# multiple homeruns for neutral only?

I work in conduit and run single wires. I have a weird one.

I have an always-hot (black) and neutral (white) pair that run from the panel passing through A to serve loads at B and C.

At C, always-hot (black) continues to a switch loop at D. The switched-hot (blue) comes back through C, B and A to go down another branch, AA, to serve the lamp.

The neutral for the lamp (white) goes from AA through A back to the service panel.

The circuit has 2 neutrals (white), one serving the receptacle loads at B and C, and the other serving the lamp load at AA. They pass like ships in the night at A, not connecting, and run alongside in the same conduit from A to the panel.

Edit: Here is a drawing. I added a second circuit (red/gray) that is unrelated, simply to illustrate that boxes B and C have a lot of other stuff going on.

Is it kosher for the circuit (black) to have two neutrals on the neutral bar?

• I admit that I am too lazy to draw this out and I have never used conduit, but the question I have is would the switched hot be an "extra" current-carrying wire in the section from C to B to A? Would this allow the instantaneous current (total of all conductors) in this section of conduit to be different from zero? – Jim Stewart May 21 '17 at 11:33
• @JimStewart great question, have already diagrammed it out confirmed that currents are equal, the star/ tree branch topology kind of enforces this anyway. Can't post drawing, onsite w/ nothing but phone. – Harper - Reinstate Monica May 21 '17 at 14:07
• No need to show the diagram; I thought this must be true even if I was not able to visualize it. Is this unusual arrangement due to the light switch being far removed from the light it controls, e.g., a switch inside a building controlling a light at the end of a long corridor or a light outside at a loading dock? – Jim Stewart May 21 '17 at 14:41
• Is there a reason that you can't connect the two neutrals at A and run a single wire back to the main panel? That'd spare you the conduit fill of the extra neutral homerun and also be a wee bit more economical on box fill at A too... – ThreePhaseEel May 21 '17 at 14:54
• @ThreePhaseEel it would add box fill, 3 wires vs 2 passthrus. Honestly this is a disentanglement projevt, and I am taking this step by step, and am suspicious the neutrals might be inappropriarely shared up in the B C complex. I am also out to eliminate all MWBCs due to no way to tie them. Jim it's a top of stairs light switch for a poured cincrete basement which is pristine, with no penetrations, so routed via attic. – Harper - Reinstate Monica May 21 '17 at 16:01

Here is the deal:

No matter how you wire things, you normally need to have two wires. One with the outgoing current and one with the return current. Otherwise you will get inductive heating of any ferrous materials.

If you split a neutral (return the outgoing current back to the source in a separate conduit) you will get inductive heating.

So, to answer your question, the neutral at point A may or may not be joined. This is a fielder's choice since all of the current leaving the panel on the black wire is returning on the two neutrals so the net current (algebraic sum of the current) in the conduit from the panel to point A is zero. However, it is bad technique to have more than one neutral return to the panel per phase conductor. It could be confusing to a future electrician.

This is the key: The algebraic sum of current at any point in the system must add to zero to avoid inductive heating.

Good luck and stay safe!

• it's an awkward situation because from A to C, I have 2 more circuits omitted from the illustration. So I am near conductor limits (9 per conduit without a derate). Since any current on blue reduces current on white, it doesn't count so I'm at 8. If (for clarity to the next guy) I doubled another neutral back from C to AA specifically for the light, blue and its new neutral would both count, putting me at 10. – Harper - Reinstate Monica Jun 27 '17 at 16:38
• Hmm not quite getting what you mean about current on blue reducing current on white. But, only current carrying conductors are counted to ampacity reduction. The Equipment Ground does not count for this calculation. If the neutral is shared between the red and black it also does not count. See 310.15(B)(5)(a) A neutral conductor that carries only the unbalanced current from other conductors of the same circuit shall not be required to be counted when applying the provisions of 310.15(B)(3)(a). – ArchonOSX Jun 27 '17 at 19:21