Solution to Question 6
Part 1: Equivalent Resistance $R_{AB}$ (Adjacent Nodes)
We use the principle of superposition of currents. Let current $I$ enter the infinite grid at node A and leave at node B. The resistance of each branch is $R$.
Step 1: Current Injection at A
If current $I$ enters at node A (sink at infinity), due to the symmetry of the hexagonal lattice (3 branches per node), the current splits equally:
$$ I_{A \to B} = \frac{I}{3} $$Step 2: Current Extraction at B
If current $I$ leaves at node B (source at infinity), current arrives equally from its 3 neighbors. The current flowing towards B from A is:
$$ I_{A \to B} = \frac{I}{3} $$Step 3: Superposition
Adding both contributions, the total current in branch AB is:
$$ I_{total(AB)} = \frac{I}{3} + \frac{I}{3} = \frac{2I}{3} $$Using Ohm’s Law ($V = IR$):
$$ V_{AB} = \left( \frac{2I}{3} \right) R \implies R_{AB} = \frac{V_{AB}}{I} = \frac{2R}{3} $$Part 2: Equivalent Resistance $R_{AC}$ (Next-Nearest Neighbors)
Node C is the second neighbor such that the path A-B-C forms an angle of $120^\circ$ (across the face of a hexagon). We find the resistance by superposing the currents for input at A and output at C.
1. Source at A (Sink at Infinity):
- Current flows $A \to B$ with magnitude $I/3$.
- At node B, this current splits into the two forward branches ($B \to C$ and $B \to \text{other}$). Due to symmetry, it splits equally.
- Current component in $B \to C$: $$ I’_{BC} = \frac{1}{2} \times \frac{I}{3} = \frac{I}{6} $$
2. Sink at C (Source at Infinity):
- By symmetry (reverse of the previous case), current flows into C from its neighbors. The current in $B \to C$ is $I/3$.
- Working backward to node B, the current supplying B comes from its neighbors (like A). The current component flowing $A \to B$ due to the sink at C is: $$ I”_{AB} = \frac{I}{6} $$
3. Superposition:
Total current in branch $A \to B$: $$ I_{AB} = I_{source} + I_{sink} = \frac{I}{3} + \frac{I}{6} = \frac{I}{2} $$ Total current in branch $B \to C$: $$ I_{BC} = I_{source} + I_{sink} = \frac{I}{6} + \frac{I}{3} = \frac{I}{2} $$
The total voltage drop from A to C is the sum of drops across AB and BC:
$$ V_{AC} = V_{AB} + V_{BC} = \left( \frac{I}{2} R \right) + \left( \frac{I}{2} R \right) = IR $$The equivalent resistance is:
$$ R_{AC} = \frac{V_{AC}}{I} = R $$