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Power & Energy Intermediate 10 min read Nov 2026

Three-Phase Power

Three-phase power delivers constant power with less conductor material. Learn the √3 formulas, star and delta connections and balanced-load calculations.

Definition

Three-phase power is the transmission and use of three alternating voltages of the same frequency and magnitude, displaced by 120° from one another. A balanced three-phase system delivers constant total power and uses conductors efficiently.

Learning objectives

  • Calculate balanced three-phase active, reactive and apparent power
  • Relate line and phase voltages and currents in star and delta
  • Explain why three-phase transmission is efficient
  • Measure three-phase power correctly
Three-Phase Power diagram Three balanced phases separated by 120 degrees L1 · L2 · L3
Three-Phase Power — schematic illustration
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Progressive explanation

Start with the plain-language idea, then move into the engineering detail. Each level stands on its own, so you can stop at the depth you need.

Beginner The core idea
A three-phase supply uses three live conductors instead of one, with waveforms spaced a third of a cycle apart. The total power delivered stays almost constant instead of pulsing to zero twice per cycle, which suits motors and heavy loads.

Homes usually receive a single phase from a three-phase distribution network. Industry uses all three phases for motors, which are smaller and smoother-running than equivalent single-phase motors.
Intermediate How it behaves in a circuit
For a balanced load, total active power is P = √3 × VL × IL × cos φ, where VL is the line-to-line voltage and IL the line current. Apparent power is S = √3 × VL × IL and reactive power Q = √3 × VL × IL × sin φ. The same power can be written per phase as P = 3 × Vph × Iph × cos φ.

In a star connection VL = √3 × Vph and IL = Iph. In a delta connection VL = Vph and IL = √3 × Iph. The total power is the same for the same load, but the voltages and currents in the windings differ.
Advanced Limits, assumptions, engineering detail
Unbalanced loads require per-phase or symmetrical-component analysis; the neutral carries the residual current and can be overloaded by triplen harmonics even when the fundamental is balanced. Sequence components separate positive, negative and zero-sequence effects, and negative sequence heats motors and generators. Power measurement uses the two-wattmeter method for a three-wire system, with the total equal to the sum of the readings. Protection, phase rotation and residual-current detection must all respect the three-phase configuration.

Formulas and variables

Balanced three-phase active power
P = √3 × VL × IL × cos φ
P Active power · W VL Line-to-line voltage · V IL Line current · A cos φ Power factor · dimensionless
400 V, 50 A and PF 0.85 give about 29.4 kW.
Star relationship
VL = √3 × Vph and IL = Iph
VL Line voltage · V Vph Phase voltage · V
A 400 V line corresponds to about 231 V per phase.
Delta relationship
IL = √3 × Iph and VL = Vph
IL Line current · A Iph Phase current · A
A 40 A phase current corresponds to about 69.3 A line current.

Units of measurement

QuantitySymbolUnit
Active power P watt (W)
Apparent power S volt-ampere (VA)
Reactive power Q volt-ampere reactive (var)
Line voltage VL volt (V)

Worked examples

Worked example

Balanced motor load

Problem. A three-phase motor draws 50 A per line at 400 V with a power factor of 0.85. Find the active, apparent and reactive power.

Solution. P = √3 × 400 × 50 × 0.85 = 1.732 × 400 × 50 × 0.85 = 29 444 W. S = √3 × 400 × 50 = 34 641 VA. Q = √(S² − P²) = √(34 641² − 29 444²) = 18 248 var.

P ≈ 29.4 kW, S ≈ 34.6 kVA and Q ≈ 18.2 kvar.
Worked example

Three single-phase heaters in star

Problem. Three 2 kW heaters are connected in star to a 400 V three-phase supply. What is the total power and the line current?

Solution. Total P = 3 × 2000 = 6000 W. Per phase Vph = 400/√3 = 231 V. Iph = 2000/231 = 8.66 A, and in star IL = Iph.

P = 6 kW and IL ≈ 8.66 A.
Worked example

Star versus delta

Problem. Three 10 Ω resistors are connected first in star and then in delta to the same 400 V supply. Compare the total power.

Solution. Star: Vph = 231 V, Iph = 23.1 A, P = 3 × 231 × 23.1 = 16.0 kW. Delta: Vph = 400 V, Iph = 40 A, P = 3 × 400 × 40 = 48.0 kW.

The delta connection delivers three times the power of the star connection for the same resistors and supply.

Practical applications

  • Industrial motor drives
  • Power distribution networks
  • Distribution and power transformers
  • Variable-frequency drives
  • Three-phase EV chargers
  • Generators and alternators

Automotive applications

Alternators generate three-phase AC that is rectified to charge the 12 V system, and electric-vehicle traction inverters produce three-phase AC for the motor. High-power AC charging can use all three phases to reduce current per conductor.

Industrial applications

Three-phase motors, distribution boards, protection relays and power-factor correction are the backbone of industrial electrical systems. Balanced loading across phases keeps neutral currents and losses low, and phase rotation must be checked before connecting motors.

Measurement procedure

  1. Use a three-phase power analyser or the two-wattmeter method for a three-wire system.
  2. Measure each line current with a clamp meter to check balance.
  3. Confirm phase rotation before connecting a motor; the wrong sequence reverses the direction.
  4. Measure phase-to-phase and phase-to-neutral voltages to verify the supply configuration.

Common mistakes

  • Using phase current instead of line current in the star line-current formula.
  • Forgetting the √3 factor and obtaining only one third of the correct power.
  • Using phase voltage in a formula that requires line voltage, or the reverse.
  • Assuming the neutral current is zero when harmonic loads are present.

Safety considerations

  • Three-phase systems commonly operate at 400 V or more; use equipment rated for the full line voltage.
  • Check phase rotation and isolation before connecting or reversing a motor.
  • Arc-flash energy is high in three-phase distribution; follow switching and PPE procedures.

Frequently asked questions

Why does the three-phase formula contain √3?
The factor arises from the 120° displacement between phases and from the relationship between line and phase quantities. Summing the three phase powers gives three times the per-phase power, which simplifies to √3 × VL × IL.
What is the difference between star and delta?
In star, one end of each winding is joined at a neutral point, so the phase voltage is lower than the line voltage by √3. In delta, the windings form a closed triangle, so the phase voltage equals the line voltage and the line current is √3 times the phase current.
Why is a neutral needed?
The neutral provides a return path for unbalanced single-phase loads and for zero-sequence current. In a perfectly balanced three-wire system it carries no fundamental current, but harmonics can still flow in it.
How can a system have both 230 V and 400 V?
The 400 V is the line-to-line voltage and the 230 V is the phase-to-neutral voltage; their ratio is √3. Both are present in the same four-wire distribution system.
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