Available fault current calculator
Free available fault current calculator: transformer kVA, %Z and voltage, then the point-to-point method along the conductor, with working shown.
Project inputs
Updates as inputs change. Unit changes preserve the same physical values.
Method & assumptions
Point-to-point method with an unlimited utility source (conservative), impedance magnitudes added, and the conductor C value from its resistance (K ÷ cmil) and the entered reactance. Single phase gives the line-to-line fault; line-to-neutral faults at a center-tapped transformer differ. Not a coordination study or an arc-flash assessment; the utility's available fault current takes precedence.
Results do not check every aspect of an electrical installation. The calculators follow US (NEC) practice only; a regional preset changes defaults such as voltage and units, and no result certifies NEC compliance.
Sources and validation →Worked example
Transformer rating: 50 kVA · Secondary voltage (line to line): 240 V · Transformer impedance (nameplate %Z): 2 % · Impedance used: Nameplate %Z · Transformer: Single phase (line-to-line fault) · Conductor size: 4/0 AWG · Conductor material and temperature: Aluminum, 25 °C (highest fault current) · Conductor length (one way): 100 ft · Conductors in parallel per phase: 1 · Conductor reactance: 0.05 Ω/1000 ft
Available fault current at the end of the run: 5,644.077 A · Fault current at the transformer terminals: 10,416.667 A · Transformer full-load current: 208.333 A · Conductor C value: 10,265.648 · f factor: 0.846 · Multiplier M: 0.542
This example uses the stated inputs; results for the entered values appear above.What available fault current is
Available fault current is the largest current that would flow at a point in an electrical system if the conductors there were connected together through a fault of negligible impedance, a bolted fault. It is highest at the transformer's secondary terminals and falls along every conductor run, because each foot of conductor adds impedance between the source and the fault.
The figure decides whether equipment can safely interrupt or withstand a fault. NEC 110.9 requires every device intended to interrupt current at fault levels to have an interrupting rating at least equal to the current available at its line terminals, and NEC 110.10 requires the equipment's short-circuit current rating and the circuit's impedance to allow protective devices to clear a fault without extensive damage. For service equipment in other than dwelling units, NEC 110.24(A) requires the maximum available fault current and the date of the calculation to be marked in the field.
How the point-to-point method works
- Transformer full-load current: FLA = kVA × 1,000 ÷ V for a single-phase transformer, or kVA × 1,000 ÷ (√3 × V) for a three-phase transformer, with V the secondary line-to-line voltage.
- Fault current at the terminals: Isc = FLA × 100 ÷ %Z. This treats the utility supply as unlimited, which gives the highest possible value. Transformer impedance has a manufacturing tolerance, so the calculation can use the nameplate %Z less 10% as a worst case.
- The f factor for the conductor run: f = k × L × Isc ÷ (C × n × V), where k is 2 for a single-phase line-to-line fault and √3 for a three-phase fault, L is the one-way length in feet, n is the number of conductors in parallel per phase and C is the conductor's constant.
- Multiplier and result: M = 1 ÷ (1 + f), and the available fault current at the end of the run is Isc × M.
The constant C is the reciprocal of the conductor's impedance per foot, C = 1,000 ÷ Z(per 1,000 ft). The calculator works it out from the conductor's resistance, K × 1,000 ÷ cmil, and the reactance entered, so it covers any size and material in the list. Published C tables, which also depend on the conduit type, give similar values; the reactance can be adjusted to match manufacturer data or NEC Chapter 9 Table 9.
Worked example
A 50 kVA, 120/240 V single-phase transformer with a nameplate impedance of 2.0% feeds a house through 100 ft of 4/0 AWG aluminum service conductors. The full-load current is 50,000 ÷ 240 = 208.3 A, and the fault current at the transformer terminals is 208.3 × 100 ÷ 2.0 = 10,417 A.
For the conductors, with aluminum at 25 °C and a reactance of 0.05 Ω per 1,000 ft, C is about 10,266. Then f = 2 × 100 × 10,417 ÷ (10,266 × 1 × 240) = 0.846, M = 1 ÷ (1 + 0.846) = 0.542, and the available fault current at the service equipment is about 5,644 A. The example transformer and conductors are illustrations; the utility's figure for an actual service takes precedence.
Table
| Length (ft) | 2 AWG copper | 1/0 AWG aluminum | 4/0 AWG aluminum | 350 kcmil aluminum |
|---|---|---|---|---|
| 0 | 10,417 | 10,417 | 10,417 | 10,417 |
| 25 | 7,608 | 7,552 | 8,599 | 9,024 |
| 50 | 5,992 | 5,923 | 7,321 | 7,960 |
| 100 | 4,206 | 4,138 | 5,644 | 6,441 |
| 150 | 3,240 | 3,179 | 4,592 | 5,409 |
| 200 | 2,635 | 2,582 | 3,871 | 4,662 |
| 300 | 1,918 | 1,876 | 2,945 | 3,653 |
Computed with the calculator above, one conductor per phase and a reactance of 0.05 Ω per 1,000 ft. The transformer and conductors are examples; at 0 ft the value is the fault current at the transformer terminals.
Limitations
This is not a short-circuit coordination study or an arc-flash assessment. It estimates one value, the bolted fault current at one point, and does not select protective devices, check their coordination or calculate incident energy.
The method assumes an unlimited utility source, adds impedance magnitudes rather than complex impedances, and gives the line-to-line fault for single-phase systems; a line-to-neutral fault at a center-tapped transformer can be higher near the transformer. Motor contribution, which adds to the fault current in systems with large running motors, is not included. Equipment ratings rely on the utility's available fault current and, where required, an engineering study.
Frequently asked questions
What is available fault current?
The largest current that a bolted short circuit could draw at a given point in the system. It is set by the source, mainly the utility transformer's rating and impedance, and by the conductors between the source and the point.
How is available fault current calculated at a service?
The transformer's fault current is its full-load current times 100 divided by its percent impedance. The point-to-point method then reduces it for the service conductors with the f factor and the multiplier M = 1 ÷ (1 + f). Utilities provide the figure for a service on request.
Why does fault current fall with conductor length?
Every foot of conductor adds resistance and reactance between the source and the fault. The longer and smaller the conductors, the larger the f factor and the lower the current that can reach the fault.
What interrupting rating do breakers need?
At least the available fault current at their line terminals, under NEC 110.9. Equipment such as panelboards also carries a short-circuit current rating that must suit the available fault current under NEC 110.10.
Is this calculator an arc-flash study?
No. It estimates the bolted fault current at one point. An arc-flash assessment and a coordination study need the full system data, protective device characteristics and an engineering analysis.
Estimating and invoicing tool
Ampxa is developing an estimating and invoicing application for electrical contractors. The list below receives notice when it opens.