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I had a problem on an exam which reads as follows: enter image description here

At the end of the day, I got that the sending end voltage is lower than the receiving end voltage, which I know isn't possible. I think I may have approached the problem the wrong way. But it is somewhat possible that the question just has incorrect/unworkable numbers. I know that in the original problem I have the sign of the third power angle wrong, but even when I change that the sending voltage is still lower than the receiving voltage. Here is my work:

enter image description here

enter image description here Any help would be greatly appreciated.

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    \$\begingroup\$ Very bad quality question, the SI system dictates spaces around all number and the correct capitalisation of units. It should read as follows: "Three loads are connected in parallel across a 4160 V RMS line supply. The loads are as follows: load one is 20 kW and 26 kvar, load two is 60 kW at 0.8 pf lagging and load three is 100 kVA at 0.9 pf leading." \$\endgroup\$ – skvery Mar 16 '17 at 18:47
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Load one is \$S= 20\$ kW and \$+j~26\$ kvar.

Load two is \$S=60\$ kW and \$+j~60\sqrt{({1\over 0.8})^2 - 1}\$ kvar.

Load three is \$S=90\$ kW and \$-j~100\sqrt{1-0.9^2}\$ kvar.

Now you can calculate the current using \$({S\over V})^* = I\$. You need to use the complex conjugate here! (If you want to know why, long story please ask in comment. :-)

The voltage drop (rise) is \$V_l = Z_l\cdot I\$, giving:

$$V_s = 4146 \angle 0^o+V_l.$$

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    \$\begingroup\$ Shouldn't it be S*/V* ? \$\endgroup\$ – user1505399 Mar 17 '17 at 14:14
  • \$\begingroup\$ No, maybe? \$S = V\cdot I^*\$ so, oops \$I^* = {S\over V}\$ my bad! I will fix it. Thanks! \$\endgroup\$ – skvery Mar 17 '17 at 14:45

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