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Three Phase Power Calculator

Convert line voltage and line current into apparent, active and reactive power for a three-phase or single-phase supply. Returns the power factor angle, phase voltage, the capacitor rating needed to reach 0.95, and the annual energy and running cost at your tariff.

S = √3 · V · I P = S · cos φ IEC 60038
🔋 Supply & Load
Line-to-line for three phase, line-to-neutral for single phase.
Used only for the P ÷ η figure — see the note under the results.
8760 is continuous running.
Display only — the tariff figure is used as entered.
Enter the supply and load data and select Calculate.
On the P ÷ η figure: when V and I are measured at the load's own terminals, the active power P is the electrical input, and the useful output is P × η. The API returns P ÷ η, which is the correct reading only if η describes the generation or supply chain upstream of the measurement point. Both figures are shown so you can pick the one that matches where your meter sits.

Three Phase Power Formulae

S = √3 · VL · IL // three phase, volt-amperes
S = V · I // single phase
P = S · cos φ  ·  Q = S · sin φ
S² = P² + Q² // the power triangle
φ = arccos( PF )  ·  Vφ = VL / √3 // star connection
Qc = P · ( tan φ₁ − tan φ₂ ) // correction kVAr, φ₂ = arccos 0.95
E = P · h / 1000  ·  Cost = E × tariff
SymbolMeaningUnit
SApparent power — sizes cables and transformersVA
PActive power — what the meter billsW
QReactive power — magnetising, no useful workVAr
VLLine-to-line voltageV
VφPhase voltage, line to neutralV
ILLine currentA
cos φPower factor, P / S
φPhase angle between voltage and current°
QcCapacitor rating to reach the target PFVAr
ηEfficiency
hOperating hours per yearh

Worked Example

415 V three-phase load drawing 100 A at 0.8 power factor

A three-phase distribution board draws 100 A from a 415 V supply at 0.8 lagging power factor. The connected equipment is 92 % efficient, runs 6000 hours a year, and energy costs 8 per kWh.

Apparent power
S = √3 × 415 × 100 = 71880.11 VA = 71.88 kVA
Active power
P = 71880.11 × 0.8 = 57504.09 W = 57.50 kW
Reactive power
sin φ = √(1 − 0.8²) = 0.6 → Q = 71880.11 × 0.6 = 43128.07 VAr
Phase angle
φ = arccos(0.8) = 36.87°
Phase voltage
Vφ = 415 / √3 = 239.60 V
P ÷ η
57504.09 / 0.92 = 62504.44 W
Correction to 0.95
Qc = 57504.09 × (tan 36.87° − tan 18.19°) = 57504.09 × 0.4213 = 24.227 kVAr
Annual energy
E = 57504.09 × 6000 / 1000 = 345024.5 kWh
Annual cost
345024.5 × 8 = 2760196.17
S = 71.88 kVA · P = 57.50 kW · Q = 43.13 kVAr · φ = 36.87° · Qc 24.227 kVAr
Correcting this load from 0.8 to 0.95 releases about 14.5 kVA of transformer and cable capacity for the same useful output, which is often the stronger argument for correction than the tariff penalty alone.

Units & Accepted Ranges

InputUnitAccepted rangeDefault
Phase systemSingle or three phaseThree phase
Line voltageV> 0415
Line currentA> 0100
Power factor> 0 and ≤ 10.80
Efficiency%> 0 and ≤ 10092
Operating hoursh/year> 0 and ≤ 87606000
Tariffper kWh≥ 08
The correction target is fixed at 0.95 and cannot be changed. Above 0.95 the calculator reports zero correction kVAr rather than a negative value, since a capacitor cannot reduce power factor.

Typical Power Factor by Load

Load typeTypical PFNatureCorrection worthwhile
Resistive heating1.00UnityNo
Incandescent lighting1.00UnityNo
LED / electronic ballast0.90–0.95Often leadingRarely
Induction motor, full load0.85–0.90LaggingMarginal
Induction motor, half load0.70–0.80LaggingYes
Induction motor, no load0.10–0.30LaggingYes — avoid idling
Welding set0.50–0.70Lagging, intermittentYes
Arc furnace0.70–0.85Lagging, harmonic-richDetuned banks only
VFD-driven load0.95–0.98Displacement near unityNo — filter harmonics instead
A motor's power factor collapses at light load while its current falls only slightly, so an oversized motor running lightly is the most common cause of poor site power factor. Right-sizing the motor usually beats adding capacitors.

Standard Supply Voltages (IEC 60038)

SystemLine voltagePhase voltageRegion
Three phase LV400 V230 VEurope, IEC standard
Three phase LV415 V239.6 VIndia, UK legacy, Australia
Three phase LV380 V219.4 VChina, parts of Asia
Three phase LV480 V277 VNorth America industrial
Three phase LV208 V120 VNorth America commercial
Three phase MV3.3 / 6.6 / 11 kV1.9 / 3.8 / 6.35 kVIndia, UK distribution
Three phase MV4.16 / 13.8 kV2.4 / 7.97 kVNorth America distribution

Frequently Asked Questions

How do I calculate three phase power?

Apparent power in volt-amperes equals √3 multiplied by the line voltage and the line current. Active power in watts is that result multiplied by the power factor. For a 415 V supply drawing 100 A, apparent power is 71880 VA or 71.88 kVA, and at a power factor of 0.8 the active power is 57504 W. The √3 appears because line current and phase voltage in a three-phase system are separated by 30°, not because there are three phases.

What is the difference between kW, kVA and kVAr?

kVA is the total power flowing, the product of voltage and current, and is what determines cable and transformer size. kW is the portion doing useful work, and is what the energy meter bills. kVAr is the portion that oscillates between the supply and the load's magnetic fields without doing work. The three form a right triangle where S² = P² + Q², and power factor is the ratio of active to apparent power.

How much capacitor kVAr do I need to correct power factor?

The capacitor rating equals the active power multiplied by the difference between the tangent of the existing phase angle and the tangent of the target angle. This calculator targets a power factor of 0.95, which is the threshold most tariffs use. A 57.5 kW load at 0.8 power factor needs about 24.2 kVAr to reach 0.95. Correcting all the way to unity is usually avoided because it risks leading power factor at light load and resonance with system harmonics.

Why is line voltage divided by root 3?

In a star or wye connection, the line voltage measured between any two phases is the vector difference of two phase voltages that are 120° apart, and that vector difference is √3 times the individual phase voltage. So a 415 V line-to-line system has 239.6 V between each phase and neutral. In a delta connection there is no neutral and the line voltage equals the phase voltage, while the line current becomes √3 times the phase current instead.

What is a good power factor?

Most utilities set the threshold at 0.95 lagging, below which a penalty applies, and many industrial tariffs offer a rebate above it. Anything below 0.85 is usually worth correcting on cost alone, because the reactive current inflates cable losses, consumes transformer capacity and often attracts a direct charge. Above 0.98 the remaining gain is small and the risk of overcorrection at light load rises, so 0.95 to 0.98 is the usual target band.

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Results are for estimation and preliminary design. Energy cost is a simple product of power and hours — real billing adds demand charges, power factor penalties and time-of-day tariffs. Verify against your tariff schedule and a qualified engineer.

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