Question

Two conductors are made of the same material and have the same length. Conductor A is a

solid wire of diameter 1.0 mm. Conductor B is a hollow tube of outside diameter 2.0 mm

and inside diameter 1.0 mm. What is the resistance ratio RA/RB, measured between their

ends?

Answer 1

The resistance in terms of the area and the length of the wire is given by:

$R = \frac{ρL}{A}$

if we have two wires, the first one is a solid wire with a diameter of $d_A=1 \times 10^{-3} \;m$ , and the second one is a hollow wire with inner diameter of $d_{B,i}=1 \times 10^{-3} \; m$ and outer diameter of $d_{B,o}=2 \times 10^{-3} \;m$ , so the cross sectional area of the first wire is:

$A_A = \pi R^2_A = \frac{\pi d^2_A}{4}$

Hence the resistance is:

$R_A = \frac{4ρL_A}{\pi d^2_A}$

The area of the second wire is:

$A_B = \pi r^2_{B,o} - \pi r^2_{B,i} = \frac{\pi}{4}(d^2_{B,o} -d^2_{B,i})$

Hence the resistance is:

$R_B = \frac{4ρL_B}{\pi(d^2_{B,o} -d^2_{B,i})}$

The ratio between the resistances:

$\frac{R_A}{R_B} = \frac{(d^2_{B,o} -d^2_{B,i})L_A}{d^2_AL_B}$

The wires have the same length, therefore:

$\frac{R_A}{R_B} = \frac{d^2_{B,o} -d^2_{B,i}}{d^2_A} \\ = \frac{(2 \times 10^{-3})^2 -(1 \times 10^{-3})^2}{(1 \times 10^{-3})^2} = 3 \\ \frac{R_A}{R_B} = 3$