Maximizing Power Delivery: Comparing 12,000V & 50,000V Electricity Transmission

[leolaw]

A power station delivers 620kW of power at 12,000 V to a factory through wires with total resistance 3.0 ohms. How much less power is wasted if the electricity is delivered at 50,000V rather than 12,000V?

I cannot quite picture what is going on in this problem. Is this a series circuit, which has 2 resistors, one is the wire, and the other one is to dissipate the power?

 

[chroot]

The power lost to heat in a wire is given by

[tex]P = I^2 R[/itex]<br /> <br /> Use Ohm's law to find how much current the wire carries at both 12 kV and 50 kV, then find how much power is lost to heat in the wire using that formula.<br /> <br /> - Warren[/tex]

 

[leolaw]

I still don't quite understand:
if I use ur method,
I first find the current through the wire (12kV): 12kV = I(3 ohms) = 4000A
and the power lost to heat: (4000)^2 * 3 = 4.8 * 10^7 W
And the current through the wire (50kV): 50kV = I (3ohms) = 16666.67A
and the power lost to heat: (16666.67)^2 * 3 = 8.3 * 10^8 W

The correct answer is 7500W, but if i substract these two numbers, they don's seem to give me the right answer

 

[leolaw]

I understand what the question is asking for now.
Thx

 

[dextercioby]

The current is not found with that "R"...It is found using the power generated and the voltage on the source.

Daniel.