In Wheatstone's bridge `P = 9 ohm`, Q = 11 ohm`, R = 4 ohm` and `S = 6 ohm`. How much resistance must be put in parallel to the resistance `S` to balance the bridge

Options

A. `24 Omega`
B. `44/ 9 Omega`
C. `26.4 Omega`
D. `18.4 Omega`

Answer

To solve the problem of finding the resistance that must be put in parallel to the resistance \( S \) in a Wheatstone bridge to achieve balance, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Wheatstone Bridge Condition**:
For a Wheatstone bridge to be balanced, the following condition must hold:
\[ \frac{P}{Q} = \frac{R}{S'} \]
where \( S' \) is the equivalent resistance of \( S \) when a resistance \( R' \) is connected in parallel with it.

2. **Identify Given Values**:
From the problem, we have:
- \( P = 9 \, \Omega \)
- \( Q = 11 \, \Omega \)
- \( R = 4 \, \Omega \)
- \( S = 6 \, \Omega \)

3. **Express the Equivalent Resistance \( S' \)**:
When a resistance \( R' \) is connected in parallel with \( S \), the equivalent resistance \( S' \) can be expressed as:
\[ \frac{1}{S'} = \frac{1}{S} + \frac{1}{R'} \]
Rearranging gives:
\[ S' = \frac{S \cdot R'}{S + R'} \]

4. **Substitute for \( S' \)**:
From the balance condition:
\[ \frac{P}{Q} = \frac{R}{S'} \]
We can substitute \( S' \):
\[ \frac{9}{11} = \frac{4}{\frac{6 \cdot R'}{6 + R'}} \]

5. **Cross Multiply and Solve for \( R' \)**:
Cross multiplying gives:
\[ 9 \cdot \frac{6 \cdot R'}{6 + R'} = 44 \]
Simplifying this:
\[ 54R' = 44(6 + R') \]
Expanding:
\[ 54R' = 264 + 44R' \]
Rearranging terms:
\[ 54R' - 44R' = 264 \]
\[ 10R' = 264 \]
Dividing by 10:
\[ R' = 26.4 \, \Omega \]

### Final Answer:
The resistance that must be put in parallel with the resistance \( S \) to balance the bridge is \( R' = 26.4 \, \Omega \).