In the $L-C-R$ circuit, shown in figure the $A C$ 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.
(a) Consider the R-L-C circuit shown in the adjacent diagram.
Given $$V=V_m \sin \omega t$$
Let current at any instant be $i$
Applying KVL in the given circuit
$$i R+L \frac{d i}{d t}+\frac{q}{C}-V_m \sin \omega t=0\quad\text{.... (i)}$$
$$\begin{gathered} \text{Now, we can write}\quad i=\frac{d q}{d t} \Rightarrow \frac{d i}{d t}=\frac{d^2 q}{d t^2} \\ \text{From Eq. (i)}\quad\frac{d q}{d t} R+L \frac{d^2 q}{d t^2}+\frac{q}{C}=V_m \sin \omega t \end{gathered}$$
$$\Rightarrow \quad L \frac{d^2 q}{d t^2}+R \frac{d q}{d t}+\frac{q}{C}=V_m \sin \omega t$$
This is the required equation of variation (motion) of charge.
(b) Let
$$\begin{aligned} q=q_m \sin (\omega t+\phi) & =-q_m \cos (\omega t+\phi) \\ i & =i_m \sin (\omega t+\phi)=q_m \omega \sin (\omega t+\phi) \\ i_m & =\frac{V_m}{Z}=\frac{V_m}{\sqrt{R^2+\left(X_C-X_L\right)^2}} \\ \phi & =\tan ^{-1}\left(\frac{X_C-X_L}{R}\right) \end{aligned}$$
When $R$ is short circuited at $t=t_0$, energy is stored in $L$ and $C$.
$$\begin{aligned} & U_L=\frac{1}{2} L i^2=\frac{1}{2} L\left[\frac{V_m}{\sqrt{\left(R^2+X_C-X_L\right)^2}}\right]^2 \sin ^2\left(\omega t_0+\phi\right) \\ & \text { and } \\ & U_C=\frac{1}{2} \times \frac{q^2}{C}=\frac{1}{2 C}\left[q^2 m \cos ^2\left(\omega t_0+\phi\right)\right] \\ & =\frac{1}{2 C}\left[\frac{V_m}{\sqrt{R^2+\left(X_C-X_L\right)^2}}\right]^2 \\ & =\frac{1}{2 C} \times\left(\frac{i_m}{\omega}\right)^2 \cos ^2\left(\omega t_0+\phi\right) \\ & =\frac{i^2 m}{2 C \omega^2} \cos ^2\left(\omega t_0+\phi\right) \\ & {\left[\because i_m=q_m \omega\right]} \\ & =\frac{1}{2 C}\left[\frac{V_m}{\sqrt{R^2+\left(X_C-X_L\right)^2}}\right]^2 \frac{\cos ^2\left(\omega t_0+\phi\right)}{\omega^2} \\ & =\frac{1}{2 C \omega^2}\left[\frac{V_m}{\sqrt{R^2+\left(X_C-X_L\right)^2}}\right]^2 \cos ^2\left(\omega t_0+\phi\right) \end{aligned}$$
(c) When $R$ is short circuited, the circuit becomes an L-C oscillator. The capacitor will go on discharging and all energy will go to $L$ and back and forth. Hence, there is oscillation of energy from electrostatic to magnetic and magnetic to electrostatic.