How this instrument works
Apparent power in kilovolt-amperes is nothing more than voltage multiplied by current, S = V·I, restated in a unit that stays indifferent to how much of that flow does useful work. Transformers, generators, UPS units, and switchgear get rated in kVA rather than kW precisely because the windings and contacts inside them heat up according to how much they carry — the I²R loss — whatever the phase relationship between that flow and the voltage happens to be. This instrument simply reverses the defining relation: multiply the kVA figure by 1000 to reach volt-amperes, then divide by the supply voltage to recover the reading, I = kVA × 1000 ⁄ V.
Because kVA already equals V·I by definition, no power-factor term enters this arithmetic at all — unlike a real-power figure in kW, which needs dividing by PF to reach an amperage, a kVA nameplate rating converts directly. That is exactly why someone sizing the output breaker or transfer switch for a backup generator reads the draw straight off that plate rather than working back through an assumed power factor: the manufacturer already built the carrying limit into that one number. The formula as written is single-phase — there is no √3 term — so it fits a single-phase generator, transformer secondary, or UPS output directly; a three-phase nameplate instead needs I = kVA × 1000 ⁄ (√3 × V_line), and reaching for the single-phase version there overstates the true per-leg figure by a factor of about 1.73.
The number this returns is a steady-state, full-load figure, not a starting or inrush current. A generator or transformer's nameplate kVA describes what it can sustain continuously; energizing an unloaded transformer can briefly draw five to ten times that figure while the core magnetizes, and a motor started directly across a generator's output can pull six or seven times its running current for a second or two. Breaker and conductor sizing built from this number still wants the usual code margins on top, and for motor loads a separate look at locked-rotor current — this arithmetic gives the nameplate rating, not the transient it survives.
- Enter the nameplate figure into Apparent power, kVA — the rating stamped on the transformer, generator, or UPS you're sizing for.
- Set Voltage to the equipment's rated single-phase output, typically 120, 230, or 240 V for portable and standby generators.
- Read Current — the full-load amperage that kVA rating is built to sustain continuously, before any code safety margin is added.
- For a three-phase nameplate, divide the Current reading by √3 (1.732) to get the true per-line draw — this formula is single-phase only.
Worked example — sizing a 10 kVA standby generator's output breaker
A 10-kilovolt-ampere standby generator carries a single-phase 230 V rating on its data plate. Feed 10 into Apparent power, kVA and 230 into Voltage: Current returns 43.478261, reading out as 43.48 A — from I = 10 × 1000 ⁄ 230 = 43.4782608696. That is the full-load draw the generator's windings, its output breaker, and the transfer switch feeding the panel all have to sustain continuously, and it is exactly the figure an installer pulls off the nameplate rather than measuring from a live load.
Scale the same generator up to 20 kilovolt-amperes at the same 230 V and the draw doubles in step, to 86.96 A — apparent power and amperage are directly proportional at fixed voltage, since V is the only thing standing between them in the formula. Apply the same arithmetic to a three-phase 10-kilovolt-ampere unit, though, and it goes wrong: three-phase equipment splits that apparent power across three conductors, so the correct figure is 10 × 1000 ⁄ (√3 × 230) = 25.1 A, not 43.48 A — a mismatch of nearly 73 percent that one missed √3 symbol will hand an installer.
Questions
Why is equipment rated in kVA instead of kW?
Because kVA is apparent power, S = V×I, and that is exactly the quantity that determines how much a winding or contact must carry — the thing that actually produces I²R heating regardless of power factor. A kW rating would additionally depend on the power factor of whatever load eventually gets connected, a number the manufacturer cannot know in advance; the apparent-power rating sidesteps that by sizing equipment on carrying capacity alone.
Does this formula work for three-phase equipment?
Not directly — it has no √3 term, so it is single-phase only. A three-phase nameplate needs I = kVA × 1000 ⁄ (√3 × V_line); applying the single-phase formula there overstates the per-line draw by a factor of √3, about 73 percent too high. Divide this instrument's result by 1.732 to get the equivalent three-phase per-line figure at the same line voltage.
Do I need to know the power factor to use this?
No, and that is the point of working from kVA rather than kW. Apparent power already equals voltage times amperage by definition, so I = kVA × 1000 ⁄ V recovers the figure with no PF term at all. Power factor only re-enters if you are asking how much real work, in kW, that flow can deliver — a different question than this instrument answers.
Is this the generator's continuous rating or its surge capacity?
Continuous. The kVA figure on a generator or transformer nameplate is what it is built to sustain indefinitely at rated voltage and temperature, and that is what this formula converts into amperes. Starting a large motor or an unloaded transformer's inrush can briefly demand several times that current; size wiring and breakers from the nameplate figure, then check the load's starting behavior separately.
How is this different from a watts-to-amps calculation?
Watts measure real power and need dividing by power factor to reach current: I = W ⁄ (V × PF). Apparent power already includes that division, since kVA equals V × I by definition, so no PF term appears here. Use this calculator from an equipment's kVA nameplate rating; reach for a watts-based formula instead when working from a real-power figure and a known power factor.
Why does entering 0 kVA give 0 A?
Because the reading is directly proportional to apparent power at a fixed voltage — set kVA to zero and there is no apparent power left for the voltage to push through anything, so the formula returns exactly zero no matter what voltage is entered. It doubles as a sanity check: halve the kVA figure and the answer halves too, since voltage sits alone in the denominator.