Watts to Amps Calculator

I = P / (V × PF) on single-phase AC, I = P / (√3 × V × PF) on three-phase, and I = P / V on DC

P

The real power the load actually converts, in watts — the figure on the nameplate, and the one that does work or makes heat. It is not the same as the volt-amperes the supply has to deliver: with a power factor of 0.9 the supply carries 2222 VA to deliver 2000 W, and the difference is reactive power that shuttles back and forth without doing anything. This page asks for the real power, which is what a nameplate states, and computes the current from it

V

The supply voltage, and on a three-phase circuit this is the line-to-line voltage, the 400 V or 380 V measured between two phases rather than the 230 V measured between a phase and neutral. This is the second place the page can go wrong without looking wrong: enter the phase voltage on a three-phase supply and the current comes out far too large. The square root of three already accounts for the relationship between the two, so only one of them belongs in the expression

PF

The power factor, the cosine of the angle by which the current lags or leads the voltage, between 0 and 1. It is what makes the current larger than the power alone suggests: at 0.9 the current is 1/0.9 = 1.111 times what a purely resistive load of the same wattage would draw, and at 0.5 it doubles. Motors, transformers and anything with a switching supply have a power factor below 1. On DC there is no phase angle, so the field is ignored — see the limitations

2

The exponent in the square root of three, and the single most important number on the page. In a balanced three-phase circuit the three phase currents share a return path, so the power is √3 times the line voltage times the current rather than three times the phase voltage times it. Dividing by 1.732 instead of nothing makes the answer 42 per cent smaller — or, seen the other way, forgetting it makes the current 73 per cent larger than it is, which is the direction that burns cable

Use it when a device is rated in watts and you need to know what it will draw: choosing a breaker, sizing a cable, deciding whether an appliance can go on an existing circuit, or working out why a generator trips when the load is supposedly within its rating. It is also the page that answers the question the power factor calculator sends you here with — entering the real power and the power factor together gives the line current directly. The one thing to settle before typing is which kind of supply you are on, because that decision changes the answer more than any other input, and getting it wrong produces a perfectly plausible number.

1. The defaults: a 2 kW appliance on a 230 V single-phase supply

   1. Real power 2000 W at 230 V with a power factor of 0.9
   2. The denominator is 230 × 0.9 = 207 V of effective driving voltage
   3. Current is 2000 / 207 = 9.662 A
   4. Apparent power is the voltage times the current: 230 × 9.662 = 2222.2 VA

   The last two rows of the panel disagree with each other on purpose, and reading why is the fastest way to understand power factor. The load converts 2000 W, but the supply has to deliver 2222 VA, and the 222 VA difference is current that flows without doing work. That is the whole reason the power factor exists as a separate input: at a power factor of 0.9, ten per cent more current is needed than the wattage alone implies, and the cable, the breaker and the generator all see that larger figure.

2. A 15 kW machine on a 400 V three-phase line

   1. Real power 15,000 W on a 400 V three-phase supply, power factor 0.85
   2. The denominator is √3 × 400 × 0.85 = 1.7320508 × 400 × 0.85 = 588.90
   3. Current is 15,000 / 588.90 = 25.471 A
   4. Apparent power is √3 × 400 × 25.471 = 17,647.1 VA

   The false answer is close enough to be dangerous. Take the square root of three out and the same load reads 44.118 A — which is exactly what this machine would draw if it were single-phase, and nothing about the number looks wrong if you did not know which supply you were on. A technician sizing from 44 A fits a larger breaker than the circuit needs, and a breaker that is too large is a protective device that no longer protects. This is why the supply type is a required field and not a default.

3. A 24 W LED strip on a 12 V battery

   1. Direct current, so neither the square root of three nor the power factor applies
   2. Current is simply 24 / 12 = 2 A
   3. In milliamps that is 2000 mA, which is the row to read for low-voltage work
   4. Apparent power equals real power, 12 × 2 = 24 VA, because there is no phase angle

   On DC the two power rows collapse into one number, and that is not a coincidence: apparent power is larger than real power only because of the phase shift between voltage and current, and there is none here. The power factor field makes no difference on this supply, and the page says so rather than making you guess — entering 0.8 here returns the same 2 A as entering 1, and a calculation that silently ignored an input you can see would be worse than one that explains it.

4. A resistive load: 2300 W at a power factor of 1

   1. A heater or an incandescent lamp: current in step with voltage, power factor 1
   2. The denominator is 230 × 1 = 230, so the expression reduces to 2300 / 230
   3. Current is 10 A, and the milliamps row is 10,000
   4. Apparent power is 230 × 10 = 2300 VA, the same number as the watts

   Power factor 1 is the ceiling, not a typical value, and it is worth seeing what it buys: the same 2300 W at a power factor of 0.9 would draw 11.111 A instead of 10. Resistive heating is the common case where the two numbers coincide, which is why a kettle or a heater can be rated in watts and amps interchangeably while a motor cannot. The field will not accept anything above 1, because a power factor greater than one is not a precision problem, it is a definition that does not exist.

Power factor is taken as a fixed number you supply, but a real motor's power factor falls as its load falls, so a motor running at half load draws more current per watt than its nameplate figure suggests. On DC the power factor field is ignored entirely, because direct current has no phase angle for it to describe. Three-phase circuits are assumed balanced, with equal currents in all three phases; an unbalanced or single-phasing supply needs a different treatment. Starting currents, which for motors run several times the running current for a fraction of a second, are not modelled at all. Nothing here accounts for the ambient temperature or the installation method, both of which decide what a cable can actually carry.

Is the current this returns the real current or the apparent one?

It is the line current itself, the one an ammeter would read and the one a breaker has to carry. The input is the real power, in watts, and the power factor is a separate field on the same screen, so the division by the power factor is already in the expression. If you entered apparent power instead — the volt-amperes from a nameplate — you would get the real current of a load with a power factor of 1, which is a different and smaller number. The result panel shows both the current and the apparent power so the two can be checked against each other.

Why does three-phase change the answer so much?

Because in a balanced three-phase circuit the three phases share a return path and the power is the square root of three times the line voltage times the line current, not three times the phase voltage times the phase current. Dividing by 1.732 rather than 1 makes the current 42 per cent smaller for the same power. Leaving the factor out is the most expensive mistake this page can make: the current comes out 73 per cent too large, the number looks entirely reasonable, and a breaker chosen from it is oversized — which means it stops protecting the circuit.

What happens to the power factor on DC?

It is ignored, and the page says so rather than pretending otherwise. Power factor measures the phase shift between voltage and current, and direct current has no phase shift, so the current is simply power divided by voltage. Entering 0.8 or 1 with the supply set to DC returns the same answer, and apparent power comes out equal to real power. The field stays visible because a control that appears on one supply and vanishes on another is harder to trust than one that is documented as inactive.

Do I enter the phase voltage or the line voltage on a three-phase supply?

The line-to-line voltage: 400 V on a European or Chinese industrial supply, 480 V in North America. That is the voltage between two phases, not the 230 V between a phase and neutral. The square root of three in the expression already carries the relationship between the two, so entering the phase voltage as well counts it twice and the current comes out about 1.7 times too large. If the only figure you have is the phase voltage, multiply it by the square root of three first.

Why is the current higher than watts divided by volts?

Because of the power factor. Real power is volts times amps times the power factor, so amps is watts divided by the product of volts and power factor; at 0.9 the current is 11 per cent higher than the naive division, and at 0.5 it is twice as high. Motors, transformers, fluorescent ballasts and switch-mode supplies all have a power factor below 1, which is why industrial installations are billed for apparent power and why a factory may be penalised for a low one.

Can I use this for a motor's starting current?

No. An induction motor draws several times its running current for a fraction of a second while it accelerates, and that inrush is what trips breakers and dims lights. It depends on the motor design and the load it is turning, and it is not a function of the steady-state numbers on this page. Size the running current here for the cable, and check the starting current against the breaker's trip curve separately — the two questions have different answers and different sources.

Does the cable have to carry this current?

Yes, and the cable is the next page along: this figure is the current input to the wire size calculator, which turns it into the smallest conductor that keeps the voltage drop within a limit you set. The current here does not depend on the cable at all — it is set by the load and the supply — but what the cable can carry continuously depends on its insulation, its cross-section and how it is installed, which is why the two questions are asked separately.

• AC power — the relationship between real power, apparent power and power factor, and why the line current is larger than the watts alone imply — Wikipedia
• Three-phase electric power — the line-to-line and line-to-neutral voltages, and where the square root of three comes from in a balanced system — Wikipedia
• OpenStax College Physics 2e, section 20.5, Alternating Current versus Direct Current — the comparison between the two supply types used on this page, including the domestic supply voltages — OpenStax