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Expand the quadratic term for resistance equation

  1. Apr 24, 2006 #1
    I said that [itex]A\ell=A_{\textrm{new}}\ell\left(1+\frac{x}{100}\right)\implies A_{\textrm{new}}=\frac{A}{1+\frac{x}{100}}[/itex]. Also, we know that [itex]R\propto\frac{\ell}{A}.[/itex]. Therefore,

    [tex]R_{\textrm{old}}\propto\frac{\ell}{A}[/tex]

    [tex]R_{\textrm{new}}\propto\frac{\ell\left(1+\frac{x}{100}\right)^{2}}{A}[/tex]

    Somehow I'm not getting as nice of an answer as I expected.
     
    Last edited: Apr 24, 2006
  2. jcsd
  3. Apr 24, 2006 #2

    Integral

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    Expand the quadradic term, you can claim that the [itex] {(\frac x {100})}^2 [/itex] term is very small and ignore it. Can you see where to go from there?
     
  4. Apr 26, 2006 #3
    So roughly [itex]2x[/itex] percent?
     
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