In a Wheatstone's bridge, three resistances P,Q and R connected in the three arms and the fourth arm is formed by two resistances `S_1 and S_2` connected in parallel. The condition for the bridge to be balanced will be

Options

A. `P/Q = (R(S_1+S_2))/(2S_1S_2)`
B. `P/Q = (R/(S_1+S_2))`
C. `P/Q = (2R)/(S_1+S_2)`
D. `P/Q = (R(S_1+S_2))/(S_1S_2)`

Explanation

To solve the problem regarding the condition for a Wheatstone bridge to be balanced with resistances P, Q, R, S1, and S2, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Configuration**: In a Wheatstone bridge, we have four resistances. Three of them are P, Q, and R, and the fourth arm consists of two resistances S1 and S2 connected in parallel.

2. **Identify the Condition for Balance**: The Wheatstone bridge is balanced when there is no current flowing through the galvanometer (or the branch connecting the two midpoints of the bridge). This occurs when the ratio of the resistances in one branch is equal to the ratio of the resistances in the other branch.

3. **Set Up the Ratios**: For the bridge to be balanced, we can express the condition as:
\[ \frac{P}{Q} = \frac{R}{S_{\text{equivalent}}} \]
where \( S_{\text{equivalent}} \) is the equivalent resistance of S1 and S2 in parallel.

4. **Calculate the Equivalent Resistance**: The equivalent resistance \( S_{\text{equivalent}} \) for two resistances S1 and S2 connected in parallel is given by the formula:
\[ S_{\text{equivalent}} = \frac{S_1 S_2}{S_1 + S_2} \]

5. **Substitute the Equivalent Resistance**: Substitute \( S_{\text{equivalent}} \) back into the balance condition:
\[ \frac{P}{Q} = \frac{R}{\frac{S_1 S_2}{S_1 + S_2}} \]

6. **Cross-Multiply to Simplify**: Cross-multiplying gives:
\[ P \cdot \frac{S_1 S_2}{S_1 + S_2} = Q \cdot R \]

7. **Rearranging the Equation**: Rearranging the equation leads to:
\[ P \cdot S_1 S_2 = Q \cdot R \cdot (S_1 + S_2) \]

8. **Final Form of the Condition**: This can be expressed as:
\[ \frac{P}{Q} = R \cdot \frac{S_1 + S_2}{S_1 S_2} \]

### Conclusion:
The condition for the Wheatstone bridge to be balanced is:
\[ \frac{P}{Q} = R \cdot \frac{S_1 + S_2}{S_1 S_2} \]