An `AC` circuit consists of an inductor of inductance `0.5H` and a capacitor of capacitance `8muF` in series. The current in the circuit is maximum when the angular frequency of `AC` source is

Options

A. 500 rad/sec
B. `2xx10^(5)`rad/sec
C. 4000 rad/sec
D. 5000 rad/sec

Answer

To solve the problem, we need to find the angular frequency at which the current in the AC circuit is maximum. The circuit consists of an inductor and a capacitor in series. The current will be maximum when the inductive reactance (XL) equals the capacitive reactance (XC).

### Step-by-Step Solution:

1. **Identify the given values:**
- Inductance (L) = 0.5 H
- Capacitance (C) = 8 µF = 8 × 10^(-6) F

2. **Write the formulas for inductive and capacitive reactance:**
- Inductive reactance: \( X_L = \omega L \)
- Capacitive reactance: \( X_C = \frac{1}{\omega C} \)

3. **Set the inductive reactance equal to the capacitive reactance:**
\[ X_L = X_C \]
\[ \omega L = \frac{1}{\omega C} \]

4. **Rearranging the equation:**
Multiply both sides by \( \omega \):
\[ \omega^2 L = \frac{1}{C} \]

5. **Solve for \( \omega^2 \):**
\[ \omega^2 = \frac{1}{LC} \]

6. **Substituting the values of L and C:**
\[ \omega^2 = \frac{1}{0.5 \times 8 \times 10^{-6}} \]

7. **Calculating the denominator:**
\[ 0.5 \times 8 \times 10^{-6} = 4 \times 10^{-6} \]

8. **Now, calculate \( \omega^2 \):**
\[ \omega^2 = \frac{1}{4 \times 10^{-6}} = 2.5 \times 10^5 \]

9. **Taking the square root to find \( \omega \):**
\[ \omega = \sqrt{2.5 \times 10^5} = 500 \, \text{rad/s} \]

10. **Conclusion:**
The angular frequency at which the current in the circuit is maximum is \( \omega = 500 \, \text{rad/s} \).

### Final Answer:
The angular frequency of the AC source when the current is maximum is \( 500 \, \text{rad/s} \).