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A resonance circuit having inductance and resistance $2 \times 10^{- 4} H$ and $6.28 Ω$ respectively, oscillates at 22.3kHz frequency. The value of quality factor of this resonator is__ ( take $π = 3.14$ ).
(Round off to nearest integer)

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answer is 4.

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Detailed Solution

Quality factor = $Q = \frac{w_{0} L}{R} = 2 π f \frac{L}{R}$

$= \frac{2 π \times 22.3 \times 10^{3} \times 2 \times 10^{- 4}}{6.28}$ $= 4.46 = 4$

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