I have 16 post lights evenly spaced on a 1000ft. run. The lights are 180 watts each on a two-pole multiwire branch circuit. I have 8 lights on each pole of the 2-pole 20 amp gfi breaker. I am using #2 aluminum USE cable to prevent unacceptable voltage drop. My question is since the load is balanced is it acceptable to use a reduced neutral such as a size 2-2-4 w/6gnd. cable. I couldn't find my situation in the code book.
reduced neutral
- Thread starter sparky59
- Start date
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- Location
- Chapel Hill, NC
- Occupation
- Retired Electrical Contractor
I am assuming these are 120 volt fixtures. This being the case I don't think you can share the neutral on a GFI circuit. It won't work. Again if it is 120 volts than why would you undersize the neutral??
- Location
- Massachusetts
You could reduce the neutral, it need be only large enough to handle the maximum imbalance.
Considering you are balancing the circuit the neutral will not carry any current until a lamp burns out.
A couple of points.
I would alternate the circuits.
I would forget the GFCI, as much as I like GFCIs it will likely be a nuisance and will not have much benefit.
You will need to run an EGC as large as your line conductors due to 250.122(B)
Considering you are balancing the circuit the neutral will not carry any current until a lamp burns out.
A couple of points.
I would alternate the circuits.
I would forget the GFCI, as much as I like GFCIs it will likely be a nuisance and will not have much benefit.
You will need to run an EGC as large as your line conductors due to 250.122(B)
- Location
- Massachusetts
Dennis Alwon said:
With a two pole GFCI this can work.
- Location
- Massachusetts
Davis9 said:
Yes, and with 15, 20 and 30 amp circuits the ratio is 1:1.
- Location
- Massachusetts
Davis9 said:
Table 250.122
Splice it down to a conductor that will fit.
Commercial type breakers, I-LInes, GE TED, TEYs etc will take a at least a 2/0 on a 15 amp breaker.
Table 250. 122?
15A 14cu 12al
20A 12cu 10al
30A 10cu 8al
Am I missing something, I'm not trying to be a pain. I took proportionately in terms of the circular mil of the ungrounded conductors. I guess it's so. But how about 250. 122 (C)? in this application. Or do we apply the 250.122(B) only? Or in addition to (C), now I'm confused.
Circuits of 15,20,30A are not neccesarily run in 14,12,10ga wire.
Tom
15A 14cu 12al
20A 12cu 10al
30A 10cu 8al
Am I missing something, I'm not trying to be a pain. I took proportionately in terms of the circular mil of the ungrounded conductors. I guess it's so. But how about 250. 122 (C)? in this application. Or do we apply the 250.122(B) only? Or in addition to (C), now I'm confused.
Circuits of 15,20,30A are not neccesarily run in 14,12,10ga wire.
Tom
- Location
- Massachusetts
sparky59 said:
That still does not alleviate you from the requirements of 250.122(B).
Besides, code wise you have one multiwire branch circuit.
Bob NH
Senior Member
- Location
- The "Live Free or Die" State
On the other hand, you might put in a 100 amp breaker because of high temperature caused by sunlight shining on the posts. Then you will be allowed to use #6 aluminum as the EGC, per Table 250.122.
Johnmcca
Senior Member
- Location
- Mid-Hudson Valley, NY
if you run a larger wire then the ground must be up-sized proportionately
- Location
- Massachusetts
Davis9 said:
It is so.
Tell me about it.
We spend a lot of our time wiring site lighting at malls, stores.
You can find us running 4 AWG CU circuit conductors with 4 AWG CU EGCs.
Tom, we would have to use 250.122(B)
This would mean if you increased a 20 amp circuit's ungrounded conductor to a #3 for VD the EGC would have to be a #3 also.
In reality the article needs some refining but as it stands right now 15,20,and 30 amp circuits must be 1-1
Roger
This would mean if you increased a 20 amp circuit's ungrounded conductor to a #3 for VD the EGC would have to be a #3 also.
In reality the article needs some refining but as it stands right now 15,20,and 30 amp circuits must be 1-1
Roger
JohnJ0906
Senior Member
- Location
- Baltimore, MD
The reason for oversizing the conductors is voltage drop, which is caused by the resistance of the wire itself. If you don't increase the size of the egc, the same resitance (of the wire itself) will reduce the fault current. It's no longer a "low impedance path"
I think I understand this better than I explain it , so if someone can jump in and explain it better, please do!
I think I understand this better than I explain it
, so if someone can jump in and explain it better, please do!- Status