Compute the magnitude of the current density

  • #1
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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


[itex]R\quad =\quad \frac { \rho L }{ A } \\ J\quad =\quad \frac { I }{ A } [/itex]

The Attempt at a Solution


[itex]\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 [/itex]

This is marked as wrong, any tips?
 

Answers and Replies

  • #2
TSny
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Your work looks correct to me, but you didn't include the units in your final answer.
 
  • #3
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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
 
  • #4
TSny
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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:
  • #5
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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!
 
  • #6
TSny
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OK. Please report back if there is a mistake in the calculation. I would like to know what I'm missing.
 
  • #7
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OK. Please report back if there is a mistake in the calculation. I would like to know what I'm missing.
Sure thing
 
  • #8
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=).
 
  • #9
SammyS
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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
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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
TSny
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OK, that's good to know. Thanks.
 

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