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