Physics E & M Question about a powerline

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Homework Help Overview

The discussion revolves around a problem related to electrical power loss in a high voltage powerline, specifically focusing on resistive power loss calculations involving voltage, current, and resistance over a specified distance.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants explore different power loss formulas, questioning the appropriateness of using V²/R versus I²R. There is a discussion about the meaning of the voltage given and its relevance to the potential drop across the powerline.

Discussion Status

Participants are actively engaging with the problem, clarifying the correct formulas to use and discussing the implications of the voltage measurement. Some have provided guidance on calculating the potential drop and have shared their results, indicating a productive exchange of ideas.

Contextual Notes

There is a noted confusion regarding the interpretation of the voltage measurement and its application in the power loss calculations. Participants are also addressing potential errors in formula usage.

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Homework Statement


A high voltage powerline operates at a 500000 V-rms and carries an rms current of 500 A. If the resistance of the cable is 0.05Ω/km, what is the resistive power loss in 200 km of the powerline.

V = 500000 V-rms
Irms = 500 A
R/x = 0.05Ω/km
x = 200 km

Homework Equations



P = I2R
P = V2/R
P = IV

The Attempt at a Solution



I attempted to use V2/R at first, and that gave me 2.5 x 1010 Watts, however the given answer is 2.5 x 106 Watts. I get that answer when I use I2R, but I was wondering why V2/R could not be used?
EDIT: Fixed to V2/R
 
Last edited:
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Hello and welcome to PF!

Your formula P = V/R is incorrect. Did you mistype the formula?

Even when you have the correct formula, you have to be careful. The 500,000 V of the powerline probably represents the voltage of the line relative to ground at one end of the line. It does not represent the potential drop as you go from one end of the line to the other end of the line.

You can calculate the potential drop V and then use the (corrected) power formula that involves V and R.
 
TSny said:
Hello and welcome to PF!

Your formula P = V/R is incorrect. Did you mistype the formula?

Even when you have the correct formula, you have to be careful. The 500,000 V of the powerline probably represents the voltage of the line relative to ground at one end of the line. It does not represent the potential drop as you go from one end of the line to the other end of the line.

You can calculate the potential drop V and then use the (corrected) power formula that involves V and R.

Sorry, I meant to type V2/R. How would one then go about calculating the potential drop V?
 
Brandon Hawi said:
Sorry, I meant to type V2/R. How would one then go about calculating the potential drop V?
You're making Georg Ohm feel neglected today.:wink:
 
TSny said:
You're making Georg Ohm feel neglected today.:wink:

Alright then, so V = IR. So I get V = 5000 V. Then I used V2/R and got 2.5 MW as my answer. Thanks ! :smile:
 
Looks good!
 

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