Find the diameter of copper wire

  1. [SOLVED] Find the diameter of copper wire

    1. The problem statement, all variables and given/known data
    Find the diameter of copper wire which has the same resistance as an aluminum wire of equal length and diameter 1.2mm. The reactivities of copper and aluminum at room temperature are 1.7x10^-8Ωm and 2.6x10^-8Ωm respectively.


    Can someone please explain to me the way of solving this question?
    Thnx in advance
     
  2. jcsd
  3. Hootenanny

    Hootenanny 9,679
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    I'm sure you mean resistivity :wink:. Anyway, this is a simple application of resistivity, so what is the equation for resistivity?
     
  4. [tex]R = \frac{Pl}{A}[/tex]
    What to do next :smile:
     
  5. Hootenanny

    Hootenanny 9,679
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    Well you're looking for when the two resistances are the same, so...
     
  6. The lengths are also the same :frown: im confused!
     
  7. Hootenanny

    Hootenanny 9,679
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    You have two equations,

    [tex]R = \frac{\rho_c \ell}{A_c}[/tex]

    And

    [tex]R = \frac{\rho_a \ell}{A_a}[/tex]

    For the copper and aluminium wire respectively. Can you see what to do next?
     
  8. [tex]\frac{P_c}{A_c} = \frac{P_a}{A_a}[/tex]

    [tex]P_c = 1.7 \times 10^{-8}\Omega{m}[/tex]
    [tex]P_a = 2.6 \times 10^{-8}\Omega{m}[/tex]

    [tex]A_a = \pi{r}^2[/tex]
    [tex]A_a = \pi \times \left(\frac{1.2 \times 10^{-3}}{2}\right)^2[/tex]
    [tex]A_a = 1.1 \times 10^{-6} m^2[/tex]

    [tex]\frac{P_c}{A_c} = \frac{P_a}{A_a}[/tex]

    [tex]\frac{1.7 \times 10^{-8}}{\pi{r}^2} = \frac{2.6 \times 10^{-8}}{1.1 \times 10^{-6}}[/tex]

    [tex]r = \sqrt{\frac{1.7 \times 10^{-8} \times 2.6 \times 10^{-8}}{2.6 \times 10^{-8}\pi}}[/tex]

    [tex]r = 7.36 \times 10^{-5}[/tex]

    Is my answer correct?
     
  9. Hootenanny

    Hootenanny 9,679
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    Nice work! Thanks for LaTeX'ing it up for me :approve:
    You're good up until this point. Your next line is wrong,
    A further word of caution: be careful with rounding errors, try not to round too early in your calculation, either leave all your calculations to the end, or store the intermediate answers in your calculator.
     
  10. Welcome Hootenanny! and thanks for the help

    [tex]r = \sqrt{\frac{1.7 \times 10^{-8} \times 1.1 \times 10^{-6}}{2.6 \times 10^{-8}\pi}}[/tex]

    [tex]r = 4.78 \times 10^{-4}m[/tex]

    Is this right?
     
    Last edited: May 9, 2008
  11. Hootenanny

    Hootenanny 9,679
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    Your method is correct, but your final answer is of by ~7x10-6 due to rounding errors.
     
  12. By rounding errors I think you meant the area of Aluminum [tex]P_a[/tex] right?

    [tex]A_a = \pi{r}^2[/tex]
    [tex]A_a = \pi \times \left(\frac{1.2 \times 10^{-3}}{2}\right)^2[/tex]
    [tex]A_a = 1.1309734 \times 10^{-6} m^2[/tex]

    [tex]r = \sqrt{\frac{1.7 \times 10^{-8} \times 1.1309734 \times 10^{-6}}{2.6 \times 10^{-8}\pi}}[/tex]

    [tex]r = 4.8 \times 10^{-6}[/tex]

    Is this right?:uhh:
     
  13. Hootenanny

    Hootenanny 9,679
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    I did indeed.
    Now you're two orders of magnitude and a rounding error off. I have 4.85...x10-4m.
     
  14. Thanks Hootenanny!! I got the final answer correct...
     
  15. Hootenanny

    Hootenanny 9,679
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    A pleasure :smile:
     
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