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.
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.
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.