NEC calculators

Available fault current calculator

Find the bolted fault current at a transformer and at the end of a feeder by the impedance method, compare it with equipment ratings, and get the number for the NEC 110.24 label.

Calculated with electrician.wiki · https://electrician.wiki/calculators/available-fault-current/

Code / standard

Source

System

Transformer rating

kVA

Nameplate impedance

Voltage at the fault

Line-to-line: 208, 240, 480 or 600. For 120/240 V use 240.

Transformer X/R

Worst case: %Z 10 % below nameplate

Utility supply at the primary

Conductor length to the fault

0 for a fault right at the source.

Parallel sets

Conductor size

Material

Raceway

Conductor resistance at

Running motors at the fault (total FLA)

Adds about 4 × FLA for the first cycles.

Check equipment rated

Result · NEC 2023

12.05 kA available at the fault

With a 500 kVA three-phase transformer at 480 V with 4.6 %Z, infinite primary through 75 ft of 2 sets of 350 kcmil copper, a bolted three-phase fault draws 12.05 kA (13.07 kA at the source): equipment there needs a rating of at least 14 kA (NEC 2023 110.9 and 110.10).

At the fault point: 12.05 kA
At the transformer secondary: 13.07 kA
Line-to-line fault: 10.44 kA
With %Z 10 % low: 13.27 kA
Transformer full-load current: 601.4 A
X/R at the fault: 2.88
Smallest common rating: 14 kA
14 kA equipment: Adequate

Nameplate %Z has a tolerance: ±10 % for UL 1561 dry-type transformers and ±7.5 % for ANSI two-winding units. With %Z 10 % low the fault current is 13.27 kA. Turn on the worst-case option to use it.
The utility is treated as an unlimited source (infinite bus), the conservative assumption. With the utility’s primary fault current the result is a little lower.
At service equipment other than in dwelling units, 110.24(A) requires a field marking of the available fault current and the date of the calculation, and 110.24(B) a recalculation when the system changes. Switchboards and panelboards outside one- and two-family dwellings are marked too (408.6).

Show the work

1. Transformer full-load current
$I_{FLA} = \frac{S}{\sqrt{3} \cdot V} = \frac{500,000 VA}{\sqrt{3} \cdot 480 V} = 601.4 A$
2. Transformer impedance in ohms
$Z_{T} = \frac{4.6}{100} \cdot \frac{{(480 V)}^{2}}{500,000 VA} = 21.2 mΩ$
X/R = 3.18 (Eaton’s typical value for a liquid-filled unit of this size; Eaton, Consulting Application Guide CA08104001E, Section 1.7 “Approximate Impedance Data” (Aug. 2017)): R = 6.359 mΩ, X = 20.22 mΩ.
3. Fault current at the secondary, infinite primary
$I_{source} = \frac{480 V}{\sqrt{3} \cdot |Z|} = \frac{480 V}{\sqrt{3} \cdot 21.2 mΩ} = 13,074 A$
The same as I_FLA × 100 ÷ %Z = 601.41 × 100 ÷ 4.6 = 13,074 A.
4. Conductor resistance: 350 kcmil copper, 2 sets, PVC raceway
$R_{C} = 0.03186 Ω/kft \cdot \frac{75 ft}{1,000 \cdot 2} = 1.195 mΩ$
Table 9 gives 0.038 Ω per 1000 ft at 75 °C; at 25 °C: × [1 + 0.00323 × (25 − 75)] = 0.03186 (Table 8 note coefficient).
5. Conductor reactance (Table 9, PVC or aluminum raceway)
$X_{C} = 0.04 Ω/kft \cdot \frac{75 ft}{1,000 \cdot 2} = 1.5 mΩ$
6. Total impedance to the fault (resistances and reactances added separately)
$|Z| = \sqrt{{(7.554 mΩ)}^{2} + {(21.72 mΩ)}^{2}} = 23 mΩ$
7. Three-phase bolted fault at the end of the run
$I_{fault} = \frac{480 V}{\sqrt{3} \cdot 23 mΩ} = 12,051 A$
X/R at the fault: 2.88.
8. Cross-check: point-to-point method (impedances added as magnitudes)
11.99 kA (−0.5 % against the impedance method).
Adding magnitudes ignores the different angles of the source and conductor impedances, so it never reads higher; the gap is largest on long runs of small conductors.

References: NEC 110.9 (2023): interrupting rating at least the available fault current · NEC 110.10 (2023): short-circuit current ratings and coordination · NEC 110.24 (2023): field marking at non-dwelling service equipment · NEC 408.6 (2023): switchboard and panelboard ratings · NEC Chapter 9, Table 9 (2023): conductor R and X · Eaton Bussmann SPD, point-to-point method: cross-check, motor and tolerance factors

Bolted (zero-impedance) fault, symmetrical RMS current: the value equipment ratings are stated in. Arcing faults are lower; the first-cycle peak is higher.
Conductor R and X from NEC Chapter 9 Table 9 (600 V, three single conductors in conduit, 60 Hz), with R corrected from 75 °C to 25 °C for the maximum fault current.
Transformer impedance from the nameplate %Z, split into R and X by the X/R ratio; the utility source, when given, is a pure reactance.
Line-to-ground faults need zero-sequence data and are not calculated. Busway, switches and breakers add a little impedance that is ignored here (conservative).

Report a problem with this result

How to use the available fault current calculator

Choose the source. For Transformer, enter the kVA rating, the nameplate impedance (%Z) and the secondary voltage, line-to-line (240 for a 120/240 V single-phase secondary). Leave the X/R ratio on the typical value unless the nameplate or test report gives %R and %X. If the utility has told you the fault current available at the primary, enter it with the primary voltage; otherwise the primary is treated as an unlimited source, which gives the highest current. For Known fault current, enter the current available at the start of the run, for example the value on a service label or in a utility letter, and its X/R if you have it.

Then describe the conductors to the fault point: length, size, copper or aluminum, the raceway (steel raises the reactance), and the number of parallel sets. Add running motors if there are any, and pick an equipment rating to check. The result gives the bolted fault current at the source and at the fault point, the line-to-line value, the worst case with the transformer impedance 10 % low, the X/R at the fault, the smallest common equipment rating, and a cross-check by the point-to-point method.

The impedance method

A bolted fault is limited only by the impedance between the source and the fault, so the job is to add those impedances up. The transformer comes first. Its %Z is the share of rated voltage that drives full-load current through a shorted secondary, so it converts to ohms on the secondary side as:

Z_{T} = \frac{%Z}{100} \cdot \frac{V^{2}}{S}

The X/R ratio splits that impedance into resistance and reactance. The utility, when its primary fault current is known, adds a source impedance of V² divided by its short-circuit power (√3 × primary volts × primary amps), taken as pure reactance. Each conductor adds the resistance and reactance per 1000 ft from NEC Chapter 9 Table 9, multiplied by the length and divided by the number of parallel sets; for a single-phase fault the conductor impedance counts twice, out and back. Resistances and reactances are added separately and combined at the end:

|Z| = \sqrt{{R_{total}}^{2} + {X_{total}}^{2}}

I_{3φ} = \frac{V}{\sqrt{3} \cdot |Z|}

At the terminals of a transformer on an infinite source this reduces to the familiar shortcut, full-load current × 100 ÷ %Z. A 75 kVA, 480 to 208 V transformer with 2 %Z carries 208.2 A at full load and can deliver 10,409 A into a bolted fault at its secondary terminals.

Worked examples

A 500 kVA service transformer and a 75 ft feeder

A 480Y/277 V pad-mount transformer (4.6 %Z, a typical value for this size) feeding the main switchboard through two sets of 350 kcmil copper in PVC. Is 14 kA equipment enough?

1. Transformer full-load current
$I_{FLA} = \frac{S}{\sqrt{3} \cdot V} = \frac{500,000 VA}{\sqrt{3} \cdot 480 V} = 601.4 A$
2. Transformer impedance in ohms
$Z_{T} = \frac{4.6}{100} \cdot \frac{{(480 V)}^{2}}{500,000 VA} = 21.2 mΩ$
3. Fault current at the secondary, infinite primary
$I_{source} = \frac{480 V}{\sqrt{3} \cdot |Z|} = \frac{480 V}{\sqrt{3} \cdot 21.2 mΩ} = 13,074 A$
4. Conductor resistance: 350 kcmil copper, 2 sets, PVC raceway
$R_{C} = 0.03186 Ω/kft \cdot \frac{75 ft}{1,000 \cdot 2} = 1.195 mΩ$
5. Conductor reactance (Table 9, PVC or aluminum raceway)
$X_{C} = 0.04 Ω/kft \cdot \frac{75 ft}{1,000 \cdot 2} = 1.5 mΩ$
6. Total impedance to the fault (resistances and reactances added separately)
$|Z| = \sqrt{{(7.554 mΩ)}^{2} + {(21.72 mΩ)}^{2}} = 23 mΩ$
7. Three-phase bolted fault at the end of the run
$I_{fault} = \frac{480 V}{\sqrt{3} \cdot 23 mΩ} = 12,051 A$
8. Cross-check: point-to-point method (impedances added as magnitudes)
11.99 kA (−0.5 % against the impedance method).

75 kVA, 480 to 208 V, 2 %Z, at the secondary terminals

The textbook case, with the panel right at the transformer.

1. Transformer full-load current
$I_{FLA} = \frac{S}{\sqrt{3} \cdot V} = \frac{75,000 VA}{\sqrt{3} \cdot 208 V} = 208.2 A$
2. Transformer impedance in ohms
$Z_{T} = \frac{2}{100} \cdot \frac{{(208 V)}^{2}}{75,000 VA} = 11.54 mΩ$
3. Fault current at the secondary, infinite primary
$I_{source} = \frac{208 V}{\sqrt{3} \cdot |Z|} = \frac{208 V}{\sqrt{3} \cdot 11.54 mΩ} = 10,409 A$
4. Cross-check: point-to-point method (impedances added as magnitudes)
10.41 kA, the same as the impedance method (one impedance only).

With a 75 kVA three-phase transformer at 208 V with 2 %Z, infinite primary, a bolted three-phase fault draws 10.41 kA: equipment there needs a rating of at least 14 kA (NEC 2023 110.9 and 110.10).

The same transformer at the low end of its impedance tolerance

UL 1561 allows ±10 % on the nameplate %Z; worst-case studies use 90 % of it.

1. Transformer full-load current
$I_{FLA} = \frac{S}{\sqrt{3} \cdot V} = \frac{75,000 VA}{\sqrt{3} \cdot 208 V} = 208.2 A$
2. Transformer impedance in ohms (%Z × 0.9, the −10 % tolerance)
$Z_{T} = \frac{1.8}{100} \cdot \frac{{(208 V)}^{2}}{75,000 VA} = 10.38 mΩ$
3. Fault current at the secondary, infinite primary
$I_{source} = \frac{208 V}{\sqrt{3} \cdot |Z|} = \frac{208 V}{\sqrt{3} \cdot 10.38 mΩ} = 11,566 A$
4. Cross-check: point-to-point method (impedances added as magnitudes)
11.57 kA, the same as the impedance method (one impedance only).

With a 75 kVA three-phase transformer at 208 V with 2 %Z (taken 10 % low), infinite primary, a bolted three-phase fault draws 11.57 kA: equipment there needs a rating of at least 14 kA (NEC 2023 110.9 and 110.10).

A subpanel 150 ft from a 42 kA service

The service label says 42 kA. The subpanel is fed by 1/0 AWG copper in PVC: how much is left at its bus?

1. Source impedance behind the known fault current
${|Z|}_{S} = \frac{480 V}{\sqrt{3} \cdot 42,000 A} = 6.598 mΩ$
2. Conductor resistance: 1/0 AWG copper, PVC raceway
$R_{C} = 0.1006 Ω/kft \cdot \frac{150 ft}{1,000 \cdot 1} = 15.09 mΩ$
3. Conductor reactance (Table 9, PVC or aluminum raceway)
$X_{C} = 0.044 Ω/kft \cdot \frac{150 ft}{1,000 \cdot 1} = 6.6 mΩ$
4. Total impedance to the fault (resistances and reactances added separately)
$|Z| = \sqrt{{(16.39 mΩ)}^{2} + {(13.07 mΩ)}^{2}} = 20.96 mΩ$
5. Three-phase bolted fault at the end of the run
$I_{fault} = \frac{480 V}{\sqrt{3} \cdot 20.96 mΩ} = 13,221 A$
6. Cross-check: point-to-point method (impedances added as magnitudes)
12.01 kA (−9.1 % against the impedance method).

With 42 kA available at 480 V through 150 ft of 1/0 AWG copper, a bolted three-phase fault draws 13.22 kA (42 kA at the source): equipment there needs a rating of at least 14 kA (NEC 2023 110.9 and 110.10).

A house fed from a 50 kVA pole transformer

120/240 V, 1.9 %Z, and an 80 ft 4/0 aluminum service lateral to the panel. Single-phase faults count the conductors out and back.

1. Transformer full-load current
$I_{FLA} = \frac{S}{V} = \frac{50,000 VA}{240 V} = 208.3 A$
2. Transformer impedance in ohms
$Z_{T} = \frac{1.9}{100} \cdot \frac{{(240 V)}^{2}}{50,000 VA} = 21.89 mΩ$
3. Fault current at the secondary, infinite primary
$I_{source} = \frac{240 V}{|Z|} = \frac{240 V}{21.89 mΩ} = 10,965 A$
4. Conductor resistance: 4/0 AWG aluminum, PVC raceway, out and back
$R_{C} = 2 \cdot 0.0835 Ω/kft \cdot \frac{80 ft}{1,000 \cdot 1} = 13.36 mΩ$
5. Conductor reactance (Table 9, PVC or aluminum raceway)
$X_{C} = 2 \cdot 0.041 Ω/kft \cdot \frac{80 ft}{1,000 \cdot 1} = 6.56 mΩ$
6. Total impedance to the fault (resistances and reactances added separately)
$|Z| = \sqrt{{(24.96 mΩ)}^{2} + {(25.12 mΩ)}^{2}} = 35.41 mΩ$
7. Line-to-line bolted fault at the end of the run
$I_{fault} = \frac{240 V}{35.41 mΩ} = 6,777 A$
8. Cross-check: point-to-point method (impedances added as magnitudes)
6.527 kA (−3.7 % against the impedance method).

With a 50 kVA single-phase transformer at 240 V with 1.9 %Z, infinite primary through 80 ft of 4/0 AWG aluminum, a bolted line-to-line fault draws 6.777 kA (10.96 kA at the source): equipment there needs a rating of at least 10 kA (NEC 2023 110.9 and 110.10).

How fast fault current falls along a feeder

Conductor impedance adds quickly. The table follows the 500 kVA example down four feeder sizes. A small feeder cuts the current in half within a couple of hundred feet; a large or paralleled feeder barely reduces it. That is why equipment near the service needs the highest ratings and why a remote panel can often use 10 kA breakers even on a large service.

Three-phase bolted fault current along copper feeders in PVC from a 500 kVA, 480 V, 4.6 %Z transformer (infinite primary, X/R 3.18, conductors at 25 °C), computed by the calculator.
Feeder	0 ft	25 ft	50 ft	100 ft	200 ft	400 ft
1/0 AWG	13.1 kA	12 kA	11 kA	9.36 kA	7.05 kA	4.62 kA
4/0 AWG	13.1 kA	12.3 kA	11.5 kA	10.3 kA	8.4 kA	6.08 kA
350 kcmil	13.1 kA	12.4 kA	11.7 kA	10.6 kA	8.95 kA	6.77 kA
2 × 350 kcmil	13.1 kA	12.7 kA	12.4 kA	11.7 kA	10.6 kA	8.95 kA

Fault current at common transformer sizes

For quick estimates, the bolted fault at the terminals depends only on the rating, the voltage and %Z. Distribution transformers range from under 2 %Z for small, efficient units to 5.75 %Z and more for large pad-mounts, so look up your nameplate rather than assuming.

Three-phase transformers, bolted fault at the secondary terminals with an infinite primary, computed by the calculator. Read across at your nameplate %Z; the current is inversely proportional to it.
kVA	208 V full load	208 V at 3 %Z	208 V at 5.75 %Z	480 V full load	480 V at 3 %Z	480 V at 5.75 %Z
45	124.9 A	4.16 kA	2.17 kA	54.13 A	1.8 kA	941 A
75	208.2 A	6.94 kA	3.62 kA	90.21 A	3.01 kA	1.57 kA
112.5	312.3 A	10.4 kA	5.43 kA	135.3 A	4.51 kA	2.35 kA
150	416.4 A	13.9 kA	7.24 kA	180.4 A	6.01 kA	3.14 kA
225	624.5 A	20.8 kA	10.9 kA	270.6 A	9.02 kA	4.71 kA
300	832.7 A	27.8 kA	14.5 kA	360.8 A	12 kA	6.28 kA
500	1,388 A	46.3 kA	24.1 kA	601.4 A	20 kA	10.5 kA
750	2,082 A	69.4 kA	36.2 kA	902.1 A	30.1 kA	15.7 kA
1,000	2,776 A	92.5 kA	48.3 kA	1,203 A	40.1 kA	20.9 kA
1,500	4,164 A	139 kA	72.4 kA	1,804 A	60.1 kA	31.4 kA
2,000	5,551 A	185 kA	96.5 kA	2,406 A	80.2 kA	41.8 kA
2,500	6,939 A	231 kA	121 kA	3,007 A	100 kA	52.3 kA

Tolerance, X/R and other conservative choices

Impedance tolerance. UL 1561 dry-type transformers of 25 kVA and up may differ from their nameplate %Z by ±10 %, and ANSI two-winding units by ±7.5 %, per Eaton’s Bussmann handbook. The worst-case option multiplies %Z by 0.9, raising the current by 11 %: the 75 kVA example goes from 10.4 kA to 11.6 kA. The calculator always shows that figure next to the nameplate result.
X/R ratio. It doesn’t change the current at the transformer terminals, but it decides how the transformer and cable impedances combine further out. Without nameplate data the calculator uses typical values: for three-phase units the higher of Eaton’s typical figures for liquid-filled and dry-type transformers of the size (3.18 at 500 kVA, 6.61 at 2500 kVA; below 112.5 kVA the smallest liquid-filled figure, 2.75, is used), and for single-phase units the values suggested in IEEE Std 242. A higher X/R gives the higher current, so the choice errs on the safe side.
Conductor temperature. Table 9 tabulates resistance at 75 °C. A fault on a lightly loaded feeder starts with cold conductors, so the default moves resistance to 25 °C with the Table 8 temperature coefficient. Eaton’s point-to-point C values correspond to the same basis: their 14 AWG copper value is the Table 8 resistance at 25 °C. On 200 ft of 2 AWG copper from the 500 kVA example the difference is 5.76 kA against 5.22 kA.
Infinite primary. Without utility data the source is unlimited, the highest possible current. Utility data lowers the answer, and utilities can change their system later, which is why the result should be rechecked when the supply changes.
Motors. Running induction motors feed a fault for a few cycles. Eaton estimates the contribution at four times their full-load current (four to six is common practice), added at every fault location.
Supply voltage. The calculation uses the nominal voltage. Eaton notes that a supply 10 % high raises the fault current by 10 %; studies that need the extreme value apply that factor too.

Comparing the result with equipment ratings

NEC 110.9 requires devices that interrupt fault current, breakers and fuses, to have an interrupting rating at least equal to the current available at their line terminals. 110.10 extends the idea to everything else on the circuit: the short-circuit current ratings of panels, controllers and equipment must be coordinated with the available fault current so a fault is cleared without extensive damage. Common ratings run 10, 14, 18, 22, 25, 35, 42, 50, 65, 100, 150 and 200 kA; the calculator names the smallest that covers the result. Where a lower-rated breaker is used downstream of a fuse or breaker it has been tested with, a listed series-rated combination is the alternative.

Breaker ratings also assume an X/R ratio. Molded-case breakers are tested at power factors that correspond to an X/R of about 1.7 for ratings of 10 kA or less, 3.2 up to 20 kA and 4.9 above 20 kA. Close to large transformers the X/R can be higher, and the breaker’s effective rating drops: at an X/R of 8, a 65 kA molded-case breaker is good for about 91 % of its rating by the derating formula in Eaton’s consulting application guide. The calculator warns when that matters.

What the NEC requires, edition by edition

110.9 and 110.10 carry the same numbers in all four editions. The 2023 edition reworded 110.9 slightly to tie it to the defined term “available fault current”.
110.24, field marking. Service equipment in other than dwelling units must be marked with the available fault current and the date of the calculation, and the calculation must be documented. When the installation changes in a way that affects it, the value is verified or recalculated and the label updated. The 2017 text said “maximum available fault current”; the 2020 edition dropped “maximum”, defined available fault current in Article 100, added a note that utilities publish fault current values, and excused supervised industrial installations from the marking. In 2026 the label has to comply with the general marking rules of 110.21(B).
408.6 arrived in 2020: switchboards, switchgear and panelboards need a short-circuit current rating not less than the available fault current, and outside one- and two-family dwellings the available fault current and the date are field marked at the point of supply. The 2026 edition turns it into a list that also requires the equipment’s short-circuit current rating to be marked, the calculation documented, the value updated after changes, and replacement breakers rated for the marked current.

Check which edition applies where you work on the NEC adoption map, and see what else changed in the 2026 NEC.

The point-to-point method as a cross-check

Eaton’s Bussmann point-to-point method is the hand calculation most electricians learn. It works on currents instead of ohms: a multiplier of 100 ÷ %Z at the transformer, then for each conductor run an “f” factor, f = 1.732 × L × I ÷ (C × n × V), and a multiplier M = 1 ÷ (1 + f). Algebraically that is the same as adding the impedance magnitudes as plain numbers, ignoring their angles. A transformer is mostly reactance and a small conductor mostly resistance, so adding magnitudes overstates the total and the point-to-point answer is never higher than the impedance method’s. Near the transformer the two agree within a few percent; on a long run of small conductors, where the source and the cable impedances are similar in size but far apart in angle, the gap can reach 10 % or more. The 150 ft subpanel example above shows it: 12 kA by point-to-point against 13.2 kA by the impedance method.

Eaton’s published System A (1500 kVA, 480 V, 3.5 %Z × 0.9, copper in steel conduit, no motors) against this calculator.
Fault point	Eaton, published	Our point-to-point, Eaton’s C values	Our impedance method, Table 9
X1: transformer secondary	57,279 A	57,279 A	57,277 A (0.0 %)
X2: 25 ft, 6 sets of 500 kcmil	55,137 A	55,139 A	54,850 A (−0.5 %)
X3: then 50 ft of 500 kcmil	38,067 A	38,068 A	36,146 A (−5.0 %)

With Eaton’s own C values our point-to-point steps reproduce the published numbers. The impedance method with NEC Table 9 data agrees within 1 % at the switchboard and reads about 5 % lower 50 ft further on, because Table 9’s reactance for conductors in steel conduit is higher than the IEEE cable data behind Eaton’s C values. Where two published methods differ by a few percent, the equipment rating decision rarely changes; when it does, use the higher figure or a full short-circuit study.

Common mistakes and limits

Confusing full-load current with fault current. At the terminals the fault current is 100 ÷ %Z times full load, roughly 17 to 60 times for common transformers.
Leaving out the tolerance on a tight result. If the nameplate result is just under a rating, the −10 % case is likely over it.
Assuming 10 kA everywhere. Residential panels near large pad-mounts, and any panel within a few feet of a commercial transformer, can see far more.
Copying a fault current from the service to a subpanel without the feeder. That is safe but expensive; the calculator shows how much a feeder takes off.
Treating the result as an arc-flash study. Arcing current is lower than bolted current, and incident energy needs a study under NFPA 70E; this calculator gives the bolted value used for equipment ratings.
Line-to-ground faults need the zero-sequence impedance and aren’t calculated here. On a 120/240 V center-tapped secondary, a line-to-neutral fault at the terminals can be about 1.5 times the line-to-line value, and the calculator flags it.

For the transformer itself, full-load currents and %Z are on the transformer calculator. IEC and BS 7671 installations use the prospective fault current calculator.

Method and verification

The calculator is a direct impedance calculation in ohms, with conductor data from NEC Chapter 9 Table 9 and the temperature coefficient from the Table 8 notes. Its tests reproduce the hand case above (10,409 A), three EC&M worked examples (a 100 kVA single-phase unit: 16,667 A against the article’s 16,680 A, which rounds full-load current to 417 A; a 2,000 kVA transformer on a 50,000 kVA utility: 26,729 A against 26,729 A; a 2,500 kVA unit: 62,975 A against 62,975 A), and all of Eaton’s published point-to-point systems, three-phase and single-phase, within the tolerances explained above. Property tests check that longer runs and smaller conductors always lower the current and that the point-to-point figure never exceeds the impedance figure.

Bolted faults on systems up to 1,000 V with Table 9 conductors (600 V cable data). Not a substitute for a short-circuit study where one is required. Updated 2026-10-14. Found a problem? Report it and we log every fix publicly.

Questions people ask

How do you calculate available fault current?

Convert the transformer %Z to ohms (%Z ÷ 100 × V² ÷ VA), add the utility source impedance if known and the conductor resistance and reactance from NEC Chapter 9 Table 9, then divide: I = V ÷ (√3 × |Z|) for a three-phase fault. At the terminals with an infinite source, it is full-load current × 100 ÷ %Z.

How much fault current can a 75 kVA transformer deliver?

At 208 V three-phase with 2 %Z, 10,409 A at the secondary terminals with an unlimited primary, or 11,566 A with the impedance 10 % low. That needs 14 kA equipment. A higher %Z gives proportionally less.

Does the NEC require fault current to be marked?

At service equipment in other than dwelling units, yes: 110.24 requires the available fault current and the date of the calculation to be field marked, and updated when the system changes. From 2020, 408.6 also requires it on switchboards and panelboards outside one- and two-family dwellings.

Why multiply transformer impedance by 0.9?

Nameplate %Z has a manufacturing tolerance: ±10 % for UL 1561 dry-type transformers (±7.5 % for ANSI two-winding units). Using 90 % of %Z gives the highest current the transformer could deliver, the worst case for equipment ratings.

Do motors add to fault current?

Yes, briefly. Running induction motors act as generators for a few cycles after a fault. A common estimate is four times their total full-load current, added at the fault point.

What is the difference between interrupting rating and SCCR?

The interrupting rating (AIC) is the current a breaker or fuse can safely interrupt (NEC 110.9). The short-circuit current rating is what equipment such as a panel or controller can withstand while its protective device clears the fault (110.10). Both must be at least the available fault current.