Power Factor Correction Calculator
Find the reactive power a capacitor bank must supply to raise the power factor of a load, the capacitance per phase for star or delta connection, and the reduction in line current.
- tan φ1 = tan(acos 0.75) = 0.8819, tan φ2 = tan(acos 0.95) = 0.3287
- Q = P × (tan φ1 − tan φ2) = 100 × 0.5532 = 55.32 kVAR
- Delta: C = Q ÷ (3 × 2πf × V²) = 340.8 µF per phase
Power Factor Correction Formulas
Delta: C = Qc ÷ (3 × 2πf × VL²) per phase
Star: C = Qc ÷ (2πf × VL²) per phase
P is active power (kW), φ1 and φ2 the angles for the existing and target power factor, f the frequency and VL the line voltage.
Worked example
A 100 kW load at pf 0.75 is corrected to 0.95 at 415 V, 50 Hz. Q = 100 × (0.882 − 0.329) = 55.3 kVAR. In delta that is about 341 µF per phase, and line current falls from 185.5 A to 146.4 A (21 % less).
Frequently Asked Questions
How do I calculate kVAR for power factor correction?
Multiply active power in kW by the difference between tan φ at the existing and target power factor: kVAR = kW × (tan φ1 − tan φ2).
Is a delta or star capacitor bank better?
Delta connection needs one third of the capacitance of star for the same kVAR, so delta banks are smaller and are standard for LV correction.
Why improve power factor?
Higher power factor reduces line current, cable and transformer losses, voltage drop and, in many tariffs, reactive energy or maximum demand charges.
Learn more on the blog
Power factor correction explained
A practical, field focused guide on instrumentationblog.in that explains the theory behind this calculator and how engineers apply it on real plants.
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