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I = P / (V × PF)

Watts to Amps Calculator

Watts to amps is the conversion behind every "will this run on a 16 A socket?" question. It needs the voltage, and on anything other than a heater it needs the power factor too.

Inputs

Supply
Current
4.38A
Apparent power1,744VA
Real power1.500kW
Over 8 hours12.00kWh

For page numbers, keep Headers and footers ticked under More settings in the print dialog.

The formula

DC circuits — the simple case, no power factor:

I = P / V

Single-phase AC:

I = P / (V × PF)

Three-phase AC:

I = P / (√3 × V × PF)

Where P is real power in watts, V is volts, PF is power factor, and I is current in amps.

Worked examples

A 1500 W heater on a 230 V single-phase supply. A heater is a resistive load, so the power factor is 1:

I = 1500 / (230 × 1) = 6.5 A

Comfortable on a 16 A socket circuit, and the reason a 6 A socket is not enough for one.

The same heater on a 120 V supply:

I = 1500 / (120 × 1) = 12.5 A

Halve the voltage and the current doubles for the same power. On a North American 15 A branch circuit, that single heater occupies 83 % of the circuit before anything else is plugged in — which is why the 12 A / 1440 W limit on portable heaters exists.

A 1500 W motor at 0.85 power factor, 230 V single-phase:

I = 1500 / (230 × 0.85) = 7.7 A

Same watts, 18 % more current than the heater, because the power factor is below 1.

Appliance chart at 230 V

Current drawn by common single-phase loads, with the power factor each type actually runs at:

Appliance Watts PF Amps at 230 V
LED bulb 9 0.9 0.04
Ceiling fan 75 0.95 0.34
Laptop charger 90 0.95 0.41
Television, 55 inch 150 0.95 0.69
Refrigerator, running 200 0.8 1.09
Desktop computer 300 0.95 1.37
Washing machine 500 0.85 2.56
Microwave oven 1200 0.95 5.49
Room heater 1500 1.0 6.52
Hair dryer 1800 1.0 7.83
1.5 ton air conditioner 1600 0.9 7.73
Electric kettle 2000 1.0 8.70
Geyser, 25 litre 2000 1.0 8.70
Induction hob 2100 0.98 9.32
Electric oven 2500 1.0 10.87
2 ton air conditioner 2200 0.9 10.63

Two things this chart does not show. Motor-driven appliances — fridges, air conditioners, washing machines — draw a starting current several times these figures for a second or so. And an inverter air conditioner varies its consumption continuously rather than cycling, so its running figure is whatever the compressor is doing at that moment, not a fixed number.

Watts, VA, and the UPS trap

This is the conversion that costs people money.

A UPS is rated in VA, apparent power. Your equipment is rated in watts, real power. They are not the same number, and the ratio between them is the power factor:

VA = W / PF

A 600 W load at 0.6 power factor needs 1000 VA of UPS. Buy a "1000 VA" UPS for a 1000 W load and it will overload immediately — that unit is designed for about 600 W.

Older UPS units state a power factor of 0.6, newer ones 0.8 to 0.9. The nameplate always shows both figures; use the watts one. And leave headroom: a UPS running at its rating runs hot, and hot is what kills the battery.

The watts to VA calculator does this conversion directly.

Three-phase, and the size of the difference

A 15,000 W load:

  • Single-phase, 230 V, PF 0.9: 15000 / (230 × 0.9) = 72.5 A
  • Three-phase, 415 V, PF 0.9: 15000 / (1.732 × 415 × 0.9) = 23.2 A

A third of the current for the same work, spread across three conductors instead of one. This is why anything much above 5 kW is supplied three-phase — the cable, the switchgear and the losses all scale with current, not with power.

Sizing the circuit around the answer

The current is where the design starts:

Continuous loads get 125 %. A load running more than three hours continuously — heating, lighting, a compressor on a duty cycle — is conventionally sized at 125 % of its current. Our 6.5 A heater wants a circuit rated at least 8.1 A, so a 10 A device.

The cable must survive the device, not the load. If the protective device is 16 A, the cable has to carry 16 A after derating for its ambient temperature and how many other cables share its route. Cable sizing handles the derating; MCB and MCCB sizing handles the device.

Volt drop over the run. On anything longer than about 20 m, volt drop rather than heating usually decides the size. Limits are 3 % for lighting and 5 % for power. The voltage drop calculator covers it.

Common mistakes

Assuming PF 1 for everything. True for heaters, kettles, filament lamps and geysers. Wrong for motors, air conditioners and anything with a switch-mode supply, and it understates the current by 10 to 40 %.

Using the wrong voltage. 230 V line-to-neutral for single-phase, 415 V line-to-line with √3 for three-phase. Mixing them up is a factor of nearly two either way.

Reading the surge rating from a nameplate. Some appliances state peak input rather than continuous. If the figure looks unusually high for the appliance, check whether it is a surge or a sustained rating.

Adding up nameplate watts for a whole installation. Nothing runs everything at once. Connected load times a diversity factor gives maximum demand — see connected load and diversity for the factors that apply.

The neutral carries the single-phase current too

On a single-phase circuit the current calculated here flows in the line conductor and returns, in full, through the neutral. That is obvious stated plainly and is routinely forgotten in two places.

The first is cable selection. A reduced-neutral cable is legitimate on a balanced three-phase circuit and is wrong on a single-phase one, where the neutral is a full current-carrying conductor and has to be the same size as the line. The second is switching and protection: the neutral must be isolated by the same device where the installation requires it, and a single-pole breaker on a circuit that needs double-pole isolation leaves the neutral live relative to earth when the circuit is supposedly dead.

Where a distribution board mixes single-phase circuits across three phases, the neutral current at the board is the vector sum rather than the arithmetic one, and on a well-balanced board it is small. That cancellation is the reason a common neutral is acceptable at all — and the reason it stops being acceptable where the loads are electronic, because triplen harmonics add in the neutral instead of cancelling.

Continuous loads and the 80 per cent rule

Protective devices are rated for the current they can carry indefinitely under standard test conditions, and those conditions are more generous than a real enclosure. Where a load runs for three hours or more at a stretch, common practice — explicit in NEC and implicit in the derating tables elsewhere — is to size the circuit at 125 per cent of the calculated current, which is the same as loading the device to no more than 80 per cent of its rating.

Water heaters, air conditioners, EV chargers, lighting circuits and anything on a process that runs a shift all qualify. A 3 kW geyser at 230 V draws about 13 A, and the circuit for it should be built around 16.3 A rather than 13 — which is the difference between a 16 A device that will spend its life near its limit and a 20 A one that will not.

The reasoning is thermal rather than bureaucratic. A breaker in a full panel, surrounded by other breakers, in an Indian summer, is running considerably hotter than the bench it was calibrated on, and its actual trip point drifts down accordingly. The margin is what stops a correctly sized circuit becoming a nuisance-tripping one in July.

Every conversion on this site runs in your browser — nothing you type is sent anywhere. See all 11 calculators.

Questions people ask

Will a 1500 W heater run on a 6 A socket?
No. A heater is resistive, so power factor is 1 and the current is simply 1500 / 230 = 6.5 A — just past a 6 A outlet and comfortable on a 16 A socket circuit. That half-amp margin is exactly why domestic wiring uses 16 A circuits for anything with an element in it, and why a 6 A point is for lighting and electronics.
Does a DC circuit need a power factor?
No — I = P / V, and that is all. Power factor exists because AC voltage and current can fall out of step, and in a DC circuit there is no phase to be out of. The same is true of any purely resistive AC load, where the power factor is 1 and the formula collapses to the DC one.
Why does the same appliance draw twice the current in North America?
Because current is power divided by voltage, and the voltage is roughly half. That 1500 W heater is 6.5 A at 230 V and 12.5 A at 120 V. It is the reason North American branch circuits are rated higher in amps for the same delivered power, and the reason a 230 V appliance brought to a 120 V country needs its circuit reconsidered rather than just its plug changed.
Why is the VA figure higher than the watts I entered?
Because volt-amps are what the wiring actually carries and watts are only the part doing work. At 0.86 power factor, 1500 W of load is 1,744 VA, and it is the 1,744 that heats the cable and loads the breaker. Size conductors and protection from the VA figure, or equivalently from the current — never from the watts alone.