In an ac circuit an alternating voltage `e = 200 sqrt(2) sin 100t` volts is connected to a capacitor of capacity 1 `muF.` The rms.value of the current in the circuit is

Options

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

Answer

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

### Step 1: Identify the given parameters
The alternating voltage is given as:
\[ e(t) = 200\sqrt{2} \sin(100t) \]
From this, we can identify:
- Peak voltage \( E_0 = 200\sqrt{2} \) volts
- Angular frequency \( \omega = 100 \) rad/s

The capacitance \( C \) 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 formula:
\[ E_{rms} = \frac{E_0}{\sqrt{2}} \]
Substituting the value of \( E_0 \):
\[ E_{rms} = \frac{200\sqrt{2}}{\sqrt{2}} = 200 \, \text{volts} \]

### 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 \( I_{rms} \) can be calculated using Ohm's law for AC circuits:
\[ I_{rms} = \frac{E_{rms}}{X_c} \]
Substituting the values we have:
\[ I_{rms} = \frac{200}{10^4} = 0.02 \, \text{A} \]

### Final Answer
The RMS value of the current in the circuit is:
\[ I_{rms} = 0.02 \, \text{A} \]

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