In a given network, each resistance has value of $6Ω$ . The point $X$ is connected to point $A$ by a copper wire of negligible resistance and point $Y$ is connected to point B by the same wire. The effective resistance between $X$ and $Y$ will be

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One end of all the three resistors are connected at the point $X$ and the other at the point $Y$ . Therefore, the equivalent circuit is given below:
The equivalent resistance is given by $\frac{1}{R} = \frac{1}{6} + \frac{1}{6} + \frac{1}{6} = \frac{3}{6} = \frac{1}{2} \Rightarrow R_{eq} = 2Ω$

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