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 the given AC circuit, we can follow these steps:

### Step 1: Identify the given values
- **Current (I_rms)**: 1.5 mA = \(1.5 \times 10^{-3}\) A
- **Angular frequency (ω)**: 100 rad/s
- **Capacitance (C)**: 0.50 µF = \(0.50 \times 10^{-6}\) F
- **Resistance (R)**: 10 kΩ = \(10 \times 10^{3}\) Ω

### Step 2: Calculate the capacitive reactance (X_c)
The capacitive reactance \(X_c\) is given by the formula:
\[ X_c = \frac{1}{\omega C} \]
Substituting the known values:
\[ X_c = \frac{1}{100 \times 0.50 \times 10^{-6}} = \frac{1}{50 \times 10^{-6}} = \frac{1}{5 \times 10^{-5}} = 2 \times 10^{4} \, \Omega \]

### Step 3: Calculate the RMS voltage across the capacitor (V_c)
The voltage across the capacitor can be calculated using Ohm's law for AC circuits:
\[ V_c = I_{rms} \times X_c \]
Substituting the values we have:
\[ V_c = (1.5 \times 10^{-3}) \times (2 \times 10^{4}) = 3 \times 10^{1} = 30 \, \text{V} \]

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

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