Question

A wire has resistance of $8\Omega$. If its length is made half by folding, find its resistance after the free ends are connected to each other.

Answer 1

Let L be the length and A be the cross sectional area

The resistance of the original wire is, $R=\rho \frac{l}{A}=8\Omega$

Now we have,

$l^{\prime }=l/2$

$A^{\prime }=A/2$

Therefore the resistance of the new wire is given by,

$R^{\prime }=\rho \frac{l^{\prime }}{A^{\prime }}$

$R^{\prime }=\rho \frac{l/2}{2A}=\frac{1}{4}\rho \frac{l}{A}=\frac{1}{4}\times 8=2\Omega$.