Question

Two conductors are made of the same material and have the same length. Conductor A is a solid wire of diameter 0.79 mm. Conductor B is a hollow tube of outside diameter 1.4 mm and inside diameter 0.91 mm. What is the resistance ratio RA/RB, measured between their ends?

Answer 1

Answer:

Explanation:

Given

$D_a=0.79 mm$

$d_{ob}=1.4 mm$

$d_{ib}=0.91 mm$

and both have same length and material

$R=\frac{\rho L}{A}$

$A_a=\frac{\pi D_a^2}{4}$

$A_b=\frac{\pi d_{ob}^2\left ( 1-\left ( \frac{d_{ib}}{d_{ob}} \right )^2\right )}{4}$

$R_a=\frac{4\rho L}{\pi D_a^2}$

so $R\propto \frac{1}{D_a^2}$

Similarly $R_b\propto \frac{1}{d_{ob}^2\left ( 1-\left ( \frac{d_{ib}}{d_{ob}} \right )^2\right )}$

Thus $\frac{R_a}{R_b}=\frac{d_{ob}^2\left ( 1-\left ( \frac{d_{ib}}{d_{ob}} \right )^2\right )}{D_a^2}$

$\frac{R_a}{R_b}=\frac{1.4^2\times 0.5575}{0.79^2}$

$\frac{R_a}{R_b}=1.813$