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(A) First,we calculate the equivalent resistance of the network.$1$. The first parallel combination of two $4 \, R$ resistors is $R_{p1} = \frac{4R

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$8$

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(A) First,we calculate the equivalent resistance of the network.$1$. The first parallel combination of two $4 \, R$ resistors is $R_{p1} = \frac{4R 	imes 4R}{4R + 4R} = 2R$.$2$. The second parallel combination of $6 \, R$ and $12 \, R$ resistors is $R_{p2} = \frac{6R 	imes 12R}{6R + 12R} = \frac{72R^2}{18R} = 4R$.$3$. The total equivalent resistance $R_{eq}$ is the sum of these in series with the other two $R$ resistors: $R_{eq} = 2R + R + 4R + R = 8R$.$4$. The power consumed is given by $P = \frac{V^2}{R_{eq}}$.$5$. Substituting the given values: $4 = \frac{16^2}{8R} \implies 4 = \frac{256}{8R} \implies 4 = \frac{32}{R}$.$6$. Therefore,$R = \frac{32}{4} = 8 \, \Omega$.

The resistive network shown below is connected to a $D.C.$ source of $16 \, V$. The power consumed by the network is $4 \, W$. The value of $R$ is ............. $\Omega$.

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