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 the resonant frequency of an LRC circuit and the capacitance. 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 \( f_0 \) when the capacitance is \( C_0 \):
\[ f_0 = \frac{1}{2\pi\sqrt{LC_0}} \]

2. **Change the capacitance**:
According to the problem, the capacitance is increased to four times its initial value:
\[ C = 4C_0 \]

3. **Calculate the new resonant frequency**:
Substitute the new capacitance into the resonant frequency formula:
\[ f = \frac{1}{2\pi\sqrt{L(4C_0)}} \]

4. **Simplify the expression**:
We can factor out the 4 from the square root:
\[ f = \frac{1}{2\pi\sqrt{L} \cdot \sqrt{4C_0}} = \frac{1}{2\pi\sqrt{L} \cdot 2\sqrt{C_0}} = \frac{1}{4\pi\sqrt{LC_0}} \]

5. **Relate the new frequency to the initial frequency**:
Notice that:
\[ \frac{1}{4\pi\sqrt{LC_0}} = \frac{1}{2} \cdot \frac{1}{2\pi\sqrt{LC_0}} = \frac{f_0}{2} \]
Therefore, the new resonant frequency \( f \) is:
\[ f = \frac{f_0}{2} \]

6. **Conclusion**:
The new resonant frequency when the capacitance is made four times the initial value is half of the original frequency:
\[ f = \frac{f_0}{2} \]

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