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Resistance of resistors

  1. Jun 12, 2010 #1
    hi guys,
    I know this sound strange but it did happen. I was working on a question asking about calculating the resistance of a resistors, using different material one is copper and another is aluminium. I got the right answer, and wrote down the formula and steps that I used. But then I only didn't put in the values that I actually used. Just formula, steps and answer. Now when I come back to look at it again, I couldn't get the same answer that I originally correct. Either could I back track how I did it, so I hope someone can point it out to me, how I did it in the first place, and did I do it right then.

    1. The problem statement, all variables and given/known data
    a 20mm long piece of copper wire (sigma = 5.8*10^7 (ohm.m)^-1) with a diameter of 0.01mm is used as a resistor. you want to design a resistor with the same resistance but the only wire avaiable to you is an aluminium wire (sigma = 3.4*10^7)ohm.m)^-1) with diameter 0.02 mm. calculate the length for the piece of aluminium wire for which the two resistances are the same

    2. Relevant equations
    R = pL/A

    3. The attempt at a solution
    the steps that I wrote down original listed below, and I haven't put in the fitted values then, but as I tries to put in all the values in again, I didn't get a consistence result.

    Rcu = pL/A
    = L/ (sigma* A)
    = (L/sigma) (4/ pi d^2)

    Values I used to reattempt to get answer:
    L = 20mm = 0.02m
    sigma = 5.8*10^7
    d = (0.01*10-3) ^2

    Rcu = (0.02/(5.8*10^7)) * (4/ (pi*(.01*10^-3)^2

  2. jcsd
  3. Jun 12, 2010 #2


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    R = ρL/A

    RCu = ρ(Cu)L1/A1 = ρ(Al)L2/A2 = R(Al)

    Substitute the values and find L2
  4. Jun 12, 2010 #3


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    I think you must be plugging it into your calculator incorrectly. I just got the same answer you got originally.
  5. Jun 12, 2010 #4
    ohh! indeed i have made a mistake at input my values. because I recalculated it everything, this time I've got a consistent answer.

    ahaha! thanks
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