SizingKitElectrical · fault current
Short circuit current calculator, with its assumptions printed
Give this page a transformer’s kVA, secondary voltage and nameplate impedance and the conductors leaving it, and it returns the bolted fault current available at both ends of that run — at the transformer terminals and at the equipment, in amperes and kiloamperes side by side. It is a point-to-point calculation for picking an interrupting rating under NEC 110.9, and it prints the five things the method leaves out and which direction each one pushes the answer. It never computes a load current, and it is not a short-circuit study.
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- No signup
- Point-to-point method
- Motor contribution included
- Every assumption printed
The source, the run and the device
Everything the point-to-point method needs, and nothing it does not: the transformer enters as an impedance, not as a current.
Off the transformer’s own plate. There is no default here because a transformer’s impedance is not predictable from its kVA.
Default 10% — A listing allowance rather than a measurement. The transformer's own factory test report carries its measured impedance to two decimals, and that figure beats any tolerance assumption. Set this to 0 when you are working from the test report.
Default 75 °C — NEC Chapter 9, Table 8 heading: 'Direct-Current Resistance at 75°C (167°F)'. A cold conductor has less resistance and passes more.
50 ft · 15.24 m
Default 4× — Varies with the motor's own subtransient reactance and decays within a few cycles. Synchronous motors contribute more and for longer. Where the motor load is large relative to the transformer, take the figure from the motor data rather than from this multiplier.
Read it off the device or its listing. This site carries no list of manufactured interrupting ratings, because no standard publishes one.
Available fault current
At the transformer terminals
At the far end of the run
| Base impedance, V² ÷ (kVA × 1000) | 0.96142 Ω |
|---|---|
| Transformer impedance at 3.15% | 0.03028 Ω |
| Conductor resistance, one way | 0.00298 Ω |
| Multiplier down the run, M | 0.91 |
| Motor feedback at 4× FLA | not included |
| Device rating you entered, minus the total | +18,390 A |
The comparison, stated as arithmetic. This calculation puts 3,610 A at the point, and the rating you entered is 22,000 A — a difference of 18,390 A in favor of the device. NEC 110.9 is the article: equipment intended to interrupt current at fault levels shall have an interrupting rating at nominal circuit voltage at least equal to the current available at its line terminals. 240.86 is the other route, where a listed series combination lets a lower-rated downstream device stand behind a higher-rated one — and it only works with the exact combination on the listing, marked on the enclosure. This page compares two numbers; it does not find that any equipment is adequately rated.
What this method leaves out, and which way each error runs:
- The primary is treated as infinite — the utility, its transformer and the primary conductors contribute no impedance at all. This raises the answer, and it is the assumption every published point-to-point procedure makes when the utility has not supplied a primary available fault current. Ask them for one and the real figure is lower.
- Conductor reactance is not included. NEC Chapter 9 Table 9 tabulates AC resistance and reactance for steel and PVC raceway and this site does not reproduce it, so only the DC resistance of Table 8 is used. Leaving reactance out lowers the modeled impedance and raises the current — safe for choosing an interrupting rating, and progressively less exact above about 1/0, which is the same size at which this engine already warns off its voltage-drop shortcut.
- The transformer and conductor impedances are added arithmetically rather than as vectors, which is what the point-to-point method does. A mostly reactive transformer impedance and a mostly resistive conductor do not add head to tail, so this overstates the total and understates the current — partly canceling the error above it.
- The fault is bolted and symmetrical. An arcing fault passes less current, and the asymmetrical first half-cycle passes more; neither is what a device's symmetrical interrupting rating is compared against.
- On a single-phase transformer with a center-tapped secondary the line-to-neutral fault can run about half again the line-to-line figure, because it sees only half a winding. That case needs the half-winding impedance, which is not on the nameplate, so this page will not compute it.
Conductor resistance comes from Computed as ρ × 1000 ÷ area in circular mils, at 75 °C, from 10.371 Ω·cmil/ft for 100% IACS copper and 17.002 Ω·cmil/ft for 61.0% IACS aluminum at 20 °C, corrected by their standard temperature coefficients. Matches the solid-conductor column of NEC Chapter 9, Table 8 (12 AWG copper 1.93, 14 AWG 3.07, 8 AWG 0.764, 12 AWG aluminum 3.18 Ω/kFT). A short-circuit study is a signed engineering document that models the utility source, every transformer and cable in the path, the motor load and the decay of its contribution, and settles arc-flash incident energy as well. This page is arithmetic over published tables and is not that document, does not replace it, and issues no finding that an installation is compliant or safe.
Worksheet for the file
Print this for the job folder. It is not the 110.24(A) field marking: that marking carries the maximum available fault current from the study of record and the date the study was performed, and 110.24(B) requires both to be revisited whenever a modification increases what is available.
| System | Three-phase |
|---|---|
| Transformer | 45 kVA, 208 V secondary, 3.5% marked impedance, 10% tolerance taken off |
| Transformer impedance | 0.03028 Ω per phase |
| Available at the terminals | 3,965 A |
| Run | 1 × 4/0 AWG copper, 50 ft (15.24 m), at 75 °C |
| Available at the point | 3,610 A |
| Motor contribution included | None |
| Device interrupting rating | 22,000 A |
| Calculated by / date |
Method: point-to-point, infinite primary. Impedance tolerance — The Bussmann point-to-point short-circuit procedure, which states that UL-listed transformers of 25 kVA and larger carry a ±10% tolerance on the nameplate impedance, and takes the negative end of it for a fault calculation because a lower impedance yields a higher available current. Motor contribution — IEEE Std 141 (the Red Book) industrial short-circuit practice: the symmetrical contribution of an induction motor to a fault is estimated at four times its rated full-load current, which is the reciprocal of an assumed 25% subtransient reactance.
How to find the fault current at a 45 kVA transformer's panel
The example the fields open on: a 45 kVA 480–208Y/120 V dry-type transformer feeding a panel 50 ft away in 4/0 copper.
Turn the transformer into an impedance
Enter the kVA, the secondary line-to-line voltage and the impedance marked on the plate. The page computes the base impedance as V² ÷ (kVA × 1000) and takes the marked percentage of it, which is why it never needs a full-load current. Leave the tolerance field at 10%: a UL-listed transformer of 25 kVA or more may come out of the factory a tenth below its marked impedance, and the low end is the one a fault study models because it passes more current.
Add the run between the transformer and the equipment
Pick the conductor metal, size, number of parallel sets per phase and the one-way length — that field takes 120, 120 ft or 35 m. Resistance comes from the NEC Chapter 9 Table 8 basis at 75 °C, and the conductor temperature is an editable default because a cold conductor has measurably less resistance and passes more. On a three-phase system only the one-way length appears; on a single-phase one the current goes out and back, so the loop is doubled.
Add the motors, then compare against the device
If motors are running downstream of the fault point, put their combined full-load amperes in — for the first few cycles a spinning induction motor is a generator, and 4 × full-load current is the standard estimate of what it feeds back. Then type the interrupting rating marked on the device you intend to use. The page states the difference between the two figures as arithmetic and cites 110.9; the decision, and the study behind it, are yours.
Technical specifications
| Method | Point-to-point with an infinite primary — the utility and everything upstream of the transformer contribute no impedance. That is the standard assumption where the utility has not published a primary available fault current, and it makes the answer higher than reality rather than lower. |
|---|---|
| Transformer model | Base impedance V² ÷ (kVA × 1000), times the nameplate percentage, times one minus the tolerance you set. A 45 kVA 208 V transformer at 3.5% marked and 10% tolerance is 0.0303 Ω per phase and 3,965 A at its own terminals. |
| Conductor model | DC resistance only, computed as 10.371 Ω·cmil/ft for copper and 17.002 for aluminum at 20 °C, corrected to the temperature you set and defaulting to 75 °C. NEC Chapter 9 Table 9 reactance is not reproduced on this site, so the model loses exactness above about 1/0 — the same boundary this engine already flags for its voltage-drop shortcut. |
| Conductor sizes | 21 sizes from 14 AWG to 1000 kcmil, in copper and aluminum, with 1 to 12 parallel sets per phase. |
| Motor contribution | An editable 4 × the combined full-load amperes of motors running downstream, the reciprocal of an assumed 25% subtransient reactance. Left out entirely if the field is empty, and it decays within a few cycles, so it matters for the interrupting duty and not for anything slower. |
| Interrupting rating input | Typed from the device label in amperes or kiloamperes. This site carries no list of manufactured interrupting ratings, because no standard publishes one — the ratings you see on shelves are a market fact, not a code table. |
| Assumptions declared | Five, each with the direction of its error: infinite primary, no conductor reactance, arithmetic rather than vector impedance addition, bolted symmetrical fault, and the line-to-neutral case on a center-tapped single-phase secondary, which this page refuses. |
| Where it runs | In the browser. Transformer nameplates and panel data are frequently a client's confidential information, and nothing typed here is uploaded, logged or retained. |
Frequently asked questions
What does an infinite primary assumption cost me?
It makes the answer high, usually by 5% to 15% on a service transformer and less on a small dry-type one deep inside a building. Treating the primary as infinite means the utility source, its transformer and the primary conductors are all modeled as zero impedance, so the only thing limiting the fault is the transformer you entered and the wire after it. Utilities will supply the actual primary available fault current on request, and putting a real number in place of infinity always lowers the result — which is why the assumption is the safe one to make while you wait for their answer.
The plate says 3.5% impedance. Why does the calculation use 3.15%?
Because the marked figure carries a tolerance and a fault study takes the end of it that passes more current. UL-listed transformers of 25 kVA and larger are allowed a ±10% spread on the nameplate impedance, so a transformer marked 3.5% may genuinely be 3.15%, and a device chosen against the marked figure would be short by that much on the day. If you have the factory test report the measured impedance is on it to two decimals, and in that case set the tolerance field to zero and use the real number.
Do motors really add to the fault current?
Yes, for the first few cycles, and it is often what pushes a panel past the rating the transformer alone would have justified. A loaded induction motor keeps turning after the voltage collapses and its trapped rotor flux drives current back toward the fault, at roughly four times its own full-load rating — the reciprocal of a 25% subtransient reactance. The contribution decays within a handful of cycles, so it is real for the interrupting duty of a device and irrelevant to anything with a longer time frame. A plant with 300 A of motors running adds about 1,200 A to whatever the transformer supplies.
My breaker is marked 10 kA but the panel label says 22 kA series rated. Which one applies?
The series rating, but only if the exact combination on that listing is what is actually installed. NEC 240.86 permits a downstream device with an interrupting rating below the available current where it is part of a tested series combination with the upstream device, and the equipment has to be legibly marked in the field to say so. The catch is that the listing names both devices by catalog number, so replacing the upstream breaker with an equivalent from another manufacturer voids it, and a motor contribution of more than 1% of the downstream device's rating disqualifies the arrangement outright.
Why is the line-to-neutral fault on a 120/240 V transformer higher than the line-to-line one?
Because it sees only half a winding, and half a winding has less than half the impedance. On a center-tapped single-phase secondary the line-to-neutral fault can run roughly one and a half times the line-to-line figure, which is the opposite of what people expect from a lower voltage. Working it properly needs the half-winding impedance, and that is not a number printed on a nameplate, so this page computes the line-to-line case and says plainly that it will not guess the other.
Does the available fault current have to be marked on the equipment?
Yes for service equipment other than at a dwelling unit — NEC 110.24(A) requires a field marking of the maximum available fault current and the date the calculation was performed, and 110.24(B) requires both to be revisited when a modification increases what is available. That marking has to reflect the study of record, not a figure produced on a phone in a switchroom, which is why the printable block at the foot of this page is labeled a worksheet for the job file rather than a label for the enclosure.
Why does 40 ft of conductor knock so much off the number?
Because at these current levels the conductor's own resistance is comparable to the transformer's impedance, and the two divide the driving voltage between them. A 45 kVA 208 V transformer at 3.15% is about 0.030 Ω, and 50 ft of 4/0 copper is about 0.003 Ω — a tenth of it, and that alone takes the available current from 3,965 A to about 3,610 A. Halve the conductor size or double the length and the effect roughly doubles, which is the single most useful lever an installer has for getting a panel under a device's rating.
About interrupting rating, and what a calculation is not
An overcurrent device carries two ratings that are easy to confuse and are about completely different events. Its ampere rating is a promise about steady-state current and is what 240.4 and Article 430 are for. Its interrupting rating is a promise about a fault: the current the device can break at its rated voltage without failing, and NEC 110.9 requires it to be at least equal to the current available at the device’s line terminals. A device asked to interrupt more than it is rated for does not simply fail to clear — it can rupture, and the energy released is the reason the marking exists. 110.10 is the related requirement on everything else in the circuit: the components have to withstand the let-through of the device ahead of them, which is a different question again and is where a current-limiting fuse class earns its price.
The arithmetic itself is short. A transformer’s impedance is a percentage of its own base impedance, so V² ÷ (kVA × 1000) times the plate percentage gives ohms per phase, and the driving voltage divided by that gives the current at its terminals. Every foot of conductor after it adds resistance in series and takes the current down. What separates a usable answer from a plausible one is being honest about the four things the short version skips — the primary source, conductor reactance, vector rather than scalar addition of impedances, and the difference between a bolted and an arcing fault — and about which way each one moves the number. Those are printed under the result rather than tucked into a disclaimer. If you need the transformer’s full-load current rather than its impedance, that is the kVA to amps calculator, and the transformer size calculator picks the rating in the first place.
This page is not a short-circuit study and cannot become one. A study models the utility source, every transformer, cable and bus in the path, the decay of the motor contribution, and the arc-flash incident energy that follows from all of it; it is signed by an engineer and it is the document 110.24 wants a marking derived from. What a calculation like this one is genuinely good for is deciding whether a device is even in the right range before you order it, and finding out that a panel at the end of a long run is comfortably below what a shorter run would have put on it. Once you know the available current, the class and rating of the device itself are on the fuse size calculator, the ampere rating for the load is on the breaker size calculator, and the conductors carrying it are sized on the wire size calculator against the run length the voltage drop calculator checks. Nothing here is a finding that any equipment is adequately rated or that an installation is safe.
Where the nameplate data goes
Every number on this page is worked out by JavaScript running in the tab you are reading it in. Nothing you type — loads, lengths, nameplate ratings, the rates your utility charges you — is uploaded, logged or kept, which is also why the calculators carry on working in a mechanical room with no signal.
That matters more here than on most pages: a transformer nameplate, a panel schedule and a run length together describe somebody’s building, and none of it is transmitted anywhere by this page.