Motor Full-Load Current (Three-Phase)

Full-load current of three-phase induction motors from 1/2 to 100 hp at 208, 230, 460 and 575 V.

When sizing wires, starters and protection for a motor, the electrical code does not use the nameplate current. It uses a table of full-load currents. This table reproduces the values for three-phase AC induction motors from NFPA 70 (the National Electrical Code) Table 430.250, for the four common supply voltages, with the motor rating in kilowatts as well.

The table

18 rows

Motor (hp)	Motor (kW)	208 V (A)	230 V (A)	460 V (A)	575 V (A)	Efficiency x power factor implied at 460 V
0.5	0.37	2.4	2.2	1.1	0.9	0.43
0.75	0.56	3.5	3.2	1.6	1.3	0.44
1	0.75	4.6	4.2	2.1	1.7	0.45
1.5	1.12	6.6	6	3	2.4	0.47
2	1.49	7.5	6.8	3.4	2.7	0.55
3	2.24	10.6	9.6	4.8	3.9	0.58
5	3.73	16.7	15.2	7.6	6.1	0.62
7.5	5.59	24.2	22	11	9	0.64
10	7.46	30.8	28	14	11	0.67
15	11.19	46.2	42	21	17	0.67
20	14.91	59.4	54	27	22	0.69
25	18.64	74.8	68	34	27	0.69
30	22.37	88	80	40	32	0.7
40	29.83	114	104	52	41	0.72
50	37.29	143	130	65	52	0.72
60	44.74	169	154	77	62	0.73
75	55.93	211	192	96	77	0.73
100	74.57	273	248	124	99	0.75

Source and notes: Full-load currents of three-phase AC induction motors as listed in NFPA 70 (National Electrical Code), Table 430.250. The code values are for sizing wiring and protection; a nameplate gives the real figure for a given motor. The last column is computed: kW out / (sqrt(3) x 460 V x listed current), which shows the table assumes a combined efficiency and power factor of about 0.45 for small motors rising to about 0.75 for 100 hp, so the code figures are on the high side of a modern motor. 1 hp = 0.7457 kW.

How to use this table

Find the motor horsepower and the supply voltage.
Read the full-load current in amperes.
Multiply by 125 percent for the minimum conductor ampacity (NEC 430.22), and use the code rules for the overload and short-circuit protection.
Use the nameplate current instead for the overload relay setting.

Worked example: a 25 hp, 460 V motor

Full-load current from the table: 34 A (the motor is 18.6 kW).
Minimum conductor ampacity = 125 percent × 34 = 42.5 A.
The same motor on 230 V draws 68 A, twice as much, so it needs a much larger cable: always check the supply voltage.

Notes and cautions

The code values are conservative averages for typical 1 750 rpm motors. A modern high-efficiency motor draws less, and a low-speed or high-speed motor can draw more: use the nameplate for overload protection.
These values are for the US code. Other countries (IEC) use the nameplate current and different rules.
Single-phase motors and DC motors have their own tables (NEC 430.248 and 430.247).

FAQ

What is the full-load current of a 5 hp motor at 460 V?

7.6 A (15.2 A at 230 V).

Why is the current half at 460 V?

Power is voltage times current: doubling the voltage halves the current for the same power.

Sources

NFPA 70, National Electrical Code, Table 430.250 (full-load current, three-phase alternating-current motors) and Section 430.22. Always check the edition in force and the authority having jurisdiction.
Full-load currents of three-phase AC induction motors as listed in NFPA 70 (National Electrical Code), Table 430.250. The code values are for sizing wiring and protection; a nameplate gives the real figure for a given motor. The last column is computed: kW out / (sqrt(3) x 460 V x listed current), which shows the table assumes a combined efficiency and power factor of about 0.45 for small motors rising to about 0.75 for 100 hp, so the code figures are on the high side of a modern motor. 1 hp = 0.7457 kW.