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Obtain the resonant frequency and $Q$-factor of a series $L C R$ circuit with $L=3.0 \mathrm{H}, C=27 \mu \mathrm{F}$, and $R=7.4 \Omega$. It is desired to improve the sharpness of the resonance of the circuit by reducing its full width at half maximum by a factor of 2 . Suggest a suitable way.
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Verified Answer
Quality factor in the given resonant circuit.
$$
Q=\frac{1}{R} \sqrt{\frac{L}{C}}=\frac{1}{7.4} \sqrt{\frac{3}{27 \times 10^{-6}}}=45
$$
We want to improve the quality factor to twice, without changing resonant frequency (without changing $L$ and C).
$$
Q^{\prime}=2 Q=90=\frac{1}{R^{\prime}} \sqrt{\frac{L}{C}}
$$
or $R^{\prime}=\frac{1}{90} \sqrt{\frac{3}{27 \times 10^{-6}}}=3.7 \Omega$
$$
Q=\frac{1}{R} \sqrt{\frac{L}{C}}=\frac{1}{7.4} \sqrt{\frac{3}{27 \times 10^{-6}}}=45
$$
We want to improve the quality factor to twice, without changing resonant frequency (without changing $L$ and C).
$$
Q^{\prime}=2 Q=90=\frac{1}{R^{\prime}} \sqrt{\frac{L}{C}}
$$
or $R^{\prime}=\frac{1}{90} \sqrt{\frac{3}{27 \times 10^{-6}}}=3.7 \Omega$
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