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(D) The given circuit can be analyzed by identifying the Wheatstone bridge structure. The resistors form a bridge between points $A$ and $B$ with node

Vedclass · Accepted Answer

$1.5$

Vedclass · Answer

(D) The given circuit can be analyzed by identifying the Wheatstone bridge structure. The resistors form a bridge between points $A$ and $B$ with nodes $C$ and $D$.Since all resistors are $3 \ \Omega$,the ratio of resistances in the arms is $\frac{3}{3} = \frac{3}{3}$,which satisfies the balanced Wheatstone bridge condition.Therefore,no current flows through the central resistor connected between $C$ and $D$. We can remove this resistor from the circuit.After removing the central resistor,the circuit simplifies to two parallel branches connected between $A$ and $B$:$1$. The upper branch consists of two $3 \ \Omega$ resistors in series: $R_{ACB} = 3 + 3 = 6 \ \Omega$.$2$. The lower branch consists of two $3 \ \Omega$ resistors in series: $R_{ADB} = 3 + 3 = 6 \ \Omega$.$3$. There is also a direct $3 \ \Omega$ resistor connected in parallel between $A$ and $B$.Now,we have three resistors of $6 \ \Omega, 6 \ \Omega$,and $3 \ \Omega$ connected in parallel.The equivalent resistance $R_{eq}$ is given by:$\frac{1}{R_{eq}} = \frac{1}{6} + \frac{1}{6} + \frac{1}{3}$$\frac{1}{R_{eq}} = \frac{1 + 1 + 2}{6} = \frac{4}{6} = \frac{2}{3}$$R_{eq} = \frac{3}{2} = 1.5 \ \Omega$.

The equivalent resistance between $A$ and $B$ in the given circuit is . . . . . . . . (in $\Omega$)

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