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Changing Rho, Finding Resistance

  1. Oct 5, 2014 #1
    1. The problem statement, all variables and given/known data
    A wire with cross-sectional radius 0.91 mm lying along the x axis from x=0 to x=0[PLAIN]http://loncapa.mines.edu/adm/jsMath/fonts/cmmi10/alpha/100/char3A.png90 [Broken] m is made of an alloy that varies with its length in such a way that the resistivity is given by [PLAIN]http://loncapa.mines.edu/adm/jsMath/fonts/cmmi10/alpha/100/char1A.png=6x^1 [Broken] [PLAIN]http://loncapa.mines.edu/adm/jsMath/fonts/cmr10/alpha/100/char0A.png[PLAIN]http://loncapa.mines.edu/adm/jsMath/fonts/cmsy10/alpha/100/char01.png [Broken] m, for x in meters. What is the resistance of this wire?

    2. Relevant equations
    dR=(rho)dL/A

    3. The attempt at a solution
    (I can't find the symbol window so bare with my explanation). Resistance is the integral of rho times length, divided by cross sectional area. I figured since L and A are constants, I could pull them out and simply do: rho, integrated from 0 to 0.9, times L/A, with everything converted to meters. Didn't work.

    Then I tried not integrated and just plugging and chugging. Didn't work.

    Then I came here.
     
    Last edited by a moderator: May 7, 2017
  2. jcsd
  3. Oct 5, 2014 #2

    Mark44

    Staff: Mentor

    What didn't work? Please show what you got and how you got it.

    Also, is there a typo in ##\rho##? It seems odd to me that it would be 6x1 ohms/m. I don't know why anyone would write x1 instead of just x.
     
    Last edited by a moderator: May 7, 2017
  4. Oct 5, 2014 #3
    So I'm on my phone, but I integrated the rho function and multipled by L/A, L being the difference in the bounds and A being the given mm amount (converted to meters).

    It's to the first power cause the numbers are random, including the exponents
     
  5. Oct 5, 2014 #4

    Mark44

    Staff: Mentor

    So what value did you get?
     
  6. Oct 5, 2014 #5

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    The notation rho*dL/A doesn't mean integrate rho and then multiply by L/A. It means integrate rho WITH RESPECT to L and then divide by A. By all means, show your work. I think you are multiplying by an extra factor L. Check the units. Remember dL has units of m. Are they ohms or something else?
     
  7. Oct 5, 2014 #6

    RUber

    User Avatar
    Homework Helper

    ##\int dR = \int_a^b \frac{\rho}{A} dL \implies R=\frac1A\int_a^b \rho dL## You should not have to multiply by length, it is included in the integral.
     
  8. Oct 6, 2014 #7
    Ok, I'm back and found out where I was going wrong. Also, the parameters change each time, so don't freak out with me inputting new numbers in my explanation.

    So yes, I don't know why I was multiplying by length afterwards . . . Good to catch it now though. Anyways, I integrated the resistivity across the bounds: the integral of 4x from 0 to 0.5. Then I divided by the AREA of the given radius. That's where I was going wrong multiple times. They gave me a radius of 0.46 MM. For some reason I was using this as "A" instead of solving for A . . . So in short, integrated my resistivity function across my wire length and divided by the cross sectional area.

    Got it instantly when I started writing out my variables here and realized it was radius, not area . . .
     
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