Question

Sol. [Key Concept: Reduce the parallel EMF-resistor branches to $E_{eq}$ and $r_{eq}$, then $I = \frac{E_{eq}}{r_{eq} + R}$.]

Solving each branch as series

$\varepsilon_{\text{eq}\text{~}} = \frac{\frac{15}{9} + \frac{15}{9} + \frac{15}{9}}{\frac{1}{9} + \frac{1}{9} + \frac{1}{9}} = 15$ volt

and $\frac{1}{r_{eq}} = \frac{1}{9} + \frac{1}{9} + \frac{1}{9}$

$$r_{eq} = 3\Omega $$

$$I_{AB} = \frac{15}{12} = 1.25\text{ }A $$