Resistors on a polyhedron

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Resistors on a polyhedron. Consider a regular tetrahedron whose six edges are identical resistors of value $R$. A potential difference is applied between any two vertices of the tetrahedron: what is the equivalent resistance seen from those two vertices? Hint: use symmetry to identify, a priori, pairs of nodes that stay at the same potential; once found, you can “short-circuit” them without altering the circuit. Optional variant: repeat for a cube whose 12 edges are resistors $R$, applying the p.d. between two diametrically opposite vertices (the classic answer is $R_\text{eq} = \tfrac{5}{6}R$). Adapted from (Povey 2015).

Solution

Key idea — equipotential nodes. If two nodes, by symmetry, are at the same potential, no current flows in the resistor connecting them: we can short-circuit them (merge them into a single node) without changing the equivalent resistance. This is the electrostatic application of the concept of equipotential surface.

Tetrahedron. Let $A$ and $B$ be the two vertices between which the p.d. is applied; let $C$ and $D$ be the other two. By symmetry (swapping $C\leftrightarrow D$ leaves the circuit identical), nodes $C$ and $D$ are at the same potential: the resistor $C$–$D$ carries no current and can be removed (or short-circuited, it makes no difference). The remaining paths are:

• the direct branch $A$–$B$: resistance $R$;
• the two branches $A$–$C$–$B$ and $A$–$D$–$B$: each $2R$, in parallel with each other giving $\tfrac{2R\cdot 2R}{2R+2R} = R$.

These two groups ($R$ direct and $R$ from the side paths) are in parallel between $A$ and $B$: $R_\text{eq} = \frac{R\cdot R}{R+R} = \frac{R}{2}$ $\ev{R_\text{eq}^{\ \text{tetraedro}} = \frac{R}{2}}$

Cube (diametrically opposite vertices). Let $A$ and $G$ be the two opposite vertices along the cube’s diagonal. Injecting current $I$ at $A$: by symmetry it splits equally among the 3 edges leaving $A$ ($I/3$ each). The 3 nodes adjacent to $A$ are equipotential; likewise the 3 nodes adjacent to $G$ are equipotential. In the middle section the current $I$ is shared among the 6 intermediate edges ($I/6$ each). Summing the potential drops along a path $A\to\text{adjacent}\to\text{adjacent}\to G$: $V = \frac{I}{3}R + \frac{I}{6}R + \frac{I}{3}R = I R\left(\frac13+\frac16+\frac13\right) = \frac{5}{6}\,I R$ From $R_\text{eq} = V/I$: $\ev{R_\text{eq}^{\ \text{cubo}} = \frac{5}{6}R}$

In both cases symmetry identifies the equipotential nodes and reduces a three-dimensional lattice to a simple series–parallel calculation. See Riferimenti bibliografici for (Povey 2015).

Links

Topics: Electrostatics Concepts: Electric potential Methods: Series-parallel simplification