How the power factor maths works
AC power has three faces. Real power P (kW) does work; reactive power Q (kVAr) sustains magnetic fields in motors, transformers and wiring; apparent power S (kVA) is the vector sum — what the wires and the utility bill demand charges actually carry:
S = P ÷ PF · Q = P × tan(acos PF)
Capacitors supply reactive power locally, cancelling the magnetising current so the source no longer carries it. The classic correction formula sizes the bank:
Qc = P × [tan(acos PF₁) − tan(acos PF₂)]
where PF₁ is the current and PF₂ the target power factor. The result is in kilovars (kVAr) — the nameplate rating you order.
Worked example
A 50 kW motor load running at PF 0.75 draws S = 50 ÷ 0.75 ≈ 66.7 kVA with Q ≈ 44.1 kVAr. Correcting to 0.95 needs Qc = 50 × (0.8819 − 0.3287) ≈ 27.7 kVAr of capacitance. Apparent power falls to 52.6 kVA — 21% of cable and transformer capacity released, lower line losses (I²R falls with the square of current), and usually the end of any utility power-factor penalty.
Why utilities care
Utilities size everything — transformers, conductors, generator capacity — on kVA, not kW. A customer at PF 0.7 forces them to deliver 1.43 kVA per kW delivered, so most commercial tariffs bill a PF penalty or a kVA demand charge below roughly 0.90–0.95. Correction to 0.95 is the economic sweet spot; pushing to unity wastes money on the last few kVAr because the required capacitance grows nonlinearly near 1.0.