Power Factor Calculator
Power factor (PF) is the ratio of real power (kW) to apparent power (kVA) in an AC circuit. PF = kW / kVA, always between 0 and 1. A PF of 1.0 (100%) means all current performs...
Formula
Source: IEEE 1459, IEEE 18 (Shunt Power Capacitors), NEC 460 | Last reviewed: July 26, 2026
Examples
100 kW
= 80 %
- kva = 125
100 kW / 125 kVA = 0.80 PF -- typical uncorrected industrial motor load
85 kW
= 85 %
- kva = 100
85 kW / 100 kVA = 0.85 PF -- minimum acceptable per most utilities
500 kW
= 95 %
- kva = 526.3
500 kW / 526.3 kVA = 0.95 PF -- target with correction capacitors
10 kW
= 100 %
- kva = 10
10 kW / 10 kVA = 1.0 PF -- pure resistive load
Quick Reference Table
| PF Old | Target PF 0.90 | Target PF 0.92 | Target PF 0.95 | Target PF 0.98 |
|---|---|---|---|---|
| 0.60 | 0.849 | 0.907 | 1.005 | 1.192 |
| 0.65 | 0.685 | 0.743 | 0.840 | 1.020 |
| 0.70 | 0.536 | 0.594 | 0.692 | 0.878 |
| 0.75 | 0.398 | 0.456 | 0.553 | 0.740 |
| 0.80 | 0.266 | 0.324 | 0.421 | 0.608 |
| 0.82 | 0.214 | 0.272 | 0.369 | 0.556 |
| 0.85 | 0.135 | 0.194 | 0.291 | 0.477 |
| 0.88 | 0.059 | 0.117 | 0.214 | 0.401 |
| 0.90 | 0.000 | 0.058 | 0.155 | 0.342 |
| 0.92 | -- | 0.000 | 0.098 | 0.284 |
| 0.95 | -- | -- | 0.000 | 0.186 |
| Equipment | Full Load PF | Half Load PF | Notes |
|---|---|---|---|
| Squirrel cage induction motor (1800 RPM) | 0.85-0.90 | 0.60-0.75 | Larger motors have better PF |
| Squirrel cage induction motor (3600 RPM) | 0.88-0.92 | 0.65-0.80 | High-speed = better PF |
| Synchronous motor (over-excited) | 1.0 (leading) | 1.0 (leading) | Can supply kVAr to system |
| Transformer (loaded) | 0.95-0.99 | 0.50-0.80 | PF improves with loading |
| Fluorescent lighting (magnetic ballast) | 0.50-0.65 | N/A | Corrected with integral capacitor |
| LED driver (switch-mode) | 0.90-0.98 | 0.85-0.92 | Depends on driver design |
| Arc furnace | 0.60-0.80 | N/A | Highly variable, requires dynamic comp. |
| Induction heating | 0.40-0.70 | N/A | Very low PF without correction |
| Resistance heating | 1.0 | 1.0 | Unity PF -- purely resistive |
| VFD input (6-pulse diode rectifier) | 0.95-0.98 | 0.93-0.96 | Displacement PF near unity; distortion PF lowers total |
Where is this used?
(1) Utility billing -- most rate schedules for industrial customers include a PF penalty.
Common structures: (a) kVA demand charges: the demand charge is based on the maximum 15-minute kVA demand rather than kW, so low PF directly increases the monthly bill even if kWh consumption is unchanged.
A facility with 5,000 kW peak demand at 0.75 PF pays demand charges on 6,667 kVA -- at $12/kVA-month, that is $80,000/year vs $63,158/year at 0.95 PF ($5,263/month vs $5,263/month -- actually the math is: 5000/0.75 = 6667 kVA x $12 = $80,004/year; at 0.95: 5000/0.95 = 5263 kVA x $12 = $63,156/year, saving $16,848/year).
(b) Explicit PF penalty: e.g., 'billable demand is increased by 1% for each 1% by which the PF is less than 0.90.' At 0.75 PF, the penalty is 15% -- billable kW = 5,000 x 1.15 = 5,750 kW, adding $9,000/year at $12/kW-month.
(c) Reactive energy charge: $0.025/kVARh for kVARh exceeding 50% of kWh (roughly PF < 0.89).
These penalties are the primary economic driver for PF correction.
(2) Transformer and generator loading -- a facility's standby generator must be sized for kVA, not kW.
A 500 kW load at 0.7 PF requires a 714 kVA generator (500 / 0.7 = 714 kVA), while the same load at 0.95 PF requires only 526 kVA (500 / 0.95 = 526 kVA).
The difference -- 188 kVA -- represents generator capacity that could serve additional loads or allow downsizing to the next smaller standard generator frame.
Generator manufacturers rate units at 0.8 PF standard, meaning a 500 kW generator can deliver 625 kVA.
A load at 0.7 PF would require derating the generator's kW output below nameplate.
(3) Capacitor bank design -- the kVAr of capacitors needed is: kVAR_needed = kW x [tan(arccos(PF_old)) - tan(arccos(PF_new))].
The existing and target PF values come directly from this calculator.
(4) Energy efficiency audits -- PF correction is consistently the fastest-payback energy conservation measure, with typical payback periods of 6-18 months.
The savings come from reduced demand charges (immediate, first month after installation) and I squared R loss reduction (ongoing).
(5) Harmonic compliance (IEEE 519) -- capacitor banks can form resonant circuits with the inductance of transformers and cables at or near harmonic frequencies produced by VFDs.
A PF correction design must include harmonic analysis: the resonant frequency f_r = f_fundamental x sqrt(kVA_SC / kVAR_cap), where kVA_SC is the short-circuit kVA at the capacitor bus.
If f_r coincides with a strong harmonic (5th, 7th, 11th), harmonic currents will be amplified, potentially damaging capacitors, overheating transformers, and causing erratic operation of electronic equipment.
Detuned capacitor banks (with series reactors tuned to 189 Hz for 60 Hz systems) shift the resonance below the 5th harmonic, protecting against amplification.
(6) Motor PF at light load -- individual motors operating at less than 50% load have notoriously low PF (0.4-0.6).
Rather than correcting at the main switchboard, it is often more cost-effective to install capacitors at the motor terminals, which also reduces losses in the motor feeder circuit.
NEC 460.8 limits the capacitor kVAr at the motor to that which raises the no-load PF to unity, preventing self-excitation and overvoltage if the motor coasts down after disconnection.
Real-World Usage Scenarios
Utility penalty audit triggers capacitor bank investment
A plastics injection molding plant with 2,800 kW average demand receives a utility bill showing a PF of 0.72. The demand charge is $14.50/kVA-month on a kVA-based rate, and the bill shows 3,889 kVA peak (2,800 / 0.72). Annual demand charges = 3,889 x $14.50 x 12 = $676,686. An engineering study shows that a 1,800 kVAr automatic capacitor bank ($42,000 installed) would raise PF to 0.96, reducing kVA demand to 2,917 (2,800 / 0.96). New annual demand charges = 2,917 x $14.50 x 12 = $507,546. Savings = $169,140/year. Additional I squared R loss savings estimated at $8,200/year. Total savings = $177,340/year on a $42,000 investment -- simple payback = 42,000 / 177,340 x 12 = 2.8 months. The project is approved in the next capital cycle, and the plant's electrical engineer specifies a 6-step automatic bank (6 x 300 kVAr) with detuning reactors (p=7%, tuned to 189 Hz) because VFD loads represent 35% of total connected load.
Standby generator loading limit forces PF correction before expansion
A hospital's essential electrical system is backed up by a 2,000 kW / 2,500 kVA diesel generator (0.8 PF rated). The facility's measured essential load is 1,680 kW at 0.76 PF = 2,211 kVA -- within the generator's 2,500 kVA limit with 11.6% margin. A planned ICU expansion adds 180 kW of load (medical equipment, HVAC, lighting). The engineer calculates: new total load = 1,860 kW; if PF remains at 0.76, kVA = 1,860 / 0.76 = 2,447 kVA -- only 2.1% margin below the 2,500 kVA rating, insufficient for code-required load growth allowance and motor starting transients. Rather than replace the $350,000 generator, the engineer specifies a 600 kVAr fixed capacitor bank on the essential bus to raise PF to 0.92. New kVA = 1,860 / 0.92 = 2,022 kVA -- 19.1% margin, well within NFPA 110 requirements. The capacitor bank costs $18,000 and is installed during a scheduled 4-hour generator test outage. Generator replacement avoided; expansion approved.
Harmonic resonance failure investigation reveals PF correction design flaw
A food processing facility installed a 1,200 kVAr fixed capacitor bank to correct PF from 0.70 to 0.96 on a 3,500 kW load. Six months later, capacitors begin failing -- bulging cases, leaking dielectric fluid, and one violent rupture. Simultaneously, the facility's 15 VFDs experience nuisance overvoltage trips. Harmonic measurements reveal: 5th harmonic current (300 Hz) is being amplified 4.7x by a parallel resonance at 282 Hz -- very close to the 5th harmonic (300 Hz). The resonance frequency f_r = 60 x sqrt(MVA_SC / MVAr_cap). The 480V bus has a short-circuit capacity of 42 MVA, and the 1.2 MVAr capacitor bank creates a resonance at 60 x sqrt(42 / 1.2) = 60 x sqrt(35) = 60 x 5.92 = 355 Hz. The 5th harmonic (300 Hz) is close enough to experience significant amplification (Q factor approximately 8 at this proximity). Solution: retrofit the capacitor bank with 7% detuning reactors (series inductors in each phase), shifting the resonant frequency to 60 x sqrt(1 / 0.07) = 60 x 3.78 = 227 Hz -- safely below the 5th harmonic. The retrofit costs $31,000. The alternative -- active harmonic filtering -- was quoted at $84,000. The facility also implements IEEE 519 harmonic limits in its equipment procurement specifications going forward.
Common Mistakes to Avoid
Treating PF as a percentage of 'efficiency' rather than a phase relationship
An operator sees PF = 0.75 and concludes the motor is '75% efficient' -- a completely different concept. Motor efficiency = mechanical output / electrical input (typically 85-96%). Power factor = kW / kVA (determined by the inductive reactance of the motor windings). A motor can be 93% efficient while having a 0.70 PF -- the two metrics describe fundamentally different energy flows. Efficiency describes how much of the electrical input becomes useful mechanical work (losses are heat in the windings, friction, windage). PF describes how much of the total current (including magnetizing current) produces real power. A 100 HP motor at full load might be 93.6% efficient (electrical input = 79.7 kW for 74.6 kW mechanical output) and have a PF of 0.87 (apparent power = 79.7 / 0.87 = 91.6 kVA). Confusing the two leads operators to believe PF correction saves kWh energy -- it does not (except for reduced I squared R losses, which is a secondary effect). PF correction primarily saves kVA demand charges, which are measured and billed in most industrial rate schedules.
Measuring only kW and assuming PF = 1.0 for cost allocation
A multi-tenant commercial building allocates electricity costs based on each tenant's kW sub-meter. Tenant A's sub-meter shows 85 kW average; Tenant B's shows 85 kW. Both pay identical shares of the master utility bill. However, Tenant A's load is primarily resistive heating (PF approximately 1.0), while Tenant B operates a large refrigeration warehouse (PF approximately 0.72). The master utility meter shows aggregate PF = 0.83, and the bill includes $4,200/month in PF penalties. Tenant B's refrigeration load is the sole cause of the penalty, but the kW-only allocation charges both tenants equally -- a $25,200/year cross-subsidy from Tenant A to Tenant B. The corrected approach: allocate the kVA demand charge based on each tenant's kVA (measured with a PF-capable meter), not kW. This reveals Tenant B's true cost and incentivizes them to install local PF correction.
Neglecting displacement PF vs distortion PF in VFD-dense facilities
A data center with UPS and VFD loads measures 'PF = 0.89' on a true RMS power analyzer, but the displacement PF (fundamental 60 Hz) is 0.98 -- nearly unity. Why the discrepancy? The UPS input rectifiers draw non-sinusoidal current with a THD of 42%. Per IEEE 1459, total PF = displacement PF / sqrt(1 + THD squared) = 0.98 / sqrt(1 + 0.42 squared) = 0.98 / 1.084 = 0.904. The low total PF is caused by harmonic distortion, not reactive power. Adding a capacitor bank would not correct this -- it could make things worse by creating a resonant circuit that amplifies the harmonics. The correct solution is an active harmonic filter, a multi-pulse rectifier (12-pulse or 18-pulse), or passive tuned harmonic filters at the 5th, 7th, and 11th harmonics. The facility manager who installs a 400 kVAr capacitor bank to 'fix the 0.89 PF' risks catastrophic equipment damage.
Industry Standards Referenced
Frequently Asked Questions
What is a good power factor?
Utilities typically require PF >= 0.85-0.90. Values below 0.85 usually incur a penalty. PF = 0.95 is a common design target for facilities with power factor correction -- it eliminates utility penalties in all jurisdictions and provides margin against degradation as equipment ages (PF naturally deteriorates as motors are downsized or lightly loaded). PF = 1.0 is ideal but rarely achieved in practice due to transformer magnetizing current (typically 1-3% of transformer rating, purely inductive) and other inherent reactive loads that cannot be perfectly cancelled without the risk of over-correction (leading PF). A PF below 0.7 indicates significant uncorrected inductive loads -- motors running at light load (low PF is characteristic of underloaded motors where the magnetizing current dominates the total current), magnetic ballast fluorescent lighting, or large transformer banks with low load factor.
How do I improve a low power factor?
Install power factor correction capacitors at the main switchboard. The kVAr rating of the capacitor bank = kW x (tan(phi_old) - tan(phi_new)). For a 500 kW load improving from 0.78 to 0.95: kVAr = 500 x (tan(arccos(0.78)) - tan(arccos(0.95))) = 500 x (0.805 - 0.329) = 238 kVAr. Select a standard capacitor bank rated approximately 250 kVAr. Automatic capacitor banks with stepped switching are preferred for variable loads -- a power factor controller monitors the bus and switches capacitor steps in/out to maintain target PF without over-correction. For facilities with many VFDs, active harmonic filters or detuned capacitor banks (with series reactors tuned to 189 Hz for 60 Hz systems) prevent harmonic resonance. Always conduct a harmonic survey before designing capacitor banks if non-linear loads exceed 25% of total load.
What causes a low power factor?
Inductive loads are the primary cause: motors (especially when underloaded -- the magnetizing current is constant regardless of load, so at light load it becomes a larger fraction of total current), transformers (magnetizing current is purely inductive, approximately 1-3% of rated current per transformer regardless of load), fluorescent lighting with magnetic ballasts, arc furnaces, and induction heaters. In industrial plants, motors typically represent 60-80% of the total electrical load, and their cumulative PF -- especially when many are lightly loaded -- drives the facility's overall PF below utility penalty thresholds. Bulk PF correction at the main service entrance is the most common solution, but point-of-use capacitors at individual large motors (>100 HP) provide the additional benefit of reducing losses in the motor feeder circuits. For new facilities, specifying premium-efficiency motors and VFDs with active front ends (which can provide unity PF and even leading PF if needed) eliminates the PF problem at the design stage.
What is the difference between displacement PF and total (true) PF?
Displacement PF (DPF) considers only the fundamental 60 Hz voltage and current. It equals cos(phi), where phi is the phase angle between fundamental voltage and fundamental current. DPF is what traditional analog PF meters and simple digital meters measure. Total PF (also called true PF) accounts for both the fundamental phase shift AND the effect of harmonic currents: Total PF = DPF / sqrt(1 + THD squared). In facilities with many VFDs, UPS systems, LED lighting, and electronic ballasts, harmonic currents (THD of 20-50% is common) significantly reduce the total PF even though the DPF may be near unity. The difference matters because: (a) capacitors correct only displacement PF -- they cannot correct distortion PF and may make it worse through resonance, and (b) utility PF penalties are increasingly based on total PF rather than displacement PF as smart meters become standard. Always use a true RMS power quality analyzer (compliant with IEEE 1459) to measure PF in modern facilities.
How does power factor affect my electric bill?
Three mechanisms: (1) kVA demand charges -- many industrial rate schedules base the monthly demand charge on the peak kVA (not kW) over a 15- or 30-minute interval. A 2,000 kW average load at 0.75 PF incurs a kVA demand of 2,667 kVA; at 0.95 PF it drops to 2,105 kVA. At $12/kVA-month, that is $32,004/month vs $25,260/month -- a $6,744 monthly difference. (2) Explicit PF penalty -- some utilities apply a multiplier to the kW demand when PF falls below a threshold (often 0.85 or 0.90). A common formula: billable demand = measured demand x (0.90 / measured PF). At 0.75 PF, billable demand = 2,000 x (0.90/0.75) = 2,400 kW -- a 20% penalty on the demand charge. (3) Reactive energy charges -- similar to kWh charges but for kVARh, typically at a lower rate ($0.02-0.04/kVARh). If kVARh exceeds 50% of kWh (equivalent to PF < 0.89), the excess kVARh is billed. Check your utility tariff to determine which mechanism applies -- many large customers have tariff choices and can select the structure most favorable after PF correction.
Reviewed for accuracy
Reviewed against IEEE 1459-2010, IEEE 18-2012 (Shunt Power Capacitors), and NEC 2023 Article 460 · Last reviewed: July 26, 2026
All calculations are for reference only. Always verify with manufacturer data and a qualified engineer for critical applications. Learn about our editorial process.