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(A) The given circuit can be simplified by identifying the series and parallel combinations of resistors.Looking at the circuit,the two $10 \Omega$ re

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$10 \Omega$

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(A) The given circuit can be simplified by identifying the series and parallel combinations of resistors.Looking at the circuit,the two $10 \Omega$ resistors on the left are in series,and the two $10 \Omega$ resistors on the right are in series.Let the top branch have two $10 \Omega$ resistors in series,giving $R_1 = 10 \Omega + 10 \Omega = 20 \Omega$.Similarly,the bottom branch has two $10 \Omega$ resistors in series,giving $R_2 = 10 \Omega + 10 \Omega = 20 \Omega$.These two branches are connected in parallel between points $A$ and $B$.Therefore,the equivalent resistance $R_{eq}$ is given by:$\frac{1}{R_{eq}} = \frac{1}{R_1} + \frac{1}{R_2} = \frac{1}{20} + \frac{1}{20} = \frac{2}{20} = \frac{1}{10}$$R_{eq} = 10 \Omega$.

Find the effective resistance between $A$ and $B$ in the given circuit.

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