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 hundredohms/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.