Electrical Tools · Free

Available Fault Current — Point to Point

NEC 110.9 requires every device that interrupts a fault to be rated for what is actually available at that point — and at a service, 110.24 requires the figure to be field-marked. This tool works it out from the transformer's impedance, then shows how much the feeder run sheds on the way to the panel.

Calculate the available fault current

Source transformer

Feeder to the panel

ft
0 ft1,000 ft

Fault current at the panel

8,626 A

Symmetrical RMS, infinite primary source

At the transformer

11,455 A

AIC there

22,000 A

Minimum interrupting rating at the panel

10,000 A

See the breakdown
Secondary FLA
Effective %Z
At the secondary
Run impedance
Source impedance
f factor
Multiplier M
At the panel

The run sheds 25% of the fault current, dropping the requirement from a 22,000 A device at the transformer to 10,000 A at the panel.

The method, explained in plain English

# Step 1 — at the transformer secondary
FLA = kVA × 1000 ÷ (V × 1.732 on three-phase)
I_sc = FLA ÷ (%Z ÷ 100)    # ×0.90 on %Z for the worst case
# Step 2 — down the run
Z_source = V_driving ÷ I_sc    # implied by the upstream current
Z_run   = √(R² + X²)    # Chapter 9, Tables 8 and 9
f = Z_run ÷ Z_source  ·  M = 1 ÷ (1 + f)
# Step 3 — the code check
device interrupting rating ≥ available fault current   # 110.9

Infinite primary is the safe assumption

Including the real utility impedance always lowers the answer. If the equipment passes on the infinite-source figure, it passes for real.

Low impedance is the worst case

Nameplate %Z carries about ±10% tolerance, and the low-impedance unit produces more fault current. For an AIC check, use the low end.

Length is your friend here

Fault current is highest at the service and falls steadily downstream. The equipment nearest the transformer needs the highest rating.

Series ratings are exact combinations

240.86 permits a tested upstream device to protect a lower-rated one — only in the listed pairing, and it has to be field-marked.

Worked examples

The same 300 kVA transformer, at three distances.

1

At the secondary terminals

FLA = 300,000 ÷ (1.732 × 480) = 360.8 A
%Z = 3.5 × 0.90 = 3.15%
I_sc = 360.8 ÷ 0.0315 = 11,455 A → needs 22 kA

Result: a 10 kA residential loadcenter would be unsafe bolted straight to this transformer. Main switchgear here needs a 22 kA rating.

2

100 ft of 4/0 copper away — the defaults

Z_run = √(0.00608² + 0.00510²) = 0.00794 Ω
Z_source = (480 ÷ 1.732) ÷ 11,455 = 0.02419 Ω
f = 0.3280 · M = 0.7530 → 8,626 A → 10 kA is adequate

Result: 100 ft of conductor sheds 25% of the fault current — enough to move the panel from a 22 kA requirement into 10 kA equipment.

3

400 ft away

M = 0.433 → 4,954 A

Result: less than half the current at the transformer. The trade-off is that the same impedance causing this reduction is also causing voltage drop — which is why long feeders get sized on drop, not ampacity.

Fault current at the secondary, by transformer size

480 V three-phase, typical impedance with the −10% tolerance applied, computed with the same helpers the calculator uses. Always use the nameplate impedance for real work.

kVA Typical %Z Secondary FLA Fault current Minimum AIC
75 2.5% 90 A 4,009 A 10,000 A
112.5 3.5% 135 A 4,296 A 10,000 A
150 3.5% 180 A 5,728 A 10,000 A
225 3.5% 271 A 8,592 A 10,000 A
300 3.5% 361 A 11,455 A 22,000 A
500 4.5% 601 A 14,850 A 22,000 A
750 4.5% 902 A 22,274 A 25,000 A
1000 5.75% 1,203 A 23,243 A 25,000 A

Sources & standards: NEC (NFPA 70) 2023 — 110.9 (interrupting rating), 110.10 (circuit impedance and other characteristics), 110.24 (field marking of available fault current at a service), 240.86 (series ratings), and Chapter 9 Tables 8 and 9 for conductor resistance and reactance. This is a planning tool for one transformer and one run using the point-to-point method with an infinite primary source; it is not a coordination or arc-flash study, and it does not model motor contribution or asymmetrical peak current. Where equipment ratings are marginal or arc-flash labelling is required, engage a licensed engineer. Local amendments override the model code, and the AHJ has final say.

Frequently asked questions

Common questions about available fault current and interrupting ratings.

What is available fault current and why does it matter?

It is the current that would flow if a bolted short circuit occurred at a given point. It matters because NEC 110.9 requires every device intended to interrupt a fault to have an interrupting rating at least equal to what is available there. A 10 kA loadcenter on a 22 kA supply can fail catastrophically rather than clearing the fault. 110.24 additionally requires the available fault current to be field-marked at a service.

Why assume an infinite primary source?

Because it is the conservative assumption and it needs no data from the utility. Treating the primary as infinite means all the impedance limiting the fault is the transformer's own, so the answer is the highest the transformer can produce. Including the real utility and primary-conductor impedance always lowers the result — so if the equipment passes on the infinite-source figure, it passes.

Why apply a −10% impedance tolerance?

Because transformer impedance is a nameplate value with a manufacturing tolerance of about ±10%, and a transformer at the low end of that band produces more fault current. For an interrupting-rating check the worst case is the low-impedance unit, so the calculator can multiply the nameplate impedance by 0.90. This is standard practice in a short-circuit study.

How does the point-to-point method work?

You take the fault current at the start of a run, add the run's own impedance, and get the reduced current at the far end. The classic form is M = 1 ÷ (1 + f), where f is the run impedance divided by the source impedance. This calculator computes the source impedance implied by the upstream current and adds the run impedance from NEC Chapter 9 Tables 8 and 9 — algebraically identical, but every number traces to the code book rather than to proprietary conductor constants.

Does conductor length really reduce fault current that much?

Yes, and it is the main reason a residential loadcenter is usually adequate. On the default 300 kVA example the current falls from 11,455 A at the transformer to 8,626 A after only 100 ft of 4/0 copper — a 25% reduction. Length is why fault current is highest at the service and drops steadily downstream, and why the equipment nearest the transformer needs the highest rating.

What interrupting ratings are available?

Residential loadcenters and their breakers are commonly rated 10 kA. Commercial panelboards run 22 kA and 65 kA, with 100 kA and 200 kA available on switchgear and current-limiting fuses. A series rating is also permitted, where a tested upstream device protects a lower-rated downstream one — but only in the exact combination the manufacturer tested and listed, and 240.86 requires it to be field-marked.

Is this a substitute for a short-circuit study?

No. This is a planning tool for a single transformer and one feeder run. A real coordination study models motor contribution, multiple sources, cable and bus impedance throughout, asymmetrical peak currents, and arc-flash incident energy — and where equipment ratings are marginal or arc-flash labelling is required, that study needs a licensed engineer.

Fault current checked? Price the equipment.

Interrupting ratings drive equipment cost more than almost anything else on a commercial job. TradesQuote turns your scope into a detailed line-item estimate with quantities and totals, checked by a built-in quality control agent.

AI line-item estimates

Quantities, unit prices, and totals generated instantly.

Knowledge base

Upload past jobs so estimates reflect your real pricing.

Shareable & signable

Clients review, accept, and sign from a public link.

No credit card required · 14-day free trial · Cancel anytime

More electrical calculators