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A resonance circuit having inductance and resistance $2 \times 10^{-4}$ H and $6.28\Omega$ respectively oscillates at 10 MHz frequency. The value of quality factor of this resonator is $[\pi = 3.14]$:
$Q = \frac{\omega L}{R} = \frac{2\pi \times 10 \times 10^6 \times 2 \times 10^{-4}}{6.28} = 2000$.
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