# Computing short circuit current of transformer

Update: My 500Va is just an example. What I wanted to compute was the short circuit current of utility pole transformers because I want to estimate how large an arc flash that can be produced.

Assuming the pole utility transformer is 100kVA, and the primary voltage is 12,000 volts. Then the computation is like this? 100kVA/12,000 volts/0.02= 416A only.. or do you use 100kVA/120 volts/0.02 = 41,666A?

original:

How do you compute for the short circuit current of a transformer? Is the following correct?

Assume the transformer is 500VA with 2% impedance.

Then 500VA/120v/0.02= 208.3A

This means if the secondary side of the 500VA transformer is short circuited. Then there will be 208.3A short circuit current produced?

How does this relate to whatever is the step down voltage?

• You are forgetting the DC resistance of the primary winding. Even though the magnetic field is 'shorted', the DC resistance of the winding is always there. 208 amps would be almost 5KW, 15 times the 500VA rating. Find the DC resistance of the primary, and how it would reduce the 208 amps.
– user105652
Feb 18, 2019 at 3:50
• Also, the secondary DC resistance is needed.
– jonk
Feb 18, 2019 at 6:12
• Measure the primary and secondary inductance and resistance. Plug it in LTSpice. As per my experience, the result is quite good. Feb 18, 2019 at 6:19
• @Sparky256 what is your estimate how it would reduce the 208A? Like would it become half or so? Also for 75kVA utility pole transformers?
– Jtl
Feb 18, 2019 at 11:04

If the primary voltage is 120 V, your calculation is correct.

This means if the secondary side of the 500VA transformer is short circuited. Then there will be 208.3A short circuit current produced?

Yes that is correct.

How does this relate to whatever is the step down voltage?

The equivalent circuit of a transformer is shown below. The impedance of the transformer as viewed from the transformer is the primary impedance (Rp + jXp) plus the parallel combination of the magnetizing branch and the secondary impedance (Rs + jXs) referred to the primary. To refer the secondary impedance to the primary, it is multiplied by the square of the turns ratio. That is how the step-down voltage is related to the percentage impedance. Stating the impedance as a single impedance percentage with the secondary impedance referred to the primary accounts for the secondary voltage, so that the actual secondary voltage does not need to be known. The transformer manufacturer has taken care of that by providing the impedance as a single impedance percentage value.

Transformer manufacturers usually state both the percentage impedance and the X/R ratio, so that other impedances in the distribution circuit can be properly added when calculating the prospective short circuit current at a given point in a distribution system.

This type of transformer specification is generally given for power transmission and distribution transformers, not for a small transformer like 500 VA. A transformer of that size would have a much higher impedance. Only a large distribution transformer would have an impedance in the 2% range,

Image from Wikimedia Commons Cblambert

Your answer is correct for secondary shorted. The secondary voltage does not matter. The impedance base (upon which that 2% impedance was determined) is 120^2/500 = 28.8 ohms. That means your 2% impedance transformer is 0.02 x 28.8 = 0.576 ohms as seen from the 120V terminals. Assuming you had a perfect 120V source connected, the fault current that would flow = 120/0.576 = 208.3. That %Z already includes the resistance (R + jX) so your answer is correct.