Among the possible box assemblies, the smallest permitted to contain the following is: Exterior MC cable connectors are used. One - 30-amp 240-volt single receptacle and One - 15-amp 120-volt single receptacle; Two - 10/2 with ground MC cables plus One - 14/2 with ground MC cable.

Among the possible box assemblies, the smallest permitted to contain the following is: Exterior MC cable connectors are used. One - 30-amp 240-volt single receptacle and One - 15-amp 120-volt single receptacle; Two - 10/2 with ground MC cables plus One - 14/2 with ground MC cable.

• A. 4 × 21/8 square box with a 3.6 in3 two-gang plaster ring
• B. two - 3 × 2 × 2 device boxes (ganged to form a 2-gang box)
• C. 4 × 11/2 square box with a 5.0 in3 two-gang plaster ring
• D. two - 3 × 2 × 21/2 device boxes (ganged to form a 2-gang box)

Answer: (C)

Box fill is the main idea here: you must have enough cubic inches inside the box for all conductors, plus the space taken by devices, grounds, and any clamps or rings. The key is counting correctly and using the volume for the largest conductor present. In this scenario the largest conductor in the box is 10 AWG (from the two 10/2 MC cables and the 14/2 MC cable). So each conductor inside the box counts as 2.5 cubic inches. Now count what goes into the box. There are three current-carrying conductors for the hot side that originate outside and terminate inside (two hots for the 240 V receptacle and one hot for the 120 V receptacle). There are three neutrals that originate outside and terminate inside (the neutrals used by the 120 V circuit plus the neutrals from the two 10/2 cables that are present in the box). Altogether that makes six current-carrying conductors. The three equipment grounding conductors are present, but all grounding conductors count as a single conductor toward box-fill. There are two devices (the two single receptacles), and each device yoke counts as one conductor. Total conductor-equivalents = 6 (current-carrying) + 1 (grounding set) + 2 (device yokes) = 9. Multiply by the largest-conductor volume: 9 × 2.5 in³ = 22.5 in³ needed. Exterior MC connectors are used, so there are no internal clamps to add. The two-gang plaster ring in the chosen assembly adds its own volume, and you need enough base box volume plus the ring to meet or exceed 22.5 in³. The smallest box option that, with its 5.0 in³ plaster ring, provides at least 22.5 in³ total is the 4 × 11/2 square box with a 5.0 in³ two-gang plaster ring. The other box options either don’t reach the required total volume even after adding their rings, or would require a larger base box, making them not the smallest permitted.

Box fill is the main idea here: you must have enough cubic inches inside the box for all conductors, plus the space taken by devices, grounds, and any clamps or rings. The key is counting correctly and using the volume for the largest conductor present.

In this scenario the largest conductor in the box is 10 AWG (from the two 10/2 MC cables and the 14/2 MC cable). So each conductor inside the box counts as 2.5 cubic inches.

Now count what goes into the box. There are three current-carrying conductors for the hot side that originate outside and terminate inside (two hots for the 240 V receptacle and one hot for the 120 V receptacle). There are three neutrals that originate outside and terminate inside (the neutrals used by the 120 V circuit plus the neutrals from the two 10/2 cables that are present in the box). Altogether that makes six current-carrying conductors. The three equipment grounding conductors are present, but all grounding conductors count as a single conductor toward box-fill. There are two devices (the two single receptacles), and each device yoke counts as one conductor.

Total conductor-equivalents = 6 (current-carrying) + 1 (grounding set) + 2 (device yokes) = 9. Multiply by the largest-conductor volume: 9 × 2.5 in³ = 22.5 in³ needed.

Exterior MC connectors are used, so there are no internal clamps to add. The two-gang plaster ring in the chosen assembly adds its own volume, and you need enough base box volume plus the ring to meet or exceed 22.5 in³.

The smallest box option that, with its 5.0 in³ plaster ring, provides at least 22.5 in³ total is the 4 × 11/2 square box with a 5.0 in³ two-gang plaster ring. The other box options either don’t reach the required total volume even after adding their rings, or would require a larger base box, making them not the smallest permitted.