(a) In a series LCR circuit connected across an ac source of variable frequency, obtain the expression for its impedance and draw a plot showing its variation with frequency of the ac source.
(b) What is the phase difference between the voltages across inductor and the capacitor at resonance in the LCR circuit ?
(c) When an inductor is connected to 200 V dc voltage, a current of 1 A flows through it.
When the same inductor is connected to a 200 V, 50 Hz ac source, only 0.5 A current flows. Explain, why ? Also, calculate the self inductance of the inductor.
OR
(a) Draw the diagram of a device which is used to decrease high ac voltage into a low ac voltage and state its working principle. Write four sources of energy loss in this device.
(b) A small town with a demand of 1200 kW of electric power at 220V is situated 20 km away from an electric plant generating power at 440 V. The resistance of the two wire line carrying power is 0.5 Q per km. The town gets the power from the line through a 4000-220 V step-down transformer at a sub-station in the town. Estimate the line power loss in the form of heat.
 

Consider an alternating e.m.f. is connected in series with an inductor, resistance R and capacitance C. Let, E and I be the instantaneous values of e.m.f. and current in the LCR circuit V_L, V_C and V_R be the instantaneous values of voltage across inductor, capacitor and resistor respectively. Then,

${\mathrm{V}}_{\mathrm{L}}={\mathrm{IX}}_{\mathrm{L}},{\mathrm{V}}_{\mathrm{C}}={\mathrm{IX}}_{\mathrm{C}}$ and ${\mathrm{V}}_{\mathrm{R}}=\mathrm{IR}$
Here, ${\mathrm{X}}_{\mathrm{L}}=\mathrm{\omega L}\text{ and }{\mathrm{X}}_{\mathrm{C}}=\frac{1}{\mathrm{\omega C}}$

In an a.c. circuit V_R and I are in same phase, ${\mathrm{V}}_{\mathrm{L}}\text{ leads }\mathrm{I}\text{ by }\mathrm{\pi }/2\text{ and }{\mathrm{V}}_{\mathrm{C}}\text{ lags }\mathrm{I}\text{ by }\mathrm{\pi }/2$ From the graph, in right angled $∆$OEA
$\begin{array}{l}\mathrm{OE}=\sqrt{{\mathrm{OA}}^2+{\mathrm{AE}}^2}=\sqrt{{\mathrm{OA}}^2+{\mathrm{OD}}^2}\\ =\sqrt{{\mathrm{V}}_{\mathrm{R}}^2+{\left({\mathrm{V}}_{\mathrm{L}}-{\mathrm{V}}_{\mathrm{c}}\right)}^2}\\ \mathrm{E}=\sqrt{(\mathrm{IR}{)}^2+{\left({\mathrm{IX}}_{\mathrm{L}}-{\mathrm{IX}}_{\mathrm{C}}\right)}^2}\\ \mathrm{E}=\mathrm{I}\sqrt{{\mathrm{R}}^2+{\left({\mathrm{X}}_{\mathrm{L}}-{\mathrm{X}}_{\mathrm{C}}\right)}^2}\\ \mathrm{I}=\frac{\mathrm{E}}{\sqrt{{\mathrm{R}}^2+{\left({\mathrm{X}}_{\mathrm{L}}-{\mathrm{X}}_{\mathrm{C}}\right)}^2}}\end{array}$
Therefore, effective resistance or impedance of LCR circuit is-

$\begin{array}{l}\mathrm{Z}=\sqrt{{\mathrm{R}}^2+{\left({\mathrm{X}}_{\mathrm{L}}-{\mathrm{X}}_{\mathrm{C}}\right)}^2}\\ =\sqrt{{\mathrm{R}}^2+{\left(\mathrm{\omega L}-\frac{1}{\mathrm{\omega C}}\right)}^2}\end{array}$

(b) At resonance
X_L = X_c
iX_L = iX_c
V_L= VC
The voltage across inductance and capacitance are equal and have a phase difference of 180° at resonance.
Since the reactance of an inductor is zero for d.c. circuit. But the inductor offers resistance to an a.c. circuit. Therefore, the current decreases for the same inductor when it is connected with an a.c. source.
When inductor is connected in a.c. circuit:
$\begin{array}{l}\mathrm{V}=200\mathrm{V}\\ \mathrm{f}=50\mathrm{Hz}\\ \mathrm{i}=0.5\mathrm{A}\\ \mathrm{i}=\frac{\mathrm{V}}{\mathrm{R}}=\frac{\mathrm{V}}{{\mathrm{X}}_{\mathrm{L}}}=\frac{\mathrm{V}}{\mathrm{\omega L}}\\ 0.5=\frac{200}{2\mathrm{\pi }\times 50\mathrm{L}}[\because \mathrm{\omega }=2\mathrm{\pi f}]\\ \mathrm{L}=\frac{200}{100\times 0.5\times 3.14}\\ =\frac{4}{3.14}=1.27\mathrm{H}\end{array}$
(a) Step-down transformer:
It is a device used for converting high alternating voltage at low current into low alternating voltage at high current and vice-versa.
The device works on the principle of mutual induction i.e., if the current or magnetic flux linked with a coil changes then an e.m.f. is induced in the other coil.
In step-down transformer Np > Ns and transformation ratio is less than 1. 

1. Copper losses: Due to resistance of winding’s in primary and secondary coils, some electrical energy is converted into heat energy.
2. Flux losses : Some of the flux produced in primary coil is not linked up with secondary coils.
3. Hysteresis losses : When the iron core is subjected to a cycle of magnetization the core gets heated up due to hysteresis known as hysteresis loss.
4. Iron losses : The varying magnetic flux produces eddy current in the iron core, which leads to the wastage of energy in the form of heat.
Length of wire line = 20 x 2 = 40 km
Resistance of wire line, r = 40 x 0.5 = 20 Ω
Power to be supplied = 1200 kW = 1200 × 103 W
Voltage at which power supplied = 4000 V
Since,     P=VI
$$\begin{gathered} \Rightarrow \mathrm{I}=\frac{\mathrm{P}}{\mathrm{V}} \\ \Rightarrow \mathrm{I}=\frac{1200 \mathrm{x} {10}^3}{4000} \\ =300 \mathrm{A} \end{gathered}$$
Therefore, line power loss
$\begin{array}{l}={\mathrm{I}}^2\times \mathrm{R}\\ =(300{)}^2\times 200\\ =36\times {10}^5\mathrm{W}\\ =360\mathrm{kW}\end{array}$
$\therefore \mathrm{Line} \mathrm{power} \mathrm{loss} \mathrm{in} \mathrm{the} \mathrm{form} \mathrm{of} \mathrm{heat} \mathrm{is} 360 \mathrm{kW}.$