In the given circuit, E_1=E_2=E_3=2 V and R_1=R_2=4 , then current...

The given circuit can be redrawn as, where, equivalent emf of the combination, $$\begin{aligned} E_{\text{eq }} & =\frac{E_1 R_1+E_2 R_2}{R_1+R_2} \\ & =\frac{2 \times 4+2 \times 4}{4+4}=2 \mathrm{ V} \end{aligned}$$ and $R_{e q}=\frac{R_1 R_2}{R_1+R_2}=\frac{4 \times 4}{4+4}=\frac{16}{8}=2 \Omega$ Net emf of the circuit, $$\begin{aligned} E^{\prime} & =E_2+E_{\mathrm{eq}} \\ & =2+2 \\ & =4 \mathrm{ V} \end{aligned}$$\therefore$Current flowing through branch$A B$,$I=\frac{E^{\prime}}{R_{e q}}=\frac{4}{2}=2 \mathrm{ A}$Clearly from figure, it is clear that direction of current is from point$A$to$B$ clockwise is loop.