CASE-007 · Teaching Example

Three-Phase Power and Power Factor

What active, apparent and reactive power does the balanced three-phase model return at 400 V, 10 A and 0.8 lagging?

Teaching example—not a real installation.

Teaching inputs

Fixed values from the versioned calculation fixture.

Each transformation stays visible.

1
Find apparent power
Balanced three-phase uses the square root of three on line voltage and line current.
S = sqrt(3) × 400 × 10 = 6928.203230276 VA
2
Apply the power factor
Active power is apparent power times the displacement power factor.
P = S × 0.8 = 5542.56258422 W
3
Find lagging reactive power
sin(acos(0.8)) is 0.6, and lagging keeps the positive sign.
Q = S × 0.6 = 4156.921938165 var

03 · Result

Teaching example—not a real installation. Results apply only to the stated fixture inputs.

Teaching Example: power factor here is P/S for this sinusoidal operating point. It is not a motor efficiency.

What the result does—and does not—support.

Harmonics, unbalance and code compliance are not evaluated.
The calculation record stores SRC-002 with url: null. The OpenStax pages are the register references, not a URL written onto that record.
This fixture stays at 400 V. It is not a converted 380 V example.

PowerEngineerLab internal method · model 1.0.0 · dataset user-input-v1

Continue exploring