Question:29
Consider the LCR circuit shown in Fig 7.6. Find the net current i and the phase of i. Show that $i=\frac{v}{Z}$. Find the impedance Z for this circuit.
Answer:
Taking Resistor andAlternating EMF circuit, $i=i_{1}+i_{2}$
$i_{2}R=V_{m}\sin \omega t\Rightarrow i_{2}=\frac{V_{m}\sin \omega t}{R}$
Taking Capacitor-Inductor and Alternating EMF circuit,
$V_{m}\sin \omega t=\frac{q_{1}}{C}+L\frac{d^{2}q}{dt^{2}}$
Assuming $q_{1}=q_{m}\; \sin (\omega t+\phi )$
$\frac{dq_{1}}{dt}=\omega q_{m}\cos (\omega t+\phi )$
$\frac{d^{2}q_{1}}{dt^{2}}=-\omega ^{2}q_{m}\sin (\omega +\phi )$
$V_{m}\sin \omega t=\frac{q_{m}\sin (\omega t+\phi )}{C}+L(-\omega ^{2}q_{m}\; \sin (\omega t+\phi ))$
$V_{m}\sin \omega t=q_{m}\left ( \frac{1}{C}-\omega ^{2}L \right )\sin (\omega t+\phi )$
$q_{m}=\frac{V_{m}}{\frac{1}{C}-\omega ^{2}L}$
$\sin \omega t=\sin (\omega t+\phi )$
$\phi =0$
$i_{1}=\frac{dq_{1}}{dt}=\omega \left ( \frac{V_{m}}{\frac{1}{C}-\omega ^{2}L} \right )\cos (\omega t)$
$i=i_{1}+i_{2}=\left ( \frac{V_{m}}{\frac{1}{\omega C}-\omega L} \right )\cos (\omega t)+\frac{V_{m}(\sin \omega t)}{R}$
$=A\; \sin (\omega t+\phi )$
$A=\sqrt{\left ( \left ( \frac{V_{m}}{\left ( \frac{1}{\omega C}-\omega L \right )} \right )^{2}+\left ( \frac{V_{m}}{R} \right )^{2} \right )}=V_{m}\sqrt{\left ( \left ( \frac{V_{m}}{\left ( \frac{1}{\omega C}-\omega L \right )} \right )^{2}+\left ( \frac{1}{R} \right )^{2} \right )}$
$\tan \phi =\frac{R}{\frac{1}{\omega C}-\omega L}$
$\phi =\tan ^{-1}\left ( \frac{R}{\frac{1}{\omega C}-\omega L} \right )$
