In the circuit shown here, $E_1 = E_2 = E_3 = 2 V$ and $R_1 = R_2 = 4\,ohms$. The current flowing between points $A$ and $B$ through battery $E_2$ is
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The given circuit can be redrawn
${E_{eq}} = \frac{{{E_1}{R_2} + {E_2}{R_1}}}{{{R_1} + {R_2}}} = \frac{{2 \times 4 + 2 \times 4}}{{4 + 4}} = 2\,V$ and
${R_{eq}} = \frac{4}{2} = 2\Omega $. Current $i = \frac{{2 + 2}}{2} = 2A$ from $A$ to $B$ through $E_2.$
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