# NCEES Power #531



## brad9m (Feb 6, 2012)

Ok I was confused once again, so hopefully you can learn from my mistakes. On NCEES, Power exam #531:




From here, I made the mistake of saying that the impedance angle was -45 and I was trying to use powere triangles to come up with 5000 MVAR. But the impedance angle is 90, so:


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## brad9m (Feb 6, 2012)

I realize know that you can't use power triangles because it is a purely reactive load. Hopefully someone else doesn't spend as much time on this problem as I did.


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## saberger_vt (Jan 31, 2013)

Someone please let me know if I did this wrong! This just seems easier:

P = (Va-Vb)^2/Z so P = (-500 + j500)^2/50&lt;90 = (707&lt;135)^2/50&lt;90 = 499990.41&lt;270/50&lt;90

This equals: 10000&lt;180 i.e. magnitude is 10,000MVAR.

I realize I did not include the "KV" in there, but it's there!

This eliminates having to figure out the current in the transmission line, since you know the impedance of the transmission line. I'm not sure if this would work for all scenarios, but anyway you cut it, you will get the same answer.

That is finding power by V*I or V^2/Z or I^2*Z etc.


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## pbo064 (Feb 7, 2013)

saberger_vt method is more direct than mine. I found I(line)=(500&lt;90-500&lt;0)/(j50)=14.14&lt;45. Then plugged this into Q=I^(2)*R=(14.14&lt;45)^(2)*j50=10000&lt;180.


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