All resistances shown in circuit are 2Ω each. The current in the resistance between D and E is

5 A 2.5 A 1 A 7.5 A

2.5 A

Due to balanced Wheatstone bridge, resistance between A and B can be removed. The current between D and E, $I_{D E}=\frac{10 V}{R_{D E}+R_{H G}}=\frac{10}{2+2}$ = 2.5 A