There are a number of techniques that can be employed to analyse in an efficient way complex passive networks by means of paper-and-pencil calculations. Of course, for a computer, matrix methods are the way to go.

Here are a few:

- For certain type of networks (e.g., series-parallel networks, ladder networks), one can apply the
*regula falsi* (*false position*): you start from a convenient resistor (e.g., the last one in a ladder network) and you assume that the current in that resistor is an arbitrary value, e.g., 1A; from this, you can walk the network up to the input terminals to find a voltage and a current whose ratio yields the equivalent resistance. This method is also useful to calculate the current in the last resistor knowing the voltage or the current at the input: you just have to take a proportion.
- Middlebrook's extra element theorem (EET) [1,2] and its extension, the $n$EET [3,4]. These are lesser known theorems of network theory which allow to find solutions in so-called "low-entropy" form, that is, a readable form evidencing the most significant terms. I gave an example of outcome from this theorem in this answer. They are not specific to the calculation of equivalent resistances -- they are much more general, in fact -- but they can be also efficiently applied in this case. In the references given below there are examples of applications.
- Bartlett's bisection theorem. This theorem, also, is not specific to the calculation of equivalent resistances, but sometimes it can be applied to this aim when the network has certain symmetries.
- Exploit the symmetry of the network, especially when you have networks with equal resistances. This is probably what you mean in your points 1 and 2. There exists also a number of works which give a systematic treatment of symmetrical networks, I'll add the references as soon as I find them in my messy folders.

[1] R. D. Middlebrook, "Null Double Injection and the Extra Element Theorem", *IEEE Trans. Edu.*, 32, 167-180, 1989 online copy

[2] V. Vorperian, *Fast Analytical Techniques for Electrical and Electronic Circuits*, Cambridge University Press, 2011.

[3] R. D. Middlebrook, V. Vorperian, J. Lindal, "The N Extra Element Theorem", *IEEE Transactions on Circuits and Systems I: Fundamental Theory and Applications*, 45, 919-935, 1998.

[4] R. W. Erickson, "The $n$ Extra Element Theorem", online.

# Some background on the handout

The scattering matrix describes how the signal at one port of a multi-port circuit is the linear sum of the signals at the other ports. In general, for a 4 port system you would need a 4x4 matrix, 16 terms. However, when you have (lots of) symmetry, then the number of independent variables quickly goes down.

In its most general form, it would be

$$S = \left[\begin{matrix}S_{11} & S_{12} & S_{13} & S_{14}\\
S_{21} & S_{22} & S_{23} & S_{24}\\
S_{31} & S_{32} & S_{33} & S_{34} \\ S_{$!} & S_{42} & S_{43} & S_{44} \end{matrix}\right]$$

And the signal $b_i$ at a given port $i$ that results from inputs $a_j$ at all ports is given by:

$$b_i = \sum S_{ij} a_j$$

(see this link fro a more in depth explanation of S parameters).

As you add symmetry, different terms end up having to have the same values. This simplifies the set of equations you have to solve to come up with the complete "transfer function" describing the behavior of the circuit. In the notes you quote, they claim (without proof) that when you have D4 symmetry, the scattering matrix reduces to 3 independent variables,

$$S = \left[\begin{matrix}A&B&B&C\\
B&A&C&B\\
B&C&A&B \\
C&B&B&A\end{matrix}\right]$$

One thing about microwave circuits (compared to simple resistor networks) is that you might not even have the most basic symmetry - if you use a *circulator* network, a signal going into port 1 might appear as an output at port 2, but a signal injected in port 2 might appear on port 3 (and not 1)... because phase shifts across the network play a major role in microwaves. That doesn't happen with a purely resistive network, but it explains why microwave circuits need more advanced analysis techniques. Which are sometimes useful even for "regular" circuit analysis.

All of that is to give you some background on the handout you were looking at. Now let's get back to your original question.

These matrix methods are probably too advanced for your students - although they are likely to do "something like this" without realizing it.

Two things in particular I think are worth noting as "tricks" to solve these problems.

# Tricks to deal with with circuits involving "left/right" and "up/down" symmetry

When you look at the symmetry of a circuit, you can divide it into "left/right" and "up/down" symmetry. I say this in reference to the direction of current, assuming that you have current flowing from "top to bottom". For symmetrical circuits, there are sometimes two branches that the current can choose, with the same current flowing in each (because of symmetry). When that happens, you can remove half the circuit, and halve the values of the resistors in the other side. Any resistors that were connecting the two nodes would have no current flowing in them (as they are at the same voltage) - and these can be removed (like in the Wheatstone bridge). This is how you handle left/right symmetry.

Regarding up/down symmetry, if your circuit looks the same when you flip it "vertically" (that is, if current flowed from A to B and you swapped A and B, could the circuit look the same?), then there will be corresponding resistors in the "top" and "bottom" half of the circuit that must have the same current in them. When you write down Kirchhoff's laws for the currents in your circuit, this means that you can use the same symbol for these two currents. Again, this reduces the number of equations you have to solve.

## Illustration of "left/right" and "up/down" symmetry

Let's take your pyramid circuit as an example, and let's solve for the resistance from A to C (because of rotational symmetry, AB and BC should give the same result). If I redraw the circuit so A is at the top and C at the bottom

I can see that B and E must be at the same potential (there is "up-down" symmetry for both the network that connects AB and BC, and the network that connects AE and EC). That means we can remove the $R_2$ resistor between B and E, and now we have a set of circuits each of which we can solve:

ABC: just two $R_1$s in series

AC: parallel of $R_1$ and a circuit consisting of AD, DF, and FC in series, where

DF: parallel of 2x $R_1$ in series, and a single $R_1$

Looking at it differently, you can say that $V_{AD} = V_{FC}$ by symmetry, and similarly $V_{AB}=V_{AE}$.

Note - these matrix equations could be used "directly" for these kinds of problems, but I'm afraid that they tend to start from the idea that you have simplified the resistor network between nodes before applying the later simplifications due to symmetry.

If this didn't help you at all, I apologize - I'm willing to try to make this clearer if you ask clarifying questions.

## Best Answer

Basically the line of symmetry tells you that resistance on both sides are equal. Since current splits inversely to the ratio, but here, the ratio is 1:1, so current splits 1:1. This means that the same amount of current goes to both sides, so they are equipotential. My explanation here probably doesn't sound very "sciency", but I hope it is understandable. There is a rule of thumb when solving circuits problem, which is to look for the line of symmetry, which can help simplify a complcated circuit(as shown in your post).