Question

PLEASE SOLVE THE QUESTIONS...
VERY URGENT

Answer 1

We select the left side AC circuit.
We write the potential difference across each network element and sum them.

$For\ capacitor\ V_C=\frac{q}{C},\ \ For\ Inductor\ induced\ voltage=L\ \frac{di}{dt},\ So\\\\L\ \frac{di}{dt}+i\ R+\frac{q}{C}=v_{in},\ \ \ v_{out}=R\ i\\\\As\ i=\frac{dq}{dt},\ \frac{di}{dt}=\frac{d^2q}{dt^2}\\\\L\ \frac{d^2q}{dt^2}+R\ \frac{dq}{dt}+\frac{q}{C}=v_{in}=v_mSin(\omega\ t),\ \ Let\ us\ say$

$Let\ q=q_mSin(\omega\ t+\theta),\ \ then\ \ \frac{dq}{dt}=q_m\omega\ Cos(\omega\ t+\theta)\\\\\frac{d^2q}{dt^2}=-q_m\omega^2Sin(\omega\ t+\theta)\\\\ Lets\ say\ X_L=impedance\ of\ Inductor=\omega\ L\\and\ X_C=impedance\ of\ capacitance=\frac{1}{\omega\ C}\\\\Hence\ v_m\ Sin(\omega\ t)=\\.\ \ \ \ q_m\ \omega(X_C-X_L)Sin(\omega\ t+\theta)+q_m\ R\ \omega\ Cos(\omega\ t+\theta)\\\\=q_m\ \omega\ [ (X_C-X_L)Sin(\omega\ t+\theta)+R\ Cos(\omega\ t+\theta) ]\\\\Let\ Cos\phi=\frac{R}{Z},\ Sin\phi=\frac{X_C-X_L}{Z}$

$and,\ \ Z=\sqrt{R^2+(X_L-X_C)^2},\ \ then\ finally\\\\v_m Sin(\omega\ t)=q_m\omega Z\ Cos(\omega\ t+\theta-\phi)\\\\\. \ \ \ =>v_m=q_m\omega Z,\ \ \ \phi-\theta=\pi/2$

$i = v_m \omega\ Cos(\omega\ t+\theta)=v_m\ \omega\ Sin(\omega\ t+\phi)\\\\i_m=\frac{v_m}{Z}=\frac{v_m}{\sqrt{R^2+(X_L-X_C)^2}}=\frac{v_m}{\sqrt{R^2+(\omega L-\frac{1}{\omega C})^2}}\\\\v_{out}=i\ R=\frac{R\ v_m}{\sqrt{R^2+(\omega L-\frac{1}{\omega C})^2}}Sin(\omega\ t+\phi),\\\\tan\phi=\frac{X_C-X_L}{R}=\frac{1-\omega^2LC}{\omega RC}$

Z = impedance in the circuit is minimum when
$X_L=X_C,\ \ \omega_o L=\frac{1}{\omega_o C},\ \ \omega_o=\frac{1}{\sqrt{LC}}$

Resonance occurs at this frequency ω₀.  Then the impedance Z = R and the output voltage 

$|v_{out} | = |v_m|, \ \ Power\ thru\ Load\ R\ is\ Maximum.$

For other values of ω, Z increases and so v_out decreases.  Hence the circuit acts as a band pass filter.  

The band width can be obtained by finding the ω₁ and ω₂, when the power is 1/2 the maximum and when v_out = v_m /√2 .

$So,\ \sqrt{R^2+(X_L-X_C)^2}=\sqrt{2R^2}\\\\X_L-X_C=R\\\\\omega\ L-\frac{1}{\omega C}=R\\\\\omega^2LC-\omega\ RC-1=0\\\\Bandwidth=\omega_1-\omega_2=\Delta\omega=\frac{\sqrt{R^2C^2+4LC}}{LC}=\frac{R}{L}\sqrt{1+\frac{4L}{R^2C}}\\\\we\ choose\ \frac{\Delta\omega}{\omega_0}<<1$

Then it works as a good band pass filter ,