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A wire with resistance of 8.0 Ω is drawn out through a die such that its new length is twice its original length. Find the resistance of the longer wire assuming that the resistivity and the density of the material of the wire are unchanged during the drawing process.

Sagot :

[tex]\begin{equation*} 16\text{ }\Omega \end{equation*}[/tex]

Explanation

the electrical resistivity is define as

[tex]\begin{gathered} \sigma=R\frac{A}{l} \\ where \\ R\text{ is the electrical resitance} \\ \text{A is the cross seccional area} \\ l\text{ is the length} \end{gathered}[/tex]

so,if we isolate R

[tex]\begin{gathered} \sigma= R\frac{A}{l} \\ R=\sigma\frac{l}{A} \end{gathered}[/tex]

hence, the ratio of the resitances is

[tex]\frac{R_1}{R_2}=\frac{\sigma\frac{l}{A}}{\sigma\frac{l_2}{A}}[/tex]

the volume of the wire is constant , therefore

[tex]A_1l_1=A_2l_2[/tex]

if the new length is twice the original

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