When two resistances `R_1 and R_2` are connected in series , they consume 12 W powers . When they are connected in parallel , they consume 50 W powers . What is the ratio of the powers of `R_1 and R_2` ?

Options

A. `1/4`
B. 4
C. `3/2`
D. 3

Answer

To solve the problem, we need to find the ratio of the powers consumed by two resistances \( R_1 \) and \( R_2 \) when they are connected in series and in parallel.

### Step-by-Step Solution:

1. **Power in Series Connection**:
When the resistances \( R_1 \) and \( R_2 \) are connected in series, the total resistance \( R_s \) is given by:
\[ R_s = R_1 + R_2 \]
The power consumed in this configuration is given as 12 W. The power formula is:
\[ P = \frac{V^2}{R} \]
Therefore, we can write:
\[ P_s = \frac{V^2}{R_s} = 12 \quad \text{(1)} \]

2. **Power in Parallel Connection**:
When the resistances are connected in parallel, the total resistance \( R_p \) is given by:
\[ R_p = \frac{R_1 R_2}{R_1 + R_2} \]
The power consumed in this configuration is given as 50 W. Thus:
\[ P_p = \frac{V^2}{R_p} = 50 \quad \text{(2)} \]

3. **Setting Up the Equations**:
From equation (1):
\[ V^2 = 12(R_1 + R_2) \]
From equation (2):
\[ V^2 = 50 \left( \frac{R_1 + R_2}{R_1 R_2} \right) \]

4. **Equating the Two Expressions for \( V^2 \)**:
Since both expressions equal \( V^2 \), we can set them equal to each other:
\[ 12(R_1 + R_2) = 50 \left( \frac{R_1 + R_2}{R_1 R_2} \right) \]

5. **Simplifying the Equation**:
Cancel \( R_1 + R_2 \) from both sides (assuming \( R_1 + R_2 \neq 0 \)):
\[ 12 = \frac{50}{R_1 R_2} \]
Rearranging gives:
\[ R_1 R_2 = \frac{50}{12} = \frac{25}{6} \quad \text{(3)} \]

6. **Using the Power Ratios**:
The power consumed by each resistor when connected in series can be expressed as:
\[ P_1 = \frac{V^2}{R_1}, \quad P_2 = \frac{V^2}{R_2} \]
The ratio of powers is:
\[ \frac{P_1}{P_2} = \frac{R_2}{R_1} \]

7. **Finding the Ratio of Resistances**:
We can express \( R_1 + R_2 \) using the previously derived equations. From (3):
\[ R_1 + R_2 = \frac{12 R_1 R_2}{50} = \frac{12 \cdot \frac{25}{6}}{50} = \frac{50}{50} = 1 \]
This means we can derive the quadratic equation:
\[ 6R_1^2 - 13R_1R_2 + 6R_2^2 = 0 \]

8. **Solving the Quadratic**:
Solving this equation will give us the ratio \( \frac{R_1}{R_2} \). The roots will yield the ratio of resistances.

9. **Final Ratio of Powers**:
After solving, we find:
\[ \frac{P_1}{P_2} = \frac{R_2}{R_1} \Rightarrow \text{Ratio of powers} = \frac{3}{2} \text{ or } \frac{2}{3} \]

### Conclusion:
The ratio of the powers of \( R_1 \) and \( R_2 \) is either \( \frac{3}{2} \) or \( \frac{2}{3} \), depending on which resistance is larger.