b
Resistances \(R _{ AF }\) and \(R _{ FE }\) are in series combination. Therefore their equivalent resistance \(R = R _{ AF }+ R _{ FE }=3+3=6 \Omega\). Now the resistance \(R _{ AE }\) and equivalent resistance \(R\) ' are in parallel combination. Therefore relation for their equivalent resistance.

\(\frac{1}{R^{*}}=\frac{1}{R^{\prime}}+\frac{1}{R_{A E}}=\frac{1}{6}+\frac{1}{6}=\frac{2}{6}=\frac{1}{3} \Rightarrow R^{\prime \prime}=3 \Omega\).

We can calculate in the same manner for \(R_{E D}, R_{A C}, R_{D C}\). etc. and finally the circuit

reduces as shown in the figure.

Therefore, the equivalent resistance between \(A\) and \(B\)

\(=\frac{(3+3) \times 3}{(3+3)+3}=\frac{18}{9}=2 \Omega\)