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(D) The given circuit consists of three identical blocks connected in series. Each block is a Wheatstone bridge-like structure where two $3\,\Omega$ r

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$9$

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(D) The given circuit consists of three identical blocks connected in series. Each block is a Wheatstone bridge-like structure where two $3\,\Omega$ resistors are in series with two other $3\,\Omega$ resistors,and these two branches are in parallel.For one block:$1$. The upper branch has two $3\,\Omega$ resistors in series: $R_1 = 3\,\Omega + 3\,\Omega = 6\,\Omega$.$2$. The lower branch has two $3\,\Omega$ resistors in series: $R_2 = 3\,\Omega + 3\,\Omega = 6\,\Omega$.$3$. These two branches are in parallel,so the equivalent resistance of one block is $R_{block} = \frac{R_1 	imes R_2}{R_1 + R_2} = \frac{6 	imes 6}{6 + 6} = \frac{36}{12} = 3\,\Omega$.Since there are three such blocks connected in series,the total equivalent resistance between $A$ and $B$ is:$R_{eq} = R_{block} + R_{block} + R_{block} = 3\,\Omega + 3\,\Omega + 3\,\Omega = 9\,\Omega$.

In the network of resistors shown in the adjoining figure,the equivalent resistance between $A$ and $B$ is ............ $ohm$.

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