neutral current

[alaskanassassin]

I have a three phase 120/208 panel and am at a total loss. A phase has 24 amps, B phase 6 amps, C phase 11 amps, and neutral has 18 amps. Circuits 1 has 9 amps and 9 amps on the neutral and circuit 4 has 7 amps and 7 amps on the neutral. When I shut off circuit 4 my main neutral goes up, I would think it would go down. I have three lights on circuit 1 and 9 fluorscent lights on circuit 4. Can someone explain what is going on? Thanks Doug

 

[brian john]

1st off clamp around A, B and C phase what do you read? This should equal the neutral current.

Calculating neutral current you should have about 16 amps on the neutral.


(A^2+B^2+C^2)-(A*B+B*C+C*A) take the square root =Neutral amps

 

[Dennis Alwon]

Here is the formula

I see Brian already posted it.

 

[alaskanassassin]

So my 16 amps in normal. I just never have checked this before. Only reason I was checking was when I disconnect a neutral on a circuit I had an arc or spark. Thanks

 

[Dennis Alwon]

The neutral current will change depending on the phase the loads you turn off are on. You say you have 6 amps on phase B but then you say you have 7 amps on cir. 4. I would assume that cir. 4 is the B phase but that cannot be.

If we assume the cir. you turn off on the C phase has 7 amps then the neutral load will increase to 19 amps from 16 amps. use the equation-- you can plug this formula into an excel spreadsheet and use cells A1 , B1 and C1 for the neutral currents and you will see what I mean

=SQRT(((A1*A1)+(B1*B1)+(C1*C1))-((A1*B1)+(B1*C1)+(C1*A1)))

 

[Dennis Alwon]

Here is another way to look at this. The neutral current is balanced when all three phases are equal. Thus is all phases had 24 amps then the neutral current is zero. Now reduce one phase to 0 amps and the neutral current shoots up to 24 amps and reducing another phase to 0 amps will still keep the neutral current at 24.

 

[GaTech04]

If there's room, you could move circuit #1 (9A) to the B phase to help even out the loads:

A: 15A
B: 15A
C: 11A

N = 4A

 

[Smart $]

...

=SQRT(((A1*A1)+(B1*B1)+(C1*C1))-((A1*B1)+(B1*C1)+(C1*A1)))

Dennis et al,

This equation's result may be inaccurate if the power factors of the line currents differ. I brought it up in this thread and at least two others.

For example, if...

Line A, 5A, pf=1
Line B, 10A, pf=1
Line C, 8.67A, pf=0.867​

...then the current on the neutral will be zero, but the formula yields a result of 4.48A.

 

[alaskanassassin]

I get it now, it has been a while since I seen the formula. Thanks to all who have responded. Doug

 

[Dennis Alwon]

Dennis et al,

This equation's result may be inaccurate if the power factors of the line currents differ. I brought it up in this thread and at least two others.

For example, if...

Line A, 5A, pf=1
Line B, 10A, pf=1
Line C, 8.67A, pf=0.867​

...then the current on the neutral will be zero, but the formula yields a result of 4.48A.

You are smart..:grin: So what formula are you using that takes into account PF for neutral current may I ask?

 

[Smart $]

You are smart..:grin: So what formula are you using that takes into account PF for neutral current may I ask?

I use vector math, but I let Excel do the computational part of it :grin:

I've posted web links to my Excel Neutral Calculator before, but it seems SkyDrive keeps changing the URL to it. So this time I've embedded it in a Word doc, so it will be on this server.

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