In the circuit shown,the readings of voltmeters $V_1, V_2$ and $V_3$ at resonance are given by

$A$ series $LCR$ circuit is connected to an ac voltage source. When $L$ is removed from the circuit,the phase difference between current and voltage is $\frac{\pi}{3}$. If instead $C$ is removed from the circuit,the phase difference is again $\frac{\pi}{3}$ between current and voltage. The power factor of the circuit is:

An inductor of inductance $2 \text{ } \mu\text{H}$ is connected in series with a resistance, a variable capacitor and an a.c. source of $10 \text{ kHz}$. The value of capacitance for which maximum current is drawn in the circuit is $\frac{1}{x} \text{ F}$, where the value of $x$ is (Take $\pi^2 = 10$).

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.

In the figure shown,three $AC$ voltmeters are connected. The circuit is at resonance.

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: