In LC oscillation resistance is 100 `Omega` and inductance and capacitance is 1 H and `10 mu F` . Find the half power of frequency .

Options

A. 266.2
B. 366.2
C. 166.2
D. 233.2

Explanation

To solve the problem of finding the half power frequency in an LC oscillation circuit with given resistance, inductance, and capacitance, we will follow these steps:

### Step 1: Identify the given values
- Resistance (R) = 100 Ω
- Inductance (L) = 1 H
- Capacitance (C) = 10 µF = 10 × 10^(-6) F

### Step 2: Calculate the resonant frequency (ω₀)
The resonant frequency (ω₀) is given by the formula:
\[ \omega_0 = \frac{1}{\sqrt{L \cdot C}} \]
Substituting the values of L and C:
\[ \omega_0 = \frac{1}{\sqrt{1 \cdot 10 \times 10^{-6}}} = \frac{1}{\sqrt{10 \times 10^{-6}}} = \frac{1}{\sqrt{10^{-5}}} = \frac{1}{\sqrt{10} \cdot 10^{-2.5}} = \frac{10^{2.5}}{\sqrt{10}} = 10^{1.5} = 31.62 \text{ rad/s} \]

### Step 3: Calculate the half power frequency (fₕ)
The half power frequency (fₕ) is given by the formula:
\[ f_h = \omega_0 - \frac{R}{2L} \]
First, we need to calculate \(\frac{R}{2L}\):
\[ \frac{R}{2L} = \frac{100}{2 \cdot 1} = \frac{100}{2} = 50 \text{ rad/s} \]

Now, substituting the values into the half power frequency formula:
\[ f_h = \omega_0 - \frac{R}{2L} = 31.62 - 50 \]
However, we need to convert the angular frequency (rad/s) to frequency (Hz) using the relationship \(f = \frac{\omega}{2\pi}\):
\[ f_h = \frac{31.62 - 50}{2\pi} \]

### Step 4: Calculate the final value
\[ f_h = \frac{-18.38}{2\pi} \approx -2.93 \text{ Hz} \]
Since frequency cannot be negative, we must check our calculations. It seems we have made a mistake in the interpretation of the half power frequency.

### Correct Calculation of Half Power Frequency
The correct formula for half power frequency in terms of frequency (f) is:
\[ f_h = f_0 - \frac{R}{2\pi L} \]
Where \(f_0\) is the resonant frequency in Hz:
\[ f_0 = \frac{1}{2\pi\sqrt{LC}} = \frac{1}{2\pi\sqrt{1 \cdot 10 \times 10^{-6}}} = \frac{1}{2\pi\sqrt{10^{-5}}} = \frac{10^{2.5}}{2\pi} \approx 15.92 \text{ Hz} \]

Now we can calculate:
\[ f_h = f_0 - \frac{R}{2\pi L} = 15.92 - \frac{100}{2\pi \cdot 1} = 15.92 - \frac{100}{6.2832} \approx 15.92 - 15.92 = 0 \text{ Hz} \]

### Conclusion
The half power frequency is approximately 0 Hz, indicating that the circuit does not oscillate at a half power frequency under the given conditions.