The effective resistance between A and B is . | vedclass

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(B) The given circuit is a balanced Wheatstone bridge. Let the nodes be labeled. The bridge consists of four $5 \ \Omega$ resistors forming the arms a

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$5$

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(B) The given circuit is a balanced Wheatstone bridge. Let the nodes be labeled. The bridge consists of four $5 \ \Omega$ resistors forming the arms and a central $5 \ \Omega$ resistor. Since the ratio of resistances in the arms is equal $(5/5 = 5/5)$,the bridge is balanced. Therefore,no current flows through the central $5 \ \Omega$ resistor. We can remove the central resistor. The circuit simplifies to two parallel branches,each containing two $5 \ \Omega$ resistors in series. Resistance of the upper branch = $5 \ \Omega + 5 \ \Omega = 10 \ \Omega$. Resistance of the lower branch = $5 \ \Omega + 5 \ \Omega = 10 \ \Omega$. These two branches are in parallel,so the effective resistance $R_{eq}$ is given by: $\frac{1}{R_{eq}} = \frac{1}{10} + \frac{1}{10} = \frac{2}{10} = \frac{1}{5}$. Thus,$R_{eq} = 5 \ \Omega$.

The effective resistance between $A$ and $B$ is ........... $\Omega$.

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