Three numbers describe every three-phase load
Any balanced three-phase load is fully described by three powers that form a right triangle. Real power (P, kilowatts) is the part that turns into torque, heat or light — the part a revenue meter bills you for. Reactive power (Q, kilovolt-amperes reactive) is the part that flows out into the magnetic fields of motor windings and transformer cores and flows back again every half cycle, doing no net work. Apparent power (S, kilovolt-amperes) is the hypotenuse: the product of the volts and amps your conductors actually carry, which is why transformers, generators, UPS units and switchgear are all rated in kVA rather than kW.
The link between them is the power factor, PF = P ÷ S = cos φ, where φ is the angle by which current lags voltage. A resistive load such as a heating element has PF = 1 and no reactive power at all. An induction motor at full load typically sits around 0.85, and the same motor at light load falls much lower, because the magnetising current it needs to hold up the rotating field barely changes while the working current collapses.
Getting these apart matters commercially as well as technically. A plant drawing 500 kW at 0.75 power factor pulls 667 kVA through its service, so its transformer, its feeders and its main breaker all have to be sized for 667 kVA of current while the meter records only 500 kW of work. That gap is the whole reason power factor correction capacitors exist.
Where the √3 actually comes from
The 1.732 in every three-phase formula is not a safety factor or a convention. It falls out of the geometry of three voltages spaced 120° apart.
Take a wye-connected source. Each winding produces a phase voltage Vph measured from its terminal to the neutral point. The voltage a meter reads between any two terminals is the vector difference of two of those phase voltages, and because they are 120° apart, that difference has magnitude 2 × Vph × cos(30°) = √3 × Vph. That is why a 480 V system has 277 V from any phase to neutral: 480 ÷ 1.732 = 277.1.
Now total the power. Each of the three windings delivers Vph × Iph × PF, so the total is 3 × Vph × Iph × PF. In a wye, line current equals winding current, and Vph = VL ÷ √3. Substitute: P = 3 × (VL ÷ √3) × IL × PF = √3 × VL × IL × PF, since 3 ÷ √3 = √3.
In a delta the roles swap. Winding voltage equals line voltage, and line current is √3 times the winding current, because each line conductor feeds two windings whose currents are 120° apart. Substituting gives exactly the same expression. That is the useful result: the line-quantity formula is identical for wye and delta, so you never need to know the internal connection to compute total power. You only need it when you are looking inside — checking the current in one leg of a delta-connected capacitor bank, or the volts across one transformer winding. This calculator reports both.
Reactive power uses the other leg of the triangle: Q = S × sin φ, where sin φ = √(1 − PF²). And rearranging the real-power formula for current gives the expression an electrician uses most often: I = P × 1000 ÷ (√3 × VL × PF). That is the current you carry into a conductor ampacity check and a voltage drop check.
Worked example: 100 A measured on a 480 V feeder at 0.85 power factor
You clamp one leg of a balanced 480 V feeder and read 100 A. The plant power meter shows 0.85 power factor. Work the triangle by hand.
- Apparent power. S = √3 × 480 × 100 ÷ 1000. First √3 × 480 = 831.38, then × 100 = 83,138 VA = 83.14 kVA.
- Real power. P = 83.14 × 0.85 = 70.67 kW.
- Phase angle. φ = arccos(0.85) = 31.79°.
- Reactive power. sin φ = √(1 − 0.85²) = √0.2775 = 0.5268, so Q = 83.14 × 0.5268 = 43.80 kVAR.
- Check the triangle. √(70.67² + 43.80²) = √(4,994 + 1,918) = √6,912 = 83.14 kVA. It closes.
- Winding quantities, wye. Each winding sees 480 ÷ 1.732 = 277.1 V and carries the full 100 A.
- Winding quantities, delta. Each winding sees the full 480 V and carries 100 ÷ 1.732 = 57.74 A.
Now read the commercial consequence. To deliver the same 70.67 kW at unity power factor you would need only 70,670 ÷ 831.38 = 85.0 A, a 15% reduction in every conductor, connection and contact in the path. The relationship is exact: at constant real power the current scales with the power factor, so 100 × 0.85 = 85 A.
Reading the result
Compare kVA with kW. Because S = P ÷ PF and PF is never above 1, apparent power is always at least as large as real power, and equal to it only at unity power factor. The ratio S ÷ P = 1 ÷ PF is the direct measure of how much extra current your system carries for the work it does. At 0.85 that is a factor of 1.176; at 0.70 it is 1.429.
Judge the power factor against the load type. Resistive heating and incandescent lighting run at essentially unity. A fully loaded induction motor lands near 0.85. Lightly loaded motors, idling machine tools and older fluorescent ballasts drag a plant into the 0.70s. Modern switch-mode power supplies with active correction sit above 0.95 but can inject harmonics that make the simple cos φ model optimistic — see the note on true power factor below.
Use kVA, not kW, for equipment ratings. A 75 kVA transformer feeding a 0.80 power factor load delivers only 60 kW. Sizing that transformer from a 75 kW load figure would overload it by 25%. The transformer kVA sizing calculator handles this conversion together with the standard rating steps.
Sanity-check line current against the service. A useful memory anchor: at 480 V three-phase, each kVA draws 1.203 A, so 100 A of line current is about 83 kVA. At 208 V each kVA draws 2.776 A. The reference table below gives the full set.
Watch the phase angle for capacitor sizing. Correcting power factor from PF₁ to PF₂ at constant real power needs Q = P × (tan φ₁ − tan φ₂) kVAR of capacitance. The phase angle output feeds straight into that calculation.
Line current per kVA at standard three-phase voltages
| Apparent power | 208 V | 240 V | 480 V | 600 V |
|---|---|---|---|---|
| 15 kVA | 41.6 A | 36.1 A | 18.0 A | 14.4 A |
| 30 kVA | 83.3 A | 72.2 A | 36.1 A | 28.9 A |
| 45 kVA | 124.9 A | 108.3 A | 54.1 A | 43.3 A |
| 75 kVA | 208.2 A | 180.4 A | 90.2 A | 72.2 A |
| 112.5 kVA | 312.3 A | 270.6 A | 135.3 A | 108.3 A |
| 150 kVA | 416.4 A | 360.8 A | 180.4 A | 144.3 A |
| 225 kVA | 624.6 A | 541.3 A | 270.6 A | 216.5 A |
| 300 kVA | 832.7 A | 721.7 A | 360.8 A | 288.7 A |
| 500 kVA | 1387.9 A | 1202.8 A | 601.4 A | 481.1 A |
| 750 kVA | 2081.9 A | 1804.2 A | 902.1 A | 721.7 A |
| 1000 kVA | 2775.8 A | 2405.6 A | 1202.8 A | 962.3 A |
These are full-load currents at the transformer rating, independent of power factor. To get real power in kW, multiply the kVA figure by the power factor.
Displacement power factor versus true power factor
The cos φ used here is the displacement power factor: it accounts only for the phase shift between fundamental voltage and fundamental current. Where the load draws non-sinusoidal current — six-pulse drives, rectifier front ends, LED drivers without correction — the harmonic current adds to the RMS amperes without adding either real or fundamental reactive power. The resulting true power factor, defined as watts divided by volt-amperes, is lower than cos φ, and a capacitor bank sized from cos φ alone will not fix it. IEEE Std 1459 sets out the power definitions that apply under those non-sinusoidal conditions.
Mistakes that produce wrong three-phase numbers
- Using phase-to-neutral voltage in the line formula. Entering 277 V instead of 480 V understates every power by a factor of 1.732. The formula wants the line-to-line voltage.
- Multiplying by 3 instead of √3. This inflates results by 73%. Three appears only when you work in winding quantities, and then the winding voltage is smaller by the same √3.
- Writing power factor as a percentage in a hand calculation. An 85 where 0.85 belongs makes real power a hundred times too large, and in the kW-to-current direction it makes the current a hundred times too small. The field here is capped at 1.0, so the slip only bites on paper.
- Assuming balance that is not there. These formulas need all three phases equal. On a heavily single-phase panel with a shared neutral, measure each leg and add the per-phase powers instead.
- Adding kVA arithmetically. Two 100 kVA loads at different power factors do not make 200 kVA. Add the kW figures, add the kVAR figures, and take the hypotenuse of the totals.
- Sizing a generator or transformer from kW. Both are limited by current and therefore by kVA. Convert first, then apply any spare-capacity allowance.
Where this fits with the other power calculations
Three-phase power is the hub that the other industrial calculations hang from. Going downstream, once you have line current you size conductors, overcurrent protection and voltage drop. Going upstream, the sum of your loads in kVA sizes the service transformer and the standby generator.
For motors specifically, work in the other direction: the nameplate gives horsepower at the shaft, and you convert to input kW by dividing by efficiency before applying these relations. The motor full load amps calculator does that chain and then compares the result with the NEC table value you are required to size from. If you are testing a running motor and want efficiency rather than current, the motor efficiency calculator takes measured volts, amps and power factor and backs out both efficiency and load percentage.
For single-phase work the same triangle applies with the √3 removed — the watts to amps calculator covers that case. And where the question is not how much current a load draws but how much reactive power to cancel, the power factor correction calculator turns the phase angle here into a capacitor bank in kVAR and microfarads.
One historical note that still shapes practice: three-phase transmission won out over Edison's DC and over two-phase systems partly because a balanced three-phase load draws constant instantaneous power. The three sinusoidal power waves, each pulsating at twice line frequency, sum to a flat line. That is why three-phase motors have no inherent torque pulsation at line frequency and why the same copper carries more power than any single-phase arrangement can.
Key terms
- Line quantity
- A voltage measured between two phase conductors, or a current measured in one phase conductor. These are what a clamp meter and a voltmeter read at a panel.
- Phase (winding) quantity
- The voltage across, or current through, one winding of the source or load. Related to line quantities by √3, in opposite directions for wye and delta.
- Power factor
- The ratio of real power to apparent power. For sinusoidal quantities it equals the cosine of the angle between voltage and current.
- kVAR
- Kilovolt-amperes reactive: the magnetising power exchanged between source and load each half cycle. It transfers no net energy but does consume conductor and transformer capacity.
- Balanced load
- A three-phase load whose three currents are equal in magnitude and 120° apart. All of the simple √3 relations depend on it.