# Question #56177

Nov 10, 2016

See proof below

#### Explanation:

The resistance of the first wire is $R = \rho \frac{l}{a}$
where $\rho =$ resistivity for this material
$l =$length
$a =$area $= \pi {r}^{2}$ where r is the radius

New length $= \frac{l}{2}$

New radius $= \frac{r}{2}$

The second wire ${R}_{1} = \rho \cdot \frac{\frac{l}{2}}{\pi {r}^{2} / 4}$

$= \frac{4 \rho l}{2 \pi {r}^{2}} = \frac{2 \rho l}{a} = 2 R$

So the new resistance is twice the other resistance