# Thread: Calculating Unbalanced Neutral Current in 3p-4w system

1. Member
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Mar 2004
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## Calculating Unbalanced Neutral Current in 3p-4w system

Is there a simple formula for determining the unbalanced neutral current in a 3p-4w system? I know how to calculate it if the line to neutral loads on two phases is equal and the line to neutral load on the third phase is different, but what if the line to neutral load is different on each phase?

In other words, if I have a 3ph-4w 208/120v panelboard and connect three circuits, one on phase A, one on phase B, and one on phase C and the corresponding 120v loads are 10A on phase A, 8A on phase B, and 6A on phase C, what would my neutral current be from the unbalanced loads which are all 120 degrees out of phase?

2. Originally Posted by stevee
Is there a simple formula for determining the unbalanced neutral current in a 3p-4w system? I know how to calculate it if the line to neutral loads on two phases is equal and the line to neutral load on the third phase is different, but what if the line to neutral load is different on each phase?

In other words, if I have a 3ph-4w 208/120v panelboard and connect three circuits, one on phase A, one on phase B, and one on phase C and the corresponding 120v loads are 10A on phase A, 8A on phase B, and 6A on phase C, what would my neutral current be from the unbalanced loads which are all 120 degrees out of phase?
SQRT (I^A + I^B + I^C) - (IA x IB) + (IB x IC) + (IC x IA)

SQRT (10 x 10 + 8 x 8 + 6 x 6) - (10 x 8) + (8 x 6) + (6 x 10)

SQRT (100 + 64 + 36) - (80) + (48) + (60)

SQRT 200 - 188 = 12

SQRT 12 = 3.46

Roger
Last edited by roger; 12-12-08 at 01:02 PM.

3. Two write it out a bit better, it's

√((IaČ+IbČ+IbČ)-(Ia*Ib)-(Ib*Ic)-(Ia*Ic))

4. Originally Posted by 480sparky
Two write it out a bit better, it's

√((IaČ+IbČ+IbČ)-(Ia*Ib)-(Ib*Ic)-(Ia*Ic))
Why do you think it's better? Might it have something to do with symbols or two?

Actually, I think one could do it fine.

Roger
Last edited by roger; 12-12-08 at 01:19 PM.

5. Originally Posted by roger
Why do you think it's better? Might it have something to do with symbols or two?

Actually, I think one could do it fine.

Roger
I^A means what? I (amps) to the power of A (?)

Also, you don't have enough parentheses nested.

Originally Posted by roger
SQRT (I^A + I^B + I^C) - (IA x IB) + (IB x IC) + (IC x IA)
.....
Roger
Appears you take the square root of the first set of parantheses, then start subtracting the other three sets.
Last edited by 480sparky; 12-12-08 at 01:23 PM.

6. Originally Posted by 480sparky
I^A means what? I (amps) to the power of A (?)

Also, you don't have enough parentheses nested.

Appears you take the square root of the first set of parantheses, then start subtracting the other three sets.

Oooops, good point should have been I^2A etc...., but I still don't know why it takes two to write it a bit better. :wink:

Roger

7. Originally Posted by roger
Oooops, good point should have been I^2A etc...., but I still don't know why it takes two to write it a bit better. :wink:

Roger
That's the problem with such a formula. Writing it out with common computer keyboard symbols makes it difficult to follow the hierarchy of the formula.

You did a good job explaining it on your first post, though.

Are you asking why it takes two sets of parentheses? To force the Square-Root function to be the last. The way you originally wrote it, it would actually work out this way:

SQRT (I^A + I^B + I^C) - (IA x IB) + (IB x IC) + (IC x IA) means the SQRT function would only apply to the first set of parentheses. You'd end up with:

SQRT (10*10 + 8*8 + 6*6) - (10*8) + (8*6) + (6*10)

SQRT (100 + 64 + 36) - (80) + (48) + (60)

SQRT (200) - (188)

At this point, the SQRT function only would aply to the 200, so you end up with

14.14 - 188 = -185.86.

It's kinda of difficult to have a negative amperage flow. :smile:

At least, this is how I remember it from school, back when dirt was invented.

Edit to add: I just noticed my first formula was wrong. I had IbČ in the first set of parenthesis twice. It should be:

√((IaČ+IbČ+IcČ)-(Ia*Ib)-(Ib*Ic)-(Ia*Ic))

And to be hard on myself, I should also add the following variables explained:
Ia = Amp flow on phase A
Ib = Amp flow on phase B
Ic = Amp flow on phase C
Last edited by 480sparky; 12-12-08 at 01:45 PM.

8. Originally Posted by 480sparky
Are you asking why it takes two sets of parentheses?
No, I'm asking why you say Two write it out a bit better. Even though I was in a hurry and answering phone calls while I was posting, I think One (even if just myself) can write it out correctly.

Roger

9. Originally Posted by roger
No, I'm asking why you say Two write it out a bit better. Even though I was in a hurry and answering phone calls while I was posting, I think One (even if just myself) can write it out correctly.

Roger
OOOOOOOOOOoooooooooooo! Sea watt hapens wen theirs know SpellCheck?

10. Originally Posted by 480sparky
Edit to add: I just noticed my first formula was wrong. I had IbČ in the first set of parenthesis twice. It should be:

If you're like me and are doing anything else or are the least little bit preoccupied, things do get missed and typo's do happen.

Roger

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