Complex Imaginary volume 4, prob 14

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cabby

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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.

 
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

 
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.

 
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.

 
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?

 
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.
 
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.

 
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

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

 
I apologize for bringing up an old thread but I have a question about this problem.  Why would I assume phase C voltage would be at an angle of 240 when the problem specifically states "On phase A (19.9 angle 0 kV) there is a 120+j65 load, on phase C (19.9 angle 0 kV) there is a 65+j20 kVA load"?  I took it to mean that phase A and C voltages were in phase.  If they wouldn't of had that voltage in parenthesis for each phase I would have assumed 19.9 kV at 0 and 240 for A and C.  Am I missing something?  Is this some notation that I'm unaware of?  Thanks.

 
I apologize for bringing up an old thread but I have a question about this problem.  Why would I assume phase C voltage would be at an angle of 240 when the problem specifically states "On phase A (19.9 angle 0 kV) there is a 120+j65 load, on phase C (19.9 angle 0 kV) there is a 65+j20 kVA load"?  I took it to mean that phase A and C voltages were in phase.  If they wouldn't of had that voltage in parenthesis for each phase I would have assumed 19.9 kV at 0 and 240 for A and C.  Am I missing something?  Is this some notation that I'm unaware of?  Thanks.
I think there may be an error in your edition of the book. 

The voltages in my copy are 19.9kV<0º for phase A and 19.9kV<240º for phase C, the loads are the same as yours.  

 
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