A 220V, 50 Hz ac source is connected to a series combination of a 3.16 micro farad capacitor , 0.63 H inductor and 600 ohm resistor.
Impedence of the circuit is

Answer

To find the impedance of the given circuit, we will follow these steps:

### Step 1: Identify the given values
- Voltage (V) = 220 V
- Frequency (f) = 50 Hz
- Capacitance (C) = 3.16 μF = 3.16 × 10^-6 F
- Inductance (L) = 0.63 H
- Resistance (R) = 600 Ω

### 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:
\[ \omega = 2\pi \times 50 = 100\pi \, \text{rad/s} \]

### Step 3: Calculate the inductive reactance (X_L)
Inductive reactance (X_L) is given by:
\[ X_L = \omega L \]
Substituting the values:
\[ X_L = 100\pi \times 0.63 = 63\pi \, \Omega \]

### Step 4: Calculate the capacitive reactance (X_C)
Capacitive reactance (X_C) is given by:
\[ X_C = \frac{1}{\omega C} \]
Substituting the values:
\[ X_C = \frac{1}{100\pi \times 3.16 \times 10^{-6}} = \frac{1}{0.000316} \approx 3164.55 \, \Omega \]

### Step 5: Calculate the impedance (Z)
The impedance (Z) of the circuit is given by the formula:
\[ Z = \sqrt{R^2 + (X_L - X_C)^2} \]
Substituting the values we calculated:
\[ Z = \sqrt{600^2 + (63\pi - 3164.55)^2} \]

### Step 6: Calculate the values inside the square root
First, calculate \(X_L - X_C\):
\[ X_L - X_C = 63\pi - 3164.55 \]
Calculating \(63\pi \approx 197.92\):
\[ X_L - X_C \approx 197.92 - 3164.55 \approx -2966.63 \]

Now, substituting back into the impedance formula:
\[ Z = \sqrt{600^2 + (-2966.63)^2} \]
Calculating \(600^2 = 360000\) and \((-2966.63)^2 \approx 8793000\):
\[ Z = \sqrt{360000 + 8793000} = \sqrt{9153000} \]

### Step 7: Final calculation
Calculating the square root:
\[ Z \approx 3025.5 \, \Omega \]

### Final Answer
The impedance of the circuit is approximately \(3025.5 \, \Omega\).