Question

A resonance circuit having inductance and resistance $2 \times 10^{-4}$ H and $6.28$ Ω respectively oscillates at 10 MHz frequency. The value of quality factor of this resonator is __________.

Hint:

The Quality Factor ($Q$) measures the sharpness of resonance. A higher $Q$ indicates a lower rate of energy loss relative to the stored energy.

Answer 1

Correct Answer: 2000

Step 1: Quality factor $Q = \frac{\omega L}{R} = \frac{2\pi f L}{R}$.
Step 2: Given $f = 10 \times 10^6$ Hz, $L = 2 \times 10^{-4}$ H, $R = 6.28$ Ω.
Step 3: $Q = \frac{2 \times 3.14 \times 10^7 \times 2 \times 10^{-4}}{6.28} = \frac{6.28 \times 10^7 \times 2 \times 10^{-4}}{6.28}$.
Step 4: $Q = 2 \times 10^3 = 2000$.