Amp to Volt Calculator

Why amps alone cannot give you volts

Amps measure current: how much charge passes a point each second. Volts measure the push that drives it. Ten amps tells you the flow, not the pressure behind the flow, so the same 10 A can travel on a 12 V wire or on a 240 V wire. Turning one number into the other needs a third quantity, and there are exactly two candidates: watts or ohms.

Both routes come out of the same pair of equations. Power is volts times amps, so P = V x I, and resistance relates the other two, so V = I x R. Rearranged, the first gives volts as watts divided by amps, and the second gives volts as amps times ohms. The OpenStax College Physics chapter on Ohm's law derives both in the same order, and it is a good place to check the algebra.

The volt itself is defined as one watt per ampere, which is why the divide route is not a trick but the definition rearranged. NIST's introduction to the ampere describes the current side of that pair, and the two units only make sense together.

A single wattage spreads across widely different currents depending on the supply, which is the whole reason this conversion needs two inputs:

Every row is 1,200 W, and the current falls as the voltage rises. If someone quotes amps without naming the voltage or the wattage, the number cannot be converted, and any answer produced from it is a guess.

Route one: divide watts by amps

This is the route to use when a label gives power and current, which is common on appliance plates and charger bricks. Divide the watts by the amps and the answer is volts.

A 1,500 W space heater that draws 12.5 A runs on 1,500 / 12.5 = 120 V. An RV refrigerator pulling 5 A on 60 W sits on 60 / 5 = 12 V. A 65 W laptop charger rated at 3.34 A works out to 65 / 3.34 = 19.46 V, which is the 19.5 V figure printed on the brick, rounded.

One caution on alternating current. The watts on a nameplate are real power, and the product of volts and amps in an AC circuit is apparent power, measured in volt-amperes. On a heater the two are equal. On a motor they are not: a 120 V motor drawing 10 A pulls 1,200 VA, and at a power factor of 0.85 it delivers 1,200 x 0.85 = 1,020 W. Divide the VA by the amps and you get the voltage back, so the divide route is safest when the label states VA rather than W.

Route two: multiply amps by ohms

When the resistance is known instead of the power, volts are amps times ohms. This route shows up in bench work, in loudspeaker and sensor circuits, and anywhere a resistor sets the operating point.

A 0.5 A current through a 240 ohm resistor develops 0.5 x 240 = 120 V across it. Two amps through 6 ohms is 2 x 6 = 12 V. A 15 A load on a 0.8 ohm element sees 15 x 0.8 = 12 V, which is how a 12 V water heater element is specified.

The three forms of Ohm's law are worth keeping in one picture, because the mistake people make is dividing when they should multiply:

Resistance in these examples is the resistance while the device is running. A heater element reads lower when cold, and a tungsten lamp reads about a tenth of its hot resistance at room temperature, so a multimeter check on a cold bulb will not reproduce the running figure.

Worked example: a 12 V circuit in an RV

Low voltage wiring is where this conversion earns its keep, because the volts at the load are often not the volts at the battery. Take a 60 W refrigerator on a 12 V bank: 60 / 12 = 5 A, and that is the number printed in the manual.

Now run that 5 A through 30 ft of 16 AWG copper, a common size for a factory RV circuit. Copper of that gauge measures about 4.016 ohms per 1,000 ft, so 30 ft one way is 0.1205 ohms. The current has to come back, so the round trip is 0.241 ohms, and the drop is 5 x 0.241 = 1.205 V. The fridge sees 12 - 1.2 = 10.8 V while the battery still reads 12 V.

That drop is not academic. At 10.8 V the fridge needs 60 / 10.8 = 5.56 A to do the same work, so the loss feeds itself a little. Rewiring the same run in 10 AWG, at about 1 ohm per 1,000 ft, gives a round trip of 0.06 ohms, a drop of 5 x 0.06 = 0.3 V, and 11.7 V at the fridge.

Battery capacity follows the same arithmetic. A 100 Ah, 12 V battery holds 100 x 12 = 1,200 Wh of energy, so a 60 W load that runs continuously draws it down in 1,200 / 60 = 20 hours on paper, and closer to 10 hours in practice once you stop at 50 percent depth of discharge.

Worked example: USB-C chargers and laptops

USB Power Delivery is a good place to see the same conversion working in both directions, because the standard negotiates a voltage and the device draws whatever current its wattage needs at that voltage.

A 65 W charger running at 20 V delivers 65 / 20 = 3.25 A. The same 65 W at 15 V needs 65 / 15 = 4.33 A, and at 9 V it needs 65 / 9 = 7.22 A. That is why the cable matters more at low voltage: the current is higher for the same power, and thin cables heat up.

The 5 V column is the reason fast charging moved up the ladder. A standard USB port is limited to 3 A at 5 V, which caps it at 15 W, and no cable can push more through that limit. Raising the voltage is the only way to move more watts through the same wire.

Breakers, nameplates and the 80 percent rule

Once volts and amps are both known, watts follow, and that is how circuit capacity is checked. A 15 A breaker at 120 V holds 15 x 120 = 1,800 W before it trips. A 20 A breaker at 120 V holds 2,400 W, and a 20 A double pole breaker at 240 V holds 20 x 240 = 4,800 W.

The National Electrical Code treats a load that runs three hours or more as continuous and caps it at 80 percent of the breaker rating. On the 15 A circuit that cap is 12 A, or 12 x 120 = 1,440 W. On the 20 A circuit it is 16 A, or 1,920 W. This is the rule behind the instruction on most 1,500 W heaters to keep them on a circuit of their own: 1,500 / 120 = 12.5 A is over the 12 A continuous limit even though it is under the 15 A trip point.

Working backwards from a nameplate is the same arithmetic. A 1,500 W appliance at 120 V draws 1,500 / 120 = 12.5 A, which fits a 20 A circuit with room and sits right at the edge of a 15 A one. OSHA's electrical safety page covers the hazard side of overloaded circuits, and it is worth a read before adding a load to a circuit that already carries a heater.

Voltage drop: why the measured volts come out lower

Every calculation on this page assumes the voltage arrives intact. Wire has resistance, so some of the push is spent moving current through the copper, and the load sees less than the panel supplies. The drop for a two wire run is V_drop = 2 x I x R_per_ft x L, where the 2 accounts for the return path.

A 20 A load on 100 ft of 12 AWG, at 1.588 ohms per 1,000 ft, drops 2 x 20 x 0.1588 = 6.35 V, so a 120 V circuit delivers 113.65 V at the far end. That is a 5.3 percent drop, and the usual guidance is to keep branch circuits under 3 percent.

Car starting is the extreme case. A starter that pulls 150 A and drags a 12.6 V battery down to 10.5 V is working through 2.1 V of internal and cable resistance, which is 2.1 / 150 = 0.014 ohms. A corroded terminal adding 0.01 ohms raises the drop by 150 x 0.01 = 1.5 V, and the engine cranks slowly for a reason that has nothing to do with the battery's state of charge.

This is also why a voltage reading taken at an outlet and a reading taken at the appliance disagree. When a calculation and a meter differ by more than a couple of percent, measure at the load before assuming the arithmetic is wrong. Our voltage drop calculator runs the wire side of this in detail.

Mistakes that flip the answer

Converting amps to volts with no third number. Ten amps is 120 V through 12 ohms and 1,200 V at 120 W. Without watts or ohms there is no answer to give, and a calculator that produces one is inventing it.

Using peak volts instead of RMS. A 120 V wall circuit peaks at 120 x 1.414 = 169.7 V. Power calculations use the RMS value, so 12.5 A at 120 V is 1,500 W, while the same current at the peak value would come out at 2,121 W. The 120 V figure is already the RMS one, and that is the one to enter.

Mixing watts and volt-amperes. Dividing VA by amps gives volts correctly. Dividing watts by amps gives volts only when the power factor is 1.0. A 10 A motor at 120 V and PF 0.85 shows 1,020 W, and 1,020 / 10 = 102 V, which is a wrong voltage produced from right numbers.

Assuming a 240 V appliance draws half the amps of the same appliance at 120 V. That holds only when the wattage is the same. A 1,500 W heater is 12.5 A at 120 V and 6.25 A at 240 V, but a 3,000 W heater at 240 V draws 12.5 A, the same current on a higher supply.

Forgetting that series halves the volts. Two identical 120 V heating elements wired in series across 240 V each see 120 V, so each produces a quarter of its rated power, not half. The current is the same through both, and the voltage divides between them.