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(B) Let the two resistors be $R_1$ and $R_2$. The equivalent resistance of two resistors in parallel is given by $\frac{1}{R_{eq}} = \frac{1}{R_1} + \

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$3$

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(B) Let the two resistors be $R_1$ and $R_2$. The equivalent resistance of two resistors in parallel is given by $\frac{1}{R_{eq}} = \frac{1}{R_1} + \frac{1}{R_2}$.Given $R_{eq} = \frac{12}{7} \ \Omega$. So,$\frac{1}{R_1} + \frac{1}{R_2} = \frac{7}{12}$.If one resistor is disconnected,the remaining resistance is simply the value of the other resistor. Given this is $4 \ \Omega$,let $R_1 = 4 \ \Omega$.Substituting $R_1$ into the parallel equation: $\frac{1}{4} + \frac{1}{R_2} = \frac{7}{12}$.$\frac{1}{R_2} = \frac{7}{12} - \frac{1}{4} = \frac{7-3}{12} = \frac{4}{12} = \frac{1}{3}$.Therefore,$R_2 = 3 \ \Omega$.

The effective resistance of two resistors in parallel is $\frac{12}{7} \ \Omega$. If one of the resistors is disconnected,the resistance becomes $4 \ \Omega$. The resistance of the other resistor is ............. $\Omega$.

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