Question

a copper wire has a diameter of 0.6 mm and resistivity og 1.7*10^-8 ohms how much does the resistence change if the diameter is doubled and if the diameter is halved

Answer 1

$$\begin{gathered} \mathrm{R}=\mathrm{\rho }\frac{\mathrm{l}}{\mathrm{A}}=\mathrm{\rho }\frac{\mathrm{l}}{{\mathrm{\pi r}}^2} \\ \mathrm{\rho }=1.7\times {10}^{-8}\mathrm{ohm}-\mathrm{m} \\ \mathrm{d}=0.6\mathrm{mm}=0.6\times {10}^{-3}\mathrm{m} \\ \mathrm{r}=0.3\times {10}^{-3}\mathrm{m} \\ \mathrm{R}=\frac{1.7\times {10}^{-8}\times \mathrm{l}}{3.14\times {\left(0.3\times {10}^{-3}\right)}^2} \\ \mathrm{R}=\frac{1.7\times {10}^{-8}\times \mathrm{l}}{0.2826\times {10}^{-6}}=6.01\times {10}^{-2}\left(\mathrm{l}\right)\mathrm{ohm} \\ 1)\mathrm{If}\mathrm{diameter}\mathrm{is}\mathrm{doubled} \\ {\mathrm{d}}_1=1.2\times {10}^{-3}\mathrm{m} \\ {\mathrm{r}}_1=0.6\times {10}^{-3}\mathrm{m} \\ {\mathrm{R}}_1=\frac{1.7\times {10}^{-8}\times \mathrm{l}}{3.14\times {\left(0.6\times {10}^{-3}\right)}^2} \\ {\mathrm{R}}_1=\frac{1.7\times {10}^{-8}\times \mathrm{l}}{1.1304\times {10}^{-6}}=1.50\times {10}^{-2}\left(\mathrm{l}\right)\mathrm{ohm} \\ 2)\mathrm{If}\mathrm{diameter}\mathrm{is}\mathrm{halved}. \\ {\mathrm{d}}_2=0.3\times {10}^{-3}\mathrm{m} \\ {\mathrm{r}}_2=0.15\times {10}^{-3}\mathrm{m} \\ {\mathrm{R}}_2=\frac{1.7\times {10}^{-8}\times \mathrm{l}}{3.14\times {\left(0.15\times {10}^{-3}\right)}^2} \\ {\mathrm{R}}_2=\frac{1.7\times {10}^{-8}\times \mathrm{l}}{0.07065\times {10}^{-6}}=24.06\times {10}^{-2}\left(\mathrm{l}\right)\mathrm{ohm} \end{gathered}$$