Three resistances P, Q and R each of resistance `5 Omega` and an unknown resistance S form the four arms of a Wheatstone bridge circuit. When a resistance of `10 Omega` is connected in parallel to S, the bridge gets balanced. The value of S is:

Options

A. `10 Omega`
B. `5 Omega`
C. `20 Omega`
D. `2 Omega`

Explanation

To solve the problem, we need to find the value of the unknown resistance \( S \) in a Wheatstone bridge circuit where three resistances \( P, Q, \) and \( R \) are each \( 5 \, \Omega \), and a \( 10 \, \Omega \) resistance is connected in parallel to \( S \) to balance the bridge.

### Step-by-Step Solution:

1. **Identify the resistances in the Wheatstone bridge:**
- Given resistances: \( P = 5 \, \Omega \), \( Q = 5 \, \Omega \), \( R = 5 \, \Omega \), and \( S \) is unknown.
- When a \( 10 \, \Omega \) resistance is connected in parallel with \( S \), the bridge becomes balanced.

2. **Use the condition for a balanced Wheatstone bridge:**
- For a balanced Wheatstone bridge, the ratio of the resistances must satisfy:
\[ \frac{P}{R} = \frac{Q}{M} \]
- Here, \( M \) is the equivalent resistance of \( S \) and \( 10 \, \Omega \) in parallel.

3. **Substitute the known values:**
- Since \( P = 5 \, \Omega \), \( R = 5 \, \Omega \), and \( Q = 5 \, \Omega \), we have:
\[ \frac{5}{5} = \frac{5}{M} \]
- This simplifies to:
\[ 1 = \frac{5}{M} \implies M = 5 \, \Omega \]

4. **Calculate the equivalent resistance \( M \):**
- The equivalent resistance \( M \) of \( S \) and \( 10 \, \Omega \) in parallel is given by:
\[ \frac{1}{M} = \frac{1}{S} + \frac{1}{10} \]
- Substituting \( M = 5 \, \Omega \):
\[ \frac{1}{5} = \frac{1}{S} + \frac{1}{10} \]

5. **Solve for \( S \):**
- Rearranging the equation gives:
\[ \frac{1}{S} = \frac{1}{5} - \frac{1}{10} \]
- Finding a common denominator (which is 10):
\[ \frac{1}{S} = \frac{2}{10} - \frac{1}{10} = \frac{1}{10} \]
- Taking the reciprocal:
\[ S = 10 \, \Omega \]

### Final Answer:
The value of \( S \) is \( 10 \, \Omega \).