How is symmetrical short-circuit current at a bus typically calculated?

• A. Isc ≈ Vphase / (Zsource + Zfault).
• B. Isc ≈ Vbase / Zpu.
• C. Isc ≈ Vphase × (Zsource + Zfault).
• D. Isc ≈ (Zsource + Zfault) / Vphase.

Answer: (A)

When a short circuit happens at a bus, you treat the network as a Thevenin source seen from the fault, with a voltage across the fault and a total impedance in the path. The current that flows is determined by Ohm’s law for that Thevenin pair: Isc equals the fault voltage across the network divided by the total impedance the current must pass through. That gives Isc ≈ Vphase / (Zsource + Zfault). This shows how the current rises as the total impedance drops and falls as the impedance rises. If the fault impedance is negligible, you can approximate Isc by Isc ≈ Vphase / Zsource, but the general form uses the sum of Zsource and Zfault.

The other options mix up the relationship: multiplying voltage by impedance would imply more current with more impedance, which isn’t correct; using Vbase with a dimensionless Zpu doesn’t yield a proper current value without converting units; and taking the reciprocal of the correct ratio would not reflect how current actually changes with impedance.