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`

Answer

To find the impedance \( Z \) of the given series circuit, we will follow these steps:

### Step 1: Identify Given Values
- Resistance \( R = 300 \, \Omega \)
- Inductance \( L = 0.9 \, H \)
- Capacitance \( C = 2.0 \, \mu F = 2.0 \times 10^{-6} \, F \)
- Angular frequency \( \omega = 1000 \, rad/s \)

### Step 2: Calculate Inductive Reactance \( X_L \)
The formula for inductive reactance is:
\[ X_L = \omega L \]
Substituting the values:
\[ X_L = 1000 \times 0.9 = 900 \, \Omega \]

### Step 3: Calculate Capacitive Reactance \( X_C \)
The formula for capacitive reactance is:
\[ X_C = \frac{1}{\omega C} \]
Substituting the values:
\[ X_C = \frac{1}{1000 \times 2 \times 10^{-6}} = \frac{1}{0.002} = 500 \, \Omega \]

### Step 4: Calculate Impedance \( Z \)
The formula for impedance in a series RLC circuit is:
\[ Z = \sqrt{(X_L - X_C)^2 + R^2} \]
Substituting the values we calculated:
\[ Z = \sqrt{(900 - 500)^2 + 300^2} \]
Calculating the values inside the square root:
\[ Z = \sqrt{(400)^2 + (300)^2} = \sqrt{160000 + 90000} = \sqrt{250000} \]
Thus, we find:
\[ Z = 500 \, \Omega \]

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

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