Cable Fill Violation Question


  #1  
Old 02-04-08, 04:13 PM
C
Member
Thread Starter
Join Date: Jan 2005
Location: Missouri
Posts: 144
Upvotes: 0
Received 1 Upvote on 1 Post
Cable Fill Violation Question

Roughing in a bathroom. Running 12/2 cable into a receptacle box, and 2 more 12/2's out to a shower light and a second receptacle. I need all these on the same circuit as it will be a GFCI breaker in the panel.

The receptacle box is 22 cu. in. Will the above result in a cable fill violation? Basically 3 x 12/2 = 9 conductors, right? Plus the outlet itself. Are there quick ways to do the math on this sort of thing?

If the above is a violation, then what are my options? Would the best thing be to put a junction box in the attic and take off the shower light and other receptacle from there? Or maybe just one of them is all I'd have to do from a junction box?

Thanks for any help.
 
  #2  
Old 02-04-08, 04:28 PM
J.D.S.'s Avatar
Member
Join Date: Jan 2008
Location: Central Minnesota
Posts: 27
Upvotes: 0
Received 0 Upvotes on 0 Posts
Are there quick ways to do the math on this sort of thing?
Check this site:
http://www.selfhelpandmore.com/homew...fill/index.htm

Use a multiplying factor of 2.25 with 12 awg wire. Grounds get a count of one.

Count:
6 for conductors, 1 for ground, 2 for device. 9 X 2.25= 20.25 CU. IN. required. The box size is adequate.
 
  #3  
Old 02-04-08, 04:43 PM
Unclediezel's Avatar
Member
Join Date: Dec 2006
Location: Northeastern PA.
Posts: 2,230
Upvotes: 0
Received 0 Upvotes on 0 Posts
Since the math usually gives me a migraine, a basic rule of thumb......

Buy the biggest box you can find that will fit in the space you have. 99 % of the time you cant go wrong. the other 1% is made up by common sense, and neatness.
 
  #4  
Old 02-04-08, 05:12 PM
C
Member
Thread Starter
Join Date: Jan 2005
Location: Missouri
Posts: 144
Upvotes: 0
Received 1 Upvote on 1 Post
So - if I read this right - the conductors always get counted separately, but the grounds from each of those NM cables coming in only get counted together as one cable. Is that correct?

3 x 3 12/2 cables = 3x2 for conductors + 1 for all grounds = 7 total wires.

I know that's what was in the formula given. I just want to confirm so going forward I can use this info.

Thanks to both for the quick response.
 
  #5  
Old 02-04-08, 05:15 PM
C
Member
Thread Starter
Join Date: Jan 2005
Location: Missouri
Posts: 144
Upvotes: 0
Received 1 Upvote on 1 Post
I 'll reply to my own question. I should have followed the link above first! It has the answers to what I just asked. All grounds count only as one.

Thanks for the link - some good stuff there.
 
  #6  
Old 02-05-08, 02:00 PM
BOA's Avatar
BOA
BOA is offline
Member
Join Date: Oct 2005
Posts: 29
Upvotes: 0
Received 0 Upvotes on 0 Posts
Don't NM clamps count as 1 also if using a metal box?
 
  #7  
Old 02-05-08, 03:15 PM
J.D.S.'s Avatar
Member
Join Date: Jan 2008
Location: Central Minnesota
Posts: 27
Upvotes: 0
Received 0 Upvotes on 0 Posts
Originally Posted by BOA
Don't NM clamps count as 1 also if using a metal box?
If they are considered "internal" cable clamps. I presumed we where talking about a single gang plastic of fiber box that has no internal clamps. A romex connector that consists of a clamp and a locknut is considered an external clamp.
 
  #8  
Old 02-06-08, 09:03 AM
BOA's Avatar
BOA
BOA is offline
Member
Join Date: Oct 2005
Posts: 29
Upvotes: 0
Received 0 Upvotes on 0 Posts
Originally Posted by J.D.S. View Post
If they are considered "internal" cable clamps. I presumed we where talking about a single gang plastic of fiber box that has no internal clamps. A romex connector that consists of a clamp and a locknut is considered an external clamp.
Gotcha. So the internal clamps are the ones with a clamp inside the box that screws down from within. Thanks. Btw, the OP never mentioned the material of the box.
 
  #9  
Old 04-02-08, 04:03 PM
C
Member
Thread Starter
Join Date: Jan 2005
Location: Missouri
Posts: 144
Upvotes: 0
Received 1 Upvote on 1 Post
Originally Posted by J.D.S. View Post
Use a multiplying factor of 2.25 with 12 awg wire. Grounds get a count of one.

Count:
6 for conductors, 1 for ground, 2 for device. 9 X 2.25= 20.25 CU. IN. required. The box size is adequate.
A late follow-up question regarding this. Would a GFCI receptacle count the same as a normal one? Or should I allow more CU. IN. for one of those?

I was originally going to use a GFCI breaker on this circuit, but after pricing them - yikes! I'm thinking a GFCI outlet would be fine. I'm trying to do several things above code, but I just don't think I can justify the cost for that breaker.
 
  #10  
Old 04-02-08, 04:08 PM
I
Forum Topic Moderator
Join Date: Feb 2005
Location: Near Lansing, Michigan
Posts: 10,941
Upvotes: 0
Received 45 Upvotes on 43 Posts
Originally Posted by cakins View Post
Would a GFCI receptacle count the same as a normal one?
Surprisingly enough, the GFCI receptacle counts the same as a standard receptacle; the fill for the device is equal to two wires.

The purpose of box fill has more to do with bending and kinking of wires than actual volume occupied. The intention of the calculation is that it requires 2.25 cu. in to put a safe bend in a #12 without damaging the insulation. Therefore, the actual size of the device is less important than the amount of available volume for the wires to fold into.
 
  #11  
Old 04-02-08, 08:07 PM
S
Member
Join Date: Nov 2007
Location: (near) Boise, ID
Posts: 441
Upvotes: 0
Received 0 Upvotes on 0 Posts
Why don't you just run the cable to the first box, then to the second, and so on, instead of branching them at one of the boxes? It will be easier now and in the future to get the devices to fit into the boxes if only 1-2 cables enter each one.
 
 

Thread Tools
Search this Thread
 
Ask a Question
Question Title:
Description:
Your question will be posted in: