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 :

Is the current distribution shown in the figure possible?

Assertion : In the purely resistive element of a series $LCR$ $AC$ circuit,the maximum value of $rms$ current increases with an increase in the angular frequency of the applied $e.m.f$.
Reason : $I_{\max} = \frac{\varepsilon_{\max}}{Z}$,where $Z = \sqrt{R^2 + (\omega L - \frac{1}{\omega C})^2}$ and $I_{\max}$ is the peak current in a cycle.

In the circuit shown,$L = 1 \mu H$,$C = 1 \mu F$,and $R = 1 k\Omega$. They are connected in series with an $a.c.$ source $V = V_0 \sin \omega t$ as shown. Which of the following options is/are correct?
[$A$] The frequency at which the current will be in phase with the voltage is independent of $R$.
[$B$] At $\omega \sim 0$,the current flowing through the circuit becomes nearly zero.
[$C$] At $\omega \gg 10^6 \text{ rad } s^{-1}$,the circuit behaves like a capacitor.
[$D$] The current will be in phase with the voltage if $\omega = 10^6 \text{ rad } s^{-1}$.

$A$ series $LR$ circuit is connected to a voltage source with $V(t) = V_0 \sin \omega t$. After a very large time,how does the current $I(t)$ behave? (Given: $t_0 \gg \frac{L}{R}$)