Start from whatever you can measure. This power factor calculator returns the real, reactive and apparent power and the power factor, on single-phase or three-phase systems, then sizes the capacitor that corrects the load to your target, in kvar and in microfarads.
RMS. Line-to-line if three-phase.
RMS. Line current if three-phase.
Lagging, above 0 up to 1.
Optional: sizes the correction capacitor.
Optional: output scales with V², so a higher-rated part delivers less at this supply.
Optional: gives the harmonic the bank resonates at. A transformer alone is about kVA ÷ per-unit impedance.
Real power P
3.6 kW
Assumes sinusoidal voltage and current and a lagging (inductive) load, the case shunt capacitors correct.
Correction to common target power factors
| Target PF | kvar per kW | Capacitor | Capacitance | Current after | I²R loss saved |
|---|---|---|---|---|---|
| 0.85 | 0.262 | 944 var | 43.5 µF | 17.6 A | 22.1 % |
| 0.90 | 0.398 | 1.43 kvar | 65.9 µF | 16.7 A | 30.6 % |
| 0.92 | 0.456 | 1.64 kvar | 75.6 µF | 16.3 A | 33.5 % |
| 0.95 | 0.553 | 1.99 kvar | 91.7 µF | 15.8 A | 37.7 % |
| 0.97 | 0.631 | 2.27 kvar | 105 µF | 15.5 A | 40.2 % |
| 0.98 | 0.679 | 2.44 kvar | 113 µF | 15.3 A | 41.4 % |
| 1.00 | 0.882 | 3.17 kvar | 146 µF | 15 A | 43.8 % |
Divide real power by apparent power. The power factor equation is PF = P ÷ S, which equals cos φ for sinusoidal waveforms. A 240 V, 20 A load drawing 3.6 kW has S = 4.8 kVA, so PF = 0.75, and it carries Q = 3.17 kvar of reactive power. Raising it to 0.95 takes a capacitor of Qc = P × (tan φ1 − tan φ2) = 1.99 kvar, which is 91.7 µF at 240 V and 60 Hz.
Real, reactive and apparent power form a right triangle. Real power P is the adjacent side, reactive power Q the opposite side, apparent power S the hypotenuse, and φ the angle between voltage and current.
A shunt capacitor supplies reactive power locally, so the source no longer has to. The capacitor needed to move from φ1 to φ2 at constant real power is the same for single-phase and three-phase systems:
The capacitance follows from Q = 2π f C V² applied to each capacitor and the voltage it sees:
At constant real power the current is inversely proportional to the power factor, and the I²R loss upstream of the capacitor follows the square of the current:
S = 240 × 20 = 4.8 kVA. P = S × 0.75 = 3.6 kW. Q = √(S² − P²) = 3.17 kvar. The angle is 41.4°.
To reach PF 0.95, the multiplier is 0.553, so Qc = 1.99 kvar and C = Qc ÷ (2π × 60 × 240²) = 91.7 µF. The current drops from 20 A to 15.8 A.
This is the example in Eaton’s guide, which reads a multiplier of 0.607 from its table. The formula gives 0.608, so Qc = 249 kvar and Eaton rounds up to a 250 kvar bank. At 480 V and 60 Hz, delta capacitors would be 956 µF per phase. The bank draws 300 A, and the plant’s line current falls from 676 A to 519 A.
Going from PF 0.8 to 0.93 takes Qc = 106 kvar (ABB’s worked example rounds tan φ2 to 0.39 and gets 108 kvar). Line current drops from 541 A to 466 A.
In delta each capacitor is 706 µF across 400 V. In star each is 2.12 mF across 231 V: three times the capacitance at a lower voltage, for the same kvar.
Multiply the load in kW by the value here to get the capacitor kvar. Each entry is tan φ1 − tan φ2, and matches published multiplier tables to within rounding.
| Present PF \ Target | 0.85 | 0.90 | 0.92 | 0.95 | 0.97 | 0.98 | 1.00 |
|---|---|---|---|---|---|---|---|
| 0.50 | 1.112 | 1.248 | 1.306 | 1.403 | 1.481 | 1.529 | 1.732 |
| 0.55 | 0.899 | 1.034 | 1.092 | 1.190 | 1.268 | 1.315 | 1.518 |
| 0.60 | 0.714 | 0.849 | 0.907 | 1.005 | 1.083 | 1.130 | 1.333 |
| 0.65 | 0.549 | 0.685 | 0.743 | 0.840 | 0.919 | 0.966 | 1.169 |
| 0.70 | 0.400 | 0.536 | 0.594 | 0.692 | 0.770 | 0.817 | 1.020 |
| 0.75 | 0.262 | 0.398 | 0.456 | 0.553 | 0.631 | 0.679 | 0.882 |
| 0.80 | 0.130 | 0.266 | 0.324 | 0.421 | 0.499 | 0.547 | 0.750 |
| 0.85 | – | 0.135 | 0.194 | 0.291 | 0.369 | 0.417 | 0.620 |
| 0.90 | – | – | 0.058 | 0.156 | 0.234 | 0.281 | 0.484 |
| 0.95 | – | – | – | – | 0.078 | 0.126 | 0.329 |
The first two columns are the capacitance across the stated voltage: single-phase, or the total of a delta bank. The delta columns are each of the three capacitors. Multiply by the kvar you need.
| Voltage | µF/kvar, 60 Hz | µF/kvar, 50 Hz | Delta µF per phase, 60 Hz | Delta µF per phase, 50 Hz | A/kvar, 1-phase | A/kvar, 3-phase |
|---|---|---|---|---|---|---|
| 120 V | 184 | 221 | – | – | 8.33 | – |
| 208 V | 61.3 | 73.6 | 20.4 | 24.5 | 4.81 | 2.78 |
| 240 V | 46.1 | 55.3 | 15.4 | 18.4 | 4.17 | 2.41 |
| 400 V | 16.6 | 19.9 | 5.53 | 6.63 | 2.5 | 1.44 |
| 415 V | 15.4 | 18.5 | 5.13 | 6.16 | 2.41 | 1.39 |
| 480 V | 11.5 | 13.8 | 3.84 | 4.61 | 2.08 | 1.2 |
| 600 V | 7.37 | 8.84 | 2.46 | 2.95 | 1.67 | 0.962 |
Starting points from ABB’s Technical Application Paper No. 8, Table 1.1. Measure your own load when you can.
| Load | Power factor |
|---|---|
| Transformer, no load | 0.1 to 0.15 |
| Motor | 0.7 to 0.85 |
| Arc welding | 0.35 to 0.6 |
| Arc welding, compensated | 0.7 to 0.8 |
| Resistance welding | 0.4 to 0.6 |
| Arc melting furnace | 0.75 to 0.9 |
| Fluorescent lamps, compensated | 0.9 |
| Fluorescent lamps, uncompensated | 0.4 to 0.6 |
| AC to DC converters | 0.6 to 0.95 |
| DC drives | 0.4 to 0.75 |
| AC drives | 0.95 to 0.97 |
| Resistive load | 1 |
The formulas assume sinusoidal waveforms and a steady load. Check these before ordering a bank.
Divide real power by apparent power: PF = W ÷ (V × A) on a single-phase circuit, and PF = W ÷ (√3 × V × A) on a three-phase circuit with line-to-line voltage and line current. A load reading 600 W at 120 V and 8 A has PF = 600 ÷ 960 = 0.625. The V, I and P mode of this PF calculator takes those three readings directly.
The apparent power formula is S = V × I on single-phase and S = √3 × V × I on three-phase. The reactive power formula is Q = √(S² − P²) = P × tan φ. With real power P they form the power triangle.
Divide kW by the power factor: S = P ÷ PF. 300 kW at PF 0.8 is 375 kVA. Choose P and PF at the top of the calculator to get this directly.
The power factor correction formula is Qc = P × (tan φ1 − tan φ2), where φ1 and φ2 are the angles of the present and target power factors. A 100 kW load going from 0.80 to 0.95 needs 42.1 kvar. Tables of "kvar per kW" multipliers are this same formula, printed.
C = Q ÷ (2π f V²) for a capacitor across voltage V. At 240 V and 60 Hz, 1 kvar is 46.1 µF. In a three-phase delta bank each capacitor sees the line voltage and supplies a third of the kvar, so 1 kvar at 480 V and 60 Hz is 3.84 µF per phase.
About 0.95 is the usual target: Eaton notes that correcting to approximately 95 % gives the maximum practical benefit. Tariffs vary, but ABB describes a typical arrangement where reactive energy is charged for power factors from 0.7 to 0.9 and nothing is due above 0.9.
Usually not. Going from 0.80 to 0.95 on 100 kW takes 42.1 kvar, while reaching 1.00 takes 75 kvar for a small further drop in current, and a fixed bank at unity turns leading as soon as the load falls.
Only for sinusoidal voltage and current. With harmonics, the total power factor P ÷ S equals cos φ of the fundamental multiplied by a distortion factor, so a drive or rectifier can have a good cos φ and a poor total power factor. Capacitors supply reactive power only, so they do not remove the distortion.
It does not change the real power the load uses. It lowers the current upstream of the capacitor, and with it the I²R loss in cables and transformers: going from 0.75 to 0.95 cuts that loss by 37.7 %. Where the tariff bills kVA demand or reactive energy, those charges fall too.
For the same kvar a star bank needs three times the capacitance per phase, but each capacitor sees only VL ÷ √3. At low voltage, delta is usually preferred because the capacitors are smaller.
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