# Sending Voltage is higher than receiving voltage: Parallel Load Problem

I had a problem on an exam which reads as follows: 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:  Any help would be greatly appreciated.

• 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." Mar 16 '17 at 18:47

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:

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

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