Volts to Watts Calculator
Watts from volts and amps — for battery, automotive 12 V, and solar-string systems where voltage is known.
Updated
You need
1,200 Wpower
120 V × 10 A
- Power
- 1,200 W
- As kilowatts
- 1.20 kW
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In short
How many watts is 12 volts at 5 amps?
Watts = volts × amps, so 12 V at 5 A is 60 W. The same product works at any system voltage: the tool default of 120 V at 10 A is 1,200 W or 1.20 kW, a 48 V solar string at 8 A is 384 W, and a 24 V trolling motor pulling 30 A is 720 W.
The product is exact for direct current and resistive loads; an alternating-current motor delivers less real power than volts × amps because its power factor is below 1.
How to use the volts to watts calculator
Enter a voltage and a current and you get power in watts, along with the same figure expressed in kilowatts. The tool is framed for the case where the voltage is the thing you already know: a 12 V automotive circuit, a 24 V truck or marine bus, a 48 V solar string, a battery bank at its nominal pack voltage.
In that world you begin from the system voltage, measure or look up the current, and want the wattage those two imply — which is exactly this direction. Starting from a known direct-current bus is a genuinely different way in.
Starting from a breaker rating instead?
It is the same relationship the amps-to-watts tool uses, but that one is written around a household breaker rating rather than a known direct-current bus.
Open the amps to watts calculator →1,200 W
120 V × 10 A
the tool default, 1.20 kW
60 W
12 V × 5 A
an ordinary automotive load
384 W
48 V × 8 A
a solar string at work
The arithmetic is watts = volts × amps, and for direct current it is exact with no caveats at all. Batteries, solar strings, automotive circuits, and USB power delivery are all direct current, so the product is the true power in every one of them. Alternating current is where the honest version needs a fourth term.
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Real power there is volts × amps × the power factor, and the power factor is the fraction of the delivered volt-amps that becomes useful work. Resistive loads such as heaters, kettles, and incandescent bulbs sit at approximately 1, so the product still holds. Motors, compressors, and switching supplies sit below 1, and for those the volt-amps you compute here are apparent power rather than real watts.
The reason this direction matters so much in direct-current work is that volts and amps trade off against each other for any fixed power. The same 1,200 W is 100 A at 12 V, 50 A at 24 V, 25 A at 48 V, 10 A at 120 V, and 5 A at 240 V.
Current is what forces conductor size and creates resistive loss, and that loss climbs with the square of the current, so quadrupling the bus voltage from 12 V to 48 V cuts the current to a quarter and the heating loss in the cable to a sixteenth. That single fact explains why solar strings, electric-vehicle packs, telecom plants, and long direct-current runs all reach for higher voltages rather than fatter cable.
Practical uses cluster around sizing. A 12 V accessory socket fused at 15 A can supply 180 W and no more, which is why a kettle in a car is a fantasy. A 12 V recovery winch pulling 40 A under load is drawing 480 W from the battery.
A 48 V string at 8 A is producing 384 W, and if you are budgeting an inverter you want that number in kilowatts, which the second output row gives you. Use the actual operating voltage and current rather than nameplate maximums where you can, because a solar string almost never sits at its rated point. And for anything permanently installed, treat these figures as a planning estimate rather than a substitute for a licensed electrician’s sizing.
Do
- Enter the nominal voltage of the system: 12 V automotive, 24 V marine, 48 V solar.
- Take the current from a clamp meter, an inline shunt, or the nameplate.
- Use the real operating figures, since a solar string rarely sits at its rated point.
- Hold the wattage against what the battery, inverter or panel string can supply.
Don't
- Expect volts alone or amps alone to give you a power figure.
- Call the product real watts on a motor, where it is apparent power in volt-amps.
- Assume more voltage means more watts unless the current stays where it is.
- Substitute this arithmetic for a licensed sizing on anything permanently installed.
Real systems where the voltage is the thing you know first, worked through volts × amps into watts and kilowatts. It shows how far apart the wattages land even when the currents look similar, which is the whole point of choosing a bus voltage.
| System | Voltage | Current | Power (volts × amps) | As kilowatts |
|---|---|---|---|---|
| USB-C power delivery, 9 V profile | 9 V | 2 A | 18 W | 0.02 kW |
| USB-C power delivery, 20 V profile | 20 V | 5 A | 100 W | 0.10 kW |
| Car accessory socket on a 15 A fuse | 12 V | 15 A | 180 W | 0.18 kW |
| 12 V LED light bar | 12 V | 4 A | 48 W | 0.05 kW |
| 12 V compressor cool box | 12 V | 4.5 A | 54 W | 0.05 kW |
| 12 V recovery winch under load | 12 V | 40 A | 480 W | 0.48 kW |
| 24 V trolling motor | 24 V | 30 A | 720 W | 0.72 kW |
| 48 V solar string | 48 V | 8 A | 384 W | 0.38 kW |
| 48 V electric bicycle controller | 48 V | 15 A | 720 W | 0.72 kW |
| North American outlet, the tool default | 120 V | 10 A | 1,200 W | 1.20 kW |
| North American 15 A branch circuit, full rating | 120 V | 15 A | 1,800 W | 1.80 kW |
| United Kingdom plug top at its fuse limit | 230 V | 13 A | 2,990 W | 2.99 kW |
| European 16 A socket circuit | 230 V | 16 A | 3,680 W | 3.68 kW |
| North American 240 V dryer circuit | 240 V | 30 A | 7,200 W | 7.20 kW |
Volts and amps trade off for a fixed power
The trade-off worth internalising is this: for a fixed power, current falls in inverse proportion to voltage. The same 1,200 W is 100 A at 12 V but only 25 A at 48 V and 5 A at 240 V.
Since resistive loss in a cable rises with the square of the current, moving a 12 V design to 48 V cuts the current to a quarter and the wasted heat to a sixteenth for the same conductor. That is why high-voltage direct-current buses exist, and it is also why the answer here is only ever a planning number: what conductor, fuse, or breaker a real installation needs is a licensed electrician’s determination, not a calculator’s.
Read it: Every bar carries identical power; the 12 V bar is what forces fat cables, big fuses and short runs, and each doubling of voltage halves it.
Computed as amps = 1,200 W ÷ volts, the same relationship the tool rearranges.
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The formula, worked line by line
One relationship underpins everything here: P = V × I, power equals voltage multiplied by current. Voltage is the electrical pressure pushing charge along, current is the rate at which that charge flows, and their product is the rate at which energy is delivered — which is what a watt measures.
Neither quantity gives you power on its own, which is why a battery labelled only in volts and a meter reading only in amps each tell you half the story.
For direct current the product is the whole truth. For alternating current, real power is P = V × I × PF, where the power factor PF measures how much of the delivered volt-amps actually does work. At PF of approximately 1, which is where resistive loads sit, the product is unchanged.
Below that, the volts × amps figure is apparent power, correctly expressed in volt-amps rather than watts, and it is what generators, inverters, and transformers are actually rated to supply. Balanced three-phase equipment carries a further factor: P = √3 × V × I × PF, using the line-to-line voltage.
watts = volts × amps
kilowatts = (volts × amps) ÷ 1000
watts = volts × amps × PF (single-phase AC, real loads)
watts = √3 × volts × amps × PF (balanced three-phase)- Supply voltage
- 120 V
- Current delivered
- × 10 A
- Power
- 1,200 W, or 1.20 kW
Drop to a 12 V automotive circuit at the same 10 A and the power falls by a factor of ten to 120 W, because the current is doing the same work against a tenth of the pressure. Push the other way to a 240 V circuit at 10 A and it doubles to 2,400 W.
A second set of examples from real direct-current systems. Each of these is a plain multiplication, exact because every one of those systems is direct current with no power factor to complicate it.
- 48 V solar string producing 8 A
- 48 × 8 = 384 W
- 24 V trolling motor pulling 30 A
- 720 W
- 12 V winch drawing 40 A under load
- 480 W
- 12 V accessory socket fused at 15 A
- 180 W
The fused socket tops out at 180 W no matter what you plug into it.
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Questions people ask
How many watts is 12 volts at 5 amps?
Sixty watts, because watts = volts × amps and 12 × 5 = 60. The same 12 V circuit at 10 A is 120 W, at 15 A it is 180 W, and a winch pulling 40 A is 480 W. Twelve-volt automotive and marine work is where this conversion comes up most often, because that is the world where you genuinely know the system voltage first and are measuring or looking up the current second, rather than the other way round.
What is the volts to watts formula?
Watts = volts × amps. You need both numbers: voltage alone or current alone cannot give you power, because power is the rate of energy transfer and that depends on how much pressure is pushing how much flow. The product is exact for direct current and for resistive alternating-current loads. For motors and switching supplies the real power is lower than the product by the power factor, so use the nameplate watts and treat volts × amps as apparent power in volt-amps.
Why is this a separate tool from amps to watts?
The arithmetic is identical — watts = volts × amps either way — but the starting point is not. The amps-to-watts tool is written around a household branch circuit, where you know the breaker rating and want the wattage budget it represents, complete with the 80 percent continuous margin. This one is written for known-voltage systems: 12 V automotive, 24 V marine, 48 V solar and battery banks, where you begin from the bus voltage and a measured current and there is no breaker rating in the picture at all.
Does it work for a solar string?
Yes, and solar is one of the cleanest cases, because it is direct current with no power factor to complicate anything. Multiply the string voltage by the string current: 48 V at 8 A is 384 W. The only caution is which numbers you feed it. Panels are rated at standard test conditions that real roofs rarely reproduce, so a string almost never sits at its nameplate maximum. Use the actual operating voltage and current from your inverter or charge controller for a figure that reflects what the array is really producing.
Does a higher voltage mean more watts?
Only if the current stays where it is. For a fixed amount of power the two move in inverse proportion: 1,200 W is 100 amps at 12 V, 25 amps at 48 V, 10 amps at 120 V, and 5 amps at 240 V. Because resistive loss in a conductor rises with the square of the current, cutting the current by four cuts the wasted heat by sixteen. That is the entire engineering argument for high-voltage direct-current buses in solar arrays, electric vehicles, and telecom plants.
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