Electrical question: a gfci AND a lighting circuit

Hey gang,
I have two circuits serving the "south 40"
"A" is a security lighting circuit currently switched from three locations with a 3-way switch at the final common work box. Power is GFCI protected at the panel
"B" is an unswitched dedicated 20A circuit with a GFCI outlet at the final common work box.
I decided after the fact (after conduit was covered) that I needed to have GFCI protected unswitched power at the light pole A. Since I had already run 12/3 WG to the light pole, and had an "extra" unused red wire, I thought that I could get a GFCI protected hot lead by connecting my "extra" red wire to the load side of my B circuit GFCI, then wired the common and ground back through the lighting circuit A.
I did it but as soon as I apply a load I throw the breaker or the GFCI (I can't remember which)
Is there a way to rewire to make this work without more wire/conduit?
That is, can I retain my switched lighting circuit "A" without creating a new branch circuit? (concerned about too much load on the circuit because I will have to walk up a steep hill to the panel in the house if the GFCI interupts or the circuit breaker throws )
If NOT, can I create an always HOT branch circuit by taking the HOT from my switched lighting circuit? (currently switched from three locations, including at the final work box)
Wouldn't that mean taking power feed from BOTH the black AND the red (traveller) at my 3-way switch at "A". I think that means the hot supply wire would change from being the black wire to the red wire every time a lighting switch was operated on circuit "A". That means the HOT would consist of a black and red cross connected. Don't think I can do that....
Any other ways to skin this cat?
TIA!
MT
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No.
GFCI's work by comparing the current on the hot and neutral wire, and tripping if the current differs by more than about 3ma.
You're pairing the neutral on "A" with the downstream GFCI'd hot from "B" to produce your outlet on pole "A". As soon as you pull 3ma or more from that outlet, _both_ GFCI's will see the exact same imbalance between their respective neutrals and hots. It's a matter of luck as to which one will trip first - might even both trip.
Secondly, since these two circuits share the same neutral, it's a no-no.
If the circuits are off the same main leg of the panel, you could fry the neutral without tripping either breaker (assuming you got around the GFCI issue).
If the circuits are off opposing legs, you won't overload the neutral, but code will not permit "haywire routing" like this (usually expressed in terms of "all conductors for a circuit need to be in the same cable"). This has to do with a notion that circuits should have "predictable behaviour" (vis-a-vis subsequent rework) without having to know what wierd way it was wired.
If you originally pulled individual wires, you _might_ get away with it code-wise, but I really don't like it - requires some additional care in connections (pigtailing) etc. But you still have the GFCI's not liking it.
While you _could_ get this to operate by converting "A" to a regular breaker (perhaps putting a outlet-less GFCI in pole "A" right on the lamp), and taking your B "hot" from the line side of the outlet (and using a GFCI outlet for "A"), you still have to contend with the shared neutral issue above, and you're not GFCI'ing anywhere near as much of the circuits as you originally were. Your other suggestions give me a headache, so I'll let someone else answer ;-)
--
Chris Lewis, Una confibula non set est
It\'s not just anyone who gets a Starship Cruiser class named after them.
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Thanks Chris. Haywired is right! I'll just tap into the lighting circuit to get power now, then I'll run more wire/conduit in the spring.
Thanks for your help.
MT

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wrote:

Can you get another wire in the conduit? Raceway systems are supposed to be installed such that wires can be installed after the conduit was installed. You may need to pull all the wire out, mate the new wires and pull it back.

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Great suggestion, but I don't think I can do it.
I never "planned" on adding the additional circuit, so I used 3/4" conduit. There are also three 90 degree bends. I'd never get the 12/3 WG out and two 12/2 WG back in. New conduit will be easier. Yesterday I capped the red I took from the load side of GFCI "B" and connected my receptacle to black from the switched hot "A". So now I have a working switched receptacle and security light on "A". Come springtime I'll run new conduit and wire for constant power at "B"
Reading my GFCI packet would have answered my question. I thought I could use any neutral to complete the circuit, since they were all interconnected at the bus on the panel. Chris Lewis explained that the GFCI works by sensing the current difference between hot and neutral. That answered my question. I was thrown off when my GFCI tester worked and showed OK, but when I applied a real load, it threw the GFCI. Tester draws almost no current.
No way to do it right without more wires. I never knew/thought about the future repercussions as Chris/Code discussed, and I like doing it "by the book". If my memory failed me on the "haywiring", I'd end up electrocuting myself servicing the circuit years from now...
Thanks!
Matt

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There is precisely one way to do it without more wire - but you lose the "unswitched" attribute on pole "A" ;-)
In short, wire the new "A" outlet in parallel with the lightbulb. You'll have a switched duplex outlet which is GFCI'd at the panel.
--
Chris Lewis, Una confibula non set est
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wrote:

Pull that stupid cable out and pull in THHN conductors. You can get a shitload in 3/4"
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