(i) The town is 10 km away, length of pair of Cu wires
used, I = 20 km = 20000 m.
Resistance of Cu wires,
\(R=\rho \frac{l}{A}=\rho \frac{l}{\pi(r)^{2}}=\frac{1.7 \times 10^{-8} \times 20000}{3.14\left(0.5 \times 10^{-2}\right)^{2}} \approx 4 \Omega\)
\(I \text { at } 220 \mathrm{~V}, V I=10^{6} \mathrm{~W} ; I=\frac{10^{6}}{220}=0.45 \times 10^{4} \mathrm{~A}\)
RI² = power loss = 4 x (0.45)² x 10⁸> 10⁶ W
Therefore, this method cannot be used for transmission.
(ii) When power, P = 10⁶ W is transmitted at 11000 V.
V \('I'\)= 10⁶ W = 11000 \(I'\)
Current drawn, \(I^{\prime}=\frac{1}{1.1} \times 10^{2}\)
Power loss \(=R I^{2}=\frac{1}{121} \times 4 \times 10^{6}=3.3 \times 10^{4} \mathrm{~W}\)
\(\therefore\)  Fraction of power loss \(=\frac{3.3 \times 10^{4}}{10^{6}}=3.3 \%\)

(i) Consider the R-L-C circuit as shown in the figure.
Given, V = V_msin \(\omega\)t
Let current at any instant be i.

Note We have to apply KVL, write the equations in the form of current and charge, double differentiate the equation with respect to time and find the required relations.
Applying KVL in the given circuit,
\(i R+L \frac{d i}{d t}+\frac{q}{C}-V_{m} \sin \omega t=0\)    ..........(i)
Now, we can write,  \(i=\frac{d q}{d t} \Rightarrow \frac{d i}{d t}=\frac{d^{2} q}{d t^{2}}\)
From Eq. (i), we get
\(\frac{d q}{d t} R+L \frac{d^{2} q}{d t^{2}}+\frac{q}{C}=V_{m} \sin \omega t\)
\(\Rightarrow \quad L \frac{d^{2} q}{d t^{2}}+R \frac{d q}{d t}+\frac{q}{C}=V_{m} \sin \omega t\)
This is-the required equation of variation motion of charge.
(ii) Let \(q=q_{m} \sin (\omega t+\phi)=-q_{m} \cos (\omega t+\phi)\)
\(i=i_{m} \sin (\omega t+\phi)=q_{m} \omega \sin (\omega t+\phi)\)
\(i_{m}=\frac{V_{m}}{Z}=\frac{V_{m}}{\sqrt{R^{2}+\left(X_{C}-X_{L}\right)^{2}}}\)
and \(\phi=\tan ^{-1}\left(\frac{X_{C}-X_{L}}{R}\right)\)
When R is short circuited at t = t_o energy is stored in Land C.
\(U_{L}=\frac{1}{2} L i^{2}=\frac{1}{2} L\left[\frac{V_{m}}{\sqrt{R^{2}+\left(X_{C}-X_{L}\right)^{2}}}\right]^{2} \sin ^{2}(\omega t+\phi)\)
\(\text { and } U_{C}=\frac{1}{2} \times \frac{q^{2}}{C}=\frac{1}{2 C} \times\left(\frac{i_{m}}{\omega}\right)^{2} \cos ^{2}\left(\omega t_{0}+\phi\right)\)
\(=\frac{i_{m}^{2}}{2 C \omega^{2}} \cos ^{2}\left(\omega t_{0}+\phi\right) \quad\left[\because i_{m}=q_{m} \omega\right]\)
\(=\frac{1}{2 C}\left[\frac{V_{m}}{\sqrt{R^{2}+\left(X_{C}-X_{L}\right)^{2}}}\right]^{2} \frac{\cos ^{2}\left(\omega t_{0}+\phi\right)}{\omega^{2}}\)
\(=\frac{1}{2 C \omega^{2}}\left[\frac{V_{m}}{\sqrt{R^{2}+\left(X_{C}-X_{L}\right)^{2}}}\right]^{2} \cos ^{2}\left(\omega t_{0}+\phi\right)\)
(iii) When R is short circuited, it becomes an L - C oscillator. The capacitor will go on discharging and all energy will go to L.
Hence, there is an oscillation of energy from electrostatic to magnetic and again to electrostatic.

Given, capacitance, \(C=2㎌=2\times { 10 }^{ -6 }F\)  
Resistance, R = 100\(\Omega\) 
Inductance, L = 8H
Peak value of voltage, E₀ = 200V
When frequency of AC source is equal to resonant frequency, then current drawn in the circuit is maximum.
\(\begin{aligned} \therefore \quad v & =v,=\frac{1}{2 \pi \sqrt{L C}} \\ \end{aligned}\)
\(\begin{aligned} & =\frac{1}{2 \times 3.14 \times \sqrt{8 \times 2 \times 10^{-6}}} \\ \end{aligned}\)
\(\begin{aligned} =\frac{1000}{8 \times 3.14}=39.8 \mathrm{~Hz} \end{aligned}\)
Peak value of current,
\(\begin{aligned} I_0 & =\frac{E_0}{R}=\frac{200}{100}=2 \mathrm{~A} \\ \end{aligned}\)
\(\begin{aligned} \because \quad X_C & =X_L=\omega L=2 \pi v L \\ \end{aligned}\)
\(\begin{aligned} & =2 \times 3.14 \times 39.8 \times 8=2000 \mathrm{~W} \end{aligned}\)
\(\begin{aligned} \Rightarrow \quad Z & =R=100 \Omega \end{aligned}\)
The voltages across inductor and resistor differ in phase by 90° and the voltages across inductor and capacitor differ in phase by 180°.

Given, inductance, L = 101.4\(\mu H\) 
\(=101.4\times { 10 }^{ -6 }H\) 
Frequency, v = 1MHz = 10⁶Hz
Since,  \(v=\frac { 1 }{ 2\pi \sqrt { LC } } \)    
\(\Rightarrow \)  \(C=\frac { 1 }{ 4{ \pi }^{ 2 }{ v }^{ 2 }L } \)  
\(=\frac { 1\times 7\times 7 }{ 4\times 22\times 22\times { \left( { 10 }^{ 6 } \right) }^{ 2 }\times 101.4\times { 10 }^{ -6 } }\)
\(=2.496\times { 10 }^{ -10 }F\)
\(C=249.6 \ ㎊\)

Given, B = 10 T, N = 50 turns
A = 100 cm² = 10^-2 m² , dt = 0.15 s
Magnetic flux link with the coil initially,
 \(\phi \) = NBA =50 x 10 x 10^-2 = 5Wb
But magnetic flux linked with the coil finally,i.e.,
(when removed from the magnetic field), \({ \phi }_{ 2 }\) = 0
Emf infuced, e = \(\frac { -d\phi }{ dt } =-(\frac { { \phi }_{ 2 }-{ \phi }_{ 1 } }{ dt } )\)
\(=-(\frac { 0-5 }{ 0.15 } )=33.33 \ v\)