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Homework Help: Copper wire resistance problem

  1. Mar 20, 2010 #1
    A copper wire is in the form of a cylinder and has a resistance R. it is stretched till its thickness reduces by half of its initial size. Find its new resistance in terms of R.
     
  2. jcsd
  3. Mar 20, 2010 #2

    Dale

    Staff: Mentor

    Re: resistance

    This should probably go into the homework section, and you will need to show your work so far.
     
  4. Mar 20, 2010 #3
    Re: resistance

    its resistance will be increased.because R is always inversely proportional to length
     
  5. Mar 20, 2010 #4
    Re: resistance

    Thanks, could you please mention by how much the resistance will rise ?
     
  6. Mar 20, 2010 #5
    Re: resistance

    I am really sorry to confuse you but this question appeared for my board exams recently . Could you please help me out . :smile:
     
  7. Mar 20, 2010 #6
    Re: resistance

    What equations do you know that relate resistance to length, and amount of substance?
     
  8. Mar 20, 2010 #7
    Last edited by a moderator: Apr 24, 2017
  9. Mar 23, 2010 #8
    Re: resistance

    it resistance will be increased by two times.
     
  10. Mar 23, 2010 #9
    Last edited by a moderator: Apr 24, 2017
  11. Mar 23, 2010 #10
    Re: resistance

    Is diameter the only thing changing?
     
  12. Mar 23, 2010 #11
    Re: resistance

    [tex] R = \rho \frac{h}{A}[/tex]

    We must first notice that both area and length change so that the volume remains constant. We assume that the stretching results in a prefect cylinder that we began with. Therefore:
    [tex] V = Ah = \pi r^2 h[/tex] and [tex] V[/tex] is a constant.

    Thickness refers to the diameter of the cylinder, so to decrease its thickness by half, you must decrease its diameter by half (or its radius by half). So the new area is:
    [tex] A' = \pi (\frac{r}{2})^2 = \pi \frac{r}{4}[/tex]

    Therefore, to destroy no matter and to keep volume constant, we must multiply [tex]h'[/tex] by a constant to keep [tex]V'[/tex] equal to [tex]V[/tex]:
    [tex] V = V' = \pi r^2 h = (\pi \frac{r^2}{4})(Kh)[/tex]
    Seen above, K must equal four to cancel out that fractional four, thus bringing both left and right equations equal and thus keeping volume constant.

    To sum:
    [tex] h' = 4h[/tex]
    [tex] A' = \pi \frac{r^2}{4} = (\pi r^2) \frac{1}{4} = \frac{A}{4}[/tex]

    We now plug in our new height and our new area into the equation for resistance:
    [tex] R' = \rho \frac{4^2h}{A}= 4^2(\rho \frac{h}{A}) = 4^2 R = 16R[/tex]
     
    Last edited: Mar 23, 2010
  13. Mar 23, 2010 #12
    Re: resistance

    Are you sure? Aren't radius and diameter directly proportional, so that if you decrease the diameter by a half, you also decrease the radius by a half?

     
  14. Mar 23, 2010 #13
    Re: resistance

    you're right
     
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