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(D) Let the two resistors be $R_1$ and $R_2$. When connected in parallel,the resultant resistance $R_{eq}$ is given by $\frac{1}{R_{eq}} = \frac{1}{R_

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$3$

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(D) Let the two resistors be $R_1$ and $R_2$. When connected in parallel,the resultant resistance $R_{eq}$ is given by $\frac{1}{R_{eq}} = \frac{1}{R_1} + \frac{1}{R_2}$,or $R_{eq} = \frac{R_1 R_2}{R_1 + R_2}$.Given $R_{eq} = \frac{6}{5} \,\Omega$.When one wire is broken,the circuit effectively becomes a single resistor. Since the new effective resistance is $2 \,\Omega$,one of the resistors must be $2 \,\Omega$. Let $R_1 = 2 \,\Omega$.Substituting these values into the parallel formula: $\frac{2 R_2}{2 + R_2} = \frac{6}{5}$.Cross-multiplying gives: $10 R_2 = 6(2 + R_2)$.$10 R_2 = 12 + 6 R_2$.$4 R_2 = 12$.$R_2 = 3 \,\Omega$.Thus,the resistance of the wire that was broken is $3 \,\Omega$.

Two resistors are joined in parallel,and their resultant resistance is $\frac{6}{5} \,\Omega$. One of the resistance wires is broken,and the effective resistance becomes $2 \,\Omega$. Then the resistance (in $\Omega$) of the wire that got broken is ..........

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