Electric Current: Reduce Wasted Power by 7500W

[mer584]

Homework Statement

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?

b]2. Homework Equations [/b]
P=IV=I^2R=V^2/R V=IR

The Attempt at a Solution

So I know you need to find the current in order to be able to do this, but I'm a bit confused about when the different formulas for P can be used. The answer is 7500W.

I tried finding the current through the wire (12kV): 12kV = I(3 ohms) = 4000A
and the power lost to heat: (4000)^2 * 3 = 4.8 x 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 x 10^8 W

but these powers don't subtract to the answer. I'm not sure I understand the problem.

 

[mer584]

OK new approach since that last one seems very wrong:
I find the current at the source using P=IV
then I find how much power that dissipates by the end using P=I^2R
then I can do that same thing for when the V is higher
and subtract the two final powers...the number i got was a bit high but I think this makes more sense

 

[cepheid]

Yeah, you know immediately that the first method is wrong because the power lost to heat is much much higher than the total power delivered!

The mistake was in assuming that the power line is the only "resistor" in the circuit. Something in the factory is acting as a load...i.e. it is making use of the electrical power in such a way that the total power consumption is 620 kW, which includes the factory power consumption and the power dissipated as heat in the power line. You can replace the factory with an equivalent load resistor that WOULD use up that much power when placed in series with the power line. However, since you don't know what this resistance IS beforehand, you can't solve for the current using knowledge of the voltage alone. You find I from P = VI, and then you find the portion of P dissipated in the power line from P_heat = I^2*R, GIVEN that that much current is flowing.

Or you could use P = V^2/R, provided that you understand that V here is NOT 12,000 V. V is the voltage DROP across the power line = IR = 155 V. However, it makes more sense just to use I^2 R, since you already have I. It saves you one step.