Question

Consider the LCR circuit shown in figure.
Find the net current $i$ and the phase of $i$.
Show that $i=\frac{W}{z}.$ Find the impendance $Z$ for this circuit.

Answer 1

Given,

Current in resistor branch (R)
$i_1=\frac{v_{m}sin\omega t}{R}$
Let charge flowing through capacitor and inductor at any time $\left(t\right),$

$q_2=q_{m}sin(\omega t+\varphi$)

Applying loop rule in the circuit containing capacitor and inductor,

$V_C+V_L=v_{m}sin\omega t$

$\frac{q_2}{C}+L\frac{d{i}_2}{dt}=v_{m}sin\omega t$

$\frac{q_2}{c}+L\frac{d{q}_2^2}{d{t}^2}=v_{m}sin\omega t...\left(i\right)$

Putting value of $q_2$ in $\left(i\right)$
$q_m(\frac{1}{C}-L{\omega }^2)sin(\omega t+\varphi )=v_{m}sin\omega t$

$\text{If}\varphi =0$

$q_m=\frac{v_m}{\frac{1}{C}=L{\omega }^2}$
$\Rightarrow \frac{1}{C}-{\omega }^{2}L>0$

Current $i_1$ and $i_2$ will be,
$i_1=\frac{v_{m}sin\omega t}{R}$
$i_2=\frac{d{q}_2}{dt}=\omega q_m\cos (\omega t+\varphi )$

Thus
$i_1\text{and}{i}_2$ will be out of phase at $\varphi =0$
Let us assume $\frac{1}{C}-{\omega }^{2}L>0$
$i_2=\omega q_{m}com\left(\omega t\right)=\omega ,\frac{v_m}{(L\omega -\frac{1}{C\omega })}\cos \omega t$

Then,
$i_1+i_2=\frac{v_m}{R}sin\omega t+\frac{v_m}{L\omega -\frac{1}{C\omega }}cos\omega t$

Let $A\cos \varphi =\frac{v_m}{R}\text{and}Asin\varphi =\frac{v_m}{L\omega -\frac{1}{C\omega }}$

$A\cos \varphi sin\omega t+Asin\varphi cos\omega t$
$=Asin(\omega t+\varphi )$

$\therefore \sqrt{(Acos\varphi {)}^2+(Asin\varphi {)}^2}=A$

Therefore,
$i_1+i_2[\frac{v_m^2}{R^2}+\frac{v{m}^2}{{[\omega L-\frac{1}{\omega C}]}^2}]sin(\omega t+\varphi )$

$\varphi ={\tan }^{-1}\left(\frac{R}{X_L-X_C}\right)$
Impendance,
$Z=\frac{V}{i}=\frac{\left(v_{m}sin\omega t\right)}{{[\frac{v{m}^2}{R^2}+\frac{v{m}^2}{{[\omega L-\frac{1}{\omega L}]}^2}]}^{\frac{1}{2}}\sin (\omega t+\varphi )}$

$\frac{1}{Z}={\{\frac{1}{R^2}+\frac{1}{(L\omega -1/\omega C{)}^2}\}}^{\frac{1}{2}}$

Final Answer:
$\frac{1}{Z}={\{\frac{1}{R^2}+\frac{1}{(L\omega -1/\omega C{)}^2}\}}^{\frac{1}{2}}$