At what frequency 1 Henry inductance offers the same impedance as `1 mu F` capacitor?

Options

A. `500/(2 pi) Hz`
B. `500/(pi) Hz`
C. `500pi Hz`
D. `1000pi Hz`

Explanation

To solve the problem of finding the frequency at which a 1 Henry inductance offers the same impedance as a 1 microfarad capacitor, we can follow these steps:

### Step-by-Step Solution:

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

2. **Understand Impedance Relationships:**
- The impedance of an inductor \( Z_L \) is given by:
\[ Z_L = j \omega L \]
- The impedance of a capacitor \( Z_C \) is given by:
\[ Z_C = \frac{1}{j \omega C} \]
- For the inductance and capacitance to have the same impedance in magnitude, we set:
\[ |Z_L| = |Z_C| \]
- This leads to:
\[ \omega L = \frac{1}{\omega C} \]

3. **Rearranging the Equation:**
- Multiplying both sides by \( \omega \):
\[ \omega^2 L C = 1 \]
- This can be rearranged to find \( \omega \):
\[ \omega^2 = \frac{1}{LC} \]
- Therefore,
\[ \omega = \frac{1}{\sqrt{LC}} \]

4. **Substituting Values:**
- Substitute \( L = 1 \, \text{H} \) and \( C = 1 \times 10^{-6} \, \text{F} \):
\[ \omega = \frac{1}{\sqrt{1 \times 1 \times 10^{-6}}} = \frac{1}{\sqrt{10^{-6}}} = 10^3 \, \text{rad/s} \]

5. **Convert Angular Frequency to Frequency:**
- The relationship between angular frequency \( \omega \) and frequency \( f \) is:
\[ \omega = 2 \pi f \]
- Therefore,
\[ f = \frac{\omega}{2 \pi} = \frac{10^3}{2 \pi} \]
- Simplifying this gives:
\[ f = \frac{1000}{2 \times 3.14} = \frac{1000}{6.28} \approx 159.15 \, \text{Hz} \]

6. **Final Calculation:**
- To express it in terms of \( \frac{500}{\pi} \):
\[ f \approx \frac{500}{\pi} \, \text{Hz} \]

### Conclusion:
The frequency at which a 1 Henry inductance offers the same impedance as a 1 microfarad capacitor is:
\[ f \approx \frac{500}{\pi} \, \text{Hz} \]