Three-phase power calculator

kW, kVA, kVAR, line current and power factor from the line voltage and any two of them.

System
V
Two quantities you know
kW
Load is

Result

Real power
100kW
Apparent power
125kVA
Reactive power, lagging
75kVAR
Line current
150.4A
Power factor, lagging
0.8
Power triangle: real power 100 kilowatts, reactive power 75 kilovars, lagging, apparent power 125 kilovolt-amperes, at 36.87 degreesP = 100.00∠0.0°. S = 125.00∠36.9°. Q = 75.00∠90.0°.φPSQ

What it calculates

The power triangle of a balanced three-phase load. Give the line-to-line voltage and any two of real power, apparent power, reactive power, line current and power factor, and it fills in the rest. Apparent power comes from the line quantities; real power is its share set by the power factor; reactive power is what is left:

S=3 VLL IS = \sqrt{3}\,V_\text{LL}\,IP=S⋅pfP = S\cdot\text{pf}Q=±S2−P2Q = \pm\sqrt{S^{2} - P^{2}}

Lagging and leading

An inductive load such as a motor draws current that lags the voltage, and its reactive power is counted as positive. A capacitive load draws current that leads the voltage, and its reactive power is negative. Reactive power is entered as a size, and the lagging or leading choice sets its sign. Power factor alone cannot say which side the current is on, so the same number can describe either.

Worked example

A 480 V load takes 100 kW at a power factor of 0.8 lagging. Apparent power, reactive power and line current follow:

S=100 kW0.8=125 kVAS = \dfrac{100\ \text{kW}}{0.8} = 125\ \text{kVA}Q=1252−1002=75 kVARQ = \sqrt{125^{2} - 100^{2}} = 75\ \text{kVAR}I=125,0003×480=150.35 AI = \dfrac{125{,}000}{\sqrt{3}\times 480} = 150.35\ \text{A}

The current lags the voltage by 36.87 degrees, since the cosine of that angle is 0.8. Correcting the power factor toward 1 lowers the kVAR and the line current while the kW stays the same.

Assumptions

  • The load is balanced and the voltages are sinusoidal.
  • Voltage is line-to-line and current is the line current, so wye and delta loads are treated alike.
  • Power factor is the displacement power factor. Harmonics are not included.

Questions

Why is there a √3 in the three-phase power formula?
Three-phase apparent power is three times the per-phase power, 3·V_LN·I. Line-to-line voltage is √3 times line-to-neutral voltage in a balanced system, so 3·V_LN·I becomes √3·V_LL·I.
Does this work for a delta-connected load?
Yes. S = √3·V_LL·I uses line quantities, which are the same whether the load is wye or delta. Only the per-phase current inside a delta differs, by a factor of √3.
Why can't I enter kVA and amps together?
Once the line voltage is known, apparent power and line current are the same information: one fixes the other. You need one more independent quantity, such as kW, kVAR or power factor.

Study areas that use this

For learning, not for design

PhasorPrep is an independent study resource. It is not affiliated with, endorsed by, or sponsored by NCEES. Questions are original and written for practice; they are not actual exam questions. Calculators are for learning, not for engineering design.

Practice questions are on the way

Original PE Power questions, each with a verified worked solution. Join the waitlist to hear when the free diagnostic opens.

See the Founding Pass