In the given circuit, $D_1$ and $D_2$ are normal diodes. The current drawn from the battery by the diode network is

1.0 A

The correct answer is Option (3) → 1.0 A

Battery = $5\ \text{V}$

$D_{1}$ is forward biased (conducting), $D_{2}$ is reverse biased (off)

Only the upper branch ($5\ \Omega$) conducts.

$I=\frac{V}{R}=\frac{5}{5}=1\ \text{A}$