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(D) $1$. The circuit consists of a resistor $r$ in series with a parallel combination of two branches.$2$. The upper branch contains three resistors o

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$5r/2$

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(D) $1$. The circuit consists of a resistor $r$ in series with a parallel combination of two branches.$2$. The upper branch contains three resistors of resistance $r$ in series,so its equivalent resistance is $R_1 = r + r + r = 3r$.$3$. The lower branch also contains three resistors of resistance $r$ in series,so its equivalent resistance is $R_2 = r + r + r = 3r$.$4$. These two branches are in parallel,so their combined equivalent resistance $R_p$ is given by $\frac{1}{R_p} = \frac{1}{R_1} + \frac{1}{R_2} = \frac{1}{3r} + \frac{1}{3r} = \frac{2}{3r}$. Thus,$R_p = \frac{3r}{2}$.$5$. The total effective resistance of the network is the sum of the series resistor $r$ and the parallel combination $R_p$: $R_{effective} = r + R_p = r + \frac{3r}{2} = \frac{5r}{2}$.

Referring to the figure below,the effective resistance of the network is

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