In an ac circuit , an alternating voltage `e = 200 sqrt2 sin 100 t` volts is connected to a capacitor of capacitance `1 mu F`. The rms value of the current in the circuit is :

Options

A. 20 mA
B. 10 mA
C. 100 mA
D. 200 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 alternating voltage is given by:
\[ e(t) = 200 \sqrt{2} \sin(100t) \]
From this, we can identify:
- The peak voltage \( E_0 = 200 \sqrt{2} \) volts.
- The angular frequency \( \omega = 100 \) rad/s.

The capacitance of the capacitor is given as:
\[ C = 1 \, \mu F = 1 \times 10^{-6} \, F \]

### Step 2: Calculate the RMS voltage
The RMS value of the voltage \( E_{RMS} \) can be calculated using the relation:
\[ E_{RMS} = \frac{E_0}{\sqrt{2}} \]
Substituting the value of \( E_0 \):
\[ E_{RMS} = \frac{200 \sqrt{2}}{\sqrt{2}} = 200 \, V \]

### Step 3: Calculate the capacitive reactance \( X_C \)
The capacitive reactance \( X_C \) is given by the formula:
\[ X_C = \frac{1}{\omega C} \]
Substituting the values of \( \omega \) and \( C \):
\[ X_C = \frac{1}{100 \times 1 \times 10^{-6}} = \frac{1}{10^{-4}} = 10^4 \, \Omega \]

### Step 4: Calculate the RMS current \( I_{RMS} \)
The RMS current can be calculated using the formula:
\[ I_{RMS} = \frac{E_{RMS}}{X_C} \]
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:
\[ I_{RMS} = 20 \, mA \]

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