Knowing what Icu and Ics mean is one thing. Knowing what fault current your board will actually see is another, and it is the number every breaking capacity decision depends on. Yet it is routinely guessed — usually by copying whatever rating was used on the last job.
Guessing fails in both directions. Underestimate and you install a breaker that cannot safely clear the fault it faces, which is a genuine safety hazard rather than a performance shortfall. Overestimate and you pay a premium across every board in the building for capacity that will never be used. This guide walks through the calculation, the adjustments that matter, and how the result maps onto a breaking capacity rating.
For a single transformer feeding a radial low-voltage system, the transformer impedance dominates everything upstream of it. The HV network is usually large enough that its contribution is small by comparison, so ignoring it gives a slightly conservative answer — which is the right kind of wrong.
The relationship is simple. Transformer impedance is quoted as the percentage of primary voltage needed to drive full-load current into a shorted secondary. A transformer with 5% impedance therefore delivers roughly 100 ÷ 5 = 20 times its full-load current into a bolted fault.
So the two-step method is:
Worked example: a 1,000 kVA transformer, 400 V secondary, 5% impedance.
That single result already matters: a 1,000 kVA transformer with 5% impedance produces close to 29 kA at its terminals. Specifying a 25 kA board immediately downstream of it is one of the more common and more serious specification errors in LV distribution.
The transformer figure is the ceiling, not the number that applies everywhere. Every metre of cable between the source and the fault adds impedance, and added impedance reduces fault current. A board 30 m from a 630 kVA transformer will see materially less fault current than the transformer terminals do.
Typical ranges in a commercial installation:
| Location | Typical prospective fault current | Why |
|---|---|---|
| Transformer LV terminals | 25–45 kA | Only transformer impedance limits it |
| Main distribution board | 20–35 kA | Short cable run from transformer |
| Sub-distribution board | 10–20 kA | 20–50 m of cable adds meaningful impedance |
| Final distribution board | 3–10 kA | Longer runs, smaller conductors |
| Socket outlet | 1–5 kA | End of a long final circuit |
This is where the money is. Specifying the same 50 kA rating at every level of a distribution system is a habit, not a calculation. A 50 kA assembly can cost substantially more than a 25 kA one of the same configuration, and across ten sub-distribution boards that difference is real capital expenditure spent on nothing.
The transformer-only formula is a sound first approximation for a single transformer feeding a radial system. It stops being adequate when:
For these cases the standard method is IEC 60909, or the equivalent national method. In practice this means power system software rather than a spreadsheet, because the moment you have parallel sources and unbalanced faults the arithmetic stops being tractable by hand.
The transformer is not the only source. Motors that are running when a fault occurs do not simply coast — for the first few cycles they act as generators and feed current back into the fault. On a board with substantial motor load, that contribution is not a rounding error.
A reasonable first approximation is to add roughly four to six times the total full-load current of the motors connected at that point, decaying within a few cycles. It matters most for the peak current and for the first-cycle duty, and it is one of the reasons the simple transformer method is treated as a floor in industrial installations rather than a final answer.
The same logic applies to on-site generation. Generators and battery-backed inverters add to the available fault current, though inverter-based sources are typically current-limited and contribute far less than a rotating machine of similar rating. Where generation is present, the source data has to come from the equipment manufacturer rather than from an assumption.
The symmetrical RMS value is what you compare against a breaking capacity. But the first half-cycle of a fault also produces a peak asymmetrical current, and that peak is what mechanically stresses busbars, supports and conductor bracing.
Ipk = κ × √2 × Isc, where κ depends on the X/R ratio of the circuit and is typically around 1.5 to 2.2 for LV systems.
Taking the 28.9 kA example with κ = 1.6 gives a peak of roughly 65 kA. A board whose busbar bracing was sized for the peak associated with a lower symmetrical rating may survive the interrupting duty and still fail mechanically. This is why assembly ratings quote both.
Once you have the prospective fault current at the point of installation, selecting the rating is a comparison — with two rules that matter more than the arithmetic:
Then step back to the device class. If the calculated level at a given point exceeds what an MCCB comfortably covers, or if the coordination study requires a declared Icw to hold a time delay, the answer at that position may be an air circuit breaker rather than a large MCCB — the trade-off we set out in MCCB vs ACB.
It is the maximum current that would flow at a given point in the installation if a zero-impedance (bolted) fault occurred there. It is determined by the source impedance and everything between the source and that point.
Calculate the transformer’s full-load current from its kVA and secondary voltage, then divide by the impedance expressed as a decimal — equivalently, multiply full-load current by 100 ÷ Z%. For a 1,000 kVA, 400 V, 5% transformer that gives about 28.9 kA at the terminals.
Yes, sometimes substantially. Cable impedance between the source and the fault reduces the fault current, which is why a sub-distribution board 30 m from the transformer sees far less than the transformer terminals do. Long runs and small conductors reduce it most.
Icu tells you the maximum the device can interrupt at all. Ics tells you the level at which it will still be serviceable afterwards. Where the circuit needs to remain operational after clearing a fault, Ics is the more appropriate and more conservative figure to check against.
The peak asymmetrical current in the first half cycle is higher than the symmetrical RMS value, typically by a factor of 1.5 to 2.2 depending on the X/R ratio. It governs the mechanical withstand of busbars and supports, so the assembly rating must cover it even when the breaker’s interrupting rating is adequate.
The fault level study is not an academic exercise — it is the input that decides whether your protection is safe and whether your switchgear budget is spent where it does something. One calculation, done properly and recorded, saves both.
If you are sizing MCCBs and want the ratings checked against your transformer data and cable runs, send us the single-line diagram. Contact our engineering team and we will confirm the breaking capacity you actually need at each level.
