Question

A copper wire has diameter 0.5mm and resistivity of $1.6\times {10}^{-8}\Omega m$. What will be the length of this wire to make its resistance $10\Omega$? How much does the resistance change if the diameter is doubled?

Answer 1

Given, diameter, $d=0.5mm$

resistivity, $\rho =1.6\times {10}^{-8}\Omega m$

Resistance, $R=10\Omega$

Let the length of wire be $l$.

$A=\pi d^2/4$

$R=\rho l/A$

$=\frac{\rho l}{\pi d^2/4}$

$⟹l=\frac{R\pi d^2}{4\rho }$

$l=\frac{10\times 3.14\times (0.5\times {10}^{-3}{)}^2}{4\times 1.6\times {10}^{-8}}$

$l=122.7m$

$R\propto 1/d^2$

If the diameter is doubled, resistance will be one-fourth.

Hence, new resistance $=2.5\Omega$