For estimation and planning use. The result is a bolted fault current from a hand method, not an arc-flash or SCCR result. A licensed engineer's short-circuit study, the equipment manufacturer's data and the authority having jurisdiction decide equipment ratings and protection. Not a substitute for a short-circuit or arc-flash study.

Maintained by BuildSolver · Updated November 5, 2026

Calculator · Short circuit

Point-to-point short circuit calculator

The point-to-point method finds the fault current available at a transformer secondary and then at the end of each feeder run, using the transformer kVA and impedance and a constant for the conductor. Bussmann's own worked example is a 300 kVA, 2 percent Z transformer at 208 V on an infinite primary. It gives 46,273 A at the secondary in the worksheet, and 39,425 A after 20 ft of 500 kcmil copper in two sets and 18,687 A after 20 ft of 2 AWG. The page carries the same steps without the rounding and returns 46,263 A, 39,419 A and 18,670 A. It treats a three-phase transformer, a single-phase transformer and the line-to-neutral fault of a center-tapped winding, and it gives a preliminary figure, not a study. To size the transformer first, use the transformer size calculator. To check the same feeder for voltage drop, use the voltage drop calculator.

Choose the phase arrangement and the secondary voltage, then enter the kVA and the nameplate percent impedance. The full-load current is 832.7 A for the 300 kVA, 208 V default, and the worst-case multiplier uses 0.9 times the impedance.
Choose from 1 to 3 runs. For each, give the length in feet, the conductor size and metal, the raceway type and the number of parallel sets. The default is Bussmann's 20 ft of 500 kcmil copper in two sets, then 20 ft of 2 AWG.
The result gives the fault current at the secondary, 46,263 A in the default case, and after each run, 39,419 A and then 18,670 A. Add a high-voltage case or a motor figure if they apply, and compare the result with equipment ratings.

Transformers of 5 to 10,000 kVA at 600 V and below, up to 3 feeder runs of 14 AWG to 1000 kcmil copper or aluminum, with an infinite primary or a known primary fault current. The form opens on Bussmann's 300 kVA example, with no motors and nominal voltage.

Transformer and fault type

Three-phase gives the three-phase bolted fault. A single-phase transformer gives the line-to-line fault, or the line-to-neutral fault of a center-tapped winding, which Bussmann starts at 1.5 times the line-to-line value.

Secondary voltage, line-to-line

The voltages of Bussmann's full-load current tables. Three-phase takes 208, 220, 240, 440, 460, 480 and 600 V, and single-phase takes 230, 240, 460 and 480 V. The conductor constants are for 600 V and below.

Transformer rating, kVA

From the nameplate. The page opens on Bussmann's 300 kVA example.

Transformer impedance, percent Z

The marked percent impedance from the nameplate. The page applies Bussmann's 0.9 factor to it for the worst case. Zero is not accepted, since it would make the fault current unlimited.

Source ahead of the transformer

Infinite is the worst case and what Bussmann's example assumes. If the utility gives the available fault current at the primary, choose the known option and enter it below.

Voltage case

Bussmann says utility voltages may vary 10 percent, so for the worst case it multiplies the secondary current by 1.1.

Total motor full-load current, A (optional)

The motors that can feed the fault. The page adds 4 times this current to the secondary and to the end of every run, and Bussmann accepts 4 to 6. Leave it blank to add nothing.

Number of feeder runs

Each run starts from the fault current at the end of the one before it. The first run starts at the transformer secondary.

Run 1 conductor metal

Run 1 conductor size

Bussmann's Table 2 covers 14 AWG to 1000 kcmil. The size is the size of each conductor, so two sets of 500 kcmil is entered as 500 kcmil with 2 sets.

Run 1 conductors and raceway

Table 2 has a column for each of the four. Steel conduit has lower C values than nonmagnetic conduit, so the current falls faster down the run. Nonmagnetic is Bussmann's heading for non-steel conduit.

Run 1 parallel sets, conductors per phase

Bussmann's n. Identical sets in parallel divide f by n. Enter 1 for a single set.

Run 1 length, ft

The one-way length of the conduit from the start of the run to the fault, in feet. A length of 0 puts the fault at the start of the run.

Run 2 conductor metal

Run 2 conductor size

Run 2 conductors and raceway

Run 2 parallel sets, conductors per phase

Run 2 length, ft

Example result for the default inputs

300 kVA transformer at 208 V, 2 feeder runs, infinite primary.

Transformer full-load current

833 A

Available fault current at the transformer secondary

46,263 A

Available fault current at the end of the last run

18,670 A

Calculation details
Transformer	300 kVA, 2 percent Z, 208 V, three-phase
Full-load current	300,000 / (208 x 1.732) = 832.7 A
Transformer multiplier	100 / (0.9 x 2) = 55.56
Fault current at the secondary	832.7 x 55.56 = 46,263 A
Run 1	20 ft, 2 sets of 500 kcmil copper, single conductors in steel conduit, C = 22,185
Run 1 factor f	1.732 x 20 x 46,263 / (22,185 x 2 x 208) = 0.174
Fault current at the end of run 1	46,263 x 0.852 = 39,419 A
Run 2	20 ft, 2 AWG copper, single conductors in steel conduit, C = 5,907
Run 2 factor f	1.732 x 20 x 39,419 / (5,907 x 1 x 208) = 1.111
Fault current at the end of run 2	39,419 x 0.474 = 18,670 A

The answer is a bolted symmetrical RMS fault current from a hand method. It is a preliminary figure for sizing and for checking equipment ratings, and it does not replace a short-circuit study, an arc-flash study or a series-rating review by a licensed engineer.
Nominal voltage is used. Bussmann says utility voltages may run 10 percent high on power services, which raises the worst-case fault current by 10 percent. Choose the high-voltage case for a worst-case figure.
No motor contribution is added. Bussmann says to add the total motor full-load current times 4 at every fault location where motors are significant, and values of 4 to 6 are commonly accepted.
This is the three-phase bolted fault. Bussmann puts a bolted line-to-line fault at about 87 percent of it, and a line-to-ground fault anywhere from 25 to 125 percent, so the ground-fault current is checked separately.
Parallel sets are taken as identical in size, length and raceway. Unequal sets or a shared conduit need a separate treatment.

For estimation purposes only. Consult a licensed engineer for final designs.

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How the available fault current is calculated

This page follows the point-to-point method that Bussmann publishes for hand calculation of the available short-circuit current. It starts at the transformer, whose full-load current and impedance fix the fault current at the secondary terminals, and then carries that current down each feeder run through a multiplier M that depends on the length, the conductor and the voltage. The answer is a bolted symmetrical RMS value at each point. Motor contribution and a high-voltage case follow Bussmann's notes, and the page does not model busway, arcing or ground faults, or any system above 600 V.

Full-load current: three-phase: I_FLA = kVA x 1000 / (E_LL x 1.732); single-phase: I_FLA = kVA x 1000 / E_LLA 300 kVA transformer at 208 V line-to-line carries 832.7 A. Bussmann's Table 3A prints 833 A for this rating, and rounding this formula reproduces every figure of its Tables 3A and 3B.
Transformer multiplier: M = 100 / (0.9 x %Z)Bussmann multiplies the marked impedance by 0.9 because the marked value of a transformer can vary 10 percent from the tested one, which gives the worst case. At 2 percent Z the multiplier is 55.56, and Bussmann's worksheet prints 55.55 for the same figure.
Fault current at the secondary, infinite primary: I_sc = I_FLA x MWith an infinite primary the transformer alone sets the secondary current. 832.7 A times 55.56 is 46,263 A, against 46,273 A in Bussmann's worksheet, which rounds the full-load current and the multiplier before it multiplies.
Known primary fault current: f = I_P x V_P x 1.732 x %Z / (100,000 x kVA), then M = 1 / (1 + f) and I_S = V_P / V_S x M x I_P; single-phase drops the 1.732When the utility gives a fault current at the primary, Bussmann's second-transformer procedure replaces the infinite-bus step. 20 kA at 13,200 V on a 500 kVA, 3.5 percent Z, 480 V transformer gives f = 32.007, M = 0.030 and 16,663 A at the secondary, against 19,093 A with an infinite primary. That procedure prints %Z as marked, so it does not carry the 0.9.
The f factor of a run: three-phase: f = 1.732 x L x I / (C x n x E_LL); single-phase line-to-line: f = 2 x L x I / (C x n x E_LL); line-to-neutral: f = 2 x L x I / (C x n x E_LN)L is the one-way length of the run in feet, I is the fault current at the start of the run, C is the constant from Bussmann's Table 2, and n is the number of conductors per phase, which adjusts C for parallel sets. Run 1 of the example is 20 ft of 2 sets of 500 kcmil copper in steel conduit, so C is 22,185 and f is 0.174. The 2 AWG run that follows has a C of 5,907 and an f of 1.111.
Multiplier M: M = 1 / (1 + f)M is the share of the current that gets through the run. An f of 0.174 leaves 0.852 of it, and an f of 1.111 leaves 0.474. Bussmann's Table 4 prints M for f from 0.01 to 100, and the page matches all 46 rows to two decimals except the 0.8 row, which prints 0.55 where the formula gives 0.56.
Fault current at the end of a run: I_end = I_start x M, and the next run starts from I_endEach run starts from the current at the end of the one before it. The 46,263 A at the secondary leaves 39,419 A after run 1 and 18,670 A after run 2, against 39,425 A and 18,687 A in the worksheet. Up to 3 runs can be chained.
Motor contribution: motor contribution = total motor full-load current x 4Bussmann says to add motor contribution at each fault location where motors are significant, and it calls 4 to 6 times the motor full-load current commonly accepted. This page uses 4 and adds the same figure to the secondary and to the end of each run, so 400 A of motors adds 1,600 A and takes the example to 20,270 A.
High utility voltage: I_sc x 1.1 at the secondaryBussmann says utility voltages may vary 10 percent on power services, so the worst case multiplies the secondary current by 1.1. That takes the example from 46,263 A to 50,890 A at the secondary and from 18,670 A to 19,381 A at the end of run 2.
Center-tapped single-phase transformer: I_L-N = 1.5 x I_L-L at the terminals, and the run factor uses E_L-NFor a center-tapped winding Bussmann puts the line-to-neutral current at the terminals at 1.5 times the line-to-line current, an approximation that varies from 1.33 to 1.67, and the run factor then uses the line-to-neutral voltage, half the line voltage. A 50 kVA, 240 V transformer with 50 ft of 2/0 copper gives 11,574 A line-to-line and 17,361 A line-to-neutral at the terminals, and 7,991 A and 7,403 A at the end of the run.

Worked examples from the Bussmann method

The default case is Bussmann's worked example, a transformer feeding two runs of cable. Start at the transformer. A 300 kVA, 2 percent Z transformer at 208 V carries 832.7 A, and the multiplier of 55.56 puts 46,263 A on the secondary. Run 1 has a C of 22,185 and an f of 0.174, which leaves 0.852 of the current and 39,419 A. Run 2 has a C of 5,907 and an f of 1.111, which leaves 0.474 and 18,670 A. That is the number to compare with a device rating.

The table below sets every number that Bussmann prints for this example beside the page's unrounded value. Bussmann rounds the full-load current to 833 A and the multiplier to 55.55 before it multiplies, so its figures differ from the page by 0.2 percent at most in every row. The gaps are small.

Bussmann printed worked example against the page
Step	Printed	This page	Difference, %
Full-load current, A	833	832.7	0.03
Transformer multiplier	55.55	55.56	0.01
Fault current at the secondary, A	46,273	46,263	0.02
Run 1 f factor	0.174	0.174	0.20
Run 1 multiplier M	0.852	0.852	0.01
Fault 1, end of run 1, A	39,425	39,419	0.02
Run 2 f factor	1.11	1.111	0.12
Run 2 multiplier M	0.474	0.474	0.08
Fault 2, end of run 2, A	18,687	18,670	0.09

Eight cases below run through the same functions the form uses. Aluminum in place of copper leaves 14,103 A after run 2, and one set of 500 kcmil in place of two takes it to 17,447 A. The 10 percent high voltage case gives 50,890 A at the secondary and 19,381 A after run 2. A 500 kVA, 480 V transformer gives 19,093 A on an infinite primary and 16,663 A when the utility gives 20 kA at 13,200 V. A 50 kVA, 240 V single-phase transformer gives 11,574 A line-to-line and 17,361 A line-to-neutral at the terminals.

Available fault current for eight example transformers
Case	Setup	Result
Bussmann example, both faults	300 kVA, 2 percent Z, 208 V three-phase; 20 ft, 2 sets of 500 kcmil copper, single conductors in steel conduit; 20 ft, 2 AWG copper, single conductors in steel conduit	46,263 A at the secondary, 39,419 A after run 1, 18,670 A after run 2
Bussmann example, fault 1 only	300 kVA, 2 percent Z, 208 V three-phase; 20 ft, 2 sets of 500 kcmil copper, single conductors in steel conduit	46,263 A at the secondary, 39,419 A after run 1
Same, aluminum conductors	300 kVA, 2 percent Z, 208 V three-phase; 20 ft, 2 sets of 500 kcmil aluminum, single conductors in steel conduit; 20 ft, 2 AWG aluminum, single conductors in steel conduit	46,263 A at the secondary, 38,380 A after run 1, 14,103 A after run 2
Same, utility voltage 10 percent high	300 kVA, 2 percent Z, 208 V three-phase, high utility voltage; 20 ft, 2 sets of 500 kcmil copper, single conductors in steel conduit; 20 ft, 2 AWG copper, single conductors in steel conduit	50,890 A at the secondary, 42,728 A after run 1, 19,381 A after run 2
500 kVA, 480 V, infinite primary	500 kVA, 3.5 percent Z, 480 V three-phase; 100 ft, 250 kcmil copper, single conductors in steel conduit	19,093 A at the secondary, 13,465 A after run 1
Same, 20 kA at a 13,200 V primary	500 kVA, 3.5 percent Z, 480 V three-phase, 20 kA at 13,200 V primary; 100 ft, 250 kcmil copper, single conductors in steel conduit	16,663 A at the secondary, 12,209 A after run 1
50 kVA, 240 V single-phase, line-to-line	50 kVA, 2 percent Z, 240 V single-phase line-to-line; 50 ft, 2/0 AWG copper, single conductors in steel conduit	11,574 A at the secondary, 7,991 A after run 1
Same, line-to-neutral	50 kVA, 2 percent Z, 240 V single-phase line-to-neutral; 50 ft, 2/0 AWG copper, single conductors in steel conduit	17,361 A at the secondary, 7,403 A after run 1

Standards and data sources

Eaton Bussmann (Cooper Bussmann), Short-Circuit Current Calculations, Point-to-Point Method - The procedure in six steps, the f factor and the multiplier M, the conductor constants of Table 2, the full-load current of Tables 3A and 3B, Table 4, and the notes on transformer impedance, utility voltage, motor contribution and center-tapped transformers. Its worked example of a 300 kVA, 2 percent Z transformer on an infinite primary is the default of this page.
Electrical Plan Review, Point-To-Point Method Of Short-Circuit Calculation (Bussmann pages in a university lecture pack) - The copy that the figures on this page were read from, because the Eaton site refused an automated download. It carries the same steps, notes, tables and worked example. Only the 600 V columns of Table 2 are used, and the busway table was not.
NFPA 70, National Electrical Code, sections 110.9, 110.10 and 240.1 - These are among the sections that Bussmann names as the reason to know the available short-circuit current. The NEC text itself was not read for this page. A figure from this page is for an engineer to compare with equipment ratings.

Limitations of this calculator

The result is a bolted symmetrical RMS fault current found by hand. It is not an arc-flash incident energy, a series-rating or SCCR determination, or a coordination study, and a licensed engineer does those from a full short-circuit study.
Conductor constants come from the 600 V columns of Bussmann's Table 2, 14 AWG to 1000 kcmil, copper and aluminum, in steel and nonmagnetic conduit. Systems above 600 V and busway runs are outside the page, because the 5 kV and 15 kV columns and the busway table were not used.
Bussmann's method rests on an infinite primary or a known primary fault current. The page does not model utility source impedance beyond that, a transformer with more than one secondary, or the grounding of a delta-wye bank.
For a center-tapped transformer the 1.5 factor on the line-to-neutral current is an approximation, and Bussmann says it varies from 1.33 to 1.67. Farther from the terminals the line-to-neutral current falls below the line-to-line current, depending on wire size.
Line-to-line and line-to-ground values are not worked out for a three-phase transformer. Bussmann puts a bolted line-to-line fault at about 87 percent of the three-phase value and a line-to-ground fault between 25 and 125 percent.
The motor contribution is a single figure of 4 times the motor full-load current, added to the secondary and to the end of each run. Bussmann accepts 4 to 6, and the page does not place motors along the feeder.
Parallel sets are taken as identical in size, length and raceway, and the length is the one-way length of the conduit to the fault. Unequal sets and mixed raceways in one run need a separate treatment.
The Table 2 constants are one over the impedance per foot, taken from the resistance and reactance values of IEEE Std 241 and 242. Cable temperature and the let-through of a current-limiting fuse or breaker are not modeled.
The transformer rating is limited to 5 to 10,000 kVA and the impedance to 0.5 to 15 percent. Bussmann's full-load tables stop at 2,000 kVA three-phase and 500 kVA single-phase, and its Table 5 lists 1 to 4 percent, so values beyond those are formula extensions.
This is a planning estimate, not a design drawing or a permit-grade study. A licensed engineer, the manufacturer's data and the authority having jurisdiction decide equipment ratings and protection.

Common short-circuit calculation mistakes

Leaving out the 0.9 on the impedance. The 0.9 matters. Bussmann multiplies the marked %Z by 0.9 to get the worst case. At 2 percent the full-load current of 832.7 A times 100 over 2 gives 41,637 A at the secondary, while the worst case gives 46,263 A. The lower figure understates the fault current by 10 percent.
Forgetting the parallel sets. Parallel sets count. Bussmann's n adjusts C for parallel runs. Entering one set of 500 kcmil in place of two sets raises f on run 1 from 0.174 to 0.347 and takes the end of the feeder from 18,670 A to 17,447 A, which understates the fault current.
Starting run 2 from the transformer value. Each run starts from the current at the end of the one before it. Starting the 2 AWG run from 46,263 A in place of 39,419 A gives 20,077 A at its end, against 18,670 A, which overstates the current by 7.5 percent.
Using the line voltage on a line-to-neutral run. A line-to-neutral run takes E_L-N, half the line voltage. With 240 V in place of 120 V the 50 ft run of 2/0 copper shows f = 0.673 and 10,380 A at its end, against f = 1.345 and 7,403 A. The error overstates the current by 40 percent.
Skipping the voltage and motor allowances. Bussmann's worst case multiplies the secondary current by 1.1 for high utility voltage and adds 4 times the motor full-load current. For the example, 400 A of motors and the high voltage take the end of the feeder from 18,670 A to 20,981 A. A device rated to the nominal figure can be under-rated.
Reading the result as an arc-flash or SCCR answer. The page gives the bolted fault current. Incident energy depends on arcing current, clearing time and working distance, and a series rating or an SCCR comes from the equipment listing. Bussmann puts a three-phase arcing fault at 89 percent of the bolted value at most, which is a different quantity from what this page returns.

Common questions about point-to-point fault current

How do you calculate available fault current by the point-to-point method?

Find the transformer full-load current, multiply it by 100 divided by 0.9 times the percent impedance, then step down each run with M = 1 / (1 + f), where f = 1.732 x L x I / (C x n x E) for three-phase. For 300 kVA at 208 V and 2 percent Z the secondary current is 46,263 A, and 20 ft of 500 kcmil copper in 2 sets leaves 39,419 A.

What is the C value in the f factor?

C is a conductor constant, one over the impedance per foot, that Bussmann prints in Table 2 by size, metal, raceway and voltage. For 500 kcmil copper single conductors in steel conduit at 600 V it is 22,185, and for 2 AWG it is 5,907. Aluminum is lower and nonmagnetic conduit is higher, so 500 kcmil aluminum is 18,756 and 500 kcmil copper in nonmagnetic conduit is 26,706.

Why is the transformer impedance multiplied by 0.9?

It builds in a margin. Bussmann says the marked impedance of a transformer can vary 10 percent from the tested value, so it multiplies %Z by 0.9 to approach the worst case. It also notes that two-winding transformers built to ANSI standards carry a 7.5 percent impedance tolerance. At 2 percent the multiplier is 55.56 with the 0.9 and 50 without it.

How do I enter parallel conductors?

Count sets, not wires. Enter the size of one conductor and the number of sets as n for the run. Bussmann's n is the number of conductors per phase and divides f, so two sets of the same size halve the factor for the same current. In the example two sets of 500 kcmil give f = 0.174, and one set would give 0.347.

What changes for a single-phase center-tapped transformer?

The full-load current uses kVA x 1000 / E with no 1.732, and each run uses 2 x L in place of 1.732 x L. For a line-to-neutral fault Bussmann starts from 1.5 times the line-to-line current at the terminals and uses the line-to-neutral voltage in f. A 50 kVA, 240 V transformer gives 11,574 A line-to-line and 17,361 A line-to-neutral at the terminals.

What if the utility gives a fault current at the primary?

Ask the utility first. Choose the known primary option and enter the current and the primary voltage. The page then uses Bussmann's second-transformer procedure, f = I_P x V_P x 1.732 x %Z / (100,000 x kVA). A 500 kVA, 480 V, 3.5 percent Z transformer on 13,200 V with 20 kA at the primary gives 16,663 A at the secondary, against 19,093 A on an infinite primary.

How much do the voltage case and motors add?

Choosing the 10 percent high voltage takes the secondary current from 46,263 A to 50,890 A. Adding 400 A of motor full-load current contributes 1,600 A at 4 times, and both together give 20,981 A at the end of the example feeder, against 18,670 A at nominal voltage without motors.

Does this replace a short-circuit study or an arc-flash analysis?

No. The page returns a bolted symmetrical RMS value from a hand method, which is the preliminary figure used to check interrupting ratings and to size equipment early. An engineer's short-circuit study, an arc-flash study and the equipment listings govern the design.

Is there a limit on how many times I can run this?

No. This calculator runs in your browser with no daily limit and no signup.

Used on the same job

Find the fault current, then compare it with the equipment

Compare, then decide. The fault current at a panel or a disconnect is the figure to set against the interrupting rating of the devices there. Take the transformer data from the nameplate, the lengths from the plans, and the final rating decision from a licensed engineer and the equipment listings. The transformer comes from the transformer size calculator, and the same feeder can be checked with the voltage drop calculator. You can also ask the assistant to go over a fault current result.

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