In the circuit shown in figure, the equivalent resistance between A and B is

 

 

 

(1)

The given network can be modified as shown in figure. Resistance of the middle branch

$\begin{array}{l}{R}^{\text{'}}=\left(\frac{8\times 2}{8+2}+\frac{4\times 6}{4+6}\right)\Omega =4\Omega \\ \therefore \frac{1}{R_{AB}}=\frac{1}{3}+\frac{1}{4}+\frac{1}{6}=\frac{4+3+2}{12}=\frac{3}{4}\Rightarrow R_{AB}=\frac{4}{3}\Omega \end{array}$