In the given circuit,& The current through battery is

In the given circuit, diode ${D}_{1}$ is reverse biased, so it will not conduct but ${D}_{2}, {D}_{3}$ are forward biased, so they will conduct, hence corresponding equivalent circuit. For the given circuit is

Now, the equivalent resistance of the circuit is

${R}_{eq} = \dfrac{\left(515\right) \times 20}{\left(5 + 5\right) + 20} + \dfrac{20}{3}Ω$

Current through battery, $I = \dfrac{10V}{\dfrac{20}{3}Ω} = 1. 5A$