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Resistance question

  1. Oct 9, 2007 #1
    1. The problem statement, all variables and given/known dataI have a question from an AS level physics paper.Calculate the resistance of a 5.46m length of lead strip,0.450cm wide and 1.25mm thick.What length of similar strip made of copper would have to be put in parallel with the lead strip for the effective resistance of the combination to be 0.100 ohms? (p of Cu and Pb are 1.72x10*-8 and 20.6x10*-800

    2. Relevant equations
    I worked the resistance of the lead to be 2 ohms using R+pL/A i know the equation needed is 1/R = 1/r1 + 1/r2 but im not sure how to find the length of copper needed to get the combine resistance?

    3. The attempt at a solution
    Last edited: Oct 9, 2007
  2. jcsd
  3. Oct 9, 2007 #2
    i would post up my working but i dont know how to get all the sighns/symbols up :redface: i have just joined.

  4. Oct 9, 2007 #3


    User Avatar

    Staff: Mentor

    Welcome to the PF, chris85. It looks like you are doing all the right things. Just keep using the equation for the resistance of a uniform bar:

    [tex]R = \frac{\rho L}{A}[/tex]

    and the equation for combining parallel resistances. If the lead is 2 Ohms (I didn't check your math), what resistance do you need to put in parallel to get down to 0.1 Ohms total? And then how long would a copper bar of the same cross-sectional area A have to be to make that resistance?

    BTW, there is a LaTex tutorial in the Tutorials forum: https://www.physicsforums.com/showthread.php?t=8997
    Last edited: Oct 9, 2007
  5. Oct 10, 2007 #4
    ok well for the resistance of the lead i did

    20.6X10^-6 X 5.46M
    __________________ = 2.00 ohms so R1 = 1/2 correct?

    Now the resistivity of copper is 1.72x10^8 but i'm having trouble finding a value of resistance for the copper that when added to the lead will give a reciprocol 0.1 for the parallel resistance can someone give me a hint where i'm going wrong?
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