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(A) The given circuit is a balanced Wheatstone bridge because the ratio of resistances in the arms is equal $(10/10 = 10/10)$.In a balanced Wheatstone

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$10$

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(A) The given circuit is a balanced Wheatstone bridge because the ratio of resistances in the arms is equal $(10/10 = 10/10)$.In a balanced Wheatstone bridge,the potential difference across the central diagonal resistor is zero,so no current flows through it.We can remove the central $10\,\Omega$ resistor from the circuit.Now,the upper two $10\,\Omega$ resistors are in series,giving an equivalent resistance of $10 + 10 = 20\,\Omega$.The lower two $10\,\Omega$ resistors are also in series,giving an equivalent resistance of $10 + 10 = 20\,\Omega$.These two branches are now in parallel between points $A$ and $B$.The total effective resistance $R_{eq}$ is given by $\frac{1}{R_{eq}} = \frac{1}{20} + \frac{1}{20} = \frac{2}{20} = \frac{1}{10}$.Therefore,$R_{eq} = 10\,\Omega$.

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

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