor resonance problems.
- Capacitance per phase assumes delta-connected capacitors at the line voltage.
- Uses the steady real power you enter. Loads that vary need automatic switched banks.
- Results are for sizing estimates, not a design for a capacitor bank installation.
Common questions
How many kVAR do I need to correct 100 kW from 0.8 to 0.95?
About 42 kVAR: 100 × (0.75 − 0.329).
Should I correct all the way to 1.0?
Rarely. Going past about 0.95–0.98 gives little extra benefit and risks leading power factor, which can overvoltage the system. Check the utility's penalty threshold first.
What does low power factor cost?
More current for the same work, so larger conductors and transformers, higher I²R losses, and often a utility demand or power-factor charge.
Can I put a capacitor across a single motor?
Yes, at the motor terminals, but the capacitor must not exceed what the motor manufacturer or code allows, and should be on the load side of the overload relay in a way the code permits. Ask a qualified electrician.
Sources
- Standard power-triangle relationships: kVAR = kW(tan θ₁ − tan θ₂); capacitor sizing C = Q/(2π f V²) per capacitor.
Updated 2026-09-30