The resonant frequency of a circuit is `f`. If the capacitance is made `4` times the initial values, then the resonant frequecy will become

Options

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

Explanation

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

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

where:
- \( L \) is the inductance,
- \( C \) is the capacitance.

### Step-by-Step Solution:

1. **Identify the Initial Resonant Frequency**:
The initial resonant frequency is given as \( f \).

2. **Understand the Effect of Changing Capacitance**:
We are told that the capacitance \( C \) is increased to \( 4C \) (i.e., it becomes 4 times its initial value).

3. **Substitute the New Capacitance into the Resonant Frequency Formula**:
The new resonant frequency \( f' \) can be expressed as:
\[ f' = \frac{1}{2\pi\sqrt{L \cdot 4C}} = \frac{1}{2\pi\sqrt{4LC}} = \frac{1}{2\pi \cdot 2\sqrt{LC}} = \frac{1}{2} \cdot \frac{1}{2\pi\sqrt{LC}} \]

4. **Relate the New Frequency to the Initial Frequency**:
Since \( \frac{1}{2\pi\sqrt{LC}} = f \), we can substitute this into our equation:
\[ f' = \frac{1}{2} f \]

5. **Conclusion**:
Therefore, when the capacitance is increased to 4 times its original value, the new resonant frequency \( f' \) becomes:
\[ f' = \frac{f}{2} \]

### Final Answer:
The new resonant frequency will be \( \frac{f}{2} \).