Motor Full Load Current Calculator

Full-load amps from a motor’s nameplate kW or HP, with efficiency and power factor in the right places — plus the starting current direct-on-line and star-delta, and the kVA it all adds up to.

Last updated 22 Sep 2026

Motor full-load current

Nameplate kW/HP → amps, running and starting

SupplyThree-phase voltage is line to line (400 V, 415 V, 480 V) and the answer is the line current.

Nameplate ratingThe shaft OUTPUT power printed on the plate, not what the motor draws.

Rating unitIndian and European plates that quote HP normally mean metric horsepower (PS); American ones mean mechanical horsepower. If the plate gives kW as well, divide to see which.

Supply voltage (V)415 V or 400 V three-phase, 230 V single-phase in India, the Gulf and Europe.

Efficiency (%)From the plate or the IE class. 90.4% is the IEC 60034-30-1 IE3 figure for a 7.5 kW 4-pole 50 Hz motor — an example, not your motor.

Power factor at full loadFrom the plate; it is not standardised the way efficiency is, so this default is an example. A lightly loaded motor is much worse than its full-load figure.

Starting current, × full-load currentFrom your motor’s data or its code letter. ABB’s Softstarter Handbook gives 6–8 times rated as usual for direct-on-line, sometimes over 10; 7 is the middle of that range and is used here as an example only.

One line of the supply, through the contactor, into the motor. The dots show the running full-load current. On three-phase this is one of the three lines and each carries the same current in a balanced motor; the starting current below is what the same line carries for the first few seconds after the contactor closes.

13.42AExample

7.5 kW, 415 V three-phase, efficiency 90.4%, power factor 0.86, starting 7 × full load

Full-load current

3-phase: I = P_out ÷ (√3 × V × η × PF) 1-phase: I = P_out ÷ (V × η × PF) I_start = I × (the multiple from the plate) I_star-delta = I_start ÷ 3

P_out: the nameplate rating — shaft power, in watts
η: efficiency as a fraction; it converts shaft power to electrical input
PF: power factor at full load, the phase angle between volts and amps
V: supply voltage; line-to-line on three-phase
√3: 1.732, because the three phases are 120° apart

Worked example

7.5 kW, 415 V three-phase, efficiency 90.4%, power factor 0.86, starting 7 × full load
Electrical input = 7,500 ÷ 0.904 = 8.296 kW
I = 8.296 kW ÷ (1.732 × 415 × 0.86) = 8,296.5 ÷ 618.17 = 13.42 A
Written the other way: I = 7,500 ÷ (1.732 × 415 × 0.904 × 0.86), because η and PF both divide — their product is 0.7774
Apparent power = 1.732 × 415 × 13.42 = 9.647 kVA
Direct-on-line start = 13.42 × 7 = 93.95 A, and star-delta a third of that, 31.32 A
796.5 W of the input never reaches the shaft

Three-phase motors at 415 V — worked with example efficiencies and power factors

PS is metric horsepower, Eff. the full-load efficiency and PF the full-load power factor. Efficiencies follow the IEC 60034-30-1 IE3 pattern where a published value exists; the power factors are plausible full-load examples and are NOT a standard — no IE class covers power factor. Every current here is the arithmetic of the row beside it, so change the efficiency or the power factor and the current changes. Use your own plate.
Rating	PS	Eff.	PF	Full-load A	Start 7×
0.37 kW	0.5 PS	70.0%	0.72	1.021 A	7.149 A
0.75 kW	1.0 PS	82.5%	0.75	1.686 A	11.8 A
1.50 kW	2.0 PS	85.3%	0.79	3.097 A	21.68 A
2.20 kW	3.0 PS	86.7%	0.81	4.358 A	30.51 A
3.70 kW	5.0 PS	88.6%	0.83	7 A	49 A
5.50 kW	7.5 PS	89.6%	0.84	10.17 A	71.16 A
7.50 kW	10.2 PS	90.4%	0.86	13.42 A	93.95 A
11.00 kW	15.0 PS	91.4%	0.86	19.47 A	136.3 A
15.00 kW	20.4 PS	92.1%	0.87	26.04 A	182.3 A
22.00 kW	29.9 PS	93.0%	0.87	37.83 A	264.8 A
37.00 kW	50.3 PS	93.9%	0.88	62.29 A	436.1 A
55.00 kW	74.8 PS	94.6%	0.88	91.91 A	643.4 A

Getting the efficiency and the power factor in the right places

This is the calculation people get wrong, and they get it wrong in a way that undersizes cables. A motor’s nameplate kW is its output: the mechanical power at the shaft. The supply has to provide that plus the motor’s own losses, so the electrical input is the rating divided by the efficiency. Only then do you turn watts into amps, which needs dividing by the voltage, by √3 if it is three-phase, and by the power factor. Both efficiency and power factor end up in the denominator, and because they multiply, their combined effect is large: 0.904 × 0.86 = 0.7774, so the real current is about 29% higher than P ÷ (√3 V) alone would give.

Check it against the plate. If the nameplate gives the output kW, the voltage and the full-load amps, then rearranging gives η × PF = P_out ÷ (√3 × V × I) directly, with no guessing. That is the honest way to fill in this page when you do not trust the efficiency and power factor separately — and it is why this page will not pretend to know your motor’s power factor. Efficiency is standardised by IE class; power factor is not.

Starting current. A cage induction motor started straight across the line is, for the first moment, a short-circuited transformer: the rotor is not turning, so the slip is 100%, and the current is limited only by the leakage impedance. ABB’s Softstarter Handbook puts it at six to eight times rated current and notes it can exceed ten. The multiple belongs to your motor — it is behind the NEMA code letter, or stated as I_p/I_n in an IEC catalogue — so this page takes it as an input rather than inventing one. At 7× the example motor pulls 93.95 A for a second or two, which is 67.53 kVA of apparent power off the supply.

Star-delta, and what it costs. Connect the windings in star for the start and each one sees the phase voltage, 1/√3 of the line voltage. Its current falls by 1/√3, and because a star connection’s line current equals its phase current where a delta’s was √3 times bigger, the line current falls by 1/√3 again: one third overall, 31.32 A here. Torque follows the square of the voltage, so it falls to one third too. That is the trade: a starter that only works if the load can be got moving on a third of the torque — a fan or an unloaded compressor, not a loaded conveyor.

Take the full-load current to the MCB size calculator and the cable size calculator, and remember that motor circuits are protected differently from fixed loads: the device has to pass the starting current without tripping while the overload relay looks after the running current. Check the drop at starting with the voltage drop calculator — that is what makes lights dim when a pump starts. For the torque and speed behind the same rating, use the motor power, torque and speed calculator; for correcting the power factor, the power factor calculator.

Frequently asked questions

How do I calculate the full load current of a 3 phase motor?

Divide the nameplate output power by the efficiency, then by √3, the line voltage and the power factor. For 7.5 kW at 415 V with η = 90.4% and PF = 0.86 that is 7,500 ÷ (1.732 × 415 × 0.904 × 0.86) = 13.42 A.

Is the motor kW on the nameplate input or output power?

Output — the mechanical power at the shaft. The input is higher by the losses. A 7.5 kW motor at 90.4% efficiency takes 8.296 kW from the supply.

How many times full load current does a motor draw when starting?

Usually six to eight times, and sometimes more than ten, according to ABB’s Softstarter Handbook. The exact figure is your motor’s: it is what the NEMA code letter encodes, or the I_p/I_n ratio in an IEC catalogue. Enter it above rather than assuming.

How much does star-delta starting reduce the current?

To one third of the direct-on-line value — and the starting torque falls to one third as well, because torque goes with the square of the voltage. For the example motor that is 93.95 A down to 31.32 A.

Why is the motor current higher than kW ÷ (√3 × V)?

Because that expression leaves out both the efficiency and the power factor. Their product here is 0.7774, so the real current is about 29% higher. Sizing a cable from the naive figure undersizes it.

Do I size the breaker from the running or the starting current?

From the running current for the thermal protection, but the device must also ride through the starting current without tripping — which is why motor circuits use a type C or D breaker, or a motor-protection circuit-breaker with a separate magnetic setting, rather than a type B. Follow your local code.

Related calculators

References

ABB. Softstarter Handbook, publication 1SFC132060M0201. Direct-on-line starting current “Usually between 6-8 times the rated current, but it can be more than 10 times the rated current”; star-delta: “The resulting current when Y-connected will be 1/3 of the current when delta connected” and the torque “ending up being 33% of the torque available when delta connected”; n = 2 × f × 60 / p; s = (n₁ − n)/n₁; T_n = 9550 × P_r/n_r.
IEC 60034-30-1 efficiency classes for line-operated AC motors (IE1 standard, IE2 high, IE3 premium, IE4 super-premium), as tabulated in ABB technical note 9AKK107319 EN 05-2018. At 50 Hz and 4 poles the nominal efficiencies are 79.6 / 82.5 / 85.7% at 0.75 kW, 85.5 / 87.7 / 90.4% at 3 kW, 88.7 / 90.4 / 92.6% at 7.5 kW and 91.6 / 93.0 / 94.5% at 22 kW for IE2 / IE3 / IE4.
National Electrical Manufacturers Association. Electric Motor Terminology and Performance Characteristics: “Synchronous Speed = 120 x Frequency / # Poles”; “% Slip = 1 − (Full Load RPM / No Load RPM) × 100”; synchronous speeds of 3600, 1800, 1200, 900 and 720 rpm at 60 Hz for 2, 4, 6, 8 and 10 poles; NEMA designs A, B, C and D, with design D full-load slip of 5–8% or 8–13% and locked-rotor current about 650% of full load for designs B, C and D.
IEC 60038:2009, IEC standard voltages: 230/400 V is the standard low-voltage three-phase system (India, the Gulf and Europe; 240/415 V is still quoted in several Gulf states, and North America uses 120/240 V and 208Y/120 V or 480Y/277 V at 60 Hz).
Hughes E, Hiley J, Brown K, Smith I M. Electrical and Electronic Technology, 12th ed. Pearson 2016: three-phase star and delta relationships, the power triangle, and the polyphase induction motor.