If a wire of resistance R is melted and recast into a wire of length 2/5th that of the original wire, what is the resistance of the new wire?

Nov 24, 2016

Let

$\text{initial length of the wire be} = {L}_{i}$

$\text{initial cross section of the wire be} = {A}_{i}$

$\text{initial resistance of the wire be} = {R}_{i}$

After melting and recast let

$\text{final length of the wire be} = {L}_{f}$

$\text{final cross section of the wire be} = {A}_{f}$

$\text{final resistance of the wire be} = {R}_{f}$

The volume of the material of the wire in both state will be same. So ${L}_{i} {A}_{i} = {L}_{f} {A}_{f} \ldots . \left(1\right)$

Again it is given that ${L}_{f} = \frac{2}{5} {L}_{i} \implies {L}_{f} / {L}_{i} = \frac{2}{5}$

Now from (1)

${L}_{i} {A}_{i} = {L}_{f} {A}_{f}$

$\implies {A}_{i} / {A}_{f} = {L}_{f} / {L}_{i} \ldots . . \left(2\right)$

Let $\rho$ be the resustivity of the material of the wire.
Then for initial wire

${R}_{i} = \rho {L}_{i} / {A}_{i} \ldots . . \left(3\right)$

And for initial wire

${R}_{f} = \rho {L}_{f} / {A}_{f} \ldots . . \left(4\right)$

Dividing (4) by (3) we get

${R}_{f} / {R}_{i} = {L}_{f} / {L}_{i} \times {A}_{i} / {A}_{f} = {\left({L}_{f} / {L}_{i}\right)}^{2} = {\left(\frac{2}{5}\right)}^{2} = \frac{4}{25}$

$\implies {R}_{f} / {R}_{i} = 0.16$

$\implies {R}_{f} = 0.16 {R}_{i}$

So the resistance of the new recast wire will be $0.16$ times the resistance of the original wire.

Nov 25, 2016

${R}_{2} = \frac{4}{25} R$
Or
${R}_{2} = 0.16 R$

Explanation:

During the recasting process the same volume of metal that was melted was used to form the new wire. For that reason we know that the volume ($V$) of both wires is equal. The resisitivity (ρ) is also the same for both wires because exactly the same material is used for both wires.

Volume is related to the length and cross-sectional area of the wire by the relationship:
$V = A L$
I will define the initial dimensions of the wire as $V$, $A$ and $L$. The dimensions of the second wire will be $V$, ${A}_{2}$ and ${L}_{2}$.

Initial resistance of the wire is: R = (ρ L)/A

The question specifies that the length is reduced to $\frac{2 L}{5}$. So ${L}_{2} = \frac{2 L}{5}$. Now use the fact that the volumes are equal to get an expression for ${A}_{2}$ in terms of $A$:
${A}_{2} = \frac{V}{L} _ 2 = \frac{V}{\left(2 L\right) / 5} = \frac{5 V}{2 L}$

Also $V = A L$ so then the equation becomes:
${A}_{2} = \frac{5 A L}{2 L} = 5 \frac{A}{2}$

Now that we have expresssions for both ${L}_{2}$ and ${A}_{2}$ we can use them in the equation for resistance to find ${R}_{2}$ in terms of $R$:
R_2 = (ρ L_2)/A_2 = (ρ (2L)//5)/((5A)/2) = (ρ 4L)/(25A) = 4/25 ((ρ L)/A)
⇒ R_2 = 4/25 R

Or
${R}_{2} = 0.16 R$