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Resistance of a Wire at varying temperatures

  1. Oct 14, 2016 #1
    1. The problem statement, all variables and given/known data
    Determine the resistance of wire at room temperature and at the melting point of the given material.

    GIVEN:
    Material = Aluminum
    Diameter of wire ##= 0.5 mm = 5.0 * 10^{-4} m##
    Length of wire ## = 0.5 m = 5.0 * 10^{-1} m##
    Temperature Coefficient: ##\alpha = 3.9 * 10^{-3}C^{-1}##
    Base Resistivity: ##\rho_{0} = 2.82 * 10^{-8} \Omega * m##
    Melting Point of Aluminum: ##T = 660.3^{\circ} C##
    Room Temperature: ## T_{0} = 20^{\circ} C##

    2. Relevant equations

    Variation of Resistivity w/ Temperature: ##\rho = \rho_{0}(1 + \alpha(\Delta T)##

    Pouillet's Law: ##R = \rho * \frac{L}{A}##

    3. The attempt at a solution

    I've already finished working the problem out, but in doing so, I had to make a few assumptions I'm not sure are correct. Actually, I'm not even sure if my methodology is correct to begin with.

    So, for the Resistance at room temperature, I assumed that the Resistivity given for Aluminum in my textbook is its resistivity at room temperature (not sure if that's right).

    The second assumption I made is that the cross-sectional area A of a wire would be given by

    $$ A = \pi r^2 = \pi *(\frac{5.0 * 10^{-4} m}{2})^{2}) = 2.0 * 10^{-7} m^{2}$$

    Which is based purely on me picturing what each "slice" would look like if I were to cut across a wire an arbitrary number of times. Again, not sure if this is valid, but a circle seemed like the closest geometric shape to what I pictured.

    If that's the case, then applying Pouillet's Law yields

    $$ R_{room} = \rho_{0} * L/A = (2.82 * 10^{-8} \Omega * m)(\frac{5.0 * 10^{-1} m}{2.0 * 10^{-7} m^{2}}) = 7.1 * 10^{-2} \Omega$$

    Then calculating the resistivity of the wire at the melting point I've got

    $$ \rho_{mp} = (2.82 * 10^{-8} \Omega * m) + (2.82 * 10^{-8} \Omega * m)(3.9 * 10^{-3}C^{-1})(660.3^{\circ} C - 20^{\circ} C) = 9.9 * 10{-8} \Omega * m$$

    So the Resistance of the wire at its melting point is

    $$R_{mp} = \rho_{mp} * L/A = (2.82 * 10^{-8} \Omega * m)(\frac{5.0 * 10^{-1} m}{2.0 * 10^{-7} m^{2}}) = 2.5 * 10^{-1} \Omega$$

    But is this correct? I can't help but feel like the Resistance of the wire should be higher at the melting point than at room temperature for some reason.
     
  2. jcsd
  3. Oct 15, 2016 #2

    haruspex

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    In your last equation line you seem to have copied in the wrong resistivity (ρ0 instead of ρmp), but I take it that's just a typo.
    The final answer looks reasonable (it is more than the room temperature resistance!)
     
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