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Three-Phase — Power Formula, Wiring & kW / Amps Reference

Three-phase electric power is the dominant industrial AC distribution scheme worldwide. This page covers the canonical power formula P = √3 × V × I × PF, the wye-versus-delta connection trade-off, three-phase wire selection, kW-to-amps conversion for conductor sizing, the open-delta variant, and a worked 50 HP motor branch example. Reviewed by a licensed PE.

Three-phase calculator (embedded)

The three-phase calculator handles all four conversion modes — kW → amps, amps → kW, V/I → kVA, kVA → kW — and returns the line / phase voltage and current relationship for both wye and delta connections.

CALC.019 3-Phase · Power · Y/Δ · Unbalanced · NEC 430 Motor
Balanced 3-phase power

For 3-phase: S = √3·V_LL·I, P = S·cos φ, Q = S·sin φ. Lagging PF (induction motors) is positive φ; leading PF (over-corrected systems) is negative.

Result
— kW
Pick a mode and enter values.
FORMULA · S = √3 · V_LL · I SOURCE · NEC 430 · IEEE STD 100 · FORTESCUE 1918

Three-phase formulas

Eq. 01 — Real power (line quantities) SI
P=3⋅VL⋅IL⋅PFP = \sqrt{3} \cdot V_L \cdot I_L \cdot PF
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P = real (active) power (W or kW)
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V_L = line-to-line voltage (V)
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I_L = line current (A)
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PF = power factor (0–1)
Eq. 02 — Apparent power and reactive power SI
S=3⋅VL⋅IL,Q=3⋅VL⋅IL⋅sin⁡ϕS = \sqrt{3} \cdot V_L \cdot I_L, \qquad Q = \sqrt{3} \cdot V_L \cdot I_L \cdot \sin \phi
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S = apparent power (VA or kVA)
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Q = reactive power (VAR or kVAR)
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PF² + sin²(φ) = 1; relate via PF = cos(φ)
Eq. 03 — Solve for line current from kW SI
IL=PkW⋅10003⋅VL⋅PFI_L = \frac{P_{kW} \cdot 1000}{\sqrt{3} \cdot V_L \cdot PF}
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Most-used direction — sizing conductors and breakers from a kW load
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√3 ≈ 1.732
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For motors use NEC Table 430.250 FLA, not computed I
Eq. 04 — Wye system relationships SI
VL=3⋅Vph,IL=IphV_L = \sqrt{3} \cdot V_{ph}, \qquad I_L = I_{ph}
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4-wire wye: line voltage 1.732 × phase (line-to-neutral) voltage
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Common: 480/277 (US industrial); 208/120 (US light commercial); 400/230 (Europe)
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Phase current equals line current
Eq. 05 — Delta system relationships SI
VL=Vph,IL=3⋅IphV_L = V_{ph}, \qquad I_L = \sqrt{3} \cdot I_{ph}
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3-wire delta: line voltage equals phase (winding) voltage
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Line current = √3 × per-winding current
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Common: 240 V corner-grounded industrial (legacy)
Eq. 06 — Open delta (V-V) capacity SI
Sopen delta=Sfull delta3=0.577⋅Sfull deltaS_{open\,delta} = \frac{S_{full\,delta}}{\sqrt{3}} = 0.577 \cdot S_{full\,delta}
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Open delta uses 2 transformers instead of 3, supplying 3-phase output
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Capacity is 57.7 % of the equivalent full-delta bank
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Used for low-load 3-phase service or as backup for 1-of-3 bank failure

Standards governing three-phase systems

DocumentScope
ANSI C84.1US standard utilization voltages (208, 240, 480, 600 V)
IEC 60038International standard voltages (230/400, 400/690 V at 50 Hz)
IEEE Std 1459-2010Power, energy, and harmonic measurement under non-sinusoidal conditions
NEC Article 430Motor branch circuits
NEC Article 408Switchboards and panelboards
NEMA MG 1Motor / generator construction — three-phase induction motor characteristics
NFPA 70 §250.20 / .26Grounding for wye-connected systems and corner-grounded delta
IEEE Std 519Harmonic limits in three-phase systems

Reference: standard three-phase voltages by region

RegionV_lineV_phaseFrequencyTypical use
US light commercial wye208 V120 V60 HzOffice, multi-family
US legacy industrial delta240 V240 V60 HzOlder industrial sites
US modern industrial wye480 V277 V60 HzStandard industrial; 277 V lighting
Canadian industrial wye600 V347 V60 HzHeavy industrial / mining
European / IEC residential wye400 V230 V50 Hz4-wire wye, all loads
European industrial wye690 V400 V50 HzHeavy industrial, large motors
UK BS 7671400 V230 V50 HzSame as Europe
Japan industrial200 V—50/60 Hz3-phase delta typical
  1. Identify the system — wye or delta Wye (Y) — 4-wire (3 hots + neutral); V_line = √3 × V_phase, I_line = I_phase. Delta (Δ) — 3-wire (3 hots only); V_line = V_phase, I_line = √3 × I_phase. Modern North American industrial supply is wye 480/277 V; legacy industrial is corner-grounded delta 240 V.
  2. Read line voltage and load current Voltmeter between any two phases gives V_line (208, 240, 400, 480, 600 V depending on tariff). Clamp ammeter on a single phase conductor gives I_line. Power factor PF from nameplate or measurement (default 0.85 for induction-motor loads, 0.95+ for power-factor-corrected industrial).
  3. Apply the three-phase power formula Real power: P = √3 × V_L × I_L × PF (in kW with V in V, I in A, then divide by 1000). Apparent power: S = √3 × V_L × I_L (kVA). Reactive power: Q = √3 × V_L × I_L × sin φ (kVAR).
  4. Solve for any unknown Three of the four (P, V, I, PF) determine the fourth. Most-used direction: kW → amps for sizing conductors. I = P × 1000 / (√3 × V × PF). Example: 100 kW at 480 V, PF 0.85 → 141 A.
  5. Size conductors and protection Apply NEC §210.20 1.25× continuous factor. Pick the smallest copper THWN-2 from NEC 310.16 75 °C column meeting the demand. For motor branches, use NEC Table 430.250 FLA, 1.25× for conductor (NEC §430.22), 2.50× for inverse-time breaker (NEC §430.52).

Worked example — 50 HP three-phase motor branch

50 HP 480 V three-phase induction motor, NEMA Design B, PF ≈ 0.85.

  1. FLA from NEC Table 430.250: 65 A (use this, not computed I, per NEC §430.6).
  2. Cross-check with formula: I = (50 × 746) / (1.732 × 480 × 0.85) = 53 A — close to NEC table 65 A; difference is the table\'s allowance for inrush + PF variation.
  3. Conductor sizing (NEC §430.22): 1.25 × 65 = 81.25 A → 4 AWG Cu THWN-2 (85 A at 75 °C).
  4. Inverse-time breaker (NEC §430.52): 2.50 × 65 = 162.5 A → 175 A (next standard).
  5. Apparent power: S = √3 × 480 × 65 / 1000 = 54 kVA.
  6. Real power: P = 54 × 0.85 = 46 kW (≈ shaft 50 HP × 0.91 efficiency).

Comparison — single-phase vs. three-phase vs. open-delta

AspectSingle-phaseThree-phase wye / deltaOpen-delta (V-V)
Conductors2 or 3 (incl. neutral)3 or 4 (incl. neutral)3
Power per conductorBaseline 1×1.73× higher1.0× (lower than full delta)
Motor self-startingNeeds starting cap or shaded poleNative (rotating field)Native — same as full delta
Best forResidential up to 5 HPIndustrial > 1 HP, commercial HVACLow-load 3-phase service, backup operation
Voltage choices120, 240 (split-phase)208/240/400/480/600 + neutral optionsSame as parent delta
Cost (per kW)Lowest equipmentLowest distribution / motorSaves 1 transformer, but 57.7 % capacity

Variants and related queries

Three-phase wire and three-phase load

A three-phase load uses three (delta) or four (wye) current-carrying conductors. For motor and large HVAC loads, the size of each conductor is set by NEC 310.16 ampacity table for the calculated I_line. A balanced 3-phase load draws equal current on each phase; an unbalanced load (mixed lighting + motor) requires a fourth wire (neutral) sized per NEC §220.61 to carry the imbalance.

Three-phase formula and three-phase calculation formula

The canonical real-power formula: P = √3 × V_L × I_L × PF. For computational shortcuts: P (kW) = (V × I × PF × 1.732) / 1000. Apparent power S (kVA) = (V × I × 1.732) / 1000. To get amps from kW: I = (P × 1000) / (V × PF × 1.732). The this calculator handles all six derivations.

Three-phase amps to kW and three-phase amperage calculator

I (A) → P (kW) on a 3-phase circuit: P = √3 × V × I × PF / 1000. At 480 V with 100 A and PF 0.85: P = 1.732 × 480 × 100 × 0.85 / 1000 = 70.6 kW. Apparent power S = 1.732 × 480 × 100 / 1000 = 83.1 kVA. The 12.5 kVAR difference is reactive power that lags 90° behind the load voltage.

Three-phase open-delta connection

Open delta (sometimes "V-V") uses two transformers connected in a V-shape to deliver three-phase output from a two-bank source. Capacity is 57.7 % of the equivalent three-bank closed delta — useful for backup operation when one bank in a 3-bank delta fails, or for low-cost low-load three-phase service in rural areas.

Three-phase load minimum requirement

"A three-phase load requires a minimum of" 3 conductors (delta) or 4 (wye). For most motor loads, 3 conductors plus an equipment grounding conductor (4 total) is the typical wiring. For wye lighting circuits with mixed loads, a 4-wire wye with a neutral that is at least the same size as each phase conductor is required by NEC §220.61.

Frequently asked questions

What is three-phase power?
Three-phase power is an AC electrical-power transmission scheme using three conductors carrying alternating currents that are 120° out of phase with each other. The instantaneous sum of the three currents is zero in a balanced system, so the neutral conductor carries no return current — meaning more power can be transferred for the same conductor cost. The 120° phase displacement also produces a rotating magnetic field, which is what makes three-phase induction motors self-starting.
What is the three-phase power formula?
For real (active) power: P = √3 × V_L × I_L × PF (kW), where V_L is line-to-line voltage in volts, I_L is line current in amperes, PF is the power factor. For apparent power: S = √3 × V_L × I_L (kVA). For reactive power: Q = √3 × V_L × I_L × sin φ (kVAR). The √3 factor (≈ 1.732) comes from the geometry of the 120° phase relationship — derive from the per-phase formula P_phase = V_phase × I_phase × PF integrated over three phases.
How do you calculate three-phase amps from kW?
I_line = P × 1000 / (√3 × V_line × PF). For 50 kW at 480 V three-phase, PF 0.85: I = 50 000 / (1.732 × 480 × 0.85) = 70.7 A. The three-phase calculator handles this in one form along with the inverse (amps → kW) and the kVA / kVAR breakdown.
What is the difference between three-phase wye and delta?
Wye (Y) connection has three windings tied together at one common neutral point; line voltage V_L = √3 × phase voltage V_ph. Delta (Δ) connection has three windings tied end-to-end in a triangle with no neutral; V_L = V_ph but I_L = √3 × I_ph. Wye gives access to two voltages (e.g. 480 V phase-phase + 277 V phase-neutral); delta is simpler with three wires only. Modern NA industrial favours wye 480/277; legacy / specialty work uses corner-grounded delta 240 V.
How much three-phase wire do I need?
A three-phase circuit needs three current-carrying conductors plus an equipment grounding conductor (4 wires) for a delta connection, or four CCCs plus EGC (5 wires) for a wye connection with neutral. Conductor size is set by NEC 310.16 ampacity for the calculated line current — for 100 A three-phase service: 3 AWG copper or 1 AWG aluminum, sized identically per phase.

Historic source — invention of the rotating magnetic field

Ferraris demonstrated in 1885 that two AC currents 90° out of phase in two perpendicular coils produce a rotating magnetic field — the foundation of the polyphase induction motor. Tesla independently reached the three-phase version, and Dolivo-Dobrovolsky\'s 1891 Lauffen-Frankfurt project demonstrated the first complete three-phase generation, transmission, and motor system. Every modern 3-phase motor descends from these three.

Galileo Ferraris — Royal Academy of Sciences of Turin → 1885 — discovery of the rotating magnetic field, the principle that makes three-phase induction motors possible

Related calculators and references

Sources and further reading

  1. ANSI C84.1 — Standard for Electric Power Systems and Equipment Voltage Ratings.
  2. IEC 60038 — IEC standard voltages.
  3. IEEE Std 1459-2010 — Standard Definitions for the Measurement of Electric Power Quantities.
  4. NFPA 70 — NEC, Article 430 (motor circuits), Article 408 (panelboards).
  5. Hayt, W. H.; Kemmerly, J. E. Engineering Circuit Analysis, 9th edition. McGraw-Hill, 2018.
  6. Ferraris, G. Atti della R. Accademia delle Scienze di Torino, 1885 — original rotating magnetic field paper.