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Let us consider a copper wire of mass $1\ \mathrm{kg}$ and length $1\ \mathrm{m}$. Let the temperature & cross-sectional area be constant.

Case 1 :

If I cut the wire with a scissor, and make it shorter from 1m to $0.8\ \mathrm{m}$, then the resistance of that $0.8\ \mathrm{m}$ wire would be the same as that of the $1\ \mathrm{m}$ copper wire.

Case 2 :

If I melt the $1\ \mathrm{m}$ & $1 \ \mathrm{kg}$ copper wire and make a $1 \ \mathrm{kg}$ copper wire of length $0.8\ \mathrm{m}$, then the resistance of the newly formed copper wire would decrease.

Is my understanding correct?

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  • $\begingroup$ No...not correct. Why do you think Case 1 at all? What is your reasoning? $\endgroup$
    – DKNguyen
    Dec 5, 2021 at 4:03

2 Answers 2

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Case 1 is definitely wrong. If it's so, then repeating this procedure will lead to a wire with vanishingly small length but finite resistance, which is non-sense.

Case 2 it's correct that the resistance will decrease, but in a more subtle way.


In general, the resistance of a material is given as (Hyperphysics):

\begin{align*} R = \rho \frac{L}{A} \end{align*}

where $L$ is the length of the material, $A$ is the cross sectional area and $\rho$ is the material dependent resistivity of the material.

This equation make sense, it basically just say the resistance double if you double the length, and halves if you increase the cross sectional area by 2.


Now for case one, the length is changed by a factor of $0.8$, so the resistance is decreased by 20%

On the other hand,for case 2, suppose the wire is still cylindrical, since the volume of the material is unchanged (i.e. $LA = \text{const}$). Making $L\to0.8L$ means you must have $A \to 0.8^{-1} A$. Putting back to the equation you can see the resistance becomes $R\to 0.8^2 R$, which is a 36% decrease of resistance

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The resistance of a wire of material with resistivity $\rho$, length $L$, and cross-section area $A$ is given by the formula $R=\rho L/A$.

So for case 1, you are reducing $L$ from 1 m to 0.8 m while keeping $A$ constant, so resistance will drop from $R=\rho *1.0/A$ to $R'=\rho *0.8/A$, i.e. resistance will drop by 20%.

And for case 2, you are reducing $L$ from 1 m to 0.8 m, but simultaneously increasing $A$ by the same proportion (as the volume $L*A=V$ remains constant). So the resistance changes to $R'=\rho *0.8/(A/0.8)$, thus it will fall by a factor $0.8*0.8=0.64$ i.e. a drop in resistance of 36%

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