In a series LCR circuit having `L=30 mH, R=8 Omega` and the resonant frequency is 50 Hz. The quality factor of the circuit is

Options

A. `0.118`
B. `11.8`
C. 118
D. `1.18`

Answer

To find the quality factor (Q) of a series LCR circuit, we can use the formula:

\[ Q = \frac{\omega_0 L}{R} \]

Where:
- \( \omega_0 \) is the angular resonant frequency in radians per second,
- \( L \) is the inductance in henries,
- \( R \) is the resistance in ohms.

### Step 1: Convert the given values into appropriate units
- Inductance \( L = 30 \, \text{mH} = 30 \times 10^{-3} \, \text{H} = 0.03 \, \text{H} \)
- Resistance \( R = 8 \, \Omega \)
- Resonant frequency \( f_0 = 50 \, \text{Hz} \)

### Step 2: Calculate the angular frequency \( \omega_0 \)
The angular frequency is given by:

\[ \omega_0 = 2 \pi f_0 \]

Substituting the value of \( f_0 \):

\[ \omega_0 = 2 \pi \times 50 \]

### Step 3: Calculate \( \omega_0 \)
Calculating \( \omega_0 \):

\[ \omega_0 = 100 \pi \, \text{rad/s} \]

### Step 4: Substitute values into the quality factor formula
Now, substitute \( \omega_0 \), \( L \), and \( R \) into the quality factor formula:

\[ Q = \frac{\omega_0 L}{R} = \frac{(100 \pi)(0.03)}{8} \]

### Step 5: Calculate \( Q \)
Now, calculate \( Q \):

\[ Q = \frac{3 \pi}{8} \]

### Step 6: Simplify \( Q \)
Using \( \pi \approx 3.14 \):

\[ Q = \frac{3 \times 3.14}{8} \approx \frac{9.42}{8} \approx 1.1775 \]

### Step 7: Round to appropriate significant figures
Rounding \( Q \) gives:

\[ Q \approx 1.18 \]

### Final Answer
The quality factor \( Q \) of the circuit is approximately **1.18**.

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