Compute the magnitude of the current density

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Homework Help Overview

The discussion revolves around calculating the magnitude of the current density for an aluminum cable in a high-voltage transmission line scenario, given specific parameters such as current and resistance per unit length.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the relationship between resistance, resistivity, and current density, with attempts to derive current density from given values. Questions arise regarding the accuracy of resistivity values and the inclusion of units in calculations.

Discussion Status

The discussion is ongoing, with participants providing feedback on calculations and questioning the resistivity values used. Some participants suggest that variations in resistivity could affect the results, while others express uncertainty about the correctness of their approaches.

Contextual Notes

There is mention of temperature dependence of resistivity, but no specific temperature information is provided in the problem statement. Additionally, participants are exploring the implications of using different resistivity values for aluminum.

Sho Kano
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Homework Statement


Copper and aluminum are being considered for a high‐voltage transmission line that must carry a current of 51.4 A. The resistance per unit length is to be 0.176 Ω/km. Compute the magnitude of the current density for an aluminum cable.

Homework Equations


R\quad =\quad \frac { \rho L }{ A } \\ J\quad =\quad \frac { I }{ A }

The Attempt at a Solution


\frac { R }{ L } \quad =\quad 0.176\frac { \Omega }{ km } \quad =\quad 1.76x10-4\frac { \Omega }{ m } \\ \\ \frac { R }{ L } \quad =\quad \frac { \rho }{ A } \quad \quad \quad \quad \quad A\quad =\quad \frac { I }{ J } \\ \\ \frac { R }{ L } \quad =\quad \frac { \rho J }{ I } \\ \\ \frac { RI }{ L\rho } \quad =\quad J\quad =\quad 1.76x10-4\quad *\quad \frac { 51.4 }{ 2.65x10-8 } \quad =\quad 3.41x10+5\quad

This is marked as wrong, any tips?
 
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Your work looks correct to me, but you didn't include the units in your final answer.
 
TSny said:
Your work looks correct to me, but you didn't include the units in your final answer.
Units of J are A/m^2
Strange, it still marks that as the wrong answer
 
Hmm. I can't see anything wrong. Different references can give somewhat different values for the resistivity of aluminum. But I guess you are using the value from your textbook or notes from class. See http://hypertextbook.com/facts/2004/ValPolyakov.shtml

Resistivity is temperature dependent, but no information about temperature is given in the statement of the problem.
 
Last edited:
TSny said:
Hmm. I can't see anything wrong. Different references can give somewhat different values for the resistivity of aluminum. But I guess you are using the value from your textbook or notes from class. See http://hypertextbook.com/facts/2004/ValPolyakov.shtml

Resistivity is temperature dependent, but no information about temperature is given in the statement of the problem.
I can't see anything wrong either, i'll just have to ask the prof. Thanks for the help!
 
OK. Please report back if there is a mistake in the calculation. I would like to know what I'm missing.
 
TSny said:
OK. Please report back if there is a mistake in the calculation. I would like to know what I'm missing.
Sure thing
 
Are you sure your resistivity value is correct? Since it is both copper and aluminum, I added the resistivity values of both copper and aluminum together and got a resistivity values of 4.5x10^-8. I am not sure if that is how you combine them, but if it is, I got a value of 2.01x10^5. I hope this helps=).
 
Quantum2323 said:
Are you sure your resistivity value is correct? Since it is both copper and aluminum, I added the resistivity values of both copper and aluminum together and got a resistivity values of 4.5x10^-8. I am not sure if that is how you combine them, but if it is, I got a value of 2.01x10^5. I hope this helps=).
It would be more appropriate to compare the two. Adding the resistivities makes little or no sense.
 
  • #10
Sorry, I didn't see that we were only looking for the resistivity of the aluminum cable. In that case, I think that the resistivity might still wrong because I looked it up, and it came up to be 2.82X10^-8. Not much of a difference, but try it. Sorry again.
 
  • #11
TSny said:
OK. Please report back if there is a mistake in the calculation. I would like to know what I'm missing.
No mistake, just a slightly different resistivity
 
  • #12
OK, that's good to know. Thanks.
 

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