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(D) The circuit is a balanced Wheatstone bridge. Let the nodes be $P$ (top) and $Q$ (bottom). The resistors are all $3 \, \Omega$. The ratio of resist

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$3 \, \Omega$

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(D) The circuit is a balanced Wheatstone bridge. Let the nodes be $P$ (top) and $Q$ (bottom). The resistors are all $3 \, \Omega$. The ratio of resistances in the arms is $\frac{3}{3} = \frac{3}{3}$,so the central $3 \, \Omega$ resistor carries no current.Removing the central resistor,we have two parallel branches. The top branch has two $3 \, \Omega$ resistors in series,giving $3 + 3 = 6 \, \Omega$. The bottom branch also has two $3 \, \Omega$ resistors in series,giving $3 + 3 = 6 \, \Omega$.These two $6 \, \Omega$ branches are in parallel. The equivalent resistance $R_{eq}$ is given by $\frac{1}{R_{eq}} = \frac{1}{6} + \frac{1}{6} = \frac{2}{6} = \frac{1}{3}$.Therefore,$R_{eq} = 3 \, \Omega$.

The equivalent resistance between points $A$ and $B$ in the following circuit is:

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