In an AC circuit , an alternating voltage e = 200√2 ``sin 100t V is connected to a capacitor of capacity `1mu F` . The rms value of the current in the circuit is

Options

A. 100 mA
B. 200 mA
C. 20 mA
D. 10 mA

Explanation

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

### Step 1: Identify the given parameters
The given alternating voltage is:
\[ e = 200\sqrt{2} \sin(100t) \, \text{V} \]
The capacitance \( C \) is:
\[ C = 1 \, \mu F = 1 \times 10^{-6} \, F \]

### Step 2: Determine the peak voltage \( E_0 \)
From the equation \( e = E_0 \sin(\omega t) \), we can identify:
\[ E_0 = 200\sqrt{2} \, V \]

### Step 3: Calculate the RMS voltage \( E_{rms} \)
The RMS voltage is given by:
\[ E_{rms} = \frac{E_0}{\sqrt{2}} \]
Substituting the value of \( E_0 \):
\[ E_{rms} = \frac{200\sqrt{2}}{\sqrt{2}} = 200 \, V \]

### Step 4: Calculate the capacitive reactance \( X_C \)
The capacitive reactance \( X_C \) is given by:
\[ X_C = \frac{1}{\omega C} \]
Where \( \omega = 100 \, \text{rad/s} \) and \( C = 1 \times 10^{-6} \, F \). Substituting these values:
\[ X_C = \frac{1}{100 \times 1 \times 10^{-6}} = \frac{1}{10^{-4}} = 10^4 \, \Omega \]

### Step 5: Calculate the RMS current \( I_{rms} \)
The RMS current in the circuit can be calculated using Ohm's law for AC circuits:
\[ I_{rms} = \frac{E_{rms}}{X_C} \]
Substituting the values we found:
\[ I_{rms} = \frac{200}{10^4} = 0.02 \, A = 20 \, mA \]

### Final Answer
The RMS value of the current in the circuit is:
\[ I_{rms} = 20 \, \text{mA} \]