Formula
Single phase: I = 746 × HP ÷ (V × PF × η)
Three phase: I = 746 × HP ÷ (√3 × VL-L × PF × η)
746 W = 1 HP (electric). η is motor efficiency as a decimal.
Worked example
A 5 HP three-phase motor at 400 V, PF 0.85, efficiency 0.9: I = 3730 ÷ (1.732 × 400 × 0.85 × 0.9) = 7.04 A full-load current.
Reference table
Quick reference (PF 0.85, η 0.9)
| HP | Amps @ 230 V (1φ) | Amps @ 400 V (3φ) |
|---|---|---|
| 0.5 | 2.12 | 0.7 |
| 1 | 4.24 | 1.41 |
| 2 | 8.48 | 2.82 |
| 3 | 12.72 | 4.22 |
| 5 | 21.2 | 7.04 |
| 10 | 42.4 | 14.08 |
Where this shows up in the real world
Motor circuits are a license-exam staple and a daily field reality: a 5 HP single-phase compressor at 240 V draws roughly 23–28 A depending on efficiency and PF, which dictates wire gauge, breaker and starter size. US practice often uses NEC Table 430.248 values for code compliance, but understanding the formula tells you *why* the table says what it says.
Common mistakes to avoid
Don't size protection from nameplate HP alone — input current depends on efficiency and PF, and code requires using table values or nameplate FLA, not back-of-envelope output math. The other classic: forgetting that direct-on-line starting pulls 5–8× full-load current, so breakers and supply must tolerate the inrush.
Frequently asked questions
Why include efficiency?
HP rates the mechanical output shaft power. The electrical input is larger by the efficiency factor, and current depends on the input.
Is this the current for sizing the starter?
It's the full-load running current. Starting (inrush) current is typically 5–8× higher for direct-on-line starts — protection devices must tolerate this.
Where do I find my motor's PF and efficiency?
On the nameplate: efficiency may appear as a percentage and PF as cos φ. Without a nameplate, 0.85 PF and 0.9 efficiency are reasonable estimates for standard induction motors.