Maximizing the conduction

Consider the star-mesh transformation that removes a valency $4$ vertex.

If the star has resistances $p,q,r,s$ as marked, then the equivalent mesh has resistances $R_{UV}$ between nodes $U$ and $V$ , where $R_{UV} \; = \; uv(a^{-1} + b^{-1} + c^{-1} + d^{-1}) \hspace{2cm} U,V = A,B,C,D\;,\; U \neq V$ Here $u = a,b,c,d$ according as $U = A,B,C,D$ respectively, and similarly for $v$ . If we place resistances $a,b,c,d,e,f,g,h,w,x,y,z$ into an octahedral array, then the effective resistance between a pair of opposite vertices is the effective resistance between $A$ and $D$ of the combination of the following three networks:

Applying the star-mesh transformation to the two "star" networks, and combining parallel resistors, we obtain that the octahedral array is equivalent to the tetrahedral array shown left in the diagram below, where

$\begin{aligned} U^{-1} & = \; \frac{1}{ad(a^{-1} + b^{-1} + c^{-1} + d^{-1})} + \frac{1}{wz(w^{-1} + x^{-1} + y^{-1} + z^{-1})} \\ V^{-1} & = \; \frac{1}{bc(a^{-1} + b^{-1} + c^{-1} + d^{-1})} + \frac{1}{xy(w^{-1} + x^{-1} + y^{-1} + z^{-1})} \\ W^{-1} & = \; \frac{1}{ab(a^{-1} + b^{-1} + c^{-1} + d^{-1})} + \frac{1}{e} + \frac{1}{wx(w^{-1} + x^{-1} + y^{-1} + z^{-1})} \\ X^{-1} & = \; \frac{1}{ac(a^{-1} + b^{-1} + c^{-1} + d^{-1})} + \frac{1}{f} + \frac{1}{wy(w^{-1} + x^{-1} + y^{-1} + z^{-1})} \\ Y^{-1} & = \; \frac{1}{bd(a^{-1} + b^{-1} + c^{-1} + d^{-1})} + \frac{1}{g} + \frac{1}{xz(w^{-1} + x^{-1} + y^{-1} + z^{-1})} \\ Z^{-1} & = \; \frac{1}{cd(a^{-1} + b^{-1} + c^{-1} + d^{-1})} + \frac{1}{h} + \frac{1}{yz(w^{-1} + x^{-1} + y^{-1} + z^{-1})} \end{aligned}$ and we can perform a $\Delta Y$ transform on the vertex $B$ , obtaining the equivalent circuit shown on the right-hand side of the above picture, where $U_1 \; = \; WY(V^{-1} + W^{-1} + Y^{-1}) \hspace{1cm} X_1 \; = \; WV(V^{-1} + W^{-1} + Y^{-1}) \hspace{1cm} Z_1 \; = \; YV(V^{-1} + W^{-1} + Y^{-1})$ Thus the effective resistance between the opposite vertices $A,D$ is $\Big[ \big(U^{-1} + U_1^{-1}\big) + \big[\big(X^{-1} + X_1^{-1}\big)^{-1} + \big(Z^{-1} + Z_1^{-1}\big)^{-1}\big]^{-1}\Big]^{-1}$ Running a computer programme to determine the least effective resistance of all the $12!$ possible arrangements of the resistances, we obtain the least resistance of $\boxed{2.2548}\;\Omega$ for the arrangement

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It is interesting to note that this optimal set-up has the four largest resistances in the equatorial ring, which has the approximate effect of insulating the longitudinal wires from each other. Thus we can approximate this optimal arrangement by a set of $3,12,48,192 \; \Omega$ resistors in parallel, making for an approximate effective resistance of $\big(3^{-1} + 12^{-1} + 48^{-1} + 192^{-1}\big)^{-1} = \tfrac{192}{85} = 2.2588 \; \Omega$ , which is a pretty close approximation to the actual value.

Apply Wheatstone bridge. And arrange parallel. Note, equivalent resistance, in parallel, is always less, than, the smallest individual resistance in parallel.