Short circuit · Transformer · Conductor run
Available Fault Current Calculator
Enter transformer kVA, secondary voltage and the nameplate impedance. See the fault current at the secondary terminals, then add cable runs to see how much each one reduces it.
By Allen Tan · Updated · How results are sourced
1Transformer
Line-to-line. Up to 600 V.
Read it from the transformer nameplate. This tool has no typical values on purpose.
2Conductor runs (optional)
Run 1
One-way distance from the transformer to the fault point.
Each run starts from the fault current at the end of the one before it. With no runs, the result is the current at the transformer secondary terminals.
3Adjustments (optional)
The handbook gives these for UL 1561 listed transformers of 25 kVA and larger (plus or minus 10% tolerance). Use the 0.9 case for equipment ratings. For an ANSI built unit (plus or minus 7.5%), enter the adjusted %Z yourself.
Handbook Note 3. Applied to the transformer terminal current before any cable run.
Adds 4 times this current as motor contribution (the handbook says 4 to 6 is accepted). Leave 0 to leave motors out.
Available fault current at the end of run 1
33,211 A
33.21 kA symmetrical, three-phase. Infinite bus on the primary.
- Transformer secondary terminals
- 34,367 A
- End of run 1
- 33,211 A
The calculation
- Full-load current
- 1,202.8 A
- Multiplier, 100 / %Z
- 28.57
- Run 1: C = 26,706, f = 0.0348
- M = 0.9663
f = 1.732 x L x I / (C x n x E), M = 1 / (1 + f), and the new current is I x M.
Share this result
https://kvcalc.com/tools/available-fault-current-calculator/
The link carries the full calculation, so anyone opening it lands on the same inputs and the same result.
How to calculate transformer fault current
The tool follows the point-to-point method in the Cooper Bussmann (now Eaton) handbook. Four steps cover a transformer and one conductor run.
- Full-load current: kVA × 1000 ÷ (1.732 × volts) for three-phase, or kVA × 1000 ÷ volts for single-phase.
- Multiplier: 100 ÷ %Z.
- Fault current at the secondary terminals: full-load current × multiplier.
- At the end of a run: f = 1.732 × L × I ÷ (C × n × E), M = 1 ÷ (1 + f), and the new current is I × M. Single-phase uses 2 in place of 1.732.
Worked example: 1000 kVA, 480 V, 3.5% Z
This is the handbook’s own case: a three-phase transformer feeding 30 ft of four parallel 500 kcmil copper conductors per phase in nonmagnetic conduit.
| Step | Result |
|---|---|
| Full-load current | 1202.8 A |
| Multiplier, 100 ÷ 3.5 | 28.57 |
| At the transformer terminals | 34,367 A |
| C for 500 kcmil copper, nonmagnetic conduit | 26,706 |
| f and M for the run | f = 0.0348, M = 0.9663 |
| At the end of the run | 33,211 A |
The handbook prints 33,215 A at the end of the run because it rounds the intermediate values. The tool does not round until the end, so it differs by a few amps.
Where the transformer impedance comes from
Read the percent impedance (%Z) from the nameplate. It is the biggest input in the answer, and it varies between designs, so this tool has no table of typical values to fall back on. If the nameplate is unreadable, ask the manufacturer for the test report, or use the value the utility gives you. For the other side of a transformer study, see the transformer fuse and breaker size calculator.
Conductor constants (C values)
Cooper Bussmann handbook Table 4, 600 V class, as used by the tool. Each C is the inverse of the impedance per foot. Single conductors and multiconductor cable are listed separately for steel and nonmagnetic conduit.
Copper
| Size | Single, steel | Single, nonmagnetic | Cable, steel | Cable, nonmagnetic |
|---|---|---|---|---|
| 14 AWG | 389 | 389 | 389 | 389 |
| 12 AWG | 617 | 617 | 617 | 617 |
| 10 AWG | 981 | 982 | 982 | 982 |
| 8 AWG | 1,557 | 1,559 | 1,559 | 1,560 |
| 6 AWG | 2,425 | 2,430 | 2,431 | 2,433 |
| 4 AWG | 3,806 | 3,826 | 3,830 | 3,838 |
| 3 AWG | 4,774 | 4,811 | 4,820 | 4,833 |
| 2 AWG | 5,907 | 6,044 | 5,989 | 6,087 |
| 1 AWG | 7,293 | 7,493 | 7,454 | 7,579 |
| 1/0 AWG | 8,925 | 9,317 | 9,210 | 9,473 |
| 2/0 AWG | 10,755 | 11,424 | 11,245 | 11,703 |
| 3/0 AWG | 12,844 | 13,923 | 13,656 | 14,410 |
| 4/0 AWG | 15,082 | 16,673 | 16,392 | 17,483 |
| 250 kcmil | 16,483 | 18,594 | 18,311 | 19,779 |
| 300 kcmil | 18,177 | 20,868 | 20,617 | 22,525 |
| 350 kcmil | 19,704 | 22,737 | 22,646 | 24,904 |
| 400 kcmil | 20,566 | 24,297 | 24,253 | 26,916 |
| 500 kcmil | 22,185 | 26,706 | 26,980 | 30,096 |
| 600 kcmil | 22,965 | 28,033 | 28,752 | 32,154 |
| 750 kcmil | 24,137 | 29,735 | 31,051 | 34,605 |
| 1000 kcmil | 25,278 | 31,491 | 33,864 | 37,197 |
Aluminum
| Size | Single, steel | Single, nonmagnetic | Cable, steel | Cable, nonmagnetic |
|---|---|---|---|---|
| 14 AWG | 237 | 237 | 237 | 237 |
| 12 AWG | 376 | 376 | 376 | 376 |
| 10 AWG | 599 | 599 | 599 | 599 |
| 8 AWG | 951 | 952 | 952 | 952 |
| 6 AWG | 1,481 | 1,482 | 1,482 | 1,482 |
| 4 AWG | 2,346 | 2,350 | 2,351 | 2,353 |
| 3 AWG | 2,952 | 2,961 | 2,963 | 2,966 |
| 2 AWG | 3,713 | 3,730 | 3,734 | 3,740 |
| 1 AWG | 4,645 | 4,678 | 4,686 | 4,699 |
| 1/0 AWG | 5,777 | 5,838 | 5,852 | 5,876 |
| 2/0 AWG | 7,187 | 7,301 | 7,327 | 7,373 |
| 3/0 AWG | 8,826 | 9,110 | 9,077 | 9,243 |
| 4/0 AWG | 10,741 | 11,174 | 11,185 | 11,409 |
| 250 kcmil | 12,122 | 12,862 | 12,797 | 13,236 |
| 300 kcmil | 13,910 | 14,923 | 14,917 | 15,495 |
| 350 kcmil | 15,484 | 16,813 | 16,795 | 17,635 |
| 400 kcmil | 16,671 | 18,506 | 18,462 | 19,588 |
| 500 kcmil | 18,756 | 21,391 | 21,395 | 23,018 |
| 600 kcmil | 20,093 | 23,451 | 23,633 | 25,708 |
| 750 kcmil | 21,766 | 25,976 | 26,432 | 29,036 |
| 1000 kcmil | 23,478 | 28,779 | 29,865 | 32,938 |
What the NEC asks for
Section 110.24 asks for service equipment in other than dwelling units to be field marked with the available fault current and the date it was calculated, and for the calculation to be documented. Section 110.9 deals with the interrupting rating of equipment and 110.10 with circuit impedance and short-circuit current ratings. These section numbers were checked against the 2023 edition, and numbering in the 2026 edition may differ, so check the edition your jurisdiction has adopted. This tool gives you the calculation, not the label.
What the result is not. It assumes an infinite source and a bolted fault, so it is an upper bound for choosing equipment ratings. It does not calculate arc flash incident energy or PPE. Do not use it to decide that a live panel is safe to work on.
Source.Cooper Bussmann (now Eaton), SPD Selection and Protection Data, "Short Circuit Current Calculations", pp. 192-198 (2005). The handbook is a manufacturer publication, not a standard. Its conductor constants are based on IEEE 241-1990 and IEEE 242-1986 data, and 3 AWG values at 5 kV and 15 kV are approximations (those columns are not used here). The tool’s results reproduce the handbook’s worked examples (three-phase with and without motors, single-phase line-to-line and line-to-neutral) within 0.03%. Checked on 2026-10-08.
Disclaimer. Method from the Cooper Bussmann handbook; NEC section numbers are those of the 2023 edition. A bolted-fault estimate for choosing equipment ratings, not an arc flash or PPE study. Not engineering advice and not a substitute for the code text. Verify against the edition adopted in your jurisdiction and with your Authority Having Jurisdiction — protection decisions affect life safety.
Questions people actually ask
What is available fault current?
It is the largest current the system can push through a short circuit at a given point, limited only by the impedance between the source and that point. Breakers, fuses and switchboards must be rated to interrupt or withstand at least that value. At a transformer secondary the main limit is the transformer impedance, which is why %Z on the nameplate drives the answer.
How do you calculate available fault current from a transformer?
Find the full-load current: kVA x 1000 divided by (1.732 x secondary volts) for three-phase, or kVA x 1000 divided by volts for single-phase. Divide 100 by the percent impedance to get the multiplier. Full-load current times the multiplier is the fault current at the secondary terminals, assuming an infinite source on the primary. A 1000 kVA, 480 V, 3.5% Z transformer gives about 1,203 A x 28.57 = 34,370 A. The tool keeps full precision and shows 34,367 A.
How do I calculate transformer fault current at the end of a cable?
Use the point-to-point method. Compute f = 1.732 x L x I / (C x n x E) for three-phase, where L is the one-way length in feet, I is the fault current at the start of the run, C is the conductor constant for the size and raceway, n is conductors per phase and E is line-to-line volts. The multiplier is M = 1 / (1 + f), and the current at the end of the run is I x M. Repeat for each run in series.
Where do I find the transformer impedance?
On the transformer nameplate, as a percentage. The Cooper Bussmann handbook this tool follows says to take %Z from the nameplate. If you only know the transformer size, ask the manufacturer or the utility, because impedance varies by design and this tool does not guess it. A wrong %Z moves the result by the same ratio: 3% instead of 6% doubles the fault current.
Does a longer cable reduce fault current?
Yes. More conductor length adds impedance between the transformer and the fault, so the current at the far end is lower. Larger conductors, or more conductors in parallel, add less impedance for the same length. In the handbook table, steel conduit gives a lower C value, meaning more impedance, than nonmagnetic conduit such as PVC for the same conductors.
What does the infinite bus assumption mean?
It assumes the utility supply on the primary side has no impedance of its own, so the transformer is the only limit. Real sources add impedance, so the true current is lower. The infinite bus value is the safe upper bound for choosing equipment ratings. For a closer value, ask the utility for the available short-circuit current at your service.
What do the impedance tolerance options do?
The handbook notes that UL 1561 listed transformers of 25 kVA and larger have a plus or minus 10% impedance tolerance, and that ANSI-built two-winding transformers have plus or minus 7.5%. To find the highest fault current from a UL-listed unit, multiply %Z by 0.9. For the lowest, multiply by 1.1. The tool applies the 0.9 and 1.1 factors from the handbook when you pick those options.
How is motor contribution added?
Running motors feed current back into a fault. The handbook says a practical estimate is the total motor current times 4, and that values of 4 to 6 are commonly accepted. Enter the total motor full-load current and the tool adds 4 times that value at every point. Motors are not added to the current that flows through the conductor run, matching the handbook examples.
What is the difference between line-to-line and line-to-neutral?
On a center-tapped single-phase transformer, the line-to-neutral fault current at the secondary terminals can be higher than the line-to-line value. The handbook uses 1.5 times the line-to-line value at the terminals as an approximation, and uses half the line-to-line voltage in the conductor calculation. Three-phase systems use the three-phase bolted fault value.
Does this calculate arc flash incident energy?
No. The result is a bolted-fault current for choosing equipment ratings. Incident energy and PPE categories depend on clearing time, working distance, electrode configuration and more. They need a study done to the method your facility follows, usually by a qualified engineer.
What does the NEC require about fault current?
Section 110.24 requires service equipment in other than dwelling units to be field marked with the available fault current and the date the calculation was made, and the calculation to be documented. Section 110.9 covers the interrupting rating of equipment, and 110.10 covers the circuit impedance and short-circuit current ratings. Read the edition your jurisdiction has adopted, and ask your inspector about local rules.
Why is the voltage limited to 600 V?
The conductor constants in the table come from the 600 V columns of the handbook, and this tool does not extrapolate to medium-voltage systems. For higher voltages use a full short-circuit study.
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