In a series resonant L-C-R circuit, the capacitance is changed from C to 3C. For the same resonant frequency, the inductance should be changed from L to

Options

A. 2L
B. L/3
C. L/4
D. 4L

Answer

To solve the problem, we need to analyze the relationship between the resonant frequency, inductance, and capacitance in a series L-C-R circuit. The resonant frequency \( f_R \) is given by the formula:

\[ f_R = \frac{1}{2\pi} \sqrt{\frac{1}{LC}} \]

### Step-by-Step Solution:

1. **Identify the Initial Conditions**:
- Let the initial capacitance be \( C \) and the initial inductance be \( L \).
- The initial resonant frequency \( f_R \) is given by:
\[ f_R = \frac{1}{2\pi} \sqrt{\frac{1}{LC}} \]

2. **Change the Capacitance**:
- The capacitance is changed from \( C \) to \( 3C \).
- We denote the new capacitance as \( C' = 3C \).

3. **Set Up the Equation for the New Resonant Frequency**:
- We want the resonant frequency to remain the same after changing the capacitance. Let the new inductance be \( L' \).
- The new resonant frequency is:
\[ f_R = \frac{1}{2\pi} \sqrt{\frac{1}{L'C'}} \]
- Substituting \( C' = 3C \):
\[ f_R = \frac{1}{2\pi} \sqrt{\frac{1}{L' \cdot 3C}} \]

4. **Equate the Two Frequencies**:
- Since the resonant frequency remains the same, we can set the two expressions for \( f_R \) equal to each other:
\[ \frac{1}{2\pi} \sqrt{\frac{1}{LC}} = \frac{1}{2\pi} \sqrt{\frac{1}{L' \cdot 3C}} \]

5. **Simplify the Equation**:
- Cancel \( \frac{1}{2\pi} \) from both sides:
\[ \sqrt{\frac{1}{LC}} = \sqrt{\frac{1}{L' \cdot 3C}} \]
- Squaring both sides gives:
\[ \frac{1}{LC} = \frac{1}{L' \cdot 3C} \]

6. **Cross Multiply**:
- Cross multiplying leads to:
\[ L' \cdot 3C = LC \]

7. **Solve for the New Inductance**:
- Dividing both sides by \( 3C \):
\[ L' = \frac{L}{3} \]

### Final Answer:
The inductance should be changed from \( L \) to \( \frac{L}{3} \).