In the circuit shown below,what will be the readings of the voltmeter and ammeter?

Given below are two statements :
Statement $I$ : In an $LCR$ series circuit,current is maximum at resonance.
Statement $II$ : Current in a purely resistive circuit can never be less than that in a series $LCR$ circuit when connected to the same voltage source.
In the light of the above statements,choose the correct option from the options given below :

$A$ direct current of $4\,A$ and an alternating current of peak value $4\,A$ flow through resistances of $3\,\Omega$ and $2\,\Omega$ respectively. The ratio of heat produced in the two resistances in the same interval of time will be.

Given below are two statements $:$ one is labelled as Assertion $(A)$ and the other is labelled as Reason $(R).$
Assertion $(A) :$ Choke coil is simply a coil having a large inductance but a small resistance. Choke coils are used with fluorescent mercury-tube fittings. If household electric power is directly connected to a mercury tube,the tube will be damaged.
Reason $(R):$ By using the choke coil,the voltage across the tube is reduced by a factor $\left(R / \sqrt{R^2+\omega^2 L^2}\right)$,where $\omega$ is the angular frequency of the supply,$R$ is the resistance,and $L$ is the inductance. If the choke coil were not used,the voltage across the tube would be the same as the applied voltage. In the light of the above statements,choose the most appropriate answer from the options given below $:$

An $LCR$ circuit is equivalent to a damped pendulum. In an $LCR$ circuit,the capacitor is charged to $Q_0$ and then connected to the $L$ and $R$ as shown below. If a student plots graphs of the square of maximum charge $(Q_{Max}^2)$ on the capacitor with time $(t)$ for two different values $L_1$ and $L_2$ $(L_1 > L_2)$ of $L$,then which of the following represents this graph correctly? (Plots are schematic and not drawn to scale.)

In an $LCR$ series circuit,when $L$ is removed from the circuit,the phase difference between voltage and current is $\frac{\pi}{3}$. If $C$ is removed from the circuit instead of $L$,the phase difference is again $\frac{\pi}{3}$. The power factor of the circuit is $(\tan 60^{\circ}=\sqrt{3})$.