Question

A copper wire has a diameter of $0.5mm$ 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

Step 1:Given data

Diameter of wire$d$=$0.5mm$=$5\times {10}^{-4}m$

The resistivity of wire$\rho$ =$1.6\times {10}^{-8}\Omega m$

Resistance $R=10\Omega$

Let the length of the wire be $l$

Let the cross-sectional area of the wire be $A$

Step 2: Calculating new resistance

$A=\frac{\pi d^2}{4}$

$$\begin{gathered} R=\frac{\rho l}{A} \\ l=\frac{\rho R}{{\displaystyle \raisebox{1ex}{$\pi d^2$}\!\left/ \!\raisebox{-1ex}{$4$}\right.}}=\frac{R\pi d^2}{\rho 4} \\ =\frac{10\times 3.14\times {(5\times {10}^{-4})}^2}{1.6\times {10}^{-8}\times 4}=\frac{5\times 31.4\times 25}{16\times 2}=122.7m \end{gathered}$$

Length to make resistance as $10\Omega is122.7m$

$R\alpha \frac{1}{d^2}$

Therefore, when the diameter is doubled, resistance will be one-fourth of the original resistance

Hence, new resistance will be $\frac{1}{4}\times 10=\frac{5}{2}=2.5\Omega$