Solution to Question 5
1. Circuit Symmetry and Nodal Potentials
The network consists of 12 edges (each $24\,\Omega$) and 6 face diagonals (each $12\,\Omega$). Let a voltage $V$ be applied at node 1 (A) and $-V$ at node 3 (B). The geometric center of the cube is at $0\,\text{V}$ potential.
We analyze the symmetry to identify equipotential nodes:
- Reflection Symmetry: The plane passing through the diagonal axis 1-3 and 5-7 divides the cube into two symmetric halves. Node 2 is symmetric to Node 4, and Node 6 is symmetric to Node 8. Thus, $V_2 = V_4$ and $V_6 = V_8$.
- Inversion Symmetry: Reversing the polarity (swapping input/output) maps Node 1 to Node 3 and Node 2 to Node 4. This implies $V_2 = -V_4$.
Combining $V_2 = V_4$ and $V_2 = -V_4$, we find that $V_2 = V_4 = 0\,\text{V}$. Similarly, $V_6 = V_8 = 0\,\text{V}$. Therefore, Nodes 2, 4, 6, and 8 are at Ground Potential (0 V). This greatly simplifies the circuit calculation, as we can analyze just the “Input Half” (Node 1 to Ground) and multiply the result by 2.
2. Analysis of Node 5 (Intermediate Node)
Node 5 is not directly grounded, but we can calculate its effective resistance to ground. Connections from Node 5:
- To Node 6 (Ground): Edge resistor $24\,\Omega$.
- To Node 8 (Ground): Edge resistor $24\,\Omega$.
- To Node 2 (Ground): Top face diagonal $12\,\Omega$.
- To Node 4 (Ground): Left face diagonal $12\,\Omega$.
- To Node 7: Back face diagonal $12\,\Omega$. By symmetry ($V_5 = -V_7$), the midpoint of this resistor is at 0V. Effectively, this is a $6\,\Omega$ resistor to ground.
Calculating the conductance $G_5$ from Node 5 to Ground:
$$ G_5 = \frac{1}{24} + \frac{1}{24} + \frac{1}{12} + \frac{1}{12} + \frac{1}{6} $$ $$ G_5 = \frac{2}{24} + \frac{2}{12} + \frac{2}{12} = \frac{1}{12} + \frac{4}{12} = \frac{5}{12}\,\Omega^{-1} $$The effective resistance from Node 5 to Ground is $R_{5g} = \frac{12}{5} = 2.4\,\Omega$.
3. Analysis of Node 1 (Input Node)
We now sum the currents leaving Node 1 for a potential $V$. The total resistance $R_{A-GND} = V / I_{total}$. Connections from Node 1:
- To Node 3 (Output): Face diagonal $12\,\Omega$. Since $V_3 = -V$, the midpoint is grounded. Effective resistance to ground is $12/2 = 6\,\Omega$.
- To Node 2 (Ground): Edge resistor $24\,\Omega$.
- To Node 4 (Ground): Edge resistor $24\,\Omega$.
- To Node 6 (Ground): Top face diagonal $12\,\Omega$.
- To Node 8 (Ground): Left face diagonal $12\,\Omega$.
- To Node 5: Edge resistor $24\,\Omega$ in series with the effective resistance of Node 5 ($2.4\,\Omega$). Total path: $24 + 2.4 = 26.4\,\Omega$.
Total Conductance $G_{total}$ at Node 1:
$$ G_{total} = \frac{1}{6} + \frac{1}{24} + \frac{1}{24} + \frac{1}{12} + \frac{1}{12} + \frac{1}{26.4} $$ $$ G_{total} = \frac{1}{6} + \frac{2}{24} + \frac{2}{12} + \frac{5}{132} \quad (\text{using } 26.4 = 132/5) $$ $$ G_{total} = \frac{1}{6} + \frac{1}{12} + \frac{1}{6} + \frac{5}{132} = \frac{2}{6} + \frac{1}{12} + \frac{5}{132} = \frac{5}{12} + \frac{5}{132} $$ $$ G_{total} = \frac{55}{132} + \frac{5}{132} = \frac{60}{132} \,\Omega^{-1} $$The resistance of the half-circuit is $R_{A-GND} = \frac{132}{60} = 2.2\,\Omega$.
4. Final Equivalent Resistance
The total equivalent resistance between A and B is the sum of the resistances of the two symmetric halves ($A \to GND$ and $GND \to B$).
$$ R_{eq} = R_{A-GND} + R_{B-GND} = 2.2 + 2.2 = 4.4\,\Omega $$