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]
Answer (integer)
2000
Solution
Given, L = 2 $\times$ 10<sup>$-$4</sup> H, R = 6.28 $\Omega$, f<sub>0</sub> = 10 MHz = 10 $\times$ 10<sup>6</sup> Hz<br/><br/>$\therefore$ Quality factor $= {\omega _0}{L \over R} = 2\pi {f_0}{L \over R}$<br/><br/>$= 2\pi \times 10 \times {10^6} \times {{2 \times {{10}^{ - 4}}} \over {6.28}}$<br/><br/>$= 2 \times {10^3} = 2000$
About this question
Subject: Physics · Chapter: Alternating Current · Topic: AC Circuits: R, L, C
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