How to determine electrical box size per NEC

Question

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.

Answer 1

Wires exiting have to be counted too. You'd need to add 4 for the two NM cables. Conduit lock nuts, NB snap and outside clamps are not counted in the fill calculations. The grounds leaving with the NM cables don't count either because you've already counted a ground.

Answer 2

You count all wires that depart the box

When dealing with box fill, you have to count all the wires that depart the box (NEC 314.16(B)(1) is phrased "Each conductor that originates outside the box and terminates or is spliced within the box..."), no matter which direction they're running in the circuit topology. Note, however, that while you need two hots and two neutrals running through the LFMC, you only need one grounding wire shared between both circuits. (Think about the case where the conduit's the grounding path.)

Note also that you only need to count 1/4 of your grounds for box fill. (It used to be that you only counted one ground out of the entire grounding bundle in the box, but that changed in 2020 to deal with boxes that had loads and loads of cables entering them.)