Calculator Inputs
Enter transformer, cable, and protective device details. Results appear above this form after calculation.
Example Data Table
These example values are illustrative and help verify your form setup before using site-specific project inputs.
| Scenario | Supply | Voltage | kVA | Z% | Length | Phase/CPC | Material | Phase Fault | Earth Fault |
|---|---|---|---|---|---|---|---|---|---|
| Site MDB feeder | Three phase | 415 V | 500 | 5.75 | 25 m | 95 / 50 mm² | Copper | 5.297 kA | 4.313 kA |
| Portable cabin board | Single phase | 240 V | 100 | 4.00 | 15 m | 25 / 16 mm² | Copper | 3.669 kA | 2.832 kA |
Formula Used
1) Full Load Current
Single phase: IFL = S / V
Three phase: IFL = S / (√3 × VLL)
2) Transformer Terminal Fault Current
Ik,tx = IFL × 100 / Z%
3) Transformer Equivalent Impedance
Ztx = Vph / Ik,tx
4) Cable Resistance at Operating Temperature
Rθ = (ρ × L / A) × [1 + α(θ − 20)]
5) Prospective Phase Fault Current
Ipf,phase = Vph / (Ztx + Rphase + Zext)
6) Prospective Earth Fault Current
Ipf,earth = Vph / (Ztx + Rphase + RCPC + Zext)
7) Design Fault Level
Idesign = max(Ipf,phase, Ipf,earth) × (1 + margin%)
How to Use This Calculator
- Select whether the circuit is single phase or three phase.
- Enter supply voltage, transformer rating, and transformer impedance.
- Add cable length, phase conductor size, and CPC size.
- Choose conductor material and operating temperature.
- Enter any known external source impedance from upstream data.
- Input breaker rating and its interrupting capacity.
- Apply a design margin if you want a conservative check.
- Press the calculate button to see results above the form.
- Review the graph, protection pass/fail status, and recommended breaking capacity.
- Download the CSV or PDF report for documentation.
FAQs
1) What is prospective fault current?
Prospective fault current is the maximum current that could flow during a fault at a chosen point. It helps engineers verify that breakers, switchgear, and cables can safely withstand and interrupt abnormal conditions.
2) Why does cable length reduce fault current?
Longer cables add more resistance to the fault path. As total impedance rises, available fault current falls. That is why remote distribution boards often show lower fault levels than points near the transformer.
3) Why is earth fault current usually lower?
Earth fault current often travels through both the phase conductor and the CPC or earth return path. The extra path resistance increases loop impedance, so the resulting current is commonly lower than a phase fault level.
4) Why is transformer impedance important?
Transformer impedance limits how much current can flow during a short circuit. Lower impedance means higher available fault current. Higher impedance reduces fault current but can also affect voltage regulation and coordination choices.
5) Should I use operating temperature or ambient temperature?
Use the expected conductor operating temperature, not just the room temperature. Hotter conductors have higher resistance, which reduces calculated fault current. This produces a more realistic estimate for loaded construction circuits.
6) Does this replace a full protection study?
No. It is a practical estimating tool. Formal studies may need utility data, reactance, motor contribution, breaker curves, cable reactance, and local code checks before approving final protection settings or equipment schedules.
7) Why compare with breaker interrupting capacity?
A breaker must interrupt the highest fault current that could appear at its location. If the available fault current exceeds the device rating, the breaker may fail during operation and create a severe safety hazard.
8) Why add a design margin?
A margin helps account for uncertainty, future changes, and conservative design practice. It is useful when upstream data is incomplete or when you want additional confidence before selecting switchgear and protective devices.