Equivalent Resistance in a Skeleton Cube

This problem involves finding the equivalent resistance of a complex network of resistors arranged in the shape of a skeleton cube. A skeleton cube has 12 edges, and each edge is represented by a wire with resistance \(r\).

Understanding the Cube Network

Imagine a cube with its corners labeled. Let's say we want to find the equivalent resistance between corner A and the diagonally opposite corner G.

• A cube has 8 corners (vertices) and 12 edges.
• Each edge is a wire with resistance \(r\).
• We need to find the resistance between two corners that are furthest apart (diagonally opposite).

Applying Symmetry for Equivalent Resistance

Due to the symmetrical nature of the cube, we can use the principle of symmetry to simplify the circuit. Let's assume a total current \(I\) enters the cube at corner A and exits at corner G.

Step-by-Step Current Distribution and Potential Levels:

1. Current entering at A: From corner A, there are three identical paths (edges) leading to adjacent corners. Due to symmetry, the current \(I\) will divide equally among these three wires.
   • Current in each wire leaving A (e.g., AB, AD, AE) = \( \frac{I}{3} \)
   The three corners connected to A (B, D, E) are equipotential points because they are symmetrically placed with respect to the input (A) and output (G) terminals. Let's call this potential level 1.
2. Current flowing from Level 1: From each of the corners B, D, E, the current \( \frac{I}{3} \) further divides into two paths (wires) each, leading to the next set of corners (C, F, H).
   • For example, from B, the current \( \frac{I}{3} \) splits into \( \frac{I}{6} \) along BC and \( \frac{I}{6} \) along BF.
   • Similarly for D (DC, DH) and E (EF, EH).
   The corners C, F, H are all equipotential points. These are the corners adjacent to the exit point G, but not G itself. Let's call this potential level 2.
3. Current entering at Level 2: Each of the corners C, F, H receives current from two paths originating from Level 1.
   • For example, corner C receives \( \frac{I}{6} \) from B and \( \frac{I}{6} \) from D. Total current entering C = \( \frac{I}{6} + \frac{I}{6} = \frac{2I}{6} = \frac{I}{3} \).
   • Similarly, F receives \( \frac{I}{3} \) and H receives \( \frac{I}{3} \).
4. Current exiting at G: From each of the corners C, F, H, the current \( \frac{I}{3} \) flows through a single wire to the exit corner G.
   • Current from C to G = \( \frac{I}{3} \).
   • Current from F to G = \( \frac{I}{3} \).
   • Current from H to G = \( \frac{I}{3} \).
   These three currents combine at G to form the total current \( \frac{I}{3} + \frac{I}{3} + \frac{I}{3} = I \), which matches the initial assumption.

Circuit Simplification and Equivalent Resistance Calculation

Based on the equipotential points, we can redraw the cube network as a series combination of three parallel resistor stages:

The total equivalent resistance between the diagonally opposite corners A and G is the sum of the equivalent resistances of these three stages, as they are effectively in series:

\( R_{total} = R_1 + R_2 + R_3 \)

\( R_{total} = \frac{r}{3} + \frac{r}{6} + \frac{r}{3} \)

To add these fractions, find a common denominator, which is 6:

\( R_{total} = \frac{2r}{6} + \frac{r}{6} + \frac{2r}{6} \)

\( R_{total} = \frac{(2 + 1 + 2)r}{6} \)

\( R_{total} = \frac{5r}{6} \)

Therefore, the equivalent resistance between the two diagonally opposite corners of the cube is \( \frac{5}{6} r \).