How to Calculate Electrical Load Current
Load current is the amount of current a piece of equipment draws from the supply. You need it to size cables, breakers, contactors and transformers, and to check that an existing circuit isn't overloaded. This guide shows the formulas, the role of power factor and efficiency, and how to read the results above.
The Core Formulas
Single-phase: I = (kW × 1000) ÷ (V × PF)
From kVA (3-phase): I = (kVA × 1000) ÷ (√3 × V)
From HP: kW = HP × 0.7457 ÷ efficiency
V is the line-to-line voltage for three-phase systems and the line-to-neutral voltage for single-phase. PF is power factor (0–1). For three-phase the √3 (≈1.732) accounts for the phase relationship between the three line currents.
Power Factor — Why It Increases Current
Power factor is the ratio of real power (kW, the useful work) to apparent power (kVA, what the supply must actually provide). Inductive loads like motors draw extra reactive current that does no work but still heats cables. A motor running at PF 0.8 draws about 25% more current than the same kW at unity power factor — so a poor power factor directly enlarges the cable and breaker you need.
Efficiency — Output vs Input Power
A motor nameplate usually states mechanical output power. The electrical input the supply must deliver is higher, because some energy is lost as heat. Input kW = output kW ÷ efficiency. A 10 kW motor at 92% efficiency actually draws about 10.9 kW of electrical power. That's why this calculator applies efficiency to kW and HP inputs — and why you should set it to 100% when entering electrical input power or kVA.
Motor Full-Load Current (NEC)
For motor circuit design, NEC requires using the table full-load current values (e.g. Table 430.250 for three-phase motors) rather than a calculated value, because manufacturers vary. The calculated figure here is excellent for estimating, transformer loading and energy studies, but for code-compliant motor protection use the NEC table current and the branch-circuit multipliers in Article 430.
Typical Full-Load Currents (3-Phase, 400 V, PF 0.85, ~92% eff.)
| Motor (kW) | Motor (HP) | Approx. FLC (A) |
|---|---|---|
| 0.75 | 1 | ~1.6 |
| 1.5 | 2 | ~3.1 |
| 3.7 | 5 | ~7.6 |
| 7.5 | 10 | ~15 |
| 15 | 20 | ~30 |
| 30 | 40 | ~59 |
| 55 | 75 | ~108 |
| 90 | 120 | ~177 |
Indicative values — actual current depends on nameplate efficiency, power factor and voltage. Verify against the manufacturer's data and the applicable code table.
Worked Example
A 22 kW (output) three-phase motor on 415 V, power factor 0.86, efficiency 93%:
- Electrical input = 22 ÷ 0.93 = 23.66 kW
- I = (23.66 × 1000) ÷ (1.732 × 415 × 0.86) ≈ 38.3 A
- Next standard device ≥ 38.3 A = 40 A (running-current basis)
Enter these values above to confirm, then click through to the cable size calculator to choose the conductor.
Common Mistakes
- Pairing single-phase voltage with a three-phase formula (the coupled system selector here prevents this).
- Forgetting efficiency on motor ratings, underestimating the real current draw.
- Applying power factor to a kVA figure twice — kVA already includes it.
- Using calculated current instead of the NEC table value for motor protection.