Short Circuit / Fault Current Calculator
Cable Segments
Fault Current Diagram
| Point | Impedance Factor | Available Fault (kA) |
|---|---|---|
| Source | -- | 31.38 |
| After Segment 1 | 0.6884 | 18.58 |
Every breaker and panel must have an interrupting rating equal to or greater than the fault current available where it is installed, as required by NEC 110.9 and documented under 110.24. This calculator uses the point-to-point method from IEEE 141 to find the available fault current at the source, walk it down each cable segment as impedance reduces it, and add motor contribution at the fault point.
Formula
Itransformer = (kVA × 1000) / (Vsec × √3 × Z%); Ipoint = Iupstream / (1 + f)
- kVA
- Transformer rating in kVA
- Vsec
- Transformer secondary line-to-line voltage (V)
- Z%
- Transformer impedance as a fraction (impedance percent ÷ 100)
- f
- Cable impedance factor for a segment, f = (√3 × L × Iupstream × Zper-ft) / V
- Ipoint
- Available fault current at the end of the segment
How it works
- Pick a source. A transformer source computes the secondary fault current from its kVA, impedance percent, and secondary voltage; a utility source uses the available fault current it provides directly.
- Add cable segments by conductor size, length, parallel conductors per phase, and material. Each segment's impedance factor f reduces the upstream fault current using the point-to-point relation Isc = Isc_upstream / (1 + f).
- Enter the running motor load and contribution percent. Motors briefly feed the fault at roughly four times their running current, and this is added to the downstream value for the total available fault current.
Worked example
A 1500 kVA transformer at 5.75% impedance, 480 V secondary, feeding 100 ft of two parallel 500 kcmil copper conductors per phase, with 800 A of motor load contributing at 100%.
- Transformer full-load amps: 1,500,000 ÷ (480 × √3) ≈ 1804 A. Source fault current: 1804 ÷ 0.0575 ≈ 31.38 kA.
- Cable impedance factor for the 500 kcmil run (Z effectively halved by two conductors per phase): f ≈ 0.1461.
- Downstream fault current: 31.38 ÷ (1 + 0.1461) ≈ 27.38 kA.
- Motor contribution: 800 A × 100% × 4 = 3200 A = 3.20 kA.
Source ≈ 31.38 kA, fault current after the cable ≈ 27.38 kA, total with motor contribution ≈ 30.58 kA.
Frequently asked questions
- Why does cable length reduce the available fault current?
- Conductors have resistance and reactance that add impedance between the source and the fault point. The longer or smaller the conductor, the larger the impedance factor f, and the point-to-point method divides the upstream fault current by (1 + f), so longer runs see proportionally lower fault current.
- Why include motor contribution?
- When a fault collapses system voltage, spinning motors momentarily act as generators and feed current back into the fault, typically about four times their running current for a few cycles. Ignoring this can understate the fault duty an interrupting device must handle.
- How does transformer impedance affect fault current?
- A transformer's impedance percent sets a ceiling on its secondary fault current: lower impedance means higher available fault current. A 5.75% transformer can deliver roughly 1/0.0575 ≈ 17 times its full-load current into a bolted secondary fault.
- Is the point-to-point method exact?
- It is an accepted simplified hand-calculation from IEEE 141 (the Red Book) that treats impedances as magnitudes rather than full complex vectors. It is excellent for sizing interrupting ratings, but a detailed study uses complex R and X and per-unit modeling for the most rigorous results.