Power Factor Correction Capacitor Calculator
Find the exact capacitor size in µF needed to correct a lagging power factor to a target value in AC electrical systems. Widely used by electrical engineers designing industrial motor circuits and utility billing compliance solutions.
Last updated: September 2026
Formula below · 2 sources (ieee.org, Wikipedia) · Updated Sep 2026
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About this calculator
In AC circuits with inductive loads (motors, transformers), current lags voltage, creating reactive power (kVAR) that increases apparent power without doing useful work. Power factor (PF) is the ratio of real power (kW) to apparent power (kVA). To improve PF from an existing value to a target, a capacitor must supply reactive power equal to Q_c = P × (tan(arccos(PF_current)) − tan(arccos(PF_target))), where P is real power in watts. The required capacitance is then C = Q_c / (2π × f × V²), expressed in farads and converted to µF by multiplying by 10⁶. This calculator handles the full derivation, requiring only real power in kW, current and target power factors (as decimals such as 0.75, or percentages such as 75), system voltage, and frequency (50 or 60 Hz). The result is the equivalent capacitance at the line voltage; for a three-phase delta-connected bank, each of the three capacitors needs one third of this value (each phase supplies a third of the kVAR at full line voltage). If the target is at or below the current power factor, no correction is needed and the result is 0.
How to use
A factory has a 100 kW inductive load at 480 V, 60 Hz, with a current power factor of 75% and a target of 95%. Q_c = 100,000 × (tan(arccos(0.75)) − tan(arccos(0.95))) = 100,000 × (0.8819 − 0.3287) = 55,320 VAR. Capacitance C = 55,320 / (2π × 60 × 480²) = 55,320 / 86,973,504 ≈ 6.36 × 10⁻⁴ F = 636 µF. Capacitor banks are normally specified in kVAR, so this is about 55 kVAR at 480 V; as a three-phase delta bank, each capacitor is about 212 µF. Choose the next standard kVAR step and avoid overcorrecting.
Frequently asked questions
Why does improving power factor reduce electricity costs for commercial and industrial facilities?
Utilities bill large commercial and industrial customers not just for real power (kWh) but also for apparent power or reactive demand (kVAR). A low power factor means the utility must supply more current than the actual work requires, stressing transmission infrastructure. Many utilities impose a power factor penalty — or reduce a discount — when PF falls below 0.90 or 0.95. Installing correction capacitors raises PF, reducing the reactive component billed and lowering peak demand charges. The payback period for capacitor banks is often less than two years in facilities with significant motor loads.
What happens if I install a capacitor that is too large for power factor correction?
Over-correction causes the power factor to go leading (capacitive) instead of lagging (inductive), which can be just as harmful as under-correction. A leading power factor can cause voltage rise on the supply network, potential resonance with transformer inductance, and in some cases utility penalties for excessive leading reactive power. It can also cause self-excitation in nearby generators or synchronous motors, leading to overvoltage damage. Always target a PF slightly below unity — typically 0.95 to 0.98 — and stage capacitor banks with automatic switching relays if the load varies significantly.
How does system frequency affect the required capacitor size for power factor correction?
Capacitive reactance (X_c) is inversely proportional to frequency: X_c = 1 / (2π × f × C). At 50 Hz (common in Europe and Asia), a given capacitor produces less reactive power than at 60 Hz (common in North America), because it charges and discharges fewer times per second. Therefore, to supply the same kVAR of correction at 50 Hz as at 60 Hz, a larger capacitance value is required. This calculator accounts for frequency directly in the formula C = Q_c / (2π × f × V²), so simply select your system frequency and the result is automatically adjusted.