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I'm trying to do a problem for revision and I'm running into problems calculating the resistance of this circuit used to approximate a twin-core wire. Full Network

It's used later in the question with the middle resistors being treated as a plastic layer, but I'm struggling with the first part of the question which is to calculate the total resistance R across the terminals on the left. What I tried is to treat the sections individually, that is:

This section has resistance \$ R_1 = 2r_a + r_b \$. I then considered adding another loop to this as essentially adding another resistor \$R_1\$ in parallel to the first loop, so this would have resistance $$\frac{1}{R_2} = \frac{1}{R_1} + \frac{1}{R_1} $$ so $$R_2 = \frac{2r_a + r_b}{2}$$

Following this logic, the resistance of \$n\$ loops would be $$R_n = \frac{2r_a + r_b}{n}$$ I don't think this makes sense since for \$n \rightarrow \infty\$, \$R_n \rightarrow 0\$. Am I missing something with this? I assume I did something wrong with the parallel/series treatment but I'm not sure.

The full question is:

The network of resistors shown in the Figure below contains a large number of identical segments and has an overall resistance \$R\$ as measured across the terminals on the left. Segments are added to the front of the network until eventually adding extra units makes no difference to \$R\$. Find \$R\$ in terms of the resistances \$r_a\$ and \$r_b\$.

1: Use the result to estimate the ultimate resistance of an infinitely-long twin-core cable made of copper wires (area 1.2mm, conductivity \$σ = 6 \cdot 10^7Ω^{−1}/m^{−1}\$). The 2mm gap between the wires is completely filled with plastic (resistivity \$ρ = 10^{13}Ω/m\$). Assume that the effective dimensions of the plastic layer are 1mm x 2mm.

2: *Roughly estimate the length of cable needed to establish this resistance.

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  • \$\begingroup\$ Are you sure that from the two left terminals one is input and one is output? In that case, yes, adding an infinite number of parallel resistors is going towards zero. But I think those two are meant to be input terminals and the "ra" are the two conductors and the rb are the contact resistances. At least that is the only thing I can currently make sense of to mean "twin core wire" in the context of that schematic \$\endgroup\$ – PlasmaHH May 24 '16 at 10:41
  • \$\begingroup\$ I'm unsure also about the question exactly, but it specifically asks for the resistance across the left terminals so I guess that it tending towards zero is the answer. I'll paste the question below. \$\endgroup\$ – bigbadpiano May 24 '16 at 10:49
  • \$\begingroup\$ ah ok, now I think the misconception here is to see all "additional loops" as being in parallel to the original one. What is really meant that they are in parallel to the r_a resistor, so for every r_a resistor you have to replace it in your calculation with the "additional loop" network, and repeat this enough times \$\endgroup\$ – PlasmaHH May 24 '16 at 10:57
  • \$\begingroup\$ So you mean that additional loops are only in parallel with the first \$r_a\$? Meaning the resistance of, say, 3 loops is \$1/R_3 = 1/r_a + 1/(r_b + r_a) + 1/R_1 + 1/R_1 \$? Sorry, I'm a physics student and circuits can really confuse me sometimes \$\endgroup\$ – bigbadpiano May 24 '16 at 11:05
  • \$\begingroup\$ Maybe redraw them in an arrangement that shows the parallel/series relationship better? Or try using an alternative notation that is somewhat in-between the real formula: ra + (rb || ( ra + (R1 || rb) + ra)) + ra \$\endgroup\$ – PlasmaHH May 24 '16 at 11:13
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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: -

schematic

simulate this circuit – 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}}\$

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  • \$\begingroup\$ Okay thanks, that seems a lot simpler than considering each segment incrementally \$\endgroup\$ – bigbadpiano May 24 '16 at 11:25
  • \$\begingroup\$ In the proof you have, the resistance \$R\$ is only on the top wire, does the presence of the second resistor on the bottom mean that in this case \$R\$ is just \$2r_a\$? \$\endgroup\$ – bigbadpiano May 24 '16 at 11:41
  • \$\begingroup\$ The series resistance in my formula is \$2r_a\$ \$\endgroup\$ – Andy aka May 24 '16 at 11:42
  • \$\begingroup\$ Not so sure that characteristics impedance equals the impedance of the cable, IMO they are two different things. Simple proof: You apply a short circuit at the end of transmission line and the whole impedance is not 50ohm + R_short. \$\endgroup\$ – Marko Buršič May 24 '16 at 12:49
  • \$\begingroup\$ @MarkoBuršič where does the 50 ohm bit come from? Anyway, one of the questions is for an infinite long cable (find its input impedance) and that will certainly be \$\sqrt{R/G}\$ at DC. \$\endgroup\$ – Andy aka May 24 '16 at 13:24

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