Power Factor Correction

How to use this calculator

This tool sizes the parallel capacitor bank that raises a load's power factor from its present value to a chosen target. Induction motors and other inductive loads draw magnetizing current that lags the voltage, so the supply must carry more amperes than the real (watt) power alone would suggest — the power factor PF = P / S = cos φ is the ratio of real to apparent power. A capacitor bank fixes this locally: it charges and discharges every half-cycle, exchanging energy with the load's magnetic field instead of forcing that energy to travel back and forth along the feeder. The required rating follows from the two power triangles at PF₁ and PF₂: Qc = P·(tanφ₁ − tanφ₂), where φₙ = acos(PFₙ). Real power stays constant throughout — correction changes only the apparent power, and therefore the current, that the supply has to carry.

Input fields

• Phase — 1-phase or 3-phase. This sets the √3 factor in the current conversion; for three-phase, enter line-to-line voltage with line current.
• Real power (P) — the load's actual watt consumption at the operating point you want to correct, in watts or kilowatts (unit selector). Take it from a meter or power analyzer, or derive it from nameplate data (mechanical output ÷ efficiency). This is P in watts, not kVA.
• Supply voltage — the nominal voltage at the correction point: line-to-line for three-phase (400 V, not 230 V), supply voltage for single-phase.
• Current power factor (PF₁) — the load's existing power factor as a decimal between 0 and 1: from the nameplate, a meter reading, or computed as kW ÷ kVA. Enter 0.75, not 75.
• Target power factor (PF₂) — the power factor you want after correction, also a decimal; it must be higher than PF₁. Values of 0.9–0.95 are the usual choice.

Steps

1. Select the phase configuration of the circuit you are correcting.
2. Enter the load's real power (setting W or kW), the supply voltage and the current power factor.
3. Enter the target power factor — 0.9 to 0.95 in most cases.
4. Read the headline capacitor bank rating in kvar; the detail rows give reactive, apparent and real power plus line current before and after correction, and the percentage current reduction.

The default example — a 15 kW three-phase load at 400 V corrected from PF 0.75 to 0.95 — gives φ₁ = acos(0.75) ≈ 41.4° (tanφ₁ ≈ 0.882) and φ₂ = acos(0.95) ≈ 18.2° (tanφ₂ ≈ 0.329), so Qc = 15 × (0.882 − 0.329) ≈ 8.3 kvar. The line current falls from I₁ = P/(√3·V·PF₁) ≈ 28.9 A to I₂ ≈ 22.8 A — a 21% reduction — while the load still consumes exactly 15 kW. Mistakes to avoid: entering power factors as percentages (75 instead of 0.75), setting the target below the current value (the calculator rejects it — that would be de-correction), sizing for today's light load rather than the maximum demand, feeding kVA into the real-power field, and applying the method to a load whose measured power factor is already leading — this tool covers lagging (inductive) loads only.

What it means in a real circuit

The motor's low power factor comes from its magnetizing current: the stator field must be established before any torque exists, and that current lags the voltage by close to 90°. A parallel capacitor is itself reactive — its current leads the voltage by 90° — so it generates precisely the kind of VARs the motor absorbs. With the bank at the load terminals, the magnetizing amperes circulate locally between capacitor and winding, and the feeder carries only the phasor sum of the two. That is why the supply current falls from I₁ to I₂ = I₁·PF₁/PF₂ while the motor itself behaves exactly as before: its own power factor, torque and speed are unchanged. Correction improves what the supply sees, not what the machine needs.

The reduced line current has a chain of practical consequences. Feeder and transformer I²R losses fall with the square of the current, so a 21% current reduction cuts that loss by about 36%. With less drop along the feeder the motor's terminal voltage rises slightly, which improves torque margin (available torque scales roughly with the square of applied voltage) and eases starting on long runs. Breakers, contactors, transformers and generators gain headroom: a circuit running at 95% of its rating can often take an additional load without an upgrade. Utilities commonly bill penalties below about 0.9 power factor — and increasingly pay incentives above it — which is why 0.9–0.95 is the standard correction target: beyond roughly 0.95 each extra kvar buys less current reduction, and a leading combined power factor can trigger its own penalty.

Where you install the bank decides how much of the benefit you capture. Per-motor correction at the terminals gives the maximum saving because every metre of feeder benefits, and it supports terminal voltage during starting; a central bank at the distribution panel is cheaper to fit for many small motors, but the feeders between panel and motor still carry uncorrected current. Over-correction — too much capacitance — pushes the combined load to leading power factor: reactive power then flows back toward the supply, terminal voltage can rise when the load switches off, and on systems with VFDs the bank can form a parallel resonance with system inductance that amplifies harmonics. Detuned reactors are specified where switching transients or harmonics make plain banks risky; this calculator sizes the ideal bank, not the protection around it.

The assumptions to keep in view: constant real power at the operating point entered, sinusoidal steady-state voltage, an ideal lossless capacitor, and a lagging load. The calculation does not model harmonic content of VFD-driven loads (which changes both measured power factor and capacitor sizing practice), bank ageing or temperature, transformer tap settings, or your utility's specific tariff thresholds. Treat the kvar figure as the engineering basis for selection — the final bank is picked from standard capacitor steps (2.5, 5, 10 kvar …) around this value.