What is the minimum size box or box assembly required to contain: Two 10/2 with ground MC cables, one 12/2 with ground MC cable, a 30-amp 240-volt single receptacle, and a 20-amp 120-volt single receptacle?

What is the minimum size box or box assembly required to contain: Two 10/2 with ground MC cables, one 12/2 with ground MC cable, a 30-amp 240-volt single receptacle, and a 20-amp 120-volt single receptacle?

• A. two - 3 × 2 × 2 device boxes (ganged)
• B. two - 3 × 2 × 2½ device boxes (ganged)
• C. 4 × 11/2 square box with a 4.1 in3 two-gang plaster ring
• D. 4 × 2 1/8 square box with a 4.1 in3 two-gang plaster ring

Answer: (D)

Box fill is about making sure there’s enough cubic inches inside a box to safely hold all conductors, devices, and any internal components. For conductors, you count each insulated conductor that originates outside and terminates or is spliced inside, using the volume per conductor based on its gauge: 10 AWG is 2.5 in³, 12 AWG is 2.25 in³, and so on. Equipment grounding conductors all together count as a single conductor of the largest size present. Device yokes or straps also add volume: each device on a box counts as one conductor of the largest conductor connected to that device. Here, you have two 10/2 with ground MC cables and one 12/2 with ground MC cable. That means six insulated conductors total (four from the two 10/2 cables and two from the 12/2). Using the gauge volumes: 4 × 2.5 in³ for the 10 AWG conductors plus 2 × 2.25 in³ for the 12 AWG conductors gives 14.5 in³ for the insulated conductors. The three equipment grounding conductors count as one 10 AWG conductor, adding 2.5 in³. So far you’re at 14.5 + 2.5 = 17.0 in³. For the devices, one receptacle is fed with 10 AWG (30-amp, 240-volt) and the other with 12 AWG (20-amp, 120-volt). Each device counts as a conductor of the size of the conductors feeding it, so add 2.5 in³ for the 30-amp device and 2.25 in³ for the 20-amp device. That’s an additional 4.75 in³, bringing the total box fill requirement to 21.75 in³. If you use a plaster ring, you must add its volume to the box’s capacity. The two-gang plaster ring in this scenario provides 4.1 in³ of additional capacity. To meet the 21.75 in³ requirement, you need a box combination that, with the 4.1 in³ ring, reaches at least 21.75 in³. A 4 × 2 1/8 square box paired with a two-gang plaster ring rated at 4.1 in³ meets this need, giving enough total capacity for all conductors and devices. The other box arrangements either don’t supply enough base volume or aren’t suitable for the two-gang, plaster-ring setup, so the described combination is the minimum that satisfies the fill calculation.

Box fill is about making sure there’s enough cubic inches inside a box to safely hold all conductors, devices, and any internal components. For conductors, you count each insulated conductor that originates outside and terminates or is spliced inside, using the volume per conductor based on its gauge: 10 AWG is 2.5 in³, 12 AWG is 2.25 in³, and so on. Equipment grounding conductors all together count as a single conductor of the largest size present. Device yokes or straps also add volume: each device on a box counts as one conductor of the largest conductor connected to that device.

Here, you have two 10/2 with ground MC cables and one 12/2 with ground MC cable. That means six insulated conductors total (four from the two 10/2 cables and two from the 12/2). Using the gauge volumes: 4 × 2.5 in³ for the 10 AWG conductors plus 2 × 2.25 in³ for the 12 AWG conductors gives 14.5 in³ for the insulated conductors. The three equipment grounding conductors count as one 10 AWG conductor, adding 2.5 in³. So far you’re at 14.5 + 2.5 = 17.0 in³.

For the devices, one receptacle is fed with 10 AWG (30-amp, 240-volt) and the other with 12 AWG (20-amp, 120-volt). Each device counts as a conductor of the size of the conductors feeding it, so add 2.5 in³ for the 30-amp device and 2.25 in³ for the 20-amp device. That’s an additional 4.75 in³, bringing the total box fill requirement to 21.75 in³.

If you use a plaster ring, you must add its volume to the box’s capacity. The two-gang plaster ring in this scenario provides 4.1 in³ of additional capacity. To meet the 21.75 in³ requirement, you need a box combination that, with the 4.1 in³ ring, reaches at least 21.75 in³. A 4 × 2 1/8 square box paired with a two-gang plaster ring rated at 4.1 in³ meets this need, giving enough total capacity for all conductors and devices. The other box arrangements either don’t supply enough base volume or aren’t suitable for the two-gang, plaster-ring setup, so the described combination is the minimum that satisfies the fill calculation.