$A$ resistor of $R=6 \Omega$,an inductor of $L=1 \text{ H}$,and a capacitor of $C=17.36 \mu \text{F}$ are connected in series with an $AC$ source. Find the $Q$ factor.

In a series $LR$ circuit,$X_{L} = R$ and the power factor of the circuit is $P_{1}$. When a capacitor with capacitance $C$ such that $X_{L} = X_{C}$ is put in series,the power factor becomes $P_{2}$. The ratio $\frac{P_{1}}{P_{2}}$ is

$A$ series $LCR$ circuit is shown in the figure. Where the inductance of $10 \ H$,capacitance $40 \ \mu F$,and resistance $60 \ \Omega$ are connected to a variable frequency $240 \ V$ source. The current at resonating frequency is (in $A$)

$A$ series $LCR$ circuit with $L=0.12\, H$,$C=480\, nF$,$R=23\, \Omega$ is connected to a $230\, V$ variable frequency supply.
$(a)$ What is the source frequency for which current amplitude is maximum? Obtain this maximum value.
$(b)$ What is the source frequency for which average power absorbed by the circuit is maximum? Obtain the value of this maximum power.
$(c)$ For which frequencies of the source is the power transferred to the circuit half the power at resonant frequency? What is the current amplitude at these frequencies?
$(d)$ What is the $Q$-factor of the given circuit?

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.

Given below are two statements:
Statement $I$: When the frequency of an $a.c.$ source in a series $LCR$ circuit increases,the current in the circuit first increases,attains a maximum value,and then decreases.
Statement $II$: In a series $LCR$ circuit,the value of the power factor at resonance is one.
In the light of the given statements,choose the most appropriate answer from the options given below: