$A$ series $LCR$ circuit with $R = 20 \ \Omega$, $L = 1.6 \ \text{H}$ and $C = 40 \ \mu\text{F}$ is connected to a variable frequency a.c. source. The inductive reactance at resonant frequency is . . . . . . $\Omega$.

An inductor of inductance $2\,\mu\text{H}$ is connected in series with a resistance,a variable capacitor,and an $AC$ source of frequency $7\,\text{kHz}$. The value of capacitance for which maximum current is drawn into the circuit is $\frac{1}{x}\text{ F}$,where the value of $x$ is $.........$. (Take $\pi = \frac{22}{7}$)

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.

For a series $LCR$ circuit,the $I$ vs $\omega$ curve is shown. Consider the following statements:
$(A)$ To the left of $\omega_{r}$,the circuit is mainly capacitive.
$(B)$ To the left of $\omega_{r}$,the circuit is mainly inductive.
$(C)$ At $\omega_{r}$,the impedance of the circuit is equal to the resistance of the circuit.
$(D)$ At $\omega_{r}$,the impedance of the circuit is $0$.
Choose the most appropriate answer from the options given below:

In an $LCR$ series circuit,$C = 2 \mu F$,$L = 1 \ mH$,and $R = 10 \ \Omega$. When the current in the circuit is maximum,what is the ratio of the energy stored in the capacitor to the energy stored in the inductor?

In a series $LCR$ resonance circuit,if we change the resistance only,from a lower to higher value: