Single & three phase formulas
| Quantity | Single phase | Three phase (line values) |
|---|---|---|
| Apparent power (VA) | V × I | √3 × VL × IL |
| Real power (W) | V × I × pf | √3 × VL × IL × pf |
| kVA | (V × I) ÷ 1000 | (√3 × VL × IL) ÷ 1000 |
| Current from power | P ÷ (V × pf) | P ÷ (√3 × VL × pf) |
Motor current from horsepower: I = (HP × 746) ÷ (V × eff × pf), and for three phase divide again by √3 → I = (HP × 746) ÷ (√3 × VL × eff × pf). One horsepower is 746 watts; efficiency and power factor are fractions between 0 and 1.
These are physics relationships, safe to state. The nameplate full-load current of a specific motor, and any code full-load-current value, come from the nameplate and your code book — not from here.
√3 ≈ 1.732. pf = power factor. eff = efficiency.
| Step | Single phase, 240 V | Three phase, 208 V line |
|---|---|---|
| Carga | 6,000 W at pf 0.8 | 6,000 W at pf 0.8 |
| Corriente | 6,000 ÷ (240 × 0.8) = 31.25 A | 6,000 ÷ (1.732 × 208 × 0.8) = 20.8 A |
| Potencia aparente | 240 × 31.25 = 7,500 VA = 7.5 kVA | 1.732 × 208 × 20.8 = 7,493 VA ≈ 7.5 kVA |
| Revisar | 7,500 × 0.8 = 6,000 W ✓ | 7,493 × 0.8 ≈ 6,000 W ✓ |
Two things to see in that table. The kVA is the same either way — a load is a load — but the three-phase line current is smaller because it is shared across three conductors. And dividing by the power factor makes the current bigger: a poor power factor means more amps for the same watts, which is the whole reason anyone cares about it.
Motor, worked: a 5 HP motor on 208 V three phase, efficiency 0.88, pf 0.85 → I = (5 × 746) ÷ (1.732 × 208 × 0.88 × 0.85) = 3,730 ÷ 269.4 ≈ 13.8 A. That is the physics estimate; the nameplate figure is what you wire to, and the code's full-load values are what you size from.
Power factor and the wye/delta line-versus-phase distinction are explained in the glossary.
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