For the NCEES power sample question #111 I have seen several discussions, but none of them outlined the KVL method in detail for finding the answer. I understand the solution was given by converting to a delta to simplify some math, but that's not the way my brain works. To me, the KVL is the more straightforward method to solve the problem.
Here is how I set it up:
Vab = VaA + VAB + VBb
or
Vab = (IaA)(Z) + VAB - (IbB)(Z)
Where Z is the line impedance (which is 11.18<63.43, given), VAB is given, and IbB equals IaA shifted by -120 degrees (because it is balanced).
So now we have:
Vab = (70<-20)(11.18<63.43) + 12.5k - (70<-140)(11.18<63.43)
Doing the math above results in:
Vab = 12,952<5.8
This method seems pretty straightforward to me, and is much more logical for me than converting to a delta and using an imaginary neutral.
Here is how I set it up:
Vab = VaA + VAB + VBb
or
Vab = (IaA)(Z) + VAB - (IbB)(Z)
Where Z is the line impedance (which is 11.18<63.43, given), VAB is given, and IbB equals IaA shifted by -120 degrees (because it is balanced).
So now we have:
Vab = (70<-20)(11.18<63.43) + 12.5k - (70<-140)(11.18<63.43)
Doing the math above results in:
Vab = 12,952<5.8
This method seems pretty straightforward to me, and is much more logical for me than converting to a delta and using an imaginary neutral.
