Question

Q). A copper wire has dimensions 0.5 mm nd resistor of $1.6\times {10}^{-8}\Omega m$. What will bet he length of this wire to make the resistance 10 $\Omega$? How much does the resistance change if the diameter is doubled?

Answer 1

Dear Student,

The resistance of the wire of length 'l', area of cross section 'A' and resistivity 'ρ' is given by,

$R=\rho \frac{l}{A}$
Here, Diameter, d = 0.5mm = 0.5 x 10^-3m , then area of cross section, A = π(d/2)², R = 10 ohm, ​ρ = 1.6 x 10^-8 ohm .m.
So, the length of the wire,
$$\begin{gathered} l=\frac{RA}{\rho }=\frac{10\times 3.14\times {\left({\displaystyle \frac{0.5\times {10}^{-3}}{2}}\right)}^2}{1.6\times {10}^{-8}} \\ =\mathbf{122}\mathbf{.}\mathbf{72}\mathbf{}\mathit{m} \end{gathered}$$


Now if the diamter of the wire is doubled than, New resistance,
$$\begin{gathered} R\text{'}=\rho \frac{l}{A\text{'}}=\frac{1.6\times {10}^{-8}\times 122.72}{3.14\times {\left({\displaystyle \frac{2\times 0.5\times {10}^{-3}}{2}}\right)}^2} \\ =\mathbf{2}\mathbf{.}\mathbf{5}\mathbf{}\mathbf{}\mathit{o}\mathit{h}\mathit{m}\mathbf{.} \end{gathered}$$

Regards