Derivation of resistance of coaxial cable

As given here, the derivation of Resistance for Coaxial Cables

Consider a coaxial cable of length $$\L\$$, consisting of a cylindrical conductor of radius a surrounded by a cylindrical conducting shell of radius $$\b\$$. The space between the conductors is filled with an insulating material.
The resistance along the length of the cable is considerably smaller than the resistance between the inner and outer cylinders. Consider current passing through a sequence of cylindrical shells of radius r and thickness dr. Each shell has a resistance $$\dR\$$ given by $$dR = \frac{\rho}{2\pi rL} dr$$ Integrating from $$\r=a\$$ to $$\r=b\$$ to find the total resistance gives: $$R = \int_a^b dR = \frac{ρ}{2\pi L}\int_a^b\frac1rdr$$ Hence $$R = \frac{ρ}{2\pi L} \ln\bigg(\frac ba\bigg)$$ Generally this resistance is several hundred ohms/m to minimize the "leakage current" that passes through the insulating material between the conductors.

The thing is, I can't seem to understand it but I do know that it did explain it completely. Maybe someone else reading would understand it. I just want to ask if someone could explain it in more detail? Like you can totally use the same variables and derivation as shown in the picture, just explain it differently. That way, I can look back on the source and kind of get what is happening from the more detailed explanations.

What I am familiar with is the basic formula for resistance
$$R = \frac {ρL}{A}$$ Based on the information, my face value thought on how the variables were represented are

• Resistance : $$\R \rightarrow dR\$$
• Resistivity : $$\\rho \rightarrow \rho\$$
• Length : $$\L \rightarrow dr\$$
• Cross Section Area : $$\A \rightarrow 2\pi rL\$$

Well I think it's wrong somewhere. I think the $$\dr\$$ is suppose to be associated with the cross section area. So I would appreciate it if the analogs from the original formula could be explained. I can take over the solving from there, since I am familiar with integral formulas and the $$\dr/r\$$ definitely results to the answer having a natural logarithm. $$\a\$$ is the lower limit, the radius of the cable and $$\b\$$ is the upper limit which is the radius of the cable including the insulation.

• The professor wants you to find the resistance of the INSULATOR between the center conductor and the outer conductor/shield. This is so unusual that it has thrown you for a loop. I had to read it three times myself. The problem is that the cross sectional area is not uniform moving outward from the inner conductor to the outer conductor. So you have to integrate using shells. Been too long since I've done calculus to bother with that. I would set up a grid in an excel spreadsheet and do it numerically if I had to figure it out. Jan 17, 2021 at 3:07
• Another approach would be to conceptually unroll the insulator into a sheet. Calculate the resistance from one plate to another passing through the sheet. For the width use the geometric average (sqrt(Ci * Co)) where Ci is inner circumference and Co is outer circumference. I think this is an approximation but it saves you from doing the integral. Jan 17, 2021 at 3:11

3 Answers

Starting from the basic formula for resistance:

R = $$\\rho L/A\$$ consider the concentric cylindrical shells, and cable length G

We want to calculate the resistance from one side to the other of a thin cylindrical shell of length G and thickness dr (current passing radially through the shell).

So L in this case is the infinitesimal radius change dr

We will use the thin slice in the radial direction because the radius changes as we go from inside to outside, and thus the resistance of a slice of the same thickness dr decreases as we go outward, and we want to integrate along that path to find the total resistance. We are integrating along the path the current follows from inside conductor to outside conductor. There is assumed negligible resistance along the center and outer conductors.

A is the length of the cable G multiplied by the circumference, which is $$\2 \pi r G\$$.

so $$\dR = \rho dr/(2\pi r G)\$$

Pulling the constants out of the integration we have:

The resistance R = $$\\frac{\rho}{2\pi G}\int_a^b\frac{1}{r}dr\$$

The definite integral we know to be ln(b)-ln(a) = ln(b/a), and we get the solution from the site.

• So let me clear this out. I'm gonna try to make a dumbed down explanation of the analog representations. Basically, for regular wires the two axis for cross section used are the ones that form a circle, and the remaining axis is used for length. In this one, only one of the original axes from the circle is used for the cross section and the other is the one originally used for the length. This is why the area seems to represent a lateral surface of a cylinder, being 2πrh or 2πrG by your representation. This also explains why radius is used. Jan 17, 2021 at 5:48
• Yes. Radius is used because the resistance of each slice varies by radius so we want to integrate along the radius to get the total. Diameter could also be used, with a few minor changes. Jan 17, 2021 at 5:51
• But I still have some questions. 1) Why bother doing the change of axis? I know how it has to do with it being a concentric cylindrical shell, but I don't get exactly why. I know you said something about current passing radially through the shell, but I'd love a more explicit explanation why doing such a drastic change from the usual axis convention done on wires. Jan 17, 2021 at 5:52
• 2) Why is the radius used a differential and not exact? Is it because the radius is not constant? Let me clarify, are getting the resistance of the insulation only right? Because I'm not sure whether the changing radius visually refers to the inner circle or the annulus formed by the insulation. I'd appreciate it if your answer was an edit from the original post because it might get lengthy if you try comparing it to original conventions in an intuitive way. Jan 17, 2021 at 5:53
• Edited. Note that if the conductivity $\rho$ changed along the path of an ordinary cylindrical wire ( $\rho(x)$) we would integrate along the length of the wire to capture that change and calculate the total resistance. Jan 17, 2021 at 6:06

The statement "Generally, this resistance is several hundred ohms/m to minimize [leakage]" is misleading in at least two ways.

First is the problem of units. The unit "ohms/m" implies there is a resistance that increases with distance. However, leakage resistance decreases with distance. The unit should either be ohm-meters, or what is more common, conductance per unit length, such as S/m.

Second, referring to the resistance as several hundred ohms/m, even if the units were correct, grossly understates the magnitude of the resistance. $$\10^{15}\$$ ohm-m or $$\10^{-15}\$$S/m would be closer to the leakage resistance of a typical coaxial cable.

As comments have pointed out, the calculation of the conductance/length from the resistivity of the insulating material requires integrating over an area, but the formula is given in the problem.

• Hmm are you sure about 1e15 ohm m ? I have the impression that the cable leakage is considered troublesome already for measurements with DUTs in the higher gigaohm range. or is it actually the cable capacitance which becomes the problem due to extremely long RC ? May 22 at 6:44
• @tobalt Many coax manufacturers don't list the leakage conductance in their specifications. HELUKABEL helukabel.com/cnen/products/… however lists a value for "minimum insulation resistance" of 10^5 Mohms-km. (The web version looks like 105, but the PDF datasheet clearly shows the 5 in an exponent position, i.e. 10^5). Which works out to 10^14 ohms-meter. What is not clear from spec is whether this refers to resistance of dielectric or outer sheath. From the context I would suspect the former, but I'm not positive. Will search more manufacturers. May 22 at 12:21

OK, so the normal formula for resistance, or the way you find it, is to integrate along the path that the current takes. If current flows in the +x direction, you integrate over x. Each section you integrate has a resistance of rho / A * dx. A is the cross-section area of the conductor. USUALLY in these problems, A is uniform. A constant. That can be pulled out in front before you do the integral. So it is barely an integral at all and you just multiply rho by the extent in x, and divide by the area. So usually Rtotal is just rho * length / area.

But in our case, we have a non-uniform cross-section so the integral is more complicated. Also, since current is not flowing along length L, we end up using L in a different place, which is also confusing.

So, what are the analogs? Instead of dx, we have dr because current is flowing outward radially. Instead of A we have circumference * length. So that is A = (2 * pi * r * L).

So now our expression takes shape. It is just:

Rshell = (rho / (2 * pi * r * L) ) dr

You are just going to integrate from r = a to r = b. Everything is constant except for dr/r, so rho / (2 * pi * r * L) gets pulled out in front. If we assume L is 1 meter, then it disappears.

Not sure if that is making any more sense for you. The key is that you are integrating along the path of the current, and you divide by the cross-sectional area of that path.

So far I have not learned mathjax. Feel free to fix it up.