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Determine resistance in a cone formed carbon resistor

  1. Sep 28, 2011 #1
    1. The problem statement, all variables and given/known data
    I have to determine the resistance of a cone shaped resistor with the following dimmensions:
    Height at the first end: 1mm
    Lengdt: 250mm
    Height at the last end: 0.5 mm

    Rho: 14.5^10-6 Ohm*m


    2. Relevant equations

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


    3. The attempt at a solution
    As the cone shaped resistor is linearly decreasing it's height across the lenght, I tried to be smart by creating a mean height to determine a fixed radius over the whole length.

    I got that when splitting the cone in two, i can put them together reversed to get a 0.75mm high and 250mm long carbon resistor.

    The crossection should be radius squared times PI, resulting in 0,44 mm^2

    Applying the formula stated above, i get a ridicoulous number of 0.000002 Ohms or something.
    The solution is 36 ohm.

    Can someone kick me in the right direction?

    If you feel that my description is inadequate, I'll upload a drawing immidiately

    -Twin
     
  2. jcsd
  3. Sep 28, 2011 #2

    berkeman

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    A drawing would help, but it sounds like you will need to use integration to solve this. Do you know how to divide the object up into small slices and integrate to get the total resistance?
     
  4. Sep 28, 2011 #3
    Give me a second and I'll make you a drawing :)
     
  5. Sep 28, 2011 #4
    see attachement for scales and my failed attempt
     

    Attached Files:

  6. Sep 28, 2011 #5

    berkeman

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    I'm not sure why you got such a small number (maybe mixing units?), but I also don't think you can use your trick to get around having to do the integration.
     
  7. Sep 28, 2011 #6
    I tried to integrate (still after cutting the rod in half)

    I dont know that the LATEX-form for integral is, but this is what I did:

    Integrated from 0-250mm for the value (0.25/250)X.
    The rectangle left underneath was added after and I still got 0.44mm^2.

    Based on the large deviation from the correct solution, i think there has to be something else that is wrong. A mean height of 0,75mm does not strike me as impossible or ilogical based on the numbers provided. Maybe the rho is wrong?
     
  8. Sep 28, 2011 #7

    berkeman

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    I have to duck out for a couple hours. Maybe start by changing the shape to a simple cylinder of about the same size, and making sure you get a reasonable resistance for that shape. In fact, you could try a cylinder that is the same size as the larger radius, and one that is the same size as the smaller radius -- make sure that you can bracket the correct answer for the cone with those two cylinders. That will help you catch any math or units errors you might be making, before doing the full integration...
     
  9. Sep 28, 2011 #8
    I'll do that.
    And let's try something else as well.
    We know that the solution is supposed to be 36,9 ohm.

    So i am gonna rearrange the equation.

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

    [tex]A = \rho \times \frac{0.25m}{36,9\Omega} = 0.00000009823[/tex]

    which looks wrong to me..
     
  10. Sep 28, 2011 #9

    berkeman

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    Remember to convert the units of the area from mm^2 to m^2. You should get reasonable values as long as you are careful with your units conversions.

    I just did the case for a 0.25m long cylinder of this material, with a diameter of 1mm (so a radius of 0.001m and an area of...), and got reasonable resistance numbers.
     
  11. Sep 29, 2011 #10

    Attached Files:

  12. Sep 29, 2011 #11

    berkeman

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    What did you get for the simple bounding cases of the large and small cylinders?
     
  13. Sep 29, 2011 #12
    Uhm, i am not sure if I understand what you mean
     
  14. Sep 29, 2011 #13

    berkeman

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    I mentioned in Post #7 that you can bound the resistance of the cone between the resistances of two cylinders. Both cylinders are the same length as the cone, but one has a diameter equal to the wide end of the cone, and the other has a diameter equal to the narrow end of the cone. The resistance of the cone has to lie between the resistances of those two rods, right?

    The calcs are simpler for the cylinders compared to the cone, and it's a good sanity check on the results that you are getting.
     
  15. Sep 29, 2011 #14
    Ah, I understand.

    For a 250mm long rod with 1mm diameter of carbon I get the following:

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

    [tex]R=\frac{14.5\times10^(-6) \times 0.25m}{0.0005^2\times \pi} = 4,6 \Omega[/tex]


    For a 250mm long rod with 0.5 mm diameter of carbon I get the following:

    [tex]R=\frac{14.5\times10^(-6) \times 0.25m}{0.00025^2\times \pi} = 18,46 \Omega[/tex]
     
  16. Sep 29, 2011 #15

    berkeman

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    Good. I got the same answer as you for the first case, and didn't do the 2nd one (but you are likely correct on that one too).

    So it would seem that the answer that you were given for the cone is incorrect, eh? Maybe show that to the instructor and ask if the answer is wrong, or if the resistivity that is given in the problem is incorrect...
     
  17. Sep 29, 2011 #16
    I fully agree.

    Thanks for giving me some perspective :)
     
  18. Sep 30, 2011 #17
    It turned out that the dimmensions I provided was not fully accurate.
    The dimmensions I used was the casing dimmension, whilst the carbon was inside like a core.

    It turned out nicely after that :)
     
  19. Sep 30, 2011 #18

    berkeman

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    Ah, thanks for the update. Makes sense now. :smile:
     
  20. Sep 30, 2011 #19

    Attached Files:

  21. Sep 30, 2011 #20

    berkeman

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    Nice. Just be sure to put a "dl" inside the integral, to show that you are integrating over the length. Good work!
     
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