In `LCR` circuit current resonant frequency is `600Hz` and half power points are at `650` and `550 Hz`. The quality factor is

Answer

To find the quality factor \( Q \) of the LCR circuit, we can follow these steps:

### Step 1: Identify the given values
- Resonant frequency \( \omega_0 = 600 \, \text{Hz} \)
- Half power points \( \omega_1 = 550 \, \text{Hz} \) and \( \omega_2 = 650 \, \text{Hz} \)

### Step 2: Calculate the bandwidth
The bandwidth \( \Delta \omega \) is calculated as the difference between the two half power points:
\[ \Delta \omega = \omega_2 - \omega_1 = 650 \, \text{Hz} - 550 \, \text{Hz} = 100 \, \text{Hz} \]

### Step 3: Use the formula for the quality factor
The quality factor \( Q \) is given by the formula:
\[ Q = \frac{\omega_0}{\Delta \omega} \]

### Step 4: Substitute the values into the formula
Substituting the values we have:
\[ Q = \frac{600 \, \text{Hz}}{100 \, \text{Hz}} = 6 \]

### Step 5: State the final answer
Thus, the quality factor \( Q \) of the LCR circuit is:
\[ \boxed{6} \]

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