Question #56177

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Question #56177

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Answer 1 · 2016-11-10T18:41:28

See proof below

Explanation:

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

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

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

The second wire $R_{1} = ρ \cdot \frac{\frac{l}{2}}{π \frac{r^{2}}{4}}$ $R_1=rho*(l/2)/(pir^2/4)$

$= \frac{4 ρ l}{2 π r^{2}} = \frac{2 ρ l}{a} = 2 R$ $=(4rhol)/(2pir^2)=(2rhol)/(a)=2R$

So the new resistance is twice the other resistance

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Question #56177

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