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Homework Help: Cylindrical bar of copper - current, resistance

  1. Jun 11, 2008 #1
    1. The problem statement, all variables and given/known data

    a cylindrical bar of copper has a resistance of 0.02 ohms

    a) if the power of the bar is 10000 watts, what is the maximum current the bar can carry?

    b) the bar is now stretched to 3 times its original length, while the total volume of the bar remains constant. what is the new resistance of the bar?

    2. Relevant equations

    power P = IV = I^2R = V^2/R where I is current, R is resistance, V is electric potential

    I = VA/L*rho = V / ((rho*L)/A) where A is area, L is length, rho is restivity

    V = IR

    3. The attempt at a solution

    i think i got part a correctly:

    P = I^2R
    10000 = I^2(0.02)
    sqrt(10000/0.02) = I
    I = 707.11 amperes

    as for part b, do i need to use the resistivity of copper, rho_copper = 1.68*10^-8 ohm-m :

    V = IR = 707.11(0.02) = 14.14 volts

    I = VA/L*rho = 14.14A/3L(1.68*10^-8) ---> holding A and L constant
    I = 14.14/(3*(1.68*10^-8)) = 2.81*10^8 ampere ---> that is massive!!

    i think i did part a correctly, very doubtful of part be though

    please tell me what needs to be fixed
  2. jcsd
  3. Jun 11, 2008 #2


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    When you stretch the wire keeping volume constant, length increases and area decreases.
    Resistance of the bar R = rho*L/A = rho*L*L/A*L = rho*L^2/V, where V is the volume of the bar.
  4. Jun 11, 2008 #3
    why is the L squared in the numerator?

    since part b states that the bar is stretched three times its original length and volume remains constant R = rho(3^2)/1 = 1.68*10^-8(9) = 1.51*10^-7 ohms

    now that's tiny, does it make sense though?
  5. Jun 11, 2008 #4


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    You don't want you use the resistivity of copper unless you assume some values for A and L that will give you a resistance of 0.02 ohms. In your treatment A and L just disappeared. If the original resistance is R=0.02 ohms=rho*L/A. Let the new length and area be L' and A'. The volume is L*A. If L'=3*L, then what is the relation between A and A' that will keep the volume constant? How will that change the value of R?
  6. Jun 11, 2008 #5
    if V = AL = A'L' where L' = 3L, then A' = A/3, correct?

    so new resistance R = rho*3L/(A/3) = rho*9L/A= rho*9 = 1.68*10^-8(9) = 1.51*10^-7 ohms

    isn't that what i got before?
  7. Jun 11, 2008 #6


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    The first part is correct. Yes, the new resistance is rho*9*L/A. That's 9*(rho*L/A) and rho*L/A is the original resistance of 0.02 ohms. What's the new resistance? Do you see why you don't actually need to know rho (or A or L)??
  8. Jun 11, 2008 #7
    oh, i see, i would've never even saw that coming if you hadn't told me, so now R = 9(0.02) = 0.18 ohms


  9. Jun 11, 2008 #8


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    Correct. It's just a ratio and proportion argument. Don't forget that trick, saves you a lot of work. Welcome.
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