Two resistances are joined in parallel whose resultant is `6//8 ohm`. One of the resistance wire is broken and the effective resistance becomes `2 Omega` . Then the resistance in ohm of the wire that got broken was

Options

A. `3//5`
B. 2
C. `6//5`
D. 3

Answer

To solve the problem step by step, we will start by defining the variables and using the formulas related to resistances in parallel.

### Step 1: Define the resistances
Let the two resistances be \( R_1 \) and \( R_2 \). According to the problem, they are connected in parallel.

### Step 2: Use the formula for equivalent resistance in parallel
The formula for the equivalent resistance \( R_{eq} \) of two resistances in parallel is given by:
\[ \frac{1}{R_{eq}} = \frac{1}{R_1} + \frac{1}{R_2} \]
From the problem, we know that:
\[ R_{eq} = \frac{6}{8} \text{ ohm} = 0.75 \text{ ohm} \]

### Step 3: Rewrite the equation
Using the equivalent resistance formula:
\[ \frac{1}{0.75} = \frac{1}{R_1} + \frac{1}{R_2} \]
This simplifies to:
\[ \frac{4}{3} = \frac{1}{R_1} + \frac{1}{R_2} \]

### Step 4: Assume one resistance is broken
When one of the resistances (let's assume \( R_2 \)) is broken, the effective resistance becomes \( R_{eq}' = 2 \text{ ohm} \). Therefore, we have:
\[ R_{eq}' = R_1 = 2 \text{ ohm} \]

### Step 5: Substitute \( R_1 \) into the equation
Now we can substitute \( R_1 \) into the equation we derived earlier:
\[ \frac{4}{3} = \frac{1}{2} + \frac{1}{R_2} \]

### Step 6: Solve for \( R_2 \)
First, we need to express \( \frac{1}{R_2} \):
\[ \frac{1}{R_2} = \frac{4}{3} - \frac{1}{2} \]
To subtract these fractions, we need a common denominator, which is 6:
\[ \frac{4}{3} = \frac{8}{6}, \quad \frac{1}{2} = \frac{3}{6} \]
Thus:
\[ \frac{1}{R_2} = \frac{8}{6} - \frac{3}{6} = \frac{5}{6} \]
Now, taking the reciprocal gives:
\[ R_2 = \frac{6}{5} \text{ ohm} \]

### Step 7: Conclusion
The resistance of the wire that got broken is:
\[ R_2 = 1.2 \text{ ohm} \]