What was the published fault current in the incident energy calculation example?

• A. 10 kA
• B. 30 kA
• C. 50 kA
• D. 100 kA

Answer: (B)

In incident energy calculations, the value you use for the fault current is the short-circuit current available at the point where the arc could occur, given the system’s sources and impedance. This published fault current is the input that drives how much energy the arc can deliver over a given arcing time and distance.

In the example you’re studying, the fault current used for the calculation is 30 kA. That choice isn’t random—it’s the specific value the example provides to illustrate how the incident energy changes with a moderate available fault current. Using 30 kA sets the calculation up so the resulting incident energy, and the corresponding PPE category and arc-flash boundary shown in the example, line up with the given numbers.

If you used a much smaller or much larger fault current (like 10, 50, or 100 kA), the calculated incident energy would be different, leading to different PPE requirements and arc-flash boundaries than those shown in the example. That’s why 30 kA is the appropriate published fault current for this particular scenario.