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Resistance of hollow copper cylinder

  1. Mar 7, 2013 #1
    1. The problem statement, all variables and given/known data
    Hey!

    We have a copper pipe, which has a outer diameter [itex]y = 4.1 mm[/itex] and inner diameter [itex]x = 4.0 mm [/itex]
    Length of the pipe is [itex]L = 120m[/itex] and resistivity of copper is [itex]ρ = 1.7 * 10^{-8}\Omega[/itex]


    2. Relevant equations
    Resistance is [itex]R = ρ * \frac{L}{A}[/itex]


    3. The attempt at a solution
    Now, I calculated the resistances for cylinders of diameter y and diameter x.
    Then I subtracted the bigger from lower one and got:

    [itex] R = 8*10^{-3} \Omega[/itex]

    But does the shape of the object matter? Don't we need just the area of the bottom of hollow cylinder?

    [itex]A = \pi * ( r_1^{2} - r_2^{2} )[/itex]

    Using the area above, I get around [itex]R = 3.2 \Omega [/itex]

    I converted diameters to radius and mm to metres.

    P.S. In back of my book the answer is [itex]R=0,8 \Omega[/itex]
     
  2. jcsd
  3. Mar 7, 2013 #2
    The answer in the book is correct. Most likely you are not using the correct units. Show the details of your calculation.
     
  4. Mar 7, 2013 #3

    SteamKing

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    I get 3.2 ohm when I calculate the resistance.

    voko: did you use 4.1 mm and 4 mm as diameters or radii?
     
  5. Mar 7, 2013 #4
    Admittedly, I (ab)used the data as radii :)
     
  6. Mar 7, 2013 #5
    First option:

    [itex]
    R = 1.7 * 10 ^{-8}\Omega m * \Big(\frac{120 m }{\pi * \big( \frac{0.004 m }{2} \big)^{2}} - \frac{120 m }{\pi * \big( \frac{0.0041 m }{2} \big)^{2}} \Big) = 7,822.. * 10^{-3} \Omega
    [/itex]

    Second option:
    [itex]
    R = 1.7 * 10 ^{-8}\Omega m * \Big(\frac{120 m }{ \pi \big( (\frac{0.0041m}{2})^{2} - (\frac{0.0040m}{2})^{2}} \Big) = 3.20667 \Omega
    [/itex]
     
    Last edited: Mar 7, 2013
  7. Mar 7, 2013 #6
    Resistance is not additive. Think about total resistance of parallel conductors.
     
  8. Mar 7, 2013 #7
    Ye, that's why option 1 seemed pretty weird to me.

    I dont see tho how option 2 is about additivity of resistance. As I just calculated the (bottom) surface area of the hollow sylinder and then proceed to calculate the resistance?
     
  9. Mar 7, 2013 #8
    Option 1 is the one one which assumes additivity of resistance. If you use resistance of parallel conductors instead, you should get the same as per option 2. I suggest that you do that symbolically.
     
  10. Mar 7, 2013 #9
    Okey,

    so we can divide the the bottom surface area to n amount of same size segments. Then they all have the same length L.

    Now as they are parallel and each has resistance [itex]R_1[/itex] we get

    [itex]R^{-1} = \big( \frac{n}{R_{1}}\big)[/itex]

    Now each segment has surface area

    [itex]A_1 = \frac{\phi}{360}\Big(\pi \big( (\frac{y}{2})^{2} - ( \frac{x}{2} ) ^{2} \big ) \Big) [/itex]

    and then

    [itex]R_1 = \frac{ρL}{\frac{\phi}{360}\Big(\pi \big( (\frac{y}{2})^{2} - ( \frac{x}{2} ) ^{2} \big ) \Big)}[/itex]

    Then as [itex]\frac{\phi}{360} = \frac{1}{n}[/itex]

    we get the

    [itex]R = \frac{ρL}{\Big(\pi \big( (\frac{y}{2})^{2} - ( \frac{x}{2} ) ^{2} \big ) \Big)} [/itex]

    ?
     
  11. Mar 7, 2013 #10
    What you did is puzzling. But the result is correct.
     
  12. Sep 14, 2016 #11
    option 1 is wrong you can not subtract the resistances of the 2 full cylinders to get the resistance of the hollow cylinder because the resistance of the large cylinder is equivalent to the hollow and small cylinders connected in parallel and not series. 1/Ry=1/Rh+1/Rx, which is your option 2
     
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