Convert the ohmic value to per-unit using base impedance: $$Z_{pu} = \frac{Z_{ohms}}{Z_{base}}$$
FLA=KVA×1000V×1.732cap F cap L cap A equals the fraction with numerator cap K cap V cap A cross 1000 and denominator cap V cross 1.732 end-fraction Use the transformer's nameplate impedance percentage (
For a fault right at the transformer's secondary side, you calculate current based on its Full Load Amps (FLA) and percentage impedance (%Z) : short ckt current calculation
A simplified version of the Ohmic method often used by contractors to calculate fault currents at various distances along a run of cable or busway. Step-by-Step Calculation (Transformer Example)
Zara pulled out her laptop and ran a proper short-circuit study per IEC 60909. Convert the ohmic value to per-unit using base
If a circuit can produce 20,000A during a fault, but your breaker is only rated for 10,000A, the breaker could melt or arc over without actually stopping the current.
Isc=FLAZ%/100cap I sub s c end-sub equals the fraction with numerator cap F cap L cap A and denominator cap Z % / 100 end-fraction Example: A 500kVA transformer at 480V with 5% impedance. Isc=FLAZ%/100cap I sub s c end-sub equals the
The "infinite bus" assumption vs. actual utility fault capacity. Conclusion
Where $K$ is a multiplication factor based on the X/R ratio of the system. A high X/R ratio (highly inductive circuit) results in a larger DC component.
