How does the sign of the phase angle $\phi$,by which the supply voltage leads the current in an $LCR$ series circuit,change as the supply frequency is gradually increased from very low to very high values?

In the circuit shown below, the ac source has voltage $V = 30 \cos(\omega t) \text{ V}$ with $\omega = 2000 \text{ rad/s}$. What will be the amplitude of the current?

An inductance of $\left(\frac{200}{\pi}\right) \text{mH}$,a capacitance of $\left(\frac{10^{-3}}{\pi}\right) \text{F}$,and a resistance of $10 \, \Omega$ are connected in series with an $AC$ source of $220 \, \text{V}, 50 \, \text{Hz}$. The phase angle of the circuit is:

In the $LCR$ circuit shown in the figure,what will be the readings of the voltmeter across the resistor $(V_R)$ and the ammeter $(A)$ if an a.c. source of $220 \ V$ and $100 \ Hz$ is connected to it?

In the following circuit,the emf of the source is $E_0 = 200 \, V$,$R = 20 \, \Omega$,$L = 0.1 \, H$,and $C = 10.6 \, F$. If the frequency is variable,then the current at frequency $f = 0$ and $f = \infty$ is:

In the circuit shown,$C = \frac{\sqrt{3}}{2} \times 10^{-3} \, F$,$R_2 = 20 \, \Omega$,$L = \frac{\sqrt{3}}{10} \, H$,and $R_1 = 10 \, \Omega$. The current in the $L-R_1$ branch is $I_1$ and in the $C-R_2$ branch is $I_2$. The voltage of the $A.C.$ source is given by $V = 200\sqrt{2} \sin(100t) \, V$. The phase difference between $I_1$ and $I_2$ is: