Question

3. A piece of wire having resistance R is cut into four equal parts.
(a) How will the resistance of each part compare with the original resistance?
(b) If the four parts are placed in parallel, how will the joint resistance compare with the resistance of the original wire?

Answer 1

Dear Student,

$$\begin{gathered} \left(\mathrm{a}\right)\mathrm{Resistance}\mathrm{of}\mathrm{a}\mathrm{piece}\mathrm{of}\mathrm{wire}\mathrm{is}\mathrm{proportional}\mathrm{to}\mathrm{its}\mathrm{length}.\mathrm{A}\mathrm{piece}\mathrm{of}\mathrm{wire}\mathrm{has}\mathrm{a}\mathrm{resistance}\mathrm{R}.\mathrm{The}\mathrm{wire}\mathrm{is}\mathrm{cut}\mathrm{into}\mathrm{four}\mathrm{equal}\mathrm{parts}. \\ \mathrm{Therefore},\mathrm{resistance}\mathrm{of}\mathrm{each}\mathrm{part}=\frac{\mathrm{R}}{4} \\ \left(\mathrm{b}\right)\mathrm{If}\mathrm{all}\mathrm{the}\mathrm{four}\mathrm{parts}\mathrm{are}\mathrm{connected}\mathrm{in}\mathrm{parallel}.\mathrm{Hence},\mathrm{equivalent}\mathrm{resistance}(\mathrm{R}’)\mathrm{is}\mathrm{given}\mathrm{as} \\ \frac{1}{\mathrm{R}\text{'}}=\frac{1}{{\displaystyle \raisebox{1ex}{$\mathrm{R}$}\!\left/ \!\raisebox{-1ex}{$4$}\right.}}+\frac{1}{{\displaystyle \raisebox{1ex}{$\mathrm{R}$}\!\left/ \!\raisebox{-1ex}{$4$}\right.}}+\frac{1}{{\displaystyle \raisebox{1ex}{$\mathrm{R}$}\!\left/ \!\raisebox{-1ex}{$4$}\right.}}+\frac{1}{{\displaystyle \raisebox{1ex}{$\mathrm{R}$}\!\left/ \!\raisebox{-1ex}{$4$}\right.}} \\ \frac{1}{\mathrm{R}\text{'}}=\frac{4}{\mathrm{R}}+\frac{4}{\mathrm{R}}+\frac{4}{\mathrm{R}}+\frac{4}{\mathrm{R}} \\ \frac{1}{\mathrm{R}\text{'}}=\frac{4+4+4+4}{\mathrm{R}} \\ \mathrm{R}\text{'}=\frac{\mathrm{R}}{16} \end{gathered}$$