In a series R-L-C circuit, the frequency of the source is half of the resonance frequency. The nature of the circuit will be

Options

A. capacitive
B. inductive
C. purely resistive
D. data unsufficient

Explanation

To determine the nature of a series R-L-C circuit when the frequency of the source is half of the resonance frequency, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Resonance Frequency**:
The resonance frequency \( f_0 \) of a series R-L-C circuit is given by the formula:
\[ f_0 = \frac{1}{2\pi\sqrt{LC}} \]

2. **Determine the Source Frequency**:
Since the frequency of the source \( f \) is half of the resonance frequency, we can express it as:
\[ f = \frac{f_0}{2} = \frac{1}{4\pi\sqrt{LC}} \]

3. **Compare Source Frequency with Resonance Frequency**:
We know that:
\[ f < f_0 \]
This indicates that the source frequency is less than the resonance frequency.

4. **Calculate Reactance**:
The inductive reactance \( X_L \) and capacitive reactance \( X_C \) can be expressed as:
\[ X_L = \omega L \quad \text{and} \quad X_C = \frac{1}{\omega C} \]
where \( \omega = 2\pi f \).

5. **Substitute the Source Frequency**:
Since \( f < f_0 \), we can substitute \( \omega \) as:
\[ \omega = 2\pi f < 2\pi f_0 \]
Therefore, we can say:
\[ \omega L < \frac{1}{\omega C} \]

6. **Rearranging the Inequality**:
This can be rearranged to show:
\[ X_L < X_C \]
This means that the capacitive reactance \( X_C \) is greater than the inductive reactance \( X_L \).

7. **Determine the Nature of the Circuit**:
Since \( X_C > X_L \), the circuit behaves in a capacitive manner. Therefore, the nature of the circuit is capacitive.

### Conclusion:
The nature of the circuit when the frequency of the source is half of the resonance frequency is **capacitive**.