Power-factor correction
Steady balanced load.
Method
Qc = P (tan φ1 − tan φ2). Worked case: 18.85 kW from 0.85 to 0.95 needs about 5.49 kVAr.
What to write beside the result
A review can only check this figure if the inputs sit next to it. Write the value, the unit and the assumption the form used. A diameter in millimetres and a diameter in metres are not the same reading. A force in kilonewtons divided as if it were newtons is a thousand-fold error.
The result is the arithmetic on this page. It is not a utilisation, not a selected product, and not a clause from a standard. If the number is going into a shared container, name the page, the date and the inputs. The file-name checker does not read the calculation.
Logic: how this figure is built.
How to read a power-factor correction
The reactive power to move a load from one power factor to another is P times the change in tangent of the two angles. 18.85 kW from 0.85 to 0.95 needs about 5.49 kVAr. That is the capacitor bank size for that change, if the load is constant.
The formula uses the angles whose cosines are the two power factors. It assumes the real power stays the same. A load that changes while you correct it is not this difference.
Leading and lagging are not distinguished. A site that is already leading does not want the same bank. Harmonics and a switched step are not in the number.
Use it to see the order of a correction before an electrical design. It is not a capacitor specification.
- Related: three-phase power
- electrical limits
Last checked 3 October 2026. See the common-calculations note and the disclaimer.