ELECTRICAL / CALCULATOR
Power Factor Correction
Estimate compensation for an inductive load and single-phase, star or delta capacitor values.
Input Parameters
Edit the example values. Results update as you type.
Enter capacitor terminal voltage for single phase and line voltage for three phase.
Results
- Required Capacitive Compensation Qc
- 42.132kvar
- Initial Phase Angle φ₁
- 36.87°
- Target Phase Angle φ₂
- 18.195°
- Initial Apparent Power
- 125kVA
- Corrected Apparent Power
- 105.263kVA
- Initial Reactive Power
- 75kvar
- Corrected Reactive Power
- 32.868kvar
- Capacitance per Phase C
- 279.394μF
- Capacitor Connection / Count
- Three-Phase Delta / 3 capacitors
Formula
φ₁ = acos(PF₁) · φ₂ = acos(PF₂) · Qc = P(tanφ₁ − tanφ₂) · Cper phase = Qc / (3 × 2πf × U²)Substitution
Qc = 100 × (tan(acos(0.8)) − tan(acos(0.95))) = 42.132 kvar；C = 42,131.589 / (3 × 2π × 50 × 400²) = 279.394 μFCalculation Steps
- Assuming constant active power, calculate initial and corrected phase angles, apparent power and inductive reactive power.
- Total compensation Qc = Q₁ − Q₂ = 42.132 kvar; convert to 42,131.589 var for capacitance.
- Calculate per phase capacitance for Three-Phase Delta; use line voltage for three-phase systems.
Result Explanation
For shunt-capacitor correction of inductive loads in sinusoidal steady state. Qc is total compensating reactive power. Capacitance is a preliminary estimate; verify equipment nameplates and site conditions.
Formulas and FAQ
Qc = P × (tanφ₁ − tanφ₂)Is the three-phase capacitance per phase or total?
The value is per phase for three capacitors. Enter line voltage. Star capacitance is Qc/(ωUᴸ²) and delta is Qc/(3ωUᴸ²). Qc is total three-phase compensation.