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(B) $(NONE)$ In the given circuit, the resistors of $3\, \Omega$ and $3\, \Omega$ connected between $A-D$ and $D-C$ are in series, so their equivalent

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$2$

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(B) $(NONE)$ In the given circuit, the resistors of $3\, \Omega$ and $3\, \Omega$ connected between $A-D$ and $D-C$ are in series, so their equivalent resistance is $3\, \Omega + 3\, \Omega = 6\, \Omega$.This $6\, \Omega$ equivalent resistance is in parallel with the $6\, \Omega$ resistor connected directly between $A$ and $C$.The equivalent resistance between $A$ and $C$ is $R_{AC} = \frac{6 	imes 6}{6 + 6} = 3\, \Omega$.Now, this $R_{AC} = 3\, \Omega$ is in series with the $3\, \Omega$ resistor connected between $C$ and $B$, giving a resistance of $3\, \Omega + 3\, \Omega = 6\, \Omega$.This $6\, \Omega$ combination is in parallel with the $3\, \Omega$ resistor connected directly between $A$ and $B$.Therefore, the total effective resistance between $A$ and $B$ is $R_{AB} = \frac{6 	imes 3}{6 + 3} = 2\, \Omega$.

The effective resistance between the points $A$ and $B$ in the figure is ............. $\Omega$.

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