What the Short-Circuit Current Calculator does
This calculator estimates the prospective short-circuit current at a board, from the transformer rating and impedance, the upstream network fault level if you have it, and the cable run between the source and the board. It shows each element's contribution to the total impedance, the fault current at the source and at the board, and compares the result with the breaking capacity of the device you intend to fit.
It is a first check rather than a fault study. Impedances are added as scalars unless you supply an X/R ratio, and an unspecified supply is treated as an infinite bus - both of which overestimate the fault current. That is the safe direction for choosing a breaking capacity and the wrong direction for proving a device will trip, which is why the page keeps the two cases apart.
How to use it
- Enter the transformer rating and its nameplate impedance, the upstream network fault level from the supplier if you have it, or both. Leaving the network out treats it as infinitely strong.
- Set the voltage factor c. IEC 60909 uses 1.05 for the maximum low-voltage fault current and 0.95 for the minimum.
- Add the cable between the source and the board. On low-voltage systems a cable run of any length is often the largest single impedance in the loop.
- Choose the conductor temperature deliberately: 20 °C for the maximum fault current, the operating temperature for the minimum.
- Enter the breaking capacity of the device you plan to fit, and read the verdict.
- Use the distance table to see how fast the fault level falls along the cable - it is often the reason a sub-board can take a much cheaper device than the main switchboard.
Reading the results
Fault current at the source is what a device mounted on the transformer terminals would have to interrupt. It is the worst case in the installation.
Fault current at the point of installation is the figure to compare with a device's Icu. Cable impedance reduces it, sometimes dramatically.
Peak current is the first asymmetrical peak, which depends on the X/R ratio and on where in the voltage cycle the fault starts. It is what a device's making capacity has to survive, and it can be over twice the rms symmetrical value.
Icu is the ultimate breaking capacity - the device interrupts the fault but may not be fit for further service. Ics is the service breaking capacity, after which it still works. Which one your design needs is a design decision, not an arithmetic one.
Worked example: a 1000 kVA transformer at 5% impedance feeding a board 60 m away
The transformer's base impedance is 400² / 1,000,000 = 0.16 ohm, so at 5% its impedance is 0.008 ohm. With c = 1.0 the fault current at its terminals is 400 / (root-three x 0.008) = 28,868 A, or 28.9 kA. The quick cross-check confirms it: full-load current is 1,443 A and dividing by the per-unit impedance of 0.05 gives the same 28.9 kA.
Now add 60 m of 95 mm² copper at 20 °C. Its resistance is 1.724 x 10⁻⁸ x 60 / 95 x 10⁻⁶ = 0.01089 ohm, and with 0.08 ohm/km of reactance it adds 0.0048 ohm of reactance, for a magnitude of 0.01190 ohm.
Total impedance is 0.008 + 0.01190 = 0.01990 ohm, and the fault current at the board falls to 400 / (root-three x 0.01990) = 11,606 A, or 11.6 kA - a 60% reduction from the transformer terminals, caused entirely by 60 m of a fairly generous cable.
That means a 25 kA device is comfortable at this board, where the main switchboard would need 36 kA. It is also a warning: shorten that cable to 10 m and the fault level jumps back to about 22 kA, so the device selection is sensitive to a routing change that might be made on site without anyone revisiting the calculation.
With an X/R ratio of 10 the peak would be around 1.75 x root-two x 11.6 = 28.7 kA, which is the figure the device's making capacity has to survive.
Formulas and scoring rules
- Transformer base impedance
Z_base = V^2 / SV is the secondary line voltage, S the rating in VA. The result is in ohms referred to the secondary.- Transformer impedance
Z_tx = (%Z / 100) x Z_base- Source impedance from a fault level
Z_source = V^2 / S_scS_sc is the network fault level in VA, as quoted by the supplier.- Cable impedance
Z_cable = sqrt(R^2 + X^2)R = rho(T) x L / A per conductor; X from the cable's reactance per kilometre.- Three-phase fault current
Isc = c x V / (root-three x Z_total)- Single-phase fault current
Isc = c x V / (2 x Z_total)Both the line and the return conductor are in the loop.- Peak current
ip = sqrt(2) x kappa x Isc, with kappa = 1.02 + 0.98 x exp(-3 / (X/R))The IEC 60909 approximation for the first asymmetrical peak.
Why the infinite-bus assumption matters
Dividing full-load current by the per-unit impedance gives the fault current the transformer alone would allow, as though the network feeding it had no impedance at all. Real networks do: a supply quoted at 250 MVA at 11 kV contributes a real impedance that reduces the low-voltage fault current, typically by 5 to 15% for a distribution transformer.
Overestimating is the right way to be wrong when you are choosing a breaking capacity, because the device will simply be stronger than it needs to be. It is the wrong way to be wrong when you are checking that a fault will trip a device within the required disconnection time, because the real fault current will be lower than your calculation and the device may be slower than you predicted. That calculation - the minimum fault current at the far end of the circuit, with a hot conductor and c = 0.95 - is a separate exercise.
What this page leaves out
A proper IEC 60909 study adds complex impedances rather than magnitudes, which matters when the source and the cable have very different X/R ratios; models the motor contribution, because large connected motors act as generators for the first few cycles and add to the fault; separates the symmetrical, peak, breaking and steady-state currents; and handles asymmetrical faults, which on some systems produce a higher current than the three-phase case.
For sizing a device on a small installation, the scalar estimate here with a conservative c factor is a reasonable first pass. For a switchboard specification, an arc-flash study or anything with significant rotating plant, it is not a substitute for the real calculation.
Limitations: what the result does not prove
- Impedances are added as magnitudes rather than as complex numbers unless an X/R ratio is given, which slightly overestimates the fault current when the angles differ.
- Motor contribution to the fault is not modelled. Large connected motors feed a fault for the first few cycles and raise both the peak and the breaking current.
- Only the balanced three-phase and the simple single-phase loop cases are covered. Line-to-earth faults in a TN system, and faults in a TT or IT system, behave quite differently.
- It gives the maximum-current case. The minimum fault current that must operate the protective device is a separate calculation with a lower c factor and a hot conductor.
- It is not an arc-flash study and says nothing about incident energy or PPE.
- Nothing here is a compliance certificate. A qualified engineer must verify the design against IEC 60909, the applicable standard and the local regulations.
Privacy: where your data goes
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Standards and sources
- IEC 60909-0 - Short-circuit currents in three-phase AC systems - checked 19 Sep 2026
- IEC 60947-2 - Low-voltage switchgear: circuit-breakers - checked 19 Sep 2026
- IEC 60076-1 - Power transformers: general (impedance voltage)
Frequently asked questions
How do I calculate short-circuit current from a transformer?
Divide the transformer's full-load current by its per-unit impedance: a 1000 kVA 400 V transformer has a full-load current of 1,443 A and at 5% impedance gives about 28.9 kA at its terminals. Equivalently, work out the base impedance as V²/S, multiply by %Z/100, and divide the voltage by root-three times that impedance.
What is the voltage factor c?
A multiplier in IEC 60909 that accounts for the supply voltage varying from nominal, transformer tap positions and load conditions. For low-voltage systems it is 1.05 when calculating the maximum fault current and 0.95 for the minimum. Using 1.0 gives the bare nominal-voltage answer.
Does cable length really reduce fault current that much?
Yes, and it is the single most useful fact in low-voltage fault calculation. Sixty metres of 95 mm² copper cuts a 28.9 kA fault to about 11.6 kA. That is why a distant sub-board can take a far cheaper device than the main switchboard - and why shortening a cable run on site without rechecking is a genuine hazard.
What is the difference between Icu and Ics?
Icu is the ultimate short-circuit breaking capacity: the device interrupts the fault safely but may not be usable afterwards. Ics is the service short-circuit breaking capacity: the device interrupts the fault and remains fit for normal service. Ics is often 50%, 75% or 100% of Icu, and which one a design must satisfy depends on the application and the standard being worked to.
Can I use a device with a lower rating than the fault current?
Only through cascading, also called back-up protection, where an upstream current-limiting device cuts the fault before the downstream one has to interrupt it fully. That is valid only for combinations the manufacturer has tested and published; it can never be inferred from the ratings of two devices from different makers.
What X/R ratio should I use?
For a low-voltage distribution transformer, 5 to 15 is typical, with larger transformers at the higher end. Cable runs have a much lower X/R because their resistance dominates, so the ratio at a distant board is smaller than at the transformer terminals. The X/R only affects the peak estimate here, not the rms symmetrical current.
Is the fault current at a socket outlet the same as at the board?
No - it is much lower, because the final circuit's cable adds substantial impedance. That is the point of the minimum fault current calculation: at the far end of a long final circuit, the earth fault current may be barely enough to operate the protective device within the required disconnection time, which is a different failure mode from a device being unable to break the current.
Does this tool do arc-flash calculations?
No. Arc-flash incident energy depends on the arcing current (not the bolted fault current), the arc duration set by the protective device's actual clearing time, the electrode configuration and the working distance, and it is calculated under IEEE 1584 or a similar method. This page gives one of the inputs to that study, not the study.
Last reviewed by the A2Z.Tools team against the sources listed above.
