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(B) The given circuit is a pentagram structure. By symmetry,we can simplify the network. $1$. The network can be reduced by identifying series and par

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$7/6\, R$

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(B) The given circuit is a pentagram structure. By symmetry,we can simplify the network. $1$. The network can be reduced by identifying series and parallel combinations of resistors. $2$. After simplifying the outer and inner loops,the circuit reduces to two parallel branches between $A$ and $B$. $3$. Each branch has an equivalent resistance of $7R/3$. $4$. The total equivalent resistance $R_{AB}$ is given by the parallel combination of these two branches: $R_{AB} = \frac{(7R/3) 	imes (7R/3)}{(7R/3) + (7R/3)} = \frac{(7R/3)^2}{2 	imes (7R/3)} = \frac{7R}{6}$.

The resistance of each wire between any two adjacent dots is $R$. Then the equivalent resistance between $A$ and $B$ as shown in the figure is:

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