I have been struggling with these questions for the past week, with not much result at all. I have to study at home, through the use of various textbooks, and with no teacher to guide me through the process. I would be extremely grateful to the person who could explain THE LAST THREE PARTS to me step-by-step, eventually reaching the answer. Thank you in advance.
The village actually gets 240V. Therefore the board will have to supply 240+PD. Assume DC as Hz not given and resistance rather than impedance is used.
Now part A
240V at village
48kW consumed
a) VI=P
.'. I=P/V
I=48000/240=200A
b)V^2/R=P
.'. R=V^2/P
R=240^2/48000=1.2 ohms
c) Voltage supplied Vs=Vpd+V Vpd is voltage drop on cables. V is voltage at village.
Vpd=IR (ohms law) Vpd=200 X 0.3= 60V
Vs=60+240=300V
d) Efficiency% for simultaneous energy consumption can be
(Output power X 100)/Input power
Pout=48kW
Pin=Vs X I=300 X 200=60 kW
eff%=48 X 100/60=80% unimpressive!
Have a go at the rest from what I have done. I will be back to solve the last bit latter. See if you can beat me to it!
Suggest you find the power loss for the cables with the higher voltage. Work back from village (CB) as the voltage will still have to be 240V there. Then step this up (X10) to find the voltage at the end of the cables. The current will be divided by 10 in the cables. Good luck!
Chris
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