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 resistors \( R_1 \) and \( R_2 \) when connected in series and parallel. Let's break down the solution step by step.

### Step 1: Understand the power formula
The power consumed in a circuit can be expressed using the formula:
\[ P = \frac{V^2}{R_{\text{eq}}} \]
where \( P \) is the power, \( V \) is the voltage across the resistors, and \( R_{\text{eq}} \) is the equivalent resistance.

### Step 2: Series Connection
When \( R_1 \) and \( R_2 \) are connected in series, the equivalent resistance is:
\[ R_{\text{eq, series}} = R_1 + R_2 \]
The power consumed in series is given as:
\[ P_1 = 12 \, \text{W} \]
Thus, we can write:
\[ P_1 \propto \frac{1}{R_1 + R_2} \]
This implies:
\[ 12 \propto \frac{1}{R_1 + R_2} \quad \text{(Equation 1)} \]

### Step 3: Parallel Connection
When \( R_1 \) and \( R_2 \) are connected in parallel, the equivalent resistance is:
\[ R_{\text{eq, parallel}} = \frac{R_1 R_2}{R_1 + R_2} \]
The power consumed in parallel is given as:
\[ P_2 = 50 \, \text{W} \]
Thus, we can write:
\[ P_2 \propto \frac{1}{\frac{R_1 R_2}{R_1 + R_2}} = \frac{R_1 + R_2}{R_1 R_2} \quad \text{(Equation 2)} \]

### Step 4: Set up the ratio of powers
From Equation 1 and Equation 2, we can set up the ratio of powers:
\[ \frac{P_2}{P_1} = \frac{R_1 + R_2}{R_1 R_2} \cdot (R_1 + R_2) \]
Substituting the known values:
\[ \frac{50}{12} = \frac{(R_1 + R_2)^2}{R_1 R_2} \]
Simplifying gives:
\[ \frac{25}{6} = \frac{(R_1 + R_2)^2}{R_1 R_2} \]

### Step 5: Cross-multiply and simplify
Cross-multiplying gives:
\[ 25 R_1 R_2 = 6 (R_1 + R_2)^2 \]
Expanding the right side:
\[ 25 R_1 R_2 = 6 (R_1^2 + 2R_1 R_2 + R_2^2) \]
Rearranging leads to:
\[ 6 R_1^2 + 6 R_2^2 - 19 R_1 R_2 = 0 \]

### Step 6: Solve the quadratic equation
This is a quadratic equation in terms of \( R_1 \) and \( R_2 \). We can factor or use the quadratic formula to find the roots.

### Step 7: Find the ratio of resistances
Assuming \( R_1 = kR_2 \), we can substitute and solve for \( k \):
\[ 6(k^2 + 1) - 19k = 0 \]
This gives us the ratio of \( R_1 \) to \( R_2 \).

### Step 8: Final ratio of powers
The ratio of the powers consumed by \( R_1 \) and \( R_2 \) can be derived from the resistances:
\[ \frac{P_1}{P_2} = \frac{R_2^2}{R_1^2} \]
Thus, the final ratio of powers is:
\[ \frac{P_1}{P_2} = \frac{3}{2} \]

### Conclusion
The ratio of the powers of \( R_1 \) and \( R_2 \) is \( \frac{3}{2} \).