Approximate short-circuit current in the transformer secondary. I_k ≈ I_n × 100 / uk It is found with; the full load current is I_n = S / (√3·U). For example, in a 1000 kVA, 400 V, UK %6 transformer, I_n ≈ 1443 A and I_k ≈ 24 kA (assuming infinite busbar, upper limit).

I_n = S / (√3·U) · I_k ≈ I_n × 100 / uk
Example calculation: 1,000 kVA, 400 V, UK %6 → I_n ≈ 1,443 A, I_k ≈ 24.1 kA (secondary terminal, upper limit). The breaking capacity of the main circuit breaker must be ≥ 24.1 kA.

Transformer Full Load & Short Circuit Current

Calculate the full-load current and approximate short-circuit current in the transformer secondary (preliminary information for panel/breaker selection).

Full load current I_n
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Approx. short circuit I_k
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Notes: I_n=S/(√3·U), I_k≈I_n·100/uk. Infinite bar assumption (Network and cable impedance neglected) → Upper limit at secondary terminal; actual fault current decreases with distance. The breaking capacity (Icu) of the circuit breaker should be selected above this value. A complete short-circuit analysis is required for precise coordination.
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How is it calculated?

  1. Enter the transformer power (kVA) and secondary voltage (V).
  2. Enter the short-circuit voltage (UK %) from the label.
  3. Press Calculate; the full load and short-circuit current will be displayed.

Frequently Asked Questions

How to find the transformer short-circuit current?
With I_k ≈ I_n·100/uk, the full load current and the transformer short-circuit voltage (uk%) are used.
What is short circuit voltage (UK)?
The percentage of voltage required to conduct the rated current when the secondary is short-circuited, as stated on the transformer label (typically %4–6).
Is this value the actual fault current?
With the assumption of an infinite busbar, the UPPER limit is at the secondary terminal; network and cable impedance reduces the actual current with distance.

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