# High Voltage line with a resistance problem

1. Feb 11, 2004

### Spectre32

Here is the question:

A high Voltage line with a resistance of .18 ohms/km carries a curent of 1525A. The line is at a potential of 1500 kV at the powerstation and caries to current to a city located 192 km away. What is the power loss due to resistance in the line. Answer in units of MW.

I throught P =IV was all i had to do. Now i'm thinking i need to calculate the total power. Then calculate the power with resistance in it and subtract the two. I'm not sure how to do the power with resisatance. I'm think i used use P=v^2/R but i think i need to conver the ohms out of ohm/km.

2. Feb 11, 2004

### turin

You want the power lost in the line. If you use the voltage that is input into the line at the station, then you will calculate the total power delivered from the station, some of which is lost, and (hopefully) most of which is delivered to the customer after the 180 km. The voltage you were given is not quite specific to the line. To have a voltage, you need two points specified. It wouldn't make much sense to specify the voltage drop from the station to the customer (well, maybe it would, but I'm going to say that it doesn't). The voltage is specified wrt ground, the same at at the customer. So what do you do?

Well, thanks to KCL, you know what the current is everywhere in the line. And you know what the resistance is per length. So this is like a really smooth series circuit. After some averaging, you get a total resistance, and you know the current. From here, you can either use Ohm's Law to get the voltage drop across the line, or you can use the one of the three power formula's that doesn't require a voltage (and is the most directly relevant/physically meaningful in this application):

P = RI2

Holy crap, that's a hell of a lot of loss.

BTW, this is why they step the voltage up to distribute it: it reduces the current for a given amount of power.

Last edited: Feb 11, 2004