What size Flexible Metal Conduit is needed for three 14/2 NM cables if each cable has a width of 372 mils and height of 170 mils?

To determine the appropriate size of the Flexible Metal Conduit (FMC) needed for three 14/2 NM cables, it's essential to consider the total space required by those cables. Each cable has a width of 372 mils (0.372 inches) and a height of 170 mils (0.170 inches).

First, we need to calculate the cross-sectional area each cable occupies. Since both dimensions are provided, we can multiply the width by the height to find the area of one cable:

Area of one cable = 0.372 inches × 0.170 inches = 0.06324 square inches.

For three cables, you would then multiply this area by three:

Total area for three cables = 0.06324 square inches × 3 = 0.18972 square inches.

Next, we must size the conduit appropriately to accommodate this total area. Electrical codes usually recommend that conduits be filled to no more than 40% of their total cross-sectional area to allow for heat dissipation and ease of pulling in the wires.

The internal cross-sectional area of the chosen FMC sizes can be derived from standard conduit size tables. A 1-inch conduit has a total cross-sectional area of approximately