In a series circuit `R=300Omega, L=0.9H, C=2.0 muF` and `omega=1000rad//sec` . The impedence of the circuit is

Options

A. `1300 Omega`
B. `900 Omega`
C. `500 Omega`
D. `400 Omega`

Explanation

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

### Step 1: Identify the given values
- Resistance (R) = 300 Ω
- Inductance (L) = 0.9 H
- Capacitance (C) = 2.0 μF = 2.0 × 10^(-6) F
- Angular frequency (ω) = 1000 rad/s

### Step 2: Calculate the inductive reactance (X_L)
The inductive reactance \(X_L\) is given by the formula:
\[ X_L = \omega L \]
Substituting the values:
\[ X_L = 1000 \, \text{rad/s} \times 0.9 \, \text{H} = 900 \, \Omega \]

### 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 the values:
\[ X_C = \frac{1}{1000 \, \text{rad/s} \times 2.0 \times 10^{-6} \, \text{F}} = \frac{1}{0.002} = 500 \, \Omega \]

### Step 4: Calculate the total reactance (X)
The total reactance \(X\) in the circuit is given by:
\[ X = X_L - X_C \]
Substituting the values:
\[ X = 900 \, \Omega - 500 \, \Omega = 400 \, \Omega \]

### Step 5: Calculate the impedance (Z)
The impedance \(Z\) of the circuit is given by:
\[ Z = \sqrt{R^2 + X^2} \]
Substituting the values:
\[ Z = \sqrt{(300 \, \Omega)^2 + (400 \, \Omega)^2} \]
Calculating:
\[ Z = \sqrt{90000 + 160000} = \sqrt{250000} = 500 \, \Omega \]

### Final Answer
The impedance of the circuit is \(Z = 500 \, \Omega\).

---