What is the smallest volume junction box that can be used for a configuration with conduits #1 and #4 each containing twenty-four 10 AWG wires, conduits #2 and #5 each containing twelve 6 AWG wires, conduits #3 and #6 each containing sixteen 8 AWG wires, and all conductors are wire-nutted inside the box with grounds equal to the circuit conductors?

What is the smallest volume junction box that can be used for a configuration with conduits #1 and #4 each containing twenty-four 10 AWG wires, conduits #2 and #5 each containing twelve 6 AWG wires, conduits #3 and #6 each containing sixteen 8 AWG wires, and all conductors are wire-nutted inside the box with grounds equal to the circuit conductors?

• A. 6" × 6" × 4" deep
• B. 8" × 8" × 4" deep
• C. 10" × 10" × 4" deep
• D. 12" × 12" × 4" deep

Answer: (C)

Box-fill calculations (NEC 314.16(B)) determine the minimum box volume by counting every conductor that enters the box and is spliced or terminated inside, adding one for all equipment grounding conductors, plus any internal clamps. In this setup, every conductor entering the box is spliced inside, so all conductors count toward fill. There are 24 wires of 10 AWG in each of two conduits (total 48), 16 wires of 8 AWG in each of two conduits (total 32), and 12 wires of 6 AWG in each of two conduits (total 24). So we have 48 of 10 AWG, 32 of 8 AWG, and 24 of 6 AWG. Use the standard fill volumes: 10 AWG = 2.5 in^3 per conductor, 8 AWG = 3.0 in^3, 6 AWG = 5.0 in^3. The current-carrying conductors contribute 48×2.5 = 120 in^3, plus 32×3.0 = 96 in^3, plus 24×5.0 = 120 in^3, totaling 336 in^3. Grounding conductors, inside the box and tied together, count as a single conductor toward fill, and use the volume of the largest grounding conductor present, which is 6 AWG (5.0 in^3). Total required fill = 336 + 5 = 341 in^3. The smallest box that meets or exceeds this is 10" × 10" × 4" (400 in^3). So that is the smallest acceptable box.

Box-fill calculations (NEC 314.16(B)) determine the minimum box volume by counting every conductor that enters the box and is spliced or terminated inside, adding one for all equipment grounding conductors, plus any internal clamps. In this setup, every conductor entering the box is spliced inside, so all conductors count toward fill. There are 24 wires of 10 AWG in each of two conduits (total 48), 16 wires of 8 AWG in each of two conduits (total 32), and 12 wires of 6 AWG in each of two conduits (total 24). So we have 48 of 10 AWG, 32 of 8 AWG, and 24 of 6 AWG.

Use the standard fill volumes: 10 AWG = 2.5 in^3 per conductor, 8 AWG = 3.0 in^3, 6 AWG = 5.0 in^3. The current-carrying conductors contribute 48×2.5 = 120 in^3, plus 32×3.0 = 96 in^3, plus 24×5.0 = 120 in^3, totaling 336 in^3.

Grounding conductors, inside the box and tied together, count as a single conductor toward fill, and use the volume of the largest grounding conductor present, which is 6 AWG (5.0 in^3).

Total required fill = 336 + 5 = 341 in^3. The smallest box that meets or exceeds this is 10" × 10" × 4" (400 in^3). So that is the smallest acceptable box.