Available Fault Current: What is it? (And How To Calculate It)

Contents

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Key learnings:

Available Fault Current Definition: Available Fault Current (AFC) is defined as the maximum current available during a fault condition, also known as available short-circuit current.
Importance of AFC Marking: AFC must be marked with a calculation date as per the 2011 NFPA 70: NEC section 110.24.
Fault Current Calculation: To calculate fault current, use the system voltage, conductor constant, and length of the service entrance conductor.
Example Calculation of AFC: In a 480V system, the AFC can be calculated using a given formula and specific parameters, resulting in 18,340A.
Reducing Fault Current: Methods like increasing cable length, using current limiting reactors, and current limiting devices can effectively reduce fault current.

What is Available Fault Current?

Available fault current (AFC), also called available short-circuit current, is the maximum prospective current at a specified point for a stated fault and set of system conditions. Its value changes with source and circuit impedance.

NFPA 70 added Section 110.24 in the 2011 National Electrical Code (NEC). The 2026 edition is the current latest version of the code at publication, but the edition adopted by a state or local authority may differ.

Section 110.24 requires applicable service equipment at locations other than dwelling units to be field-marked with available fault current and the calculation date. The calculation must be documented, and relevant installation changes require review and an updated marking where necessary.

The marked AFC is an installation calculation, not an equipment rating. It states the prospective fault current at the equipment’s line terminals under the calculation assumptions.

Short-circuit current rating (SCCR) is the fault current an equipment assembly can withstand under its rating conditions. Where an SCCR is required, it must be at least the AFC at the equipment terminals unless an approved series or current-limiting combination establishes a suitable rating.

AFC allows designers and installers to compare interrupting ratings and SCCR with the prospective current addressed by NEC Sections 110.9 and 110.10.

Available Fault Current Formula

A short-circuit study starts with source data, usually the available current or source impedance at the utility transformer secondary. NEC 110.24 marking does not apply to dwelling-unit service equipment, although equipment ratings still must suit the available current under the adopted code.

The utility may provide source fault-current data or transformer impedance for the serving transformer. Do not assume that a transformer-terminal value is already marked or that it remains valid after system changes.

Calculate AFC at each point from the source data and intervening circuit impedance. Motors and generators can also contribute fault current, so a complete study may need more than a utility-source calculation.

The following Eaton Bussmann point-to-point steps approximate a three-phase bolted fault through one conductor run. Use the conductor constant for the exact material and size, and use a method that matches the system phase and fault type:

Find line-to-line system voltage ( $E_{L-L}$ ).
Find conductor constant C for the conductor material and size.
Find one-way conductor length L in the units required by the constant table.
Use source current I and the following equations to calculate factor F and multiplier M. In practice, multiply I by M for the endpoint estimate. Do not assume the source value is labelled at the transformer.

$F = \frac{1.73 \times L \times I}{C \times E_{L-L}}$

$Multiplier\ (M) = \frac{1}{1+F}$

To find the available fault current at premises, this multiplier (M) is multiplied by the available fault current labeled at the secondary terminal of the utility transformer.

How to Calculate Available Fault Current

This example applies the stated point-to-point assumptions.

Assume a 480 V three-phase system and conductor constant C of 13,900 for the selected conductor.

Source AFC at the transformer secondary is 35,000 A, and the one-way conductor length is 100 ft.

E_L-L = 480V

C = 13,900

I = 35,000A

L = 100ft

Substitute the values:

$F = \frac{1.73 \times L \times I}{C \times E_{L-L}}$

$F = \frac{1.73 \times 100 \times 35,000}{13,900 \times 480}$

$F = 0.9075$

$Multiplier (M) = \frac{1}{1+F}$

$M = \frac{1}{1+0.9075}$

$M = 0.524$

$I_{sc} = I \times M$

$I_{sc} = 35,000 \times 0.524$

$Available \, Fault \, Current \, I_{sc} = 18,340A$

How to Reduce Available Fault Current

If AFC exceeds an equipment rating, redesign the system or use an evaluated protection method. A qualified engineer should confirm the study, equipment ratings, coordination, voltage drop and arc-flash implications.

Common sources of added impedance or current limitation include:

Conductor impedance already present in the designed route
Current-limiting reactors
Listed current-limiting fuses, circuit breakers or evaluated combinations

Increase Cable Length

Longer conductors add impedance and therefore reduce calculated AFC. Adding unnecessary cable solely for this purpose is poor practice because it also increases voltage drop, losses and installation cost.

If the actual designed route in the example were 200 ft rather than 100 ft, keep the other stated assumptions and recalculate the endpoint AFC.

The point-to-point estimate is about 12.4 kA, subject to rounding and the selected conductor constant. Verify voltage drop and protective-device performance for the route.

Within this approximation, F is proportional to conductor length L.

Using Current Limiting Reactor

A properly engineered reactor adds inductive impedance at a chosen point in the distribution system. It also causes normal-operation voltage drop and losses, so placement, rating, insulation and protection require study. Cost depends on the installation.

A current-limiting reactor is connected in series with the affected feeder or bus section.

Using Current Limiting Devices

Current-limiting fuses and circuit breakers restrict peak current and let-through energy by clearing within their current-limiting range. They are not generic two-state devices that change from zero resistance to high resistance.

During normal operation, the protective device carries load current within its continuous rating and has a finite voltage drop and power loss.

During a high fault, an appropriate device opens quickly enough to limit peak and let-through current. Manufacturer tables and an evaluated combination must show that the downstream equipment rating is adequate; adding a current-limiting label alone does not establish compliance.

current limiting devices — Current Limiting Devices