What the Three-Phase Power Calculator does
This calculator converts between three-phase kW, kVA, kVAr and line current for a balanced load, and shows the line and phase quantities separately for both star and delta connections. Enter the line voltage and any one of the three - power, apparent power or current - and the rest follows from the root-three relationship.
Almost every mistake in three-phase arithmetic comes from mixing a line quantity with a phase quantity, so the result is laid out as a table with a column for each. It also shows the same load in the other connection, because star versus delta is the question people are really asking most of the time. Everything is computed in your browser.
How to use it
- Enter the line voltage - the voltage between any two lines, which is what a nameplate and a voltmeter across two phases both give you.
- Enter one of kW, kVA or line current, and the power factor. Leave the other two blank so the calculator knows which one to solve for.
- Pick star or delta. This does not change the total power or the line current; it changes what each of the three windings of the load sees.
- Read the line column for cable and breaker sizing and the phase column for anything about the windings themselves - a motor's winding current, a capacitor's voltage rating, a heater element's resistance.
- Use the comparison block to see the same load in the other connection, and export the figures with copy, CSV or JSON.
Reading the results
Line current is the figure that sizes cables and protective devices. It is the same in star and delta for a given total power.
Phase (winding) current is what flows inside the load. In delta it is the line current divided by root-three, which is why a star-delta starter's contactors are rated at 58% of full-load current.
Phase voltage in star is the line voltage divided by root-three: 230 V from a 400 V system, 240 V from a 415 V system. In delta the winding sits across the full line voltage.
Power per phase is a third of the total, because the load is assumed balanced. The moment the three phases differ, this assumption fails and each phase has to be worked separately.
Worked example: a 30 kW motor on a 400 V supply at 0.88 power factor
Apparent power is 30 / 0.88 = 34.091 kVA. Line current is 34,091 / (root-three x 400) = 34,091 / 692.82 = 49.21 A.
Reactive power is the third side of the triangle: the square root of 34.091 squared minus 30 squared, which is 16.20 kVAr.
Connected in delta, each winding sits across the full 400 V and carries 49.21 / root-three = 28.41 A. Connected in star, each winding sees 400 / root-three = 230.9 V and carries the full 49.21 A. Either way the supply delivers 30 kW and the cable carries 49.21 A.
That difference is the whole basis of star-delta starting: running the windings in star on a 400 V supply puts only 230.9 V across each one, so the winding current, the line current and the torque all fall to a third of their direct-on-line values.
Note that 30 kW here is the input power. A motor nameplate quotes shaft output, so a 30 kW motor at 93% efficiency actually draws 32.26 kW and 52.9 A - use the motor full-load current calculator when the 30 kW is a nameplate figure.
Formulas and scoring rules
- Apparent power
S = root-three x V_line x I_lineIn volt-amperes, for a balanced three-phase load.- Real power
P = root-three x V_line x I_line x pfEquivalently P = 3 x V_phase x I_phase x pf, which gives the same answer in either connection.- Line current
I_line = S / (root-three x V_line) = P / (root-three x V_line x pf)- Star (wye)
V_phase = V_line / root-three ; I_phase = I_lineRoot-three is 1.7320508; a 400 V system gives 230.9 V to neutral.- Delta
V_phase = V_line ; I_phase = I_line / root-three1/root-three is 0.5774, the 58% figure used for star-delta contactors.- Reactive power
Q = sqrt(S^2 - P^2)Shown as a magnitude; the calculator does not know whether the load is lagging or leading.
Why root-three and not three
Each of the three phase voltages has the same magnitude but sits 120 degrees apart in time. Subtracting one from another is a vector subtraction, not an arithmetic one, and the result has a magnitude of 2 x sin(60 degrees) = root-three times a single phase voltage. That is where 400 V comes from on a system whose phase-to-neutral voltage is 230 V.
The factor then reappears in the power formula, but for a different reason: total power is three times the per-phase power, and substituting V_line / root-three for V_phase turns 3 into 3 / root-three, which is root-three. The two appearances are related but not the same step, which is why it is worth writing the derivation out rather than memorising the constant.
Balanced, and what happens when it is not
Every figure on this page assumes the three phases carry equal current at equal power factor. Distribution boards with single-phase circuits never quite manage that, and the imbalance has consequences: a neutral current that the formula does not predict, extra losses, and on a motor, negative-sequence current that heats the rotor out of proportion to its size.
For a board, work each phase separately - the load schedule builder does exactly that and reports the imbalance. For a motor, a voltage imbalance of even 2% is worth investigating; NEMA MG 1 requires derating above it.
Limitations: what the result does not prove
- Balanced loads only. Unbalanced systems need per-phase analysis and, strictly, symmetrical-component analysis; nothing here does either.
- Sinusoidal waveforms are assumed. Drives, rectifiers and switch-mode supplies distort the current, and their true power factor is lower than the displacement power factor this arithmetic implies.
- Neutral current is not calculated. With balanced linear loads it is nearly zero; with single-phase non-linear loads the third-harmonic components add rather than cancel and the neutral can carry more than a phase.
- The result does not tell you whether a cable, a breaker or a transformer can supply the load. Use the cable, breaker and transformer calculators for that, and have the design verified by a qualified engineer against the applicable standard.
Privacy: where your data goes
Everything you paste, type or drop is processed in this browser tab. It is not uploaded, logged, stored or sent to analytics. Session recording and tag-manager scripts are switched off on this page.
Standards and sources
- IEC 60364-5-52 - Selection and erection of wiring systems - checked 19 Sep 2026
- NFPA 70 - National Electrical Code - checked 19 Sep 2026
- IEC 60038 - IEC standard voltages
Frequently asked questions
How do I convert kW to amps on a three-phase supply?
Divide the power in watts by root-three times the line voltage times the power factor: I = P / (1.732 x V x pf). At 400 V and 0.9 power factor, 30 kW draws 48.1 A; at 415 V the same load draws 46.4 A, because the higher voltage carries the same power with less current.
Is the line current different in star and delta?
No. For the same load drawing the same power from the same supply, the line current is identical. What differs is what happens inside the load: in delta each winding carries 58% of the line current at full line voltage, and in star each winding carries the full line current at 58% of the line voltage.
What is the difference between 400 V and 415 V systems?
They are the same kind of system at slightly different nominal voltages: 400 V gives 230 V to neutral and 415 V gives 240 V. IEC 60038 standardised on 230/400 V, but plenty of equipment and plenty of countries still work to 240/415 V, and the difference is enough to matter in a current calculation.
Why is my measured current higher than this calculation?
The three usual reasons are a nameplate kW that is shaft output rather than input, a real power factor worse than the one entered, and an unbalanced load so that the phase you clamped is carrying more than its share. Harmonic current from electronic loads is a fourth: a true-rms clamp meter reads it, this arithmetic does not.
Can I use this for a single-phase load on a three-phase board?
No - a single-phase circuit fed from one line and neutral uses the phase voltage and has no root-three in it. Use the single-phase calculator for that circuit, and the load schedule builder when you want to see how several such circuits sit across the three phases.
What power factor should I assume for a whole board?
There is no safe default, but 0.85 to 0.95 covers most mixed commercial boards, with motor-heavy industrial loads at the lower end and boards dominated by electronics or corrected loads at the upper end. If a kVA demand charge or a transformer size depends on the answer, measure it over a working week rather than assuming.
Does the calculator handle a three-wire supply with no neutral?
Yes - a three-wire delta or corner-earthed supply has the same line quantities, and the star column simply describes a notional star point that may not physically exist. What changes is the earthing arrangement and the fault behaviour, which are separate questions from the power arithmetic.
How much more power can three phases carry than one?
For the same conductor size and the same voltage to earth, a three-phase circuit carries about three times the power of a single-phase one using only 1.5 times the copper, because the three currents partly cancel in the return path. That economy, not the power itself, is the reason distribution is three-phase.
Last reviewed by the A2Z.Tools team against the sources listed above.
