How this instrument works
Apparent power is the product of the RMS voltage across a circuit and the RMS current flowing through it, scaled down by 1000 to land in the kilo unit written on nameplates: S = V × I ⁄ 1000. Multiplying the two this way ignores the timing relationship between the voltage and current waveforms entirely — it treats every volt and every amp as pulling together, which is exactly the load a generator winding or transformer coil has to physically endure no matter what the attached equipment does with that current afterward.
The base unit here is the volt-ampere, VA, deliberately not the watt, even though volts times amps looks identical to a watt calculation on paper. The distinction matters: watts measure real power, the portion of current that actually performs work once its phase lag against the voltage is accounted for, while volt-amps measure the total current the supply conductors must be sized to carry regardless of that lag. Dividing by 1000 turns an ugly figure like 4600 VA into the 4.6 kVA a switchboard label actually shows.
This particular formula covers a single-phase circuit — one live conductor, one voltage, one current. A three-phase supply needs a factor of root three worked into the product, because its three line currents sit 120 degrees apart rather than adding directly, so plugging single-phase readings into a three-phase nameplate rating understates the true figure by a wide margin. It is also worth remembering that S sets a ceiling, not a promise: a 4.6 kVA machine can carry up to that much apparent demand, but how much becomes usable kilowatts still depends on the power factor of whatever gets plugged in.
- Enter the circuit's RMS voltage into the Voltage field — 230 V for a typical single-phase supply, or the line voltage measured at the board.
- Enter the RMS current the load draws into the Current field, read from a clamp meter or taken from the equipment's nameplate amperage.
- Read the result in the Apparent power, kVA field — the figure to match against a generator, UPS, or transformer's kVA rating, not its kW rating.
- To find real power instead, multiply the kVA reading by the load's power factor (cos φ) — this instrument reports apparent power only and assumes no power factor for you.
Worked example — sizing a generator for a 230 V, 20 A board
A site electrician is choosing a generator for a temporary distribution board on a single-phase 230 V supply. Clamping the feeder while the lighting towers and power tools run shows a steady 20 A draw. Apparent power works out to S = 230 × 20 ⁄ 1000 = 4.6 kVA, meaning the board's supply conductors, and whatever generator feeds them, must be rated to carry at least 4.6 kVA continuously.
That 4.6 kVA is the figure to check against the generator's nameplate, not against a kW rating. If the tools on the board run at a typical induction-motor power factor of 0.8, the real power consumed is only 4.6 × 0.8 = 3.68 kW — a smaller number describing fuel burn, not conductor heating. A generator chosen by matching 3.68 kW instead of 4.6 kVA would run its windings well past their rated current, which is precisely why kVA, not kW, is the number stamped on the machine.
Questions
Why are generators and transformers rated in kVA rather than kW?
Because the heat a winding produces depends on the current passing through it, not on how much of that current turns into useful work. A machine rated 4.6 kVA must survive 20 A at 230 V continuously regardless of the attached load's power factor, so manufacturers state the current-carrying limit — kVA — rather than a real-power figure that would change with every different load plugged in.
What is the difference between kVA and kW?
kVA is apparent power, the plain product of RMS voltage and RMS current; kW is real power, the portion that actually does work once the phase lag between voltage and current is accounted for. The two are equal only when the power factor is exactly 1, as with a purely resistive load such as a heating element. Any motor, transformer, or ballast pulls current at least slightly out of step with the voltage, so its kW figure sits below its kVA figure.
How do I convert kVA to kW?
Multiply the kVA figure by the load's power factor: kW = kVA × cos φ. A 4.6 kVA reading at a power factor of 0.8 gives 3.68 kW of real power; at a power factor of 1.0 the two figures match exactly. Power factor is not something this formula supplies on its own — it depends on the specific equipment connected, typically 0.8 to 0.9 for induction motors and close to 1.0 for resistive heaters.
Does this formula apply to three-phase circuits?
No. This calculator's S = V × I ⁄ 1000 is the single-phase form, with one voltage and one current. A balanced three-phase circuit needs S = root three × line voltage × line current ⁄ 1000 instead, since the three phase currents sit 120 degrees apart rather than combining directly. Using the single-phase formula on three-phase readings understates the true apparent power by roughly 42 percent.
Why does apparent power multiply volts by amps instead of using real power's formula?
Because apparent power deliberately ignores timing. Real power multiplies the instantaneous voltage and current together and averages the result, which cancels out any portion where the two waveforms disagree in direction; apparent power instead multiplies the RMS magnitudes only, keeping that cancelled portion in the total. The gap between the two figures is the reactive power that transformer and motor windings still have to carry even though it delivers no work.
Can I enter current in milliamps or voltage in kilovolts?
Yes. The Voltage field accepts mV, V, or kV, and the Current field accepts mA or A; the instrument converts everything to volts and amps before dividing by 1000. Enter 230 V and 20 A, or the equivalent 0.23 kV and 20000 mA, and the readout is the same 4.6 kVA either way.