Power-factor correction
Size the capacitor bank that lifts a load from its present power factor to a target, and see the line current it saves. Real power never changes — only the reactive component the supply has to carry.
| Quantity | Before | After |
|---|---|---|
| Power factor | 0.720 | 0.950 |
| Phase angle | 43.9° | 18.2° |
| Apparent power | 347.2 kVA | 263.2 kVA |
| Reactive power | 241.0 kVAr | 82.2 kVAr |
| Line current | 501.2 A | 379.8 A |
| Capacitor bank | 158.8 kVAr | 3159 µF per phase |
The bank supplies reactive power locally, so the utility no longer has to deliver it through the same cables — which is why line current falls without the real load changing.
Q = P·tan(acos pf) · per phase quantities scale identically
Q_c = P(tan φ₁ − tan φ₂) · the classic correction formula
C_ph = (Q_c/3)/(2πf·V_ph²) · three-phase: Q_c split equally across phases; V_ph = V_L/√3, so this equals Q_c/(2πf·V_L²)
I = P/(√3·V_L·pf) · three-phase, line quantities
A clear path from inputs to results.
Correction capacitors supply reactive power locally, so the utility no longer has to deliver it through the same cables. Real power never changes — only the current that carries it, which is why correction releases capacity without touching the load.
How to use
- Enter the real power of the load.
- Set the present power factor and the target.
- Enter the supply voltage, frequency and phase count.
- Read the kVAr required and the capacitance per phase.
- Check the current relief and the apparent power released.
The formula
Reference: IEEE 18 (shunt power capacitors); IEC 60831.
Worked example
250 kW at 0.72 corrected to 0.95 on a 400 V system
- Real power 250 kW.
- Power factor 0.72 → 0.95.
- 400 V, 50 Hz, three phase.
- 1
Q before: 250 × tan(acos 0.72) = 240.9 kVAr.
- 2
Q after: 250 × tan(acos 0.95) = 82.2 kVAr.
- 3
Bank: 241.0 − 82.2 = 158.8 kVAr.
- 4
Capacitance per phase: 158.8e3 / (3 × 2π × 50 × 230.9²) = 3,159 µF.
- 5
Line current falls from 501 A to 380 A.
158.8 kVAr bank · 3,159 µF per phase · line current down 121 A (24%).
Interpreting the result
The current reduction is where the money is: lower cable losses, released transformer capacity, and no utility penalty. But correction is a one-time capital cost against a permanent saving, so the payback depends on load hours and the tariff structure. Aim for 0.95–0.98, not unity — over-correction at light load gives a leading power factor, which some tariffs penalise too.
Assumptions & limitations
- A linear, sinusoidal load with no significant harmonic content.
- The load runs at a steady power factor; varying loads need switched or automatic banks.
- Voltage is constant and the capacitor is rated for it; capacitors on higher-voltage systems need series reactors.
- No resonance study is performed, which is essential where variable-speed drives are present.
Frequently asked questions
Why not correct to unity power factor?
Because the last few kVAr buy almost no current relief — the relationship is tangent-driven and flattens near unity — while risking a leading power factor at light load, which some utilities penalise.
Can capacitors cause problems with drives?
Yes. Capacitors and drive harmonics can resonate, amplifying harmonic voltages. Detuned reactors or active filters are required, and a harmonic study should precede the installation.
What is the payback on correction?
Typically one to three years for industrial loads running many hours. The saving comes from removed kVA demand charges, lower losses and avoided penalties.
How do I size the capacitor for a single-phase load?
Use the phase voltage in C = Q_c/(2πfV²) and set the phase count to one. The formula is the same; only the voltage basis changes.
Methodology last reviewed 2026-10-18. Results are informational and not financial or engineering advice — see our disclaimer.