Using Kirchhoff's loop rule for the circuit: The total EMF in the loop is $10 - 5 = 5$ V. The total resistance is $2 + 1 + 7 = 10 \Omega$. Current $I = V/R = 5 / 10 = 0.5$ A. Since the 10V battery is stronger, the current flows from A to B through the branch containing the 5V battery.