Difference between voltage and voltage drop?


by VenaCava
Tags: difference, voltage
VenaCava
VenaCava is offline
#1
Apr26-09, 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?
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Feldoh
Feldoh is offline
#2
Apr26-09, 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.
n.karthick
n.karthick is offline
#3
May18-10, 03:53 AM
P: 241
Quote Quote by VenaCava View Post

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 [itex] P_{loss}[/itex] in line. You have use the voltage drop across the line to find [itex] P_{loss}[/itex].
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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