Consider the circuit shown in the figure, where E _1 = 3 V , E _2 =...

Using Millman's theorem, the equivalent EMF of the parallel combination of the three branches gives the potential difference between nodes A and B. Taking the potential at B to be higher than A (due to the orientation of the batteries):

$V_B - V_A = \dfrac{\dfrac{\mathcal{E}_1}{r_1} + \dfrac{\mathcal{E}_2}{r_2} + \dfrac{\mathcal{E}_3}{r_3}}{\dfrac{1}{r_1} + \dfrac{1}{r_2} + \dfrac{1}{r_3}}$

Substituting the given values:

$V_B - V_A = \dfrac{\dfrac{3}{1} + \dfrac{2}{1} + \dfrac{1}{1}}{\dfrac{1}{1} + \dfrac{1}{1} + \dfrac{1}{1}} = \dfrac{6}{3} = 2 \text{ V}$

The potential difference between A and B is $2 \text{ V}$.

Let the currents flowing from A to B in the three branches be $i_1$, $i_2$, and $i_3$ respectively. Applying Kirchhoff's Voltage Law for each branch from A to B:

$V_A - i_1 r_1 + \mathcal{E}_1 = V_B \Rightarrow i_1 = \dfrac{\mathcal{E}_1 - (V_B - V_A)}{r_1}$

$i_1 = \dfrac{3 - 2}{1} = 1 \text{ A}$

$V_A - i_2 r_2 + \mathcal{E}_2 = V_B \Rightarrow i_2 = \dfrac{\mathcal{E}_2 - (V_B - V_A)}{r_2}$

$i_2 = \dfrac{2 - 2}{1} = 0 \text{ A}$

$V_A - i_3 r_3 + \mathcal{E}_3 = V_B \Rightarrow i_3 = \dfrac{\mathcal{E}_3 - (V_B - V_A)}{r_3}$

$i_3 = \dfrac{1 - 2}{1} = -1 \text{ A}$

Thus, the potential difference is $2 \text{ V}$ and the currents are $i_1 = 1 \text{ A}$, $i_2 = 0 \text{ A}$, and $i_3 = -1 \text{ A}$.

Answer: $2 \text{ V}$; $i_1 = 1 \text{ A}$, $i_2 = 0 \text{ A}$, $i_3 = -1 \text{ A}$