In the following circuit diagram, the current through the battery is

From the figure, It is clear that diodes ${D}_{1}$ and ${D}_{3}$ are reverse biased, therefore they will not conduct any current and the resistance of $11Ω$ and $5Ω$ will be not in use. The diode ${D}_{2}$ is forward biased, therefore the current flowing& through the diode ${D}_{2}$ is same as the current through the battery.

On redrawing the circuit as& shown, we can obtain the value of current as

$i = \dfrac{20}{7 + 3} = 2A$