# How much power is lost in heating the wires of a power line

1. Sep 22, 2007

### gamesandmore

A generating station is producing 1.4e6 W of power that is to be sent to a small town located 7.0 km away. Each of the two wires that comprise the transmission line has a resistance per kilometer of length of 5.0e-2 $$\Omega$$/km.

(a) Find the power lost in heating the wires if the power is transmitted at 1100 V.

(b) A 70:1 step-up transformer is used to raise the voltage before the power is transmitted. How much power is now lost in heating the wires?

Heres what I did:
a) P = 1.4e6 W
L = 7.0 km
R/L = 5.0e-2 ohm/km
Vp = 1100 V

R = (5.0e-2 ohm/km)*(7.0km)
R = .35 ohm
I = P/V = 1.4e6W/1100V = 1272.73A
P = I^2 * R = (1272.73)^2 * (.35ohm)

P = 5.7e5 W

But it is wrong....

2. Sep 22, 2007

### Staff: Mentor

Please note that there are two power lines, not one. Each has a resistance of 0.05 ohms/km. Do they represent parallel resistances, or simply divide the power?

3. Sep 22, 2007

### gamesandmore

Aren't they in series, so I could just then do:
5.0e-2 ohm/km * .35 for each one and then add them,
wouldn't that still come out to .35 ohm though?

4. Sep 22, 2007

### mgb_phys

No they are in parallel - assume the electricity goes down both of them to the town.

5. Sep 22, 2007

### gamesandmore

Alright, so:
1/Rt = 1/.35 + 1/.35
Rt = .175 ohm,
using this resistance still gives me the incorrect answer.

6. Sep 22, 2007

### PhanthomJay

The wires are in parallel, but you're using the wrong approach. How does the current that you have calculated divide between the 2 wires? Then find the heat loss in each, and the total of the losses.

7. Oct 19, 2009

### vatoburg

Re: Powerlines

The wires are in series, you just multiply the resistance/km (.05) times the distance (7km) to get .35ohm, but then multiply that product by two to account for the other wire, so instead of .35ohm use .7ohm and you should get it. (At least that's what I did and I got the right answer according to webassign)