Consider the L-C-R circuit. Applying KVL for the loop, we can write

$$\begin{aligned} &\Rightarrow \quad L \frac{d i}{d t}+\frac{q}{C}+i R=V_m \sin \omega t\quad\text{.... (i)}\\ &\text { Multiplying both sides by } i \text {, we get }\\ &L i \frac{d i}{d t}+\frac{q}{C} i+i^2 R=\left(V_m i\right) \sin \omega t=V i\quad\text{..... (ii)} \end{aligned}$$

where

$$\begin{aligned} & L i \frac{d i}{d t}=\frac{d}{d t}\left(\frac{1}{2} L i^2\right)=\text { rate of change of energy stored in an inductor. } \\ & R i^2=\text { joule heating loss } \\ & \frac{q}{C} i=\frac{d}{d t}\left(\frac{q^2}{2 C}\right)=\text { rate of change of energy stored in the capacitor. } \end{aligned}$$

$V i=$ rate at which driving force pours in energy. It goes into (i) ohmic loss and (ii) increase of stored energy.

Hence Eq. (ii) is in the form of conservation of energy statement. Integrating both sides of Eq. (ii) with respect to time over one full cycle $(0 \rightarrow T)$ we may write

$$\begin{aligned} &\begin{aligned} \int_0^T \frac{d}{d t}\left(\frac{1}{2} L i^2+\frac{q^2}{2 C}\right) d t+\int_0^T R i^2 d t & =\int_0^T V i d t \\ \Rightarrow\quad 0+(+v e) & =\int_0^T V i d t \end{aligned}\\ &\Rightarrow \int_0^T V i d t>0 \text { if phase difference between } V \text { and } i \text { is a constant and acute angle. } \end{aligned}$$

(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.