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Equipment is rated in different units, kW on one nameplate, kVA on another, amps on a third, and the panel schedule wants amps. Sooner or later you have one number and need another, and the single-phase habit of dividing watts by volts gives the wrong answer on a three-phase supply.

This guide shows the arithmetic for both supplies, where power factor changes the answer, and how to check a conversion against the nameplate. It converts quantities; it does not size wire or breakers.

The two formulas

Real power is what does the work and is counted in watts or kilowatts; kilo is the SI prefix for 10³, so 1 kW is 1,000 W (NIST). DOE's fact sheets define power factor as the fraction of power actually delivered compared with what the same voltage and current would deliver without the phase shift caused by motor windings, so a resistive load such as a heating element, with no phase shift, has a power factor of 1. The physics texts say the same: in a DC circuit power is volts × amps (HyperPhysics), and in an AC circuit the average power picks up a factor of cos φ, where φ is the phase angle between voltage and current; that factor is the power factor, and it is 1 for a purely resistive circuit (HyperPhysics). Rearranged for current, the single-phase arithmetic is volts × amps × PF = watts:

  • Single-phase: amps = kW × 1,000 ÷ (volts × PF)
  • Three-phase: amps = kW × 1,000 ÷ (√3 × volts × PF)

The three-phase form is the rearrangement of the equation in the DOE motor-load fact sheet: kW = V × I × PF × √3 ÷ 1,000, where V is the rms voltage measured line to line (the mean of the three phases), I is the rms current (mean of the three phases), and PF is a decimal. The √3 is 1.7320508…; our calculator uses the exact value, so results can differ from a hand calculation with 1.73 in the third digit.

Apparent power in kVA ignores the power factor, so those conversions drop it:

  • Single-phase: amps = kVA × 1,000 ÷ volts
  • Three-phase: amps = kVA × 1,000 ÷ (√3 × volts)

DOE defines power factor as real power divided by apparent power (DOE), so kW = kVA × PF.

Line-to-line voltage

On a three-phase supply there are two ways to measure voltage: between two of the three lines, or between a line and neutral. The DOE fact sheet says service and motor voltages are quoted line to line unless noted otherwise, and that line-to-line readings are 1.73 times line-to-neutral readings.

That is why the formula uses the line-to-line number. At 208 V line to line, line to neutral works out to 208 ÷ √3 = 120.09 V; at 480 V it is 277.13 V. Plug a line-to-neutral value into the three-phase formula and your current comes out about 73% too high. The formula is exact only for a balanced load, the same current on all three lines. DOE handles a slightly unbalanced real reading by averaging the three phases, which makes the result an approximation.

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Worked example: the same 10 kW on three supplies

Take a 10 kW resistive load, so PF is 1.0. The loads in these examples are made up to show the arithmetic.

SupplyCalculationAmps
240 V single-phase10,000 ÷ 24041.67
208 V three-phase10,000 ÷ (√3 × 208)27.76
480 V three-phase10,000 ÷ (√3 × 480)12.03

The power is identical, but the current falls with voltage and falls again when three phases share the load. For the same kW, voltage and power factor, a single-phase circuit carries √3 (1.732) times the amps of a three-phase one: at 240 V, 10 kW is 41.67 A single-phase and 24.06 A three-phase.

Fig. 1 One load, three suppliesDrawn to scale: amps
The same 10 kW load at power factor 1.0 as current on one scale: 240 V single-phase 41.67 amps; 208 V three-phase 27.76 amps; 480 V three-phase 12.03 amps. The square root of 3 is 1.732.240 V single-phase41.67 A208 V three-phase27.76 A480 V three-phase12.03 AAmps
Fig. 1 — The same 10 kW resistive load at power factor 1.0 on three supplies. The current falls from 41.67 A on 240 V single-phase to 27.76 A on 208 V three-phase and 12.03 A on 480 V three-phase.

Lower current also helps the other end of the job. The volts lost in a wire scale with current, and a higher supply voltage makes that loss a smaller percentage; the voltage drop guide works that out.

What power factor does

Resistive loads, such as heating elements, have a power factor of 1. Motors and other inductive loads do not: DOE says the inductance in motor windings delays the current, and its motor-load fact sheet shows power factor falling off at part load and differing by motor size. We found no source that gives one typical value for motors, compressors or drives, so we don't offer one: use the nameplate, the equipment data sheet or a measurement.

Here is a hypothetical 30 kW load on 208 V three-phase at three power factors:

Power factorCalculationAmps
1.0030,000 ÷ (√3 × 208 × 1.00)83.27
0.8530,000 ÷ (√3 × 208 × 0.85)97.97
0.7030,000 ÷ (√3 × 208 × 0.70)118.96

The kilowatts never changed, yet the current rose by 17.6% at 0.85 and by 42.9% at 0.70. DOE's power-factor sheet describes the same effect: low power factor reduces the capacity of the distribution system by increasing current flow and causing voltage drops, and it causes power losses in that system. DOE adds that many utilities charge extra when a customer's power factor falls below 0.95. Rates differ by utility, so that is a reason to ask yours, not a number to plan around.

kVA, kW and the nameplate

kVA is apparent power. When a transformer or generator nameplate gives kVA, converting is the same square-root arithmetic without PF:

  • A 75 kVA three-phase source at 208 V: 75,000 ÷ (√3 × 208) = 208.18 A. At 480 V: 90.21 A.
  • Real power from that 75 kVA depends on the load. At PF 0.85 it is 75 × 0.85 = 63.75 kW; at 0.70, 52.5 kW.
  • Going the other way, a 63.75 kW load at PF 0.85 on 208 V draws 208.18 A, the full 75 kVA.

Reading it back from measurements works the same. The DOE fact sheet's own example measures 469.7 V line to line, 37 A and a power factor of 0.763. That is √3 × 469.7 × 37 × 0.763 ÷ 1,000 = 22.97 kW (the sheet prints 22.9). The apparent power is 30.10 kVA, and 22.97 ÷ 30.10 gives back the 0.763.

When does the nameplate win? Whenever it lists amps. A nameplate current is the equipment maker's rated value for that unit, so it needs no assumed power factor from you. DOE notes it applies only at the rated voltage, so a current measured on a different supply voltage will differ. The calculated number is a cross-check, or a stand-in when all you have is a kW or kVA rating. Horsepower is the worst starting point. NIST lists 1 electric horsepower as 746 W, but that is mechanical output: turning hp into input kW needs efficiency (DOE's formula divides by it) and into amps needs power factor too, and both change with load.

Where the math stops

  • A current is not a conductor or breaker size. Wire and overcurrent sizing follow the electrical code, the equipment's listing and its installation instructions. Our tool does not apply them; your local authority having jurisdiction decides what applies on your job. This is general information, not code advice.
  • Only sinusoidal, balanced loads. The sources we read cover rms values on balanced three-phase systems. Variable-speed drives, LED drivers and other nonlinear loads can behave differently.
  • Single-phase loads on a three-phase supply. A load fed from two lines is a single-phase calculation at that line-to-line voltage. A load wired line to neutral uses the line-to-neutral voltage, the line-to-line value ÷ √3 (our arithmetic from DOE's 1.73 ratio).
  • Work on live equipment only if you are qualified, and follow the equipment maker's instructions.

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Next step

Enter your kW or kVA, the line-to-line voltage and the nameplate power factor in the Ohm's law and power calculator, and compare the answer with the nameplate current. If the run to the equipment is long, check the loss with the voltage drop calculator. For the heating side of a water heater job, the kW arithmetic is in the water heater sizing guide.

Sources

  1. Determining Electric Motor Load and Efficiency (Motor Challenge fact sheet)U.S. Department of Energy
  2. Reducing Power Factor Cost (Motor Challenge fact sheet)U.S. Department of Energy
  3. Electric PowerHyperPhysics, Georgia State University
  4. Power Factor for AC PowerHyperPhysics, Georgia State University
  5. NIST Guide to the SI, Appendix B.9: Conversion Factors for PowerNational Institute of Standards and Technology
  6. NIST Guide to the SI, Chapter 4: The Two Classes of SI Units and the SI PrefixesNational Institute of Standards and Technology

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