EGC Fill

[Newtechemmanuel]

Quick Question!

Article 314.16 (b)(5) 2020 NEC states that where up to four EGC enter a box , that a single volume allowance in accordance with Table 314.16(B) shall be made based on the largest conductor.

I am confused about something. Lets say I have 5 - # 14 AWG CU EGC and 5 - #12 AWG CU. EGC. When doing the calculation do I just forget about the # 14AWG conductors or do I add them to the number 12AWG, so instead of having 5 # 12 AWG CU I would have 10 # 12 AWG.

 

[wwhitney]

Lets say I have 5 - # 14 AWG CU EGC and 5 - #12 AWG CU. EGC. When doing the calculation do I just forget about the # 14AWG conductors or do I add them to the number 12AWG, so instead of having 5 # 12 AWG CU I would have 10 # 12 AWG.

The last sentence of (2020) 314.16(B)(5) says "A 1/4 volume allowance shall be made for each additional equipment grounding conductor or equipment bonding jumper that enters the box, based on the largest equipment grounding conductor or equipment bonding conductor."

So you have 10 EGCs. 4 are in the first allowance. 6 are extra, so you need 6/4 = 1.5 extra allowances. What size allowance? It's "based on the largest equipment grounding conductor or equipment bonding conductor", so those are all #12 allowances. The total EGC allowance is (2.5 allowances) * (2.25 in^3/allowance) = 5.6 in^3.

Cheers, Wayne

 

[Dennis Alwon]

Yes, as you can see by Wayne's post all the equipment grounding conductor's become #12, in your case, and get counted together

 

[Newtechemmanuel]

The last sentence of (2020) 314.16(B)(5) says "A 1/4 volume allowance shall be made for each additional equipment grounding conductor or equipment bonding jumper that enters the box, based on the largest equipment grounding conductor or equipment bonding conductor."

So you have 10 EGCs. 4 are in the first allowance. 6 are extra, so you need 6/4 = 1.5 extra allowances. What size allowance? It's "based on the largest equipment grounding conductor or equipment bonding conductor", so those are all #12 allowances. The total EGC allowance is (2.5 allowances) * (2.25 in^3/allowance) = 5.6 in^3.

Cheers, Wayne

 

[Newtechemmanuel]

Thank you both and just so I am clear on my understanding all the conductors would be based on #12 AWG CU

Check my math for me please

5 # 12 AWG = 2.25
5 # 14 AWG = 2.00

Since #12 is largest the answer is
The first four equipment grounding conductors = 2.25 in.3
The next 6 equipment grounding conductors = 2.8125 in.3 (2.25 in.3 X ¼ = .5625 in.3) (5 more wires X .5625 in.3 = 2.8125 in.3)
Total equipment grounding conductor fill =5.0625

Am I correct in my arithmetic?

 

[Dennis Alwon]

You did something incorrectly. It should be 5.625 cu.in maybe a typo

The easiest way to do this is to look at how many wire counts you have
First 4 equipment grounding conductor = 1 allowance
Rest 6 x 1/4" = 1.5

Total is 2.5 allowance
2.5 x 2.25 = 5.625 in cu

 

[wwhitney]

The next 6 equipment grounding conductors = 2.8125 in.3 (2.25 in.3 X ¼ = .5625 in.3) (5 more wires X .5625 in.3 = 2.8125 in.3)

That's 6 more wires 6 * .5625 = 3.375.

Cheers, Wayne

 

[Newtechemmanuel]

That's 6 more wires 6 * .5625 = 3.375.

Cheers, Wayne

Thanks Wayne. How did get the original .5625? Did you do 2.25 x .25 = .5625 ( One conductor). I'm just confused as to how come you re-added the conductor from which you got the .5625. I would think it would be subtracted since it was already used in finding the original .5625.

 

[Newtechemmanuel]

Thanks Wayne. How did get the original .5625? Did you do 2.25 x .25 = .5625 ( One conductor). I'm just confused as to how come you re-added the conductor from which you got the .5625. I would think it would be subtracted since it was already used in finding the original .5625.

Disregard I understand what you are saying. I get it!

 

[Newtechemmanuel]

That's 6 more wires 6 * .5625 = 3.375.

Cheers, Wayne

I know I'm bombarding you with emails. But mistake came when I didn't add back in the .5625

So my calculation was wrong because I did know to add .5625 back into the original equation.

2.8125 in.3 (2.25 in.3 X ¼ = .5625 in.3) (5 more wires X .5625 in.3 = 2.8125 in.3)= .5625 + 2.8125 = 3.375 or like you said 6 x .5625. I will adopt your way of doing solving these problems, it's faster and easier to check.

 

[wwhitney]

Even easier: Find largest EGC, get unit allowance: #12 largest means 2.25 in^3 unit allowance.

3 or fewer EGCs: just use one allowance
4 or more EGCS: use unit allowance * #EGCs / 4.

Cheers, Wayne