Prospective Fault Current Calculator

Inputs

All fields support decimals. Units are shown in labels.

System and fault

System

Single‑phase uses the L‑N voltage for calculations.

Fault type

L‑N uses an approximate 2× loop length.

Line voltage (L‑L), V *

Phase voltage (L‑N), V *

Typical: 230 V or 120 V depending on region.

Source X/R ratio

Used to split source impedance into R and X.

Motor contribution factor

Use 1.00 if no motor contribution is expected.

Safety factor

Optional multiplier for conservative design checks.

Source model

Method

Transformer rating, kVA *

Transformer impedance, %Z *

Nameplate impedance. Typical range: 4–8%.

Supply fault level, kA *

Use the declared prospective short‑circuit current.

Cable and location

Cable length, meters

Parallel runs

Temperature, °C

Material

Conductor size, mm²

Manual R and X per km

Manual entries override size and material tables.

R (Ω/km)

X (Ω/km)

What you get

Prospective fault current at the selected point.
Total impedance breakdown (source + cable).
Fault level in kVA and peak-current estimate.
CSV and PDF results for documentation.

Quick guidance

Select system and fault type.
Choose the source method and fill values.
Enter cable details to the fault location.
Press Calculate and review the results card.

Typical checks

Device breaking capacity ≥ adjusted fault current.
Energy withstand and cable adiabatic limits.
Disconnection time requirements for the earthing system.

Formula used

Transformer method

Full‑load current (3‑phase): I_FL = (kVA×1000)/(√3×V_LL)
Prospective at terminals: I_SC = I_FL×(100/%Z)

Impedance approach

Source impedance magnitude: |Z_s| = V/(√3×I) (3‑phase)
Cable impedance: R = (RΩ/km)×L×m, X = (XΩ/km)×L×m
Total: |Z| = √(R² + X²)
3‑phase fault current: I = V_LL/(√3×|Z|)
L‑N fault current: I = V_LN/|Z|

This calculator uses typical conductor impedance values and a simplified X/R split for the source. For final designs, use verified data and a formal short‑circuit study.

How to use this calculator

Pick the system: choose three‑phase or single‑phase and select the fault type.
Choose a source model: transformer (%Z) or declared supply fault level (kA).
Enter voltages: use L‑L for three‑phase; include L‑N when needed.
Describe the run: cable length, parallel runs, material, size, and temperature.
Optional precision: switch to manual R and X per km if you have data.
Review results: compare adjusted fault current to device ratings and limits.
Export: use CSV/PDF buttons in the results card for records.

Example data table

Scenario	System	Source	Cable	Result (approx.)
Small panel near transformer	3‑phase 400 V	500 kVA, 5.75%Z	30 m Cu 50 mm²	≈ 7–10 kA (depends on X/R and temperature)
Long feeder to sub‑board	3‑phase 400 V	10 kA supply level	120 m Cu 35 mm²	Lower kA due to higher cable impedance
Socket circuit L‑N fault	1‑phase 230 V	5 kA supply level	40 m Cu 6 mm² (loop 2×)	Compare with disconnection time requirements

Example rows are illustrative and vary by installation details.

Professional article

Why prospective fault current matters on sites

Prospective fault current is the available short-circuit current at a point. It sets breaking capacity, busbar withstand, and incident energy. Temporary boards, long leads, and changing supplies during construction can shift values across the site. Use it early to confirm device ratings on sites.

Source impedance from transformers and utility

The tool models the source from transformer percent impedance or a declared supply level. Example: a 500 kVA transformer at 400 V has about 721 A full-load. With 5.75% impedance, terminals can see roughly 12.5 kA before cable losses. Utility services often fall between 6 and 25 kA.

Using X/R to estimate peak making duty

Peak making duty depends on the network X/R ratio. Many low-voltage systems fall around X/R 5 to 15. Higher X/R raises the first peak, so check both symmetrical RMS and peak duties for breakers, contactors, and switchgear.

Cable length, parallel runs, and temperature impacts

Cables add resistance and reactance that reduce fault current with distance. Doubling length roughly doubles impedance, lowering kA at remote boards. Parallel runs divide impedance and can increase current. Copper resistance rises about 0.39% per °C above 20°C.

Comparing three-phase and line-to-neutral faults

Three-phase faults often give the highest RMS current for breaking checks. Line-to-neutral faults depend on the full return path and earthing. This calculator uses an approximate two-times loop length for L-N faults; verify loop impedance for compliance work.

Interpreting results for device selection

Compare the adjusted fault current to interrupting ratings at the installation point. If the value is too high, use higher-rated devices, current-limiting fuses, or added impedance. Also review coordination so downstream protection clears first where practical. Document any assumed margins used in the safety factor.

Documentation and commissioning checks

Capture assumptions and export CSV or PDF for records. During commissioning, confirm transformer impedance, cable lengths, and the utility’s declared fault level. Any change in supply or routing should trigger a quick recalculation and an updated panel schedule. Recheck after adding parallel feeders or large motors.

Limits and when to run a detailed study

This calculator uses typical impedances and a simplified source split. Large motors, generators, multiple transformers, or complex earthing need a formal short-circuit study with verified manufacturer data and network modeling. Treat outputs as estimates and validate critical points. Always align results with local standards and project specifications.

FAQs

What is prospective fault current?

It is the maximum current that could flow during a short circuit at a location, before protective devices operate. It depends on source strength and the impedance of cables, transformers, and connections.

Which input method should I use?

Use the transformer option when you know kVA and percent impedance. Use the supply-level option when the utility provides a prospective short-circuit current at the service point or main switchboard.

Why does cable length reduce the current?

Longer conductors add resistance and reactance, increasing total impedance. Because fault current is inversely related to impedance, more impedance means fewer amperes available at the fault point.

How should I pick the X/R ratio?

If you have a study, use its X/R. Otherwise, a typical low-voltage assumption of 5–15 is common. Higher values increase the peak current estimate and can affect making-duty checks.

What does the motor factor do?

It applies an optional multiplier for motor contribution to the initial fault current, often relevant near large motors. Set it to 1.00 if motors are not significant at the selected point.

Is the line-to-neutral result code compliant?

It is an approximation. Real L-N values depend on the earthing system and the full return path. For compliance and disconnection times, confirm loop impedance using verified methods and measurements.

How do I use the outputs for equipment selection?

Compare the adjusted fault current to the interrupting rating of breakers, fuses, and switchgear at that point. Also check busbar withstand and coordination so upstream protection supports safe, selective operation.

Practical tips

Use manufacturer data for cable impedance when available.
Account for earthing arrangement and return-path impedance.
Check both symmetrical RMS and peak making duties.

Always verify results against local codes and studies carefully.