A series combination of a `0.2 k omega` resistor and `1.5 muF` capacitor is given. If, this circuit is connected across a 220 V, 50 Hz AC source what will be the impedance of the circuit

Options

A. `250 omega`
B. `268 omega`
C. `29.15 omega`
D. `291.5 omega`

Explanation

To solve the problem of finding the impedance of a series combination of a resistor and a capacitor connected to an AC source, we can follow these steps:

### Step 1: Identify the given values
- Resistance (R) = 0.2 kΩ = 200 Ω
- Capacitance (C) = 1.5 µF = 1.5 × 10^(-6) F
- Frequency (f) = 50 Hz

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

### Step 3: Calculate the capacitive reactance (Xc)
The capacitive reactance (Xc) is calculated using the formula:
\[ X_c = \frac{1}{\omega C} \]
Substituting the values of ω and C:
\[ X_c = \frac{1}{100 \pi \times 1.5 \times 10^{-6}} \]
Calculating this gives:
\[ X_c \approx \frac{1}{4.71 \times 10^{-4}} \]
\[ X_c \approx 2120.3 \, \Omega \]

### Step 4: Calculate the impedance (Z)
The impedance (Z) in a series circuit with a resistor and capacitor is given by:
\[ Z = \sqrt{R^2 + X_c^2} \]
Substituting the values of R and Xc:
\[ Z = \sqrt{(200)^2 + (2120.3)^2} \]
Calculating this gives:
\[ Z = \sqrt{40000 + 4490564.09} \]
\[ Z = \sqrt{4530564.09} \]
\[ Z \approx 2688.4 \, \Omega \]

### Step 5: Final result
Thus, the impedance of the circuit is approximately:
\[ Z \approx 2688.4 \, \Omega \]