## 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,
$V_1-V_2=IR=\left(\frac{11}{60}\:\text{A}\right)\left(1\:\Omega\right)=\frac{11}{60}\:\text{V}$.
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
$V_1-V_2=\left(\frac{11}{60}\:\text{A}\right)\left(1\:\Omega\right)=\frac{11}{60}\:\text{V}$.

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 $\frac{11}{12}\:\text{A}+\frac{1}{12}\:\text{A}=1\:\text{A}$ 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
$R_{eq}=\frac{V_1-V_2}{I}=\frac{\frac{11}{30}\:\text{V}}{1\:\text{A}}=\frac{11}{30}\:\Omega$.

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
$V_1-V_2=\frac{19}{60}\:\text{V}+\frac{19}{60}\:\text{V}=\frac{19}{30}\:\text{V}$,
and thus an equivalent resistance of
$R_{eq}=\frac{V_1-V_2}{I}=\frac{\frac{19}{30}\:\text{V}}{1\:\text{A}}=\frac{19}{30}\:\Omega$.

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?