Three-phase power is how electricity is generated, transmitted and delivered to most commercial and industrial loads. Three alternating voltages, each a third of a cycle apart, deliver power smoothly and need less conductor material than single-phase for the same load. This calculator finds real, apparent and reactive power from line voltage, current and power factor — or the line current for a given kW — and breaks the result down per phase for wye and delta connections, with a power triangle drawn to scale.
How to use the three-phase power calculator
- Choose Power from volts & amps or Current from kW.
- Enter the line-to-line voltage, the voltage between any two phase conductors.
- Enter the line current or the real power in kW.
- Enter the power factor. Resistive heaters are 1; induction motors at full load are usually 0.8 to 0.9.
- Choose wye or delta to see the voltage and current in each phase of the load.
Three-phase power formulas
P is real power in watts, S apparent power in volt-amperes and Q reactive power in VAR. The phase angle φ = arccos(PF) is the angle between voltage and current, and the angle at the corner of the power triangle.
Worked example
480 V, 50 A, power factor 0.85, wye (the default)
P = 1.7321 × 480 × 50 × 0.85 = 35.33 kW (about 47.4 hp).
S = 1.7321 × 480 × 50 = 41.57 kVA, and Q = √(41.57² − 35.33²) = 21.9 kVAR. The phase angle is arccos 0.85 = 31.79°.
Each wye phase sees 480 ÷ 1.732 = 277.1 V and carries 50 A, delivering 11.78 kW. In delta, each phase would see 480 V and carry 28.87 A.
Current from power. Switch to Current from kW and enter 35 kW: I = 35,000 ÷ (1.732 × 480 × 0.85) = 49.53 A.
Common three-phase voltages in North America
| System | Line-to-line | Line-to-neutral | Typical use |
|---|---|---|---|
| 208Y/120 V | 208 V | 120 V | Offices, retail, apartment buildings |
| 240 V delta | 240 V | 120 V on two legs if high-leg | Older shops and farms |
| 480Y/277 V | 480 V | 277 V | Industrial plants, large HVAC, lighting at 277 V |
| 600Y/347 V | 600 V | 347 V | Canadian industrial systems |
Because current falls as voltage rises, the same 35 kW motor that draws about 50 A at 480 V would draw about 114 A at 208 V, needing much larger conductors.
Why three-phase is efficient
In a balanced three-phase system, the instantaneous power is constant instead of pulsing twice per cycle as it does in single-phase. Motors therefore run with steadier torque and less vibration, and three-phase motors start without the auxiliary windings or capacitors single-phase motors need. Three conductors carry √3 times the power of a single-phase pair at the same voltage and current — about 15% more power per conductor, which adds up to a real saving in copper on long feeders.
To size a transformer or generator in kVA, see the kVA calculator; to improve a low power factor, the power factor calculator. The voltage drop calculator checks long three-phase feeders.
Three-phase systems carry hazardous voltages and fault currents. These calculations are for estimation only; follow your local electrical code and have a licensed electrician perform all work.
Frequently asked questions
What is the formula for three-phase power?
P = √3 × VL × IL × PF, where VL is the line-to-line voltage, IL the line current and PF the power factor. At 480 V, 50 A and PF 0.85, P = 1.732 × 480 × 50 × 0.85 ≈ 35.3 kW.
Why is there a square root of 3 in three-phase formulas?
The three phase voltages are 120° apart, so the voltage between two lines is √3 times the voltage of one phase. Writing total power in terms of line voltage instead of phase voltage turns the factor of 3 (three phases) into √3.
What is the difference between wye and delta connections?
In a wye (star) connection each load sits between a line and neutral, so it sees VL ÷ √3 (277 V on a 480 V system) and carries the full line current. In a delta connection each load sits between two lines, so it sees the full line voltage but carries IL ÷ √3. Total power is the same either way.
How do I calculate three-phase current from kW?
I = P ÷ (√3 × VL × PF). A 35 kW load at 480 V and PF 0.85 draws 35,000 ÷ (1.732 × 480 × 0.85) ≈ 49.5 A per line.
What happens if the load is not balanced?
The √3 formula assumes equal current in all three phases. With unbalanced loads, calculate each phase separately as V(phase) × I × PF and add the three results. The neutral then carries the imbalance current in a wye system.