1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Temperature Coefficient of Resistivity

  1. Oct 7, 2012 #1
    Hi

    I was wondering if anyone could help. My teacher isnt very good at explaining things and hes given us a question without going through the reasoning behind the equation.

    Aluminium wire has a resistance of .250Ω at 0°C. Give that the temperature coefficient of resistivity at 0°C for aluminium is 0.004 find the resistance of the wire at 200°C

    My answer is Rt = .250 x (1+.004x200) = 0.45Ω

    What I dont get however is why isnt it just .250Ω + (0.004 x 200) which would equal 1.05Ω

    I thought the temperature coefficient was how much the resistance increases per increased degree, which you then add to the original resistance.

    If anyone could help explain ild be very grateful :)
     
  2. jcsd
  3. Oct 7, 2012 #2

    berkeman

    User Avatar

    Staff: Mentor

    Welcome to the PF.

    The only error that you've made is that the coefficient of resistivity applies to differences in temperature. So if you start at 25C and go to 200C, what is the delta temperature?

    Oh, and it's best to put units on everything in your calculations. What are the units on the coefficient of resistivity?
     
  4. Oct 7, 2012 #3

    berkeman

    User Avatar

    Staff: Mentor

    Actually, there may be one more error, but I'm not sure. Please write out the full corrected equation and show the answer so we can check it. Thanks.
     
  5. Oct 7, 2012 #4
    hi, thanks for the welcome :)

    The question I have is:

    A long thin aluminium wire has a resistance of 0.250Ω at 0°C. Given that the temperature coefficient of resistivity at 0°C for aluminium is 4x10-3/0°C find the resistance of the wire at 200°C

    Ive simply transported my values into this equation:

    RT = Rref [1 + ɑ(T-Tref)]

    R200°C = R0°C [1 + ɑ(200-0)]

    R200°C = 0.250Ω (1 + 4x10-3 x 200)

    R200°C = 0.45Ω

    But thats just blindly following the formula. What i dont get is that i thought the temperature coefficient was just the measure of change in resistance as the temperature increases.

    So an increase of 200°C, would simply mean you multiply the coefficient (4x10-3) by 200 and add that to your original resistance (0.250Ω)

    Basically what Im asking is, how come the formula is:

    R200°C = 0.250Ω (1 + 4x10-3 x 200)

    And not simply

    R200°C = 0.250Ω + (4x10-3 x 200)

    I mean why do you have to multiply everything in the brackets by 0.250Ω?
     
  6. Oct 7, 2012 #5
    The temp coeff of 0.004 means that the resistance increases (changes) by 0.004 of the resistance at 0C for each change of 1C in temp.
    That is what the equation R = R0(1 + αt) means
    R = R0 + R0αt
     
  7. Oct 7, 2012 #6
    sorry i still dont understand why you have to multiply it by R0 :/
     
  8. Oct 8, 2012 #7

    berkeman

    User Avatar

    Staff: Mentor

    Since the increase is just a percentage, you need to multiply that percentage increase by the original resistance to find out what the full change in resistance is. If you have a 1% increase in resistance due to temperature, that will result in twice as big a change in resistance for a 200 Ohm resistor versus a 100 Ohm resistor.

    Does that help?
     
  9. Oct 8, 2012 #8
    arh, the coefficient is a percentage of the original resistance at 0°C

    So to find the actual resistance increase per degree, I need to find out what 0.004% of .250 is

    To do that I have to times my original resistance of .250 x the coefficient of 0.004 = 0.001Ω

    this means that an increase in each degree causes a change of 0.001Ω in the material

    Because I have a 200°C increase i need to multiply 0.001Ω by 200 = 0.2Ω

    Add that to my original resistance 0.25Ω + 0.2Ω = 0.45Ω

    Many thanks everyone, I very much appreciate the help youve given me :)
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook