Question

Consider the $LCR$ circuit shown in Fig. Find the net current $i$ and the phase of $i$. Show that $i=\frac{V}{Z}$. Find the impedance $Z$ for this circuit.

Answer 1

Total current $i$ from the source $V_m\sin \omega t$ is divided at $B$ in two part, $i_1$ through capacitor and inductor and part $i_2$ through resistance.
Potential across $R=$ Potential of source
$P.D.$ across $R=V_m\sin \omega t$
$i_{2}R=V_m\sin \omega t$
$i_2=\frac{V_m\sin \omega t}{R}....\left(I\right)$
$q_1$ is charge on the capacitor at any time $t$, then for series combination of $C,L$ applying Kirchhoff's voltage law in loop $ABEFA$.
$V_C+V_L=V_M\sin \omega t$
$\frac{q_1}{C}+L\frac{d{i}_1}{dt}=V_m\sin \omega t$
$\frac{q_1}{C}+L\frac{d^2{q}_1}{d{t}^2}=V_m\sin \omega t....II$
Let $q_1=q_m\sin (\omega t+\varphi )....III$
$i_1=\frac{d{q}_1}{dt}=q_m\omega \cos (\omega t+\varphi )...IV$
$\frac{d^2{q}_1}{d{t}^2}=-q_m{\omega }^2\sin (\omega t+\varphi )...V$
Substitute the values of equations (III) and (V) in equation (II)
$\frac{q_m\sin (\omega t+\varphi )}{C}-L{q}_m{\omega }^2\sin (\omega t+\varphi )=V_m\sin \omega t$
$q_m\sin (\omega t+\varphi )[\frac{1}{C}-L{\omega }^2]=V_m\sin \omega t$
at $\varphi =0$,
$q_m\sin (\omega t+\varphi )[\frac{1}{C}-L{\omega }^2]=V_m\sin \omega t$
$q_m[\frac{1}{C}-L{\omega }^2]\sin \omega t=V_m\sin \omega t$
$q_m[\frac{1}{C}-L{\omega }^2]=V_m$
$q_m=\frac{V_m}{\omega \left[\frac{1}{C\omega -L\omega }\right]}....\left(VI\right)$
Applying Kirchhoff's junction rule as junction $B,i=i_1+i_1$ using relation $I,IV$
$i=\frac{V_m\sin \omega t}{R}+q_m\omega \cos (\omega t+\varphi )$
Now using relation VI for $q_m$ and at $\varphi =0$
$i=[\frac{V_m\sin \omega t}{R}+\frac{V_m\omega \cos \omega t}{\omega [\frac{1}{\omega C}-\omega L]}]$
$i=\frac{V_m}{R}\sin \omega t+\frac{V_m}{(\frac{1}{\omega C}-\omega L)}\cos \omega t$
Let
$A=\frac{V_m}{r}=C\cos \varphi ....\left(VII\right)$
$B=\frac{V_m}{\frac{1}{\omega C}-\omega L}=C\cos \varphi ....\left(VIII\right)$
$i=C\cos \varphi \sin \omega t+C\sin \varphi .\cos \omega t$
$=C[\cos \varphi \sin \omega t+\sin \varphi \cos \omega t]$
$i=C\sin (\omega t+\varphi )$
Squaring and adding (VII), (VIII)
$A^2+B^2=C^2{\cos }^2\varphi +C^2{\sin }^2\varphi$
$=C^2[{\cos }^2\varphi +{\sin }^2\varphi )$
$A^2+B^2=C^2$
or $C=\sqrt{A^2+B^2}$
$\varphi ={\tan }^{-1}\frac{B}{A}={\tan }^{-1}\frac{\frac{V_m}{\frac{1}{\omega C}-\omega L}}{\frac{V_m}{R}}$
$\therefore \tan \varphi =\frac{R}{(\frac{1}{\omega C}-\omega L)}$
$\because C^2=A^2+B^2=\frac{V_m^2}{R^2}+\frac{V_m^2}{{(\frac{1}{\omega C}-\omega L)}^2}$
$C={[\frac{V_m^2}{R^2}+\frac{V_m^2}{{(\frac{1}{\omega C}-\omega L)}^2}]}^{\frac{1}{2}}$
$\therefore i={[\frac{V_m^2}{R^2}+\frac{V_m^2}{{(\frac{1}{\omega C}-\omega L)}^2}]}_2\sin (\omega t+\varphi )$
$i=V_m{[\frac{1}{R^2}+\frac{1}{{(\frac{1}{\omega C}-\omega L)}^2}]}^{\frac{1}{2}}\sin (\omega t+\varphi )....\left(IX\right)$
And $\varphi ={\tan }^{-1}\frac{R}{(\frac{1}{\omega C}-\omega L)}$
$\because I=\frac{V}{R}$ or $i=\frac{V}{Z}$
For ac $i=\frac{V}{Z}\sin (\omega t+\varphi )....\left(X\right)$
Comparing (IX) and (X)
So, $\frac{1}{Z}={[\frac{1}{R^2}+\frac{1}{{(\frac{1}{\omega C}-\omega L)}^2}]}^{\frac{1}{2}}$
This is the impedance $Z$ for the circuit.