In a series LCR circuit, operated with an ac of angular frequency `omega` , the total impedance is

Options

A. `[R^(2)+(Lomega-Comega)^(2)]^(1//2)`
B. `[R^(2)+(Lomega+1/(Comega))^(2)]^(1//2)`
C. `[R^(2)+(Lomega-1/(Comega))^(2)]^(1//2)`
D. `[(Romega)^(2)+(Lomega-1/(Comega))^(2)]^(1//2)`

Answer

To find the total impedance in a series LCR circuit operated with an AC source of angular frequency \( \omega \), we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Components of the LCR Circuit**:
- The series LCR circuit consists of an inductor (L), a capacitor (C), and a resistor (R) connected in series with an AC source.

2. **Identify the Reactances**:
- The inductive reactance \( X_L \) is given by the formula:
\[ X_L = \omega L \]
- The capacitive reactance \( X_C \) is given by the formula:
\[ X_C = \frac{1}{\omega C} \]

3. **Write the Formula for Total Impedance**:
- The total impedance \( Z \) in a series LCR circuit is given by:
\[ Z = \sqrt{R^2 + (X_L - X_C)^2} \]

4. **Substitute the Values of Reactances**:
- Substitute \( X_L \) and \( X_C \) into the impedance formula:
\[ Z = \sqrt{R^2 + \left(\omega L - \frac{1}{\omega C}\right)^2} \]

5. **Final Expression for Impedance**:
- Therefore, the total impedance \( Z \) can be expressed as:
\[ Z = \sqrt{R^2 + \left(\omega L - \frac{1}{\omega C}\right)^2} \]

### Conclusion:
The total impedance in a series LCR circuit operated with an AC of angular frequency \( \omega \) is:
\[ Z = \sqrt{R^2 + \left(\omega L - \frac{1}{\omega C}\right)^2} \]