Available fault current (point-to-point)

A planning aid. The point-to-point method gives a reasonable worst-case bolted fault current for checking interrupting ratings; it is not an arc-flash or coordination study. Based on the NEC 2023 — check the edition your jurisdiction has adopted and its amendments; the inspector decides.

How to use it

Source. Choose how the fault current starts:

Transformer — the utility transformer’s kVA and nameplate impedance (%Z), with an infinite primary: the usual worst case. Leave 0.9 × %Z ticked: UL-listed transformers may be up to 10 % below their marked impedance, which raises the fault current.
Transformer + utility — the same, but with the primary voltage and the fault current the utility says is available on the primary. That lowers the result a little.
Utility figure — the available fault current the utility gives you at the service, used as is.

Pick three-phase or single-phase and the secondary voltage. For a 120/240 V centre-tapped transformer you can ask for the line-to-neutral fault, which is about 1.5 × the line-to-line current at the terminals. Enter the running motor load if motors could feed a fault: the method adds 4 × their full-load current at every point (4 to 6 is accepted).

Points. Add each step from the source down: a run of conductors (copper or aluminium, size, steel or nonmagnetic conduit or cable, length in feet and parallel sets), busway, or a second transformer. Give each the interrupting rating or SCCR of the equipment at its end, in kA.

Results. The fault current at each point, a one-line diagram, a table with f and M, and a check mark or cross against each rating. Show the formulas prints every step with the numbers put in. The marking box gives the text for the 110.24 label on service equipment (other than dwelling units): the maximum available fault current and the date. Worked example loads the method’s own example — 300 kVA, 2 %Z, 208 V, with 20 ft of two sets of 500 kcmil to the main panel and 20 ft of 2 AWG to a branch panel — which gives 46,273 A, 39,425 A and 18,687 A.

How it is worked out

Transformer FLA     = kVA × 1000 ÷ (√3 × E)        (single-phase: ÷ E)
Multiplier          = 100 ÷ %Z                      (0.9 × %Z for the worst case)
Isc at secondary    = FLA × multiplier
f (three-phase)     = √3 × L × I ÷ (C × n × E)
f (single-phase)    = 2 × L × I ÷ (C × n × E)
M                   = 1 ÷ (1 + f)
Isc at the point    = I × M
Second transformer  f = I(P) × Vp × √3 × %Z ÷ (100 000 × kVA);  Isc(S) = Vp/Vs × M × I(P)

L is the length in feet, I the fault current at the start of the run, n the number of parallel sets and C the conductor constant — one over the impedance per foot. The C values are the 600 V columns of the published point-to-point tables (Bussmann, now Eaton), which are based on resistance and reactance from IEEE Std 241 and 242. They are facts about conductors, used here as data.

110.9 asks that a device meant to interrupt a fault be rated for the current available at its line terminals; 110.10 that equipment’s short-circuit current rating not be exceeded. If a panel’s breakers fall short, the answer is higher-rated equipment, a tested series combination (240.86), or current-limiting protection ahead of it.

Related tools: transformers and three-phase for kVA, amps and protection, wire size and voltage drop, grounding and bonding, panel schedule, conduit bending and the electrician’s quick reference.

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