A) 2.02
B) 2.5434
C) 20.54
D) 200.54
Correct Answer: B
Solution :
Key Idea Quality factor defines sharpness of tuning at resonance. The Q-factor of series resonant circuit is defined as the ratio of the voltage developed across the inductance or capacitance at resonance to the impressed voltage, which is the voltage applied across R. i.e., \[Q=\frac{voltage\text{ }across\text{ }L\text{ }or\text{ }C}{~applied\text{ }voltage\text{ (}=\text{ }voltage\text{ }across\text{ }R)}\] \[Q=\frac{({{\omega }_{r}}L)i}{Ri}=\frac{{{\omega }_{r}}L}{R}\] or \[Q=\frac{(1/{{\omega }_{r}}C)i}{Ri}=\frac{1}{RC{{\omega }_{r}}}\] Using, \[{{\omega }_{r}}=\frac{1}{\sqrt{LC}},\] we get \[Q=\frac{L}{R}.\frac{1}{\sqrt{LC}}\] or \[Q=\frac{1}{R}.\sqrt{\frac{L}{C}}\] Here, \[L=8.1mH,\,C=12.5\mu F,\,R=10\Omega \] \[f=500Hz.\] \[\therefore \] \[Q=\frac{{{\omega }_{r}}L}{R}=\frac{2\pi fL}{R}\] \[=\frac{2\times \pi \times 500\times 8.1\times {{10}^{-3}}}{10}=\frac{8.1\pi }{10}\] \[=2.5434\]
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