Question

A wire of resistance $1ohm$ is stretched to double its length. What is the new resistance?

Answer 1

Given: a wire of resistance $1\Omega$is stretched to double its length.

To find the new resistance

Solution:

Here,

$R=1\Omega ,\text{Length}=L,\text{Cross sectional area, CSA}=A$

The formula used is,

$R=\rho \frac{L}{A}⟹\rho \frac{L}{A}=\mathrm{1..........}\left(i\right)$

The new length, $L^{\prime }=2L$

And new CSA = $A^{\prime }$

So we have,

Volume of the wire= $AL$

Now after the act of stretching the volume remains constant.

Therefore, $AL=A^{\prime }{L}^{\prime }$

$⟹AL=2L{A}^{\prime }⟹A^{\prime }=\frac{A}{2}$

Now the new resistance of the wire becomes

$R_{new}=\rho \frac{L^{\prime }}{A^{\prime }}⟹R_{new}=\rho \frac{2L}{\frac{A}{2}}⟹R_{new}=4\rho \frac{L}{A}$

$R_{new}=4\times 1$ by using equation (i)

Therefore the new resistance is $4\Omega$