In an LCR series circuit the capacitance is changed from `C` to `4C` For the same resonant fequency the inductance should be changed from `L` to .

Options

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

Explanation

To solve the problem, we need to understand the relationship between the inductance (L), capacitance (C), and resonant frequency (f) in an LCR series circuit. The resonant frequency is given by the formula:

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

### Step 1: Write the formula for resonant frequency
The resonant frequency \( f \) is defined as:
\[ f = \frac{1}{2\pi\sqrt{LC}} \]

### Step 2: Set up the equation for the initial condition
Initially, we have:
\[ f = \frac{1}{2\pi\sqrt{L \cdot C}} \]

### Step 3: Change the capacitance
Now, the capacitance is changed from \( C \) to \( 4C \). We need to find the new inductance \( L_1 \) that maintains the same resonant frequency.

### Step 4: Set up the equation for the new condition
For the new condition with capacitance \( 4C \), the resonant frequency can be expressed as:
\[ f = \frac{1}{2\pi\sqrt{L_1 \cdot 4C}} \]

### Step 5: Equate the two expressions for frequency
Since the resonant frequency remains the same, we can set the two equations equal to each other:
\[ \frac{1}{2\pi\sqrt{L \cdot C}} = \frac{1}{2\pi\sqrt{L_1 \cdot 4C}} \]

### Step 6: Simplify the equation
Removing the common terms \( 2\pi \) from both sides, we have:
\[ \sqrt{L \cdot C} = \sqrt{L_1 \cdot 4C} \]

### Step 7: Square both sides
Squaring both sides gives us:
\[ L \cdot C = L_1 \cdot 4C \]

### Step 8: Cancel \( C \) from both sides
Assuming \( C \neq 0 \), we can cancel \( C \) from both sides:
\[ L = 4L_1 \]

### Step 9: Solve for \( L_1 \)
Rearranging gives:
\[ L_1 = \frac{L}{4} \]

### Conclusion
Thus, the new inductance \( L_1 \) should be changed from \( L \) to \( \frac{L}{4} \).

### Final Answer
The inductance should be changed from \( L \) to \( \frac{L}{4} \).

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