Difference between voltage and voltage drop?

[VenaCava]

Homework Statement

A power station delivers 440 kW of power through 3 ohm lines. How much power is wasted if it is delievered at 12000v?

Homework Equations

v=IR
P=I^2R
P=IV

The Attempt at a Solution

I believe you are supposed to solve it like this but I do not understand why:

I=P/V = 440000/12000=36.67 A
P lost =I^2R=(36.67)^2 (3) = 4033 W

But my gut instict tells me to do this which I believe is wrong from what I've read:

P=V^2/R = (12000)^2/3 = 4.8 x 10^7 W
P lost = P - Pused = (4.8 x 10^7 - 440000) =4.756 x 10^7W

I think I'm getting confused with what "V" is. I keep googling it and all I can tell is that I don't understand the different between voltage and voltage drop. I'm not clear what either is. Could anyone please explain?

Why does P=I^2R give you the power lost rather than the original power (440 kW) or the power used?

 

[Feldoh]

You cannot talk about voltage at a single point, you can only talk about a change in voltage. You'll notice that V = Ed, this is the difference in potential between the distance d.


Power is a change in energy and obvious way to look at it is to say that if there is a change in energy it has to be a negative since the resistor will heat up.

 

[n.karthick]

But my gut instict tells me to do this which I believe is wrong from what I've read:

P=V^2/R = (12000)^2/3 = 4.8 x 10^7 W
P lost = P - Pused = (4.8 x 10^7 - 440000) =4.756 x 10^7W

No you are wrong. You cannot use V=12000V for calculating power loss $P_{loss}$ in line. You have use the voltage drop across the line to find $P_{loss}$.
Since we don't know that, (It is equal to difference between voltage at supply end ($V_{s}$) and voltage at load end (12000V). Since we don't know $V_{s}$, but we know current through the line, we can find power lost directly from the relation
[itex] P_{loss}=I^{2}R[/tex]