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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, $[π = 3.14]$

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2000

200

1000

answer is C.

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Detailed Solution

$Q = \frac{x_{L}}{R} = \frac{ω L}{R} = \frac{2 π f L}{R}$
$Q = \frac{2 π \times 10^{6} \times 10 \times 2 \times 10^{- 4}}{6.28} = 2000$
$Q = 2000$

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