ElectroHub

Power Factor Correction Calculator

Inductive loads — motors, chokes, welding sets — draw lagging reactive current that the supply must carry but nobody is paid for. A capacitor bank supplies that reactive power locally, cutting the current in the cable and, in most industrial tariffs, the penalty on the bill.

Power-Factor Correction — Q_c = P(tan φ₁ − tan φ₂)

Size the shunt capacitor that lifts a load's power factor from cos φ₁ to a better cos φ₂. Add the supply voltage and frequency to get the capacitance in farads (single-phase).

Reactive power before Q₁
Reactive power after Q₂
Capacitor rating Q_c
Apparent power before S₁
Apparent power after S₂
Capacitance C
Power triangle before correction (φ₁)
P = 100 kWQ₁ = 102.0204 kVARφ=45.6°S₁ = 142.8571 kVA
  1. 1.Given: P = 100 kW, cos φ₁ = 0.7, cos φ₂ = 0.95
  2. 2.Formulaφ = cos⁻¹(pf), then reactive power Q = P·tan φSubstituteQ₁ = 100000 × tan(45.573°), Q₂ = 100000 × tan(18.1949°)ResultQ₁ = 102.0204 kVAR, Q₂ = 32.8684 kVAR
  3. 3.FormulaCapacitor VARs Q_c = Q₁ − Q₂ = P(tan φ₁ − tan φ₂)SubstituteQ_c = 102020.4061 − 32868.4105ResultQ_c = 69.152 kVAR
  4. 4.FormulaC = Q_c ÷ (2πf·V²)SubstituteC = 69151.9956 ÷ (2π × 50 × 400²)ResultC = 1375.7352 µF
Common trap: The capacitor only cancels the reactive power — real power (kW) is unchanged, so the load does exactly the same work. What drops is the current the supply must push, which is why correcting to unity would need a huge capacitor for almost no extra benefit past ~0.95.

The formula

Q_c = P × (tan φ₁ − tan φ₂), where φ = arccos(pf)

P
real power of the load, unchanged by correction (watts)
pf₁, pf₂
power factor before and after correction
Q_c
reactive power the capacitor must supply (VAr)

Worked example

A 10 kW load at 0.7 lagging, to be corrected to 0.95.

  1. φ₁ = arccos(0.7) = 45.6°, so tan φ₁ = 1.020
  2. φ₂ = arccos(0.95) = 18.2°, so tan φ₂ = 0.329
  3. Q_c = 10 × (1.020 − 0.329)

Q_c ≈ 6.9 kVAr of capacitance, which drops the line current by about 26 %.

Where you'll use it

Industrial energy audits and the standard exam question on capacitor bank sizing. Correcting all the way to unity is rarely done — the last few percent needs disproportionately large capacitors and risks leading power factor at light load.

The laws behind it

Parts this applies to

Resistor (fixed)Capacitor (aluminium electrolytic)Capacitor (tantalum)LED (light-emitting diode)MOSFETRectifier DiodeSchottky DiodeZener Diode

Related calculators

Power Triangle CalculatorS² = P² + Q²Three-Phase Power CalculatorS = √3 × V_L × I_LCapacitive and Inductive Reactance CalculatorX_C = 1 / (2πfC)RLC Impedance CalculatorZ = √(R² + (X_L − X_C)²)