In ac circuit contains a pure capacitor, across which an ac emf
`e = 100 sin (1000t)`, volt
is applied. If the peak value of the current is 200 mA, then the value of the capacitor is

Options

A. `2mu`
B. `20 muF`
C. `5 muF`
D. `500 muF`

Answer

To find the value of the capacitor in the given AC circuit, we can follow these steps:

### Step-by-step Solution:

1. **Identify the given parameters:**
- The AC voltage is given as \( e = 100 \sin(1000t) \) volts.
- The peak current \( I_0 = 200 \, \text{mA} = 200 \times 10^{-3} \, \text{A} = 0.2 \, \text{A} \).

2. **Calculate the peak voltage \( V_0 \):**
- The peak voltage \( V_0 \) from the given equation is \( 100 \, \text{V} \).

3. **Determine the capacitive reactance \( X_C \):**
- The relationship between peak voltage and peak current in a capacitor is given by:
\[ X_C = \frac{V_0}{I_0} \]
- Substituting the values:
\[ X_C = \frac{100}{0.2} = 500 \, \Omega \]

4. **Use the formula for capacitive reactance:**
- The capacitive reactance \( X_C \) is also given by:
\[ X_C = \frac{1}{\omega C} \]
- Where \( \omega \) is the angular frequency. From the voltage equation, we have \( \omega = 1000 \, \text{rad/s} \).

5. **Set the two expressions for \( X_C \) equal to each other:**
\[ 500 = \frac{1}{1000 C} \]

6. **Solve for the capacitance \( C \):**
- Rearranging the equation gives:
\[ C = \frac{1}{500 \times 1000} \]
- Calculating this yields:
\[ C = \frac{1}{500000} = 2 \times 10^{-6} \, \text{F} = 2 \, \mu\text{F} \]

7. **Conclusion:**
- The value of the capacitor is \( 2 \, \mu\text{F} \).

### Final Answer:
The value of the capacitor is \( 2 \, \mu\text{F} \).

---