HP to Amps Calculator

The formula#

Three phase:   I = HP × 746 / (√3 × V × PF × η)
Single phase:  I = HP × 746 / (V × PF × η)

Where η is motor efficiency and PF is the power factor at the load the motor is actually running.

Three terms in the denominator, and only one of them appears in a unit conversion. That is why "5 HP is 3.73 kW, so divide by 415 and √3" gives an answer roughly a quarter too low.

Why both terms are needed#

Efficiency accounts for the losses between the supply and the shaft. The horsepower on the plate is what comes out; the supply has to deliver that plus the copper, iron, friction and windage losses. At 88 % efficiency, 5 HP of shaft power needs 4.24 kW of electrical input.

Power factor accounts for the magnetising current. An induction motor draws current to establish its rotating field whether or not it is doing work, and that current is out of phase with the voltage. It does no work, but it flows in the cable and it heats it exactly as any other current does.

Together they are the difference between a plausible number and a correct one:

5 HP, 415 V three-phase, PF 0.85, η 88 %

Shaft power       = 5 × 0.7457      = 3.73 kW
Electrical input  = 3.73 / 0.88     = 4.24 kW
Full load current = 4240 / (1.732 × 415 × 0.85) = 6.9 A

Ignore both terms and the same motor appears to draw 5.2 A. That is 25 % low, before any derating is applied.

Power factor falls when the motor is lightly loaded#

This is the part that surprises people. An induction motor's power factor is worst at no load and improves as it is loaded up — a motor running at a quarter of its rating can sit near 0.4 rather than the 0.85 on its plate.

The magnetising current is roughly constant, so as the working current falls the ratio between them worsens. Which means an oversized motor is a double penalty: it costs more, and it draws more current per unit of useful work than a correctly sized one would.

The nameplate power factor applies at full load. If the machine habitually runs at half load, use a lower figure.

Starting current is a different question#

The calculator also reports around six times full load current, which is what a direct-on-line induction motor draws until it reaches speed.

That figure sizes the protection, not the cable. A cable can carry a several-hundred-percent overload for a couple of seconds without harm, because heating takes time — so cables are sized on full load current, and the starting transient is handled by choosing a breaker curve that does not trip on it. The motor starter calculator covers the protection side, and cable sizing the conductor.

If the starting current is the problem rather than the protection — lights dimming, a generator stumbling, an inverter tripping — the answer is a different starting method. Star-delta, a soft starter or a VFD each reduce it, and choosing between them depends on the load rather than the motor.

Typical full load currents at 415 V, three phase#

At 0.85 power factor and 88 % efficiency:

Use these to sanity-check a measurement, not to design from. Efficiency and power factor both vary with size, speed and load, and the nameplate figures for the actual machine are always better than a table.