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

Step-by-step 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=(\rho L)/(A)`

`A_a=(\pi D_a^2)/(4)`

`A_b=(\pi d_(ob)^2\left ( 1-\left ( (d_(ib))/(d_(ob)) \right )^2\right ))/(4)`

`R_a=(4\rho L)/(\pi D_a^2)`

so
`R\propto (1)/(D_a^2)`

Similarly
`R_b\propto (1)/(d_(ob)^2\left ( 1-\left ( (d_(ib))/(d_(ob)) \right )^2\right ))`

Thus
`(R_a)/(R_b)=(d_(ob)^2\left ( 1-\left ( (d_(ib))/(d_(ob)) \right )^2\right ))/(D_a^2)`

`(R_a)/(R_b)=(1.4^2* 0.5575)/(0.79^2)`

`(R_a)/(R_b)=1.813`