Single-phase versus three-phase
See the 120° phase relationship and compare single-phase and balanced three-phase load-current calculations.
Learning goal: Choose the correct power equation and recognize what the square-root-of-three factor represents.
Prepared by Power Stations Worldwide. Sources checked on 30 September 2026; not independently engineering peer reviewed.
The essentials
One waveform or three
Single-phase AC uses one alternating voltage waveform across a load. A balanced three-phase set has three equal sinusoidal voltages separated by 120 electrical degrees. Phase names identify the waveforms, not three independent energy sources.
Specify the voltage basis
Single-phase apparent power is S = VI. Total balanced three-phase apparent power is S = √3 VLL IL, using line-to-line voltage VLL and line current IL. For real power, multiply apparent power by power factor.
Balance matters
With sinusoidal, balanced loading, the three phase-power pulsations sum to constant total instantaneous power. An unbalanced or nonlinear load requires more detailed analysis. A lower current in the example below does not establish a universal current ratio: the supply voltages differ.
A balanced set is separated by 120°
Worked example
The same 5 kW load on two hypothetical supplies
Given
- Real power P = 5 kW = 5,000 W
- Power factor PF = 1 in both cases
- Single-phase V = 230 V RMS; three-phase VLL = 400 V RMS
Formula toolbox
- Single-phase: P = VI × PF
- Balanced three-phase total: P = √3 VLL IL × PF
- I₁φ = P / (V × PF); IL,₃φ = P / (√3 VLL × PF)
Calculate step by step
- Single-phase current: I = 5,000 / (230 × 1) ≈ 21.74 A.
- Balanced three-phase line current: IL = 5,000 / (√3 × 400 × 1) ≈ 7.22 A.
- The 5 kW is total three-phase power, not 5 kW per phase.
21.74 A single-phase · 7.22 A per three-phase line
The result compares the stated voltage bases at the same total real power. It is not a cable-size, breaker-size or supply-conversion recommendation.
Results are rounded only for display. Reproduce the calculations using unrounded intermediate values.
Quick quiz
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- 120°. Three equal phase voltages are separated by 120 electrical degrees.
- About 6.93 kW. P = √3 × 400 × 10 / 1,000 ≈ 6.93 kW total.
Curriculum connection: MIT 6.061 (Spring 2011). Its readings include 2007 notes; some equations use peak rather than RMS amplitudes. This lesson states its own value conventions and uses original diagrams, numbers and quiz questions. No MIT affiliation or endorsement is claimed.
Textbook references
Suggested technical background for this topic. These books are further reading, not proof that every explanation was checked against every edition. See the full bibliography and review scope.
- Fundamentals of Electric CircuitsSinusoidal AC and three-phase circuit foundations.
- Electrical Machines, Drives and Power SystemsThree-phase circuits and their connection to machines.