The current in resistance $R_3$ in the given circuit is $\frac{2}{x}A$. Find the value of $x$.

 

Points A and B are at same potential. Between A&D and between B&C, the battery emf (2V) is same. Hence, D and C are also at same potential. No current flows through $R_1$ .
Between D&E and between C&F, the battery emf (3V) is same. Hence, E and F are also at same potential. No current flows through $R_2$ . $R_1\&R_2$can be disconnected.  
The effective circuit reduces to

$i=\frac{4-2}{3}=\frac{2}{3}A$