In the circuit shown here, ${\mathrm{E}}_1={\mathrm{E}}_2={\mathrm{E}}_3=2\mathrm{V}\mathrm{and}{\mathrm{R}}_1\mathrm{and}{\mathrm{R}}_2=4\mathrm{ohms}.$ The current flowing between points A and B through battery E₂ is

The given circuit can be redrawn as:

${\mathrm{E}}_{\mathrm{eq}}=\frac{{\mathrm{E}}_1{\mathrm{R}}_1+{\mathrm{E}}_2{\mathrm{R}}_2}{{\mathrm{R}}_1+{\mathrm{R}}_2}=\frac{2\times 4+2\times 4}{4+4}=2\mathrm{V}\mathrm{and}$

${\mathrm{R}}_{\mathrm{eq}}=\frac{4}{2}=2\mathrm{\Omega }$

Current, $\mathrm{i}=\frac{2+2}{2}=2\mathrm{A}$ from A to B through E₂.