A resonant `AC` circuit contains a capacitor of capacitance `10^(-6)F` and an inductor of `10^(-4)H`. The frequency of electrical oscillation will be

Options

A. `105 Hz`
B. `10 Hz`
C. `(10^(5))/(2pi) Hz`
D. `10/(2pi)Hz`

Explanation

To find the resonant frequency of the given AC circuit containing a capacitor and an inductor, we can use the formula for the resonant frequency \( f \) in a LC circuit:

\[ f = \frac{1}{2\pi \sqrt{LC}} \]

Where:
- \( L \) is the inductance in henries (H)
- \( C \) is the capacitance in farads (F)

### Step-by-Step Solution:

1. **Identify the Values**:
- Given:
- Capacitance \( C = 10^{-6} \, \text{F} \)
- Inductance \( L = 10^{-4} \, \text{H} \)

2. **Substitute the Values into the Formula**:
- Plugging in the values into the formula for resonant frequency:
\[ f = \frac{1}{2\pi \sqrt{(10^{-4})(10^{-6})}} \]

3. **Calculate the Product \( LC \)**:
- Calculate \( LC \):
\[ LC = 10^{-4} \times 10^{-6} = 10^{-10} \]

4. **Calculate \( \sqrt{LC} \)**:
- Now, calculate the square root:
\[ \sqrt{LC} = \sqrt{10^{-10}} = 10^{-5} \]

5. **Substitute Back into the Frequency Formula**:
- Now substitute \( \sqrt{LC} \) back into the frequency formula:
\[ f = \frac{1}{2\pi \times 10^{-5}} \]

6. **Calculate the Frequency**:
- Finally, calculate the frequency:
\[ f = \frac{1}{2\pi} \times 10^{5} \approx \frac{10^{5}}{6.2832} \approx 15915.5 \, \text{Hz} \]

### Final Answer:
The frequency of electrical oscillation in the resonant AC circuit is approximately \( 15915.5 \, \text{Hz} \).