Q.A battery of $10\ \text{V}$ and negligible internal resistance is connected across the diagonally opposite corners of a cubical network consisting of 12 resistors each of resistance $1\ \Omega$ (Fig. 3.16). Determine the equivalent resistance of the network and the current along each edge of the cube.

The network is symmetric about the body diagonal from the entry corner $A$ to the opposite corner $G$, and all 12 edges are $1\ \Omega$.

Symmetry currents. Let the total current be $I$. It splits equally among the three edges at $A$ (current $I/3$ each). Each of these reaches a vertex where it divides into two of the six middle edges ($I/6$ each). At the far end three middle edges feed each of the three edges into $G$ ($I/3$ each).

Equivalent resistance. Add the potential drops along a path $A \to \text{(adjacent)} \to \text{(middle)} \to G$:

With $V = 10\ \text{V}$: $\dfrac{5}{6}I = 10 \Rightarrow I = 12\ \text{A}$, so

Edge currents. $I/3 = 4\ \text{A}$ on each of the three edges at $A$ and the three at $G$; $I/6 = 2\ \text{A}$ on each of the six middle edges.

Setting up the symmetry. The battery is across the body diagonal, from corner $A$ (entry) to the opposite corner $G$ (exit). Three edges leave $A$; by symmetry they are indistinguishable, so the total current $I$ splits equally: $I/3$ in each. Each such edge ends on a vertex adjacent to $A$; from there two edges continue toward $G$, so by symmetry the $I/3$ splits into $I/6$ in each of these six 'middle' edges. Finally three edges arrive at $G$, each carrying $I/3$ (pairs of middle edges of $I/6$ merging).

Potential drop along a diagonal path. Follow $A \to B \to F \to G$ (adjacent $\to$ middle $\to$ exit):

Equivalent resistance. This drop equals the battery voltage, $\dfrac{5I}{6} = 10\ \text{V}$, and by definition $V = I R_\text{eq}$, so

Total and branch currents.

• Three edges at $A$ and three at $G$: $I/3 = 4\ \text{A}$ each.
• Six middle edges: $I/6 = 2\ \text{A}$ each.

Method: Exploiting Symmetry to Reduce Resistor Networks

This method solves resistor networks with multiple identical branches (e.g. a cube, wire mesh, or ladder network) fed between two symmetric terminals, without writing individual Kirchhoff equations for every branch.

Steps

Step 1: Identify equivalent points via symmetry

Look at the geometry of the network relative to the entry and exit terminals. If several branches are indistinguishable from the source's point of view (same distance, same connectivity, mirror images of one another), by symmetry they must carry equal currents and their far ends must sit at equal potentials.

Step 2: Group vertices into equipotential "shells"

Points that are equidistant from the entry terminal along equivalent paths are at the same potential. This lets you assign a single current variable to each symmetric group of edges, rather than one variable per edge.

Step 3: Distribute the total current using symmetry counts

If $n$ identical edges leave a junction and, by symmetry, must carry equal current, each carries $I/n$ of whatever current arrives there. Track how the current $I$ from the source splits and recombines through each successive shell of the network.

Step 4: Sum potential drops along ONE representative path

Every direct path from the entry to the exit terminal must produce the same total potential drop $V$ (the applied voltage). Pick the simplest such path and add up the $IR$ drop across each segment along it:

Solve this single equation for the overall current $I$.

Step 5: Applying to this problem

Once $I$ is known, $R_\text{eq} = V/I$, and the current in any individual edge follows directly from the symmetry grouping set up in Step 2 — e.g. edges nearest the entry/exit terminals carry a larger fractional share of $I$ than the "equatorial" edges further from either terminal.