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(D) To find the equivalent resistance between $A$ and $B$,we simplify the circuit step-by-step.By labeling the nodes,we observe that the circuit can b

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$19$

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(D) To find the equivalent resistance between $A$ and $B$,we simplify the circuit step-by-step.By labeling the nodes,we observe that the circuit can be reduced by identifying parallel branches.The resistors $10 \Omega$ and $5 \Omega$ are in series $(15 \Omega)$,$4 \Omega$ and $11 \Omega$ are in series $(15 \Omega)$,and the central vertical branch also effectively forms a $15 \Omega$ path between nodes $C$ and $D$.These three $15 \Omega$ branches are in parallel between nodes $C$ and $D$.The equivalent resistance of these three parallel branches is $R_p = \frac{15 \Omega}{3} = 5 \Omega$.Now,the circuit simplifies to a series combination of $6 \Omega$,$5 \Omega$,and $8 \Omega$.Therefore,the total equivalent resistance is $R_{eq} = 6 \Omega + 5 \Omega + 8 \Omega = 19 \Omega$.

The equivalent resistance between $A$ and $B$ is: (in $\Omega$)

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