The $i -
u$ curve for an anti-resonant circuit is:

What is the reason we prefer $A.C.$ voltage over $D.C.$ voltage for power transmission?

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 $\omega = 10^6 \text{ rad } s^{-1}$.

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 :

You are given many resistances,capacitors and inductors. These are connected to a variable $DC$ voltage source (the first two circuits) or an $AC$ voltage source of $50 \ Hz$ frequency (the next three circuits) in different ways as shown in Column $II$. When a current $I$ (steady state for $DC$ or rms for $AC$) flows through the circuit,the corresponding voltage $V_1$ and $V_2$ (indicated in circuits) are related as shown in Column $I$. Match the two.

In the shown $AC$ circuit,the phase difference between currents $I_1$ and $I_2$ is: