Question

Two conductors are made of the same material and have the same length. Conductor $A$ is a solid wire of diameter $1mm$. Conductor $B$ is a hollow tube of outer diameter $2mm$ and inner diameter $1mm$.
Find the ratio of resistance $R_A$ to $R_B$.

Answer 1

$\text{Find resistance of wire}A.$

$\text{Formula used}:R=\frac{\rho l}{A}$

Let the conductor is made up of a material of resistivity $\rho$ and length of the conductor be $l$.

Diameter of the wire $A,d=1mm$

Radius of the wire, $r=\frac{d}{2}=0.5\times {10}^{-3}m$

Resistance of the wire $A$,

$R_A=\frac{\rho l}{\pi (0.5\times {10}^{-3}{)}^2}$

$\text{Find resistance of wire}B$.

$\text{Formula used}:R=\frac{\rho l}{A}$

Let the conductor is made up of a material of resistivity $\rho$ and length of the conductor be $l$.

Inner diameter of the wire $B,d_1=1mm$

Inner radius of the wire,

$r_1=\frac{d_1}{2}=0.5\times {10}^{-3}m$


Outer diameter of the wire $B$,

$d_2=2mm$

Outer radius of the wire,

$r_2=\frac{d_2}{2}=1\times {10}^{-3}m$

Resistance of the wire $B$,

$R_B=\frac{\rho l}{\pi \left(\right(r_2{)}^2-\left(r_1{)}^2\right)}$

$R_B=\frac{\rho 1}{\pi \left(\right(1\times {10}^{-3}{)}^2-(0.5\times {10}^{-3}{)}^2)}$


$\text{Find ratio of}{R}_A\text{and}{R}_B.$

Ratio of $R_A$ and $R_B$,

$\frac{R_A}{R_B}=\frac{\left(\right(1\times {10}^{-3}{)}^2-(0.5\times {10}^{-3}{)}^2)}{(0.5\times {10}^{-3}{)}^2}$

$\frac{R_A}{R_B}=\frac{3}{1}$

$\text{Final answer}:3:1$