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Electric pipe

  1. May 17, 2007 #1
    A cylindrical aluminum pipe of length 1.36 m has an inner radius of 1.80 10-3 m and an outer radius of 3.00 10-3 m. The interior of the pipe is completely filled with copper. What is the resistance of this unit? (Hint: Imagine that the pipe is connected between the terminals of a battery and decide whether the aluminum and copper parts of the pipe are in series or in parallel.)

    Here is what I did:
    L = 1.36 m
    ri = 1.80e-3m
    ro = 3.00e-3 m
    R = ? ohms
    R = pL/A
    pcopper = 1.72e-8 ohm*m
    palum = 2.82e-8 ohm*m

    In = copper
    Out = aluminum

    Ai = pi*ri^2 = 1.01788e-5 m^2
    Ao= pi*ro^2 = 2.827e-5 m^2

    Ri = pcopper * L/Ai = .002298 ohms
    Ro = palum * L/Ao = .0013566 ohms

    Then I did
    1 / ( (1/.002298ohms) + (1/.0013566 ohms) )
    and got 8.53e-4 Ohms, but it was wrong...
     
  2. jcsd
  3. May 17, 2007 #2

    nrqed

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    The cross sectional area for the aluminum cylinder is not ][itex] \pi r_0 ^2 [/itex], you must use only the area of the outside cyliner which wil be [itex] \pi(r_0^2 - r_i^2) [/itex]
     
  4. May 17, 2007 #3
    So Ao = pi*(ro^2 - ri^2) ?
     
  5. May 17, 2007 #4

    nrqed

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    Yes, that's what I wrote.
     
  6. May 17, 2007 #5
    Thanks, worked perfect :)
     
  7. May 17, 2007 #6

    nrqed

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    Great. Am glad I could help. I hope you see why this is the correct equation to use for the cross sectional area of the aluminum pipe.

    good luck
     
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