For the network shown in fig, the value of n current is

The circuit is a balanced Wheatstone bridge. Therefore, the resistance 4Ω is  and can be eliminated. The resistances 4 Ω and 2Ω are in series and their equivalent resistance is 6Ω. Similarly, the equivalent resistance of 6Ω and 3Ω resistances is 9 Ω The equivalent resistance of the circuit is given byReq=6×96+9=185∴ i=VReq=V(18/5)=5V18