(i) The town is 10 km away, length of pair of Cu wires used, $L=20 \mathrm{~km}=20000 \mathrm{~m}$.

Resistance of Cu wires,

$$\begin{aligned} R & =\frac{l}{A}=\frac{l}{\pi(r)^2} \\ & =\frac{1.7 \times 10^{-8} \times 20000}{3.14\left(0.5 \times 10^{-2}\right)^2}=4 \Omega \end{aligned}$$

$$\begin{aligned} I \text { at } 220 \mathrm{~V} \quad V I & =10^6 \mathrm{~W} ; I=\frac{10^6}{220}=0.45 \times 10^4 \mathrm{~A} \\ R I^2 & =\text { power loss } \\ & =4 \times(0.45)^2 \times 10^8 \mathrm{~W} \\ & >10^6 \mathrm{~W} \end{aligned}$$

Therefore, this method cannot be used for transmission.

(ii) When power $P=10^6 \mathrm{~W}$ is transmitted at 11000 V .

$$\begin{aligned} V^{\prime} I^{\prime} & =10^6 \mathrm{~W}=11000 I^{\prime} \\ \text { Current drawn, } I^{\prime} & =\frac{1}{1.1} \times 10^2 \\ \text { Power loss } & =R I^2=\frac{1}{1.21} \times 4 \times 10^4 \\ & =3.3 \times 10^4 \mathrm{~W} \\ \text { Fraction of power loss } & =\frac{3.3 \times 10^4}{10^6}=3.3 \% \end{aligned}$$

In the given figure $i$ is the total current from the source. It is divided into two parts $i_1$ through $R$ and $i_2$ through series combination of $C$ and $L$.

So, we can write $i=i_1+i_2$

$$\begin{aligned} \text{As,}\quad V_m \sin \omega t & =R i_1 \quad\text{[from the circuit diagram] }\\ \Rightarrow\quad i_1 & =\frac{V_m \sin \omega t}{R}\quad\text{... (i)} \end{aligned}$$

If $q_2$ is charge on the capacitor at any time $t$, then for series combination of $C$ and $L$. Applying KVL in the Lower circuit as shown,

$$\begin{aligned} &\begin{aligned} & \frac{q_2}{C}+\frac{L d i_2}{d t}-V_m \sin \omega t=0 \\ \Rightarrow\quad & \frac{q_2}{C}+\frac{L d_2 q_2}{d t^2}=V_m \sin \omega t \quad\left[\because i_2=\frac{d q_2}{d t}\right] \ldots (ii) \end{aligned}\\ &\begin{aligned} \text{Let}\quad q_2 & =q_m \sin (\omega t+\phi) \quad\text{... (iii)}\\ \therefore\quad\frac{d q_2}{d t} & =q_m \omega \cos (\omega t+\phi) \\ \Rightarrow\quad\frac{d^2 q_2}{d t^2} & =-q_m \omega^2 \sin (\omega t+\phi) \end{aligned} \end{aligned}$$

Now putting these values in Eq. (ii), we get

$$q_m\left[\frac{1}{C}+L\left(-\omega^2\right)\right] \sin (\omega t+\phi)=V_m \sin \omega t$$

If $\phi=0$ and $\left(\frac{1}{C}-L \omega^2\right)>0$, then

$$q_m=\frac{V_m}{\left(\frac{1}{C}-L \omega^2\right)}\quad\text{.... (iv)}$$

$$\begin{aligned} \text{From Eq. (iii), }\quad i_2 & =\frac{d q_2}{d t}=\omega q_m \cos (\omega t+\phi) \\ \text{using Eq. (iv),}\quad i_2 & =\frac{\omega V_m \cos (\omega t+\phi)}{\frac{1}{C}-L \omega^2} \\ \text { Taking } \phi & =0 ; i_2=\frac{V_m \cos (\omega t)}{\left(\frac{1}{\omega C}-L \omega\right)}\quad\text{.... (v)} \end{aligned}$$

From Eqs. (i) and (v), we find that $i_1$ and $i_2$ are out of phase by $\frac{\pi}{2}$.

Now, $$i_1+i_2=\frac{V_m \sin \omega t}{R}+\frac{V_m \cos \omega t}{\left(\frac{1}{\omega C}-L \omega\right)}$$

Put $\quad \frac{V_m}{R}=A=C \cos \phi$ and $\frac{V_m}{\left(\frac{1}{\omega C}-L \omega\right)}=B=C \sin \phi$

$\therefore \quad i_1+i_2=C \cos \phi \sin \omega t+C \sin \phi \cos \omega t$ $=C \sin (\omega t+\phi)$

where $C=\sqrt{A^2+B^2}$

and $$\phi=\tan ^{-1} \frac{B}{A} C=\left[\frac{V_m^2}{R^2}+\frac{V_m^2}{\left(\frac{1}{\omega C}-L \omega\right)}\right]^{1 / 2}$$

and $$\phi=\tan ^{-1} \frac{R}{\left(\frac{1}{\omega C}-L \omega\right)}$$

Hence, $$i=i_1+i_2=\left[\frac{V_m^2}{R^2}+\frac{V_m^2}{\left(\frac{1}{\omega C}-L \omega\right)^2}\right]^{1 / 2} \sin (\omega t+\phi)$$

or $$\frac{i}{V_m}=\frac{1}{Z}=\left[\frac{1}{R^2}+\frac{1}{\left(\frac{1}{\omega C}-L \omega\right)^2}\right]^{1 / 2}$$

This is the expression for impedance $Z$ of the circuit.

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.