An alternating voltage `E (in volts) = 200sqrt2` sin (100t) is connected to a 1 muF capacitor through an a.c. ammeter. The reading of the ammeter shall be

Options

A. 10 mA
B. 20 mA
C. 40 mA
D. 80 mA

Answer

To solve the problem, we need to find the reading of the AC ammeter connected to a capacitor in an AC circuit. The given alternating voltage is \( E(t) = 200\sqrt{2} \sin(100t) \) and the capacitance \( C = 1 \mu F \).

### Step-by-Step Solution:

1. **Identify the Peak Voltage**:
The peak voltage \( E_0 \) is given as \( 200\sqrt{2} \) volts.

2. **Calculate the RMS Voltage**:
The RMS (Root Mean Square) voltage \( E_{rms} \) can be calculated using the formula:
\[ E_{rms} = \frac{E_0}{\sqrt{2}} \]
Substituting the value of \( E_0 \):
\[ E_{rms} = \frac{200\sqrt{2}}{\sqrt{2}} = 200 \text{ volts} \]

3. **Determine the Angular Frequency**:
The angular frequency \( \omega \) is given in the voltage equation as \( 100 \) rad/s.

4. **Calculate the Capacitive Reactance**:
The capacitive reactance \( X_c \) is calculated using the formula:
\[ X_c = \frac{1}{\omega C} \]
Here, \( C = 1 \mu F = 1 \times 10^{-6} F \). Thus,
\[ X_c = \frac{1}{100 \times 1 \times 10^{-6}} = 10^4 \text{ ohms} \]

5. **Calculate the RMS Current**:
The RMS current \( I_{rms} \) through the capacitor can be calculated using Ohm's law for AC circuits:
\[ I_{rms} = \frac{E_{rms}}{X_c} \]
Substituting the values:
\[ I_{rms} = \frac{200}{10^4} = 0.02 \text{ A} = 20 \text{ mA} \]

6. **Final Answer**:
The reading of the ammeter will be \( 20 \text{ mA} \).