Electrical calculators

Three-phase power calculator

Enter the line-to-line voltage and either the line current or the load in kW or kVA. You get kW, kVA, kVAR and line current, with the working.

Calculated with electrician.wiki · https://electrician.wiki/calculators/three-phase-power/

Result

P = 70.67 kW · S = 83.14 kVA

100 A per line at 480 V three-phase with power factor 0.85 is 70.67 kW of real power, 83.14 kVA apparent and 43.8 kVAR reactive.

Real power (P): 70.67 kW
Apparent power (S): 83.14 kVA
Reactive power (Q): 43.8 kVAR
Line current: 100 A
Phase voltage, wye (VLN): 277.1 V
Phase current, delta: 57.74 A

If your utility charges for low power factor: raising it from 0.85 to 0.95 would take about 20.57 kVAR of capacitors and cut the line current from 100 A to 89.47 A for the same 70.67 kW. Where drives or other harmonic sources are present, use detuned capacitor banks.

Show the work

1. Apparent power
$S = \sqrt{3} \cdot V_{LL} \cdot I_{L} = \sqrt{3} \cdot 480 V \cdot 100 A = 83.14 kVA$
2. Real power
$P = S \cdot PF = 83.14 kVA \cdot 0.85 = 70.67 kW$
3. Reactive power (power triangle)
$Q = \sqrt{S^{2} - P^{2}} = \sqrt{{83.14}^{2} - {70.67}^{2}} = 43.8 kVAR$
S in kVA and P in kW give Q in kVAR.
4. Phase values for wye and delta loads
$V_{LN} = \frac{V_{LL}}{\sqrt{3}} = \frac{480 V}{\sqrt{3}} = 277.1 V, I_{ph} = \frac{I_{L}}{\sqrt{3}} = \frac{100 A}{\sqrt{3}} = 57.74 A$
Wye load: each element sees 277.1 V and carries the full 100 A. Delta load: each element sees the full 480 V and carries 57.74 A.

References: IEEE Std 141 (Red Book): three-phase power relationships

Balanced load: the same current and power factor in every phase.
Voltage is line-to-line; voltage and current are RMS values.
Power factor is the true power factor (kW ÷ kVA). With distorted current, √(S² − P²) includes distortion power as well as reactive power, and capacitors alone will not remove it.

New to this? Power factor explained

Report a problem with this result

How to use the three-phase power calculator

This three-phase power calculator works in both directions. From current takes the line-to-line voltage, the current in each line and the power factor, and returns real power (kW), apparent power (kVA) and reactive power (kVAR). From power takes a load in kW or kVA and returns the line current. Both modes also give the phase voltage of a wye load and the phase current of a delta load, and every step is shown.

Enter the voltage between two phases: 208, 400, 480 or 600 V, not the voltage to neutral. If you only have a line-to-neutral reading, the amps to watts calculator accepts it directly. For the power factor, use 1 for resistance heating, about 0.8 to 0.9 for induction motors at full load, and the measured value for a whole building or panel.

Three-phase power formulas

For a balanced load, everything follows from the apparent power. The real power is the share of it set by the power factor, and the reactive power is the remaining side of a right triangle.

S = \sqrt{3} \cdot V_{LL} \cdot I_{L}

P = S \cdot PF = \sqrt{3} \cdot V_{LL} \cdot I_{L} \cdot PF

Q = \sqrt{S^{2} - P^{2}}

PF = \frac{P}{S} = \cos (φ)

I_{L} = \frac{P}{\sqrt{3} \cdot V_{LL} \cdot PF} = \frac{S}{\sqrt{3} \cdot V_{LL}}

The power triangle

Real power P does the work: heat, light, shaft torque. Reactive power Q builds the magnetic fields in motors and transformers and flows back and forth without being used up. The cable, breaker and transformer have to carry both, and their combined size is the apparent power S. The angle φ between P and S is the phase angle between voltage and current, and its cosine is the power factor.

The power triangle for the default result, drawn to scale: 100 A per line at 480 V and power factor 0.85 (cos φ).

Line and phase values in wye and delta

Line values are what you measure at the supply: the voltage between two phases and the current in each phase conductor. Phase values are what each element of the load sees. A wye load puts each element between a line and the star point, so it sees 277.1 V on a 480 V system and carries the full line current. A delta load connects each element across two lines, so it sees the full 480 V and carries the line current divided by √3. The total power is the same either way, which is why the line-value formulas above work without knowing the connection.

Wye load: voltage across each element

V_{LN} = \frac{V_{LL}}{\sqrt{3}}

Delta load: current through each element

I_{ph} = \frac{I_{L}}{\sqrt{3}}

Line and phase values for 100 A per line at 480 V, power factor 0.85.
Per load element	Wye (Y) load	Delta (Δ) load
Voltage across it	277.1 V (VLL ÷ √3)	480 V (VLL)
Current through it	100 A (IL)	57.74 A (IL ÷ √3)
Real power in it	23.56 kW	23.56 kW
Total, three elements	70.67 kW	70.67 kW

Worked examples

480 V, 100 A per line at power factor 0.85

A motor control center feeder reading 100 A per line on a 480Y/277 V system.

1. Apparent power
$S = \sqrt{3} \cdot V_{LL} \cdot I_{L} = \sqrt{3} \cdot 480 V \cdot 100 A = 83.14 kVA$
2. Real power
$P = S \cdot PF = 83.14 kVA \cdot 0.85 = 70.67 kW$
3. Reactive power (power triangle)
$Q = \sqrt{S^{2} - P^{2}} = \sqrt{{83.14}^{2} - {70.67}^{2}} = 43.8 kVAR$
4. Phase values for wye and delta loads
$V_{LN} = \frac{V_{LL}}{\sqrt{3}} = \frac{480 V}{\sqrt{3}} = 277.1 V, I_{ph} = \frac{I_{L}}{\sqrt{3}} = \frac{100 A}{\sqrt{3}} = 57.74 A$

100 A per line at 480 V three-phase with power factor 0.85 is 70.67 kW of real power, 83.14 kVA apparent and 43.8 kVAR reactive.

A 208 V panel drawing 60 A per line

A 208Y/120 V panelboard in a small commercial building, with a measured power factor of 0.9.

1. Apparent power
$S = \sqrt{3} \cdot V_{LL} \cdot I_{L} = \sqrt{3} \cdot 208 V \cdot 60 A = 21.62 kVA$
2. Real power
$P = S \cdot PF = 21.62 kVA \cdot 0.9 = 19.45 kW$
3. Reactive power (power triangle)
$Q = \sqrt{S^{2} - P^{2}} = \sqrt{{21.62}^{2} - {19.45}^{2}} = 9.422 kVAR$
4. Phase values for wye and delta loads
$V_{LN} = \frac{V_{LL}}{\sqrt{3}} = \frac{208 V}{\sqrt{3}} = 120.1 V, I_{ph} = \frac{I_{L}}{\sqrt{3}} = \frac{60 A}{\sqrt{3}} = 34.64 A$

60 A per line at 208 V three-phase with power factor 0.9 is 19.45 kW of real power, 21.62 kVA apparent and 9.422 kVAR reactive.

Line current for a 22 kW load on 400 V

A 22 kW electrical input at power factor 0.86 on a 230/400 V supply.

1. Line current
$I_{L} = \frac{P}{\sqrt{3} \cdot V_{LL} \cdot PF} = \frac{22,000 W}{\sqrt{3} \cdot 400 V \cdot 0.86} = 36.92 A$
2. Apparent power
$S = \frac{P}{PF} = \frac{22 kW}{0.86} = 25.58 kVA$
3. Reactive power (power triangle)
$Q = \sqrt{S^{2} - P^{2}} = \sqrt{{25.58}^{2} - 22^{2}} = 13.05 kVAR$
4. Phase values for wye and delta loads
$V_{LN} = \frac{V_{LL}}{\sqrt{3}} = \frac{400 V}{\sqrt{3}} = 230.9 V, I_{ph} = \frac{I_{L}}{\sqrt{3}} = \frac{36.92 A}{\sqrt{3}} = 21.32 A$

22 kW at 400 V three-phase with power factor 0.86 draws 36.92 A per line (25.58 kVA apparent power).

Line current for a 150 kVA load on 480 V

A load given in kVA, as generator and UPS loads often are. The current does not depend on the power factor.

1. Line current
$I_{L} = \frac{S}{\sqrt{3} \cdot V_{LL}} = \frac{150,000 VA}{\sqrt{3} \cdot 480 V} = 180.4 A$
2. Real power
$P = S \cdot PF = 150 kVA \cdot 0.9 = 135 kW$
3. Reactive power (power triangle)
$Q = \sqrt{S^{2} - P^{2}} = \sqrt{150^{2} - 135^{2}} = 65.38 kVAR$
4. Phase values for wye and delta loads
$V_{LN} = \frac{V_{LL}}{\sqrt{3}} = \frac{480 V}{\sqrt{3}} = 277.1 V, I_{ph} = \frac{I_{L}}{\sqrt{3}} = \frac{180.4 A}{\sqrt{3}} = 104.2 A$

150 kVA at 480 V three-phase draws 180.4 A per line; at power factor 0.9 that is 135 kW of real power.

Quick multipliers for common three-phase systems

For estimates in the field, multiply kVA by the amps-per-kVA figure, or scale the per-100 A values. To size cable from the current, continue to the wire size calculator or the IEC cable size calculator; for a transformer, the transformer calculator gives full-load amps for standard kVA ratings.

Quick multipliers for balanced loads. Amps per kVA hold at any power factor; the last column assumes 0.85.
System	Line-to-neutral	Amps per kVA	kVA per 100 A	kW per 100 A, PF 0.85
208Y/120 V	120.1 V	2.776	36.03	30.62
220/380 V	219.4 V	1.519	65.82	55.95
230/400 V	230.9 V	1.443	69.28	58.89
240/415 V	239.6 V	1.391	71.88	61.1
480Y/277 V	277.1 V	1.203	83.14	70.67
600Y/347 V	346.4 V	0.9623	103.9	88.33

Power factor correction in brief

Capacitors supply reactive power locally, so less of it travels through the supply. Raising the default load from power factor 0.85 to 0.95 needs about 20.57 kVAR of capacitance, from $Q_{C} = P \cdot (\tan (φ_{1}) - \tan (φ_{2}))$ . The real power stays at 70.67 kW, while the apparent power falls to 74.39 kVA and the line current from 100 A to 89.47 A. That frees capacity in the feeder and transformer and reduces losses. Where drives or other harmonic sources are present, use detuned capacitor banks, because plain capacitors can resonate with the supply.

Mistakes to avoid

Using the line-to-neutral voltage. 277 V in place of 480 V understates every result by a factor of √3. The calculator warns when it sees a common line-to-neutral value.
Averaging unbalanced currents. The formulas assume equal currents. For a four-wire wye with unequal loading, add the phases: each contributes VLN × I × PF. On three-wire systems a power meter is the reliable answer.
Treating single-phase loads as three-phase. A 208 V or 480 V load connected across two phases is single-phase: P = V × I × PF with no √3.
Assuming VLN = VLL ÷ √3 on a high-leg delta. On a 240 V four-wire delta, the high leg is about 208 V to neutral and must be identified, usually in orange (NEC 110.15). The wye phase values here do not apply to that supply.
Mixing up kW and kVA. Generators and transformers are limited by current, so they are rated in kVA. A 100 kW load at power factor 0.85 needs 117.6 kVA of capacity, not 100 kVA.
Trusting nameplate cos φ on drives. A drive’s displacement power factor can be close to 1 while its true power factor, which includes harmonics, is lower. Use the true figure from a power analyzer.

Method and verification

The calculator computes S = √3 × VLL × IL (or the line current from a known kW or kVA), then P = S × PF and Q = √(S² − P²), in double precision. Phase values are VLL ÷ √3 for a wye load and IL ÷ √3 for a delta load. Results are displayed to four significant figures; nothing is rounded between steps.

The tests check six cases worked by hand, among them 480 V at 100 A and power factor 0.85 (70.67 kW, 83.14 kVA, 43.8 kVAR), 208 V at 60 A, a 22 kW load on 400 V and a 150 kVA load on 480 V. They also confirm that S² = P² + Q², that converting current to power and back returns the same current, that wye and delta phase values deliver the same total power, and that the answers match the amps to watts and watts to amps calculators exactly.

Assumes a balanced load, RMS values and the true power factor. Relationships as in IEEE Std 141. Updated 2026-10-14. Found a problem? Report it and we log every fix publicly.

Questions people ask

How do you calculate three-phase power?

Multiply √3 (about 1.732) by the line-to-line voltage, the line current and the power factor: P = √3 × VLL × I × PF. At 480 V, 100 A and power factor 0.85 that is 70.67 kW.

What is the difference between kW, kVA and kVAR?

kW is real power, the part that does work. kVAR is reactive power, which builds magnetic fields and flows back and forth. kVA is the combination the wiring has to carry: kVA² = kW² + kVAR². The ratio kW ÷ kVA is the power factor.

Why is √3 used in three-phase calculations?

The voltage between two lines is the difference of two phase voltages 120° apart, which works out to √3 times either phase voltage. Writing three-phase power with the line-to-line voltage brings in that √3: 3 × VLN equals √3 × VLL.

How many amps is 100 kW at 480 V three-phase?

120.3 A per line at power factor 1, and 141.5 A at power factor 0.85. Divide the watts by √3 × 480 V × the power factor.

Do wye and delta loads use different power formulas?

Not for line values: P = √3 × VLL × IL × PF holds for both. They differ inside the load. Wye elements see VLL ÷ √3 and carry the line current; delta elements see VLL and carry IL ÷ √3. The total power is identical, as the wye and delta table on this page shows.