A resistor of `50 Omega`, an inductor of `(20//pi)H` and a capacitor of `(5//pi) mu F` are connected in series to a voltage source 230 V, 50 Hz. Find the impedance of the circuit.

Answer

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

### Step 1: Identify the given values
- Resistance (R) = 50 Ω
- Inductance (L) = \( \frac{20}{\pi} \) H
- Capacitance (C) = \( \frac{5}{\pi} \) μF = \( \frac{5}{\pi} \times 10^{-6} \) F
- Frequency (f) = 50 Hz

### Step 2: Calculate the inductive reactance (X_L)
The inductive reactance \( X_L \) is given by the formula:
\[ X_L = \omega L \]
where \( \omega = 2\pi f \).

Calculating \( \omega \):
\[ \omega = 2\pi \times 50 = 100\pi \text{ rad/s} \]

Now substituting \( L \):
\[ X_L = 100\pi \times \frac{20}{\pi} = 100 \times 20 = 2000 \, \Omega \]

### Step 3: Calculate the capacitive reactance (X_C)
The capacitive reactance \( X_C \) is given by the formula:
\[ X_C = \frac{1}{\omega C} \]

Substituting \( \omega \) and \( C \):
\[ X_C = \frac{1}{100\pi \times \frac{5}{\pi \times 10^6}} = \frac{1}{\frac{500}{10^6}} = \frac{10^6}{500} = 2000 \, \Omega \]

### Step 4: Calculate the impedance (Z)
The impedance \( Z \) of the circuit is given by:
\[ Z = \sqrt{R^2 + (X_L - X_C)^2} \]

Substituting the values:
\[ Z = \sqrt{50^2 + (2000 - 2000)^2} \]
\[ Z = \sqrt{2500 + 0} = \sqrt{2500} = 50 \, \Omega \]

### Final Answer
The impedance of the circuit is \( Z = 50 \, \Omega \).

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