When two resistances are connected in series. The effective resistance is found to be `48Omega `. Their resistances if their values are in the ratio 3:1.

Options

A. `24 Omega and 24 Omega `
B. `36 Omega and 12 Omega `
C. `12 Omega and 48 Omega `
D. `36 Omega and 36 Omega `

Answer

To solve the problem of finding the resistances when two resistances are connected in series and their values are in the ratio of 3:1, we can follow these steps:

### Step-by-Step Solution

1. **Understand the Problem**: We have two resistances, \( R_1 \) and \( R_2 \), connected in series. The effective resistance \( R_{eff} \) is given as \( 48 \Omega \). The resistances are in the ratio \( 3:1 \).

2. **Set Up the Ratios**: Since the resistances are in the ratio \( 3:1 \), we can express them as:
\[ R_1 = 3x \quad \text{and} \quad R_2 = x \]
where \( x \) is a common factor.

3. **Write the Equation for Effective Resistance**: The effective resistance for resistors in series is given by:
\[ R_{eff} = R_1 + R_2 \]
Substituting the expressions for \( R_1 \) and \( R_2 \):
\[ R_{eff} = 3x + x = 4x \]

4. **Set Up the Equation**: We know that the effective resistance is \( 48 \Omega \), so we can write:
\[ 4x = 48 \]

5. **Solve for \( x \)**: To find \( x \), divide both sides by 4:
\[ x = \frac{48}{4} = 12 \]

6. **Find the Values of \( R_1 \) and \( R_2 \)**:
- Now that we have \( x \), we can find \( R_1 \) and \( R_2 \):
\[ R_2 = x = 12 \Omega \]
\[ R_1 = 3x = 3 \times 12 = 36 \Omega \]

7. **Conclusion**: The values of the resistances are:
\[ R_1 = 36 \Omega \quad \text{and} \quad R_2 = 12 \Omega \]

### Final Answer
The resistances are \( 36 \Omega \) and \( 12 \Omega \).

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