Transmission & Grid calculator
Power Factor Correction Calculator
How much capacitor kvar raises your power factor to a target, and how much current and kVA you free up.
GridBasis: IEEE 1036 (shunt capacitors) · IEC 60831
Results
- Enter values to calculate
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Formula
QC = P · [tan(arccos PF1) − tan(arccos PF2)]
S1 = P / PF1, S2 = P / PF2; current falls in the same ratio as S.
Worked example
500 kW at 0.75 PF corrected to 0.95: QC = 500 × (0.882 − 0.329) = 276.6 kvar. Apparent power falls from 666.7 kVA to 526.3 kVA, cutting current by 21 %.
Assumptions and standards
- Real power stays constant after correction.
- Choose the next standard capacitor bank size up, and check harmonics: capacitors can resonate with system inductance, so detuned banks may be needed where drives or rectifiers are present.
- Avoid over-correction to a leading power factor at light load; switched or automatic banks help.
Formula and data last verified: October 2026.
Frequently asked questions
Why correct power factor?
Many utilities charge penalties below about 0.9–0.95 PF. Correction also frees transformer and cable capacity and reduces I²R losses.
Where should capacitors be installed?
At the main switchboard for billing correction, or at large motors to also relieve the feeders. Motor-applied capacitors must not exceed the motor’s magnetizing kvar, to avoid self-excitation.
Can I use these results for final design?
Use them for preliminary design and checking. Final designs must be verified by a qualified engineer against the code edition and standards adopted where the work is installed.
Is my data sent anywhere?
No. Every calculation runs in your browser. Nothing you type is uploaded or stored on our servers.