The frequency at which the inductive reactance of `2 H` inductance will be equal to the capactive reactance of `2 mu F` capactiance (nearly)

Options

A. `80 Hz`
B. `40 Hz`
C. `60 Hz`
D. `20 Hz`

Explanation

To find the frequency at which the inductive reactance of a 2 H inductor is equal to the capacitive reactance of a 2 µF capacitor, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Reactance Formulas**:
- The inductive reactance (\(X_L\)) is given by:
\[ X_L = \omega L = 2\pi f L \]
- The capacitive reactance (\(X_C\)) is given by:
\[ X_C = \frac{1}{\omega C} = \frac{1}{2\pi f C} \]

2. **Set the Reactances Equal**:
- We want to find the frequency (\(f\)) at which \(X_L = X_C\):
\[ 2\pi f L = \frac{1}{2\pi f C} \]

3. **Rearranging the Equation**:
- Multiply both sides by \(2\pi f C\):
\[ (2\pi f)^2 L C = 1 \]
- This simplifies to:
\[ 4\pi^2 f^2 LC = 1 \]

4. **Solve for Frequency**:
- Rearranging gives:
\[ f^2 = \frac{1}{4\pi^2 LC} \]
- Taking the square root:
\[ f = \frac{1}{2\pi \sqrt{LC}} \]

5. **Substituting Values**:
- Here, \(L = 2 \, \text{H}\) and \(C = 2 \, \mu\text{F} = 2 \times 10^{-6} \, \text{F}\).
- Substitute these values into the equation:
\[ f = \frac{1}{2\pi \sqrt{2 \times (2 \times 10^{-6})}} \]

6. **Calculating the Square Root**:
- Calculate \(LC\):
\[ LC = 2 \times 2 \times 10^{-6} = 4 \times 10^{-6} \]
- Now calculate \(\sqrt{LC}\):
\[ \sqrt{LC} = \sqrt{4 \times 10^{-6}} = 2 \times 10^{-3} \]

7. **Final Calculation of Frequency**:
- Substitute back into the frequency formula:
\[ f = \frac{1}{2\pi (2 \times 10^{-3})} \]
- This simplifies to:
\[ f = \frac{1}{4\pi \times 10^{-3}} \approx \frac{1000}{4\pi} \approx 79.6 \, \text{Hz} \]

8. **Conclusion**:
- The frequency at which the inductive reactance equals the capacitive reactance is approximately \(80 \, \text{Hz}\).