A conductor of capacity 1pF is connected to an `A.C` source of `220 V` and `50 Hz` frequency. The current flowing in the circuit will be

Options

A. `6.9 xx 10^(-8) A`
B. `6.9 A`
C. `6.9 xx 10^(-6) A`
D. zero

Explanation

To find the current flowing in the circuit when a capacitor of capacity 1 pF is connected to an AC source of 220 V and 50 Hz, we can follow these steps:

### Step-by-Step Solution:

1. **Identify Given Data**:
- Voltage (Vrms) = 220 V
- Frequency (f) = 50 Hz
- Capacitance (C) = 1 pF = \(1 \times 10^{-12}\) F

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

3. **Calculate Capacitive Reactance (Xc)**:
- The capacitive reactance (Xc) is given by:
\[ X_c = \frac{1}{\omega C} \]
- Substituting the values of ω and C:
\[ X_c = \frac{1}{100\pi \times 1 \times 10^{-12}} = \frac{1}{100\pi \times 10^{-12}} \approx \frac{1}{3.14 \times 10^{-10}} \approx 3.18 \times 10^{9} \, \Omega \]

4. **Calculate the Current (Irms)**:
- The RMS current (Irms) in the circuit can be calculated using:
\[ I_{rms} = \frac{V_{rms}}{X_c} \]
- Substituting the values:
\[ I_{rms} = \frac{220}{3.18 \times 10^{9}} \approx 6.9 \times 10^{-8} \, \text{A} \]

5. **Final Answer**:
- The current flowing in the circuit is approximately:
\[ I_{rms} \approx 6.9 \times 10^{-8} \, \text{A} \]