Determine the potential difference V_a - V_b for the electrical...

Let the potential at node b be $V_b = 0$. The potential difference to be found is $V_a - V_b = V_a$.

Both circuits (a) and (b) consist of three parallel branches connected between nodes a and b. The first branch contains a battery $\mathcal{E}_1$ and a resistor $R_1$, with the positive terminal of $\mathcal{E}_1$ connected to node a. The second branch contains only a resistor $R_3$. The third branch contains a battery $\mathcal{E}_2$ and a resistor $R_2$, with the positive terminal of $\mathcal{E}_2$ connected to node a.

Applying Kirchhoff's Current Law at node a, the sum of currents leaving the node must be zero:

$\dfrac{V_a - \mathcal{E}_1}{R_1} + \dfrac{V_a - 0}{R_3} + \dfrac{V_a - \mathcal{E}_2}{R_2} = 0$

Grouping the terms with $V_a$ on one side:

$V_a \left( \dfrac{1}{R_1} + \dfrac{1}{R_2} + \dfrac{1}{R_3} \right) - \dfrac{\mathcal{E}_1}{R_1} - \dfrac{\mathcal{E}_2}{R_2} = 0$

$V_a \left( \dfrac{1}{R_1} + \dfrac{1}{R_2} + \dfrac{1}{R_3} \right) = \dfrac{\mathcal{E}_1}{R_1} + \dfrac{\mathcal{E}_2}{R_2}$

$V_a = \dfrac{\dfrac{\mathcal{E}_1}{R_1} + \dfrac{\mathcal{E}_2}{R_2}}{\dfrac{1}{R_1} + \dfrac{1}{R_2} + \dfrac{1}{R_3}}$

This result can also be directly written using Millman's theorem.

Answer: $\dfrac{\dfrac{\mathcal{E}_1}{R_1} + \dfrac{\mathcal{E}_2}{R_2}}{\dfrac{1}{R_1} + \dfrac{1}{R_2} + \dfrac{1}{R_3}}$