The Resistor Cube

Consider twelve resistors of equal value arranged along the edges of a unit cube, connected to one another only at the corners. This looks like

The task is to determine equivalent resistance across different elements of the cube. There are three distinct cases: resistance along an edge, resistance across a face diagonal and resistance across a body diagonal.

Currents in the resistors will be labeled with a numeric subscript indicating the initial and final corners. Starting from the origin and omitting zero for there, these will be designated

i₁	move along x-axis
i₂	move along y-axis
i₃	move along z-axis
i₁₂	first along x-axis then y-axis
i₁₃	first along x-axis then z-axis
i₂₁	first along y-axis then x-axis
i₂₃	first along y-axis then z-axis
i₃₁	first along z-axis then x-axis
i₃₂	first along z-axis then y-axis
i₃₊	finally along z-axis, i₁₂ / i₂₁ coterminus
i₂₊	finally along y-axis, i₁₃ / i₃₁ coterminus
i₁₊	finally along x-axis, i₂₃ / i₃₂ coterminus

Overlaying on wires without resistor bodies one has

Conservation of charge at each corner gives Kirchoff’s first law, which says that the sum of currents at each circuit node is zero. Without attaching signs to currents, one can write schematically

$\begin{array}{r} i_{1} + i_{2} + i_{3} \approx 0 \\ i_{1} + i_{12} + i_{13} \approx 0 \\ i_{2} + i_{21} + i_{23} \approx 0 \\ i_{3} + i_{31} + i_{32} \approx 0 \end{array} \begin{array}{r} i_{12} + i_{21} + i_{3+} \approx 0 \\ i_{13} + i_{31} + i_{2+} \approx 0 \\ i_{23} + i_{32} + i_{1+} \approx 0 \\ i_{1+} + i_{2+} + i_{3+} \approx 0 \end{array}$

These at first appear to be eight equations for the twelve unknowns, but some caution is in order. With directional signs properly assigned, these eight equations are not all independent. If current is introduced into the cube at the origin, the equation for which it exits the cube is equivalent to the first equation and must be omitted. This has the added benefit of simplifying the overall calculation, as will be made clear.

The remaining independent equations come from Kirchoff’s second law, which says the sum of voltages around a closed circuit is zero. Since all resistors are equal, by Ohm’s law $V = I R$ all voltage drops are proportional to currents. Without attaching signs one can again write schematically

$\begin{array}{r} i_{1} + i_{2} + i_{12} + i_{21} \approx 0 \\ i_{1} + i_{3} + i_{13} + i_{31} \approx 0 \\ i_{2} + i_{3} + i_{23} + i_{32} \approx 0 \end{array} \begin{array}{r} i_{12} + i_{13} + i_{2+} + i_{3+} \approx 0 \\ i_{21} + i_{23} + i_{1+} + i_{3+} \approx 0 \\ i_{31} + i_{32} + i_{1+} + i_{2+} \approx 0 \end{array}$

This is as far as one can go without further specification of the circuit.

And now the fun begins! The third case above, resistance across a body diagonal, is actually the easiest due to the symmetry of the cube. The problem will not be evaluated using this, but instead with a method that directly produces answers for the other cases as well.

With current entering at the origin, the point of exit for the third case will be [1,1,1]. Adding graphical indicators of the directions of circuits one has

The circuit equations including directional signs are then

$\begin{array}{r} i_{1} + i_{2} + i_{3} = i \\ i_{1} - i_{12} - i_{13} = 0 \\ i_{2} - i_{21} - i_{23} = 0 \\ i_{3} - i_{31} - i_{32} = 0 \\ i_{1} - i_{2} + i_{12} - i_{21} = 0 \\ i_{1} - i_{3} + i_{13} - i_{31} = 0 \\ i_{2} - i_{3} + i_{23} - i_{32} = 0 \end{array} \begin{array}{r} i_{12} + i_{21} - i_{3+} = 0 \\ i_{13} + i_{31} - i_{2+} = 0 \\ i_{23} + i_{32} - i_{1+} = 0 \\ i_{1+} + i_{2+} + i_{3+} = i \\ i_{12} - i_{13} - i_{2+} + i_{3+} = 0 \\ i_{21} - i_{23} - i_{1+} + i_{3+} = 0 \\ i_{31} - i_{32} - i_{1+} + i_{2+} = 0 \end{array}$

In the first group of eight, adding together the second through seventh equations indicates the first and eighth equations are identical, so the latter must be omitted. In the second group of second, order around the loop does not matter as long as the signs are consistent.

If the currents are collected in a vector in the order labeled,

$i = [i_{1}, i_{2}, i_{3}, i_{12}, i_{13}, i_{21}, i_{23}, i_{31}, i_{32}, i_{3+}, i_{2+}, i_{1+}]$

along with a mostly-zero current source vector,

$I = [i, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0]$

and a matrix for the body diagonal case is formed using coefficients from the seven equations of the first group and five from the second,

then the currents satisfy the simple matrix equation

$M_{body} i = I$

whose solution is straightforwardly

$i = M_{body}^{- 1} I$

The inverse matrix can be evaluated numerically using Math, and given the structure of the matrix equation only the first column of this inverse is relevant. That first column is

When evaluated exactly, the values here are either one third or one sixth. The above mentioned symmetry of the cube is why the first three entries, as well as the last three, are identically equal, and similarly for the six intermediate entries.

To obtain the effective resistance across the body diagonal, equate the input voltage to the sum of voltage drops along any path. For example

$\begin{array}{l} i R_{body} = i_{1} R + i_{12} R + i_{3+} R = (\frac{1}{3} + \frac{1}{6} + \frac{1}{3}) i R = \frac{5}{6} i R \\ \Rightarrow R_{body} = \frac{5}{6} R \end{array}$

One gets the same result for any three sequential voltage drops, again evidence of the symmetry of this case. This is immediately clear if the current labels are replaced by values from the column matrix in the graphic visualization:

In fact one gets the same result over any path whatsoever around the cube, as long as one includes signs of currents from their directions.

For the other two cases, proceed in the same manner. The point of exit of current for the face diagonal is either of two upper corners, so choose [1,0,1] for simplicity. The graphic visualization of currents is then

where only one current, $i_{2+}$ , is reversed. The graphic corresponds to the circuit equations

$\begin{array}{r} i_{1} + i_{2} + i_{3} = i \\ i_{1} - i_{12} - i_{13} = 0 \\ i_{2} - i_{21} - i_{23} = 0 \\ i_{3} - i_{31} - i_{32} = 0 \\ i_{1} - i_{2} + i_{12} - i_{21} = 0 \\ i_{1} - i_{3} + i_{13} - i_{31} = 0 \\ i_{2} - i_{3} + i_{23} - i_{32} = 0 \end{array} \begin{array}{r} i_{12} + i_{21} - i_{3+} = 0 \\ i_{13} + i_{31} + i_{2+} = i \\ i_{23} + i_{32} - i_{1+} = 0 \\ i_{1+} - i_{2+} + i_{3+} = 0 \\ i_{12} - i_{13} + i_{2+} + i_{3+} = 0 \\ i_{21} - i_{23} - i_{1+} + i_{3+} = 0 \\ i_{31} - i_{32} - i_{1+} - i_{2+} = 0 \end{array}$

In the first group of eight, adding together all equations with zero right-hand sides indicates that this time the first and sixth equations are identical, so the latter must be omitted. The matrix of coefficients is now

the first column of whose inverse is

This time an exact evaluation would give values of one quarter, one eighth or three eighths. The effective resistance along the shortest path is

$\begin{array}{l} i R_{face} = i_{1} R + i_{13} R = (\frac{3}{8} + \frac{3}{8}) i R = \frac{3}{4} i R \\ \Rightarrow R_{face} = \frac{3}{4} R \end{array}$

but one can evaluate the same value using any path that avoids the two edges for which the current is zero. This is again immediately clear if the current labels are replaced by values from the column matrix in the graphic visualization:

Mathematically one also finds the same effective resistance using paths with edges having zero values of current. This is naturally physically contradictory, since it essentially corresponds to removal of a resistor from the cube.

For the third case the point of exit of current for a single edge is either of two lower corners, so choose [1,0,0] for simplicity. The graphic visualization of currents is now

with four currents reversed. The graphic corresponds to the circuit equations

$\begin{array}{r} i_{1} + i_{2} + i_{3} = i \\ i_{1} + i_{12} + i_{13} = i \\ i_{2} - i_{21} - i_{23} = 0 \\ i_{3} - i_{31} - i_{32} = 0 \\ i_{1} - i_{2} - i_{12} - i_{21} = 0 \\ i_{1} - i_{3} - i_{13} - i_{31} = 0 \\ i_{2} - i_{3} + i_{23} - i_{32} = 0 \end{array} \begin{array}{r} i_{12} - i_{21} - i_{3+} = 0 \\ i_{13} - i_{31} - i_{2+} = 0 \\ i_{23} + i_{32} - i_{1+} = 0 \\ i_{1+} - i_{2+} - i_{3+} = 0 \\ i_{12} - i_{13} - i_{2+} + i_{3+} = 0 \\ i_{21} - i_{23} - i_{1+} - i_{3+} = 0 \\ i_{31} - i_{32} - i_{1+} - i_{2+} = 0 \end{array}$

In the first group of eight, adding together the third through eighth equations indicates that the first and second equations are identical, so the latter must be omitted. The matrix of coefficients is then

the first column of whose inverse is

This time an exact evaluation would give values that are not immediately identifiable:

$\frac{7}{12} \approx .5833, \frac{5}{24} \approx .2083, \frac{1}{6} \approx .1667, \frac{1}{12} \approx .0833, \frac{1}{24} \approx .0417$

The effective resistance along the lower right edge is

$i R_{edge} = i_{1} R = \frac{7}{12} i R \Rightarrow R_{edge} = \frac{7}{12} R$

and one can again evaluate the same value using any path through the cube. The graphic with current labels are replaced by values from the column matrix is

Adding up fractional values along the paths takes a bit more effort this time, but the results are as expected.

And that covers all three cases, quod erat quaerendum.