# To reduce cost and weight, power transmission lines are made

• g98
In summary, the conversation discusses determining the resistance of a 30 km long wire, transferring 1000 MW of power at a potential difference of 500 kV. The current through the wire is calculated to be 2000A, but the value for wire resistance is incorrect. The information about the wire being aluminum with a diameter of 4cm is mentioned, but the specific resistivity value is not given. The units for the area in the formula R=ρ(l/A) can be either m^2 or mm^2, as long as they are consistent.

## Homework Statement

Determine the resistance of a 30 km long wire. b. Assume that the line transfers 1000 MW of power at a potential difference of 500 kV. What is the current through the wires? c. How much power is lost during transmission? What fraction of the transmitted power is lost?

## Homework Equations

R=p*l/A P=I*deltaV P=I^2*R

## The Attempt at a Solution

so, for the first one i determined the resistance to be 4,23*10^14 ohm, the current through the wire is 2000A . I am not quite sure for the above values and because of that, i get weird values for the lost power in c) I would very much appreciate a bit of help with the exercise

The wire resistance can't be 10^14 ohms. You have done something wrong. Show us how you did the calculations of the wire resistance.

I think there must be some important information we're missing, probably given in a "part a" portion of the problem?

• CWatters
g98 said:
How much power is lost during transmission? What fraction of the transmitted power is lost?
That can't be determined unless the resistance of the line is specified.

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gneill said:
I think there must be some important information we're missing, probably given in a "part a" portion of the problem?
The thing i didn't mention is that the wire is aluminium and the diameter is 4cm

phyzguy said:
The wire resistance can't be 10^14 ohms. You have done something wrong. Show us how you did the calculations of the wire resistance.
I was wondering for the formula R=p(l/A) (p being the specificity) what should be the units of the area? i though it should be m^2 but i checked in internet and in many places i saw mm^2 and i got slightly confused

g98 said:
I was wondering for the formula R=p(l/A) (p being the specificity) what should be the units of the area? i though it should be m^2 but i checked in internet and in many places i saw mm^2 and i got slightly confused

OK, You're right that the resistance is given by R=ρ(l/A). You can use any units you want as long as you are consistent. If your value for resistivity is in Ohm-m, then you would want to use m for the length and m^2 for the area. Why don't you set up this calculation and show us the values you are using and the results.