Quality factor of a series $L-C-R$ circuit decreases from 3 to 2 . Resonant frequency is $600 \mathrm{~Hz}$.…
Quality factor of a series $L-C-R$ circuit decreases from 3 to 2 . Resonant frequency is $600 \mathrm{~Hz}$. Change in bandwidth is
zero
$100 \mathrm{~Hz}$ increase
$100 \mathrm{~Hz}$ decrease
$300 \mathrm{~Hz}$ 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
$\Delta \beta=\beta_{2}-\beta_{1}=\frac{f_{0}}{Q_{2}}-\frac{f_{0}}{Q_{1}}=f_{0}\left(\frac{1}{Q_{2}}-\frac{1}{Q_{1}}\right)$
$=600\left(\frac{1}{2}-\frac{1}{3}\right)=100 \mathrm{~Hz}$
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