# NCEES #511 (Power Depth)



## Kuku (Aug 13, 2008)

Can someone walk me through this problem. Thanks.


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## Flyer_PE (Aug 13, 2008)

There is a pretty good discussion of this one here. Let us know if you need more.


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## Kuku (Aug 13, 2008)

Flyer_PE said:


> There is a pretty good discussion of this one here. Let us know if you need more.


Thanks for the link. I actually did search for this topic prior to posting, but the search returned absolutely nothing for me. Go figure.


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## Flyer_PE (Aug 13, 2008)

No problem. There's a lot of good stuff on this board. I also had the advantage of knowing almost exactly where to look.


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## jproctor6 (Aug 13, 2008)

Kuku said:


> Can someone walk me through this problem. Thanks.


This may not make much sense without a picture of the line and load but here goes:

VAB = 12.5 &lt;0 degrees (kV)

so

VAN = 12.5/sqrt3 &lt;-30 degrees (kV) = 7.22 &lt;-30 degrees (kV)

now

Van = VAN + IaA(5+j10) = (7.22 &lt;-30degrees) + (70/1000 &lt;-20degrees)(5+j10) = 7.48 &lt;-24degrees

|Vab| = sqrt3*|Van| = (sqrt3)(7.48) = 12.95

Basically, you take the phase-to-phase load voltage "VAB" given to find the phase-to-neutral voltage "VAN" seen at load point "A".

Then you determine the phase-to-neatral voltage "Van" seen at the source by adding the load voltage "VAN" to the line voltage drop given by ohm's law = line current * line impedance.

Now the magnitude of the phase-to-phase voltage "Vab" seen at the source is just the phase-to-neutral voltage you just calculated "Van" multiplied by the square root of 3.

I hope that makes sense.

Sorry if my explaination or method is not helpful.


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