Calculate the capacitance of a capacitro connected across an alternating power supply with the reactance equal to the reactance of an inductor coil of 1H. The frequency of power supply is 200 Hz.

Answer

To calculate the capacitance of a capacitor connected across an alternating power supply, where the capacitive reactance is equal to the inductive reactance of a 1 H inductor at a frequency of 200 Hz, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Given Values:**
- Inductance \( L = 1 \, \text{H} \)
- Frequency \( f = 200 \, \text{Hz} \)

2. **Calculate the Angular Frequency (\( \omega \)):**
\[ \omega = 2 \pi f \]
Substituting the value of \( f \):
\[ \omega = 2 \pi \times 200 = 400 \pi \, \text{rad/s} \]

3. **Calculate the Inductive Reactance (\( X_L \)):**
The inductive reactance is given by:
\[ X_L = \omega L \]
Substituting the values:
\[ X_L = (400 \pi) \times 1 = 400 \pi \, \Omega \]

4. **Set the Capacitive Reactance (\( X_C \)) Equal to Inductive Reactance (\( X_L \)):**
Since \( X_C = X_L \), we have:
\[ X_C = \frac{1}{\omega C} \]
Therefore:
\[ \frac{1}{\omega C} = 400 \pi \]

5. **Rearranging to Find Capacitance (\( C \)):**
Rearranging the equation gives:
\[ C = \frac{1}{\omega X_C} = \frac{1}{400 \pi} \]

6. **Substituting \( \omega \):**
Now substituting \( \omega = 400 \pi \):
\[ C = \frac{1}{(400 \pi)(400 \pi)} = \frac{1}{160000 \pi^2} \]

7. **Calculating the Value of Capacitance:**
Using \( \pi \approx 3.14 \):
\[ C = \frac{1}{160000 \times (3.14)^2} \approx \frac{1}{160000 \times 9.8596} \approx \frac{1}{1577593.6} \approx 6.33 \times 10^{-7} \, \text{F} \]

8. **Convert Capacitance to Microfarads:**
\[ C \approx 6.33 \times 10^{-7} \, \text{F} = 0.633 \, \mu\text{F} \]

### Final Answer:
The capacitance of the capacitor is approximately \( 0.63 \, \mu\text{F} \).