The magnitude and direction of the current in the following circuit is:

The net EMF in the loop is $10\text{V}-5\text{V}=5\text{V}$ (since they oppose each other).The total resistance is $2+7+1=10\Omega$.Current $I=\frac{E_{net}}{R_{total}}=\frac{5}{10}=0.5\text{A}$.The direction is determined by the stronger $10\text{V}$ battery, which pushes current from its positive terminal (point A) towards point B through the resistor branch containing E.