How many amps is 1,500 watts?

12.5 A at 120 V, 6.25 A at 240 V (resistive). With a motor at PF 0.85, 14.7 A at 120 V.

What power factor should I use?

1 for heaters, ranges, incandescent lamps and EV chargers; 0.8–0.9 for motors and older fluorescent ballasts; the nameplate often states it. If unsure, 0.8 is conservative.

Why √3 for three-phase?

With a balanced load the power is shared by three lines and the line-to-line voltage is √3 times the line-to-neutral voltage; the two effects combine into P = √3 × VLL × I × PF.

Watts to amps calculator

Current · single-phase12.50AI = P ÷ (V × PF)1,500 W at 120 V, PF 1.00

Power

From the nameplate

Voltage

120, 240, or a DC value

Phase

Power factor

1 for heaters and resistive loads; 0.8–0.9 for motors

Apparent powerwatts ÷ PF; what the conductor and breaker see: 1,500 VA
At 125 % for a continuous load210.19(A)(1): conductors and OCPD sized at 125 % of a continuous load: 15.63 A
Kilowatts: 1.50 kW

Three-phase current is per line with a balanced load; the voltage is line-to-line (208, 480). For single-phase 240 V loads use 240, not 120.

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Watts and volts give amps. For DC and resistive loads it is P ÷ V; for AC with a power factor below 1 the current is higher, P ÷ (V × PF); for balanced three-phase, divide by √3 as well and use the line-to-line voltage. The tally adds the apparent power in VA, which is what the conductor and breaker actually see, and the 125 % figure for a continuous load.

Worked examples

1,500 W heater at 120 V: 12.5 A. At 125 % for a continuous load, 15.6 A, which is why a 1,500 W heater wants a 20 A circuit to itself. Open this run
5,000 W (5 kW) at 240 V single-phase: 20.8 A. A 5 kW water heater element on a 30 A circuit with 10 AWG. Open this run
10 kW three-phase motor at 480 V, PF 0.8: I = 10,000 ÷ (√3 × 480 × 0.8) = 15.0 A per line; apparent power 12.5 kVA. Open this run
100 W solar panel at 18 V (DC): 5.6 A. The 12 V wire size calculator sizes the run. Open this run

How it is worked out

DC: I = P ÷ V. Single-phase AC: I = P ÷ (V × PF). Three-phase AC (balanced): I = P ÷ (√3 × V × PF), V line-to-line.

Apparent power S = P ÷ PF in VA. Continuous-load figure = I × 1.25 (210.19(A)(1) and 210.20(A) size conductors and overcurrent devices at 125 % of a continuous load).

The figures behind it

Power formulas: P = V × I × PF and P = √3 × V × I × PF; arithmetic, no table. 125 % continuous-load rule from NEC 210.19(A)(1) / 210.20(A) (2017 text). Source

Questions people ask

How many amps is 1,500 watts?: 12.5 A at 120 V, 6.25 A at 240 V (resistive). With a motor at PF 0.85, 14.7 A at 120 V.
What power factor should I use?: 1 for heaters, ranges, incandescent lamps and EV chargers; 0.8–0.9 for motors and older fluorescent ballasts; the nameplate often states it. If unsure, 0.8 is conservative.
Why √3 for three-phase?: With a balanced load the power is shared by three lines and the line-to-line voltage is √3 times the line-to-neutral voltage; the two effects combine into P = √3 × VLL × I × PF.

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