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(D) The circuit consists of several branches connected in parallel between points $A$ and $B$.$1$. The first branch (leftmost) has two $5 \Omega$ resi

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$2$

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(D) The circuit consists of several branches connected in parallel between points $A$ and $B$.$1$. The first branch (leftmost) has two $5 \Omega$ resistors in series,giving $R_1 = 5 \Omega + 5 \Omega = 10 \Omega$.$2$. The second branch has a $10 \Omega$ resistor connected directly between $A$ and $B$,so $R_2 = 10 \Omega$.$3$. The third branch has two $5 \Omega$ resistors in series,giving $R_3 = 5 \Omega + 5 \Omega = 10 \Omega$.$4$. The fourth branch has a $10 \Omega$ resistor connected directly between $A$ and $B$,so $R_4 = 10 \Omega$.$5$. The fifth branch has a $10 \Omega$ resistor connected directly between $A$ and $B$,so $R_5 = 10 \Omega$.All these branches are in parallel. The equivalent resistance $R_{eq}$ is given by:$\frac{1}{R_{eq}} = \frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_3} + \frac{1}{R_4} + \frac{1}{R_5}$$\frac{1}{R_{eq}} = \frac{1}{10} + \frac{1}{10} + \frac{1}{10} + \frac{1}{10} + \frac{1}{10} = \frac{5}{10} = \frac{1}{2}$$R_{eq} = 2 \Omega$

The equivalent resistance between points $A$ and $B$ in the given network is: (in $\Omega$)

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