Suppose you have 30 identical 1 Ω resistors connected to form the edges of a regular icosahedron? What, then, is the effective resistance between adjacent vertices? What about the effective resistance between adjacent vertices of a regular dodecahedron formed of 30 identical 1 Ω resistors?
For both, we can make heavy use of the symmetries of the polyhedra. First, for the icosahedron, let us number two adjacent vertices as vertex 1 and vertex 2. Now, let us consider the situation where a current of 
A enters the icosahedron through vertex 1, and a current of 
A leaves the icosahedron through each of the other 11 vertices (why these values will become apparent later). Then, we note that since the icosahedron has five-fold symmetry about an axis through opposite vertices, we see this setup has five-fold symmetry about vertex 1, and thus the five resistors connected to vertex 1 must each carry the same amount of current; and thus a current of =\frac{11}{60}\:\text{A})
flows through each of these resistors, one of which is the one connecting vertices 1 and 2. Thus, the voltage difference between vertices one and two is, via Ohm’s law,
\left(1\:\text{\Omega}\right)=\frac{11}{60}\:\text{V})
.
Next, consider a second situation, with a current of A leaving via vertex 2, and a current of A entering through each of the other 11 vertices. Again, due to the same symmetry, a current of 
A flows through each of the five resistors connecting at vertex 2, and so the voltage difference between vertices 1 and 2 is again
\left(1\:\text{\Omega}\right)=\frac{11}{60}\:\text{V})
.
We can then superimpose these two setups. Adding, we see that the currents entering and leaving cancel for all vertices except 1 and 2, leaving us with a current of 
entering through vertex 1, leaving through vertex 2, and a potential difference of 
.
Now, since 
, we see
.
For the dodecahedron, we can use a similar setup. First, let 
flow in through vertex 1 and 
flow out through each of the other 19 vertices. There are three edges meeting at each vertex of a dodecahedron, and due to the symmetry, the three resistors connected to vertex 1 will each carry a current of =\frac{19}{60}\:\text{A})
, which gives a voltage difference of \left(1\:\text{\Omega}\right)=\frac{19}{60}\:\text{V})
. Next, we consider flowing out through vertex 1 and flowing in through each of the other 19 vertices. As before, we have 
through each resistor connecting to vertex 2, and thus a voltage difference of 
. Superimposing, we obtain a situation with 1 A entering vertex 1, 1 A leaving vertex 2, no current entering or leaving the dodecahedron through the other vertices, and a potential difference between vertices 1 and 2 of
,
and thus an equivalent resistance of
.
Note that in both cases, if you label the number of edges E and the number of vertices by V, then the equivalent resistance across an edge is 
. What then, might the equivalent resistance be across one edge of a cube of 1 Ω resistors? A tetrahedron?
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Tags: Dodecahedron, Equivalent Resistance, Friday Physics, Icosahedron, Ohm's Law, physics
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