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(B) The circuit is symmetric about the axis $xy$. The nodes above and below the horizontal axis are at the same potential due to symmetry.By simplifyi

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$2R/5$

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(B) The circuit is symmetric about the axis $xy$. The nodes above and below the horizontal axis are at the same potential due to symmetry.By simplifying the circuit using symmetry,the resistors connected in parallel can be combined.The equivalent resistance of the network between $x$ and $y$ is calculated as follows:$\frac{1}{R_{xy}} = \frac{1}{2R} + \frac{1}{2R} + \frac{1}{2R} + \frac{1}{2R} + \frac{1}{2R} = \frac{5}{2R}$Therefore,$R_{xy} = \frac{2R}{5}$.

Identical resistors of resistance $R$ are connected as shown in the figure. Find the equivalent resistance between $x$ and $y$.

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