Why is this called a 'worst-case' or 'infinite source' estimate?

It assumes the utility source behind the transformer has zero impedance (infinite fault capacity) and ignores all downstream conductor/cable impedance. Both of those real-world factors add impedance to the fault current path and REDUCE the actual available fault current below this estimate, so this method gives an upper-bound screening number, not the precise available fault current.

Why does a lower transformer impedance percentage increase the fault current?

Impedance is what limits fault current in this simplified method — the available fault current is the full-load amps divided by the impedance percentage (as a decimal), so a smaller impedance percentage produces a larger calculated fault current.

Can the actual fault current ever come out higher than this figure?

Yes — and it is worth being precise about what "worst case" means here. This figure is an upper bound on what the transformer itself contributes, expressed as a symmetrical RMS current; it is not an upper bound on everything that arrives at the bus. Two things sit outside it. Motors running on the secondary feed current back into a fault for the first few cycles and add to what the transformer delivers; there is no motor input on this page, so none of that contribution is in the number. And because a fault in an inductive circuit carries a DC offset, the peak in the first half cycle is higher again than the symmetrical RMS value shown here. Screen with this, then let the real study work from the utility's stated available fault current at the primary, the connected motor load, and the actual conductor runs.

What do I enter for a 480Y/277 V transformer, and does this work on a single-phase unit?

Line-to-line, always: 480 rather than 277, 208 rather than 120. The value is used as the line-to-line voltage in the three-phase full-load-amps formula, kVA × 1000 ÷ (V × √3), so entering 277 V for a 500 kVA transformer at 5.5% returns about 18,950 A against the correct 10,930 A — 73% high. Single-phase transformers are outside this tool: the √3 is fixed in the formula and there is no phase selector, so a 75 kVA 240 V single-phase unit at 3% impedance comes back near 6,010 A when the transformer-only figure is closer to 10,400 A, roughly 42% low — an error in the direction that makes equipment look better rated than it is. For a single-phase unit, work it by hand: kVA × 1000 ÷ secondary volts, then ÷ impedance as a decimal. That hand figure is the full-winding case — a 120 V line-to-neutral fault on a centre-tapped 120/240 V unit runs through half the secondary winding and is a separate calculation this formula does not give.

The nameplate shows two kVA ratings and the transformer on site isn't the one I specified — which figures do I use?

Both figures come off the unit actually standing there. Percent impedance belongs to one specific transformer rather than to a model line, and it is generally referenced to the self-cooled base kVA — so on a fan-cooled unit enter the base rating, not the forced-air uprating, or the result is overstated. A value pulled from a catalogue page, a spec section or the transformer originally ordered can easily sit a quarter of a point away from what got delivered, and that is not academic: 5.5% against 5.75% on a 500 kVA 480 V transformer is a difference of roughly 475 A. Substitutions get made at procurement, and the figure often never gets re-run afterwards. When you record the result, record the three inputs beside it — a bare ampere number tells the next person nothing about which transformer or which impedance produced it.

Materials & Quantities

Transformer-Limited Short-Circuit Current Calculator

Estimate the maximum available short-circuit current at a transformer's secondary, limited by transformer impedance alone.

Answers as you typeNo submit button
Every formula citedNamed published standards
Calculated in your browserNo account, nothing uploaded

SettingsSettings for this calculationUS

Market

Imperial · sales tax

Transformer Rating (kVA)

The transformer's nameplate kVA rating.

Use the transformer's nameplate apparent power rating in kVA.

Secondary Voltage (V, Line-to-Line)

The three-phase line-to-line secondary voltage.

Line to line, not line to neutral — the two differ by the square root of three, and entering the line-to-neutral figure inflates the available fault current by about 73%. Take it from the transformer nameplate rather than from a measurement: fault current is computed from the nominal secondary and the impedance, and a measured no-load voltage runs above nominal.

Transformer Impedance (%, from Nameplate)

The transformer's nameplate percent impedance.

A lower impedance percentage allows more fault current to flow, since impedance is what limits the fault current in this simplified method.

Estimated maximum available fault current

11,000 A

Low confidence

This is a transformer-only screening approximation (the 'infinite source' method), and it is NOT a worst case in either direction. It ignores upstream utility source impedance and downstream conductor impedance, which reduce the current available at a point further from the transformer. It also ignores two things that push the real figure ABOVE it: running motors feed a fault for the first few cycles at roughly four to six times their own full-load current, and nameplate impedance carries an ANSI/IEEE tolerance of plus or minus 7.5%, so a transformer at the low end of that band delivers proportionally more. Do not select an interrupting rating from this number. This is NOT a substitute for a complete short-circuit study (per IEEE 141/242 or equivalent software) required for proper overcurrent protective device rating, selective coordination, and arc-flash hazard analysis, which must be performed by a qualified electrical engineer.

Transformer full-load amps: 601.41 A

Then change the inputs to see how far the answer moves.

Show calculation logic

How this was calculated

Formula source(s)

Simplified transformer-only fault current estimate: full-load amps = (transformer kVA × 1000) ÷ (voltage × √3) for three-phase; available fault current = full-load amps ÷ (transformer impedance percentage ÷ 100) — this is the classical 'infinite source' approximation used for preliminary screening. It ignores upstream utility source impedance and downstream conductor impedance, which reduce the current available at a given point — and it also ignores motor contribution and the tolerance on nameplate impedance, which raise it

Inputs used

Transformer Rating (kVA): 500
Secondary Voltage (V, Line-to-Line): 480
Transformer Impedance (%, from Nameplate): 5.5

Intermediate steps

Transformer full-load amps: 601.41 A

Final result10,934.66 A

Confidence note: This is a transformer-only screening approximation (the 'infinite source' method), and it is NOT a worst case in either direction. It ignores upstream utility source impedance and downstream conductor impedance, which reduce the current available at a point further from the transformer. It also ignores two things that push the real figure ABOVE it: running motors feed a fault for the first few cycles at roughly four to six times their own full-load current, and nameplate impedance carries an ANSI/IEEE tolerance of plus or minus 7.5%, so a transformer at the low end of that band delivers proportionally more. Do not select an interrupting rating from this number. This is NOT a substitute for a complete short-circuit study (per IEEE 141/242 or equivalent software) required for proper overcurrent protective device rating, selective coordination, and arc-flash hazard analysis, which must be performed by a qualified electrical engineer.

What this calculation does not cover

The secondary voltage is used once, as the nominal line-to-line figure in (kVA × 1000) ÷ (volts × √3), and no input asks what the bus was actually sitting at in the moment before the fault, so a system running a few percent above nominal drives proportionally more current than this arithmetic returns — and typing that elevated voltage into the box moves the answer the wrong way, because the term sits in the denominator.
Entries outside 15 to 2,500 kVA, 120 to 600 V or 2 to 8% impedance are pulled back to the nearest bound and answered there, so a 3,000 kVA unit or a 1.8% nameplate produces a figure lower than the transformer in front of you would actually deliver.
One division by the impedance decimal serves for every kind of fault, and no input describes the winding connection or the grounding arrangement, so a line-to-ground fault — whose return path those two things govern — is not the event this number describes.
The transformer full-load amps carried in the breakdown is the dividend of that same division, derived from nameplate kVA, so it is the working shown rather than an independent check, and it is rated capacity rather than the load actually connected to the secondary.
Nothing in the arithmetic has a time dimension: the output is a magnitude at the instant of the fault, with no duration behind it, so it says nothing about how long the transformer or the bus it feeds can hold that current.

Add the equipment this sizes

This result is a specification — 11,000 A — not a quantity. Put the thing it sizes into your project: how many, what you call it, and your supplier’s price.

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Computed in your browser — nothing you enter is uploaded. Presented in US customary units and US trade terminology. Where a formula follows a published standard, that standard and its edition are cited beside it on this page; where none governs, the page says so. Local amendments override model codes — verify against the code in force where you build.

Sources checked 2026-09-05 · in the site-wide review of 2026-09-06 · v1.0.1

Regulatory standards & verification citations1

Simplified transformer-only fault current estimate: full-load amps = (transformer kVA × 1000) ÷ (voltage × √3) for three-phase; available fault current = full-load amps ÷ (transformer impedance percentage ÷ 100) — this is the classical 'infinite source' approximation used for preliminary screening. It ignores upstream utility source impedance and downstream conductor impedance, which reduce the current available at a given point — and it also ignores motor contribution and the tolerance on nameplate impedance, which raise it

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These guides cover the work this quantity is for — the first ones run this calculator inside the section that raises the question.

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Wiring a Motor Circuituses this calculator
A motor branch circuit carries four separate duties, and the device that guards against a short circuit is deliberately too big to catch an overload.

Called something else where you work? Residual current and ground-fault protection — the term in each market, how close the equivalence really is, and the standard that governs it.

Still deciding? Breaker Trip Rating vs Interrupting Rating — the factors that actually differ, with no invented prices.

How to calculate transformer-limited short-circuit current in 4 steps

Transformer Rating (kVA)The transformer's nameplate kVA rating.
Secondary Voltage (V, Line-to-Line)The three-phase line-to-line secondary voltage.
Transformer Impedance (%, from Nameplate)The transformer's nameplate percent impedance.
Estimated maximum available fault currentThe tool computes the estimated maximum available fault current from those figures and shows the formula, its sources, and a confidence rating alongside it.

Estimated maximum available fault current by transformer rating (kVA)

Page defaults, not your figures above.

Transformer Rating (kVA)	Estimated maximum available fault current (A)
50	1,093
100	2,187
200	4,374
500	10,935
1,000	21,869
2,000	43,739

Frequently asked questions

Why is this called a 'worst-case' or 'infinite source' estimate?: It assumes the utility source behind the transformer has zero impedance (infinite fault capacity) and ignores all downstream conductor/cable impedance. Both of those real-world factors add impedance to the fault current path and REDUCE the actual available fault current below this estimate, so this method gives an upper-bound screening number, not the precise available fault current.
Can I use this result to select and rate my overcurrent protective devices?: No. This is NOT a substitute for a complete short-circuit study (per IEEE 141/242 or equivalent software) required for proper overcurrent protective device rating, selective coordination, and arc-flash hazard analysis. That analysis must be performed by a qualified electrical engineer.
Why does a lower transformer impedance percentage increase the fault current?: Impedance is what limits fault current in this simplified method — the available fault current is the full-load amps divided by the impedance percentage (as a decimal), so a smaller impedance percentage produces a larger calculated fault current.
Can the actual fault current ever come out higher than this figure?: Yes — and it is worth being precise about what "worst case" means here. This figure is an upper bound on what the transformer itself contributes, expressed as a symmetrical RMS current; it is not an upper bound on everything that arrives at the bus. Two things sit outside it. Motors running on the secondary feed current back into a fault for the first few cycles and add to what the transformer delivers; there is no motor input on this page, so none of that contribution is in the number. And because a fault in an inductive circuit carries a DC offset, the peak in the first half cycle is higher again than the symmetrical RMS value shown here. Screen with this, then let the real study work from the utility's stated available fault current at the primary, the connected motor load, and the actual conductor runs.
What do I enter for a 480Y/277 V transformer, and does this work on a single-phase unit?: Line-to-line, always: 480 rather than 277, 208 rather than 120. The value is used as the line-to-line voltage in the three-phase full-load-amps formula, kVA × 1000 ÷ (V × √3), so entering 277 V for a 500 kVA transformer at 5.5% returns about 18,950 A against the correct 10,930 A — 73% high. Single-phase transformers are outside this tool: the √3 is fixed in the formula and there is no phase selector, so a 75 kVA 240 V single-phase unit at 3% impedance comes back near 6,010 A when the transformer-only figure is closer to 10,400 A, roughly 42% low — an error in the direction that makes equipment look better rated than it is. For a single-phase unit, work it by hand: kVA × 1000 ÷ secondary volts, then ÷ impedance as a decimal. That hand figure is the full-winding case — a 120 V line-to-neutral fault on a centre-tapped 120/240 V unit runs through half the secondary winding and is a separate calculation this formula does not give.
The nameplate shows two kVA ratings and the transformer on site isn't the one I specified — which figures do I use?: Both figures come off the unit actually standing there. Percent impedance belongs to one specific transformer rather than to a model line, and it is generally referenced to the self-cooled base kVA — so on a fan-cooled unit enter the base rating, not the forced-air uprating, or the result is overstated. A value pulled from a catalogue page, a spec section or the transformer originally ordered can easily sit a quarter of a point away from what got delivered, and that is not academic: 5.5% against 5.75% on a 500 kVA 480 V transformer is a difference of roughly 475 A. Substitutions get made at procurement, and the figure often never gets re-run afterwards. When you record the result, record the three inputs beside it — a bare ampere number tells the next person nothing about which transformer or which impedance produced it.

Preliminary estimate, not certified engineering. This tool produces an indicative quantity calculation for planning purposes only — it is not a certified structural analysis, a guaranteed material takeoff, or a substitute for building department approval. Always verify measurements on-site and have a licensed contractor or structural engineer review any load-bearing, code-sensitive, or safety-critical work before purchasing materials or starting construction. Spotted an arithmetic or standards error? Report it to contact@craftquantities.com with your inputs — a confirmed fix gets a permanent check of its own, so the same mistake cannot come back.