Question

Two conductors are made of the same material and have the same length. Conductor A is a solid wire of diameter 1 m. Conductor B is a hollow tube of inside diameter 1 m and outside diameter 2 m. The ratio of their resistance, RA/RBRA/RB, is:

Answer 1

Ra / Rb = 3

Step-by-step explanation:

The resistance is defined by the expression

r = ρ L / A

where A is the area of ​​the wire and L is the length

let's find the area of ​​each wire

a) solid wire (A) of d₁ = 1 m

r = d₁ / 2 = 0.5 m

A = π r²

A = π 0.5²

R_a =
`(\rho \ L)/(\pi \ 0.5^2)`

R_a = 4 \frac{\rho \ L}{\pi }

b) hollow tube (B) of d₁ = 1 m and d₂ = 2m

the radios are

r₁ = 0.5 m

r₂ = 1 m

the area is

A = π r₂² - π r₁²

A = π (r₂² - r₁²)

A = π (1² - 0.5²)

A = π 0.75

resistance is

R_b =
`( \rho \ L )/(\pi \ 0.75)`

R_b = 1.33 \frac{ \rho \ L }{\pi }

therefore the relationship between these two resistances is

Ra / Rb = 4 / 1.33

Ra / Rb = 3