Question

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

Answer 1

We know that the resistance of wire is `R = ρ l/A` where A is cross-sectional area of conductor and ρ is the specific resistance or resistivity and L is the length of conductor.

The resistance of first conductor

`R_A = (ρl)/(pi(10^-3 xx 0.5)^2`

The resistance of second conductor,

`R_B = (ρl)/(pi[(10^-3)^2 - (0.5 xx 10^-3)^2]`

Now, the ratio of two resistors is given by

`R_A/R_B = ((10^-3)^2 - (0.5 xx 10^-3)^2)/(0.5 xx 10^-3)^2` = 3:1