An alternating voltage `E=200sqrt(2)sin(100t)` is connected to a `1` microfarad capacitor through an AC ammeter. The reading of the ammeter shall be

Options

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

Explanation

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

### Step-by-Step Solution:

1. **Identify the RMS Voltage**:
The given voltage is in the form \( E(t) = V_{max} \sin(\omega t) \), where \( V_{max} = 200\sqrt{2} \) and \( \omega = 100 \).
The RMS (Root Mean Square) voltage is given by:
\[ V_{RMS} = \frac{V_{max}}{\sqrt{2}} = \frac{200\sqrt{2}}{\sqrt{2}} = 200 \, V \]

**Hint**: Remember that the RMS value of a sinusoidal voltage is the peak voltage divided by \(\sqrt{2}\).

2. **Calculate the Capacitive Reactance**:
The capacitive reactance \( X_C \) is given by the formula:
\[ X_C = \frac{1}{\omega C} \]
Here, \( C = 1 \, \mu F = 1 \times 10^{-6} \, F \) and \( \omega = 100 \).
Substituting the values:
\[ X_C = \frac{1}{100 \times 1 \times 10^{-6}} = \frac{1}{0.0001} = 10000 \, \Omega \]

**Hint**: Capacitive reactance decreases with increasing frequency and capacitance.

3. **Calculate the RMS Current**:
The RMS current \( I_{RMS} \) through the capacitor can be calculated using Ohm's law for AC circuits:
\[ I_{RMS} = \frac{V_{RMS}}{X_C} \]
Substituting the values we found:
\[ I_{RMS} = \frac{200}{10000} = 0.02 \, A = 20 \, mA \]

**Hint**: The current through a capacitor in an AC circuit is directly proportional to the voltage and inversely proportional to the reactance.

4. **Final Result**:
The reading of the ammeter will be \( 20 \, mA \).

### Conclusion:
The reading of the AC ammeter connected to the capacitor is \( 20 \, mA \).