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 solve the problem step by step, we will follow these calculations:

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

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

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

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

### Conclusion
The RMS value of the current in the circuit is:
\[ \boxed{20 \, mA} \]

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