- Location
- New Jersey
- Occupation
- Journeyman Electrician
The suggestion was type AC cable - you would save 2.25 cu in because there is no EGC.
Glad that you can read. :grin:
The suggestion was type AC cable - you would save 2.25 cu in because there is no EGC.
I very often turn a single gang on it's side in between drawers or above.
Ah you got me on the AC/MC thing.Glad that you can read. :grin:
....The blue carlon box someone stated earlier, how do you cut that in and mount it????Those are just a square box with no flanges.....
Cut it in very tight, and use some hot-melt glue to keep it in place. Works like a charm!![]()
I hope you don't mean the 8cu.in. box with the little triangular KO's. I don't believe they're approved for electrical devices. Besides, a single 12-2 and a receptacle require 10 cu.in.I use the carlon blue pastic shallow box.
This one is 2" deep and 17 cu in.
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Tell the cabinet guys what needs to be done and work out a solution. Very rarely do I not find a legal option.
I see Scott beat me to it. Oh, well. One thing I like about this box is that the flat half pops off for easier wiring before you install it. I don't know if it's supposed to, but it does.I like this box, kinda big but maybe it will work. I'll check into this one....
Hot melt: the listed installation method for all carlon boxes, I'm sure :wink:
Ok what am I doing wrong here on calculation---12-2 nm + device......2 conductors + ground + device= 2+1+2 x 2.25=11.25 cu.in + the strap thingy in the metal box (per inspector) +1 x 2.25= total 13.5 cu.in.I hope you don't mean the 8cu.in. box with the little triangular KO's. I don't believe they're approved for electrical devices. Besides, a single 12-2 and a receptacle require 10 cu.in.
Often there's enough room for a standard-depth box behind the drawer, even when it's pushed all the way in, or, if the drawer isn't too tall, you can mount the box just below it.
For another option, take a look at this Carlon box:
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Besides, a single 12-2 and a receptacle require 10 cu.in.
Unless I'm mistakenly using 2 cu.in. per conductor. :roll: Oops!:-?.......I come up with 11.25 cu. in.....assuming you are talking about NM-B