All tools Print with PE stamp box Designed for sealed engineering submittals — print drops PE stamp + signature block at the end.

Three-Phase Power

Real (active), apparent, and reactive power for three-phase and single-phase AC circuits. Compute kW from kVA and PF, kVAR for capacitor sizing, line current from kW, or PF from kW and kVA.

V (L-L for 3-ph; L-N for 1-ph)
A
— (cos φ; positive = lagging)
kVA
kW
kVAR
degrees (lagging)
kVAR (capacitor needed)

Defaults: 480V three-phase, 100 A, PF = 0.85 lagging — typical industrial motor service. Lagging PF means inductive load (motors, transformers); leading PF means capacitive (over-corrected, generators).

Three-phase apparent power:
$$ S = \sqrt{3} \, V_{LL} \, I \quad \text{(volt-amperes)} $$
Single-phase: $S = V \cdot I$
Real and reactive components:
$$ P = S \cos\phi, \qquad Q = S \sin\phi $$
Power triangle: $S^2 = P^2 + Q^2$, $\text{PF} = \cos\phi = P/S$
S apparent power (kVA) · P real power (kW), what does the work · Q reactive power (kVAR), magnetizes inductors and capacitors · PF power factor = real / apparent · φ phase angle between voltage and current.

Why √3 in three-phase

The √3 factor appears because line-to-line voltage is √3 times line-to-neutral voltage in a balanced three-phase system. Per-phase apparent power = V_LN × I_phase, and total = 3 × V_LN × I_phase = √3 × V_LL × I_line. So total three-phase power = √3 V_LL I_line, the standard formula.

Power factor — what it costs you

Apparent power (kVA) sizes the wire, transformer, and switchgear. Real power (kW) is what you pay for and what does useful work. A 100 kW motor at PF = 0.7 needs a 143 kVA service; the same motor at PF = 1.0 needs only 100 kVA. The "extra" 43 kVA is wasted on charging and discharging the motor's magnetizing inductance.

Utilities penalize low PF in two ways:

Adding shunt capacitors at the customer side raises PF. The required capacitor kVAR to correct from PF1 to PF2:

kVAR_cap = P (tan φ_1 − tan φ_2)

Lagging vs leading

Inductive loads (motors, transformers) draw lagging current — current peaks after voltage. PF is positive but the load is "behind" voltage in time. Capacitive loads (overcompensated systems, harmonic generators, synchronous motors operated leading) push PF to lead — current peaks before voltage. Most industrial and commercial loads are lagging; correction adds capacitors to bring PF closer to 1.0.

Real vs apparent in motor sizing

Motors are nameplate-rated by horsepower (mechanical real power). Convert to kVA: kVA = (hp × 0.746) / (η_motor × PF). For a 100 hp motor at η = 0.92 and PF = 0.85: 100 × 0.746 / (0.92 × 0.85) = 95 kVA at full load. The contributor wire and starter must handle this kVA, not just kW.

Three-phase vs single-phase distribution

Higher voltage at the same kW means lower current, smaller wire, less voltage drop. Industrial plants distribute at medium voltage and step down at use point.

Reference: IEEE Std 141 (Red Book): Recommended Practice for Electric Power Distribution for Industrial Plants. NEC (NFPA 70-2023). Bergen, A.R., Vittal, V. (2000). Power Systems Analysis, 2nd ed., Pearson.

Related tools

Monthly engineering case studies

One real stormwater or hydraulics design problem per month, with the math worked out and the gotchas called out. No tutorials, no fluff.

Free. Unsubscribe anytime. Privacy.

Engineer of Record — Stamp & Signature
APPLY PE STAMP HERE
Engineer Name
License No.
State
Signature
Date
Project / Sheet
By stamping and signing, the Engineer of Record certifies that the inputs, formulas, and applicability of this calculation have been reviewed for the specific design context. PE-Calc tools provide computational support only — the engineer is responsible for verifying results, applying engineering judgment, and complying with applicable codes and standards.
Calculation generated at pe-calc.com