A wire of reistance R is connected in series with an inductor of reactance `omega` L. Then quality factor of `RL` circuit is

Options

A. `(R)/(omegaL)`
B. `(omegaL)/(R)`
C. `(R)/(sqrt(R^(2)+omega^(2)L^(2)))`
D. `(omegaL)/(sqrt(R^(2)+omega^(2)L^(2)))`

Answer

To find the quality factor (Q) of an RL circuit consisting of a resistor (R) and an inductor (L) in series, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Components**: We have a resistor with resistance \( R \) and an inductor with inductive reactance \( X_L \). The inductive reactance is given by the formula:
\[ X_L = \omega L \]
where \( \omega \) is the angular frequency and \( L \) is the inductance.

2. **Define the Quality Factor**: The quality factor \( Q \) for an RL circuit is defined as the ratio of the voltage across the inductor \( V_L \) to the voltage across the resistor \( V_R \):
\[ Q = \frac{V_L}{V_R} \]

3. **Relate Voltages to Current**: The voltage across the inductor \( V_L \) can be expressed as:
\[ V_L = I \cdot X_L \]
and the voltage across the resistor \( V_R \) can be expressed as:
\[ V_R = I \cdot R \]
where \( I \) is the current flowing through the circuit.

4. **Substitute the Expressions**: Now, substituting these expressions into the quality factor formula gives:
\[ Q = \frac{I \cdot X_L}{I \cdot R} \]

5. **Cancel Out Current**: Since the current \( I \) is the same through both components in a series circuit, we can cancel \( I \) from the numerator and denominator:
\[ Q = \frac{X_L}{R} \]

6. **Substitute for Inductive Reactance**: Now, substitute \( X_L = \omega L \) into the equation:
\[ Q = \frac{\omega L}{R} \]

7. **Final Expression**: Therefore, the quality factor \( Q \) for the RL circuit is:
\[ Q = \frac{\omega L}{R} \]

### Conclusion:
The quality factor of the RL circuit is given by:
\[ Q = \frac{\omega L}{R} \]