CSLB C-10 Electrical ContractorWiring Methods and MaterialsMedium
A two-gang metal device box contains one duplex receptacle and one single-pole switch (each on its own yoke), plus two 12/2 NM-B cables entering the box (4 current-carrying 12 AWG conductors and grounding conductors). Using the box fill allowances of NEC 314.16, what is the minimum required box volume?
- A11.25 cubic inches
- B20.25 cubic inches
- C22.5 cubic inches
- D15.75 cubic inches
Show answer & explanationAnswer & explanation
Correct answer: B. 20.25 cubic inches
Box fill units: 4 current-carrying conductors = 4 units; all grounding conductors count as 1 unit; each device yoke counts as 2 units, and there are 2 yokes = 4 units. Total = 4 + 1 + 4 = 9 volume allowances. At 2.25 in³ per 12 AWG conductor, the minimum box volume is 9 × 2.25 = 20.25 in³.
Why the other options are wrong
- A. This undercounts either the devices or the grounding allowance.
- C. This overcounts the grounding conductors as more than one allowance.
- D. This omits one of the device yoke allowances.
Box Fill with Multiple Devices
Box fill calculations count conductors, one allowance for all grounding conductors combined, and two volume units per device yoke.
- Each current-carrying conductor = 1 unit
- All grounds together = 1 unit (based on largest conductor)
- Each device yoke = 2 units
- 12 AWG = 2.25 in³ per unit
Memory trick: Conductors count once, grounds count once total, but devices count double