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Current and Painted Tube

  1. Feb 19, 2007 #1
    1. The problem statement, all variables and given/known data

    A plastic tube 25.0 m long and 4.00 cm in diameter is dipped into a silver solution, depositing a layer of silver 0.100 mm thick uniformly over the outer surface of the tube. If this coated tube is then connected across a 12.0-V battery, what will be the current?

    2. Relevant equations

    R = (p*L)/A

    V = I*R, where I = V/R

    3. The attempt at a solution

    I’m a bit uncertain about this problem since there is both a semiconductor (Ag) and plastic involved.

    A = 2*pi*(r1 + r2)*L – 2*pi*(r1)*L = 2*pi*r2* L = 2*pi*(0.100* 10^-3 m)*25 m = 0.0157 m^2???

    R = (p_Ag*L)/A = [(1.47*10^-8 ohm*m)*(25 m)]/(0.0157 m^2) = 2.34*10^-5 ohm???

    I = 12 V/(2.34*10^-5 m^2 ohm) = 5.129*10^5 A ???

    If the above is incorrect, please direct me. Any help is appreciated.

    Last edited: Feb 19, 2007
  2. jcsd
  3. Feb 19, 2007 #2


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    1) Ag is not a semiconductor - it's a conductor. I would assume the plastic is an insulator.

    2) A is the cross-sectional area of the silver coating. It doesn't involve L. I would rethink that part of the calculation.

    3) The units for R are just ohms. Take another look at that line.
  4. Feb 19, 2007 #3
    A = pi*(d + r)^2 - pi*r^2, where r is radius of insulator section and d is thickness of Ag layer

    A = pi*[0.02 m + (0.1*10^-3 m)] - pi(0.02 m)^2
    = 0.001269 m^2 - 0.001257 m^2
    = 1.256*10^-5 m^2

    R = [25 m*(1.47*10^-8 ohm*m)]/[1.256*10^5 m^2]
    = 0.02952 ohm

    I = (12 V)/(0.02952 ohm) = 410.2 A ????
  5. Feb 19, 2007 #4


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    Looks good to me!
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