Power Loss in 400kV Cable Carrying 1GW of Power

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Discussion Overview

The discussion revolves around calculating power loss in a 400kV cable carrying 1GW of power, specifically focusing on resistive losses due to line resistance. The context includes a homework problem with multiple-choice answers and attempts to derive the correct power loss using relevant equations.

Discussion Character

  • Homework-related
  • Mathematical reasoning

Main Points Raised

  • One participant presents a problem involving a 400kV line with a specified line resistance and power load, seeking help to calculate power loss.
  • Another participant suggests checking the equation relating power and current, implying a potential error in the initial approach.
  • A different participant suspects that the original poster may be using only the line resistance instead of the total resistance, which includes load resistance, to compute the current.
  • One participant calculates the current using I=P/V and finds it to be 2500A, then applies the power loss formula P=I^2/R to find a power loss of 1.25x10^6 W.
  • Another participant echoes this calculation, suggesting that the original poster made a similar mistake in their approach.
  • A later reply acknowledges the mistake in the calculation method, indicating a realization of the error in applying the power loss formula.

Areas of Agreement / Disagreement

There is no consensus on the correct power loss value, as participants are discussing different approaches and calculations. Some participants agree on the calculation method, while others point out potential errors in the initial reasoning.

Contextual Notes

Participants have not provided complete work for all calculations, which limits the ability to identify specific errors or assumptions made in the problem-solving process.

JordanHood
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Homework Statement


A 400kV line is carrying 1GW of power and experiences a line resistance of 5 ohms. What is the power loss in the cable due to resistive losses?

Homework Equations

The Attempt at a Solution


You are given 4 potential answers to this question
a) 3.2x10^7 W
b) 78 W
c) 5000 W
d) 2x10^8 W

I have tried using I=P/V and then substituting this value into P=I^2/R but this doesn't seem to give any of the answers. Can anyone help me out?
 
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You might want to check your equation relating power and current.
 
I suspect you are probably using the LINE resistance, when you should be using the line resistance plus the load resistance, to computer the current. Since you did not bother to post any work, I'm just guessing here. Please post your work when you ask such questions. How can we tell where you've gone wrong when we don't see what you did?
 
I used I=P/V to get 1x10^9/400x10^3 = 2500A. I then substituted this into Power loss= I^2/R to get 2500^2/5 = 1.25x10^6 W.
 
JordanHood said:
I used I=P/V to get 1x10^9/400x10^3 = 2500A. I then substituted this into Power loss= I^2/R to get 2500^2/5 = 1.25x10^6 W.
So you are agreeing with me; you made exactly the mistake I thought you probably had made. Think about it.
 
phinds said:
So you are agreeing with me; you made exactly the mistake I thought you probably had made. Think about it.
ah yes, gosh that was stupid P=i^2*R not divided like I was doing
 

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