Obtain the resonant frequency and $Q$-factor of a series $LCR$ circuit with $L=3.0\; H$,$C=27\; \mu F$,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.

(A) Inductance,$L = 3.0\; H$
Capacitance,$C = 27\; \mu F = 27 \times 10^{-6}\; F$
Resistance,$R = 7.4\; \Omega$
At resonance,the angular frequency $\omega_r$ is given by $\omega_r = \frac{1}{\sqrt{LC}}$.
$\omega_r = \frac{1}{\sqrt{3.0 \times 27 \times 10^{-6}}} = \frac{1}{\sqrt{81 \times 10^{-6}}} = \frac{1}{9 \times 10^{-3}} = 111.11\; rad/s$.
The $Q$-factor is given by $Q = \frac{\omega_r L}{R}$.
$Q = \frac{111.11 \times 3.0}{7.4} \approx 45.04$.
The 'full width at half maximum' (bandwidth) is given by $\Delta \omega = \frac{R}{L}$.
To reduce the bandwidth by a factor of $2$ while keeping $\omega_r$ constant,we must reduce the resistance $R$ by a factor of $2$.
New resistance $R' = \frac{R}{2} = \frac{7.4}{2} = 3.7\; \Omega$.