Resistance calculation -- Comparing coax cable insulator resistances

In summary, the conversation discusses the calculation of resistance between two coaxial cylinders with the same material and length. The formula used for the calculation is R = ρ*L/A, where ρ is the resistivity of the material. The correct calculation is R = ρ*ln(b/a)/2π, which results in R1 = R2. However, there is no matching choice among the given answers, leading to a discussion about the correct formula and calculation method.
  • #1
hidemi
208
36
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?

The answer is R.
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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  • #2
hidemi said:
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.
 
  • #3
haruspex said:
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.
 
  • #4
hidemi said:
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?
 
  • #5
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.
 
  • #6
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.
 
  • #7
hidemi said:
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.
 
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  • #8
Thanks for confirming my calculation!
 

1. What is resistance calculation?

Resistance calculation is the process of determining the resistance of a material, such as a coax cable insulator, to the flow of electrical current. It is typically measured in ohms (Ω) and is an important factor in determining the efficiency and performance of electronic systems.

2. Why is it important to compare coax cable insulator resistances?

Comparing coax cable insulator resistances allows us to determine which cable is more efficient in terms of minimizing signal loss. This is important in electronic systems, as higher resistance can lead to weaker signals and poor performance.

3. How do you calculate the resistance of a coax cable insulator?

The resistance of a coax cable insulator can be calculated using the formula R = ρ * (L/A), where R is resistance, ρ is the resistivity of the material, L is the length of the cable, and A is the cross-sectional area of the cable. The lower the resistance value, the more efficient the cable is at conducting electrical current.

4. What factors can affect the resistance of a coax cable insulator?

The resistance of a coax cable insulator can be affected by the material it is made of, the length and diameter of the cable, and the temperature. Higher temperatures can increase resistance, while different materials and sizes can have varying resistivities.

5. How can comparing coax cable insulator resistances help in selecting the right cable for a specific application?

By comparing coax cable insulator resistances, we can determine which cable is best suited for a specific application based on its efficiency in conducting electrical current. A lower resistance value indicates a more efficient cable, which can be beneficial in high-performance electronic systems where signal loss must be minimized.

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