Resistances P, Q, S and R are arranged in a cyclic order to form a balanced Wheatstone's network. The ratio of power consumed in the branches (P+Q) and (R+S) is

Options

A. `1:1`
B. `R:P`
C. `P^(2):Q^(2)`
D. `P^(2):R^(2)`

Explanation

To solve the problem regarding the ratio of power consumed in the branches of a balanced Wheatstone network, we can follow these steps:

### Step 1: Understand the Wheatstone Bridge Configuration
In a balanced Wheatstone bridge, we have four resistances P, Q, R, and S arranged in a cyclic order. The branches of the bridge are (P + Q) and (R + S).

### Step 2: Write the Expression for Power
The power consumed in any branch can be expressed using the formula:
\[ P = \frac{V^2}{R} \]
where \( V \) is the potential difference across the branch and \( R \) is the equivalent resistance of that branch.

### Step 3: Identify the Branches
For the two branches:
- The first branch (P + Q) has a power \( P_1 \):
\[ P_1 = \frac{V^2}{P + Q} \]
- The second branch (R + S) has a power \( P_2 \):
\[ P_2 = \frac{V^2}{R + S} \]

### Step 4: Calculate the Ratio of Powers
The ratio of the powers consumed in the two branches can be expressed as:
\[ \frac{P_1}{P_2} = \frac{\frac{V^2}{P + Q}}{\frac{V^2}{R + S}} \]
This simplifies to:
\[ \frac{P_1}{P_2} = \frac{R + S}{P + Q} \]

### Step 5: Use the Condition of a Balanced Bridge
In a balanced Wheatstone bridge, the condition is:
\[ \frac{P}{R} = \frac{Q}{S} \]
From this, we can derive:
\[ \frac{P}{Q} = \frac{R}{S} \]
Let’s denote this ratio as \( K \).

### Step 6: Express the Resistors in Terms of K
Using the ratio \( K \):
- We can express \( S \) in terms of \( R \) and \( K \):
\[ S = \frac{R}{K} \]
- Similarly, express \( Q \) in terms of \( P \):
\[ Q = \frac{P}{K} \]

### Step 7: Substitute Back into the Power Ratio
Now substitute \( S \) and \( Q \) back into the power ratio:
\[ \frac{P_1}{P_2} = \frac{R + \frac{R}{K}}{P + \frac{P}{K}} \]
This can be simplified:
\[ \frac{P_1}{P_2} = \frac{R(1 + \frac{1}{K})}{P(1 + \frac{1}{K})} \]
The \( (1 + \frac{1}{K}) \) terms cancel out:
\[ \frac{P_1}{P_2} = \frac{R}{P} \]

### Conclusion
Thus, the final ratio of power consumed in the branches (P + Q) and (R + S) is:
\[ \frac{P_1}{P_2} = \frac{R}{P} \]