1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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

    rl.bhat

    User Avatar
    Homework Helper

    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

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    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

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    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

    correct?

    thanks
     
  9. Jun 11, 2008 #8

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    Correct. It's just a ratio and proportion argument. Don't forget that trick, saves you a lot of work. Welcome.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Cylindrical bar of copper - current, resistance
Loading...