A resistor `R`, an inductor `L` and a capacitor `C` are connected in series to an oscillator of frequency `n`. If the resonant frequency is `n_r,` then the current lags behind voltage, when

Options

A. n=0
B. `n lt n_(f)`
C. `n lt n_(r)`
D. `n gt n_(r)`

Answer

To solve the problem, we need to analyze the behavior of a series RLC circuit connected to an oscillator and determine under what conditions the current lags behind the voltage.

### Step-by-Step Solution:

1. **Understanding the Circuit**:
- We have a resistor (R), an inductor (L), and a capacitor (C) connected in series to an AC voltage source (oscillator) with frequency \( n \).
- The resonant frequency \( n_r \) of the circuit is given by the formula:
\[ n_r = \frac{1}{2\pi\sqrt{LC}} \]

2. **Reactance of Components**:
- The inductive reactance \( X_L \) is given by:
\[ X_L = \omega L = 2\pi n L \]
- The capacitive reactance \( X_C \) is given by:
\[ X_C = \frac{1}{\omega C} = \frac{1}{2\pi n C} \]

3. **Condition for Current Lagging Voltage**:
- In a series RLC circuit, the current lags behind the voltage when the inductive reactance is greater than the capacitive reactance:
\[ X_L > X_C \]
- Substituting the expressions for \( X_L \) and \( X_C \):
\[ 2\pi n L > \frac{1}{2\pi n C} \]

4. **Rearranging the Inequality**:
- Multiply both sides by \( 2\pi n C \) (assuming \( n \) and \( C \) are positive):
\[ (2\pi n)^2 L C > 1 \]
- This can be rearranged to:
\[ n^2 > \frac{1}{4\pi^2 LC} \]

5. **Expressing in Terms of Resonant Frequency**:
- Recognizing that \( \frac{1}{4\pi^2 LC} = n_r^2 \):
\[ n^2 > n_r^2 \]
- Taking the square root of both sides gives:
\[ n > n_r \]

6. **Conclusion**:
- Therefore, the current lags behind the voltage when the frequency of the oscillator \( n \) is greater than the resonant frequency \( n_r \):
\[ n > n_r \]

### Final Answer:
The current lags behind the voltage when \( n > n_r \).