Resistance of twin core wire
Regard it as a transmission line problem and from that, we know that the characteristic impedance is: -

\$Z_0 = \sqrt{\dfrac{R + j\omega L}{G + j\omega C}}\$

It's not too tricky to prove this - see my answer here

So for DC this reduces to \$Z_0 = \sqrt{\dfrac{R}{G}}\$

R is series resistance per unit length and G is parallel conductivity per unit length. For example if R is 1 ohm per metre and G is 1 micro siemen per metre then the characteristic impedance is 1000 ohms.



Looks like I'll have to do the proof. Imagine one section of the line having R series ohms and 1/G parallel Mohms. If this section is repeated to infinity the impedance looking into the 1st section is the same impedance as if the 1st section is discarded and you looked into the 2nd section.

From this simple and straightforward observation you can say: -

\$Z_{IN} = R + \dfrac{1}{G} || Z_{IN}\$. In other words this: -







simulate this circuit &ndash; Schematic created using CircuitLab

So, \$Z_{IN} = R + \dfrac{\frac{Z_{IN}}{G}}{Z_{IN} + \frac{1}{G}} \$

\$Z_{IN} = R + \dfrac{Z_{IN}}{1+Z_{IN}\cdot G}\$

\$Z_{IN} + Z_{IN}^2\cdot G = R + Z_{IN}\cdot G\cdot R + Z_{IN} = R(1 + Z_{IN}\cdot G) + Z_{IN}\$

As the sections of cable are made infinitely small, \$Z_{IN}\cdot G\$ becomes an insignificant term (right hand side of the equation) hence we are left with: -

\$Z_{IN}^2\cdot G = R\$ or \$Z_{IN} = \sqrt{\dfrac{R}{G}}\$