Transformer Fault Current & AIC Calculator
Enter the transformer, upstream source, feeder, and device rating. The calculator estimates symmetrical bolted-fault current at the transformer secondary and at the feeder end, then checks the entered interrupting rating.
Fault current at feeder end4,810.69
- Interrupting-rating check
- PASS
- Device AIC margin
- 107.87%
- Transformer full-load current
- 312.27
- Infinite-source terminal fault current
- 5,430.76
- Finite-source terminal fault current
- 5,430.76
Impedance referred to transformer secondary
| Component | Resistance (ohms) | Reactance (ohms) |
|---|---|---|
| Transformer | 0.01 | 0.02 |
| Primary source | 0 | 0 |
| Feeder | 0 | 0 |
Currents are RMS symmetrical amperes for a bolted fault at the selected point.
Primary fault current of zero is treated as an infinite upstream source.
Conductor values use a compact NEC Chapter 9 Table 9 lookup at 75 C conditions.
How to use this calculator
- Enter the transformer phase, kVA, voltages, percent impedance, and transformer X/R ratio.
- Enter the available primary fault current, or leave it at zero to model an infinite primary source.
- Choose the phase-conductor material, size, raceway type, feeder length, and number of parallel sets.
- Enter the downstream device interrupting rating in kA RMS symmetrical.
- Compare the feeder-end fault current with the AIC check and margin.
How the fault current is calculated
This calculator treats the fault path as a series impedance on the transformer secondary side. It begins with transformer full-load current, then converts transformer percent impedance, finite upstream source impedance, and feeder impedance into resistance and reactance before calculating the symmetrical RMS current.
Let S be transformer kVA times 1,000, g be 1 for single-phase or sqrt(3) for three-phase, Vs be secondary voltage, and zpu be transformer percent impedance divided by 100. Full-load current is:
IFL = S / (g x Vs)
The infinite-source terminal current is:
Iinf = IFL / zpu
The transformer impedance magnitude referred to the secondary is:
|Zt| = zpu x Vs^2 / S
Each impedance magnitude is split with its X/R ratio: R = |Z| / sqrt(1 + q^2) and X = q x R. If the entered primary fault current is greater than zero, the primary source impedance is Vp / (g x Ip) and is referred to the secondary by multiplying by (Vs / Vp)^2. If primary fault current is zero, the source impedance is set to zero.
The selected conductor size supplies resistance and reactance in ohms per 1,000 ft. Feeder impedance is multiplied by length divided by 1,000 and divided by the number of parallel sets. Single-phase two-wire faults include both the outgoing and return conductor, so feeder resistance and reactance are doubled.
The terminal and feeder-end fault currents are then:
I = Vs / (g x sqrt(Rtotal^2 + Xtotal^2))
What moves the result most
Transformer percent impedance usually has the largest effect near the transformer terminals. A lower percent impedance or larger transformer kVA raises available fault current. Farther downstream, conductor length, conductor size, raceway reactance, and parallel sets can materially reduce the current available at the device.
A finite utility or upstream source also lowers the terminal current. That is why the infinite-source value is useful as a quick upper bound, but it can overstate the available current when the upstream system is weak or the transformer primary voltage is high.
What this calculator leaves out
The estimate is for a bolted, symmetrical RMS short circuit. It does not include arcing impedance, motor contribution, generator contribution, current-limiting fuses, transformer manufacturing tolerance, temperature changes, enclosure details, or protective-device time-current behavior. Interrupting ratings also depend on voltage, equipment class, series ratings, and the actual installation.
Worked example
For the default three-phase example, a 112.5 kVA transformer at 208 V secondary has a full-load current of 312.27 A. With 5.75% impedance, the infinite-source secondary-terminal fault current is 5,430.76 A.
The default primary source is infinite, so the finite-source terminal current is also 5,430.76 A. A 50 ft run of one set of 4/0 copper conductors in PVC or aluminum raceway adds 0.00305 ohms resistance and 0.00205 ohms reactance per phase. The feeder-end fault current is 4,810.69 A. A 10 kA interrupting rating is therefore above the calculated duty, with a margin of 107.87%.
Common questions
How do I calculate fault current from transformer kVA and percent impedance?
For an infinite primary source, divide transformer full-load current by transformer per-unit impedance. Full-load current is transformer VA divided by secondary voltage, and by sqrt(3) for three-phase systems. This calculator also adds finite source and feeder impedance when you enter those details.
Why is actual secondary fault current lower than the infinite-bus value?
The infinite-bus value assumes the upstream system has no impedance, so only the transformer limits the current. Real utility systems, service conductors, and upstream transformers add impedance. That extra impedance reduces the current available at the transformer secondary.
Does feeder length reduce available fault current?
Yes. A longer feeder adds more resistance and reactance between the transformer and the fault. Larger conductors and more parallel sets reduce that added impedance, while a short or zero-length feeder makes the feeder-end value approach the terminal value.
What is the difference between available fault current and AIC?
Available fault current is the current the electrical system can deliver at a point during a short circuit. AIC, or interrupting rating, is the marked current a protective device is rated to interrupt under specified conditions. The rating should be at least the available fault current at that device.
Should single-phase fault calculations include both conductors?
For a two-wire single-phase bolted fault, current leaves on one conductor and returns on the other. This calculator therefore doubles the feeder resistance and reactance for single-phase feeder runs. The transformer and source impedances remain referred to the secondary side.
Does this replace an arc-flash study?
No. This is a preliminary bolted-fault and interrupting-duty estimate. Arc-flash studies require additional equipment data, protective-device clearing times, arcing-current calculations, working distances, and applicable standard methods.