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Math Help - temperature coefficients

  1. #1
    Super Member bigwave's Avatar
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    temperature coefficients

    Temperature coefficients of resistance for copper and platinum are respectively
    0.00427^oC^{-1}, and 0.00357^oC^{-1}
    Resisters made of copper and platinum have respective resistences 17\Omega and 18\Omega at 20^oC.
    Calculate the temperature at which the two resistances will be equal. Ans. 165.4^oC

    I am very new to this but knowing that

    R_t=R_o\Left(1+ \alpha T \Right)

    using this simultaneously is the way to solve it. however apparently it assumes some things that I don't know to set this up....

    thanks ahead for any help
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  2. #2
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    Couldn't you just substitute your values for copper and platinum individually to find R0 for both metals? (i.e. for copper sub Rt = 17, a = 0.00427, T = 20)

    Then you could just eqaute R0(copper)(1+a(copper)T) = R0(platinum)(1+a(platinum)T) and solve for T.
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  3. #3
    Super Member bigwave's Avatar
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    here is an example problem

    Quote Originally Posted by Gusbob View Post
    Couldn't you just substitute your values for copper and platinum individually to find R0 for both metals? (i.e. for copper sub Rt = 17, a = 0.00427, T = 20)

    Then you could just eqaute R0(copper)(1+a(copper)T) = R0(platinum)(1+a(platinum)T) and solve for T.
    not sure if I follow just how this is plugged in. but here is an example problem to show somewhat how this is supposed to be solved... however I don't know how they got \alpha  and R_0 from this

    Electrical conductivity, and hence resistence, is a slowly varying function of temperature. To a good approximation, the Resistance R_T at temperature T(^o C) is related to the resistance  R_o at  0^o by:

    R_T=R_o(1+\alpha T)

    where \alpha is a constant known as temperature coefficient of resistence of the material. If a coil has a resistence of 4.0\Omega at 20^o C and 4.52 \Omega at  80^o C calculate (a) the temperature coefficient of resistance of the material; (b)the resistance of the coil at 100^o C
    (a)
    R_{20}=R_o(1+20\alpha)=4.00\Omega
    R_{80}=R_o(1+80\alpha)=4.52\Omega

    Solving simultaneously, \alpha = 2.27 \times 10^{-3}^o C^{-1} and R_o = 3.38\Omega

    (b)
    R_{100}=(3.83)[1+10092.27\times10^-3] = 4.69\Omega

    thanks ahead for help explaining this.
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