A resonance circuit having inductance $$2 \times 10^{-4}$$ H and resistance 6.28 $$\Omega$$ respectively oscillates at 10 MHz frequency. The value of quality factor of this resonator is ______. $$\pi = 3.14$$

We have a resonance circuit with inductance $$L = 2 \times 10^{-4}$$ H, resistance $$R = 6.28$$ $$\Omega$$, and resonant frequency $$f = 10$$ MHz $$= 10 \times 10^6$$ Hz.

The quality factor of a resonant circuit is given by $$Q = \frac{\omega L}{R}$$, where $$\omega = 2\pi f$$ is the angular frequency.

Calculating $$\omega$$: $$\omega = 2\pi \times 10 \times 10^6 = 2 \times 3.14 \times 10^7 = 6.28 \times 10^7$$ rad/s.

Now substituting into the quality factor formula: $$Q = \frac{6.28 \times 10^7 \times 2 \times 10^{-4}}{6.28}$$.

$$Q = \frac{6.28 \times 2 \times 10^3}{6.28} = 2 \times 10^3 = 2000$$

So, the answer is $$2000$$.