Transformer kVA to Current Calculator
Transformer kVA to Current Calculator
Full-load current on both windings from a transformer’s kVA rating — or the kVA from a measured current — plus the short-circuit current its percentage impedance allows at the terminals.
Transformer kVA ⇄ current
100 kVA, 11,000 V / 415 V three-phase, 4.5% impedance
Rating, current and fault level
- kVA
- the rating, the same on both windings
- V
- the winding’s rated voltage; line-to-line on three-phase
- %Z
- impedance voltage: the percentage of rated voltage that drives rated current into a shorted secondary
- Isc
- prospective short-circuit current at the terminals, with an infinitely stiff supply
Worked example
100 kVA, 11,000 V / 415 V three-phase, 4.5% impedance
Secondary: I = 100,000 ÷ (1.732 × 415) = 139.1 A
Primary: I = 100,000 ÷ (1.732 × 11,000) = 5.249 A — the same kVA, 26.5 times the voltage, so 26.5 times less current
Short circuit: 139.1 A × 100 ÷ 4.5 = 3.092 kA at the secondary terminals
As apparent power that is 100 kVA ÷ 0.045 = 2.222 MVA
Allowing the usual 10% margin for the impedance tolerance gives 3.401 kA — the figure to compare against a breaker's breaking capacity
Distribution transformers at 11 kV / 415 V, three-phase
| Rating | HV line current at 11 kV | LV line current at 415 V | Fault current at 4.5% Z |
|---|---|---|---|
| 25 kVA | 1.312 A | 34.78 A | 772.9 A |
| 63 kVA | 3.307 A | 87.65 A | 1.948 kA |
| 100 kVA | 5.249 A | 139.1 A | 3.092 kA |
| 160 kVA | 8.398 A | 222.6 A | 4.947 kA |
| 250 kVA | 13.12 A | 347.8 A | 7.729 kA |
| 400 kVA | 20.99 A | 556.5 A | 12.37 kA |
| 630 kVA | 33.07 A | 876.5 A | 19.48 kA |
| 1,000 kVA | 52.49 A | 1.391 kA | 30.92 kA |
From the rating plate to the current and the fault level
A transformer’s nameplate gives a rating in kVA, not in kW, and that is deliberate: what heats the windings is current, and what the insulation has to stand is voltage, so the product of the two is the honest limit. The load’s power factor decides how many of those volt-amperes turn into watts, but the transformer neither knows nor cares. Divide the rating by the voltage — and by √3 as well on a three-phase winding, because the nameplate voltage is measured between two lines — and you have the full-load current that winding is designed to carry continuously.
Both windings, same volt-amperes. A 100 kVA 11 kV/415 V transformer carries 5.249 A on the high-voltage side and 139.1 A on the low-voltage side. The ratio between them is the voltage ratio, so the HV cable is a thin thing and the LV busbar is a slab of copper. Size each side from its own current.
What percentage impedance is actually telling you. Short the secondary, then wind the primary voltage up from zero until rated current flows. The voltage you needed, as a percentage of rated, is the impedance voltage — the %Z on the plate. It says the whole internal impedance of the transformer is that fraction of the base impedance V²/S. Turn it round and you get the number that matters for protection: with full voltage applied and the secondary bolted short, the current is the full-load current divided by that fraction. At 4.5% the example transformer can deliver 3.092 kA into a fault at its terminals, which is 22.2 times its own full-load current.
The assumptions behind that number, plainly. It takes the supply network above the transformer as having no impedance of its own — the “infinite bus” assumption that Bussmann’s short-circuit method states explicitly. That is not true, so the real fault current is a little lower, which makes this the safe way round for choosing breaking capacity. It applies at the transformer terminals only: every metre of cable downstream adds impedance and lowers it. And %Z itself has a tolerance — IEC 60076-1 allows ±7.5% on a declared impedance of 10% or more and ±10% below that — which is why the standard practice is to multiply by 1.1 for the worst case, giving 3.401 kA here. Motor contribution during the first cycles adds more; that is a study, not a one-line calculation.
Take the secondary full-load current to the cable size calculator for the cable and the MCB size calculator for the device, check the drop with the voltage drop calculator, and use the short-circuit figure to check that device’s breaking capacity. For the ratio and winding side of the same transformer, see the transformer turns ratio calculator; for what the kVA becomes in kilowatts at a given power factor, the kVA to kW calculator.
Frequently asked questions
How do I calculate the full-load current of a transformer?
Divide the rating in volt-amperes by the voltage, and by √3 as well for three-phase. A 100 kVA transformer with a 415 V three-phase secondary gives 100,000 ÷ (1.732 × 415) = 139.1 A.
What is percentage impedance on a transformer nameplate?
The percentage of rated primary voltage needed to drive rated current through a short-circuited secondary. It sets both the voltage regulation under load and the fault current the transformer can deliver. A lower %Z means better regulation and a bigger fault current.
How do I work out short-circuit current from percentage impedance?
Divide the full-load current by the impedance expressed as a fraction: Isc = Ifl × 100 ÷ %Z. For the example above, 139.1 A × 100 ÷ 4.5 = 3.092 kA. That is the worst case at the terminals with an infinitely stiff supply behind the transformer.
Why are transformers rated in kVA and not kW?
Because the limits are thermal and dielectric, not mechanical. Current heats the windings and voltage stresses the insulation, whatever the phase angle between them. The power factor belongs to the load, so the transformer is rated in the product of volts and amps.
Does the primary or the secondary carry more current?
The low-voltage winding, always. The same volt-amperes pass through both, so the currents are in inverse proportion to the voltages.
Is the voltage ratio the same as the turns ratio?
Only on a star-star or a single-phase transformer. On a three-phase transformer the nameplate voltages are line-to-line, so on a delta-star (Dyn) unit the ratio of the two windings’ turns differs from the ratio of the two nameplate voltages by √3. This page reports the voltage ratio, which is what the plate gives; for turns, work from the phase voltage of each winding.
Related calculators
References
- Eaton / Bussmann. Electrical Formulas and Short-Circuit Current Calculations: transformer full-load current = kVA × 1000 ÷ E single-phase and kVA × 1000 ÷ (E × 1.73) three-phase; “Multiplier = 100 / (%Z transformer)” and “IS.C. = Transformer F.L.A. × Multiplier”, stated under an “Available Utility Infinite Assumption”, with “for highest short circuit conditions, multiply values as calculated in step 3 by 1.1”.
- IEC 60076-1:2011, Power transformers — Part 1: General. Table 1 tolerances: voltage ratio at the principal tapping, the lesser of ±0.5% of the declared ratio and ±1/10 of the actual percentage impedance; short-circuit impedance at the principal tapping ±7.5% of a declared value of 10% or more and ±10% below that. Definitions 3.4.6 rated power, 3.4.4 rated voltage ratio, 3.7.1 short-circuit impedance, 3.6.2 no-load current.
- IEC 60038:2009, IEC standard voltages: 230/400 V is the standard low-voltage three-phase system (India, the Gulf and Europe; 240/415 V is still quoted in several Gulf states, and North America uses 120/240 V and 208Y/120 V or 480Y/277 V at 60 Hz).
- Hughes E, Hiley J, Brown K, Smith I M. Electrical and Electronic Technology, 12th ed. Pearson 2016: three-phase star and delta relationships, the power triangle, and the polyphase induction motor.