formula.expert

Transformer Full-Load Current

Also known as transformer secondary amps · kVA to amps · FLA from kVA · transformer primary current · rated current single and three phase

IFL=Sk VI_{FL} = \frac{S}{k \, V}

Enter your known values, leave one input blank, and solves for the missing one. Try different units for next level excitement!

See also
Three-Phase Apparent Power — The three-phase case written out with its √3 in place, if you never need the single-phase form.Available Short-Circuit Current from Percent Impedance — Turns this rated current into the available fault current at the secondary terminals.Derated Conductor Ampacity — The ampacity this current has to fit inside once the corrections are applied.

Learning zone

A transformer's nameplate gives kilovolt-amperes, and kVA is the honest rating: the windings care about amps, the core cares about volts, and neither knows anything about power factor. Turning kVA into amps is just division — by the voltage for a single-phase unit, and by √3 times the line-to-line voltage for a three-phase one. This page carries that choice as a single factor k: enter 1 for single-phase, 1.732 for three-phase. A 75 kVA three-phase transformer on a 208 V secondary is rated 75 000 / (1.732 × 208) ≈ 208 A, which is a pleasing coincidence and nothing more.

Do this on both sides. The same kVA at a higher voltage means proportionally fewer amps, which is the whole argument for distribution voltage: that 75 kVA unit fed at 600 V pulls only 72 A on the primary. Getting the sides mixed up — using the primary voltage with the secondary conductors — is the classic error, and it always errs in the dangerous direction on the low-voltage side. If you ever solve this page for k as a check and get something between 1 and 1.732, that is the diagnosis: your voltage, your current and your kVA are not all from the same winding.

Full-load current is where transformer protection and conductor sizing start, but it is only the start. Primary and secondary overcurrent protection follow the code's own percentage rules, which deliberately allow the device to sit well above this current so that magnetizing inrush does not trip it on every energization — and those percentages differ by voltage class, by whether secondary protection is provided, and by edition. Look them up. What this page gives you is the rated current the rest of that arithmetic hangs from.

Transformer Full-Load Current
IFL=Sk VI_{FL} = \frac{S}{k \, V}
IFLVSk
Where
  • IFLI_{FL}= Full-load current (A)
  • SS= Transformer rating (VA) (W)
  • VV= Winding voltage (V)
  • kk= Phase factor
Missing one of these? Work it out first, then come back