The calculation, without hidden assumptions
Estimate the symmetrical RMS fault current available at a transformer secondary terminal when transformer impedance is the only modeled limiting impedance. The calculator stays deliberately narrow: it does not pretend a terminal infinite-bus estimate is a downstream coordination or arc-flash study.
How to use this calculator
Enter the transformer nameplate kVA and secondary RMS voltage.
Choose single- or balanced three-phase to match the transformer/output being screened.
Enter the actual tested/nameplate percent impedance; enter 5.75 for 5.75%, not 0.0575.
Read the kA result together with rated current, 100/Z multiplier and fault MVA cross-check.
For any downstream panel, protection selection, coordination or arc-flash work, continue with a full study including all source/path impedances and applicable contributions.
Where people use it
- •Screen transformer-secondary prospective fault current before a detailed study.
- •Cross-check a nameplate %Z against the expected current multiplier.
- •Compare how candidate transformer impedances change terminal fault-current magnitude.
- •Provide a transparent hand-calculation check for electrical design review.
Example: 1000 kVA, 415 V, 5% Z
Rated current is about 1,391.2 A. A 5% impedance is 0.05 pu, so the transformer-only terminal estimate is about 27.82 kA and the corresponding fault level is 20 MVA.
What the result does not assume
- •Use actual nameplate/tested %Z whenever available; generic typical impedance values are not design inputs.
- •The model assumes an infinite upstream source and a bolted fault at the transformer terminals.
- •Cable, busway, utility/source impedance and connections can reduce downstream available current; rotating machines can add contribution.
- •Single-line-to-ground and other unbalanced faults require sequence impedances and grounding data; this owner does not invent them.
- •Peak/asymmetrical current requires X/R and an applicable short-circuit method; it is intentionally not inferred from %Z alone.
- •Do not use this simplified result as an arc-flash incident-energy calculation or automatic breaker/SCCR approval.
Frequently asked questions
Why divide rated current by Z%/100?+
Percent impedance is the per-unit transformer impedance on its own base. Under the simplified infinite-bus terminal model, current in per unit is the reciprocal of that impedance, so a 5% transformer corresponds to about 20 times rated current.
Is this the fault current at a downstream panel?+
No. The result is for the modeled transformer secondary terminal. Downstream conductors, busway and connections add impedance, while motors and other sources can contribute current.
Can I use a typical %Z?+
Only for rough exploration. For engineering decisions use the transformer manufacturer/nameplate tested impedance and the project short-circuit model.
Does this calculate line-to-ground faults?+
No. Ground faults require grounding and positive-, negative- and zero-sequence network data. A kVA/%Z shortcut cannot establish those quantities safely.
Why is fault MVA useful?+
It is an independent scale check: transformer kVA multiplied by 100/Z% gives the transformer-limited short-circuit apparent-power level under the same simplified assumptions.
Does a lower %Z increase fault current?+
Yes in this model. Because Isc is inversely proportional to per-unit impedance, halving %Z doubles the transformer-limited terminal fault current.
Semantic next steps
Continue the calculation
These links move to a different input, formula or project stage rather than a keyword variation of this page.
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