Q169 / 180 · COMEDK 2016 · Physics

Quality factor of a series $L-C-R$ circuit decreases from 3 to 2 . Resonant frequency is $600 \mathrm{~Hz}$. Change in bandwidth is

  • A
    zero
  • Correct answer
    B
    $100 \mathrm{~Hz}$ increase
  • C
    $100 \mathrm{~Hz}$ decrease
  • D
    $300 \mathrm{~Hz}$ increase

Solution

Given, $f_{0}=600 \mathrm{~Hz}, Q_{1}=3, Q_{2}=2$ The bandwidth in $L-C-R$ circuit, $\beta=\frac{f_{0}}{Q}$ As, quality factor decreases, bandwidth increases. This increase in bandwidth is given by $\Delta \beta=\beta_{2}-\beta_{1}=\frac{f_{0}}{Q_{2}}-\frac{f_{0}}{Q_{1}}=f_{0}\left(\frac{1}{Q_{2}}-\frac{1}{Q_{1}}\right)$ $=600\left(\frac{1}{2}-\frac{1}{3}\right)=100 \mathrm{~Hz}$

Hence, the correct answer is option: 'B'.

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