The quality factor of LCR circuit having resistance ( R ) and inductance ( L ) at resonance frequency `(omega )` is given by

Options

A. `("CR")/(omega_(@))`
B. `(omega_(@)L)/(R )`
C. `(omega_(@)R)/(L)`
D. `(R )/((omega_(@)C )`

Answer

To find the quality factor (Q) of an LCR circuit at resonance frequency (ω), we can follow these steps:

### Step-by-Step Solution:

1. **Understanding the Circuit**:
- We have an LCR circuit which consists of an inductor (L), a capacitor (C), and a resistor (R) connected to an AC source.

2. **Resonance Condition**:
- At resonance frequency (ω), the inductive reactance (X_L) is equal to the capacitive reactance (X_C). However, for the quality factor, we focus on the inductor and resistor.

3. **Voltage Across Inductor**:
- The voltage across the inductor (V_L) can be expressed as:
\[ V_L = I \cdot X_L \]
- Where \(X_L\) is the inductive reactance given by:
\[ X_L = \omega L \]

4. **Voltage Across Resistor**:
- The voltage across the resistor (V_R) is given by:
\[ V_R = I \cdot R \]

5. **Quality Factor Definition**:
- The quality factor (Q) is defined as the ratio of the voltage across the inductor to the voltage across the resistor:
\[ Q = \frac{V_L}{V_R} \]

6. **Substituting the Expressions**:
- Substituting the expressions for \(V_L\) and \(V_R\) into the quality factor formula:
\[ Q = \frac{I \cdot X_L}{I \cdot R} \]
- The current (I) cancels out:
\[ Q = \frac{X_L}{R} \]

7. **Substituting Inductive Reactance**:
- Now substituting \(X_L = \omega L\):
\[ Q = \frac{\omega L}{R} \]

8. **Final Result**:
- Thus, the quality factor of the LCR circuit at resonance frequency is given by:
\[ Q = \frac{\omega L}{R} \]