In the following circuit, the current through the resistor R(=2Ω) is I Amperes. What is the value of I?

Due to balanced wheatstone bridge ABCE, ABE=8Ω can be taken out. RAC=(1+2)(2+4)/(1+2)+(2+4)=18/9=2Ω. Now, ACFGHA is a balanced Wheatstone bridge, hence HC=10Ω can be taken out. RAG=6*18/(6+18)=108/24=4.5Ω. Rtotal=2+4.5=6.5Ω. Hence, current i=V/Rtotal=6.5/6.5=1A