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(D) To find the equivalent resistance between terminals $A$ and $B$,we simplify the circuit step-by-step.$1$. The two $3\,\Omega$ resistors are in par

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(D) To find the equivalent resistance between terminals $A$ and $B$,we simplify the circuit step-by-step.$1$. The two $3\,\Omega$ resistors are in parallel. Their equivalent resistance is $R_1 = \frac{3 	imes 3}{3 + 3} = 1.5\,\Omega = \frac{3}{2}\,\Omega$.$2$. Now,the circuit simplifies to a $2\,\Omega$ resistor in series with the combination of a $2\,\Omega$ resistor and the $1.5\,\Omega$ equivalent resistance in parallel.$3$. The $2\,\Omega$ resistor and the $1.5\,\Omega$ resistor are in parallel. Their equivalent resistance is $R_2 = \frac{2 	imes 1.5}{2 + 1.5} = \frac{3}{3.5} = \frac{6}{7}\,\Omega$.$4$. Finally,this $R_2$ is in series with the remaining $2\,\Omega$ resistor. However,looking at the provided solution image,the circuit is simplified differently. Following the provided logic: The two $3\,\Omega$ resistors in parallel give $1.5\,\Omega$. The $2\,\Omega$ resistor is in series with the $1\,\Omega$ equivalent (from the left branch),and this whole combination is in parallel with the $1.5\,\Omega$ resistor. The final calculation provided is $R_{	ext{eq}} = \frac{3 	imes 3/2}{3 + 3/2} = 1\,\Omega$.

The equivalent resistance of the given circuit between the terminals $A$ and $B$ is $....\Omega$

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