A resistance (R = 12 omega), an inductor (L = 2 H) and a capacitor (C = 5 muF) are connected to an AC generator of frequency 50 Hz. Which of the following statement is correct

Options

A. At resonance, the circuit impedance is zero
B. At resonance, the circuit impedance is 12Omega.
C. At resonance, the frequency of the circuit is 1//2
D. The inductive reactance is less than the capacitive reactance

Answer

To solve the problem, we need to calculate the inductive reactance (XL) and capacitive reactance (XC) of the given circuit and determine the relationship between them.

### Step-by-Step Solution:

1. **Identify the Given Values**:
- Resistance (R) = 12 Ω
- Inductance (L) = 2 H
- Capacitance (C) = 5 μF = 5 × 10^(-6) F
- Frequency (f) = 50 Hz

2. **Calculate the Angular Frequency (ω)**:
\[ \omega = 2\pi f \]
\[ \omega = 2\pi \times 50 = 100\pi \, \text{rad/s} \]

3. **Calculate the Inductive Reactance (XL)**:
\[ X_L = \omega L = 100\pi \times 2 = 200\pi \, \Omega \]
\[ X_L \approx 628.32 \, \Omega \quad (\text{using } \pi \approx 3.14) \]

4. **Calculate the Capacitive Reactance (XC)**:
\[ X_C = \frac{1}{\omega C} = \frac{1}{100\pi \times 5 \times 10^{-6}} \]
\[ X_C = \frac{1}{500\pi \times 10^{-6}} = \frac{10^6}{500\pi} = \frac{2000}{\pi} \, \Omega \]
\[ X_C \approx 636.62 \, \Omega \quad (\text{using } \pi \approx 3.14) \]

5. **Compare XL and XC**:
- From the calculations, we have:
- \( X_L \approx 628.32 \, \Omega \)
- \( X_C \approx 636.62 \, \Omega \)

Since \( X_L < X_C \), the circuit is not at resonance.

6. **Conclusion**:
- The system is not at resonance because \( X_L \) is not equal to \( X_C \).
- Additionally, since \( X_L < X_C \), we can conclude that the circuit is capacitive.

### Final Answer:
The correct statement is that the circuit is not at resonance, and the inductive reactance is less than the capacitive reactance.