Irodov 3.150: Symmetry Folds the Resistor Cube – Amit & Deepali Sinha Foundation

Problem Irodov 3.150

A wire frame has the shape of a cube, and each of its twelve edges has resistance $R$ . On one face the vertices are numbered $1$ (bottom left), $2$ (top left), $3$ (top right) and $4$ (bottom right); on the opposite face $5$ , $6$ , $7$ and $8$ lie directly behind $1$ , $2$ , $3$ and $4$ . Find the resistance of the frame measured between the vertices

$1$ and $7$ , the ends of a body diagonal
$1$ and $2$ , the ends of an edge
$1$ and $3$ , the ends of a face diagonal

The wire cube. Every edge has resistance R. Vertices 5 to 8 lie behind 1 to 4. — The wire cube. Every edge has resistance $R$ . Vertices $5$ to $8$ lie behind $1$ to $4$ .

A current $I$ is fed in at one vertex and taken out at the other; the resistance is the potential difference between them divided by $I$ . In each part a symmetry of the cube shows which vertices share a potential, and vertices at one potential can be joined without changing any current.

Given

$R$ on each of the $12$ edges

Find

(a)

R_{17}

, (b)

R_{12}

, (c)

R_{13}

Solution

Step 1(a) Three layers of equal potential

So vertices $2$ , $4$ , $5$ share one potential and $3$ , $6$ , $8$ share another. The current divides equally at each stage: $I/3$ in each of the three edges leaving $1$ , $I/6$ in each of the six edges joining the two layers, and $I/3$ in each of the three edges entering $7$ .

The cube for terminals 1 and 7, with the equal-potential vertices joined: three, six and three edges in parallel, every resistor R. — The cube for terminals $1$ and $7$ , with the equal-potential vertices joined: three, six and three edges in parallel, every resistor $R$ .

The potential difference adds up along any path from $1$ to $7$ , for example $1 \to 2 \to 3 \to 7$ :

\varphi_1 - \varphi_7 = R\,\frac{I}{3} + R\,\frac{I}{6} + R\,\frac{I}{3} = \frac{5}{6}\,IR

Dividing by $I$ :

R_{17} = \frac{5}{6}\,R

(1)

Step 2(b) The mirror plane of an edge

Measuring potentials from the mean of the terminals, the reflection therefore sends every potential to its negative: mirror vertices have opposite potentials, and every point on the plane is at zero. The plane cuts four edges ( $1$ - $2$ , $4$ - $3$ , $5$ - $6$ , $8$ - $7$ ) at their midpoints, so each splits into two halves of $R/2$ , and the four midpoints, all at zero, can be joined into one node $M$ . The network falls into two equal halves in series:

R_{12} = 2R_{1M} \tag{2}

The half of the cube on the side of vertex 1 for terminals 1 and 2. The four edges cut by the mirror plane leave halves R/2 ending on the common node M. — The half of the cube on the side of vertex $1$ for terminals $1$ and $2$ . The four edges cut by the mirror plane leave halves $R/2$ ending on the common node $M$ .

Step 3(b) Reducing the half network

Joined, $4$ and $5$ become one node, and the pairs of equal resistors attached to it combine in parallel: $1$ - $4$ with $1$ - $5$ , $4$ - $M$ with $5$ - $M$ , and $4$ - $8$ with $5$ - $8$ .

R_{1,45} = \frac{R}{2}, \qquad R_{45,M} = \frac{R}{4}, \qquad R_{45,8} = \frac{R}{2}

From the joined node to $M$ the direct pair $R/4$ is in parallel with the route through $8$ , which is $R/2 + R/2 = R$ :

\frac{1}{R_{45 \to M}} = \frac{4}{R} + \frac{1}{R} = \frac{5}{R}

In series with $R_{1,45}$ this gives the route from $1$ to $M$ by way of $4$ and $5$ , and that route is in parallel with the half edge $R/2$ running straight from $1$ to $M$ :

\begin{aligned} R_{1 \to 45 \to M} &= \frac{R}{2} + \frac{R}{5} = \frac{7R}{10}\\[2pt] \frac{1}{R_{1M}} &= \frac{10}{7R} + \frac{2}{R} = \frac{24}{7R} \end{aligned}

Doubling $R_{1M} = 7R/24$ as (2) says:

R_{12} = \frac{7}{12}\,R

(3)

Step 4(c) The mirror plane of a face diagonal

So $2$ , $4$ , $6$ , $8$ are all at zero potential and can be joined into one node $M$ . The edges $2$ - $6$ and $4$ - $8$ then run from $M$ to $M$ and carry no current, so they drop out. What is left is two equal halves in series, each made of the edges $1$ - $2$ and $1$ - $4$ straight to $M$ , in parallel with $1$ - $5$ followed by the pair $5$ - $6$ , $5$ - $8$ :

The cube for terminals 1 and 3, with vertices 2, 4, 6, 8 joined. Edges 2-6 and 4-8 carry no current and are omitted; every resistor is R. — The cube for terminals $1$ and $3$ , with vertices $2$ , $4$ , $6$ , $8$ joined. Edges $2$ - $6$ and $4$ - $8$ carry no current and are omitted; every resistor is $R$ .

R_{1M} = \left(\frac{2}{R} + \frac{1}{R + R/2}\right)^{-1} = \left(\frac{2}{R} + \frac{2}{3R}\right)^{-1} = \frac{3}{8}\,R

The half on the side of $3$ is the mirror image, so the total is twice this:

R_{13} = \frac{3}{4}\,R

(4)

Check

Foster’s theorem for a network of $N$ nodes states that the effective resistances across the edges, each divided by that edge’s own resistance, sum to $N - 1$ . The cube has $N = 8$ and twelve equivalent edges, so each edge must give $7/12$ of $R$ , which is (3) exactly. The three answers also grow with distance on the cube, $7/12 < 3/4 < 5/6$ , and each lies below the single shortest path ( $R$ , $2R$ , $3R$ ), as it must with other routes in parallel.

Remark

The two tools used here cover most symmetric networks. A symmetry that keeps the terminals in place makes points equipotential, as in (a); a symmetry that swaps the terminals makes the plane between them a zero of potential, as in (b) and (c), and any resistor lying in that plane, like $2$ - $6$ and $4$ - $8$ , carries no current, exactly as the bridge arm of a balanced bridge. The resistance grows far more slowly than the path length, $0.58R$ , $0.75R$ , $0.83R$ against $R$ , $2R$ , $3R$ , because the number of parallel routes grows with the distance too.