Expand the quadratic term for resistance equation

1. Apr 24, 2006

amcavoy

I said that $A\ell=A_{\textrm{new}}\ell\left(1+\frac{x}{100}\right)\implies A_{\textrm{new}}=\frac{A}{1+\frac{x}{100}}$. Also, we know that $R\propto\frac{\ell}{A}.$. Therefore,

$$R_{\textrm{old}}\propto\frac{\ell}{A}$$

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

Somehow I'm not getting as nice of an answer as I expected.

Last edited: Apr 24, 2006
2. Apr 24, 2006

Integral

Staff Emeritus
Expand the quadradic term, you can claim that the ${(\frac x {100})}^2$ term is very small and ignore it. Can you see where to go from there?

3. Apr 26, 2006

amcavoy

So roughly $2x$ percent?