# How to determine electrical box size per NEC

Here is a quick exercise:

• Wires for two circuits are entering a junction box through one conduit.
• On the other end, the circuits split into two branches
• Two neutrals entering (current count = 2)
• Two hots entering (current count = 4)
• Two grounds entering (current count = 5)
• Wire gauge is 12, so total volume is `5 * 2.25 = 12.25 cu inches`

Does this calculation have to include wires that are exiting? I will have two 12/2 Romex. If this figures into calculation, 2x the volume is needed (24.5 cu inches).

Edit 1: I should clarify that this will be a junction box without any devices on it. On one end, a 1/2 inch LFMC is entering with a plastic connector, and on the other end, two 12/2 Romex wires will exit individually through a snap on connector each.

• What size conduit? Aug 24, 2020 at 22:23
• Wires exiting have to be counted too. You'd need to add 4 for the two NM cables and possibly 1 for the cable clamps depending on type.
– JACK
Aug 24, 2020 at 22:59
• A cable is several wires bundled. Aug 25, 2020 at 2:43

• The box I am using is 21 cubic inches in size. With 5 cables entering and 4 cables exiting, the minimum volume comes out to `9 * 2.25 = 20.25 cubic inches`. I guess I am running pretty tight, but I will let the inspector decide if he wants to fail me on that count. Second, I had a discussion with someone who basically told me that using the same ground raises the risk of overheating and fire. Not sure how likely that is. Aug 24, 2020 at 23:49