Three-Phase Systems
Line and phase quantities, star and delta, balanced loads, power and current and why industry uses three phases.
Why three phases
A three-phase supply carries three sinusoidal voltages of the same size, 120° apart. The total power of a balanced three-phase load is constant, not pulsating as in single-phase; a three-phase motor starts by itself with a rotating field (see Induction Motors and Drives); and for the same power and current density it uses about half the copper of a single-phase supply. Industry and every power grid use it for these reasons.
Star (wye) and delta
The three windings of a motor, a transformer or a load can be connected in two ways:
- Star (Y or wye): one end of each winding goes to a common point (the star point or neutral); the other ends go to the three lines. Each winding sees the phase voltage, Vph = VL ÷ √3, and the line current equals the winding current.
- Delta (Δ): the windings are connected in a triangle, each across two lines. Each winding sees the full line voltage, and the line current is √3 times the winding current, IL = √3 Iph.
Star: VL = √3 Vph, IL = Iph Delta: VL = Vph, IL = √3 Iph
| System | Line-to-line | Line-to-neutral | Where |
|---|---|---|---|
| 230/400 V, 50 Hz | 400 V | 230 V | Europe, much of Asia, Africa, Australia |
| 120/208 V, 60 Hz | 208 V | 120 V | US and Canada commercial buildings |
| 277/480 V, 60 Hz | 480 V | 277 V | US industrial and large commercial |
| 347/600 V, 60 Hz | 600 V | 347 V | Canada industrial |
| 6.6 kV, 11 kV, 33 kV | distribution | — | utility feeders, large motors |
Line and phase quantities
On a 400 V supply the star-point voltage of each phase is 400 ÷ √3 = 230.9 V. The factor √3 = 1.732 comes from adding two 230 V phasors 120° apart, and it is the number to remember. A common mistake is to use the line voltage with the phase current or vice versa: always check which one a formula expects.
Three-phase power
P = √3 VL IL cos φ S = √3 VL IL Q = √3 VL IL sin φ
To find the line current from the power: IL = P ÷ (√3 VL cos φ). With kW and volts: I (A) = 1 000 × P (kW) ÷ (1.732 × V × cos φ) = 0.577 × P (kW) ÷ (V (kV) × cos φ).
Worked example: the motor on the 400 V supply
Worked example: the 11 kW press motor
- Input power 12.09 kW at a power factor of 0.85 on 400 V: IL = 12 088 ÷ (1.732 × 400 × 0.85) = 20.5 A.
- Check: √3 × 400 × 20.5 × 0.85 = 12 088 W.
- Star-connected the winding carries the line current, 20.5 A, at 230.9 V: each phase impedance is 11.25 Ω.
- Delta-connected the winding carries 11.85 A at 400 V: 33.8 Ω per winding, three times the star value. A motor plate “400 V Δ / 690 V Y” is the same motor; on 400 V it must be connected in delta.
- On a single-phase 230 V supply the same load would draw 61.8 A, 3.0 times as much. The conductors carry 61.6 A (three wires of 20.5) against 123.7 A (two wires of 61.8): about 50 percent of the copper.
- Star or delta for a heater? Three 30 Ω elements draw 16.0 kW in delta and a third of that, 1.78 kW, in star: switching the connection is a simple way to get two heat ratings.
Unbalance and the neutral
If the three currents are equal and 120° apart, they cancel at the star point and the neutral carries nothing. Single-phase loads spread over the phases are never perfectly balanced, so the neutral carries the difference. For line currents of 20, 15 and 10 A (unity power factor) the neutral current is √(Ia² + Ib² + Ic² − IaIb − IbIc − IcIa) = 8.7 A. A voltage imbalance of only 2 percent raises a motor’s current imbalance by 6 to 10 times and its heating by about 8 percent; keep the imbalance under 1 to 2 percent.
Beware of harmonics: the third harmonic (150 Hz) currents of computers, LED lighting and drives are in phase in all three lines, so they add in the neutral instead of canceling. A neutral may then carry up to 1.7 times the phase current (about 18.5 A here from a 30 percent third harmonic per phase); size the neutral for it in offices and data rooms (Power Factor, Harmonics and Energy).
The rotating magnetic field
Three coils 120° apart carrying three-phase currents create a magnetic field of constant strength that rotates at the synchronous speed ns = 120 f ÷ p (rpm), with p the number of poles. At 50 Hz: 2 poles 3 000 rpm, 4 poles 1 500 rpm, 6 poles 1 000 rpm, 8 poles 750 rpm; at 60 Hz the speeds are 20 percent higher (1 800 rpm for 4 poles). Swapping any two lines reverses the direction of rotation: the first thing to check when a new motor runs the wrong way.
Related
- Previous: AC Power Basics. Next: Transformers and Distribution.
- Three-Phase Power and Power Factor Calculator.
- Wire Ampacity and Voltage Drop Calculator.
- Electrical Power: the whole guide.
FAQ
Why 1.732?
It is √3, from the geometry of three vectors 120° apart: the voltage between two lines is the difference of two phase voltages, 2 × cos 30° = 1.732 times the phase value.
Do I need the neutral?
For single-phase loads (230 V sockets, lighting) yes. A pure three-phase motor load needs only three conductors and earth.
What is phase sequence?
The order in which the phases peak (L1, L2, L3 or L1, L3, L2). It sets the direction of a motor; a phase-sequence meter confirms it before connecting.
Can a three-phase motor run on single phase?
Not directly. A phase-converter or a variable-speed drive with a single-phase input (derated) makes three-phase from single-phase; a capacitor-start motor is the simple option for small loads.
References
- Wildi, T., Electrical Machines, Drives and Power Systems, Pearson; Glover, J. D., et al., Power System Analysis and Design, Cengage.
- IEC 60038 (standard voltages), IEC 60034-1 (rotating machines: rating and performance) and IEC 60364 (low-voltage installations).
- Fitzgerald, A. E., Kingsley, C., Umans, S. D., Electric Machinery, McGraw-Hill.
Last reviewed: October 2026. Voltages are standard values; motor data are typical.