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(B) To find the equivalent resistance between points $A$ and $B$,we analyze the circuit structure.Looking at the circuit,we can see that it consists o

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$\frac {5R}{8}$

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(B) To find the equivalent resistance between points $A$ and $B$,we analyze the circuit structure.Looking at the circuit,we can see that it consists of five resistors,each of resistance $R$.By redrawing the circuit,we can identify the series and parallel combinations.Two resistors are in series with each other,forming a branch of resistance $2R$. This branch is in parallel with another resistor $R$,resulting in an equivalent resistance of $\frac{R 	imes 2R}{R + 2R} = \frac{2R}{3}$.This combination is in series with another resistor $R$,giving a total resistance of $R + \frac{2R}{3} = \frac{5R}{3}$.Finally,this entire combination is in parallel with the remaining resistor $R$ connected across $A$ and $B$.The total equivalent resistance $R_{eq}$ is given by $\frac{1}{R_{eq}} = \frac{1}{R} + \frac{1}{5R/3} = \frac{1}{R} + \frac{3}{5R} = \frac{5+3}{5R} = \frac{8}{5R}$.Therefore,$R_{eq} = \frac{5R}{8}$.

The equivalent resistance between $A$ and $B$ is

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