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TOPIC-02 · Engineering guide

Balanced Three-Phase Power and Current

For a balanced three-phase load, apparent power is √3 × line voltage × line current. At 400 V, 10 A and 0.8 lagging, active power is 5542.56258422 W, apparent power is 6928.203230276 VA, and lagging reactive power is 4156.921938165 var.

Formula, variables and conditions

k = sqrt(3) for balanced three-phase, otherwise 1
S = k × V × I
P = S × pf
Q = +S × sqrt(1 − pf²) when lagging, and the opposite sign when leading
From active power, I = P / (k × V × pf).

Variables and units

  • V: phase voltage for single-phase, line-to-line voltage for balanced three-phase. Unit V.
  • I: line current in A.
  • pf: displacement power factor, greater than 0 and at most 1.
  • P, S, Q: active watts, apparent volt-amperes, and reactive vars.

Conditions

  • Sinusoidal steady state. The three-phase path assumes balance.
  • Motor-measurement mode is a different question: it divides measured input watts by the apparent power from RMS voltage and current. This example is from-current, not motor-measurement.
  • The calculation record stores SRC-002 with url: null. The project source register names the OpenStax sections below as previously checked. This guide does not write a URL onto the calculation record.

Worked example

Locked TC-004 fixture: balanced-three, from-current, lagging.

Line voltage
400 V
Line current
10 A
Power factor
0.8 lagging
Active power P
5542.56258422 W
Apparent power S
6928.203230276 VA
Reactive power Q
4156.921938165 var

S = sqrt(3) × 400 × 10. P = S × 0.8. Q = S × 0.6 because sin(acos(0.8)) = 0.6 for a lagging load. Line current stays 10 A in from-current mode.

Boundaries and common mistakes

Boundaries

  • Harmonics, unbalance and code compliance are not evaluated.
  • A leading load uses a negative Q. The 400 V example above is lagging.
  • Three-phase fault current is a different calculator and stays in Wiring, circuits & protection.

Common mistakes

  • Dropping √3 and treating 400 V × 10 A as the three-phase apparent power.
  • Using a phase voltage and √3 together.
  • Reading power factor as P/S and then applying it to shaft power. This mode does not know efficiency.

Sources