Calculadora de Corriente de Cortocircuito Gratis
Calcula la corriente de cortocircuito en un punto del sistema eléctrico. Esencial para protección y coordinación.
Prospective Short-Circuit Current
14,434 A
Prospective Short-Circuit Current vs Transformer Rating
Fórmula
## Transformer Short-Circuit Current The prospective short-circuit current at the transformer secondary is limited by the transformer impedance. ### Formulas **I_FL = kVA x 1000 / (sqrt(3) x V)** (full load current) **I_SC = I_FL / (Z%/100)** (short-circuit current) where Z% is the transformer impedance expressed as a percentage. This is the maximum fault current available at the transformer terminals, ignoring cable impedance downstream.
Ejemplo Resuelto
A 500 kVA transformer, 400 V secondary, 5% impedance.
- 01I_FL = 500,000 / (1.732 x 400) = 721.7 A
- 02I_SC = 721.7 / 0.05 = 14,434 A = 14.43 kA
- 03Switchgear must have a breaking capacity above 14.43 kA.
Preguntas Frecuentes
Why does transformer impedance matter?
Higher impedance limits short-circuit current (reducing switchgear cost) but increases voltage regulation (more voltage drop under load). Typical impedance: 4-6% for distribution transformers.
Does cable impedance reduce fault current?
Yes. Cable resistance and reactance add to the total impedance, reducing the fault current at points downstream of the transformer. Use I_fault = V / Z_total for accurate calculations.
What is the asymmetric peak current?
The first half-cycle of fault current includes a DC offset that can increase the peak to 1.8-2.5 times the RMS value. Equipment must withstand this momentary peak, expressed as the peak making capacity.
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