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(A) The given circuit is a balanced Wheatstone bridge-like structure. Due to the symmetry of the circuit,the potential at the nodes connected by the v

Vedclass · Accepted Answer

$\frac{9}{2} \, \Omega$

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(A) The given circuit is a balanced Wheatstone bridge-like structure. Due to the symmetry of the circuit,the potential at the nodes connected by the vertical $3 \, \Omega$ resistors is the same. Therefore,no current flows through these vertical resistors.We can remove these vertical resistors from the circuit.Now,the top branch consists of three $3 \, \Omega$ resistors in series,giving a total resistance of $3 + 3 + 3 = 9 \, \Omega$.Similarly,the bottom branch consists of three $3 \, \Omega$ resistors in series,giving a total resistance of $3 + 3 + 3 = 9 \, \Omega$.These two branches are in parallel between points $A$ and $B$.The equivalent resistance $R_{AB}$ is given by $\frac{1}{R_{AB}} = \frac{1}{9} + \frac{1}{9} = \frac{2}{9}$.Thus,$R_{AB} = \frac{9}{2} \, \Omega$.

Calculate the equivalent resistance between $A$ and $B$.

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