In the figure shown, the current in the

$\mathrm{E}_{\mathrm{eq}}=\frac{20 \times 10}{17}=\frac{200}{17}$

and $\mathrm{R}_{\text {eq }}=\frac{7 \times 10}{17}=\frac{70}{17}$

$\therefore \quad I=\frac{\frac{20}{17}-10}{4+\frac{70}{17}}=0.21 \mathrm{~A}$

from +ve to -ve terminal