Two resistance wires on joining in parallel the resultant resistance is ohms `(6)/(5)` ohms. One of the wire breaks, the effective resistance is 2 ohms . The resistance of the broken wire is

Options

A. `(3)/(5)Omega`
B. `2Omega`
C. `(6)/(5)Omega`
D. `3Omega`

Answer

To solve the problem, we need to find the resistance of the broken wire given the conditions of the circuit. Let's denote the two resistances as \( R_1 \) and \( R_2 \).

### Step 1: Write the formula for equivalent resistance in parallel
The formula for the equivalent resistance \( R \) of two resistors \( R_1 \) and \( R_2 \) connected in parallel is given by:
\[ \frac{1}{R} = \frac{1}{R_1} + \frac{1}{R_2} \]
Given that the resultant resistance when both wires are connected in parallel is \( \frac{6}{5} \) ohms, we can write:
\[ \frac{1}{\frac{6}{5}} = \frac{1}{R_1} + \frac{1}{R_2} \]
This simplifies to:
\[ \frac{5}{6} = \frac{1}{R_1} + \frac{1}{R_2} \]

### Step 2: Rearranging the equation
Rearranging the equation gives:
\[ \frac{1}{R_1} + \frac{1}{R_2} = \frac{5}{6} \]
Multiplying through by \( R_1 R_2 \) gives:
\[ R_2 + R_1 = \frac{5}{6} R_1 R_2 \]

### Step 3: Consider the case when one wire breaks
When one of the wires breaks, let's assume \( R_1 \) breaks. The effective resistance then becomes \( R_2 \), which is given as 2 ohms. Therefore:
\[ R_2 = 2 \text{ ohms} \]

### Step 4: Substitute \( R_2 \) back into the equation
Substituting \( R_2 = 2 \) into the earlier equation:
\[ 2 + R_1 = \frac{5}{6} R_1 \cdot 2 \]
This simplifies to:
\[ 2 + R_1 = \frac{10}{6} R_1 \]
or
\[ 2 + R_1 = \frac{5}{3} R_1 \]

### Step 5: Solve for \( R_1 \)
Now, rearranging gives:
\[ 2 = \frac{5}{3} R_1 - R_1 \]
This can be rewritten as:
\[ 2 = \left(\frac{5}{3} - \frac{3}{3}\right) R_1 \]
\[ 2 = \frac{2}{3} R_1 \]
Multiplying both sides by \( \frac{3}{2} \):
\[ R_1 = 3 \text{ ohms} \]

### Conclusion
Thus, the resistance of the broken wire \( R_1 \) is \( 3 \) ohms.

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