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Electric Currents and Resistance I NEED HELP!

  1. Mar 27, 2004 #1
    Our physics teacher gaves us online homework, and he was gone on friday and it is due by midnight tonight. Here is an example problem that we are having trouble with:

    A copper cable is designed to carry a current of 400A with a power loss of 4 W/m. What is the required radius of this cable?
     
  2. jcsd
  3. Mar 27, 2004 #2

    Integral

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    Start with P=I2R

    You have the information you need to compute the resistance of 1m of wire.

    Use

    [tex] R= \frac {\rho L } A [/tex]

    where [tex] \rho [/tex] is the resistivity of copper.

    Can you finish it?
     
    Last edited: Mar 27, 2004
  4. Mar 27, 2004 #3
    All i know what to do is to plug in the amperage, but i don't know where the other numbers are substituted for what letter, and i have no idea what Latex is for this equation.
     
  5. Mar 27, 2004 #4

    Integral

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    P = power = 4 w
    R = resistance
    A = area
    L = lenght (use 1m)
    [tex] \rho [/tex] is the resistivity be sure your value is in ohms/m

    Edit:
    Oh yeah.... I = current
     
  6. Mar 27, 2004 #5
    Is the answer from those numbers going to be the diameter or the radius? And if I used 1.56e-8 for the resisitivity, would the answer be in m or cm?
     
  7. Mar 27, 2004 #6
    Power, as Integral said, is given by:
    [tex]P = I^2R[/tex]
    Subtitute R and you get:
    [tex]P = \frac{I^2\rho L}{A} = \frac{I^2\rho L}{\pi r^2}[/tex]
    Solve for r:
    [tex]r = \sqrt{\frac{I^2\rho L}{\pi P}}[/tex]
    Since the question says the power per length unit is 4W/m, L/P is 0.25m/W. The rest you should have.
     
  8. Mar 27, 2004 #7
    The answer that I came up with was .01409, now is that in "m" or "cm", and is it the radius?
     
  9. Mar 27, 2004 #8

    Integral

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    Chen, we really in courage people to the complete solution on their own. this is simply doing his homework for him. While we appreicate your efforts please let them do some of the work :)

    BTW:
    Note how he completely solved the problem in terms of the the variables before plugging in a single number. This is the right way to do physics problems.
     
  10. Mar 27, 2004 #9

    Integral

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    If you solve for the diameter then your answer will be the diameter, if you solve for the radius then your answer will be the radius.

    What are the units of the value you used for resistivity? That will determine the units of your answer.
     
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