# Resistance calculation -- Comparing coax cable insulator resistances

hidemi
Homework Statement:
A high-resistance material is used as an insulator between the conductors of a length of coaxial cable. The resistance material, which forms a hollow tube, has an inner radius a and an outer radius b, and the insulator provides a resistance R between the conductors. If a second insulator, made of the same material and having the same length, is made with double both the inner radius and the outer radius of the first, what resistance would it provide between the conductors?

Relevant Equations:
R = density*(Length/Area)
Since both cylinders have the same material, this implies the density is the same. Length is also the same based on the question. We only need to compare the radii for resistance.
R2 = pi ((2b)^2 - (2a)^2)
R1 = pi (b^2 - a^2)
R2/R1 = 4, so my answer is C. But, the given answer is A.
Where did I do wrong?

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R2 = pi ((2b)^2 - (2a)^2)
That's the area, but it is not the resistance.
If you don't have a formula then you will need to do an integration. Or cheat and look up resistance of an annulus.

hidemi
That's the area, but it is not the resistance.
If you don't have a formula then you will need to do an integration. Or cheat and look up resistance of an annulus.
Sorry I missed a "L"
Here's the correct calculation:
R2 = density * L / ((2b)^2 - (2a)^2)
R1 = density * L / (b^2 - a^2)
R2/R1 = 1/4
So, there is in fact no matching choice.

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Sorry I missed a "L"
Here's the correct calculation:
R2 = density * L / ((2b)^2 - (2a)^2)
R1 = density * L / (b^2 - a^2)
R2/R1 = 1/4
So, there is in fact no matching choice.
I think you have misunderstood either the question or the formula.
In R=Resistivity x length/area, the length is the length of the resistor in the direction of the current and the area is the cross sectional area, i.e. normal to the current. So you have calculated the resistance parallel to the coaxial core.
You are asked for the resistance between the core and the outer, i.e. for a current flowing radially.
You cannot simply apply that same formula because the cross section to the flow is a cylinder. Its area changes as its radius changes from core to outer. Consider a cylindrical shell, radius a<r<b, thickness dr. What is its area? What is its resistance to a radial flow?

Homework Helper
In the formula R = ρ*L/A, ρ is the resistivity, not the density. (It doesn't matter here as it cancels out, but it will matter in other situations.) The same symbol is used for both, as there are a lot more physical quantities than letters, but it's important to understand which is meant in which situation.

hidemi
Thanks for the reminder!
I am not sure if the following calculation is right?

lambda: density of the resistance material
dR = lambda * dr/ (2pi*r)
R1 = lambda * ln(2b/2a) / 2pi (because integrate from radius 2a to 2b)
R2 = lambda * ln(b/a) / 2pi (because integrate from radius a to b)
Therefore, R1 = R2.

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2022 Award
Thanks for the reminder!
I am not sure if the following calculation is right?

lambda: density of the resistance material
dR = lambda * dr/ (2pi*r)
R1 = lambda * ln(2b/2a) / 2pi (because integrate from radius 2a to 2b)
R2 = lambda * ln(b/a) / 2pi (because integrate from radius a to b)
Therefore, R1 = R2.
Yes, except that as @mjc123 posted, you mean resistivity, not density.

• hidemi
hidemi
Thanks for confirming my calculation!