Fault Current at the End of a Feeder

1. Transformer impedance in ohms = %Z × V² ÷ kVA = 0.0575 × 480² ÷ 500,000 = 0.02650 Ω.
2. At the terminals: I = V ÷ (√3 × Z) = 480 ÷ (1.732 × 0.02650) = 10,459 A, the value from the earlier problem.
3. Feeder (0.15 thousand feet): R = 0.027 × 0.15 = 0.00405 Ω and X = 0.048 × 0.15 = 0.00720 Ω.
4. Add resistances and reactances separately: R = 0.00405 Ω and X = 0.02650 + 0.00720 = 0.03370 Ω.
5. Total impedance = √(0.00405² + 0.03370²) = 0.03394 Ω.
6. At the panel: I = 480 ÷ (1.732 × 0.03394) = 8,166 A.

Answer

What it means

• Every device must be rated for the fault current at its own location. A 10 kA-rated device would be enough at this panel but not at the transformer secondary, where 10.5 kA is available.
• The drop is only 22% because a short feeder adds little impedance next to a large transformer. A long feeder or small conductors lower the fault current a lot more.
• This still assumes an infinite bus and ignores motor contribution. A full study includes the utility source and motors, which raise the value.

Common mistakes

• Using the transformer fault current for every panel downstream, which can lead to over-sized, costlier equipment.
• Adding the impedance magnitudes instead of adding resistance and reactance separately. Here that gives 8.0 kA, a little low.
• Doubling the feeder length as if it were a single-phase loop. For a balanced three-phase fault the impedance is per phase, so the one-way length is used.
• Using a feeder impedance value from a different conduit material without checking.