An `LCR` series circuit contains `L = 8 H, C = 0.5 muF` and `R =100 Omega` The resonant frequency of the circuit is .

Options

A. `(1000)/(pi)Hz`
B. `(500)/(pi)Hz`
C. `(250)/(pi)Hz`
D. `(125)/(pi)Hz`

Explanation

To find the resonant frequency of an LCR series circuit, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the given values:**
- Inductance, \( L = 8 \, \text{H} \)
- Capacitance, \( C = 0.5 \, \mu\text{F} = 0.5 \times 10^{-6} \, \text{F} \)
- Resistance, \( R = 100 \, \Omega \) (not needed for resonant frequency calculation)

2. **Write the formula for resonant frequency:**
The resonant frequency \( f_R \) of an LCR circuit is given by the formula:
\[ f_R = \frac{1}{2\pi\sqrt{LC}} \]

3. **Substitute the values into the formula:**
\[ f_R = \frac{1}{2\pi\sqrt{8 \times 0.5 \times 10^{-6}}} \]

4. **Calculate the product inside the square root:**
\[ 8 \times 0.5 = 4 \]
Thus,
\[ f_R = \frac{1}{2\pi\sqrt{4 \times 10^{-6}}} \]

5. **Simplify the square root:**
\[ \sqrt{4 \times 10^{-6}} = 2 \times 10^{-3} \]
Therefore,
\[ f_R = \frac{1}{2\pi \times (2 \times 10^{-3})} \]

6. **Calculate the denominator:**
\[ 2 \times 10^{-3} = 0.002 \]
Thus,
\[ f_R = \frac{1}{4\pi \times 10^{-3}} \]

7. **Calculate the resonant frequency:**
\[ f_R = \frac{1}{4\pi} \times 10^{3} \text{ Hz} \]

8. **Final calculation:**
To express this in a more usable form, we can approximate \( \pi \approx 3.14 \):
\[ f_R \approx \frac{1000}{4 \times 3.14} \approx \frac{1000}{12.56} \approx 79.58 \text{ Hz} \]

### Final Answer:
The resonant frequency of the circuit is approximately \( 79.58 \, \text{Hz} \).

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