Determine the potential difference V_a - V_b between the points a...

In the circuit labeled (a), let the potential at the bottom wire be $V_b = 0$.

The $12 \text{ V}$ battery fixes the potential at the node directly above it to $12 \text{ V}$.

Let the potential at the node between the $12 \text{ V}$ battery and node $a$ be $x$.

Applying the principle of charge conservation at node $x$ yields $2(x - 12) + 4(x - 0) + 2(x - V_a) = 0$, which simplifies to $8x - 24 - 2V_a = 0$ or $4x - V_a = 12$.

Applying charge conservation at node $a$ gives $2(V_a - x) + 4(V_a - 0) = 0$, resulting in $6V_a - 2x = 0$, or $x = 3V_a$.

Substituting $x = 3V_a$ into the first equation gives $4(3V_a) - V_a = 12$, so $11V_a = 12$ and $V_a = \dfrac{12}{11} \text{ V}$.

Therefore, the potential difference is $V_a - V_b = \dfrac{12}{11} - 0 = \dfrac{12}{11} \text{ V}$.