ElectroCalculators
Free electronics & electrical engineering calculators

Motor FLA Calculator

Full load current from motor kW or hp, including efficiency — the term most calculators drop, which understates the current by 10–15%.

Inputs

Supply
Rating in

The nameplate figure — mechanical power at the shaft, not electrical input.

V

%

IE3 motors are 89–94% depending on size. From the nameplate where possible.

1.15 is common on North American motors; 1.0 elsewhere.

Results

Full load amps14.15 AI = P / (√3 · V · η · cos φ)
Size conductors for
17.69 A125% of FLA
Starting current
84.9 A
Electrical input power
8.333 kW
Apparent power
9804
Heat lost in motor
833.3 W

Size branch-circuit conductors for 17.7 A — NEC 430.22 requires 125% of full load current for a continuous-duty motor.

Direct-on-line starting draws roughly 84.9 A (6× FLA). Protection must tolerate this inrush without tripping, which is why motor circuits use time-delay devices rather than ordinary breakers.

0.833 kW is lost as heat inside the motor. That is what the cooling fan has to remove, and what an enclosure has to dissipate.

For NEC conductor and overload sizing in the United States, use the table values in NEC 430.248–430.250 rather than this calculated figure or the nameplate. The tables are what the code requires; the nameplate is only used for overload protection.

Diagram

Motor circuit with running and starting currentA 7.5 kW motor on 400 V draws 14.2 A running and about 84.9 A on starting.3Ø 400 VMPCBM7.5 kWRunning14.2 AStarting84.9 ASize conductors for 17.7 A (125% of FLA)

Worked examples

7.5 kW three-phase, 400 V

A standard industrial motor — pump, fan or conveyor drive.

14.15 A FLA, size conductors for 17.7 A, 85 A starting surge

10 hp, 460 V North American

The same job in imperial units on a US industrial supply.

11.9 A FLA — but always check NEC Table 430.250 for the code figure

2.2 kW single-phase, 230 V

A workshop machine on a domestic supply. Note the current compared to three-phase.

14.1 A — near the limit of a 16 A circuit before inrush is considered

Same motor on a VFD

A drive removes the inrush problem entirely, which often lets you use smaller protection.

Starting current drops from 85 A to 21 A

How to calculate motor full load amps

When to use this: sizing cable and protection for a motor circuit, checking whether an existing supply can take another machine, or verifying that a motor is drawing what it should rather than something worrying.

Full load current is the motor's rated output divided by everything it takes to get there:

Three-phase: I = P_out / (√3 × V × η × cos φ)
Single-phase: I = P_out / (V × η × cos φ)

P_out is the nameplate rating — mechanical power at the shaft. η is efficiency and cos φ is power factor, both also on the nameplate.

The efficiency term people drop

This is the most common mistake in the calculation. The nameplate says 7.5 kW, but that is what the motor delivers at the shaft. To deliver it, the motor must draw more, because some is lost as heat in the windings, iron and bearings.

Electrical input = 7.5 kW / 0.90 = 8.33 kW

Leaving efficiency out understates the current by 10–15%. On a circuit sized close to its limit, that is the difference between a cable that runs warm and one that runs hot.

The NEC trap

If you are working under the National Electrical Code in the United States, do not size conductors from this calculation or from the nameplate.

NEC 430.6(A)(1) requires branch-circuit conductor and short-circuit protection sizing to use the current values in Tables 430.248 through 430.250 — tables of typical currents by horsepower and voltage. The nameplate current is used only for overload protection sizing under 430.32.

The table values are deliberately generic and often differ from a specific motor's actual nameplate. That is intentional: it means a replacement motor of the same horsepower will always fit the existing circuit. This calculator gives you the engineering figure, which is the right number outside NEC jurisdictions and for sanity-checking — but the code number is what an inspector will check.

Starting current is the real design constraint

An induction motor at standstill looks electrically like a short-circuited transformer. Until it turns, there is no back-EMF opposing the supply, so it draws enormous current — typically six times full load, for a few seconds.

Starting methodInrushStarting torqueSuits
Direct-on-line~6× FLA100%Small motors, stiff supplies
Star-delta~2× FLA33%Fans, centrifugal pumps — loads that start unloaded
Soft starter~3× FLAAdjustableConveyors, anything where mechanical shock matters
VFD~1.5× FLAFull, from zero speedAnything, if you also want speed control

Protection has to ride through this surge. That is why motor circuits use time-delay fuses, motor-rated circuit breakers or motor protection relays rather than ordinary breakers — an ordinary breaker sized for the running current would trip on every start.

Worked examples

7.5 kW three-phase on 400 V

Input = 7500 / 0.90 = 8333 W
I = 8333 / (1.732 × 400 × 0.85) = 14.15 A
Conductors at 125% = 17.7 A
DOL starting ≈ 85 A for a few seconds

A 2.5 mm² cable handles 17.7 A comfortably. The protection needs to be motor-rated to tolerate the 85 A surge — a standard 20 A breaker would trip instantly on start.

2.2 kW single-phase on 230 V

Input = 2200 / 0.85 = 2588 W
I = 2588 / (230 × 0.80) = 14.1 A

Almost the same current as the 7.5 kW three-phase motor, for less than a third of the mechanical output. That is the case for three-phase in one line: for the same power, the current is √3 times lower, so conductors, contactors and switchgear are all smaller.

Why oversizing a motor is expensive

The magnetising current that establishes the rotating field is roughly constant regardless of load. The working current scales with load. At light load the magnetising component dominates, so power factor collapses:

LoadTypical efficiencyTypical power factor
100%90%0.85
75%89%0.81
50%86%0.71
25%75%0.47

A motor at 25% load draws current at 0.47 power factor — the supply carries roughly double the current per useful watt compared to full load. Specifying a motor "with margin" costs money continuously, not once.

Reading the nameplate

  • Output rating (kW or hp). Mechanical power at the shaft. Never electrical input.
  • FLA / FLC. The manufacturer's own full load current for this specific motor, at the stated voltage. Used for overload sizing under the NEC.
  • Efficiency and cos φ. At full load. Both fall off at partial load.
  • Service factor. Continuous overload capability — 1.15 means it tolerates 15% above rating. Headroom, not a target.
  • IE class. IE2 through IE5 under IEC 60034-30-1. Higher is more efficient; IE3 is the minimum legal class for most new industrial motors in the EU.
  • Duty type (S1–S9). S1 is continuous. Intermittent duties allow higher loading because the motor cools between runs.

How to use this calculator

  1. Enter the nameplate rating

    In kW or hp. This is shaft output — the mechanical power the motor delivers, not what it draws.

  2. Add efficiency and power factor

    Both are on the nameplate. Efficiency is what converts shaft output into electrical input, and dropping it is the most common error in this calculation.

  3. Pick the starting method

    Direct-on-line draws about six times FLA for a few seconds. That surge, not the running current, is what protection has to tolerate.

  4. Size conductors at 125%

    NEC 430.22 requires branch-circuit conductors rated for 125% of full load current on a continuous-duty motor.

Frequently asked questions

How do I calculate motor full load amps?

For three-phase: I = P_output / (√3 × V × efficiency × power factor). For single-phase, drop the √3. A 7.5 kW motor at 400 V, 90% efficient with 0.85 power factor draws 7500 / (1.732 × 400 × 0.9 × 0.85) = 14.15 A.

Why does efficiency appear in the formula?

The nameplate kW is mechanical power at the shaft, but current is drawn to supply electrical input, which is larger by the efficiency. A 90% efficient 7.5 kW motor draws 8.33 kW electrically. Leaving efficiency out understates the current by 10–15%, which is enough to undersize a cable.

Should I use the calculated FLA or the nameplate value?

Neither, for code compliance in the United States. NEC 430.6(A)(1) requires conductor and short-circuit protection sizing to use the table values in 430.248–430.250, not the nameplate and not a calculated figure. The nameplate current is used only for overload protection sizing. This calculator is for engineering estimates and non-NEC jurisdictions.

What is starting current and why does it matter?

An induction motor started direct-on-line draws roughly six times its full load current for a few seconds until it reaches speed. Protection must ride through that surge without tripping, which is why motor circuits use time-delay fuses or motor-rated breakers rather than ordinary ones.

How much does star-delta starting reduce current?

To about a third — roughly 2× FLA instead of 6×. The trade-off is that starting torque falls by the same factor, so it only suits loads that start unloaded, such as fans and centrifugal pumps. A conveyor or compressor may not accelerate at all.

What is service factor?

A multiplier showing how far above nameplate rating the motor can run continuously without damage. A 1.15 service factor on a 10 hp motor means it will tolerate 11.5 hp. It is headroom for occasional overload, not a licence to run there permanently — sustained operation in the service factor range shortens insulation life.

Why does a lightly loaded motor have poor power factor?

The magnetising current that establishes the rotating field is roughly constant regardless of load, while the working current scales with load. At light load the magnetising component dominates, so power factor collapses — often to 0.4 or lower. This is the main argument against oversizing motors.

How do I convert horsepower to kW?

One mechanical horsepower is 745.7 W, so multiply hp by 0.7457 to get kW. A 10 hp motor is 7.46 kW. Note that motors are rated in output power in both systems, so the conversion applies to the shaft rating directly.

Sources and further reading

Last reviewed .

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