An alternating current of 1.5 mA with angular frequency 100 rad/s flows through a `10 k Omega` resistor and a `0.50 mu F` capacitor in series. The rms potential drop across the capacitor is

Options

A. 4.8 V
B. 30 V
C. 15 V
D. 190 V

Answer

To find the rms potential drop across the capacitor in a series circuit with a resistor and a capacitor, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Given Values**:
- Current, \( I = 1.5 \, \text{mA} = 1.5 \times 10^{-3} \, \text{A} \)
- Angular frequency, \( \omega = 100 \, \text{rad/s} \)
- Resistance, \( R = 10 \, \text{k}\Omega = 10 \times 10^{3} \, \Omega = 10^{4} \, \Omega \)
- Capacitance, \( C = 0.50 \, \mu\text{F} = 0.50 \times 10^{-6} \, \text{F} \)

2. **Calculate the Capacitive Reactance (\(X_C\))**:
The capacitive reactance is given by the formula:
\[ X_C = \frac{1}{\omega C} \]
Substituting the values:
\[ X_C = \frac{1}{100 \times 0.50 \times 10^{-6}} = \frac{1}{50 \times 10^{-6}} = \frac{10^{6}}{50} = 20,000 \, \Omega = 2 \times 10^{4} \, \Omega \]

3. **Calculate the rms Voltage across the Capacitor (\(V_C\))**:
The rms voltage across the capacitor can be calculated using:
\[ V_C = I \times X_C \]
Substituting the values:
\[ V_C = (1.5 \times 10^{-3}) \times (2 \times 10^{4}) = 3 \times 10^{1} = 30 \, \text{V} \]

4. **Conclusion**:
The rms potential drop across the capacitor is \( V_C = 30 \, \text{V} \).

### Final Answer:
The rms potential drop across the capacitor is **30 V**.