Why is it called a rejector circuit?

 

The readings of ammeter and voltmeter in the following circuit are respectively

The amplitude of the current in series LCR circuit connected to an AC frequency $"\omega "$is given by,

${\mathrm{i}}_{\mathrm{m}}=\frac{{\mathrm{v}}_{\mathrm{m}}}{\sqrt{{\mathrm{R}}^2+{\left({\mathrm{X}}_{\mathrm{L}}-{\mathrm{X}}_{\mathrm{C}}\right)}^2}}$.

Where ${\mathrm{X}}_{\mathrm{L}} \mathrm{and} {\mathrm{X}}_{\mathrm{C}}$ are inductive and capacitive reactants respectively and $"{\mathrm{v}}_{\mathrm{m}}"$ is the amplitude of the voltage. Starting from this equation show that the sharpness of resonance the circuit is equal to the quality factorof the circuit.

In the given circuit, the resonant frequency is

In the above circuit, $C=\frac{\sqrt{3}}{2}\mu F,R_2=20\mathrm{ \Omega },L=\frac{\sqrt{3}}{10}H$ and $R_1=10\mathrm{ \Omega }.$ Current in $L\text{-}{R}_1$ path is $I_1$ and in $C-R_2$ path it is $I_2.$ The voltage of $$AC source is given by, $V=200\sqrt{2}\sin \left(100t\right)$ volts. The phase difference between $I_1$ and $I_2$ is:

A series LCR circuit contains a pure inductor of inductance $5.0 \mathrm{H}$, a capacitor of capacitance $20 \mu \mathrm{F}$ and a resistor of $40 \mathrm{\Omega }$. Calculate the Quality factor (Q-factor) of the circuit. 

A parallel plate capacitor in series with a resistance of $100 \mathrm{\Omega },$ an inductor of $20 \mathrm{mH}$ and an $AC$ voltage source of variable frequency shows resonance at a frequency of $\frac{1250}{\pi }\mathrm{Hz}$. If this capacitor is charged by a $DC$ voltage source to a voltage $25 \mathrm{V},$ what amount of charge will be stored in each plate of the capacitor?

Find the peak current and resonant frequency of the following circuit (as shown in figure).