An alternating emf = `220 sqrt(2) sin ` 100 t V is applied to a capacitor `1 mu F ` . The current flowing through the capacitor is

Options

A. 22mA
B. 12mA
C. 32mA
D. 42mA

Explanation

To find the current flowing through the capacitor when an alternating emf is applied, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Given Values**:
- The alternating emf is given as \( E(t) = 220\sqrt{2} \sin(100t) \) volts.
- The capacitance \( C = 1 \mu F = 1 \times 10^{-6} F \).

2. **Determine the Peak Voltage \( E_0 \)**:
- From the equation \( E(t) = E_0 \sin(\omega t) \), we can identify:
- \( E_0 = 220\sqrt{2} \) volts.

3. **Calculate the RMS Voltage \( E_{rms} \)**:
- The RMS (Root Mean Square) voltage is calculated using the formula:
\[ E_{rms} = \frac{E_0}{\sqrt{2}} \]
- Substituting the value of \( E_0 \):
\[ E_{rms} = \frac{220\sqrt{2}}{\sqrt{2}} = 220 \text{ volts} \]

4. **Determine the Angular Frequency \( \omega \)**:
- From the given emf, we have:
- \( \omega = 100 \) rad/s.

5. **Calculate the Capacitive Reactance \( X_C \)**:
- The capacitive reactance is given by the formula:
\[ X_C = \frac{1}{\omega C} \]
- Substituting the values of \( \omega \) and \( C \):
\[ X_C = \frac{1}{100 \times 1 \times 10^{-6}} = \frac{1}{10^{-4}} = 10^4 \text{ ohms} \]

6. **Calculate the RMS Current \( I_{rms} \)**:
- The RMS current can be calculated using Ohm's law for AC circuits:
\[ I_{rms} = \frac{E_{rms}}{X_C} \]
- Substituting the values of \( E_{rms} \) and \( X_C \):
\[ I_{rms} = \frac{220}{10^4} = 0.022 \text{ A} = 22 \text{ mA} \]

### Final Answer:
The current flowing through the capacitor is \( 22 \text{ mA} \).

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