CSLB C-10 Electrical ContractorWiring Methods and MaterialsMedium
A single-gang metal device box contains two 12/2 NM-B cables (4 current-carrying 12 AWG conductors, 2 grounds) with no internal cable clamps, and will hold one duplex receptacle. Using NEC box fill rules where each 12 AWG conductor counts as 2.25 cubic inches, what is the minimum required box volume?
- A15.75 cubic inches
- B18.0 cubic inches
- C13.5 cubic inches
- D20.25 cubic inches
Show answer & explanationAnswer & explanation
Correct answer: A. 15.75 cubic inches
Conductors: 4 × 2.25 = 9.0. Grounding conductors (all count as one, based on largest): 1 × 2.25 = 2.25. Device yoke (counts as two volume allowances, based on largest conductor): 2 × 2.25 = 4.5. Total = 9.0 + 2.25 + 4.5 = 15.75 cubic inches.
Why the other options are wrong
- B. 18.0 cu in overstates the required volume by double-counting an allowance.
- C. 13.5 cu in omits either the grounding or device yoke allowance.
- D. 20.25 cu in incorrectly adds an extra conductor allowance not present.
Box Fill with Device Yoke
Per NEC 314.16(B), each strap/yoke supporting a device counts as two conductor-volume allowances based on the largest conductor connected to it.
- Conductors: 1 volume allowance each based on size.
- All grounds together: 1 allowance based on largest ground.
- Device yoke: 2 allowances based on largest conductor connected.
Memory trick: 'Grounds unite, devices double' — grounds count once, yokes count twice.