An inductor opposes flow of current through it by developing a back emf according to Lenz's law. The induced voltage has a polarity so as to maintain the current at its present value. If the current is decreasing, the polarity of the induced emf will be so as to increase the current and vice -versa.

Since, the induced emf is proportional to the rate of change of current, it will provide greater reactance to the flow of current if the rate of change is faster, i.e., if the frequency is higher. The reactance of an inductor, therefore, is proportional to the frequency. Mathematically, the reactance offered by the inductor is given by $X_L=\omega L$.

$$\begin{aligned} & \text { Given, power drawn }=P=2 \mathrm{~kW}=2000 \mathrm{~W} \\ & \tan \phi=-\frac{3}{4}, I_M=I_0=\text { ?, } R=\text { ?, } X_C-X_L=\text { ? } \\ & V_{\mathrm{rms}}=V=223 \mathrm{~V} \\ & \text { Power } P=\frac{V^2}{Z} \\ \Rightarrow\quad& Z=\frac{V^2}{P}=\frac{223 \times 223}{2 \times 10^3}=25 \end{aligned}$$

$$\begin{aligned} \text { Impedance } Z & =25 \Omega \\ \text { Impedance } Z & =\sqrt{R^2+\left(X_L-X_C\right)^2} \\ \Rightarrow\quad 25 & =\sqrt{R^2+\left(X_L-X_C\right)^2} \\ \text{or}\quad 625 & =R^2+\left(X_L-X_C\right)^2 \quad\text{... (i)}\\ \text{Again}\quad\tan \phi & =\frac{X_L-X_C}{R}=\frac{3}{4} \\ \text{or}\quad X_L-X_C & =\frac{3 R}{4}\quad\text{.... (ii)} \end{aligned}$$

$$\begin{aligned} &\text { From Eq. (ii), we put } X_L-X_C=\frac{3 R}{4} \text { in Eq. (i), we get }\\ &625=R^2+\left(\frac{3 R}{4}\right)^2=R^2+\frac{9 R^2}{16} \end{aligned}$$

or $625=\frac{25 R^2}{16}$

(a) Resistance $R=\sqrt{25 \times 16}=\sqrt{400}=20 \Omega$

(b) $X_L-X_C=\frac{3 R}{4}=\frac{3}{4} \times 20=15 \Omega$

(c) Main current $I_M=\sqrt{2} I=\sqrt{2} \frac{V}{Z}=\frac{223}{25} \times \sqrt{2}=12.6 \mathrm{~A}$

As $R, X_C, X_L$ are all doubled, $\tan \phi$ does not change. $Z$ is doubled, current is halved. So, power is also halved.

(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

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