Power Factor Correction Capacitor
Required kVAR and capacitance to raise existing PF to target.
Example
You enter
- Real power (kW) 100
- Existing PF (0-1) 0.75
- Target PF (0-1) 0.95
- System voltage (V) 480
- Phase three
You get
- Required kVAR 55.32
- Capacitance per leg 636.93
Checked against published worked examples from: Eaton Corporation (Cutler-Hammer).
Details, formula, and sources
Formula or tablekVAR = kW × (tan(acos(pf₁)) − tan(acos(pf₂))); μF from Q = V² × 2π f × C at 60 Hz; three-phase per leg, wye C = kVAR / (3 × 2π f × V_LN²) and delta C = kVAR / (3 × 2π f × V_LL²), one third of wye.
Source and editionClassical AC theory; IEEE 141 by name.
Free accessPrinciples free at IEEE-USA outreach.
What governsEstimate. AHJ and licensed electrician govern. Verify against the NEC edition adopted in your jurisdiction.
API field nameskW, pf1, pf2, system_V, phase, kVAR, capacitance_uF
- Frequency 60 HzANSI C84.1