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Joined: Jan 2004
Posts: 1,507
Article 310 Section 310.15 (B)(3) talkes about adjustment factors for more than 3 current carrying conductors in a raceway or cable.
The question that was given to me is that if you have 6 two wire circuits each with it's own neutral in the same conduit, can he use 310.15(B)(5)(a) and not count the neutrals. I said "No" because crudely put (a) is really talking about a Delta system and (b) and (c) are talking about a Wye system. Plus for his example none of the circuits shared the same neutral wire as it would if they were multi-wire circuits.

The code references are from the '11 NEC but it's worded the same in the '08 NEC, they just moved it around.

George Little
2017 / 2014 NEC & Related Books and Study Guides
Joined: Mar 2011
Posts: 98

(6) two wire circuits? So, 12 conductors + 6 neutrals?

In that situation it doesn't matter whether you're counting the neutrals or not since table 310.15(B)(3)(a) gives an adjustment factor or 50% for 10-20 conductors and either way (12 or 18 conductors) the factor is 50%.

Unless I'm misunderstanding your question.

Joined: Jul 2004
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George, there are several ways to explain it but I agree with the answer.

Greg Fretwell
Joined: Apr 2002
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6 x 2 wire = 12 conductors. 6 current carrying; 80% derated. Without adjustments for ambient temperature, if required.

Joined: Jul 2004
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Why aren't the neutrals current carrying? Dr Kirchoff tells us all of the current going out, has to come back and I2R losses are the same going both ways. The only time you don't have to count neutrals is if they only carry the unbalanced current in a MWBC.

Greg Fretwell
Joined: Oct 2000
Posts: 5,389
i never could figure how we 'balance' a multitude of non-simultaneous loads, or what this unbalanced current could be calculated at....?


Last edited by sparky; 07/21/11 08:59 AM.
Joined: Oct 2000
Posts: 2,722
Broom Pusher and

The question that was given to me is that if you have 6 two wire circuits each with it's own neutral in the same conduit, can he use 310.15(B)(5)(a) and not count the neutrals.

In this case, there are no "Neutrals" - only Grounded Conductors and Ungrounded Conductors.

George's scenario states "6 Two-Wire Circuits", which results in 12 Current Carrying Conductors.

Derating Factor will be 50%.

Installing (12) #10 THHN CU. Conductors with 20 Amp OCPDs will comply.

BTW, these Two-Wire Circuits could be any of the following flavors;

  • One Ungrounded Conductor + One Grounded Conductor,
  • Two Ungrounded Conductors.

In either case, there will be Two Current Carrying Conductors per Circuit.

The time a Grounded Conductor may be determined as a Common "Neutral" is when it is part of a Multiwire Branch Circuit.

For the Common Grounded "Neutral" Conductor of a Multiwire Branch Circuit to become a "Balanced Load Carrying Conductor", the System needs to be derived from a Single Phase 3 Wire Transformer - either via the Center Transformer of a 4 Wire Delta System, or from a "Stand-Alone" 1 Phase 3 Wire Transformer.

The "Center-Tapped" Secondary Winding will have only the Imbalanced Load flowing, when Two L-N Connections are made across the entire length of the Coil.

Simply stated, using L-N-L (AKA: 1 Phase 3 Wire Multiwire Circuit) with a 10 Amp Load between the "A" Line Output and the Center Tap, and a 15 Amp Load between the "B" Line Output and the Center Tap, the Center Tap Load will be the imbalanced Load value - which is 5 Amps.

This occurs due to the "Common Load" flow running across the entire length of the Secondary Coil - as if the connection was a 10 Amp, 240V Load; and the imbalanced "Half-Winding Section Load" flow being circulated between the Center Tap and one half of the entire Secondary Coil is 5.0 Amps.

In this case, we will say the Right-Hand side of the Secondary Winding's end will be Line "B", so the Winding Section between the Center Tap and Terminal "B" will have 5.0 Amps circulating, and the entire Winding will have 10 Amps circulating.

Let's say the Primary of this Transformer is 480V; the Primary Load will be 6.25 Amps - or 3.0 KVA


With a Wye System, the Common Grounded Conductor will only "Balance" under certain conditions - mainly when the connected Loads are equal and at Unity Power Factor.
Three equally sized, equally loaded 1 Phase Motors across a 3 Phase 4 Wire Wye Multiwire Circuit will result in a low Percentage of L-N Load Current flowing.

Other than these Load types, the Common Star Point Derived Grounded Conductor of a Wye System will draw close to the same Load Amps as the Highest L-N Load value.

With Reflective Loads (Multiple Harmonic Frequency Load Currents Reflected back into the Power System), the Common Grounded Conductor will carry the highest L-N Load Amperes (Fundamental Amperes), along with 3 times the Percentage of Reflected Harmonic Amperes.

THD = 10%.

  1. Load's Fundamental Amperes = 10.0 Amps,
  2. Load's Harmonic Reflected Amperes = 1.0 Amps,
  3. Total Load Amperes = 11.0 Amps.
  4. (3) of these Loads connected L-N (Phase A-N, Phase B-N, Phase C-N),
  5. Load on "N" = 10.0 Amps Fundamental, plus 3 times the Harmonic Reflections (1.0 Amps x 3),
  6. Total Load on "N" = 13.0 Amps.

Using a 1 Phase 3 Wire Multiwire Circuit from the 4 Wire Wye creates an "Open Wye" Polyphase System, so once again the Common Grounded Conductor will be a Current Carrying Conductor - functioning similar to a 3 phase 3 Wire Branch Circuit.


Lastly, not all Grounded Conductors are the same as a "Neutral", and vice-verse...

The Grounded Conductor of a 3 Phase 3 Wire Corner Grounded Delta is not a "Balanced Neutral" at all.
This Grounded Conductor is one of the three system phases. The only difference is it is intentionally ground bonded. Otherwise it functions exactly the same as any of the remaining Two Ungrounded Conductors on the System.

Will stop here and wait for feedback.

-- Scott

Scott " 35 " Thompson
Just Say NO To Green Eggs And Ham!
Joined: Jan 2005
Posts: 5,316
Cat Servant
Let's look at this issue by taking a few examples, and see how the results are affected.

In one example, you have a perfectly balanced multi-wire circuit. Three wires, two of them 'hot.' The third- the neutral- doesn't carry any current. So, from this one circumstance, we get the assertion that 'neutrals are not current carrying conductors.' OK, I'll buy that- for this example ONLY.

In a similar arrangement, the two circuits are not perfectly balanced. Therefore, the neutral does carry SOME current. I don't see how you can claim it's not a current carrying conductor; you have to count all three wires.

Now, let's forget the MWBC, and run two hots with two neutrals. Each neutral is now carrying exactly as much current as the hot wire- so of course they're current carrying conductors.

If you 'share' neutrals, you're reducing the total number of wires, and that alone will improve the derating situation. To eliminate the neutrals from consideration completely is not only technically wrong - because the neutrals ARE carrying current - but you're repeating the 'discount' you got by combining neutrals in the first place.

Joined: Apr 2002
Posts: 7,283
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Pardon me for being confused. George said two wire, Tesla came up with 18 conductors, I only can 'see' 12.

To me, a 2 wire is a hot & a neutral; ie: black & white. What did I miss?? (Besides a George Question)

Joined: Jul 2004
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2 wire is also 2 current carrying conductors, no matter how you do it. If you duplicate that 6 times you have 12 current carrying conductors. Tesla was pointing out, the only way the grounded conductor is not considered current carrying is if it is only handling the unbalanced current of a unity PF multiwire circuit.
It still carries current but only the current that is not flowing in one of the ungrounded legs thus it nulls the overall I2R heating.

Greg Fretwell
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