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(B) The circuit can be simplified by observing the symmetry and series-parallel combinations.$1$. The resistors along $AF$ and $FE$ are in series. The

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$2 \; \Omega$

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(B) The circuit can be simplified by observing the symmetry and series-parallel combinations.$1$. The resistors along $AF$ and $FE$ are in series. Their equivalent resistance is $R_1 = 3 \; \Omega + 3 \; \Omega = 6 \; \Omega$.$2$. This $R_1$ is in parallel with the $6 \; \Omega$ resistor connected along $AE$. The equivalent resistance $R_{AE}'$ is given by $\frac{1}{R_{AE}'} = \frac{1}{6} + \frac{1}{6} = \frac{2}{6} = \frac{1}{3} \Rightarrow R_{AE}' = 3 \; \Omega$.$3$. Similarly, the resistors along $ED$ and $DC$ are in series with the $6 \; \Omega$ resistor along $AD$ in parallel, leading to an equivalent resistance of $3 \; \Omega$ between $A$ and $C$.$4$. Finally, the circuit reduces to three resistors of $3 \; \Omega$ each connected between $A$ and $B$ (one directly, and two through path $C$).$5$. The equivalent resistance $R_{eq}$ is the parallel combination of the direct $3 \; \Omega$ resistor and the series combination of the other two $3 \; \Omega$ resistors $(3+3=6 \; \Omega)$.$6$. $R_{eq} = \frac{3 	imes 6}{3 + 6} = \frac{18}{9} = 2 \; \Omega$.

Six resistors of $3 \; \Omega$ each are connected along the sides of a hexagon and three resistors of $6 \; \Omega$ each are connected along $AC$, $AD$, and $AE$ as shown in the figure. The equivalent resistance between $A$ and $B$ is equal to

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