Worked problem

Short-Circuit Current at a Transformer Secondary

Estimate the available fault current on the secondary of a 500 kVA, 480 V transformer with 5.75% impedance, assuming an infinite bus.

Electrical & Power · Last reviewed

Given

500 kVA, 3-phase transformer, 480 V secondary, nameplate impedance 5.75%. Assume the utility source is an infinite bus (it does not limit the current), which gives a conservative upper estimate.

Find

The available three-phase fault current at the secondary terminals, and what it means for equipment ratings.

Solution

  1. Full-load current = 500,000 ÷ (√3 × 480) = 601 A.
  2. Fault current = full-load current ÷ per-unit impedance = 601 ÷ 0.0575 = 10,459 A, about 10.5 kA.

Answer

about 10.5 kA at the transformer secondary terminals

What it means

  • Every device on this bus must have an interrupting or short-circuit current rating at or above the available fault current. The next standard breaker ratings above 10.5 kA are commonly 14 kA, 18 kA, 22 kA or 25 kA.
  • The value falls with distance, because conductor impedance adds to the transformer impedance. Calculate it again at each panel.
  • The utility source, motor contribution and conductor impedance change the result. A real study accounts for them.

Common mistakes

  • Using the load current instead of the transformer rated current.
  • Entering impedance as 5.75 instead of 0.0575.
  • Forgetting that the available fault current is a property of the supply, not of the load that is connected.

Try two variations

  1. A 1,000 kVA, 480 V transformer at 5.75%: full-load current = 1,203 A, so the fault current is 1,203 ÷ 0.0575 = 20.9 kA.
  2. A 500 kVA transformer with a 208 V secondary at 5.75%: full-load current = 1,388 A, so the fault current is 24.1 kA.

The second case shows that the same kVA at a lower voltage gives a higher current, and so a higher fault current.

Sanity check

With the infinite-bus assumption, the fault current is the full-load current times 100 ÷ (impedance in percent). A 5.75% transformer gives about 17.4 times full-load current. If your answer is not near that multiple, check the impedance entry.

Related tools and labs

Calculators, labs and simulators are part of the All-Access Pass.

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Questions

Why assume an infinite bus?

It needs no utility data and always gives a result at or above the real value, so it is a safe first estimate. A real study uses the utility's available fault level.

Does this apply to single-phase transformers?

The same idea applies, with the single-phase full-load current: kVA × 1000 ÷ V.

What is the difference between available fault current and an AIC rating?

The available fault current is what the system can deliver at a point. The AIC (interrupting) rating is what a device can safely interrupt. The rating must be at or above the available current.

How does motor contribution change the result?

Running motors act as generators for the first few cycles of a fault and add current. A full study includes them.

What does a short-circuit study include?

Utility source data, transformer and cable impedances, motor contributions, and device ratings, followed by a check that every device is rated for the current at its location. A coordination and arc-flash study often follows.

Educational material. Examples use simplified, hypothetical inputs and are not a design, a construction document or professional engineering advice. Confirm requirements with the codes and standards adopted where the work is built and with a licensed engineer.

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