Available fault current is the largest amount of current that can travel in a circuit during a fault condition—usually a short circuit between conductors, a line-to-earth fault, or a line-to-neutral fault. It is based upon how much electrical energy the source can supply and how much impedance is between the source and the point of fault.

The lower the circuit impedance, the greater the fault current. That’s why faults at or near transformers or service points are more likely to be serious than those at the end of a long distribution feeder.

Why Fault Current Matters

Knowing what fault current is available is critical when choosing protective devices. Each device has an interrupting rating—a highest fault current it can interrupt safely. If the fault current to be interrupted is higher than this rating, the device can fail to interrupt the circuit safely.

In switchboards, fault withstand ratings also exist for the internal parts—copper busbars and supports, for example. When the fault current goes beyond these ratings, catastrophic physical damage can happen. That’s why it’s not all about design compliance: knowing your available fault current is an issue of safety.

Step 1: Determine the Source of Supply

Your reference point is the point of supply. Most commercial and industrial buildings are fed by a utility transformer. You’ll usually need to be aware of:

The transformer’s rated kVA or MVA
The voltage level (e.g., 400 V, 11 kV)
The transformer’s impedance (typically stated as a percentage)

Assume you have a 1000 kVA, 400 V transformer with an impedance of 6%. The first thing to calculate is its full-load current (FLA):

FLA = (kVA × 1000) / (√3 × Voltage) = (1000 × 1000) / (√3 × 400) ≈ 1443 A
  

Now calculate the available fault current at the transformer terminals:

AFC = FLA / Impedance fraction = 1443 / 0.06 ≈ 24,050 A
  

That’s your worst-case, 3-phase bolted fault current at the secondary terminals of the transformer.

Step 2: Compensate for the Cable Impedance

The fault current at the true switchboard or load point will typically be lower than at the transformer because of cable and conductor impedance in between. You must determine the voltage drop (or, better, the impedance drop) resulting from the run of cable between the transformer and the point of fault.

Let’s say you’ve run 20 metres of 3-core 240 mm² copper XLPE cable to a main switchboard. The typical impedance (R + jX) for this cable is around 0.12 Ω/km. Over 20 metres, that’s:

Z = 0.12 Ω/km × 0.02 km = 0.0024 Ω
  

You now compute fault voltage on this cable impedance and the transformer’s internal impedance. Add both impedances (in ohms) to determine the total impedance presented to the fault, then use Ohm’s Law:

Total Z = Z_transformer + Z_cable = (400² / AFC) + 0.0024 ≈ 6.65 mΩ + 2.4 mΩ = 9.05 mΩ
New AFC at MSB = (400 × √3) / 0.00905 ≈ 76,000 V / 0.00905 = 8,400 A
  

That’s a more realistic fault current at your main switchboard, adjusted for the cable run.

Step 3: Add Motor Contributions (if needed)

If you’ve got large motors on-site, especially induction motors, they can contribute additional current to the fault—sometimes up to 4 to 6 times their full-load current for a brief period (about 100 milliseconds).

Assume you have 90 kW motor with full-load current of 160 A. Its contribution may be approximately:

160 A × 4 = 640 A
  

If this motor is located near the switchboard and electrically close to the fault, you’ll need to add its fault contribution to the utility-based fault current.

Step 4: Adjust for Downstream Points

As you go further downstream from the transformer or switchboard, fault current available continues to decrease. Increasing cable runs, smaller conductors, or multiple connections will contribute to impedance in the system.

Each panelboard or sub-board should have its available fault current recalculated based on the impedance from the source. This is particularly important if you’re selecting circuit breakers for a final distribution board. A breaker with a 6 kA interrupting rating might be fine at the far end of a warehouse run, but it would be undersized if used at the MSB fed directly from a 1 MVA transformer.

You can perform these downstream calculations manually from impedance tables (such as those in AS/NZS 3008), or you can utilize modelling programs such as PowerCAD, AMTECH, or ETAP for larger installations.

Step 5: Use Conservative Assumptions

If you don’t know the actual transformer impedance or you can’t obtain the complete system data, it’s customary practice to use conservative assumptions—like assuming minimum impedance or using the worst-case transformer rating your DNSP will permit your site size.

But don’t be too conservative if you’re near the fault rating capacities of your equipment. Overestimating will cause you to oversize equipment unnecessarily.

Step 6: Check With the Utility (if necessary)

A few DNSPs will quote fault current on request—particularly for larger or more sensitive connections. This is typically given in the form of a fault level at the point of supply, in MVA. You can convert this to current with:

I_fault = (Fault level (MVA) × 10⁶) / (√3 × Voltage)
  

For example, if the fault level is 25 MVA at 400 V:

(25 × 10⁶) / (√3 × 400) ≈ 36,083 A
  

Use this as your starting point, then de-rate accordingly.