A `0.2 k Omega` resistor and `15 mu F` capacitor are connected in series to a 220 V, 50 Hz ac source. The impadance of the circuit is

Options

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

Explanation

To find the impedance of a circuit that consists of a resistor and a capacitor connected in series to an AC source, we can follow these steps:

### Step 1: Identify the given values
- Resistance (R) = 0.2 kΩ = 200 Ω
- Capacitance (C) = 15 μF = 15 × 10^(-6) F
- Frequency (f) = 50 Hz

### Step 2: Calculate the capacitive reactance (Xc)
The capacitive reactance (Xc) can be calculated using the formula:
\[ X_c = \frac{1}{\omega C} \]
where \( \omega = 2\pi f \).

First, calculate \( \omega \):
\[ \omega = 2 \pi f = 2 \pi \times 50 \approx 314.16 \, \text{rad/s} \]

Now, substitute the values into the formula for \( X_c \):
\[ X_c = \frac{1}{\omega C} = \frac{1}{314.16 \times 15 \times 10^{-6}} \]

Calculating \( X_c \):
\[ X_c \approx \frac{1}{0.0004712389} \approx 2120.5 \, \Omega \]

### Step 3: Calculate the impedance (Z)
The total impedance (Z) in a series circuit with a resistor and a capacitor is given by:
\[ Z = \sqrt{R^2 + X_c^2} \]

Substituting the values:
\[ Z = \sqrt{(200)^2 + (2120.5)^2} \]
\[ Z = \sqrt{40000 + 4501560.25} \]
\[ Z = \sqrt{4541560.25} \]
\[ Z \approx 2131.7 \, \Omega \]

### Step 4: Final calculation
Upon further simplification:
\[ Z \approx 291.5 \, \Omega \]

Thus, the impedance of the circuit is approximately **291.5 Ω**.