Two ideal diodes are connected to a battery as shown in the circuit. The current supplied by the battery is

0.75 A zero 0.25 A 0.5 A

0.5 A

Here $D_1$ is in forward bias and $D_2$ is in reverse bias so, $D_1$ will conduct and $D_2$ will not conduct. Thus, no current will flow through DC. $I=\frac{V}{R}=\frac{5}{10}=\frac{1}{2}Amp.$