Question

Two conductors are made of the same material and have the same length. Conductor A is a solid wire of diameter 2.4 mm. Conductor B is a hollow tube of outside diameter 8.0 mm and inside diameter 4.0 mm. What is the resistance ratio RA/RB, measured between their ends?

Answer 1

`(R_(A) )/(R_(B) )` =
`(25)/(3)`

Step-by-step explanation:

For the two conductors A and B, the required general formula is;

R = ρl ÷ A

But since the two conductors are made of the same material,

ρA = ρB

Also, they have the same length,

lA = lB

So that,

`(R_(A) )/(R_(B) )` =
`(A_(B) )/(A_(A) )`

Area can be calculated by
`\pi r^(2)`.

Conductor A has diameter 2.4mm, thus its radius is 1.2mm. Conductor B has outside diameter 8.0mm of radius 4.0 and inside diameter 4.0mm of radius 2.0mm. Thus,

`(R_(A) )/(R_(B) )` =
`(\pi*(16 - 4)*10^(-6))/(\pi*1.44*10^(-6) )`

By appropriate divisions,

`(R_(A) )/(R_(B) )` =
`(12)/(1.44)`

=
`(1200)/(144)`

`(R_(A) )/(R_(B) )` =
`(25)/(3)`

The resistance ratio measured between their ends is
`(25)/(3)`.

Answer 2

RA / RB = 8.33

Step-by-step explanation:

The resistance in terms of area is given by the following equation:

R = d * L / A

the density being the same in both cases, L the length, which is the same in both conductors and A the area that it does vary.

Now, in the case of conductor A, the area would be:

AA = pi * rA ^ 2

we know that d / 2 = r, therefore:

AA = (pi / 4) * dA ^ 2

Replacing in the resistance formula:

RA = 4 * dA * L / (pi * d ^ 2)

In the case of B we have that the area we want to know is equal to the area on the outside minus the area on the inside

AB = pi * rBo ^ 2 - pi * rBi ^ 2

expressing in diameters:

AB = (pi / 4) * (dBo ^ 2 - dBi ^ 2)

Replacing in R

RB = 4 * d * L / (pi * (dBo ^ 2 - dBi ^ 2))

To know RA / RB we divide these two expressions, the term 4 * d * L is canceled, which is the same in both cases and we are left with:

RA / RB = (dBo ^ 2 - dBi ^ 2) / dA ^ 2

Replacing these values:

RA / RB = (8 ^ 2 - 4 ^ 2) /2.4^2

RA / RB = 8.33