kVA Calculator — Apparent Power from Volts and Amps, kW, or a Nameplate Rating

Background.

kVA is the size of an electrical supply. kW is how much of it does work. The gap between them is why a 100 kW load will not run on a 100 kVA generator, and why the transformer feeding your workshop is rated in a unit most people never meet anywhere else. This calculator moves between the two in every direction, on single-phase and balanced three-phase supplies.

Apparent power, written S and measured in volt-amperes, is simply the product of the voltage and the current the supply has to deliver: S = V × I on a single-phase circuit, and S = √3 × V × I on a balanced three-phase one, where V is the line-to-line voltage. Nothing about the load's behaviour enters that product. Conductors heat up according to the current they carry and insulation is stressed by the voltage across it, so the volt-ampere product is what actually determines the size of every piece of equipment in the path — the cable, the switchgear, the transformer, the generator, the UPS. That is why they are all sold in kVA.

Real power, written P and measured in watts, is the part of that apparent power that leaves the circuit as motion, heat or light. On an AC supply the voltage and current do not necessarily peak at the same instant; the cosine of the angle between them is the power factor, and P = S × PF. A resistive load such as an immersion heater has a power factor of 1 and consumes every volt-ampere it draws. An induction motor at 0.8 lagging draws 25 percent more current than its wattage suggests, because a quarter of the apparent power is spent building and collapsing magnetic fields rather than turning the shaft. That circulating component is reactive power, Q, measured in kilovars, and it is the third side of a right triangle whose other two sides are P and S.

The practical consequences show up the moment you size something. Convert a kW load to kVA by dividing by the power factor, never by assuming they are the same. Convert a kVA rating to amps to check it against the cable and the protective device. Convert measured volts and amps to kVA to see how much headroom a transformer has left. All four of those moves are on this page, as a dropdown, along with the kW and kvar that accompany each answer.

Two warnings worth reading before you trust a number. In three-phase mode the voltage must be the line-to-line value — the source this page's √3 comes from specifies "RMS voltage, mean line-to-line of 3 phases", and quietly using line-to-neutral instead is the most common way this calculation goes wrong, by a factor of 1.73. And the whole model assumes a balanced, sinusoidal, steady-state supply. Variable-frequency drives, switch-mode power supplies and LED drivers draw distorted current whose true apparent power is higher than √3 × V × I predicts, and an unbalanced three-phase load is not described by a single line current at all.

Finally, this is an electrical identity, not a code compliance calculation. Conductor ampacity, overcurrent protection, continuous-load derating and transformer sizing margins are governed by the electrical code adopted where you are, and local amendments vary. Use the number here to understand the physics and to sanity-check a quote; have a licensed electrician or engineer sign off before any work proceeds.

What is kva calculator?

kVA stands for kilovolt-ampere, a unit of apparent power — the product of the root-mean-square voltage across a circuit and the root-mean-square current through it, divided by a thousand. It measures the total electrical burden a supply has to carry, without regard to how much of that burden ends up doing useful work.

Apparent power exists as a separate quantity because AC voltage and current are not necessarily in step. In a purely resistive circuit they rise and fall together and every volt-ampere becomes a watt. Add inductance — a motor winding, a transformer core, a fluorescent ballast — and the current lags the voltage, so part of the energy flowing into the load each cycle flows straight back out again. That returned energy still had to travel down the cable and through the transformer, still heated the conductors on the way, and still counts against the equipment's rating, but it never turned into anything. Apparent power counts it. Real power does not.

The three quantities form the power triangle: S² = P² + Q², where S is apparent power in kVA, P is real power in kW, and Q is reactive power in kvar. The power factor is the ratio P ÷ S, which is also the cosine of the phase angle between voltage and current. On a single-phase supply S = V × I. On a balanced three-phase supply each phase carries V_phase × I_line, and because the line-to-line voltage is √3 times the phase voltage, the total works out to S = √3 × V_line-to-line × I_line — the origin of the √3 that appears in every three-phase formula.

How to use this calculator.

1. 01Choose the supply type. Three phase applies the √3 factor and expects the line-to-line voltage; single phase does not.
2. 02Choose what you want to solve for: kVA from measured volts and amps, kVA from a kW load, amps from a kVA rating, or volts from a kVA rating.
3. 03Enter the line voltage. On three phase this is the line-to-line value — 208 V, 400 V, 415 V or 480 V on common systems, not the 120 V or 230 V you measure to neutral.
4. 04Enter whichever of current, kVA rating or kW load your chosen mode asks for. Fields the mode does not read are ignored and can be left at zero.
5. 05Enter the power factor as a decimal. Use 1 for a purely resistive load; read it off the nameplate for a motor or a drive. It does not affect the kVA figure computed from volts and amps, but it does set the kW and kvar that go with it.
6. 06Read the apparent power in kVA as the headline, then check the line current against your cable and protective device, and the kW against what the load actually needs.
7. 07Round up to the next standard rating on the manufacturer's list before ordering anything, and have the installation signed off by a licensed electrician or engineer.

The formula.

Start with the single-phase case, because the three-phase one is built from it. Apparent power is the plain product of the rms voltage and the rms current: S = V × I. There is no phase angle in that expression, which is exactly the point — it describes what the supply has to carry, not what the load does with it. Real power carries the angle: P = V × I × cos φ, and cos φ is the power factor. OpenStax University Physics Volume 2 gives both forms in section 15.4, as Equation 15.12 for the general case and Equation 15.13 for a resistor, and names the factor: "In engineering applications, cos(ϕ) is known as the power factor." Dividing the general expression by the resistive one isolates V × I as the power-factor-free product, which is apparent power.

On a balanced three-phase supply, each of the three phases delivers V_phase × I_line, so the total is 3 × V_phase × I_line. Three-phase equipment is labelled with the line-to-line voltage rather than the phase voltage, and in a balanced system those differ by √3, so substituting V_phase = V_LL ÷ √3 turns 3 × V_phase × I into √3 × V_LL × I. That is where the √3 comes from — it is a consequence of measuring voltage between lines instead of to neutral, not a fudge factor.

The US Department of Energy states the resulting relation directly. Fact sheet DOE/GO-10097-517, Determining Electric Motor Load and Efficiency, gives Equation 1 as "Pi = V x I x PF x √3 / 1000", with V defined as "RMS voltage, mean line-to-line of 3 phases" and I as "RMS current, mean of 3 phases". Setting PF = 1 in that equation leaves the apparent power in kVA. This page uses the exact √3 = 1.7320508075688772935…, carried at forty significant digits, and rounds once at the very end. That matters more than it looks: the fact sheet's own worked example prints 22.9 kW for 469.7 V, 37 A and PF 0.763, while the exact √3 gives 22.9671681259 kW. Substituting the schoolbook √3 ≈ 1.73 reproduces the printed figure at 22.94 kW, so the difference is the source's hand-calculation rounding rather than a disagreement about physics. Both values are recorded, and a test on this calculator asserts the exact one and explicitly rejects 22.94.

The reactive power follows from the power triangle. If P = S cos φ then Q = S sin φ, and since sin φ = √(1 − cos²φ), the calculator computes Q = S × √(1 − PF²) without ever needing the angle itself. It reports the magnitude only: a single power-factor number does not record whether the current lags the voltage (an inductive load, the usual case) or leads it (a capacitive one), so signing the result would be inventing information the input never carried.

Every mode on the page is a rearrangement of the same identity. Amps from a rating is I = S ÷ (√3 × V). Volts from a rating is V = S ÷ (√3 × I). kVA from a kW load is S = P ÷ PF, and once S is known the current follows from the voltage. Nothing is solved numerically and nothing is approximated; each answer is a closed-form inverse of the one relation.

A worked example.

Frequently asked questions.

References& sources.

1. [1]US Department of Energy, Office of Energy Efficiency and Renewable Energy, fact sheet DOE/GO-10097-517, "Determining Electric Motor Load and Efficiency", January 1997 (OSTI report numbers DOE/GO--10097-517; 319). Primary source for the three-phase relation. Equation 1 reads "Pi = V x I x PF x √3 / 1000", with "Pi = Three-phase power in kW", "V = RMS voltage, mean line-to-line of 3 phases", "I = RMS current, mean of 3 phases", "PF = Power factor as a decimal". Setting PF = 1 isolates the apparent power. RECORDED CONFLICT: the same document's worked example prints "469.7 x 37 x 0.763 x √3 / 1000 = 22.9 kW", while the exact √3 gives 22.9671681259 kW; the printed figure is reproduced by √3 ≈ 1.73, which yields 22.94 kW. This page uses the exact √3 and a test asserts 22.9671681259 and explicitly rejects 22.94. Access: open. Retrieved and text-verified 2026-07-29.
2. [2]OpenStax (Rice University), University Physics Volume 2, §15.4 "Power in an AC Circuit". Primary source for the power factor and the rms voltage-current product. Equation 15.12 gives "Pave = (1/2)I₀V₀cos(ϕ)" and Equation 15.13 gives "Pave = (1/2)I₀V₀ = IrmsVrms = Irms²R"; the section states "In engineering applications, cos(ϕ) is known as the power factor, which is the amount by which the power delivered in the circuit is less than the theoretical maximum of the circuit due to voltage and current being out of phase", and defines the rms values as "Irms = (1/√2)I₀ and Vrms = (1/√2)V₀". Access: open. Retrieved 2026-07-29.
3. [3]OpenStax (Rice University), University Physics Volume 2, §9.5 "Electrical Energy and Power". Secondary check on the DC power identities underlying the AC forms: Equation 9.12 "P = IV" and Equation 9.13 "P = I²R = V²/R", plus "The energy unit on electric bills is the kilowatt-hour (kW·h), consistent with the relationship E = Pt" and "1 kW·h = 3.6 × 10⁶ J". Access: open. Retrieved 2026-07-29.
4. [4]International Electrotechnical Commission, IEC 60050 International Electrotechnical Vocabulary, part 131, entry 131-11-41 "apparent power". Cited by IEV reference number for the standardised definition of the term and its unit, the volt-ampere. ACCESS: GATED — electropedia.org returned HTTP 403 to the retrieval used for this page, so nothing is quoted from the entry and no figure on this page rests on it. Declared as unopened rather than presented as verified. Retrieval attempted 2026-07-29.
5. [5]NIST, Special Publication 811, "Guide for the Use of the International System of Units (SI)", 2008 edition (DOI 10.6028/NIST.SP.811e2008). Used only for unit symbols and typography — V, A, W, VA — not for any electrical relation. NIST's own landing page notes that SP 811 "has not yet been updated to reflect the changes in the SI that came into effect on May 20, 2019"; none of those changes affect the symbols used here. Access: open. Retrieved 2026-07-29.