Problem 13: Equivalent Resistance
This problem can be solved using Nodal Analysis. Due to the specific values of the resistors, the circuit has a skew-symmetry. The top branch has resistances $1\Omega, 2\Omega, 3\Omega$, and the bottom has $3\Omega, 2\Omega, 1\Omega$.
Let us apply a potential $V = 100$ V at terminal A and $0$ V at terminal B. Let $V_C$ be the potential after the first top resistor ($1\Omega$) and $V_D$ be the potential after the first bottom resistor ($3\Omega$).
Using Kirchhoff’s Current Law (KCL) at node C:
$$ \frac{V_C – 100}{1} + \frac{V_C – V_E}{2} + \frac{V_C – V_D}{2} = 0 $$Using KCL at node D:
$$ \frac{V_D – 100}{3} + \frac{V_D – V_F}{2} + \frac{V_D – V_C}{2} = 0 $$Due to symmetry, the potentials satisfy $V_E = 100 – V_D$ and $V_F = 100 – V_C$. Solving these linear equations gives:
$$ V_C = 75 \text{ V}, \quad V_D = 62.5 \text{ V} $$The total current $I_{total}$ entering A splits into the top and bottom branches:
$$ I_{top} = \frac{100 – 75}{1} = 25 \text{ A} $$ $$ I_{bottom} = \frac{100 – 62.5}{3} = 12.5 \text{ A} $$ $$ I_{total} = 37.5 \text{ A} $$Finally, the equivalent resistance is:
$$ R_{eq} = \frac{V}{I_{total}} = \frac{100}{37.5} = \frac{8}{3} \, \Omega $$Answer: $8/3 \, \Omega$