How to use this calculator
- Enter the transformer rating in kVA, the secondary line-to-line voltage and the impedance voltage uk (Z%) from the nameplate.
- Optionally enter the load loss Pk in watts so the impedance is split into resistance and reactance.
- Leave c at 1.0 for the plain formula, or enter cmax from IEC 60909-0 for a maximum-current basis. Optionally enter the upstream fault level in MVA.
- To find the current at a distribution board, tick the cable option and enter the cable length, size, material, parallel runs and reactance.
- Read the current at the transformer terminals and at the cable end, and compare them with the breaking capacity of the devices.
How it works: the formula
The impedance voltage is the percentage of rated voltage that drives rated current through the shorted transformer. With the full voltage applied, the current is therefore In divided by Z in per unit. Converting the impedance to ohms on the secondary side lets you add the cable and the upstream network as vectors, R with R and X with X. The cable is taken at 20 °C because its lowest resistance gives the highest fault current.
Variables and units
| Symbol | Meaning | Unit |
|---|---|---|
| S | Transformer rated power | kVA |
| V | Secondary line-to-line voltage | V |
| Z%, uk | Short-circuit impedance (impedance voltage) | % |
| Pk | Load loss at rated current | W |
| c | Voltage factor (IEC 60909-0) | — |
| Ssc | Upstream short-circuit power | MVA |
| L, A, n | Cable route length, conductor size and parallel runs | m, mm², — |
| x | Cable reactance per metre | mΩ/m |
Worked example
1000 kVA, 400 V, Z = 5 %, infinite source:
With an assumed load loss of 10 500 W and 50 m of 240 mm² copper (X = 0.08 mΩ/m) to a main board:
Assumptions
- Three-phase symmetrical bolted fault on the LV side.
- Infinite upstream source unless you enter its fault level, which is then treated as pure reactance.
- Without the load loss, the transformer impedance is treated as purely reactive.
- Cable resistance at 20 °C from resistivity (Cu 0.017241, Al 0.028264 Ω·mm²/m).
Limitations
- Motor contribution, which adds to the initial fault current, is ignored.
- Arc resistance, which reduces real fault currents, is ignored.
- Single-phase and earth-fault currents (needed for disconnection checks) are not calculated.
- Equipment ratings (Icu, Icw, peak withstand) should be based on a short-circuit study to IEC 60909-0 with the real network data, including the peak current ip and the DC component.
- Transformers in parallel multiply the fault current; enter their combined data only if they are identical and share the load.
References
Standards and texts that cover this topic. Use the edition adopted in your jurisdiction.
- IEC 60909-0, Short-circuit currents in three-phase AC systems — Calculation of currents (voltage factor c, impedance method).
- IEC 60076-5, Power transformers — Ability to withstand short circuit.
- IEC 60076-1, Power transformers — General (impedance voltage, load loss and tolerances).
- IEC 60364-4-43, Protection against overcurrent (breaking capacity at least the prospective fault current).
- IEC 60364-4-41, Protection against electric shock (disconnection times using fault currents).
- BS 7671, Requirements for Electrical Installations — Regulation 434.1 (determination of prospective fault current).
Frequently asked questions
Why is the infinite-source result the usual first estimate?
Ignoring the upstream network gives the highest possible current at the terminals, so it is on the safe side for choosing breaking capacity. Real networks have some source impedance, which reduces the current slightly; enter the fault level from the utility to include it.
What is the voltage factor c?
IEC 60909-0 multiplies the nominal voltage by a factor c to cover voltage variations. For maximum currents in low-voltage systems cmax is 1.05 or 1.10 depending on the voltage tolerance. The plain formula uses c = 1.
Does the load loss change the terminal fault current?
No. The terminal current depends only on |Z|. Splitting Z into R and X matters once a cable is added, because cable impedance is mostly resistive and adds to the transformer's R and X separately.
Why is the cable taken at 20 °C?
A cold cable has its lowest resistance, which gives the highest fault current. That is the right basis for checking breaking capacity. For disconnection-time checks the minimum fault current with the cable at its operating temperature is needed instead.
How fast does the current fall along a cable?
Quickly for small cables. In the example, 50 m of 240 mm² copper reduces 28.87 kA to 17.84 kA. A long run of a small cable can cut the fault current to a few kiloamperes, which may then be too low for fast magnetic tripping.