There are many programs to determine the available fault current at a given location but most just provide the final AFC value, not the math to back it up. What would be the available fault current at the service if the service entrance conductors are only 80 feet instead of 95?

Available Fault Current (AFC) calculations are most often performed by using fault current calculating software or spreadsheets that automatically calculate the final amount of AFC after entering specific values. Long before the software became available, the calculation was done manually with a few basic formulas that are commonly found in many of the electrician’s pocket guides like the “Ugly’s Electrical References” guide. These types of guides provide the most common electrical references and formulas used by the electrician. It is important to understand how the basic formula works even if the plan is to use software for the calculation.

Below is a 3-step formula to calculate three-phase AFC, also called the available short circuit current (ISC) at the end of a run of wire:

**Step 1:**

** F = (1.732 X L X I) ÷ (C X E_(L_L))**

**Step 2:**

** Multiplier (M) = 1 ÷ (1 + F)**

**Step 3:**

** Available Short-Circuit Current (ISC) = I X M**

Here is an example using realistic data for the below electrical installation:

**3 conductors in a raceway****3/0 Copper conductors****PVC conduit**- 3-phase
**480**Volt system - AFC
**35,000**amps - Length of conductor
**95 feet**

**Step 1:**

**F = (1.732 X 95 X 35,000) ÷ (13,923 X 480) F = .861**

**Step 2:**

**Multiplier (M) = 1 ÷ (1 + .861) M = .537**

**Step 3:**

**ISC = 35,000 X .537 = 18,795 amps (18,795 is the available fault current at the end of the run of conductor).**

**In the calculations shown above, which of the following is the available fault current at the beginning of the run of conductor?**

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