# Complex Imaginary volume 4, prob 14

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### #1 cabby

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Posted 09 April 2012 - 02:42 AM

Has anyone worked this problem yet? 34.5kv, 3phase, 4 wire, Y-connected, distrobution panel is experiencing unbalanced load conditions. Phase A load=(120+65j)kva @ (19.9 angle 0) kv. Phase C load=(65+20j)kva @ (19.9 angle 0)kv. Calculate the neutral current.

for the life of me, I do not understand the solution nor do I think the problem is set up correctly. I thought the voltage for phase C would be 19.9 at angle 240 or something to that affect. If anyone has some examples of unbalanced load problems it would be great.

### #2 nmh0408

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Posted 09 April 2012 - 04:24 AM

Since it doesn't specify where the load is connected and it is asking for In you would assume the loads are connected between each leg and the neutral and therefore In=load A/(V/root3)angel 0 + load B/(V/root3)angel 120+ load B/(V/root3)angel 240

### #3 stinkycheese

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Posted 09 April 2012 - 12:59 PM

I was just looking at this one! I'm taking the week off as final prep. The Phase C voltage should be <240. However, I don't think their answer is right. I get that IA = 6.86<-28.44 and IC = 3.42<137.1 (I think they got confused with conjugates). IN = -(IA+IC) gives me IN = 3.65<165 for an answer of 3.65A.

### #4 nmh0408

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Posted 09 April 2012 - 02:19 PM

Since it doesn't specify where the load is connected and it is asking for In you would assume the loads are connected between each leg and the neutral and therefore In=load A/(V/root3)angel 0 + load B/(V/root3)angel 120+ load C/(V/root3)angel 240

Note the correction.

### #5 nmh0408

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Posted 09 April 2012 - 02:21 PM

I was just looking at this one! I'm taking the week off as final prep. The Phase C voltage should be <240. However, I don't think their answer is right. I get that IA = 6.86<-28.44 and IC = 3.42<137.1 (I think they got confused with conjugates). IN = -(IA+IC) gives me IN = 3.65<165 for an answer of 3.65A.

If you have three unbalanced loads, why you are not adding up Ib to the neutral current?

### #6 stinkycheese

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Posted 09 April 2012 - 02:28 PM

If you have three unbalanced loads, why you are not adding up Ib to the neutral current?

There is no load on phase B.

### #7 nmh0408

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Posted 09 April 2012 - 03:38 PM

If you have three unbalanced loads, why you are not adding up Ib to the neutral current?

Since I do t have there book, what was the answer?

Hence, the question is giving you C@0 degrees, therefore you start your sequence with C, I solved it based on CBA sequence and I got 5.23angel 119?
There is no load on phase B.

### #8 stinkycheese

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Posted 09 April 2012 - 09:39 PM

Since I do t have there book, what was the answer?

Hence, the question is giving you C@0 degrees, therefore you start your sequence with C, I solved it based on CBA sequence and I got 5.23angel 119?

The CI answer is 6.6. As I wrote above, I got 3.65A.

### #9 stbtigerr

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Posted 10 April 2012 - 04:01 PM

What is their math? I don't have that volume.

### #10 stbtigerr

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Posted 10 April 2012 - 05:00 PM

This is what I did using 240 for the Phase C voltage:

Ia=(120+65i)/(19.9 @ 0) = 6.86@28 => 6.86@-28

Ic=(65+20i)/(19.9@240) = 3.42@137 => 3.42@-137

In= 3.42@-137 + 6.86@-28 = 6.59@-57A

### #11 stinkycheese

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Posted 10 April 2012 - 07:34 PM

It looks like I was the one confused about conjugates. Argh. Thanks!

### #12 cabby

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Posted 10 April 2012 - 10:05 PM

Thank you everyone. The answer to the problem was 6.6 @ -58 degrees. The magnitude therefore is 6.6A for the neutral current.

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