Three resistance `P, Q, R` each of `2 Omega` and an unknown resistance `S` from the four amrs of a Wheatstone's bridge circuit. When a resistance of `6 Omega` is connected in parallel to `S` the bridge gets balanced. What is the value of `S` ?

Options

A. `2Omega`
B. `3Omega`
C. `6Omega`
D. `1Omega`

Answer

To solve the problem, we need to find the value of the unknown resistance \( S \) in a Wheatstone bridge circuit where three resistances \( P, Q, R \) are each \( 2 \, \Omega \) and a \( 6 \, \Omega \) resistance is connected in parallel with \( S \) to balance the bridge.

### Step-by-Step Solution:

1. **Understand the Wheatstone Bridge Balance Condition**:
The Wheatstone bridge is balanced when the ratio of the resistances in one arm is equal to the ratio of the resistances in the other arm. Mathematically, this can be expressed as:
\[ \frac{P}{Q} = \frac{R}{S_{\text{eq}}} \]
where \( S_{\text{eq}} \) is the equivalent resistance of \( S \) in parallel with the \( 6 \, \Omega \) resistor.

2. **Identify the Known Values**:
From the problem, we have:
- \( P = 2 \, \Omega \)
- \( Q = 2 \, \Omega \)
- \( R = 2 \, \Omega \)
- The resistance \( S \) is unknown.
- A \( 6 \, \Omega \) resistor is connected in parallel with \( S \).

3. **Calculate the Equivalent Resistance \( S_{\text{eq}} \)**:
The equivalent resistance \( S_{\text{eq}} \) of \( S \) in parallel with \( 6 \, \Omega \) is given by the formula:
\[ S_{\text{eq}} = \frac{S \cdot 6}{S + 6} \]

4. **Substitute Values into the Balance Condition**:
Substituting the known values into the balance condition:
\[ \frac{2}{2} = \frac{2}{S_{\text{eq}}} \]
This simplifies to:
\[ 1 = \frac{2}{S_{\text{eq}}} \]
Therefore:
\[ S_{\text{eq}} = 2 \, \Omega \]

5. **Set Up the Equation for \( S \)**:
Now we can set up the equation using the expression for \( S_{\text{eq}} \):
\[ \frac{S \cdot 6}{S + 6} = 2 \]

6. **Cross Multiply and Solve for \( S \)**:
Cross multiplying gives:
\[ S \cdot 6 = 2(S + 6) \]
Expanding this:
\[ 6S = 2S + 12 \]
Rearranging the equation:
\[ 6S - 2S = 12 \]
\[ 4S = 12 \]
Dividing both sides by 4:
\[ S = 3 \, \Omega \]

### Final Answer:
The value of the unknown resistance \( S \) is \( 3 \, \Omega \).