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Help (calculating resistance)

  1. Dec 8, 2003 #1

    eee

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    Can anyone help me with this resistance problem?

    Total Resistance in a series = R1+R2+...

    Using this principle, estimate the resistance of a cylindrical aluminum block (resistivity = 2.65 x 10^-8 ohm-meter)that has a radius w/c lineary increases from 5cm to 10 cm.

    Then divide block into 3 cylinders w/ equal thickness and uniform radius. State how you assumed the radii of the 3 cylinders.

    If the 3 cylinders are cut to very thin disks, the resistance may be calculated w/ an integral. Set-up the integral.
     
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  3. Dec 8, 2003 #2

    Integral

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    We do not do homework for you on this forum. Show us something you have tried. I will give you something to get you started.


    You should know the expression for resistance.

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

    The second phrase seems to be a hint of how to accomplish the first part. Use that to get an estimate ( it this case it should be a correct value)
     
    Last edited: Dec 8, 2003
  4. Dec 8, 2003 #3

    HallsofIvy

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    The question isn't very clear. Are you not told the length of the cylindrical?

    Also the "then" makes me think you were expected to do find the resistance of the whole block directly and "then" do the rest of the problem to find the resistance by integrating- and I don't see how to find the resistance without integrating!

    I see no way, nor reason, to divide this into 3 cylinders and then divide those 3 cylinders into "very thin disks".


    Finally, resistivity is given as 2.65 x 10^-8 ohm-meter: that is, it's per meter- length, not volume. Your formula says nothing about how that varies with cross-section area.
     
  5. Dec 8, 2003 #4
    I believe there is no need of length of cylinder

    as
    therefore

    r = A+Bx
    where x is the distance from one end of cylinder.
    A & B are to be founded by boundary conditions for r
    also r is the radius of the disc at distance x. therefore area=[tex]\pi r ^2[/tex].

    Now apply differential form for [tex]dR= \frac {\rho dx} A [/tex]
     
  6. Dec 8, 2003 #5

    Integral

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    To get a numeric answer you will need a length. As the proplem now reads the resistance can only be expressed in terms of the unknown length. We are only given 2 diameters and the fact that it changes lineraly, this is not enough to completly solve the problem.
     
  7. Dec 8, 2003 #6
    sorry we will need length
     
  8. Dec 8, 2003 #7

    eee

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    Thank you for replying to my post. :)

    I did not meant for anyone to solve the problem for me, I just want to know how to solve it.

    I know the formula: [tex]R= \frac {\rho L} A [/tex]

    I'm just confuse on how to use it on a radius that lineary increases.

    Should I try to estimate a Resistance total b4 getting the 3 separated resistance or the other way around? Although I think I should be doing the former for this problem...hmm

    Well I assumed that 1 of my radii would be 0.05 meters since it's the starting radius and that the constant length for the 3 divided cylinders would be perhaps 0.33 meters..I dunno since uh 1/3? ehehe.

    But then again, how would I get the other 2 radii if I don't know the total Resistance?

    I guess the question really here is how do you solve for the resistance of something that has a radius that lineary increases? Is it possible or you would just have to settle an estimate? If so, what should you do to get the best estimate?

    The thing w/ the integral, well that was just a bonus question. I really don't know how to solve that because we haven't tackled it yet in our calculus. I don't even get the question.

    Hmm..I'm still trying to solve this though I am of need of assistance. :>
     
  9. Dec 8, 2003 #8

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    Ack I guess I forgot to say that L = 1m.
    Sorry.
     
  10. Dec 9, 2003 #9

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    Well I already sort of answered this an hour after my last post.

    Mostly, I did it by trial and error.

    I used .06 m as the radius for the Total Resistance.

    And I got .05, .065, and .08 as the radii for the 3 blocks, the length I used was .33 m.

    If anyone has a better solution, I would like to know. This homework was already passed so it doesn't really matter but still, I'd really like to know :smile:
     
  11. Dec 9, 2003 #10

    HallsofIvy

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    Okay, given that [tex]R= \frac {\rho L} A [/tex], that the length is 1 meter, and that the radius increases linearly from 5 cm. to 10 cm. , then, the radius at distance x (in meters) from the small end is .05x+ 0.05 meters. The area of a cross section is π(0.05x+ 0.05)2. Thinking of the entire length as being divided into think disk of height dx (thin enough so that the radius does not change significantly), the resistance of each disk is [tex]dR= \frac {\rho dx} A [/tex], just as himansu121 said.

    integrate that from 0 to 1.
     
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