Question

A piece of wire of resistance 20 Ω is drawn out so that its length is increased to twice its original length. Calculate the resistance of the wire in the new situation.

Answer 1

We know that the resistance of a conductor is given by:
$\mathrm{R}=\rho \frac{\mathrm{l}}{\mathrm{A}}=20\mathrm{\Omega }$
where ρ = resistivity
l = length of the conductor
A = area of cross-section of the conductor
Now the length is increased to twice the original length. Then let the new resistance be denoted by R'.

$$\begin{gathered} {\mathrm{R}}^{,}=\mathrm{\rho }\frac{2\mathrm{l}}{{\displaystyle \frac{\mathrm{A}}{2}}}=4\mathrm{\rho }\frac{\mathrm{l}}{{\displaystyle \mathrm{A}}} \\ {\mathrm{R}}^{,}=4\mathrm{R}=4\times 20=80\mathrm{\Omega } \end{gathered}$$

Thus, the new resistance will become four times.