Kilowatts to Volts Calculator
The junction box has no label. Someone stripped the tag years ago, or it was never printed in the first place, but the load itself isn't a mystery: a spec plate deeper in the equipment reads 4.8 kW, and a clamp meter on the feeder reads 20 A. That's enough to work backward to the one number missing from the box: the voltage the circuit is actually running at.
Solved voltage
240.0V
V = (4.8 × 1000) / (20 × 1) = 240.0 V
Why kilowatts alone can't give you the voltage
Converting kilowatts to horsepower needs nothing else, because both units describe the same quantity, real power, just scaled differently; divide by 0.7457 and the answer falls straight out. Voltage doesn't work that way. Power is the product of voltage and current, so one wattage figure by itself matches an unlimited number of voltage-and-current pairs that all multiply to the same result. A 4.8 kW load could be 240 V at 20 A, 480 V at 10 A, or 48 V at 100 A, the kilowatt reading alone cannot distinguish between any of them.
Solving for the missing voltage needs a second known quantity to pin down which of those pairs is real, and current is usually the one already sitting on hand: read off a clamp meter, a breaker's rated draw, or a feeder schedule. That's why the calculator above asks for three known values (kilowatts, current, and power factor) rather than the two a straight unit conversion would need.
That third value is what separates this page from a plain unit conversion. Converting kW to horsepower or BTU is a fixed-ratio rescale: the same number always maps to the same answer. Solving for voltage is closer to solving a small system of equations: kilowatts and current each constrain the answer, and only together do they leave exactly one voltage that satisfies both. Drop either one and the voltage becomes indeterminate again, no matter how precisely the other is known.
Why voltage needs a third known value
Watt's Law states power equals voltage times current: P = V × I. For an AC circuit, only the power-factor fraction of the current is doing real work, so P = V × I × PF, and solving that for voltage gives V = P / (I × PF). Swapping in kilowatts for watts adds the ×1000 conversion factor.
Three-phase circuits divide by √3 × A × PF (line-to-line) or 3 × A × PF (line-to-neutral) instead of just A × PF, for the same reason the kW-to-amps formula adds that factor in the other direction. Picking the wrong circuit type is enough to turn a correct-looking calculation into a wrong voltage, so it's worth confirming the phase configuration before trusting the result.
A DC circuit is the simplest case to check first, since power factor drops out of the equation entirely: V = (kW × 1000) / A, with nothing left to assume. Anything running on AC needs a real power-factor figure, ideally read from a nameplate rather than guessed, since a wrong PF assumption is one of the most common ways this calculation drifts off the circuit's real voltage.
Worked example: the unlabeled feeder
4.8 kW load per the spec plate, 20 A measured on the feeder, power factor 1.0 (resistive load).
- 1
Convert kW to watts and apply the formula
V = (4.8 × 1000) / (20 × 1.0)
- 2
Solve
V = 4800 / 20 = 240 V
- 3
Compare against a standard rail
240 V → matches 240 V single-phase
This is general reference arithmetic for diagnosing a figure that's missing on paper, not a substitute for a direct meter reading. A licensed electrician should confirm any voltage on an energized circuit before it's treated as verified.
Known kW and A pairs, solved for voltage
Single-phase, power factor 1.0, each row shows a kilowatt-and-amp pair landing on a standard voltage rail once solved with V = (kW × 1000) / (A × PF).
| Power | Current | Solved voltage |
|---|---|---|
| 1.2 kW | 10 A | 120 V |
| 3.12 kW | 15 A | 208 V |
| 4.8 kW | 20 A | 240 V |
| 4 kW | 10 A | 400 V |
| 6.925 kW | 25 A | 277 V |
| 14.4 kW | 30 A | 480 V |
Reference estimates only, not measured values. A real feeder's power factor and meter tolerance can shift a calculated voltage a little off these clean numbers.
Solving for the other circuit values
Questions
Kilowatts to volts FAQ
Diagnostic questions that come up when a voltage has to be worked out rather than read off a label.
Can voltage be calculated from kilowatts alone, without a current reading?
No. A single kilowatt figure matches infinitely many voltage-and-current pairs that all multiply to the same power, so there is no unique voltage to solve for until a second measured quantity narrows it down. 4.8 kW could be 240 V at 20 A, 480 V at 10 A, or 48 V at 100 A. The wattage alone cannot tell those apart. A measured current is what pins down which pair actually describes the circuit.
What if the measured current includes a reactive component I can't isolate?
Assuming a power factor of 1.0 for a load that isn't purely resistive will pull the calculated voltage lower than the circuit's real voltage. For example, a 240 V circuit drawing 20 A at an actual power factor of 0.8 delivers 3.84 kW of real power. Feed 3.84 kW and 20 A into the formula assuming PF = 1.0 and it solves to 192 V, 48 V short of the true 240 V. A nameplate power-factor value, not an assumed one, is what keeps a reverse-calculated voltage accurate on anything with a motor or a switching supply.
When would I calculate voltage instead of just reading it off a multimeter?
A meter across the terminals is always the more direct answer when the terminals are safely accessible. Calculating is the fallback for when they are not. A sealed enclosure, a de-energized panel being scoped from a spec sheet, or a remote piece of equipment with only a wattage rating and a clamp-meter current logged from its feeder are all cases where working backward from kW and A is the only option on hand. It's also a useful cross-check against a meter reading that looks questionable.
Should the calculated voltage match a standard value like 120, 208 or 240 V exactly?
It should land close to a standard rail, but an exact match is not guaranteed. A few volts of drift is normal and does not necessarily mean the inputs are wrong. Rounded nameplate kW, a clamp-meter reading with its own tolerance, an assumed rather than measured power factor, and ordinary voltage sag under load can all shift the result a little off a clean number. A result that lands 15-20 V or more off any standard rail is worth rechecking, rather than one that is off by a volt or two.
Does selecting single-phase vs. three-phase change the answer for the same kW and A?
Yes, the current-type selection changes the multiplier in the denominator, so the same 4.8 kW and 20 A numbers solve to a different voltage depending on which one is picked. Single-phase gives 240 V, as in the example above, but three-phase line-to-line at the same kW and A solves to roughly 138.6 V instead, because the formula divides by √3 × A × PF rather than just A × PF. Picking the wrong circuit type is one of the fastest ways to get a plausible-looking but wrong answer.
What is the most common reason a reverse-calculated voltage comes out clearly wrong?
Usually one of the three inputs was actually something else: an apparent-power (kVA) reading used in place of real power (kW), a current logged on the wrong leg or phase, or an assumed power factor that was never confirmed. Because the formula has no way to sanity-check its own inputs. A swapped or misread figure still produces a confident-looking number; it just happens to be wrong. Cross-checking against a nameplate, a spec sheet, or a second measurement is the only real safeguard.