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

Background.

How many amps can a 500 kVA transformer deliver on a 415 V three-phase supply? About 695.6 A per line. How large a generator does 100 kW of load need if its power factor is 0.8? At least 125 kVA. Both answers come out of one relation between voltage, current and apparent power, and both go wrong in the same two ways: dropping the √3 that a three-phase supply carries, or treating kVA and kW as interchangeable.

Real power, in kW, is the share of apparent power that leaves as shaft work, heat or light, and the ratio of the two is the power factor. A motor whose current runs partly out of step with the voltage still loads its supply with every ampere it draws, which is why transformers, generators and UPS units carry kVA ratings. At a power factor of 0.8, each kW of load needs 1.25 kVA of supply, so equipment sized on kW alone comes up 20 percent short.

On three phase, the V in S = √3 × V × I is the voltage between two lines. Type the line-to-neutral figure instead and the answer shrinks by a factor of √3: the 480 V, 100 A feeder in the worked example carries 83.1384 kVA, but entered as 277 V it reports 47.9778 kVA, an undersize of more than 40 percent.

The relations assume a balanced, sinusoidal, steady supply. Variable-speed drives, switch-mode power supplies and LED drivers draw distorted current, so √3 × V × I worked from a fundamental-frequency current reading understates their apparent power, and an unbalanced three-phase load has to be worked phase by phase. This is the physics, not a wiring-code calculation: cable ampacity, breaker sizing and derating follow the electrical rules adopted where you are, and a licensed electrician or engineer has to approve real work.

What is kva calculator?

Apparent power, S, is the product of the rms voltage and the rms current, measured in volt-amperes; divide by 1,000 for kVA. On an AC circuit it splits into two parts at right angles: real power P, in kilowatts, which is consumed, and reactive power Q, in kilovars, which is stored in magnetic or electric fields and handed back to the supply every cycle. They form the power triangle, S² = P² + Q², and the power factor is P ÷ S, the cosine of the angle between voltage and current for a sinusoidal supply. On a balanced three-phase supply each phase carries the phase voltage times the line current, and because the line-to-line voltage is √3 times the phase voltage, the total 3 × V_phase × I becomes √3 × V_line × I.

How to use this calculator.

1. 01Choose single phase or three phase. In three-phase mode the voltage must be the line-to-line value, for example 208, 400, 415 or 480 V, not the 120, 230 or 277 V measured to neutral.
2. 02Choose what to solve for: kVA from measured volts and amps, kVA from a kW load and its power factor (sizing a generator or UPS), amps from a kVA rating, or volts from a rating and a current. Fields the chosen mode does not use are ignored.
3. 03Enter the power factor as a decimal (0.85, not 85), from the motor, drive or UPS nameplate where you can; use 1 only for resistive loads such as heaters. It does not change kVA worked out from volts and amps, but it sets the kW and kvar beside it, and in the kW mode it sets the answer.
4. 04Round the kVA up to the next rating the manufacturer actually sells, add the margin your application needs, and have conductor and protection sizes checked against your local wiring rules.

The formula.

Single phase: S = V × I. Balanced three phase: S = √3 × V_LL × I, with V_LL the line-to-line voltage and I the current in one line. Real power is P = S × PF and reactive power is Q = S × √(1 − PF²), which follows from the power triangle without ever needing the phase angle itself. Every mode is the same identity rearranged, with k = 1 for single phase and √3 for three phase: I = S ÷ (k × V), V = S ÷ (k × I) and S = P ÷ PF. The √3 is not a correction factor but bookkeeping: three phases each carrying V_LL ÷ √3 sum to 3 ÷ √3 = √3 times V_LL × I. The calculator uses its exact value, 1.7320508…, rather than 1.73. Reactive power is reported as a magnitude because a single power-factor number cannot say whether the current lags the voltage, as in motors and transformers, or leads it, as with capacitor banks. Of all the inputs, the voltage reference is the one most often wrong: the line-to-neutral value in three-phase mode returns only 57.7 percent of the true kVA.

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, giving 22.9671681259 rather than 22.94. 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". 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". 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. Retrieved 2026-07-29.