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Homework Help: 2 resistance problems,

  1. Feb 11, 2007 #1
    2 resistance problems, plz help

    1. The problem statement, all variables and given/known data
    Um so I have two problems, they should be fairly easy after explanation. So the first problem is that there's a rod, 65 cm long. 25 cm of it is a material with resistivity a and the rest of it is another material with resistivity b. cross section is same throughout and I'm suppose to find resistance.
    The second problem is a temperature problem, so we have an initial temp of 60C and the measured current on copper wire is 1.3A the same voltage is applied when it is -88C with the same wire. We're suppose to find the new current.

    3. The attempt at a solution
    So for the first problem I don't understand how to do this at all. I think integration might be involved but I'm not sure. For the second problem I've set up an equation with the formula they provided in text book: Rf=Ri(1+alpha(Tf-Ti)) alpha is temp coeficient of resistivity which is in book. So my equation is V/If=(V/1.3A)(1+.0039(-88-60)) because same voltage is applied voltage cancels out. I then solved for I final but the answer is not correct. I don't know what I'm doing wrong. Any advice will help. Thank you
  2. jcsd
  3. Feb 11, 2007 #2


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    Question 2: Check your formula for resistance.

    The constant .0039 will only apply when Ti is a fixed reference temperature. From your number 0.0039, I would guess the reference temp is either 20 degC or 300 degK - check in your book.

    Assuming the reference temperature was 20C: if the resistance is R at 20C, then it is R(1 + .0039(60-20)) at 60C and R(1 + .0039(-88-20)) at -88C. The value of R wll cancel out, like the voltage did.

    Question 1: You should have a formula for the resistance of a rod in terms of its resistivity, length, and cross section area. The two parts of the rod act like two resistors in series.
    Last edited: Feb 11, 2007
  4. Feb 11, 2007 #3
    oh! i didn't realize that there's a reference temperature thing. Thanks. But for question 1, I tried that approach and its not correct :( k so the resistivity is 6.00x10^-3, one side is 25 cm long, then other is 40 cm long. the cross section is 3mm. I figured that since both sides have same resistivity i can just use R= p(l/A) to get resistance but apparently that's not the right way. So I have no idea how to approach this
  5. Feb 11, 2007 #4
    According to your first post, the resistivities are different because the materials are different. The total resistance would then be the sum of the individual resistances
  6. Feb 11, 2007 #5
    yes, you see, this homework system is on the internet and numbers are randomized. I however, ended up with same resistivity for both material so the equation should just be resistivity multiply by total length of rod and divided by cross section area correct? But when I enter this answer into the homework system, it says my answer is not correct. I also tried doing the resistance separately and I end up with the same answer which is 1.3 ohms so I'm not really sure what I should do now
  7. Feb 12, 2007 #6
    nvm i got it, i misread something, the supposedly cross section area is only one side, I had to square it. thanks for all the help!!!
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