A capacitor of `1muF` is connected to source of ac having emf given by equation `E = 200 sin(120pit)`. Find the value of rms current through the capacitor.

Explanation

To find the value of the RMS current through a capacitor connected to an AC source, we can follow these steps:

### Step 1: Identify the given values
- Capacitance \( C = 1 \, \mu F = 1 \times 10^{-6} \, F \)
- The EMF of the AC source is given by \( E(t) = 200 \sin(120 \pi t) \).

### Step 2: Determine the angular frequency \( \omega \)
From the equation of the EMF, we can identify the angular frequency \( \omega \):
\[ \omega = 120 \pi \, \text{rad/s} \]

### Step 3: Calculate the peak voltage \( E_0 \)
The peak voltage \( E_0 \) is given as:
\[ E_0 = 200 \, V \]

### Step 4: Calculate the RMS voltage \( E_{rms} \)
The RMS voltage is related to the peak voltage by the formula:
\[ E_{rms} = \frac{E_0}{\sqrt{2}} = \frac{200}{\sqrt{2}} \approx 141.4 \, V \]

### Step 5: 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:
\[ X_C = \frac{1}{120 \pi \times 1 \times 10^{-6}} \approx \frac{1}{0.00037699} \approx 2654.9 \, \Omega \]

### Step 6: Calculate the RMS current \( I_{rms} \)
The RMS current through the capacitor can be calculated using Ohm's law:
\[ I_{rms} = \frac{E_{rms}}{X_C} \]
Substituting the values:
\[ I_{rms} = \frac{141.4}{2654.9} \approx 0.0533 \, A \approx 0.05 \, A \]

### Final Answer
The value of the RMS current through the capacitor is approximately:
\[ I_{rms} \approx 0.05 \, A \]

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