# Resistance and How It Varies With Diameter

1. Mar 12, 2013

### Bashyboy

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

So, if I were to double the diameter, the cross sectional area would become $A = \pi \frac{(2d)^2}{4} \rightarrow A = \pi d^2$ Does this mean that the cross sectional area quadruples? further implying that the resistance decreases by 1/4?

2. Mar 12, 2013

### TSny

Yes. That's correct.