Almost every electrical sizing decision starts with one number: the current a load draws. Cables, breakers, contactors, transformers and generators are all chosen by amperes, but loads are described in kilowatts or horsepower. Getting from one to the other is simple arithmetic, provided you know which formula applies to your supply and what the power factor is doing.
DC: the easy case
In a DC circuit power is voltage times current, full stop: P = V × I. A 600 W load on a 48 V battery bank draws 600 ÷ 48 = 12.5 A. There is no power factor and no phase relationship to worry about, which is one reason DC distribution is attractive for solar and battery systems, and one reason its high currents at low voltage make voltage drop such a problem.
Single-phase AC: power factor enters
In an AC circuit the voltage and the current are both sine waves. If the load is purely resistive, a heater or an incandescent lamp, they rise and fall together and the power is again V × I. If the load has inductance, a motor, a transformer, a fluorescent ballast, the current lags behind the voltage. Part of each cycle the voltage is pushing one way while the current is still flowing the other, and that part delivers no net energy. The useful power is:
where φ is the angle by which the current lags and cos φ is the power factor. The product V × I without the power factor is the apparent power S, measured in volt-amperes (VA), and it is what the wiring and the supply actually have to carry. The part that does no work is the reactive power Q in volt-amperes reactive (var). The three are related as the sides of a right triangle: S² = P² + Q².
A 2 kW single-phase heater at 230 V draws 2 000 ÷ 230 = 8.7 A. A 2 kW single-phase motor at 0.8 power factor draws 2 000 ÷ (230 × 0.8) = 10.9 A, a quarter more current for the same useful power, and its apparent power is 2.5 kVA.
Three-phase AC: where the √3 comes from
A three-phase supply has three line conductors whose voltages are 120° apart. Between any line and neutral the voltage is 230 V; between any two lines it is √3 × 230 ≈ 400 V. A balanced three-phase load takes equal current from each line, and the total power is three times the single-phase power of one line-to-neutral circuit:
The second form, with the line-to-line voltage and the factor √3 ≈ 1.732, is the one everyone uses, because 400 V is the number on the nameplate and the meter. Rearranged for current:
A 15 kW three-phase load at 400 V and 0.85 power factor draws 15 000 ÷ (1.732 × 400 × 0.85) = 25.5 A in each line. The power and current calculator does this and also returns the kVA and kVAr.
The commonest error is to put 230 V into the three-phase formula, which overstates the current by √3 and leads to cables and breakers nearly twice as large as needed; the second commonest is to put 400 V into the single-phase formula for a load that is actually connected line-to-neutral.
kW, kVA and kVAr: what each one is for
- kW (real power) is what does the work and what the energy meter integrates into kWh. Loads are rated in it; engines are rated in it.
- kVA (apparent power) is voltage times current. Transformers, generators and UPS units are rated in kVA because their limit is current (heating in windings) regardless of what the load does with it. A 100 kVA transformer can deliver 100 kW to a resistive load but only 80 kW to a load at 0.8 power factor.
- kVAr (reactive power) is the circulating part. It heats cables and occupies capacity without doing work, which is why utilities penalise it and why capacitor banks are installed to cancel it. The power factor guide covers this.
Converting: kVA = kW ÷ power factor; kVAr = √(kVA² − kW²). At unity power factor the three collapse to one number, which is only true for heaters and modern electronics with active power-factor correction.
Motors: horsepower, efficiency and starting current
A motor nameplate gives the mechanical output at the shaft, in kW or in horsepower (1 hp = 745.7 W; some nameplates use the metric horsepower of 735.5 W, 1.4 % less). The electricity it draws is that output plus its own losses, so:
A 10 hp motor that is 90 % efficient has a shaft output of 7.46 kW and an input of 8.29 kW; at 400 V and 0.85 power factor it draws 14.1 A. Small motors are less efficient (75–85 %) and have lower power factor; large ones are better on both counts. The nameplate full-load current, when it is given, is the number to use, since it already includes both effects.
Starting is another matter. An induction motor started direct-on-line draws five to eight times its full-load current for the second or two it takes to reach speed. Breakers and fuses are chosen with curves that ride through this; generators must be able to supply it (see the generator sizing calculator); and cables must not drop so much voltage during the start that the motor stalls.
Unbalanced loads and the neutral
The √3 formula assumes the three lines carry equal current. In a building, single-phase loads (lighting, sockets, small equipment) are spread across the three phases, never perfectly, and the difference flows in the neutral. When designing a distribution board, calculate each phase separately from the loads connected to it and then balance them as well as you can; the incoming cable is sized for the most heavily loaded phase, not the average.
Electronic loads add a twist: their distorted current contains a third harmonic that does not cancel in the neutral but adds, so a neutral serving mostly computers or LED drivers can carry more current than any phase. Standards require a full-size or oversized neutral in that case.
Two worked examples from the field
A flour mill's 75 kW three-phase motor at 400 V, 0.88 power factor, 94 % efficiency: input = 75 ÷ 0.94 = 79.8 kW; current = 79 800 ÷ (1.732 × 400 × 0.88) = 131 A per line; apparent power 90.7 kVA. The feeder cable, the starter and the generator are all sized from 131 A and 91 kVA, not from 75 kW.
An office's single-phase air conditioner rated 1.5 ton, drawing 1.8 kW at 230 V and 0.9 power factor: current = 1 800 ÷ (230 × 0.9) = 8.7 A running, with an inverter type starting gently and a fixed-speed type surging to 30–40 A for a moment. A 16 A circuit is normal for the running load; the starting surge is why several fixed-speed units on one generator need a much larger set than their kilowatts suggest.
References
- IEC 60038:2009, IEC standard voltages
- IEEE Std 1459-2010, Standard Definitions for the Measurement of Electric Power Quantities Under Sinusoidal, Nonsinusoidal, Balanced, or Unbalanced Conditions
- IEC 60034-1:2022, Rotating electrical machines — Part 1: Rating and performance
- IEC 60364-5-52:2009, Low-voltage electrical installations — Selection and erection of electrical equipment — Wiring systems — neutral conductor sizing with harmonic currents, Annex E