In the given circuit The current through the battery is:

In the given circuit, diode $D_1$ is reverse biased, so it will not conduct. Diodes $D_2$ and $D_3$ are forward biased, so they will conduct. The corresponding equivalent circuit is a shown in the figure.

The equivalent resistnce of the circuit is

$R_{eq}=\frac{(5+5)×20}{(5+5)+20}=\frac{10×20}{10+20}=\frac{200}{30}=\frac{20}{3}Ω$

Current through the battery, $I=\frac{10V}{\frac{20}{3}Ω}=1.5A$