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?

$A$ capacitance of $\left(\frac{10^{-3}}{2 \pi}\right) F$,an inductance of $\left(\frac{100}{\pi}\right) mH$,and a resistance of $10 \Omega$ are connected in series with an $AC$ voltage source of $220 V, 50 Hz$. The phase angle of the circuit is (in $^{\circ}$)

In a series $LCR$ circuit,if the current leads the source voltage,then

In an $LCR$ series $ac$ circuit,the voltage across each of the components,$L, C$ and $R$ is $50\,V$. The voltage across the $LC$ combination will be........$V$.

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:

$A$ resistor of resistance $30 \Omega$, an inductor of reactance $10 \Omega$, and a capacitor of reactance $10 \Omega$ are connected in series to an $AC$ voltage source $V = 300 \sqrt{2} \sin(\omega t)$. The current in the circuit is . . . . . . (in $\text{ A}$)