In an A.C. Circuit containing an inductance and a capacitance in series, the current is found to be maximum, when `L=0.5H` and `C=8 muF`. Then the angular frequency of the applied alternating e.m.f. will be

Options

A. `25 rad//sec`
B. `500 rad//sec`
C. `750 rad//sec`
D. `1000 rad//sec`

Explanation

To solve the problem, we need to find the angular frequency of an AC circuit containing an inductance (L) and a capacitance (C) in series, given that the current is maximum.

### Step-by-Step Solution:

1. **Identify Given Values**:
- Inductance, \( L = 0.5 \, H \)
- Capacitance, \( C = 8 \, \mu F = 8 \times 10^{-6} \, F \)

2. **Use the Formula for Angular Frequency**:
The angular frequency \( \omega \) in an LC circuit is given by the formula:
\[ \omega = \frac{1}{\sqrt{LC}} \]

3. **Substitute the Values into the Formula**:
Substitute the values of \( L \) and \( C \) into the formula:
\[ \omega = \frac{1}{\sqrt{0.5 \times 8 \times 10^{-6}}} \]

4. **Calculate the Product Inside the Square Root**:
First, calculate \( L \times C \):
\[ L \times C = 0.5 \times 8 \times 10^{-6} = 4 \times 10^{-6} \, H \cdot F \]

5. **Calculate the Square Root**:
Now, calculate the square root:
\[ \sqrt{4 \times 10^{-6}} = 2 \times 10^{-3} \]

6. **Calculate Angular Frequency**:
Now substitute back into the formula for \( \omega \):
\[ \omega = \frac{1}{2 \times 10^{-3}} = 500 \, rad/s \]

7. **Conclusion**:
The angular frequency of the applied alternating e.m.f. is:
\[ \omega = 500 \, rad/s \]

### Final Answer:
The angular frequency of the applied alternating e.m.f. is \( 500 \, rad/s \).

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