For a set involving thirty-four 8 AWG, twenty 12 AWG, nine 6 AWG, and one 10 AWG THHN/THWN conductors in EMT between two junction boxes, which is the correct minimum size?

For a set involving thirty-four 8 AWG, twenty 12 AWG, nine 6 AWG, and one 10 AWG THHN/THWN conductors in EMT between two junction boxes, which is the correct minimum size?

• A. 2 in.
• B. 2-1/2 in.
• C. 3 in.
• D. 3-1/2 in.

Answer: (A)

Conduit fill is the key idea here: you size the raceway by how much space all the conductors need inside, and you must stay within the allowed fill percentage of the conduit’s inner cross‑section. For EMT, the typical rule used in practice is a 40% fill, though the exact table allows slightly more as the number of conductors grows. The important part for this problem is comparing the total cross‑sectional area of all insulated conductors (in circular mils, cmil) to the raceway’s available area times the permitted fill. First, sum the cross‑sectional areas of every insulated conductor in the run: - 8 AWG each ≈ 16,510 cmil; 34 of them → about 561,340 cmil - 12 AWG each ≈ 6,530 cmil; 20 of them → about 130,600 cmil - 6 AWG each ≈ 26,240 cmil; 9 of them → about 236,160 cmil - 10 AWG (THHN/THWN) ≈ 10,380 cmil; 1 → about 10,380 cmil Total ≈ 938,480 cmil. Convert to square inches (since conduit interior area is in square inches): 1 cmil ≈ 7.85398×10^-7 in^2, so total area ≈ 0.736 in^2. Now compare to the allowed fill. A 1.5 inch EMT has interior cross‑section around π(0.75)^2 ≈ 1.767 in^2; 40% of that is about 0.707 in^2, which is slightly less than the 0.736 in^2 needed, so 1.5 inches isn’t enough. A 2 inch EMT has interior area around π(1.0335)^2 ≈ 3.35 in^2; 40% of that is about 1.34 in^2, which comfortably exceeds 0.736 in^2. Therefore, the smallest EMT size that meets the fill requirement is 2 inches.

Conduit fill is the key idea here: you size the raceway by how much space all the conductors need inside, and you must stay within the allowed fill percentage of the conduit’s inner cross‑section. For EMT, the typical rule used in practice is a 40% fill, though the exact table allows slightly more as the number of conductors grows. The important part for this problem is comparing the total cross‑sectional area of all insulated conductors (in circular mils, cmil) to the raceway’s available area times the permitted fill.

First, sum the cross‑sectional areas of every insulated conductor in the run:

• 8 AWG each ≈ 16,510 cmil; 34 of them → about 561,340 cmil

• 12 AWG each ≈ 6,530 cmil; 20 of them → about 130,600 cmil

• 6 AWG each ≈ 26,240 cmil; 9 of them → about 236,160 cmil

• 10 AWG (THHN/THWN) ≈ 10,380 cmil; 1 → about 10,380 cmil

Total ≈ 938,480 cmil.

Convert to square inches (since conduit interior area is in square inches): 1 cmil ≈ 7.85398×10^-7 in^2, so total area ≈ 0.736 in^2.

Now compare to the allowed fill. A 1.5 inch EMT has interior cross‑section around π(0.75)^2 ≈ 1.767 in^2; 40% of that is about 0.707 in^2, which is slightly less than the 0.736 in^2 needed, so 1.5 inches isn’t enough. A 2 inch EMT has interior area around π(1.0335)^2 ≈ 3.35 in^2; 40% of that is about 1.34 in^2, which comfortably exceeds 0.736 in^2.

Therefore, the smallest EMT size that meets the fill requirement is 2 inches.