A capacitor of capacitance `2 mu F` is connected in the tank circuit of an oscillator oscillating with a frequency of 1 kHz. If the current flowing in the circuit is `2 mA`, the voltage across the capacitor will be

Options

A. 0.16V
B. 0.32V
C. 79.5V
D. 159V

Explanation

To find the voltage across the capacitor in the given tank circuit, we can follow these steps:

### Step-by-Step Solution:

1. **Identify Given Values:**
- Capacitance \( C = 2 \, \mu F = 2 \times 10^{-6} \, F \)
- Frequency \( f = 1 \, kHz = 1 \times 10^{3} \, Hz \)
- Current \( I = 2 \, mA = 2 \times 10^{-3} \, A \)

2. **Calculate the Angular Frequency \( \omega \):**
\[ \omega = 2 \pi f = 2 \pi \times 1 \times 10^{3} \approx 6283.19 \, rad/s \]

3. **Calculate the Capacitive Reactance \( X_c \):**
The formula for capacitive reactance is:
\[ X_c = \frac{1}{\omega C} \]
Substituting the values:
\[ X_c = \frac{1}{6283.19 \times 2 \times 10^{-6}} \approx \frac{1}{0.0000125664} \approx 7957.75 \, \Omega \]

4. **Calculate the Voltage Across the Capacitor \( V \):**
Using Ohm's law for AC circuits, the voltage across the capacitor is given by:
\[ V = I \times X_c \]
Substituting the values:
\[ V = (2 \times 10^{-3}) \times 7957.75 \approx 15.91 \, V \]

5. **Final Result:**
The voltage across the capacitor is approximately \( 15.91 \, V \).