Physics Friday 105

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 \frac{11}{12} A enters the icosahedron through vertex 1, and a current of \frac1{12} 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 \frac15\left(\frac{11}{12}\:\text{A}\right)=\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,
Next, consider a second situation, with a current of \frac{11}{12} A leaving via vertex 2, and a current of \frac1{12} A entering through each of the other 11 vertices. Again, due to the same symmetry, a current of \frac{11}{60} A flows through each of the five resistors connecting at vertex 2, and so the voltage difference between vertices 1 and 2 is again

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 \frac{11}{12}\:\text{A}+\frac{1}{12}\:\text{A}=1\:\text{A} entering through vertex 1, \frac{11}{12}\:\text{A}+\frac{1}{12}\:\text{A}=1\:\text{A} leaving through vertex 2, and a potential difference of V_1-V_2=\frac{11}{60}\:\text{V}+\frac{11}{60}\:\text{V}=\frac{11}{30}\:\text{V}.
Now, since V_1-V_2=IR_{eq}, we see

For the dodecahedron, we can use a similar setup. First, let \frac{19}{20}\:\text{A} flow in through vertex 1 and \frac{1}{20}\:\text{A} 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 \frac13\left(\frac{19}{20}\:\text{A}\right)=\frac{19}{60}\:\text{A}, which gives a voltage difference of V_1-V_2=\left(\frac{19}{60}\:\text{A}\right)\left(1\:\Omega\right)=\frac{19}{60}\:\text{V}. Next, we consider \frac{19}{20}\:\text{A} flowing out through vertex 1 and \frac{1}{20}\:\text{A} flowing in through each of the other 19 vertices. As before, we have \frac{19}{60}\:\text{A} through each resistor connecting to vertex 2, and thus a voltage difference of V_1-V_2=\frac{19}{60}\:\text{V}. 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 \frac{V-1}{E}\:\Omega. What then, might the equivalent resistance be across one edge of a cube of 1 Ω resistors? A tetrahedron?


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