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(D) $1$. Observe the bridge structure in the upper part of the circuit. The four $3\,\Omega$ resistors form a balanced Wheatstone bridge because the r

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$9$

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(D) $1$. Observe the bridge structure in the upper part of the circuit. The four $3\,\Omega$ resistors form a balanced Wheatstone bridge because the ratio of resistances in opposite arms is equal $(3/3 = 3/3)$.$2$. In a balanced Wheatstone bridge,no current flows through the central branch. Thus,the two $3\,\Omega$ resistors in the upper arms are in series $(3 + 3 = 6\,\Omega)$,and the two $3\,\Omega$ resistors in the lower arms are also in series $(3 + 3 = 6\,\Omega)$.$3$. These two $6\,\Omega$ branches are in parallel,so their equivalent resistance is $R_p = (6 	imes 6) / (6 + 6) = 36 / 12 = 3\,\Omega$.$4$. Now,the total circuit consists of this $3\,\Omega$ equivalent resistance in series with the two $3\,\Omega$ resistors connected to terminals $A$ and $B$ at the bottom. The total resistance is $R_{AB} = 3 + 3 + 3 = 9\,\Omega$.

The equivalent resistance between points $A$ and $B$ is ............ $\Omega$.

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