A `1mu F` capacitor is connected to 220 V - 50 Hz a.c. source Find the virtual value of current through the circuit. What is the peak voltage across the capacitor ?

Answer

To solve the problem step by step, we will follow the outlined procedure to find the virtual value of current through the circuit and the peak voltage across the capacitor.

### Step 1: Identify the given values
- Capacitance (C) = 1 µF = \(1 \times 10^{-6}\) F
- RMS Voltage (V_RMS) = 220 V
- Frequency (f) = 50 Hz

### Step 2: Calculate the angular frequency (ω)
The angular frequency (ω) is given by the formula:
\[ \omega = 2\pi f \]
Substituting the value of frequency:
\[ \omega = 2 \times \pi \times 50 \approx 314.16 \, \text{rad/s} \]

### 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 ω and C:
\[ X_C = \frac{1}{314.16 \times 1 \times 10^{-6}} \approx 3183.1 \, \Omega \]

### Step 4: Calculate the RMS current (I_RMS)
The RMS current (I_RMS) can be calculated using Ohm's law for AC circuits:
\[ I_{RMS} = \frac{V_{RMS}}{X_C} \]
Substituting the values:
\[ I_{RMS} = \frac{220}{3183.1} \approx 0.069 \, \text{A} \]

### Step 5: Calculate the peak voltage (V_0)
The peak voltage (V_0) across the capacitor can be calculated using the relationship:
\[ V_0 = \sqrt{2} \times V_{RMS} \]
Substituting the value of V_RMS:
\[ V_0 = \sqrt{2} \times 220 \approx 311.13 \, \text{V} \]

### Final Results
- The virtual value of current (I_RMS) through the circuit is approximately **0.069 A**.
- The peak voltage (V_0) across the capacitor is approximately **311 V**.