For a 12' run of 21/2 inch FMC, what is the maximum area that can be occupied by the circuit conductors?

The maximum area that can be occupied by circuit conductors within a conduit is calculated based on the inner diameter of the conduit and the area required by the conductors. In this case, for 2½ inch Flexible Metal Conduit (FMC), we can use standard industry guidelines to determine the conduit fill.

The internal diameter of 2½ inch FMC is approximately 2.500 inches. The area of a circle is calculated using the formula A = πr², where r is the radius. The radius in this instance would be half of the internal diameter, resulting in a radius of approximately 1.250 inches.

Calculating the area:

1. Use the radius (1.250 inches):

A = π (1.250)² = π (1.5625) ≈ 4.9088 sq. in.

The National Electrical Code (NEC) provides fill requirements for conduits. For the standard fill calculations, a maximum fill of 40% is allowed for more than two conductors, which keeps conductors from overheating due to insufficient space for the wires.

By applying the 40% fill rule, we have:

Max area for conductors = 0.40 x 4.