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(A) To find the equivalent resistance across $A$ and $B$,we simplify the circuit step-by-step:$1$. The two $4 \,\Omega$ resistors on the left are in p

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$2$

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(A) To find the equivalent resistance across $A$ and $B$,we simplify the circuit step-by-step:$1$. The two $4 \,\Omega$ resistors on the left are in parallel,giving an equivalent resistance of $R_1 = (4 	imes 4) / (4 + 4) = 2 \,\Omega$.$2$. The two $4 \,\Omega$ resistors on the right are also in parallel,giving an equivalent resistance of $R_2 = (4 	imes 4) / (4 + 4) = 2 \,\Omega$.$3$. Now,the circuit simplifies to a bridge-like structure where we have two branches in parallel between the input and output nodes. Each branch consists of two $2 \,\Omega$ resistors in series.$4$. The resistance of the upper branch is $2 \,\Omega + 2 \,\Omega = 4 \,\Omega$.$5$. The resistance of the lower branch is $2 \,\Omega + 2 \,\Omega = 4 \,\Omega$.$6$. Finally,these two $4 \,\Omega$ branches are in parallel,so the total equivalent resistance is $R_{eq} = (4 	imes 4) / (4 + 4) = 2 \,\Omega$.

The equivalent resistance across $A$ and $B$ as shown in the figure is ............. $\Omega$.

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