In the circuit of figure, treat the diodes as ideal. Current in the 4-ohm resistor is (In A)

With battery polarity, $D_2$  is reverse biased and no current flows through $3\Omega$ resistor.

$D_1$ is in forward biasing means $2 \Omega$ and $4 \Omega$ are in series.

 So, $I=\frac{12}{4+2}=2A$