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Difference between voltage and voltage drop? 
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#1
Apr2609, 09:27 PM

P: 20

1. The problem statement, all variables and given/known data
A power station delivers 440 kW of power through 3 ohm lines. How much power is wasted if it is delievered at 12000v? 2. Relevant equations v=IR P=I^2R P=IV 3. 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? 


#2
Apr2609, 10:01 PM

P: 1,345

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. 


#3
May1810, 03:53 AM

P: 241

Since we don't know that, (It is equal to difference between voltage at supply end ([itex] V_{s}[/itex]) and voltage at load end (12000V). Since we dont know [itex] V_{s}[/itex], but we know current through the line, we can find power lost directly from the relation [itex] P_{loss}=I^{2}R[/tex] 


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