Derivation of resistance of coaxial cable

Question

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.

Answer 1

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.

Answer 2

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.

Answer 3

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.