Power Factor Correction Calculator

Capacitor kVAR and microfarads needed to raise a load’s power factor to a target, with line current and kVA before and after correction.

Measured kW drawn by the load (not kVA). For a motor, use electrical input power.
Lagging (inductive) power factor as a decimal, e.g. 0.75.
0.95 is a common target; correcting to 1.00 costs more capacitors and risks over-correction at light load.
V
Single-phase: line voltage. Three-phase: line-to-line voltage.
Hz
60 Hz in North America, 50 Hz in most other countries.
Capacitor rating required
55.32 kvar
Total for all three phases
Capacitance per phase, delta-connected
212.3 µF
Wye-connected: 636.9 µF per phase
Reactive power before
88.19 kvar
Reactive power after
32.87 kvar
Apparent power before
133.33 kVA
Apparent power after
105.26 kVA
Line current before
160.4 A
Line current after
126.6 A
21.1% less current

Estimate generated with SortedSums: https://sortedsums.com/power-factor-correction-calculator/. For informational purposes only; verify before relying on it.

Estimate only. This tool is for informational and educational purposes. Results depend on your inputs and simplifying assumptions, and are not a substitute for professional engineering, design or financial advice. Always verify with a qualified professional and applicable codes before purchasing, building or making decisions.

How it’s calculated

Real power P (kW), reactive power Q (kvar) and apparent power S (kVA) form a right triangle, with power factor pf = P/S = cos φ. A capacitor supplies leading reactive power that cancels part of the load’s lagging (inductive) kvar, so the real power stays the same while S and the line current fall.

Qc = P × (tan φ₁ − tan φ₂) tan φ = √(1 − pf²) / pf Single-phase: C = Qc / (2π f V²) Three-phase: C per phase (delta) = Qc / (3 × 2π f V_LL²) Line current: I = S / V (1φ) or S / (√3 V_LL) (3φ)

Example (Kuphaldt, Electric Circuits II §11.4): a 240 V, 60 Hz single-phase load draws 1.5 kW at pf 0.65 (S = 2.308 kVA, Q = 1.754 kvar). Correcting fully to pf 1.0 needs Qc = 1.754 kvar, so C = 1754 ÷ (2π × 60 × 240²) = 80.8 µF.

Three-phase example: 100 kW at pf 0.75 → target 0.95. tan φ₁ = 0.8819, tan φ₂ = 0.3287, so Qc = 100 × 0.5532 = 55.3 kvar. At 480 V the line current falls from 160.4 A to 126.6 A.

Assumes a steady, sinusoidal load. Harmonic-rich loads (drives, rectifiers) can resonate with capacitors; capacitor banks need proper rating, protection and discharge resistors. Choose the next standard kvar size.

Frequently asked questions

Why not correct to a power factor of 1.0?

The last few percent need disproportionately more kvar, and when the load drops the bank can over-correct to a leading power factor. Most utilities’ penalty thresholds are 0.90–0.95.

Does power factor correction reduce my kWh?

Not directly: real power is unchanged. It reduces current, kVA demand charges and I²R losses in cables and transformers upstream of the capacitors.

Delta or wye capacitors?

For the same kvar, delta-connected capacitors need one third of the capacitance of wye-connected ones because each sees the full line-to-line voltage, but they must be rated for that voltage.

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Sources

Formulas are taken from the free public references above. Results are provided “as is” for informational and educational purposes only. See our disclaimer.

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