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#173114 01/02/08 11:09 PM
Joined: Apr 2001
Posts: 109
Hello from South Dakota!!!!

I need a little help.
My mind does not seem to be working any more.

If you have a single phase service, 35a on leg #1, 25a on leg #2, your neutral should carry the 10a unballanced load.

Now if you have the same situation on a three phase 208/120 service, but you are only using two legs on a 2pole breaker, what would you expect to see on your neutral. I always thought it would be the same.

I have a service that we have just installed that has a very high number on the grounded conductor. I will stop by in the morning and get the exact numbers with my clamp meter. Will post later.


Jon Niemeyer
Joined: Jul 2007
Posts: 1,335
The neutral will carry the unbalanced loads of all three legs so the reading on the neutral will likely be different then you think. I'm not sure how to calculate it though.

"Live Awesome!" - Kevin Carosa
sparkyinak #173118 01/03/08 01:31 AM
Joined: Jul 2004
Posts: 9,766
Likes: 13
If you have reactive loads like PCs and electronic ballasts throw your calculator back in your truck. The neutral load will be bigger than it computes because of harmonics

Greg Fretwell
gfretwell #173122 01/03/08 02:42 AM
Joined: Oct 2000
Posts: 2,722
Broom Pusher and
Typically, the Common Grounded Conductor (AKA "Neutral") of a 3Ø 4 Wire Wye system will carry between 70% and 100% of the highest L-N load value.

Most commonly, the Grounded Conductor will be carrying 100% of the highest L-N load Ampere rating.

If there are extreme Harmonics across all 3 L-N circuits, expect upto 173% of the L-N load Amperes flowing on the Grounded Conductor (this would be very rare, and would exist if all 3 L-N circuits were carrying the same load Amperes, and had the same Harmonic Distortion values on the connected loads).

If the loads are connected as an "Open Wye" (1Ø 3 Wire), the Grounded Conductor will carry 100% of the highest L-N load Amperes.

If the connected loads happen to be _ALL_ Pure Resistance types (1.0 PF / < 0.5% Reactive / 99.99999% True Power / etc.), then the load Amperes on the common Grounded Conductor may be found using Vectors.
In these cases, the load Amperes will be lower than the highest L-N load Ampere value.

With unbalanced loads, the results will be different per L-N load values; and with connected loads made across 1Ø 3 Wire circuitry - vs - 3Ø 4 Wire circuitry.

If a 3Ø 4 Wire multiwire branch circuit was driving 3 L-N connected _PURE RESISTANCE_ loads of the exact same characteristics
(total circuit Impedance ((Z)) is exactly the same for all 3 connected loads),
there will be nearly zero load on the common Grounded Conductor - provided the Voltage across each load is exactly the same as the other two L-N loads.

FYI: "Pure Resistance" loads = things like:
  • Incandescent Lamps,
  • Quartz-Halogen Lamps (in normal operation),
  • Electric Resistance Heating Elements for Stoves and Ovens,
  • Electric Water Heater Elements.

To me, Incandescent Lamps and especially Quartz-Halogen Lamps have enough Inductive Reactance to qualify them as 95% Pure Resistance (0.98 PF).
This is due to the coiled filaments.

Along with this, many tightly wound Resistance Elements would have similar characteristics.

As for 1Ø 3 Wire, from a Single Phase source (derived from a single center tapped coil), the "Center Tapped Grounded Conductor (AKA "Neutral") will carry the "Unbalanced" load Amperes from two L-N load connections - as connected between the end of the coil leads to the center tap.

This would describe the typical 120/240V 1Ø 3 Wire Center Tapped Transformer, found in most Residential settings;
The center Tapped winding of a 240/120V 3Ø 4 Wire Delta Transformer arrangement.

208/120V 3Ø 4 Wire Open Delta "Tee" Transformer setups, may have odd resulting L-N Unbalanced load values found on the Grounded Conductor -
may be whatever the unbalanced load is,
may be 100% of the highest L-N load,
may be 141% of the balanced L-L-L-N load.

Good luck!!!


Scott " 35 " Thompson
Just Say NO To Green Eggs And Ham!
Scott35 #173141 01/03/08 04:26 PM
Joined: Nov 2000
Posts: 2,148
Without the effects of harmonics you can find the neutral current using the following:
=square root of [(A^2 + B^2 + C^2)-(A x B)-(B x C)-(A x C)]
My experience does not show the amount of harmonics that all of the information says there should be. In my opinion that is because this information is provided by people that have a vested economic interest in solving the harmonic problem.

resqcapt19 #173148 01/03/08 07:34 PM
Joined: Apr 2001
Posts: 109
Scott and Don,

Thank you very much for your replies. That is exactly what I was looking for.

I was unable to get the readings today but I will make sure I get them posted before Monday.

God, I love this site!!!


Jon Niemeyer
Joined: Mar 2005
Posts: 1,213
Don, I've never seen nuetral current above line current in any of our data centers. Harmonics coming off most quality UPSs, VFDs and computer power supplies are filtered down so that they're really not an issue at all. Heck, a lot of them are simply straight delta and don't even have a neutral connection.

OTOH, I've heard stories from the old timers about a fractional neutral in an old buildings quite literally glowing cherry red from the harmonics...

Based on our load profile, I always specify 100% neutral.

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