Power Factor Correction — Capacitor kVAR Sizing
Inductive loads draw magnetising current that does no work but still has to travel through your conductors and transformer. Capacitors supply that reactive current locally instead. The sizing formula is Qc = P × (tan φ₁ − tan φ₂) — and the practical target is 0.95, not unity, because over-correction causes its own problems.
Size the capacitor bank
From your utility bill or a power-quality meter
0.95 is the practical target — avoid over-correction
Leave at 0 if your tariff bills kW demand only — correction then saves capacity, not money.
Capacitance required
Correcting 0.75 → 0.95 cuts apparent power and line current.
Standard bank
60 kVAR
Line current freed
33.76 A
See the breakdown
Where VFDs or other harmonic sources are present, use a detuned or harmonic-filtered bank — plain capacitors can resonate with the supply inductance and amplify harmonic currents.
The formula, explained in plain English
Correction doesn't reduce kW
Real power is unchanged — the load still does the same work. What falls is apparent power and therefore current, which frees capacity and cuts a kVA demand charge.
Check the tariff first
If the utility bills kW demand with no PF penalty, correction saves no money on the bill. It is a capacity project then, not an energy project — still valuable, but justified differently.
Location changes the benefit
At the motor, correction unloads the branch circuit too. At the service, only the utility side benefits and internal conductors still carry the reactive current.
Harmonics complicate it
Capacitors plus supply inductance form a resonant circuit. If that resonance sits near a harmonic a VFD produces, currents amplify. Detuned banks exist for this.
Worked examples
A standard correction, a tariff with a demand charge, and the diminishing return near unity.
100 kW load, 0.75 → 0.95 at 480 V
The defaults above.
Qc = 100 × (0.8819 − 0.3287) = 55.32 kVAR → 60 kVAR bank
kVA: 133.33 → 105.26 · amps: 160.4 → 126.6
line current freed: 33.76 A (21.1%)
Result: 21% of the feeder's current capacity recovered without touching a conductor. On a service running near its limit, that is often cheaper than an upgrade.
Same load with an $18/kVA demand charge
A typical commercial kVA demand tariff.
28.07 × $18 = $505/month → $6,063/year
Result: a 60 kVAR bank typically installs for a few thousand dollars, so payback here is well under a year. This is the case that makes correction an easy sell — but it exists only because the tariff bills kVA.
Chasing unity — 0.95 → 1.00
The same 100 kW load, going the last step.
0.95 → 1.00 needs a further 32.87 kVAR and saves only 5.26 kVA
59% more capacitance for 19% more benefit
Result: the tangent curve steepens sharply near unity, so the last 0.05 is poor value — and fixed capacitance that reaches unity at full load will over-correct into leading power factor whenever the load drops. That is why 0.95 is the standard target.
kVAR multiplier table
Multiply your load in kW by the figure below to get the required capacitor kVAR. These are the tan φ₁ − tan φ₂ values, computed with the same function the calculator uses.
| Current PF | Target 0.90 | Target 0.95 | Target 1.00 |
|---|---|---|---|
| 0.65 | 0.6848 | 0.8404 | 1.1691 |
| 0.70 | 0.5359 | 0.6915 | 1.0202 |
| 0.75 | 0.3976 | 0.5532 | 0.8819 |
| 0.80 | 0.2657 | 0.4213 | 0.7500 |
| 0.85 | 0.1354 | 0.2911 | 0.6197 |
| 0.90 | — | 0.1556 | 0.4843 |
Sources & standards: Qc = P × (tan φ₁ − tan φ₂) is standard power-engineering practice. NEC (NFPA 70) 2023 — Article 460 capacitors, 460.8 conductor sizing at 135% of capacitor rated current, 460.9 overcurrent protection, Article 430 Part VIII for capacitors on motor circuits. IEEE 519 covers harmonic limits relevant to capacitor resonance. Utility demand-charge structures vary — confirm the tariff before promising a saving.
Frequently asked questions
Common questions about power factor, capacitor sizing, and over-correction.
How do I calculate capacitor kVAR for power factor correction?
Qc = P × (tan φ₁ − tan φ₂), where P is real power in kW, φ₁ = arccos(current PF) and φ₂ = arccos(target PF). Correcting a 100 kW load from 0.75 to 0.95 needs 100 × (0.8819 − 0.3287) = 55.32 kVAR of capacitance, so you would install a 60 kVAR bank.
What is power factor and why is it low?
Power factor is the ratio of real power to apparent power — how much of the current is actually doing work. It drops below 1.0 whenever inductive loads are present, because they draw magnetising current that lags the voltage. Motors, transformers, welders, and magnetic ballasts are the usual culprits. A lightly loaded motor is worse than a fully loaded one, so oversized motors are a common cause of poor plant power factor.
What power factor should I target?
0.95 is the usual target. It captures most of the available benefit and stays clear of the risk of over-correction, where too much capacitance pushes power factor leading and can cause voltage rise, resonance with harmonics, and self-excitation problems on motors. Chasing 0.99 costs disproportionately more capacitance for very little extra gain — the tangent curve steepens sharply near unity.
How much money does correction actually save?
It depends entirely on your tariff. If the utility bills on kW demand only and applies no power-factor penalty, correction saves nothing directly on the bill — though it still frees conductor and transformer capacity. If the tariff has a kVA demand charge or a power-factor penalty below some threshold (commonly 0.90 or 0.95), the savings can be substantial and pay back a bank in one to three years. Enter your kVA demand rate above to see the figure.
Where do the capacitors go?
Three options with different trade-offs. At the motor is most effective because it unloads the branch circuit as well as the service, but needs care not to over-correct at light load and must be switched with the motor. At the panel is a good compromise. At the service corrects the utility bill but leaves internal conductors carrying the full reactive current. Automatic switched banks are used where load varies widely.
Can capacitors cause problems?
Yes, and this is why 0.95 rather than unity is the target. Fixed capacitance on a variable load can over-correct at light load, producing a leading power factor and voltage rise. Capacitors also form a resonant circuit with the supply inductance, and if that resonance lands near a harmonic frequency present in the system — very common where there are VFDs — currents can amplify badly. Harmonic-filtered or detuned banks exist for exactly this case.
Correction sized? Quote the install in seconds.
Capacitor bank, disconnect, conductors, and labor — TradesQuote turns the scope into a detailed line-item estimate with quantities, unit prices, and totals, validated by a built-in quality control agent.
AI line-item estimates
Quantities, unit prices, and totals generated instantly.
Knowledge base
Upload past jobs so estimates reflect your real pricing.
Shareable & signable
Clients review, accept, and sign from a public link.
No credit card required · 14-day free trial · Cancel anytime