How to calculate resistivity of coaxial cylinder

In summary, the conversation discusses the calculation of the final resistivity along the axis of two tightly-attached coaxial cylinder, specifically in the example of a copper core wire coated with a layer of aluminum. The method involves using the cross-sectional area, length, and conductivity of both the core and the shield in a formula to calculate the resistance, and then adding together the resistance of the core and the shield. There is some debate about whether to add the resistances or the conductances, but it is determined that the two are in parallel. The conversation also touches on the difference between resistance and resistivity, and clarifies the setup of the coaxial cable.
  • #1
jacobier
2
0
Is it any model on resisivity of two tightly-attached coaxial cylinder? For example, a copper core wire is coated with layer of aluminum. How to calculate the final resistivity along axis?
 
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  • #2
There is a method on this but that's related to the radial resistance ( the insulator is consider imperfect).

The resistance along axis seems much simpler to me. Just plug in the cross-sectional area, length and conductivity of both the core and the shield in the related formula ( [itex]R=\frac{l}{σS}[/itex]), and add together the resistance of the core and the shield. Hopefully I'm not missing something!
 
  • #3
Hassan2 said:
add together the resistance of the core and the shield.

I don't think you mean add the resistances together. These are in parallel here, so it's the conductances that add.

Btw, in the OP, you mean resistance, not resistivity. Resistivity is a property of the material, independent of shape and size.
 
  • #4
Thank you very much. Actually I mean "resistance". I have been looking for solution for this for a long time. Probably the problem is not as complicated as I thought, no one have interest to talked about that.
 
  • #5
haruspex said:
I don't think you mean add the resistances together. These are in parallel here, so it's the conductances that add.

Actually what I had in mind is that the two , with a load connected to the end form a series circuit, that's why i added the " resistances" together. (OP asked about resistance along the axis.)
 
  • #7
Hassan2 said:
Actually what I had in mind is that the two , with a load connected to the end form a series circuit, that's why i added the " resistances" together. (OP asked about resistance along the axis.)
I get the feeling you have the wrong model for the set-up. There's a copper core and an Al coating. The axial current will consist of some current in each, in parallel. At each end there may be some radial flow, but I'm assuming we can ignore that.
 
  • #8
Haruspex is correct that the conductances add. To do it with resistance instead, separately calculate the resistance of a length of the copper core and the same length of hollow Al tube. The total resistance is the parallel combination of the two, using the usual formula for resistors in parallel.
 
  • #9
marcusl and haruspex,

In the attached figure, aren't the series resistances of the core and the shield added together to give the cable resistance? Sorry I understand that this is very simple question but I would like to know what I am not getting.

Thanks.
 

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  • #10
Ah - it was me that had the wrong set-up in mind. Yes, they're in series.
 
  • #11
Hassan2 said:
marcusl and haruspex,

In the attached figure, aren't the series resistances of the core and the shield added together to give the cable resistance? Sorry I understand that this is very simple question but I would like to know what I am not getting.

Thanks.

But this is a coaxial cable and not what is described in the OP.
The copper core is not "coated with layer of aluminum" but they are separated by insulator. The OP describes a two tightly attached coaxial cylinders.
 
  • #12
You are right nasu. I got it wrong from the beginning. They are parallel then. Many thanks.
 

Related to How to calculate resistivity of coaxial cylinder

1. What is the formula for calculating the resistivity of coaxial cylinders?

The formula for calculating the resistivity of coaxial cylinders is ρ = (ln(b/a))/(2πL), where ρ is the resistivity, b is the outer radius of the cylinder, a is the inner radius of the cylinder, and L is the length of the cylinder.

2. What are the units of resistivity for coaxial cylinders?

The units of resistivity for coaxial cylinders are ohm-meters. This is derived from the formula ρ = RA/L, where R is the resistance, A is the cross-sectional area, and L is the length of the cylinder. Therefore, the units of resistivity are ohms (Ω) divided by meters (m).

3. How do I measure the inner and outer radii of a coaxial cylinder?

The inner and outer radii of a coaxial cylinder can be measured using a caliper or a ruler. Place the measuring tool directly on the surface of the cylinder and take the measurement in either millimeters or centimeters. Be sure to measure the inner radius from the inside edge of the cylinder and the outer radius from the outside edge.

4. Can I use the same formula to calculate the resistivity of any coaxial cylinder?

Yes, the formula ρ = (ln(b/a))/(2πL) can be used to calculate the resistivity of any coaxial cylinder, regardless of the material. This formula is based on the dimensions of the cylinder and does not depend on the material properties.

5. How accurate is the calculated resistivity for coaxial cylinders?

The calculated resistivity for coaxial cylinders can be very accurate if the measurements of the inner and outer radii are precise. However, there may be slight variations due to imperfections in the cylinder's shape or material properties. It is always best to take multiple measurements and calculate the average resistivity for more accurate results.

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