Calculating Resistance in a Parallel Circuit

This problem involves calculating the individual resistances in a circuit where two resistances are initially connected in parallel. We are given the equivalent resistance when both are present and the effective resistance when one of them is broken.

Understanding the Problem

We have two resistors, let's call their resistances $R_1$ and $R_2$.

Case 1: Both resistors are connected in parallel. The equivalent resistance ($R_{eq}$) is given as $6/5 \, \Omega$.

Case 2: One of the resistors breaks. Let's assume resistor with resistance $R_1$ breaks. The circuit now effectively has only the resistor with resistance $R_2$. The problem states that the effective resistance in this case becomes $2 \, \Omega$. This means the resistance of the remaining wire ($R_2$) is $2 \, \Omega$.

Applying the Formula for Parallel Resistance

The formula for the equivalent resistance ($R_{eq}$) of two resistors ($R_1$ and $R_2$) connected in parallel is:

$\frac{1}{R_{eq}} = \frac{1}{R_1} + \frac{1}{R_2}$

Solving for the Unknown Resistance

From Case 2, we know that the resistance of the wire that did NOT break is $R_2 = 2 \, \Omega$.

From Case 1, we know that the equivalent resistance when both were in parallel was $R_{eq} = 6/5 \, \Omega$.

Now, we can substitute these values into the parallel resistance formula:

$\frac{1}{6/5} = \frac{1}{R_1} + \frac{1}{2}$

Simplify the left side:

$\frac{5}{6} = \frac{1}{R_1} + \frac{1}{2}$

To find $\frac{1}{R_1}$, subtract $\frac{1}{2}$ from both sides:

$\frac{1}{R_1} = \frac{5}{6} - \frac{1}{2}$

To subtract the fractions, find a common denominator, which is 6:

$\frac{1}{R_1} = \frac{5}{6} - \frac{3}{6}$

Perform the subtraction:

$\frac{1}{R_1} = \frac{5 - 3}{6} = \frac{2}{6}$

Simplify the fraction:

$\frac{1}{R_1} = \frac{1}{3}$

To find $R_1$, take the reciprocal of both sides:

$R_1 = 3 \, \Omega$

Thus, the resistance of the wire that got broken was $3 \, \Omega$. We assumed $R_1$ was the broken wire, and we found $R_1 = 3 \, \Omega$. If we had assumed $R_2$ was the broken wire, then $R_1 = 2 \, \Omega$ would be the remaining resistance, and we would solve for $R_2$ using the same method, getting $R_2 = 3 \, \Omega$. The result is consistent.

Conclusion

The resistance of the wire that got broken is $3 \, \Omega$.