Transformer current calculator

Primary and secondary full-load current, and the largest secondary fault current its impedance allows.

Transformer
Nameplate
kVA
%
kV
kV

Result

Secondary full-load current
1,804A
Primary full-load current
62.76A
Largest secondary fault current (infinite source)
31.38kA

What it calculates

The current each winding carries at rated load, and the largest fault current the transformer can pass to a short circuit on its secondary. Full-load current follows from the kVA rating and the voltage on that side. For a three-phase unit, use line-to-line voltage:

IFL=S3 VLLI_\text{FL} = \dfrac{S}{\sqrt{3}\,V_\text{LL}}

For a single-phase unit, divide by the winding voltage:

IFL=SVI_\text{FL} = \dfrac{S}{V}

Why %Z sets the fault current

Percent impedance is the fraction of rated voltage that drives full-load current through the transformer when its secondary is shorted. Applying full voltage to the same impedance gives a current larger by the reciprocal of that fraction:

Isc=IFL%Z/100I_\text{sc} = \dfrac{I_\text{FL}}{\%Z/100}

Worked example

A three-phase 1,500 kVA transformer steps 13.8 kV down to 480 V with 5.75% impedance. The calculator opens on this case. The secondary full-load current and the largest fault current are:

IFL=1,5003×0.48=1,804.2 AI_\text{FL} = \dfrac{1{,}500}{\sqrt{3} \times 0.48} = 1{,}804.2\ \text{A}Isc=1,804.20.0575=31.38 kAI_\text{sc} = \dfrac{1{,}804.2}{0.0575} = 31.38\ \text{kA}

The primary full-load current is 62.76 A.

Assumptions

  • The source is infinite: it holds the primary voltage constant, with no impedance of its own. A real source only lowers the fault current. To include it, use the fault current calculator.
  • The fault is bolted, with zero fault impedance, at the secondary terminals.
  • The result is the symmetrical current, from the nameplate %Z. It does not include motor contribution or tolerance on the impedance.
  • Three-phase voltages are line-to-line. The transformer connection does not change the line currents calculated here.

Questions

Why is the fault current limited by %Z?
Percent impedance is the share of rated voltage needed to drive full-load current through a shorted secondary. At full voltage the current is therefore full-load current divided by %Z/100. Any source impedance only lowers it, so this is the upper bound.
Should I use the nameplate %Z or a tolerance-adjusted value?
Nameplate impedance can differ from the as-built unit by a few percent. For the highest fault duty, many engineers reduce %Z by the standard manufacturing tolerance; for voltage drop, they use the nameplate value or higher.
Is the secondary voltage line-to-line?
Yes, for three-phase units enter line-to-line kV on both sides. For single-phase units enter the winding voltage, for example 0.24 kV for a 120/240 V secondary.

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