In the circuit shown below, D_1 and D_2 are silicon diodes....

The positive terminal of the battery is connected to the p-side of diode $D_1$ and the n-side of diode $D_2$.

Therefore, $D_1$ is forward-biased and $D_2$ is reverse-biased.

The branch containing $D_2$ acts as an open circuit, so no current flows through $R_3$.

The active circuit is a series loop with the battery, $R_1$, $D_1$, and $R_2$.

The net voltage driving the current is the battery voltage minus the barrier potential of the forward-biased diode $D_1$.

$V_{net} = 3.7 - 0.7 = 3.0$ V

The equivalent resistance of the active circuit is the sum of the series resistors $R_1$ and $R_2$.

$R_{eq} = R_1 + R_2 = 5 + 5 = 10\ \Omega$

The current supplied by the battery is calculated using Ohm's law.

$I = \dfrac{V_{net}}{R_{eq}} = \dfrac{3.0}{10} = 0.30$ A

Answer: $0.30\text{ A}$