In the following circuit, the current through the resistor $R(=2\Omega )$ is I amperes. The value of I is_____ 

Due to balanced wheatstone bridge  $ABCEABE=8\Omega$ can be taken out
 

$R_{AC}=\frac{(1+2)(2+4)}{(1+2)+(2+4)}=\frac{18}{9}=2$
Now, ACFGHA is a balanced Wheatstone bridge, hence HC=10Ω can be taken out

$R_{AG}=\frac{6\times 18}{6+18}=\frac{108}{24}=4.5\Omega$

Hence, current $i=\frac{V}{R_{total}}=\frac{6.5}{6.5}=1A$