What the running amps and startup load calculator does
Motors and compressors draw two very different currents. While they run steadily they pull the running current (also called full-load amps, or FLA). At the instant they start the rotor is still and the winding behaves almost like a short circuit, so they briefly pull a much larger starting current (locked-rotor amps, or LRA). This calculator turns those two amp figures, together with the supply voltage, into the electrical load they place on a circuit, inverter, UPS or generator, giving both the steady running power and the momentary startup surge.
How to use it
Enter the supply voltage (for example 120 V or 240 V in North America, or 230 V in Europe), the appliance's running current in amps, and its starting or locked-rotor current in amps. Both amp ratings are usually printed on the motor nameplate or the appliance data plate. Optionally set the power factor (a value between 0 and 1) so the running real power in watts is accurate; if you leave it blank a typical motor value of 0.9 is used. The calculator then reports the running power, the running apparent power, the startup apparent power, and the ratio between the starting and running current.
The formula explained
For a single-phase supply the apparent power is simply voltage multiplied by current, and the real power adds the power factor:
$$S_{run} = V \times I_{run}$$ $$P_{run} = V \times I_{run} \times PF$$The startup surge uses the much larger starting current, and because inrush is mostly reactive it is sized by apparent power in volt-amps:
$$S_{start} = V \times I_{start}$$The surge ratio shows how many times the running current the appliance draws at the moment it starts:
$$\text{Surge ratio} = I_{start} \div I_{run}$$Here V is the supply voltage in volts, I_run is the running current in amps, I_start is the starting current in amps, and PF is the power factor.
Worked example
Suppose a well pump runs at 8 A on a 120 V circuit and its nameplate lists a locked-rotor current of 40 A. The running apparent power is 120 x 8 = 960 VA, and with a 0.9 power factor the running real power is 120 x 8 x 0.9 = 864 W. At startup the apparent load jumps to 120 x 40 = 4,800 VA, a surge ratio of 40 divided by 8 = 5. So although the pump only needs about 864 watts to keep running, the power source must be able to supply roughly 4,800 VA for a fraction of a second to get it spinning.
Frequently asked questions
Why is the startup load shown in VA instead of watts? Starting inrush is largely reactive current, so its real-watt content is small but its apparent power (volt-amps) is large. Generators, inverters and UPS units are limited by apparent power, so VA is the correct figure to compare against a device's surge rating.
What if I only know the running amps? Many induction motors draw a starting current of roughly 3 to 8 times their running current, so a rough estimate is 5 to 6 times the running amps. For accurate sizing use the locked-rotor amps printed on the nameplate rather than a guess.
Does this work for three-phase motors? No. These formulas are for single-phase supplies. For a three-phase load multiply the apparent power by the square root of three, so S equals root-three times V times I using the line-to-line voltage.
