An `AC` source having a frequency of `50Hz` and voltage supply of `300v` is applied directly to the condenser of capacity `100muF`. The peak and `rms` values of displacement current are

Options

A. `9.42 A`, `(9.42)/(sqrt(2))A`
B. `(9.42)/(sqrt(2))A`, `9.42sqrt(2)A`
C. `9.42sqrt(2)A`, `9.42 A`
D. `9.42A`, `9.42 A`

Explanation

To solve the problem, we need to find the peak and RMS values of the displacement current when an AC voltage is applied to a capacitor. Here’s a step-by-step solution:

### Step 1: Identify Given Values
- Frequency (f) = 50 Hz
- Voltage (E) = 300 V
- Capacitance (C) = 100 µF = 100 × 10^-6 F

### Step 2: Calculate Angular Frequency (ω)
The angular frequency (ω) is given by the formula:
\[ \omega = 2\pi f \]
Substituting the given frequency:
\[ \omega = 2\pi \times 50 \, \text{Hz} = 100\pi \, \text{rad/s} \]

### Step 3: Calculate Capacitive Reactance (X_C)
The capacitive reactance (X_C) is calculated using the formula:
\[ X_C = \frac{1}{\omega C} \]
Substituting the values of ω and C:
\[ X_C = \frac{1}{100\pi \times 100 \times 10^{-6}} = \frac{1}{0.01\pi} = \frac{100}{\pi} \, \Omega \]

### Step 4: Calculate RMS Current (I_RMS)
The RMS current (I_RMS) can be calculated using the formula:
\[ I_{RMS} = \frac{E}{X_C} \]
Substituting the values of E and X_C:
\[ I_{RMS} = \frac{300}{\frac{100}{\pi}} = 300 \times \frac{\pi}{100} = 3\pi \, \text{A} \approx 9.42 \, \text{A} \]

### Step 5: Calculate Peak Current (I_peak)
The peak current (I_peak) is related to the RMS current by the formula:
\[ I_{peak} = I_{RMS} \times \sqrt{2} \]
Substituting the value of I_RMS:
\[ I_{peak} = 9.42 \times \sqrt{2} \approx 9.42 \times 1.414 \approx 13.33 \, \text{A} \]

### Final Results
- **RMS Current (I_RMS)**: Approximately 9.42 A
- **Peak Current (I_peak)**: Approximately 13.33 A