A resonance circuit having inductance and resistance `2 xx 10^(-4)` H and 6.28 `Omega ` respectively oscillates at 10 MHz frequency. The value of quality factor of this resonator is_____.

Answer

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

### Step 1: Identify the given values
- Inductance (L) = \(2 \times 10^{-4} \, \text{H}\)
- Resistance (R) = \(6.28 \, \Omega\)
- Frequency (f) = \(10 \, \text{MHz} = 10 \times 10^6 \, \text{Hz} = 10^7 \, \text{Hz}\)

### Step 2: Calculate the angular frequency (ω)
The angular frequency (ω) is given by the formula:
\[ \omega = 2\pi f \]
Substituting the value of f:
\[ \omega = 2\pi \times 10^7 \]

### Step 3: Substitute values into the quality factor formula
The quality factor (Q) is given by the formula:
\[ Q = \frac{\omega L}{R} \]
Substituting the values of ω, L, and R:
\[ Q = \frac{(2\pi \times 10^7) \times (2 \times 10^{-4})}{6.28} \]

### Step 4: Simplify the expression
Calculating the numerator:
\[ \text{Numerator} = 2\pi \times 10^7 \times 2 \times 10^{-4} = 4\pi \times 10^3 \]
Now substituting this back into the equation for Q:
\[ Q = \frac{4\pi \times 10^3}{6.28} \]

### Step 5: Calculate the value of Q
Using the approximation \(\pi \approx 3.14\):
\[ Q = \frac{4 \times 3.14 \times 10^3}{6.28} \]
Calculating:
\[ Q = \frac{12.56 \times 10^3}{6.28} = 2 \times 10^3 = 2000 \]

### Final Answer
The value of the quality factor (Q) of this resonator is \(2000\).

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