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 \[Q-\] resonant circuit is denned 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. ie, \[Q=\frac{\text{voltage}\,\text{across}\,\text{L}\,\text{or}\,\text{C}}{\text{applied}\,\text{voltage}\,\text{(=voltage}\,\text{actross}\,\text{R)}}\] \[=\frac{({{\omega }_{r}}L)i}{Ri}=\frac{{{\omega }_{r}}L}{R}\] or \[=\frac{(1/{{\omega }_{r}}C)i}{Ri}=\frac{1}{RC{{\omega }_{r}}}\] Using, \[{{\omega }_{r}}=\frac{1}{\sqrt{LC}},\]we get \[Q=\frac{L}{R}.\frac{1}{\sqrt{LC}}=\frac{1}{R}\sqrt{\frac{L}{C}}\] or \[=\frac{1}{R}\sqrt{\frac{L}{C}}\] Here, \[L=8.1\,mH,C=12.5\,\mu F,R=10\,\Omega ,\] \[f=500\,Hz.\] \[\therefore \] \[Q=\frac{{{\omega }_{r}}L}{R}=\frac{2\pi fL}{R}\] \[=\frac{2\times \pi \times 500\times 8.1\times {{10}^{-3}}}{10}=\frac{8.1\pi }{10}\] \[=2.5434\]
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