A resistance of `50Omega`, an inductance of `20//pi` henry and a capacitor of `5//pimuF` are connected in series with an A.C. source of 230 volt and 50Hz. The impedance of circuit is –

Options

A. `5Omega`
B. `50Omega`
C. `5KOmega`
D. `500Omega`

Answer

To find the impedance of the given series circuit with a resistance, inductance, and capacitance connected to an AC source, we can follow these steps:

### Step-by-Step Solution:

1. **Identify Given Values**:
- Resistance (R) = 50 Ω
- Inductance (L) = \( \frac{20}{\pi} \) H
- Capacitance (C) = \( \frac{5}{\pi} \) μF
- Frequency (f) = 50 Hz

2. **Calculate Angular Frequency (ω)**:
\[ \omega = 2\pi f \]
Substituting the frequency:
\[ \omega = 2\pi \times 50 = 100\pi \text{ rad/s} \]

3. **Calculate Inductive Reactance (Xl)**:
\[ X_l = \omega L \]
Substituting the values of ω and L:
\[ X_l = 100\pi \times \frac{20}{\pi} = 2000 \text{ Ω} \]

4. **Calculate Capacitive Reactance (Xc)**:
\[ X_c = \frac{1}{\omega C} \]
First, convert capacitance to farads:
\[ C = \frac{5}{\pi} \times 10^{-6} \text{ F} \]
Now substitute the values:
\[ X_c = \frac{1}{100\pi \times \frac{5}{\pi} \times 10^{-6}} = \frac{1}{100 \times 5 \times 10^{-6}} = \frac{10^6}{500} = 2000 \text{ Ω} \]

5. **Calculate Impedance (Z)**:
The formula for impedance in an RLC series circuit is:
\[ Z = \sqrt{R^2 + (X_l - X_c)^2} \]
Substituting the values:
\[ Z = \sqrt{50^2 + (2000 - 2000)^2} = \sqrt{2500 + 0} = \sqrt{2500} = 50 \text{ Ω} \]

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

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