• Product
  • Suppliers
  • Manufacturers
  • Solutions
  • Free tools
  • Knowledges
  • Experts
  • Communities
Search


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

Electrical4u
Field: Basic Electrical
0
China
what is available fault current

What is Available Fault Current?

Available fault current (AFC) is defined as the largest amount of current available during a fault. It is the maximum amount of current that can be delivered to the electrical equipment under a fault condition. The available fault current is also known as the available short-circuit current.

The term ‘Available Fault Current’ was introduced in the 2011 NFPA 70: National Electric Code (NEC) in section 110.24 (latest version of the code).

According to this section, it is compulsory to mark the maximum amount of available fault current with a date of fault current calculation that was performed.

The rating marked as available fault current is not an equipment rating. But it is the maximum amount of unwanted current that will flow on equipment in case of fault occurs.

The term short-circuit current rating (SCCR) is different from available fault current. For all equipment or circuits must not have SCCR less than the AFC.

The reason behind marking AFC in equipment is that the electrician can take that rating and use it to select the proper equipment rating to comply with other code sections like NEC 110.9 and 110.10.

Available Fault Current Formula

According to NEC 110.24, labeling of available fault current is necessary. But before calculating the available fault current of equipment within dwellings, we require the rating of available fault current at secondary terminals on the utility transformer feeding said dwelling.

In most cases, the rating of available fault current is provided by the utility and it is labeled at the secondary terminal of the utility transformer.

According to this rating, the available fault current is calculated for all equipment. The calculation for all equipment is different as it depends on the circuit impedance.

Follow the below steps to calculate the available fault current;

  1. Find the system voltage (E_{L-L})

  2. Find the conductor constant (C) from the table

  3. Find the length of the service entrance conductor (L)

  4. Now, using the above values, calculate the value of the multiplier (M) using the below equations.


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



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


  1. 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

Let’s take an example to understand how to calculate available fault current.

For that we consider a three-phase system having 480V line-line voltage. And the conductor constant C for this system is 13900.

The available fault current at a secondary winding of the utility transformer is 35000A, and the length of the service entrance conductor is 100ft.

EL-L = 480V

C = 13,900

I = 35,000A

L = 100ft

Now, put these values in the above equation.


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



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



  \[ F = 0.9075 \]


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



  \[ M = \frac{1}{1+0.9075} \]



  \[ M = 0.524 \]



  \[ I_{sc} = I \times M \]



  \[ I_{sc} = 35,000 \times 0.524 \]



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


How to Reduce Available Fault Current

In a low-resistance circuit, the current that passes through the circuit increases drastically. And it may damage the circuit. Therefore, we need to use other concepts or equipment to reduce the short circuit current or available fault current.

Mainly there are three methods to reduce available fault current:

  1. Increasing cable length

  2. Using current limiting reactors

  3. Using current limiting devices

Increase Cable Length

We can reduce the fault current by increasing the length of the cable.

In the above example, we have taken the length of the cable as 100ft. Instead of that, let’s calculate the fault current with a 200ft length cable and other parameters remaining as it is.

By calculating the above steps, you can find the available fault current is 12460A. Hence, it is reduced by increasing the length of the cable.

Observing the relationship between F and L, we can see that they are directly proportional to one another.

Using Current Limiting Reactor

The reactors can be used at any place in the distribution system to reduce the fault current. However, it will degrade the voltage profile during normal operation. But it is the cheapest method to reduce the fault current.

In most cases, a current limiting reactor is connected in series with the circuit or system.



current limiting reactor
Current Limiting Reactor



Using Current Limiting Devices

This is a non-linear device that is used in two states; conducting (zero resistance) and limiting (high resistance).

When the circuit is operating in normal conditions, this device simply conducts and there is no change in the voltage profile.

But as the current increases, this device increases the resistance in the circuit to limit the magnitude of the fault current.



current limiting devices
Current Limiting Devices



Statement: Respect the original, good articles worth sharing, if there is infringement please contact delete.

Give a tip and encourage the author!

Recommended

Main Transformer Accidents and Light Gas Operation Issues
1. Accident Record (March 19, 2019)At 16:13 on March 19, 2019, the monitoring background reported a light gas action of No. 3 main transformer. In accordance with the Code for Operation of Power Transformers (DL/T572-2010), operation and maintenance (O&M) personnel inspected the on-site condition of No. 3 main transformer.On-site confirmation: The WBH non-electrical protection panel of No. 3 main transformer reported a Phase B light gas action of the transformer body, and the reset was ineff
02/05/2026
Faults and Handling of Single-phase Grounding in 10kV Distribution Lines
Characteristics and Detection Devices for Single-Phase Ground Faults1. Characteristics of Single-Phase Ground FaultsCentral Alarm Signals:The warning bell rings, and the indicator lamp labeled “Ground Fault on [X] kV Bus Section [Y]” illuminates. In systems with a Petersen coil (arc suppression coil) grounding the neutral point, the “Petersen Coil Operated” indicator also lights up.Insulation Monitoring Voltmeter Indications:The voltage of the faulted phase decreases (in
01/30/2026
Neutral point grounding operation mode for 110kV~220kV power grid transformers
The arrangement of neutral point grounding operation modes for 110kV~220kV power grid transformers shall meet the insulation withstand requirements of transformer neutral points, and shall also strive to keep the zero-sequence impedance of substations basically unchanged, while ensuring that the zero-sequence comprehensive impedance at any short-circuit point in the system does not exceed three times the positive-sequence comprehensive impedance.For 220kV and 110kV transformers in new constructi
01/29/2026
Why Do Substations Use Stones, Gravel, Pebbles, and Crushed Rock?
Why Do Substations Use Stones, Gravel, Pebbles, and Crushed Rock?In substations, equipment such as power and distribution transformers, transmission lines, voltage transformers, current transformers, and disconnect switches all require grounding. Beyond grounding, we will now explore in depth why gravel and crushed stone are commonly used in substations. Though they appear ordinary, these stones play a critical safety and functional role.In substation grounding design—especially when multiple gr
01/29/2026
Send inquiry
+86
Click to upload file
Download
Get the IEE Business Application
Use the IEE-Business app to find equipment, obtain solutions, connect with experts, and participate in industry collaboration anytime, anywhere—fully supporting the development of your power projects and business.