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Homework Help: Resistivity and Power

  1. Feb 17, 2006 #1
    A potential difference V is applied to a wire of cross section A, length L, and resistivity p. You want to change the applied potential difference and stretch the wire so that the energy dissipation rate is multiplied by 35 and the current is multiplied by 3. What should be the new values of L and A [in relationship to the old values]?

    I know that R = pL/A and P = IV (etc.), but it doesn't make sense that they'd want specific answers for both L and A. Can't either one change without the other and still affect the power?
  2. jcsd
  3. Feb 17, 2006 #2


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    HINT: The volume of the metal does not change upon stretching.
  4. Feb 17, 2006 #3
    So then volume V = AL. And if you're stretching it then A gets smaller and L gets bigger. But what does volume have to do with power?
    Last edited: Feb 17, 2006
  5. Feb 17, 2006 #4
    Fine. It doesn't matter anymore. Thanks.
  6. Feb 17, 2006 #5


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    Resistance can be found from L, A, and rho. Power can be found with P = IV = V^2/R.

    - Warren
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