What is the right box fill for my 1x30A circuit and 2x20A circuit case?

Question

Box fill calculations are so frustratingly ambiguous and confusing. I have the following setup:

• A 30A dryer circuit along with 2x20A circuits comes via FMC from panel
• The dryer (3x8AWG) passes through (even without wire nuts)
• The dryer neutral is not current carrying (since it only carries the imbalance between the hots)
• The two 20A circuits (washer and bathroom) also pass through but will have a wire nut because they switch from THHN to romex
• Box is Steel City 72171-1 (42in^3) and no internal clamps
• A total of 4 grounds are needed to go in/out but additionally, the box itself has to be bonded as well
• It is not clear if the allowances count on each individual AWG or if I need to take the factor of the largest conductor.

With all of these ambiguities, I arrive at anywhere between 17.25in^3 to 45.75in^3:

To make things worse, the Steel City 72171-1 has dimensions 4-11/16 times 4-11/16 times 2-1/8 ... which amounts to 46.6919cu^3. Yet the advertised volume is only 42in^3. Why?

Summary:

1. When a wire passes through a junction box, does it count as one allowance or two allowances (1 for in, 1 for out)?
2. Does this change when the pass-through is broken and connected via wire nut?
3. Does the ground box for the box itself count as allowance?
4. Do only current-carrying conductors count as allowance? (and not inactive circuits for example)
5. Is the neutral of a MBWC or 240V circuit (such as my 30A dryer circuit) defined as current-carrying or not?
6. Do all the allowances need to be multiplied with the worst case in^3/allowance factor (in my case 3) or can I add the allowances for the AWG#12 circuits with factor 2.25?

EDIT: I believe the actual volume is 33in^3 (box bonding is pigtail and does not count, 240V neutral does not count, unbroken AWG#8 only count once and everything uses the worst case factor, 3) but I'd really like to confirm all these points.

Answer 1

The shared ground needs to be #10 since one circuit has wire rated for 21-60 amps.

For a 30A circuit, #10 conductors will suffice unless it is aluminum.

Wires which pass thru a box only count as 1 wire count, unless you are leaving more than 12” of wire in the box for a future cut-and-splice loop.

Thus I get

• three #8 wires @ 3 ci = 9 ci
• pigtails are free
• two #10 and two #12 grounds. Grounds are 4 for the pruce of 1 so 1 @ 2.5 c.i.
• eight #12 wires @ 2.25 c.i. = 18 c.i.

Total 29.5 cu.in.

When a wire passes through a junction box, does it count as one allowance or two allowances (1 for in, 1 for out)?

Pass thrus count as 1 unless they are 12” or longer, then 1 wire count per 12”.

Does this change when the pass-through is broken and connected via wire nut?

When there is a splice each participating wire takes 1 wire count... except for pigtails, which are free.

Does the ground box for the box itself count as allowance?

No, pigtails are free.

Do only current-carrying conductors count as allowance? (and not inactive circuits for example)

For box fill, all count except pigtails........ and grounds, which are "4 for the price of 1".

Is the neutral of a MBWC or 240V circuit (such as my 30A dryer circuit) defined as current-carrying or not?

Well if you have a 240V supply (not 208), the MWBC or 240V neutral does not count as a CCC for thermal calculations. But it does count as wire fill.

The only ones that don't count are pigtails (free) or grounds (BOG3).

Do all the allowances need to be multiplied with the worst case in^3/allowance factor (in my case 3) or can I add the allowances for the AWG#12 circuits with factor 2.25?

No, each wire counts for its own actual size. Except for grounds, where it follow normal BOGO rules and you must pay for the most expensive of the 4.

Answer 2

You have:

3 #8 conductors passing through the box.

8 #12 conductors entering the box and spliced there.

1 #12 ground wire.

For which my calculator says you need a 29.3 CI box.