Power, current and voltage calculator
Estimates for planning and study. Fixed wiring must be designed and installed to the wiring regulations that apply where you are, by someone competent to do it. The 125% rows follow the NEC continuous-load rule and apply in the United States; BS 7671 and IEC 60364 size the device against the cable instead.
Power is volts times amps. On AC, real power is V × I × pf single phase and √3 × V × I × pf three phase, so a three-phase 400 V supply at 16 A with a 0.9 power factor delivers 9.98 kW real out of 11.09 kVA apparent. Give any two of volts, amps, ohms and watts to get the rest.
How to calculate electrical power
The gap between watts and volt-amps is the single most misunderstood thing in electrical sizing. Watts is the power actually doing work; volt-amps is the product of the voltage and the current the supply has to deliver. For a resistive load they are equal. For a motor, current lags voltage and the two diverge by the power factor, so a 10 kW motor at 0.8 might pull 12.5 kVA. Cables, breakers and generators all have to be sized for the apparent power, because the current is real whether it is doing useful work or not.
Going the other way, from a load rating to a current, is where domestic limits become concrete. A 230 V circuit protected at 16 A tops out near 3.7 kW; a 120 V circuit at 15 A stops at 1.8 kW. That is why showers, hobs and EV chargers need their own circuits rather than a socket, and why a 7.4 kW charger is 32 A on single phase but only 10.7 A per line at 400 V three phase. The 125% row is the NEC’s answer to a load that runs three hours or more, which would otherwise sit at the protective device’s thermal limit with nothing in reserve; BS 7671 and IEC 60364 reach the same safety a different way, by requiring the design current, the device rating and the cable’s derated capacity to fall in that order, with no uplift applied to the current itself. Either way a motor’s inrush of six or seven times its running current for a second or two is a separate matter, handled by the device’s trip curve rather than by its rating.
At 12 and 24 volts the same arithmetic turns thick. A 500 W inverter load is 41.7 A at 12 V against 2.2 A at 230 V, so vehicle and van wiring is far heavier than mains wiring for the same power, and doubling the system voltage to 24 V halves the current and quarters the loss in the cable. It is also why the fuse belongs close to the battery: the fuse protects the cable, and unprotected cable between battery and fuse is the dangerous part.
Ohm’s law itself is one equation, V = I × R, with P = V × I as a separate one, and between them any two of the four quantities determine the other two — the twelve rearrangements of the familiar wheel. The law describes ohmic components, which is a real restriction. A resistor is ohmic: double the voltage and the current doubles. A filament lamp is not, because its resistance rises sharply as it heats, which is why bulbs almost always fail at the moment you switch them on. Diodes, LEDs, motors and anything semiconductor are non-ohmic too, so the law gives a number that is right only at one operating point.
What people use it for
- Sizing a generator or UPS in kVA
- Converting a motor nameplate figure into supply current
- Checking a circuit will carry an appliance
- Converting between kW and horsepower
- Sizing a fuse or breaker for a new load
- Working out the supply voltage a component needs
- Sizing a supply for three-phase machinery
- Checking a heater against a household circuit
- Converting an appliance wattage into current draw
- Turning a clamp meter reading into watts
- Sizing a circuit for an EV charger or heat pump
- Finding the voltage across a resistor at a known dissipation
- Fusing a 12 V accessory in a van or car
- Comparing a 12 V and a 24 V system design
- Checking headroom on a 230 V socket circuit
- Solving an Ohm’s law problem from any two values
- Finding a resistor value and the rating it needs
- Checking total demand against the main supply fuse
- Comparing the 120 V and 230 V versions of the same appliance
- Working out the DC current an inverter pulls from a 12 V or 24 V bank
- Comparing a single-phase and a three-phase supply for the same machine
- Checking a motor nameplate against a clamp meter reading
Questions
P = V × I. DC stops there, because direct current has no power factor. AC adds one: P = V × I × pf on single phase, and P = √3 × V × I × pf on three phase.
Divide watts by volts. On AC with a power factor, divide by volts times the power factor as well. Amperage is the older word for the current you get.
12.5 A at 120 V, 6.5 A at 230 V, or 125 A at 12 V. Voltage decides it, so there is no fixed conversion.
1,840 W at 230 V or 960 W at 120 V, assuming a power factor of 1.
kW is real power doing work; kVA is apparent power the supply must deliver. They are equal only when the power factor is 1.
Use 1 for heating, incandescent lighting and resistive loads. Motors are typically 0.8 to 0.9; check the nameplate.
Divide by 745.7 for mechanical horsepower. A 3 kW motor is about 4 hp of output, before efficiency losses.
Because line-to-line voltage and phase current are not in phase with each other. The √3 falls out of the vector sum of three balanced phases.
In a star connection, yes. In delta the line current is √3 times the winding current, which is what catches people out on a motor nameplate: the current in the winding and the current in the supply lead are two different numbers.
32 A at 230 V single phase, or 10.7 A per line at 400 V three phase, at unity power factor.
It is American. Under the NEC a load running three hours or more is continuous, and the branch circuit and its protective device must be rated at least 125% of that load current, so a 32 A continuous draw needs a 40 A circuit. There is no equivalent uplift under BS 7671 or IEC 60364: those ask only that the design current, the device rating and the cable’s current-carrying capacity fall in that order, with the derating done on the cable side. Read the 125% row if you are wiring to the NEC and ignore it if you are not.
Nominally 230 V across most of Europe, at a tolerance of ±10%, so anything from 207 to 253 V is a compliant supply. Harmonising on 230 V was what let the old 220 V and 240 V countries keep their transformers and their equipment, and the width of the band is why a real measurement wanders: a house close to its substation reads high, one at the end of a long rural feed reads low, and both are in specification. Use 230 V for planning and measure if the margin is tight.
No. It is more dangerous per contact, because the current through a body is higher at the same skin resistance. It needs half the current for the same power, though, which means thinner conductors and less heat in the wiring, so the fire risk goes the other way. That trade is the whole argument between the two standards.
A stationary motor looks almost like a short circuit until it spins up. Inrush of six to seven times running current for a second or two is normal.
About 2,990 W on a 13 A plug at 230 V, which is where the informal 3 kW ceiling for a plug-in appliance comes from, and 3,680 W on a 16 A circuit. Keep anything that runs for hours below roughly 80% of either.
8.33 A. At 24 V the same load is 4.17 A, and at 230 V only 0.43 A.
Half the current for the same power, and a quarter of the cable loss. Cable gets thinner and cheaper.
Only approximately. A filament’s resistance rises sharply when hot, which is why bulbs fail at switch-on: the cold inrush current is large.
No. Power is volts times amps, so a second quantity is needed: V = P ÷ I with a current, or V = √(P × R) with a resistance. That is the rule for the whole page. Any two of voltage, current, resistance and power fix the other two, and in the Ohm’s law mode you leave the two you do not know empty.
Yes. Voltage is always a difference between two points; a single point has no voltage of its own without a reference to measure it against, which is what a ground or a battery negative provides.
Yes, by exactly I × R. In a series chain the individual drops always add back up to the supply voltage, which is the quickest way to check a divider by hand.
Almost always a unit slip: milliamps entered as amps, or kilohms as ohms. Most of the modes here work a voltage against a current, and P = V × I carries the error straight through: enter 500 mA as 500 A and the watts land a thousand times out. It compounds only where the current or the voltage gets squared, which is the Ohm’s law branch that has to reach the answer through a resistance — P = I² × R or V² ÷ R — where the same thousandfold slip lands a million times out. Either way the wattage is the quickest check on the page, because a household number and an absurd one are easy to tell apart.
For a resistive AC load, yes, using RMS values, which is what a multimeter and a nameplate both quote. Anything with significant inductance or capacitance needs impedance rather than resistance, and that is what the power factor field stands in for here.