Question

In the $LCR$ circuit shown in figure, the AC driving voltage is $v=v_{m}sin\omega t.$


A) Write down the equation of motion for q(t).

B) At $t=t_0,$ the voltage source stops, and $R$ is short circuited. Now write down how much energy is stored in each of $L$ and $C$.

C) Describe subsequent motion of charges.

Answer 1

A)
From Kirchhoff's voltage rule:
$iR+L\frac{di}{dt}+\frac{q}{c}-V_{m}sin\omega t=\mathrm{0...}\left(i\right)$

Since, $i=\frac{dq}{dt}$

$\Rightarrow \frac{dt}{dt}=\frac{d^{2}q}{d{t}^2}$

Substituting in equation (i),

$\frac{dq}{dt}R+L\frac{d^{2}q}{d{t}^2}+\frac{q}{C}=V_{m}sin\omega t$

Rearranging

$L\frac{d^{2}q}{d{t}^2}+R\frac{dq}{dt}+\frac{q}{C}=v_{m}sin\omega t$

Final Answer:
$L\frac{d^{2}q}{d{t}^2}+R\frac{dq}{dt}+\frac{q}{C}=v_{m}sin\omega t$


B) Energy stored in inductor

$U_L=\frac{1}{2}L{i}^2$

$U_L=\frac{1}{2}L{\left[\frac{V_m}{\sqrt{(R^2+(X_C-X_L{)}^2)}}\right]}^{2}si{n}^2(\omega t_0+\varphi )$

Energy stored in capacitor

$U_C=\frac{q^2}{2C}$

Putting value of $q$

$U_C=\frac{1}{2C}\left[q{m}^{2}co{s}^2\right(\omega t_0+\varphi \left)\right]$

$U_C=\frac{1}{2C}\times {\left(\frac{i_m}{\omega }\right)}^{2}co{s}^2(\omega t_0+\varphi )$

$U_C=\frac{1}{2C{\omega }^2}{\left[\frac{V_m}{\sqrt{R^2+(X_C-X_L{)}^2}}\right]}^{2}co{s}^2(\omega t_0+\varphi )$

Final Answer:

$U_L=\frac{1}{2}L{i}^2$

$=\frac{1}{2}L{\left[\frac{V_m}{\sqrt{(R^2+(X_C-X_L{)}^2}}\right]}^{2}si{n}^2(\omega t_0+\varphi )$

$U_C=\frac{q^2}{2C}$

$=\frac{1}{2C{\omega }^2}{\left[\frac{V_m}{\sqrt{(R^2+(X_C-X_L{)}^2}}\right]}^{2}si{n}^2(\omega t_0+\varphi )$


C) When $R$ is short circuited it becomes $L-C$ circuit. Now, capacitor is fully charged, so it will transfer energy to oscillator, so the subsequent motion of charges happen between capacitor and inductor and no energy gets wasted.

In $LC$ oscillations the energy in the system oscillates between the capacitor and the inductor.

Final Answer: Capacitor will go on discharging and all energy will go to L and back and forth.