Just The Series-Parallel Combinations

The possible combinations that are "series" on the highest level are $\begin{cases} R_1 = a+b+c+d & 1 \\ R_2 = (a|b)+c+d & 6 \\ R_3 = (a|b|c)+d & 4 \\ R_4 = (a|b)+(c|d) & 3 \\ R_5 = ((a+b)|c)+d & 12 \end{cases}$ with + for series, and | for parallel. The numbers indicate how many alternatives can be made be permuting $a, b, c, d$ . By creating the dual circuits (interchanging series and parallel) we double the number of possibilities. This accounts for $2\times 26 = 52$ different arrangements.

The bigger challenge is to prove that the 52 circuits can all have different values. This can be shown as follows.

Step 1 : Suppose all four resistors have the same value, $a = b = c = d = r$ . Then the resistance depends only on the general geometry of the circuit. For the five patterns given above we get $R_1 = 4r;\ \ \ R_2 = 2\tfrac12r;\ \ \ R_3 = 1\tfrac13r;\ \ \ R_4 = r;\ \ \ R_5 = 1\tfrac23r.$ The resistances of the dual circuits follow immediately: $R_1^\star = \tfrac14 r;\ \ \ R_2^\star = \tfrac25 r;\ \ \ R_3^\star = \tfrac34r;\ \ \ R_4^\star = r;\ \ \ R_5^\star = \tfrac35 r.$ Thus the different shapes of circuits have different resistors except those of the form $R_4 = (a|b)+(c|d) = r$ and $R_4^\star = (a+b)|(c+d) = r$ . However, an infinitesimal change in $a$ will guarantee that they are different. If we set $a = r+\epsilon$ , the resistances become $R_4 = (a|b)+(c|d) = \frac{4+3\epsilon}{4+2\epsilon}r;\ \ \ R_4^\star = (a+b)|(c+d) = \frac{4+2\epsilon}{4+\epsilon}r,$ which are clearly different.

Step 2 : Now make slight changes in the values so that $a < b < c < d$ , but small enough so that the overall resistance changes by less than $\tfrac16r$ . This guarantees that the conclusion from Step 1 is still valid.

Using $x|(y+z) < y|(x+z)\ \ \text{and}\ \ x+(y|z) < y+(x|z)\ \ \text{iff}\ \ x < y,$ we see that e.g. $a+b+(c|d) < a+c+(b|d) < a+d+(b|c), b+c+(a|d) < b+d+(a|c) < c+d+(a|b),$ and similar comparisons apply to the other geometries. In this example, we have no way to compare $a+d+(b|c)$ and $b+c+(a|d)$ ; however, when increasing $a$ infinitesimally to $a+\epsilon$ , the former value increases by $\epsilon$ while the latter increases by $d^2/(a+d)^2\epsilon < \epsilon$ . Thus we can always make small adjustments in the value of $a$ to guarantee the two are different.

In my investigation, there are $10$ ways to wire four resistors in different series-parallel combinations. Somehow, it is related to the number of partitions of 4, and each partition having two variants. I will be adding the pictures later as I am on my mobile device as of the moment.

4 has 5 partitions

4

3+1

2+2

2+1+1

1+1+1+1

If we treat the individual numbers as the number of resistors in a parallel subnetwork and the addition operation as placing these networks in series, we get 5 different patterns. If we interchange the treatments, we get another 5 patterns.

In each way, there are different number of permutations allowing for different values given differently valued resistors.

To lessen confusion, let us name the arbitrary resistors in each combination as $p, q, r$ and $s$ .

Way # 1. All resistors in series (form p+q+r+s.)

Here, no matter what arrangement of the resistors $A, B, C,$ and $D$ are, the equivalent resistance is the same.

= 1 combination

Way # 2. One parallel pair in series with the others (form p+q + r||s)

Here, we take into account the number of possible pairs from {A,B,C,D} which will be the parallel pair. That is $_4C_2 = 6$ . We no longer take the arrangements of the remaining resistors in series, the arrangement of the resistors in parallel, and the position in the parallel pair as it does not give a change of value in the net resistance.

= 6 combinations

Way #3. One resistor in series with a parallel network of two resistors in series and a resistor ( form p+ (q+r)||s)

There are 4 different ways of having the sole resistor in parallel, and with the remaining 3 resistors, there are $_3C_2 = 3$ ways of selecting the two resistors in series to be paralleled with the firstly chosen resistor. That gives us $3 \times 4 = 12$ combinations.

= 12 combinations.

Way #4. 3 resistors series in parallel with a resistor (form (p+q+r)||s)

There are 4 ways of selecting the sole resistor in the parallel pair. Thus there are four distinguishable resistance networks of this form with differing values.

= 4 combinations

Way #5. Two resistors in series in parallel with the other two resistors in series (form (p+q)||(r+s))

Here, there are just three configurations, as if we count $_4C_2$ ways of pairing the resistor in one branch of the network (the other being automatic), then each configuration is counted twice. Thus there are only 3 combinations of this form.

= 3 combinations.

Way #6. Three resistors in parallel with one another, in series with the fourth resistor (form (p||q||r) + s)

This is the direct complement of Way #4. (Notice how the "parallel" and "series" operators are swapped). Surprisingly, this also gives us just 4 ways, following the analysis given in Way #4.

= 4 combinations.

Way #7. A resistor in parallel with a parallel pair and a resistor in series [ form p||((q||r) + s) ]

Again, this is a direct complement of Way #3 . Counting the same way, we also get 12 combinations.

= 12 combinations.

Way #8. Three parallel networks (two resistors individually, and two resistors in series). (form p||q||(r+s) )

You could've guessed it. This is complementary to Way #2 . Again, there are 6 ways to have differently valued networks in this form.

= 6 combinations.

Way #9 . Two parallel networks in series. ( form (p||q) + (r||s) )

Similar to Way #5 , there are only three ways to have this form.

= 3 combinations.

Way #10. All in parallel with one another.

Well, this also is similar to Way #1 . No matter what order the resistors are, it gives the same net value, thus providing us just 1 combination.

= 1 combination.

So, summing it up, we get $\boxed {52}$ .