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What happens with the resistance of a wire with twice the diameter

  1. Nov 16, 2010 #1
    hi everybody

    Question:
    a length of uniform wire has a R of 2 ohms. calculate the R of a wire of the same metal and original length but twice the diameter.

    I thought as the wire gets twice as thick the R must halve to 1ohms. But the solution is 0.5 ohms.
    what did i do wrong?


    thanks in advance
     
  2. jcsd
  3. Nov 16, 2010 #2

    cepheid

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    Can you list for me the parameters that resistance depends upon? (Hint: it does not depend *directly* on the diameter.) Which one of these parameters is relevant here?
     
  4. Nov 16, 2010 #3
    is it that the R decreases when the A increases?
     
  5. Nov 16, 2010 #4

    cepheid

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    Yes, cross-sectional area is one of the parameters on which resistance depends. (I had asked you to list all of the parameters, but this one is the relevant one here). So, what does changing the diameter do to the area? What does that resulting change in area do to the resistance?
     
  6. Nov 16, 2010 #5
    is it that when you double the diameter you double the A?
     
  7. Nov 16, 2010 #6
    as the area increases the R decrease
     
  8. Nov 16, 2010 #7

    cepheid

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    This is false. What is the area of a circle in terms of its diameter (or radius)?
     
  9. Nov 16, 2010 #8
    r ^2 times pi
     
  10. Nov 16, 2010 #9

    cepheid

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    Right. And since the diameter is just twice the radius, what this means is that the area depends on the SQUARE of the diameter. So, if you double the diameter, you don't double the area. If you double the diameter, the area increases by a factor of _____?
     
  11. Nov 16, 2010 #10
    sorry i don't get what you mean with the A depends on the square of diameter.
    can you show me an example with numbers please so that i can visualise this problem?
     
  12. Nov 16, 2010 #11

    cepheid

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    Alright look. Let's call the original diameter d1. Now:

    [tex] A_1 = \pi r_1^2 [/tex]

    But the radius is just half the diameter, so:

    [tex] r_1 = d_1 / 2 [/tex]

    Agreed? Therefore:

    [tex] A_1 = \pi \left(\frac{d_1}{2}\right)^2 = \pi \frac{d_1^2}{4} [/tex]

    This is how the area of a circle depends upon its diameter. As you can see, area is equal to a constant times the diameter squared.

    Now, what happens if we change the diameter by doubling it? Let's call the new diameter d2 so that:

    [tex] d_2 = 2d_1 [/tex]

    We can plug this new diameter into the formula for the area in order to find the new area:

    [tex] A_2 = \pi \frac{d_2^2}{4} = \pi \frac{(2d_1)^2}{4} = 4 \pi \frac{d_1^2}{4} = 4A_1 [/tex]

    So we have the result that A2 = 4A1. After doubling the diameter, the new area is equal to FOUR times the original area. This is because the area of a circle is proportional to its diameter SQUARED. So if you double the diameter, you quadruple the area. If you triple the diameter, you increase the area by a factor of nine. If you quadruple the diameter, the area increases by a factor of 16. Now do you understand?
     
  13. Nov 16, 2010 #12
    wow I'm impressed with that.
    Yes I understand now. Thanks a lot for your answer :)
     
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