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Resistance and How It Varies With Diameter

  1. Mar 12, 2013 #1
    So, the relationship of resistance is [itex]R= \rho \frac{l}{A}[/itex]; assuming the resistor is cylindrical, the cross-sectional area of the resistor is [itex]A = \pi r^2[/itex]. The relationship between the radius and diameter is [itex]r = \frac{d}{2}[/itex]. Substituting in this relationship, [itex]A = \pi (\frac{d}{2})^2[/itex]

    So, if I were to double the diameter, the cross sectional area would become [itex]A = \pi \frac{(2d)^2}{4} \rightarrow A = \pi d^2[/itex] Does this mean that the cross sectional area quadruples? further implying that the resistance decreases by 1/4?
     
  2. jcsd
  3. Mar 12, 2013 #2

    TSny

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    Yes. That's correct.
     
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