To reduce the resonant frequency in an `LCR` series circuit with a generator

Options

A. the generator frequency should be reduced
B. another capacitor should be added in parallel to the first
C. the iron core of the inductor should be removed
D. dielectric in the capacitor should be removed

Explanation

To reduce the resonant frequency in an LCR series circuit with a generator, we can follow these steps:

### Step-by-Step Solution:

1. **Understanding Resonant Frequency**:
The resonant frequency \( f_0 \) of an LCR series circuit is given by the formula:
\[ f_0 = \frac{1}{2\pi\sqrt{LC}} \]
where \( L \) is the inductance and \( C \) is the capacitance.

2. **Identifying the Relationship**:
From the formula, we can see that the resonant frequency is inversely proportional to the square root of the product of inductance and capacitance. Thus, to reduce the resonant frequency, we need to either increase the inductance \( L \) or increase the capacitance \( C \).

3. **Analyzing Options**:
- **Option 1**: Reducing the generator frequency.
- This option is incorrect because the resonant frequency \( f_0 \) is independent of the generator frequency.

- **Option 2**: Adding another capacitor in parallel to the first.
- When we add another capacitor in parallel, the equivalent capacitance \( C_{eq} \) increases (since \( C_{eq} = C_1 + C_2 \)). An increase in capacitance leads to a decrease in resonant frequency. Therefore, this option is correct.

- **Option 3**: Removing the iron core of the inductor.
- Removing the iron core decreases the inductance \( L \). A decrease in inductance results in an increase in resonant frequency, making this option incorrect.

- **Option 4**: Removing the dielectric from the capacitor.
- Removing the dielectric decreases the capacitance \( C \). A decrease in capacitance results in an increase in resonant frequency, making this option incorrect as well.

4. **Conclusion**:
After analyzing all options, the only correct method to reduce the resonant frequency in the LCR circuit is by adding another capacitor in parallel to the existing one.

Thus, the correct answer is **Option 2**.