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(A) Let the two resistances be $R_1$ and $R_2$. When connected in parallel,the equivalent resistance $R_{eq}$ is given by $\frac{R_1 R_2}{R_1 + R_2} =

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$3$

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(A) Let the two resistances be $R_1$ and $R_2$. When connected in parallel,the equivalent resistance $R_{eq}$ is given by $\frac{R_1 R_2}{R_1 + R_2} = \frac{6}{5} \, \Omega$. When one resistance is removed,the circuit consists of only the remaining resistance. Given that the effective resistance becomes $2 \, \Omega$,we can assume $R_1 = 2 \, \Omega$. Substituting $R_1 = 2 \, \Omega$ into the parallel combination formula: $\frac{2 R_2}{2 + R_2} = \frac{6}{5}$ $10 R_2 = 6(2 + R_2)$ $10 R_2 = 12 + 6 R_2$ $4 R_2 = 12$ $R_2 = 3 \, \Omega$. Thus,the resistance of the wire removed is $3 \, \Omega$.

When two resistances are connected in parallel,the equivalent resistance is $\frac{6}{5} \, \Omega$. When one of the resistances is removed,the effective resistance is $2 \, \Omega$. The resistance of the wire removed will be ................ $\Omega$.

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