The magnitude and direction of the current in the circuit shown will be

Since E₁(10V) > E₂(4V)
So current in the circuit will be clockwise.

Applying Kirchhoff's voltage law,

$\begin{array}{l}-1\times \mathrm{i}+10-4-2\times \mathrm{i}-3\mathrm{i}=0\Rightarrow \mathrm{i}=1\mathrm{A}\left(\mathrm{a}\mathrm{to}\mathrm{b}\mathrm{via}\mathrm{e}\right)\\ \therefore \mathrm{Current}=\frac{\mathrm{V}}{\mathrm{R}}=\frac{10-4}{6}=1.0\mathrm{ampere}\end{array}$