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Resistance and Temperature

  1. Oct 13, 2009 #1
    1. The problem statement, all variables and given/known data

    An engineer needs a resistor with zero overall temperature coefficient of resistance at 20°C. She designs a pair of circular cylinders, one of carbon and one of Nichrome.

    The device must have an overall resistance of R1 + R2 = 15.0 Ω independent of temperature and a uniform radius of r = 1.40 mm. Determine the lengths l1 and l2 of the carbon and Nichrome segments respectively. You may ignore thermal expansion of the cylinders and assume both are always at the same temperature.

    2. Relevant equations
    R1+R2=15
    alpha1+alpha2=0
    R=pl/A
    pcarbon = 3.5e-5
    pnichrome = 1.5e-6



    3. The attempt at a solution
    I dont really know where to begin. The problem states that the 15ohms is independent of temperature. I know what that means but does it have any relevance to the problem? Any help getting started would be appreciated. Thanks
     
  2. jcsd
  3. Oct 13, 2009 #2

    berkeman

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    Staff: Mentor

    One of those materials would have to have a negative temperature coefficient of resistance. Double check the rho numbers -- is one negative?
     
  4. Oct 13, 2009 #3
    Rho is the resistivity. Alpha is the temperature coefficient. Both values of resistivity are positive as written. Carbons temperature coefficient is positive and nichromes is negative.
     
  5. Oct 13, 2009 #4

    berkeman

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    Staff: Mentor

    Ah, thanks for the catch. My bad.
     
  6. Oct 13, 2009 #5
    Could still use some help. Thanks.
     
  7. Oct 13, 2009 #6

    ideasrule

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    Homework Helper

    The two alpha's don't need to add up to zero; R1 and R2 just need to sum to 15--not 15+0.2T or anything, just 15. All the other relevant equations are correct. Why not plug them all into R1+R2=15 and see what you get?
     
  8. Oct 13, 2009 #7
    Plugging into the two formulas I got.

    3.5e-5(l1) + 1.5e-6(l2) = 9.2363e-5

    but this gives me 1 equation and 2 unknowns
     
  9. Oct 13, 2009 #8
    Can someone please help me out?
     
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