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Electrical resistance

  1. Aug 5, 2010 #1
    This problem is about a cylinder, which its length is 'L', its radius 'a' and its resistivity 'p'.

    We pierce a hole in the cylinder (this hole has a radius of a/2)

    I have now to

    a) prove that the resistance equation(between the ends) of this hollow cylinder is

    R= 4/3 pL/(pi)a^2

    b) prove that the total resistance of the same hollow cylinder filled with a new resistivity material p2 is:

    R= 4/3 pL/(pi)a^2 X p2/p2+p

    I tried to prove A but it gives me R=pL/pi(b^2 - a^2)....which is not what I want. I have NO idea where does the 4/3 from the equation of the question comes from. If someone could help me with a hint or something, it would be much appreciated.
     
  2. jcsd
  3. Aug 5, 2010 #2

    nrqed

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    By [itex] b^2 - a^2 [/itex] I think you mean the outer radius squared minus the inner radius squared, right? So you mean [itex] a_{max}^2 - a_{min}^2 [/itex]. Then just replace a_max by "a" and a_min by "a/2". You will see the 4/3 appear.
     
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