Why short-circuit current matters
Every protective device has to survive the fault current that can appear at its terminals. If the prospective short-circuit current is higher than a device's rated breaking capacity, the device may fail to clear the fault safely. The calculation is therefore not academic: it decides which equipment is permitted on that part of the installation.
Two questions are answered at once. The first is the magnitude of the prospective fault current at each point of the network. The second is whether the protective devices can break that current within the requirements of IEC 60364.
The same calculation supports several later decisions: it confirms the breaking capacity of breakers and fuses, feeds the selectivity study between upstream and downstream devices, and shows where the fault level is high enough to demand attention in the layout.
The quantities behind a short-circuit
A short-circuit is a fault of near-zero impedance. The current is limited only by the impedance of the path from the source to the fault and back. Three quantities define the problem:
| Quantity | Symbol | Notes |
|---|---|---|
| Voltage | U | Line voltage for three-phase, phase voltage for single-phase |
| Total impedance | Z | Sum of source, transformer and conductor impedance |
| Fault current | Ik | Derived from voltage and impedance |
For a three-phase fault the current is Ik = U / (√3 · Z), where U is the line voltage. For a single-phase fault it is Ik = U / Z, with U the phase voltage and Z the impedance of the phase-and-protective loop.
Higher impedance means lower current. That is why the fault level falls as you move away from the source: every metre of conductor adds impedance, and the lowest fault current is usually found at the end of the longest circuit.
Step 1: Establish the source impedance
The path starts at the supply. Take the prospective fault current from the supply authority, or derive the source impedance from the upstream transformer data.
Transformer impedance
For a transformer of rated apparent power Sr and per-unit impedance z%, the impedance referred to the secondary is Zt = (z / 100) · (Us² / Sr), where Us is the secondary line voltage.
Upstream network
Do not discard the impedance of the network ahead of the transformer. If it is ignored, the calculated fault current is too high and the equipment looks more stressed than it is.
Where the supply data is unknown, request it rather than assume the worst case. A calculation built on an invented source impedance is worse than no calculation, because it looks precise while resting on a guess.
Step 2: Add the conductor impedance
The conductors between the source and the fault add resistance and reactance. For short runs reactance is small, so resistance dominates and the impedance is close to the cable resistance.
Copper resistance at 20 °C can be estimated as R ≈ 17.5 / S mΩ per metre, where S is the cross-section in mm²:
| Cross-section (Cu, mm²) | R (mΩ/m) |
|---|---|
| 1.5 | 11.7 |
| 2.5 | 7.0 |
| 4 | 4.4 |
| 6 | 2.9 |
| 10 | 1.75 |
| 16 | 1.1 |
| 25 | 0.7 |
For a single-phase fault, the protective conductor is part of the loop, so its impedance is added as well. Remember that resistance rises with temperature; IEC calculations use the operating-temperature value, which is higher than the 20 °C figure and gives a more conservative result.
Long runs with large cross-sections may also carry meaningful reactance. For the kind of final circuits most engineers check, however, the resistive term is dominant and a resistance-only approximation is close enough to make the right equipment decision.
Step 3: Calculate the fault current — worked example
Take a 400 V three-phase supply from a 630 kVA transformer with z = 4 %, feeding a 25 mm² copper cable 40 m long. Only the transformer and the cable are considered.
- Transformer impedance: Zt = (4 / 100) · (400² / 630000) = 0.0102 Ω, or 10.2 mΩ.
- Cable resistance: 0.7 mΩ/m × 40 m = 28.0 mΩ.
- Total impedance: Z = 10.2 + 28.0 = 38.2 mΩ.
- Fault current: Ik = 400 / (√3 · 0.0382) ≈ 6.0 kA.
The prospective fault current at the far end of the cable is about 6.0 kA. A protective device at that point must therefore have a breaking capacity of at least 6 kA, so a device rated 10 kA is a safe choice.
Run the same example without the cable, at the transformer terminals, and the current is far higher — close to 22 kA. The two values show why every point on the network is checked separately rather than once for the whole installation.
Step 4: Verify the protective device against IEC
The calculation is only useful if it is compared with the equipment. Check the following against IEC 60364:
- The device's rated breaking capacity is at least the prospective fault current at its point of installation.
- The device disconnects within the required time — 0.4 s for 230 V final circuits, up to 5 s for distribution circuits.
- The fault current lies above the instantaneous trip threshold, or the time-current curve confirms acceptable disconnection.
- Both the highest fault current (near the source) and the lowest (at the end of a long circuit) are checked.
Current-limiting devices may be verified against manufacturer curves rather than a single current value. Where the calculated current exceeds the device rating, move the device upstream, choose a higher-rated one, or reduce the loop impedance.
The lowest fault current deserves as much attention as the highest. If it is too small to trip the device within the required time, the circuit is not adequately protected even though the breaker has plenty of breaking capacity for a larger fault.
Common mistakes to avoid
- Ignoring the upstream supply impedance, which makes the fault current look larger than it is.
- Using 20 °C conductor resistance when the standard expects the operating-temperature value.
- Forgetting the protective conductor in a single-phase loop impedance.
- Comparing the device's rated current with the fault current instead of its breaking capacity.
- Assuming the highest fault current always gives the fastest disconnection without checking the curve.
- Entering the transformer's rated current in place of its actual impedance.
Each of these errors tends in one direction — toward a fault current that looks either too large or too small — and both directions lead to the wrong equipment choice.
Checklist and next steps
Before releasing a calculation, confirm the following:
- Source and transformer impedance established from real data.
- Conductor impedance added at the operating temperature.
- Three-phase and single-phase currents calculated where both apply.
- Highest and lowest fault currents identified across the network.
- Every protective device rated at or above the local fault current.
- Disconnection times verified against IEC 60364.
GorkyCAD runs the short-circuit, heating and voltage-drop checks from one model and flags devices whose breaking capacity does not cover the fault current, so the verification follows directly from the design.
FAQ
What is prospective short-circuit current?
The current that would flow at a point if a zero-impedance fault occurred there. It is calculated from the supply voltage and the total impedance from the source to that point.
Does cable length change the fault current?
Yes. Longer runs add impedance and reduce the current, so the lowest fault current usually occurs at the end of the longest circuit, not at the board.
What breaking capacity does a breaker need?
At least the prospective fault current at its point of installation. If the calculated current is 6 kA, a device rated 6 kA or higher, such as 10 kA, is suitable.
Why use operating-temperature resistance?
IEC calculations use conductor resistance at the expected operating temperature. It is higher than the 20 °C value and gives a more conservative, safe result.