question_answer

A wire of uniform cross-section and length l has a resistance of \[16\,\Omega .\] It is cut into four equal parts. Each part is stretched uniformly to length l and all the four stretched parts are connected in parallel. Calculate the total resistance of the combination so formed, Assume that stretching of wire does not cause any change in the density of its material.

Answer:

Given, \[R=16\Omega ,\] when length = l When the wire is cut into 4 equal parts, then length of each part = 1/4. As         \[R\propto l\]                              \[\left[ \because R=\rho \frac{l}{A} \right]\] The resistance of each part \[R'=\frac{R}{4}=\frac{16}{4}=4\,\Omega \] If a wire is stretched to n times its original length, then its new resistance is given by             \[{{R}_{new}}={{n}^{2}}{{R}_{original}}\] Here each part of the wire is stretched to length l i.e.                        n = 4 \[\therefore \] The resistance of each part \[={{n}^{2}}\times R'={{4}^{2}}\times 4=64\,\Omega \] When 4 such resistances are connected in parallel, the effective resistance is             \[\frac{1}{{{R}_{eff}}}=\frac{1}{64}+\frac{1}{64}+\frac{1}{64}+\frac{1}{64}\]             \[\frac{1}{{{R}_{eff}}}=\frac{4}{64}\] \[\therefore \]           \[{{R}_{eff}}=\frac{64}{4}=16\,\Omega \]