# NCERT Solutions Class 12 Physics Chapter 7

## NCERT Solutions Class 12 Physics Chapter  7 Alternating Current

The NCERT Solutions Class 12 Physics Chapter 7 will guide students to understand the concepts and solve questions appropriately. Class 12 Physics Chapter 7 NCERT Solutions are prepared by the subject experts at Extramarks. All answers are explained clearly in simple language while preparing the solutions, and all the guidelines given by CBSE have been kept in mind

CBSE or Central Board of Secondary Education is the most preferred educational board among schools. It covers all topics required to understand the concept clearly and thoroughly. The  NCERT Solutions Class 12 Physics Chapter 7 act as a helping hand for students to learn the subject effectively.

### Key Topics Covered In NCERT Solutions Class 12 Physics Chapter 7

Chapter 7 Physics Class 12 explains the concept of Alternating current, its properties and use cases, a vital sub-topic under the Electromagnetic Induction unit. Students can expect questions on this topic in first term examinations. To score good marks, students may refer to NCERT Solutions Class 12 Physics Chapter 7.

Following are the topics covered in NCERT Solutions Class 12 Physics Chapter 7

 Topic Number Topic Name 7.1 Introduction 7.2 AC Voltage Applied to resistor 7.3 AC Current, voltage by rotating vectors 7.4 AC voltage to inductor 7.5 AC Voltage to capacitor 7.6 AC Voltage to Series LCR 7.7 Power factor 7.8 Oscillations 7.9 Transformers

A brief of the topics covered in NCERT Solutions Class 12 Physics Chapter 7  is below.

Topic 7.1 Introduction

Alternating current (AC) is an electric powered current that periodically reverses direction and adjusts its magnitude constantly with time, in contrast to direct current (DC), which flows best in a single direction.

Topic 7.2 AC Voltage Is Applied to a Resistor

To have a sinusoidal alternating current, we want to have an AC voltage supply because the current is at once proportional to voltage. An AC generator or AC dynamo may be used as an AC voltage supply.

Voltage V(t) is applied across resistance R. V(t) is sinusoidal voltage with peak Vm and time period T.

T=1f=2

Where f is frequency and ω is the angular frequency. This kind of circuit is a purely resistive circuit. According to Kirchhoff’s law –

v(t)=Ri(t)

i(t)=v(t)R

i(t)=Vmsin(t)R

im=VmR

i(t)=imsin(t)

Here voltage and current have equal frequency, and each is in a similar phase.

The average current value may be located out with the aid of summing over the whole change inside the voltage and dividing it using the number of instances we do the measurements.

I.e iavg=0

Over a cycle, the average value of AC current is 0 because, in the 1st half of the time, the current is positive and in the second half, the current is negative.

Root Mean Square Value of Current, i.e. irms=im2

Topic 7.3 Representation of AC Current and Voltage through Rotating Vectors or Phasors

The projection of the phasor at the vertical axis represents the value of the quantity. For example, inside the case of a current or a vector phasor, the projection of the phasor at the vertical axis, given via means of vmsinωt and imsinωt, respectively, offers a current value or voltage value.

Topic 7.4 AC Voltage Applied to an Inductor

It implies an AC circuit composed only of an inductor of inductance L linked to an AC supply. The AC voltage throughout the supply is V = Vm sin(t). The converting current output of the AC supply offers a returned emf inside the coil of magnitude that’s given via the means of VL = L di/dt.

Topic 7.5 AC Voltage Applied to a Capacitor

When an alternating voltage is carried out throughout a capacitor, the current leads the voltage via a phase angle of ninety degrees. This approach oscillates 1 / 4 of the cycle before the voltage.

Topic 7.6 AC Voltage Applied to a Series LCR Circuit

A chain LCR circuit is connected to an AC supply. The voltage of an AC supply is V=Vmsinωt. Potential distinction throughout the ends of capacitor =qC and potential distinction throughout the resistance ends =IR.

Topic 7.7 Power in AC Circuit: The Power Factor

Power factor (PF) is the ratio of running power, measured in kilowatts (kW), to apparent power, measured in kilovolt amperes (kVA). Apparent power, additionally called demand, is the degree of the quantity of power used to run equipment and devices all through a certain period. It is observed with the aid of using multiplying (kVA = V x A).

Topic 7.8 LC Oscillations

The LC Oscillator employs a tank circuit (comprising an inductor and a capacitor) to offer tremendous vital comments to preserve oscillations in a circuit going.

Working of an LC Oscillator:

• When a completely powered capacitor is connected to a de-energised inductor, the capacitor gets all the circuit’s energy while the inductor receives none. Let’s name the energy saved inside the capacitor (electrical energy) UE and the magnetic energy saved inside the inductor (UB).
• Current flows from the capacitor to the inductor, energizing the inductor and discharging the capacitor. The inductor’s energy starts to evolve to rise simultaneously as the capacitor’s energy falls. The bars underneath the circuit diagram display that 1/2 of the energy saved in an inductor is equal to 1/2 of the energy saved in a capacitor, indicating that the capacitor has transferred 1/2 of its energy to the inductor.
• All of the capacitor’s energy will now be transmitted to the inductor as quickly because the capacitor is drained. As a result, a vast quantity of electrical energy is converted to magnetic energy.
• Because the capacitor has been absolutely drained and the inductor has been absolutely energized, the inductor will now start to charge the capacitor in the same direction because of the current. The present state of the circuit is depicted in the fourth diagram. As a result, 1/2 of the inductor’s energy has been transferred to the capacitor.
• Finally, the capacitor may be absolutely charged once more, and the inductor may be activated. On the alternative hand, the capacitor will now have its polarity reversed. As a result, if the current inside the circuit begins to move once more from the capacitor, it’ll flow in the opposite direction. We might also say that the primary 1/2 of the AC cycle has ended, and the second 1/2 has started because the current inside the circuit now has an opposing flow of current.
• As a result, each capacitor and the inductor may be charged two times for the duration of the cycle.

Topic 7.9 Transformers

These are used to increase or decrease the voltage of alternating currents. A transformer includes coils of wire wound on a metallic core. An alternating voltage is carried out to at least one coil (the first coil). This causes a changing (alternating) magnetic area to be installed inside the core.

### List of NCERT Solutions Class 12 Physics Chapter 7 Alternating current

Click on the below links to view NCERT Solutions Class 12 Physics Chapter 7

• NCERT Solutions Class 12 Physics Chapter 7: Exercise 7.1
• NCERT Solutions Class 12 Physics Chapter 7: Exercise 7.2
• NCERT Solutions Class 12 Physics Chapter 7: Exercise 7.3
• NCERT Solutions Class 12 Physics Chapter 7: Exercise 7.4
• NCERT Solutions Class 12 Physics Chapter 7: Exercise 7.5
• NCERT Solutions Class 12 Physics Chapter 7: Exercise 7.6
• NCERT Solutions Class 12 Physics Chapter 7: Exercise 7.7
• NCERT Solutions Class 12 Physics Chapter 7: Exercise 7.8
• NCERT Solutions Class 12 Physics Chapter 7: Exercise 7.9

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### Key Features of NCERT Solutions Class 12 Physics Chapter 7

The key features of the NCERT Solutions Class 12 Physics Chapter 7 offered by Extramarks include

• Detailed explanations for each topic under chapter 7 Physics class 12
• Curated by Subject Experts in respective fields
• Easy to understand and simple language
• Answers are written by giving importance to the weightage.

To help students in exams and the learning process, Extramarks provides expert solutions, notes and explanatory articles for all subjects and classes.

Q.1 A 100 Ω resistor is connected to a 220 V, 50 Hz ac supply.
(a) What is the rms value of current in the circuit?
(b) What is the net power consumed over a full cycle?

Ans.

$\begin{array}{l}\text{Here, resistance of the resistor, R=100 Ω}\\ \text{Supply voltage, V=220 V}\\ \text{Frequency,}\mathrm{\nu }\text{= 50 Hz}\\ \left(\text{a}\right)\text{The rms value of current in the circuit is given by,}\\ \text{I =}\frac{\text{V}}{\text{R}}\text{=}\frac{\text{220 V}}{\text{100 Ω}}\\ \text{= 2.20 A}\\ \left(\text{b}\right)\text{The net power consumed over a full cycle is given by,}\\ \text{P = VI}\\ \text{= 220 V × 2.2 A}\\ \text{= 484 W}\end{array}$

Q.2 (a) The peak voltage of an ac supply is 300 V. What is the rms voltage?
(b) The rms value of current in an ac circuit is 10 A. What is the peak current?

Ans.

$\begin{array}{l}\left(\text{a}\right)\text{Here, peak voltage of the ac supply},{\text{V}}_{0}=\text{3}00\text{V}\\ \text{Rms voltage is given by,}\\ \text{V=}\frac{{\text{V}}_{0}}{\sqrt{2}}\\ =\frac{300\mathrm{V}}{\sqrt{2}}=212.1\text{ }\mathrm{V}\\ \left(\text{b}\right)\text{The rms value of current is given by,}\\ \text{I}=\text{1}0\text{A}\\ \therefore {\text{Peak current, I}}_{\text{0}}=\sqrt{2}\mathrm{I}=\text{1}0\sqrt{2}\text{}\mathrm{A}=14.1\text{ }\mathrm{A}\end{array}$

Q.3 A 44 mH inductor is connected to 220 V, 50 Hz ac supply. Determine the rms value of the current in the circuit.

Ans.

$\begin{array}{l}{\text{Here, inductance of inductor, L = 44 mH = 44 × 10}}^{\text{-3}}\text{H}\\ \text{Supply voltage, V = 220 V}\\ \text{Frequency,}\mathrm{\nu }\text{= 50 Hz}\\ \text{Angular frequency, ω = 2}\mathrm{\pi \nu }\text{}\to \text{(i)}\\ {\text{Inductive reactance, X}}_{\text{L}}\text{= ω L}\to \text{(ii)}\\ \text{From equation (i) and (ii), we have}\\ {\text{X}}_{\text{L}}\text{= 2}\mathrm{\pi \nu }\text{L}\\ {\text{X}}_{\text{L}}\text{= 2}\mathrm{\pi }{\text{× 50 Hz × 44 ×10}}^{\text{-3}}\text{ Ω}\\ \text{Rms value of current, I =}\frac{\text{V}}{{\text{X}}_{\text{L}}}\\ \text{I =}\frac{\text{220 V}}{{\text{2π×50 Hz × 44×10}}^{\text{-3}}\text{Ω}}\text{=15.92 A}\\ \therefore \text{The rms value of current in the circuit is 15.92 A.}\end{array}$

Q.4 A 60 μF capacitor is connected to a 110 V, 60 Hz ac supply. Determine the rms value of the current in the circuit.

Ans.

$\begin{array}{l}{\text{Here, capacitance of capacitor, C = 60 μF = 60×10}}^{\text{-6}}\text{F}\\ \text{Supply voltage, V = 110 V}\\ \text{Frequency,}\mathrm{\nu }\text{= 60 Hz}\\ \text{Angular frequency, ω = 2}\mathrm{\pi \nu }\\ \text{Capacitive reactance is given by,}\\ {\text{X}}_{\text{c}}\text{=}\frac{\text{1}}{\text{ωC}}\text{=}\frac{\text{1}}{\text{2}\mathrm{\pi \nu }\text{C}}\\ \text{=}\frac{\text{1}}{{\text{2× 3.14 × 60 Hz × 60 ×10}}^{\text{-6}}\mathrm{F}}{\text{\hspace{0.17em}\hspace{0.17em}Ω}}^{\text{-1}}\\ \text{Rms value of current is given by, }\\ \text{I =}\frac{\text{ν}}{{\text{X}}_{\text{c}}}\\ {\text{= 110 × 2 × 3.14 × 60 Hz × 60 × 10}}^{\text{-6}}\text{}\mathrm{F}\\ \text{= 2.49 A}\\ \therefore \text{The rms value of current is 2.49 A.}\end{array}$

Q.5 In Exercises 7.3 and 7.4, what is the net power absorbed by each circuit over a complete cycle. Explain your answer.

Ans.

$\begin{array}{l}\text{In Exercise 7.3,}\\ \text{Rms value of the current, I = 15.92 A}\\ \text{Rms value of the voltage, V = 220 V}\\ \text{The net power absorbed can be calculated by the relation,}\\ \text{P = VI cos}\mathrm{\varphi }\\ \text{Here,\hspace{0.17em}\hspace{0.17em}}\mathrm{\varphi }\text{= Phase difference between V and I}\\ \text{In a pure inductive circuit, the phase difference}\\ \text{between alternating voltage and current is 90° i.e.,}\\ \mathrm{\varphi }\text{= 90°}\\ \therefore \text{P = VI cos 90° = 0}\\ \therefore \text{The net power is zero.}\\ \text{In Exercise 7.4,}\\ \text{Rms value of the current,}\\ \text{I = 2.49 A}\\ \text{Rms value of the voltage,}\\ \text{V = 110 V}\\ \text{The net power absorbed can be obtained as:}\\ \text{P = VI cos}\mathrm{\varphi }\\ \text{In a pure capacitive circuit, the phase difference}\\ \text{between alternating voltage and current is 90° i.e.,}\\ \mathrm{\varphi }\text{= 90°.}\\ \therefore \text{P = VI cos 90° = 0}\\ \therefore \text{The net power is zero.}\end{array}$

Q.6 Obtain the resonant frequency ωr of a series LCR circuit with L = 2.0 H, C = 32 μF and R = 10 Ω. What is the Q-value of this circuit?

Ans.

$\begin{array}{l}\text{Here, inductance, L = 2.0 H}\\ {\text{Capacitance,C = 32 μF = 32 × 10}}^{\text{-6}}\text{F}\\ \text{Resistance, R = 10 Ω}\\ \text{Resonant frequency can be obtained by the relation,}\\ {\text{ω}}_{\text{r}}\text{=}\frac{\text{1}}{\sqrt{\text{LC}}}\\ \text{=}\frac{\text{1}}{\sqrt{{\text{2 H × 32 × 10}}^{\text{-6}}\text{F}}}\text{=}\frac{\text{1}}{{\text{8×10}}^{\text{-3}}}{\text{s}}^{\text{-1}}{\text{= 125 s}}^{\text{-1}}\\ \text{Now, Q-value of the circuit is given by:}\\ \text{Q =}\frac{\text{1}}{\text{R}}\sqrt{\frac{\text{L}}{\text{C}}}\\ \text{=}\frac{\text{1}}{\text{10}}\sqrt{\frac{\text{2H}}{{\text{32 × 10}}^{\text{-6}}\text{F}}}\\ \text{=}\frac{\text{1}}{\text{10}}\text{×}\frac{\text{1}}{{\text{4×10}}^{\text{-3}}}\text{= 25}\\ \therefore \text{The Q-Value of this circuit is 25.}\end{array}$

Q.7 A charged 30 μF capacitor is connected to a 27 mH inductor. What is the angular frequency of free oscillations of the circuit?

Ans.

$\begin{array}{l}{\text{Here, capacitance, C = 30 μF = 30 ×10}}^{\text{-6}}\text{ F}\\ {\text{Inductance, L = 27 mH = 27 ×10}}^{\text{-3}}\text{H}\\ \text{Angular frequency is given by the relation,}\\ {\text{ω}}_{\text{r}}\text{=}\frac{\text{1}}{\sqrt{\text{LC}}}\text{=}\frac{\text{1}}{\sqrt{{\text{27 × 10}}^{\text{-3}}{\text{H ×30 × 10}}^{\text{-6}}\text{F}}}\\ {\text{ω}}_{\text{r}}\text{=}\frac{\text{1}}{{\text{9×10}}^{\text{-4}}}{\text{= 1.11×10}}^{\text{3}}{\text{ rads}}^{\text{-1}}\\ \therefore \text{The angular frequency of free oscillations of the}\\ {\text{circuit is 1.11 × 10}}^{\text{3}}{\text{rads}}^{\text{-1}}\text{.}\end{array}$

Q.8 Suppose the initial charge on the capacitor in Exercise 7.7 is 6 mC. What is the total energy stored in the circuit initially? What is the total energy at later time?

Ans.

$Here, capacitance of capacitor, C = 30 μF = 30×10 -6 F Inductance of inductor, L = 27 mH = 27×10 -3 H Charge on capacitor, Q = 6 mC = 6 × 10 -3 C Total energy stored in the capacitor can be obtained by the relation, E= 1 2 Q 2 C = 1 2 × 6 × 10 -3 C 2 30×10 -6 F = 6 10 J = 0.6 J At later time, total energy will remain the same because energy is shared between the capacitor and the inductor.$

Q.9 A series LCR circuit with R = 20 Ω, L = 1.5 H and C = 35 μF is connected to a variable frequency 200 V ac supply. When the frequency of the supply equals the natural frequency of the circuit, what is the average power transferred to the circuit in one complete cycle?

Ans.

$When the frequency of the supply power equals the natural frequency of the given LCR circuit, resonance takes place. Here, resistance, R = 20 Ω Inductance, L = 1.5 H Capacitance, C = 35 μF= 35 × 1 0 −6 F AC supply voltage to the circuit, V = 200 V Impedance of the circuit is given by, Z = R 2 + ωL – 1 ωC 2 At resonance, ωL = 1 ωC ∴Z=R =20 Ω Current in the circuit, I = V Z = 200 V 20 Ω =10 A ∴The average power transferred to the circuit in one complete cycle = VI = 200 V × 10 A = 2000 W$

Q.10 A radio can tune over the frequency range of a portion of MW broadcast band: (800 kHz to 1200 kHz). If its LC circuit has an effective inductance of 200 μH, what must be the range of its variable capacitor?

[Hint: For tuning, the natural frequency i.e., the frequency of free oscillations of the LC circuit should be equal to the frequency of the radiowave.]

Ans.

$Here, lower tuning frequency, ν 1 = 800 kHz = 800 × 10 3 Hz Upper tuning frequency, ν 2 = 1200 kHz = 1200 ×10 3 Hz Effective inductance of LC circuit, L = 200 μH = 200 ×10 -6 H Capacitance of the variable capacitor for ν 1 is given by: C 1 = 1 ω 1 2 L Here,ω 1 = Angular frequency for capacitor C 1 ω 1 = 2π ν 1 = 2π × 800 ×10 3 rads -1 ∴ C 1 = 1 2π × 800 × 10 3 rads -1 2 × 200 ×10 -6 H C 1 = 1 .9809×10 -10 F = 198.1 pF Capacitance of variable capacitor for ν 2 is given by, C 2 = 1 ω 2 2 L Here,ω 2 = Angular frequency for capacitor C 2 ω 2 = 2π ν 2 = 2π ×1200×10 3 rads -1 ∴ C 2 = 1 2π ×1200 ×10 3 rads -1 2 ×200×10 -6 H C 2 = 88.04 pF ∴The range of the variable capacitor is from 88.04 pF to 198.1 pF.$

Q.11 Figure 7.21 shows a series LCR circuit connected to a variable frequency 230 V source. L= 5.0 H, C = 80μF, R = 40 Ω
(a) Determine the source frequency which drives the circuit in resonance.
(b) Obtain the impedance of the circuit and the amplitude of current at the resonating frequency.
(c) Determine the rms potential drops across the three elements of the circuit. Show that the potential drop across the LC combination is zero at the resonating frequency.

Ans.

$\begin{array}{l}\text{Here, }\mathrm{i}\text{nductance of inductor},\text{L}=\text{5}.0\text{H}\\ \text{Capacitance of capacitor},\text{C}=\text{8}0\text{}\mathrm{\mu F}\text{}\\ \mathrm{C}=\text{8}0×\text{1}{0}^{-\text{6}}\text{F}\\ \text{Resistance of resistor},\text{R}=\text{4}0\text{}\mathrm{\Omega }\\ \text{Potential of variable voltage source},\text{V}=\text{23}0\text{V}\\ \left(\text{a}\right)\text{Resonance angular frequency is given by}:\\ {\mathrm{\omega }}_{\mathrm{R}}=\frac{1}{\sqrt{\mathrm{LC}}}\\ =\frac{1}{\sqrt{\text{5}.0\text{}\mathrm{H}×\text{8}0×\text{1}{0}^{-\text{6}}\text{}\mathrm{F}}}\\ =\frac{{10}^{3}}{20}\text{}{\mathrm{rads}}^{-1}\\ =50\text{ }{\mathrm{rads}}^{-1}\\ \therefore \text{The circuit will come in resonance for a source}\\ \text{frequency of 5}0{\text{rads}}^{\text{-1}}.\\ \left(\text{b}\right)\text{Impedance of the circuit is given by},\\ \mathrm{Z}=\sqrt{{\mathrm{R}}^{2}+{\left(\mathrm{\omega L}-\frac{1}{\mathrm{\omega C}}\right)}^{2}}\\ \text{At resonance},\\ \mathrm{\omega L}=\frac{1}{\mathrm{\omega C}}\\ \therefore \mathrm{Z}=\mathrm{R}\\ =40\text{ }\mathrm{\Omega }\\ \text{Amplitude of current at resonating frequency},\text{ }{\mathrm{I}}_{\mathrm{o}}=\frac{{\mathrm{V}}_{\mathrm{o}}}{\mathrm{Z}}\\ \text{Here},{\text{V}}_{0}=\text{Peak voltage}\\ \text{=}\sqrt{2}\mathrm{V}\\ \therefore \text{ }{\mathrm{I}}_{\mathrm{o}}=\frac{\sqrt{2}\mathrm{V}}{\mathrm{Z}}\\ =\frac{\sqrt{2}×230\text{}\mathrm{V}}{40\text{}\mathrm{\Omega }}\\ =8.13\text{ }\mathrm{A}\\ \therefore \text{At resonance},\text{the impedance of the circuit is 4}0\text{}\mathrm{\Omega }\text{}\\ \text{and the amplitude of the current is 8}.\text{13 A}.\\ \left(\text{c}\right)\text{Across the inductor, }\mathrm{r}\text{ms potential drop},\\ {\left({\text{V}}_{\text{L}}\right)}_{\text{rms}}\text{}=\text{I}×\text{}{\mathrm{\omega }}_{\text{R}}\text{L}\\ \text{Here},\text{I}=\text{rms current}\\ \text{=}\frac{{\mathrm{I}}_{\mathrm{o}}}{\sqrt{2}}\\ =\frac{\sqrt{2\text{}}\mathrm{V}}{\sqrt{2}\text{}\mathrm{Z}}\\ =\frac{230}{40}\text{ }\mathrm{A}\\ {\left({\text{V}}_{\text{L}}\right)}_{\text{rms}}\text{}=\frac{230}{40}\text{A}×50\text{}{\mathrm{rads}}^{-1}×5\text{H}\\ =1437.5\text{ }\mathrm{V}\\ \text{Potential drop across capacitor},\\ {\left({\text{V}}_{\mathrm{C}}\right)}_{\text{rms}}=\text{I}×\frac{1}{{\mathrm{\omega }}_{\text{R}}\mathrm{C}}\\ =\frac{230\text{}}{40\text{}}\mathrm{A}×\frac{1}{50\text{}{\mathrm{rads}}^{-1}×80×{10}^{-6}\text{F}}\\ =1437.5\text{ }\mathrm{V}\\ \text{Potential drop across resistor},\\ {\left({\text{V}}_{\text{R}}\right)}_{\text{rms}}=\text{IR}\\ =\frac{230}{40}\mathrm{A}×\text{4}0\text{}\mathrm{\Omega }\text{}\\ =\text{23}0\text{V}\\ \text{Potential drop across LC combination},\\ {\mathrm{V}}_{\mathrm{LC}}=\mathrm{I}\left({\mathrm{\omega }}_{\text{R}}\text{L}-\frac{1}{{\mathrm{\omega }}_{\text{R}}\mathrm{C}}\right)\\ \text{At resonance},\\ {\mathrm{\omega }}_{\text{R}}\text{L =}\frac{1}{{\mathrm{\omega }}_{\text{R}}\mathrm{C}}\\ \therefore {\text{V}}_{\text{LC}}=0\\ \therefore \text{It is proved that the potential drop across the LC}\\ \text{combination is zero at resonating frequency}.\end{array}$

Q.12 An LC circuit contains a 20 mH inductor and a 50 μF capacitor with an initial charge of 10 mC. The resistance of the circuit is negligible. Let the instant the circuit is closed be t =0.

(a) What is the total energy stored initially? Is it conserved during LC oscillations?

(b) What is the natural frequency of the circuit?

(c) At what time is the energy stored?
(i) Completely electrical (i.e., stored in the capacitor)?
(ii) Completely magnetic (i.e., stored in the inductor)?

(d) At what times is the total energy shared equally between the inductor and the capacitor?

(e) If a resistor is inserted in the circuit, how much energy is eventually dissipated as heat?

Ans.

$\begin{array}{l}{\text{Here, inductance of inductor,L = 20 mH = 20×10}}^{\text{-3}}\text{ H}\\ {\text{Capacitance of capacitor, C = 50 μF = 50 × 10}}^{\text{-6}}\text{ F}\\ {\text{Initial charge on capacitor,Q = 10 mC = 10 × 10}}^{\text{-3}}\text{ C}\\ \text{(a) Total energy stored initially in the circuit is given by the relation,}\\ \text{E =}\frac{\text{1}}{\text{2}}\frac{{\text{Q}}^{\text{2}}}{\text{C}}\\ \text{=}\frac{{\left({\text{10×10}}^{\text{-3}}\text{C}\right)}^{\text{2}}}{{\text{2×50×10}}^{\text{-6}}\text{F}}\text{= 1 J}\\ \text{Since there is no resistor connected in the given LC circuit, therefore total }\\ \text{energy stored initially in the given LC circuit is conserved during oscillations.}\\ \text{(b) The relation for natural frequency of the circuit is given as:}\\ \mathrm{\nu }\text{=}\frac{\text{1}}{\text{2}\mathrm{\pi }\sqrt{\text{LC}}}\\ \mathrm{\nu }\text{=}\frac{\text{1}}{\text{2}\mathrm{\pi }\sqrt{{\text{20×10}}^{\text{-3}}{\text{H × 50×10}}^{\text{-6}}\text{}\mathrm{F}}}\\ \mathrm{\nu }\text{=}\frac{{\text{10}}^{\text{3}}}{\text{2}\mathrm{\pi }}\text{= 159.24 Hz}\\ \text{Natural angular frequency is given as:}\\ {\text{ω}}_{\text{r}}\text{=}\frac{\text{1}}{\sqrt{\text{LC}}}\\ \text{=}\frac{\text{1}}{\sqrt{{\text{20×10}}^{\text{-3}}{\text{H× 50 ×10}}^{\text{-6}}}\mathrm{F}}\text{=}\frac{\text{1}}{\sqrt{{\text{10}}^{\text{-6}}}}\\ {\text{=10}}^{\text{3}}{\text{ rads}}^{\text{-1}}\\ \therefore {\text{Natural angular frequency of the given circuit is 10}}^{\text{3}}{\text{ rads}}^{\text{-1}}.\\ \text{(c) (i) At any instant t,total charge on the capacitor is given as:}\\ \text{Q’ = Q cos}\frac{\text{2}\mathrm{\pi }}{\text{T}}\text{t}\\ \text{For electrical energy stored,we can write,}\\ \text{Q’ = Q}\\ \therefore \text{It can be concluded that the energy stored in the}\\ \text{capacitor is completely electrical at time,}\\ \text{t = 0,}\frac{\text{T}}{\text{2}}\text{,T,}\frac{\text{3T}}{\text{2}}\text{,}.\dots ..\\ \text{}\left(\text{ii}\right)\text{Magnetic energy is the maximum when electrical energy,Q’ = 0.}\\ \therefore \text{It can be concluded that the energy stored in the}\\ \text{capacitor is completely magnetic at time,}\\ \text{t =}\frac{\text{T}}{\text{4}}\text{,}\frac{\text{3T}}{\text{4}}\text{,}\frac{\text{5T}}{\text{4}}\text{,}.\dots ..\\ \left(\text{d}\right){\text{Q}}^{\text{1}}\text{= Charge on capacitor when total energy is equally shared between}\\ \text{the capacitor and the inductor at time t.}\\ \text{When total energy is equally shared between the inductor and capacitor,}\\ \text{Energy stored in the capacitor=}\frac{\text{1}}{\text{2}}\left(\text{maximum energy}\right)\\ ⇒\frac{\text{1}}{\text{2}}\frac{{\left({\text{Q}}^{\text{1}}\right)}^{\text{2}}}{\text{C}}\text{=}\frac{\text{1}}{\text{2}}\left(\frac{\text{1}}{\text{2}}\frac{{\text{Q}}^{\text{2}}}{\text{C}}\right)\text{=}\frac{\text{1}}{\text{4}}\frac{{\text{Q}}^{\text{2}}}{\text{C}}\text{ }\\ \left({\text{Here, Q}}^{\text{1}}\text{=}\frac{\text{Q}}{\sqrt{\text{2}}}\right)\\ {\text{From Q}}^{\text{1}}\text{= Q cos}\frac{\text{2}\mathrm{\pi }}{\text{T}}\text{t, we have}\\ \frac{\text{Q}}{\sqrt{\text{2}}}\text{= Q cos}\frac{\text{2}\mathrm{\pi }}{\text{T}}\text{t}\\ \mathrm{cos}\frac{2\mathrm{\pi }}{\mathrm{T}}\mathrm{t}=\frac{1}{\sqrt{2}}=\mathrm{cos}\left(2\mathrm{n}+1\right)\frac{\mathrm{\pi }}{4}\\ \mathrm{Where}\text{ }\mathrm{n}=0,1,2,3,...\\ \therefore \mathrm{t}=\left(2\mathrm{n}+1\right)\frac{\mathrm{T}}{8}\\ \therefore \text{Total energy is equally shared between the inductor and the capacity at time},\\ \mathrm{t}=\frac{\mathrm{T}}{8},\frac{3\mathrm{T}}{8},\frac{5\mathrm{T}}{8},.\dots ..\\ \left(\text{e}\right)\text{The presence of a resistor in the circuit involves},\text{dissipation of energy as}\\ \text{heat energy in the circuit}.\text{The resistance damps out the LC oscillations and }\\ \text{they disappear when total energy of 1 J is dissipated in the form heat.}\end{array}$

Q.13 A coil of inductance 0.50 H and resistance 100 Ω is connected to a 240 V, 50 Hz ac supply.

(a) What is the maximum current in the coil?

(b) What is the time lag between the voltage maximum and the current maximum?

Ans.

$Inductance of inductor, L = 0.50 H Resistance of resistor, R = 100 Ω Potential of supply voltage, V = 240 V Frequency of the supply voltage, ν = 50 Hz a Peak voltage is given by the relation, V 0 = 2 V = 2 × 240 V = 339.41 V Angular frequency of the supply voltage is given as: ω = 2 πν = 2π × 50 Hz = 100 π rads -1 Maximum current in the circuit, I 0 = V 0 R 2 + ω 2 L 2 = 339.41 V 100 Ω 2 + 100π rads -1 2 0.50 H 2 =1.82 A b Voltage is given by the relation, V = V 0 cos ωt Current is given by the relation, I = I 0 cos ωt – ϕ Here,ϕ = Phase difference between voltage and current At time, t = 0 V= V 0 voltage is maximum For ωt – ϕ =0 i.e.,at time t = ϕ ω , I = I 0 current is maximum ∴The time lag between maximum voltage and maximum current = ϕ ω The relation for the phase angle ϕ is given as: tan ϕ = ωL R = 2π ×50 Hz ×0.5 H 100 Ω =1.57 ∴ϕ = 57.5°= 57.5ϕ 180 rad ∴ωt = 57.5π 180 ∴t = 57.5π 180 × 2π ×50 Hz = 3 .19 ×10 -3 s = 3.2 ms ∴The time lag between maximum voltage and maximum current is 3.2 ms.$

Q.14 Obtain the answers (a) to (b) in Exercise 7.13 if the circuit is connected to a high frequency supply (240 V, 10 kHz). Hence, explain the statement that at very high frequency, an inductor in a circuit nearly amounts to an open circuit. How does an inductor behave in a dc circuit after the steady state?

Ans.

$Here, inductance of inductor, L = 0.50 H Resistance of resistor, R = 100 Ω Potential of supply voltages, V = 240 V Frequency of supply,ν=10 kHz =1 0 4 Hz Angular frequency of supply, ω=2πν =2π×1 0 4 rads -1 a Peak voltage is given by the relation, V 0 = 2 ×V =240 2 V Maximum current is given by, I 0 = V 0 R 2 + ω 2 L 2 = 240 2 V 100 Ω 2 + 2π× 10 4 rads -1 2 × 0.50 H 2 =1.1× 10 −2 A b The relation for phase difference Φ is given as: tanϕ= ωL R = 2π× 10 4 rads -1 ×0.50 H 100 Ω =100π ϕ= 89.82 o = 89.82π 180 rad ωt= 89.82π 180 rad ∴t= 89.82π rad 180×2π× 10 4 rads -1 =25 μs We observe that, I 0 is very small in this case. ∴At high frequencies, the inductor amounts to an open circuit. In a dc circuit, after a steady state is reached, ω=0. ∴ X L = ωL = 0 ∴Inductor L behaves like a pure conducting object.$

Q.15 A 100 μF capacitor in series with a 40 Ω resistance is connected to a 110 V, 60 Hz supply.

(a) What is the maximum current in the circuit?

(b) What is the time lag between the current maximum and the voltage maximum?

Ans.

$Here, capacitance of capacitor, C = 100 μF = 100 ×10 -6 F Resistance of resistor, R = 40 Ω Supply voltage, V = 110 V a Here, frequency of oscillations, ν = 60 Hz Angular frequency,ω = 2πν = 2π × 60 rads -1 In a RC circuit, the relation for impedance is given as: Z= R 2 + 1 ω 2 C 2 Peak voltage is given as, V 0 = V 2 = 110 2 V Maximum current is given by, I 0 = V 0 Z = V 0 R 2 + 1 ω 2 C 2 = 110 2 V 40 Ω 2 + 1 120π rads -1 2 10 -4 F 2 = 110 2 V 1600 Ω 2 + 10 8 120π 2 Ω 2 = 3.24 A b In the RC circuit, the voltage lags behind the current by phase angle ϕ. Phase angle is given by the relation: tanϕ = 1 ωC R = 1 ωCR = 1 120π rads -1 ×10 -4 F× 40 Ω = 0.6635 ∴ϕ = tan -1 0.6635 = 33 .56 o = 33.56π 180 rad ∴Time lag = ϕ ω = 33.56π 180 × 120π = 1 .55 × 10 -3 s = 1.55 ms ∴The time lag between maximum current and maximum voltage is 1.55 ms.$

Q.16 Obtain the answers to (a) and (b) in Exercise 7.15 if the circuit is connected to a 110 V, 12 kHz supply? Hence, explain the statement that a capacitor is a conductor at very high frequencies. Compare this behaviour with that of a capacitor in a dc circuit after the steady state.

Ans.

$Here, capacitance of capacitor, C = 100 μF = 100×10 -6 F Resistance of resistor, R = 40 Ω Supply voltage,V = 110 V Frequency of supply voltage, ν = 12 kHz =12×10 3 Hz Angular Frequency, ω = 2 πν = 2 × π × 12 ×10 3 rads -1 = 24π ×10 3 rads -1 Peak voltage is given by the relation, V 0 = V 2 = 110 2 Maximum current is given by, I 0 = V 0 R 2 + 1 ω 2 C 2 = 110 2 40 Ω 2 + 1 24π ×10 3 rads -1 2 100×10 -6 F 2 = 110 2 1600+ 10 24π 2 = 3.9 A In an RC circuit, the voltage lags behind the current by phase angle ϕ. The relation for phase angle ϕ is given as: tanϕ ​​= 1 ωC R = 1 ωCR = 1 24π ×10 3 rads -1 × 100 ×10 -6 F × 40 Ω tanϕ = 1 96π ∴ϕ≃0.2°= 0.2π 180 rad ∴Time lag = ϕ ω = 0.2π 180× 24π ×10 3 rads -1 = 1 .55×10 -3 s = 0.04 μs ∴ϕ tends to become zero at high frequencies. ∴At very high frequencies, capacitor C acts as a conductor of negligible capacitive reactance. In a dc circuit, after the steady state, ω = 0. ∴ X C = 1 ωC →∞ ∴Capacitor C amounts to an open circuit.$

Q.17 Keeping the source frequency equal to the resonating frequency of the series LCR circuit, if the three elements, L, C and R are arranged in parallel, show that the total current in the parallel LCR circuit is minimum at this frequency. Obtain the current rms value in each branch of the circuit for the elements and source specified in Exercise 7.11 for this frequency.

Ans.

$\begin{array}{l}\text{An inductor}\left(\text{L}\right),\text{a capacitor}\left(\text{C}\right),\text{and a resistor}\left(\text{R}\right)\text{is}\\ \text{connected in parallel with each other in a circuit where},\\ \text{Inductance of inductor L}=\text{5}.0\text{H}\\ \text{Capacitance of capacitor C}=\text{8}0\text{}\mathrm{\mu }\text{F}\\ =\text{8}0×\text{1}{0}^{-\text{6}}\text{F}\\ \text{Resistance of resistor R}=\text{4}0\text{}\mathrm{\Omega }\\ \text{Potential of voltage source},\text{}\\ \text{V}=\text{23}0\text{V}\\ \text{The relation for }\mathrm{i}\text{mpedance}\left(\text{Z}\right)\text{of the given parallel}\\ \text{LCR circuit is given as}:\\ \frac{1}{\mathrm{Z}}=\sqrt{\frac{1}{{\mathrm{R}}^{2}}+{\left(\frac{1}{\mathrm{\omega L}}-\mathrm{\omega C}\right)}^{2}}\\ \text{Here},\\ \mathrm{\omega }=\text{Angular frequency}\\ \text{At resonance},\\ \frac{1}{\mathrm{\omega L}}-\mathrm{\omega C}=0\\ \therefore \mathrm{\omega }=\frac{1}{\sqrt{\mathrm{LC}}}\\ =\frac{1}{\sqrt{5\text{H}×80×{10}^{-6}\text{}\mathrm{F}}}\\ =50\text{ }{\mathrm{rads}}^{-1}\\ \therefore \text{Z is the maximum at 5}0{\text{rads}}^{\text{-1}}.\text{}\\ \therefore \mathrm{T}\text{otal current is minimum}.\\ \text{Rms current flowing through inductor L is given by,}\\ {\text{I}}_{\text{L}}\text{=}\frac{\mathrm{V}}{\mathrm{\omega L}}\\ =\frac{230\text{}\mathrm{V}}{50{\text{rads}}^{\text{-1}}×5\text{H}}\\ =0.92\text{ }\mathrm{A}\\ \text{Rms current flowing through capacitor C is given by,}\\ {\text{I}}_{\text{C}}=\frac{\mathrm{V}}{\frac{1}{\mathrm{\omega C}}}\\ =\mathrm{\omega CV}\\ =50{\text{rads}}^{\text{-1}}×80×{10}^{-6}\text{F}×230\text{V}\\ =0.92\text{ }\mathrm{A}\\ \text{Rms current flowing through resistor R is given by,}\\ {\text{I}}_{\mathrm{R}}=\frac{\mathrm{V}}{\mathrm{R}}\\ =\frac{230\text{}\mathrm{V}}{40\text{}\mathrm{\Omega }}\\ =5.75\text{ }\mathrm{A}\end{array}$

Q.18 A circuit containing a 80 mH inductor and a 60 μF capacitor in series is connected to a 230 V, 50 Hz supply. The resistance of the circuit is negligible.
(a) Obtain the current amplitude and rms values.
(b) Obtain the rms values of potential drops across each element.
(c) What is the average power transferred to the inductor?
(d) What is the average power transferred to the capacitor?
(e) What is the total average power absorbed by the circuit? [‘Average’ implies ‘averaged over one cycle’.]

Ans.

$\begin{array}{l}\text{Here, Inductance},\text{L}=\text{8}0\text{mH}\\ =\text{8}0×\text{1}{0}^{-\text{3}}\text{H}\\ \text{Capacitance},\text{C}=\text{6}0\text{}\mathrm{\mu }\text{F}\\ =\text{6}0×\text{1}{0}^{-\text{6}}\text{F}\\ \text{Supply voltage},\text{V}=\text{23}0\text{V}\\ \text{Frequency},\text{}\mathrm{\nu }=\text{5}0\text{Hz}\\ \text{Angular frequency},\text{}\mathrm{\omega }=\text{2}\mathrm{\pi \nu }\\ =\text{1}00\text{}\mathrm{\pi }{\text{rads}}^{\text{-1}}\\ \text{Peak voltage is given by},{\text{V}}_{0}=\text{V}\sqrt{2}\\ =\text{23}0\sqrt{2}\text{ V}\\ \left(\text{a}\right)\text{Maximum current is given by the relation,}\\ {\text{I}}_{\text{0}}=\frac{{\mathrm{V}}_{0}}{\left(\mathrm{\omega L}-\frac{1}{\mathrm{\omega C}}\right)}\\ =\frac{\text{23}0\sqrt{2}}{\left(\text{1}00\mathrm{\pi }{\text{rads}}^{\text{-1}}×\text{8}0×\text{1}{0}^{-\text{3}}\mathrm{H}-\frac{1}{\text{1}00\text{}\mathrm{\pi }×\text{6}0×\text{1}{0}^{-\text{6}}\mathrm{L}}\right)}\\ =\frac{\text{23}0\sqrt{2}\text{V}}{\left(8\mathrm{\pi }-\frac{1000}{6\mathrm{\pi }}\right)\text{}\mathrm{\Omega }}\\ =-11.63\text{ }\mathrm{A}\\ \text{Negative sign appears because }\mathrm{\omega L}<\frac{1}{\mathrm{\omega C}}\\ \text{Amplitude of the maximum current},\left|{\text{I}}_{\text{0}}\right|=11.63\text{ }\mathrm{A}\\ \therefore \text{Rms value of current},\text{ }\mathrm{I}=\frac{{\text{I}}_{\text{0}}}{\sqrt{2}}\\ =\frac{-11.63\text{A}}{\sqrt{2}}\\ =-\text{ }8.22\text{ }\mathrm{A}\\ \left(\text{b}\right)\text{Potential difference across the inductor is given by},{\text{\hspace{0.17em}\hspace{0.17em}V}}_{\text{L}}=\text{I}×\mathrm{\omega }\text{L}\\ =\text{8}.\text{22 A}×\text{1}00\text{}\mathrm{\pi }{\text{rads}}^{\text{-1}}×\text{8}0×\text{1}{0}^{-\text{3}}\text{H}\\ =\text{2}0\text{6}.\text{61 V}\\ \text{Potential difference across the capacitor is given by},\\ {\text{V}}_{\mathrm{C}}=\text{I}×\frac{1}{\mathrm{\omega C}}\\ =8.22×\frac{1}{\text{1}00\text{}\mathrm{\pi }{\text{rads}}^{\text{-1}}×\text{6}0×\text{1}{0}^{-6}\text{F}}\\ =436.3\text{ }\mathrm{V}\\ \left(\text{c}\right)\text{Average power transferred over a complete cycle }\\ \text{by the source to inductor is zero as actual voltage}\\ \text{leads the current by}\frac{\mathrm{\pi }}{2}\text{ }\mathrm{in}\text{ }\mathrm{the}\text{ inductor }.\\ \left(\text{d}\right)\text{Average power transferred over a complete cycle }\\ \text{by the source to the capacitor is zero as voltage lags}\\ \text{current by}\frac{\mathrm{\pi }}{2}\text{ }\mathrm{in}\text{ }\mathrm{the}\text{ }\mathrm{capacitor}.\\ \left(\text{e}\right)\text{The }\mathrm{t}\text{otal average power absorbed by the circuit is}\\ \text{zero}.\end{array}$

Q.19 Suppose the circuit in Exercise 7.18 has a resistance of 15 Ω. Obtain the average power transferred to each element of the circuit, and the total power absorbed.

Ans.

$Here, inductance of the inductor, L = 80 mH = 80 × 10 -3 H Capacitance of the capacitor, C = 60 μF =60×10 -6 F Resistance of the resistor, R = 15 Ω Potential of the voltage supply, V = 230 V Frequency of the signal, ν = 50 Hz Angular frequency of the signal, ω = 2πν =2π× 50 Hz =100π rads -1 Average power transferred to resistor=788.44 W Average power transferred to capacitor=0 W Total power absorbed by circuit=788.44 W As elements are connected in series to each other, ∴Impedance of the circuit is given as: Z= R 2 + ωL− 1 ωC 2 = 15 2 + 100π 80 × 1 0 −3 H − 1 100π×60×1 0 −6 F 2 = 15 Ω 2 + 25.12 Ω−53.08 Ω 2 =31.728 Ω Current flowing in the circuit is given by, I= V Z = 230 V 31.728 Ω =7.25 A Average power transferred to resistance is given by the relation, P R = I 2 R = 7.25 A 2 ×15 Ω =788.44 W Average power transferred to capacitor, P C =0 Average power transferred to inductor, P L = 0 Total power absorbed per cycle by the circuit: = P R + P C + P L =788.44+0+0 =788.44 W Thus, the total power absorbed by the circuit is 788.44 W.$

Q.20 A series LCR circuit with L = 0.12 H, C = 480 nF, R = 23 Ω is connected to a 230 V variable frequency supply.
(a) What is the source frequency for which current amplitude is maximum. Obtain this maximum value.
(b) What is the source frequency for which average power absorbed by the circuit is maximum. Obtain the value of this maximum power.
(c) For which frequencies of the source is the power transferred to the circuit half the power at resonant frequency? What is the current amplitude at these frequencies?
(d) What is the Q-factor of the given circuit?

Ans.

$\begin{array}{l}\text{Here, Inductance},\text{L}=0.\text{12 H}\\ \text{Capacitance},\text{C}=\text{48}0\text{nF}=\text{48}0×\text{1}{0}^{-\text{9}}\text{F}\\ \text{Resistance},\text{R}=\text{23}\mathrm{\Omega }\\ \text{Supply voltage},\text{V}=\text{23}0\text{V}\\ \text{Peak voltage is given by, }\\ {\text{V}}_{0}=\sqrt{2}\text{V =}\sqrt{2}×\text{23}0\text{V}\\ =\text{325}.\text{22 V}\\ \left(\text{a}\right)\text{Current flowing in the circuit is given by},\\ {\mathrm{I}}_{0}=\frac{{\mathrm{V}}_{0}}{\sqrt{{\mathrm{R}}^{2}+{\left(\mathrm{\omega L}-\frac{1}{\mathrm{\omega C}}\right)}^{2}}}\\ \text{Here},{\text{I}}_{0}\text{}=\text{maximum at resonance}\\ \text{At resonance},\\ {\mathrm{\omega }}_{\mathrm{R}}\mathrm{L}-\frac{1}{{\mathrm{\omega }}_{\mathrm{R}}\mathrm{C}}=0\\ \text{Here},{\mathrm{\omega }}_{\text{R}}\text{}=\text{Resonance angular frequency}\\ \therefore {\mathrm{\omega }}_{\mathrm{R}}=\frac{1}{\sqrt{\mathrm{LC}}}\\ =\frac{1}{\sqrt{0.12\text{}\mathrm{H}×480×{10}^{-9}\mathrm{F}}}\\ =4166.67\text{ }{\mathrm{rads}}^{-1}\\ \therefore \text{Resonant frequency is given by the relation,}\\ {\mathrm{\nu }}_{\text{R}}\text{=}\frac{{\mathrm{\omega }}_{\mathrm{R}}}{2\mathrm{\pi }}\\ =\frac{4166.67\text{}{\mathrm{rads}}^{-1}}{2×3.14}\\ =663.48\text{ }\mathrm{Hz}\\ \text{Maximum current is given by the relation,}\\ {\left({\mathrm{I}}_{0}\right)}_{\mathrm{max}}=\frac{{\mathrm{V}}_{0}}{\mathrm{R}}\\ =\frac{325.22\mathrm{V}}{23\text{}\mathrm{\Omega }}\\ =14.14\text{ }\mathrm{A}\\ \left(\text{b}\right)\text{Maximum average power absorbed by the circuit}\\ \text{is given by the relation,}\\ {\left({\mathrm{P}}_{\mathrm{av}}\right)}_{\mathrm{max}}=\frac{1}{2}{{\left({\mathrm{I}}_{0}\right)}^{2}}_{\mathrm{max}}\mathrm{R}\\ =\frac{1}{2}×{\left(14.14\mathrm{A}\right)}^{2}×23\mathrm{\Omega }\\ =2299.3\text{ }\mathrm{W}\\ \therefore \text{Resonant frequency}\left({\mathrm{\nu }}_{\text{R}}\right)\text{is }663.48\text{ }\mathrm{Hz}.\\ \left(\text{c}\right)\text{Power transferred to the circuit is half at resonant}\\ \text{frequency, when}\\ \mathrm{\Delta \omega }\text{=}\frac{\mathrm{R}}{2\mathrm{L}}\\ =\frac{23\text{}\mathrm{\Omega }}{2×0.12\text{H}}\\ =95.83\text{ }{\mathrm{rads}}^{-1}\\ \therefore \text{Frequencies at which power transferred is half}\\ \text{=}{\mathrm{\omega }}_{\text{r}}±\mathrm{\Delta \omega }\\ =4166.7\text{}{\mathrm{rads}}^{-1}±95.83\text{}{\mathrm{rads}}^{-1}\\ =4262.53\text{ }{\mathrm{rads}}^{-1}\text{}\mathrm{and}\text{ }4070.87\text{ }{\mathrm{rads}}^{-1}\\ \mathrm{At}\text{ }\mathrm{these}\text{ }\mathrm{frequencies},\text{ }\mathrm{current}\text{ }\mathrm{amplitude}\text{ }\mathrm{is}\text{ }\mathrm{given}\text{ }\mathrm{by},\\ \mathrm{I}‘=\frac{1}{\sqrt{2}}×{\left({\mathrm{I}}_{0}\right)}_{\mathrm{max}}\\ =\frac{14.14\text{}\mathrm{A}}{\sqrt{2}}\\ =10\text{ }\mathrm{A}\\ \left(\text{d}\right)\text{Q}-\text{factor of the given circuit is given by the }\\ \text{relation},\mathrm{Q}=\frac{{\mathrm{\omega }}_{\mathrm{R}}\mathrm{L}}{\mathrm{R}}\\ =\frac{4166.7\text{}{\mathrm{rads}}^{-1}×0.12\text{H}}{23\text{}\mathrm{\Omega }}\\ =\text{21}.\text{74}\\ \therefore \text{The Q}-\text{factor of the given circuit is 21}.\text{74}.\end{array}$

Q.21 Obtain the resonant frequency and Q-factor of a series LCR circuit with L = 3.0 H, C = 27 μF, and R = 7.4 Ω. It is desired to improve the sharpness of the resonance of the circuit by reducing its ‘full width at half maximum’ by a factor of 2. Suggest a suitable way.

Ans.

$Here, inductance, L = 2.0 H Capacitance,C = 32 μF = 32 × 10 -6 F Resistance, R = 10 Ω Resonant frequency can be obtained by the relation, ω r = 1 LC = 1 2 H × 32 × 10 -6 F = 1 8×10 -3 s -1 = 125 s -1 Now, Q-value of the circuit is given by: Q = 1 R L C Q = 1 10 2H 32 × 10 -6 F = 1 10 × 1 4×10 -3 = 25 ∴The Q-Value of this circuit is 25.$

Q.22 A power transmission line feeds input power at 2300 V to a step-down transformer with its primary windings having 4000 turns. What should be the number of turns in the secondary in order to get output power at 230 V?

Ans.

$Here, Input voltage, V i = 2300 V Number of turns in primary coil, n p = 4000 Output voltage, V o = 230 V Let number of turns in secondary coil = n s The relation between voltage and number of turns is given as: V i V o = n p n s ∴ 2300 V 230 V = 4000 n s ∴ n s = 4000 × 230 2300 = 400 ∴There are 400 turns in the secondary coil.$

(a) In any ac circuit, is the applied instantaneous voltage equal to the algebraic sum of the instantaneous voltages across the series elements of the circuit? Is the same true for rms voltage?

(b) A capacitor is used in the primary circuit of an induction coil.

(c) An applied voltage signal consists of a superposition of a dc voltage and an ac voltage of high frequency. The circuit consists of an inductor and a capacitor in series. Show that the dc signal will appear across C and the ac signal across L.

(d) A choke coil in series with a lamp is connected to a dc line. The lamp is seen to shine brightly. Insertion of an iron core in the choke causes no change in the lamp’s brightness. Predict the corresponding observations if the connection is to an ac line.

(e) Why is choke coil needed in the use of fluorescent tubes with ac mains? Why can we not use an ordinary resistor instead of the choke coil?

Ans.

(a) Yes; the statement is not true for rms voltage.

It is true that in any ac circuit, the applied instantaneous voltage is equal to the algebraic sum of the instantaneous voltages across the series elements of the circuit. But, this is not true for rms voltage because such voltages across different elements may not be in same phase.

(b) The capacitor is used in the primary circuit of an induction coil. This is because, when a circuit is broken, a high induced voltage is used to charge the capacitor to avoid sparking.

(c) The dc signal will appear across capacitor C because for dc signals, the impedance of an inductor (L) is very low while the impedance of a capacitor (C) is very high (almost infinite). The capacitor blocks the dc signal. Therefore, a dc signal appears across C. For an ac signal of high frequency, the impedance of L is high and that of C is negligible. The inductor blocks the ac signal of high frequency. Therefore, an ac signal of high frequency appears across L.

(d) On a dc line, choke offers no impedance. Therefore, the lamp shines brightly and the insertion of an iron core in the choke causes no change in the lamp’s brightness. However, on an ac line, both the choke coil and the iron core increase the impedance of the circuit. Therefore, if an iron core is inserted in the choke coil (which is in series with a lamp connected to the ac line), the lamp will glow dimly.

(e) A choke coil is needed in the use of fluorescent tubes with ac mains because it reduces ac across the tube without much loss of power. We cannot use an ordinary resistor instead of a choke coil for this purpose because it wastes power in the form of heat.

Q.24 At a hydroelectric power plant, the water pressure head is at a height of 300 m and the water flow available is 100 m3s–1. If the turbine generator efficiency is 60%, estimate the electric power available from the plant (g = 9.8 ms–2).

Ans.

$\begin{array}{l}\text{Here, height of water pressure head, h = 300 m}\hfill \\ {\text{Volume of the water flow per second, V = 100 m}}^{\text{3}}{\text{s}}^{\text{-1}}\hfill \\ \text{Efficiency of the turbine generator, η = 60%}\hfill \\ \text{= 0}\text{.6}\hfill \\ \text{Acceleration due to gravity, g = 9}{\text{.8 ms}}^{\text{-2}}\hfill \\ {\text{Density of the water, ρ = 10}}^{\text{3}}{\text{kgm}}^{\text{-3}}\hfill \\ \text{Hydroelectric power =}\frac{\text{work}}{\text{time}}\hfill \\ \text{=}\frac{\text{Force × displacement}}{\text{time}}\hfill \\ \text{= Force × Velocity}\hfill \\ \text{=}\left(\text{Pressure × Area}\right)\text{× Velocity}\hfill \\ {\text{As, area×velocity = volumesec}}^{\text{-1}}\hfill \\ \text{= V}\hfill \\ \therefore \text{Hydroelectric power = P × V}\hfill \\ \text{=}\left(\text{hρg}\right)\text{× V}\hfill \\ \text{= hρgV}\hfill \end{array}$

Q.26 A small town with a demand of 800 kW of electric power at 220 V is situated 15 km away from an electric plant generating power at 440 V. The resistance of the two wire line carrying power is 0.5 Ω per km. The town gets power from the line through a 4000-220 V step-down transformer at a sub-station in the town.
(a) Estimate the line power loss in the form of heat.
(b) How much power must the plant supply, assuming there is negligible power loss due to leakage?
(c) Characterise the step up transformer at the plant.

Ans.

$\begin{array}{l}{\text{Here, total requirement of electric power, P = 800 kW = 800×10}}^{\text{3}}\text{W}\\ \text{Supply voltage, V = 220 V}\\ \text{Voltage at which the electric plant is generating power, V’ = 440 V}\\ \text{Distance between power generating station and the town, d =15 km}\\ \text{Resistance of the two wire lines carrying power = 0.5 Ω/km}\\ \text{Total resistance of the two wires, R = 2×15 km × 0.5Ω/km =15 Ω}\\ \text{As supply is through a step-down transformer of}\\ \text{rating 4000 – 220 V,}\\ \therefore {\text{Input voltage, V}}_{\text{i}}\text{= 4000 V}\\ {\text{Output voltage, V}}_{\text{o}}\text{= 220 V}\\ \text{Rms current in the wire lines, I =}\frac{\text{P}}{{\text{V}}_{\text{i}}}\text{=}\frac{{\text{800×10}}^{\text{3}}\text{}\mathrm{W}}{\text{4000 V}}\text{= 200 A}\\ \left(\text{a}\right){\text{\hspace{0.17em}\hspace{0.17em}Line power loss = I}}^{\text{2}}\text{R}\\ \text{=}{\left(\text{200A}\right)}^{\text{2}}\text{×15}\mathrm{\Omega }{\text{= 600×10}}^{\text{3}}\text{W = 600 kW}\\ \text{}\left(\text{b}\right)\text{\hspace{0.17em}\hspace{0.17em}Assuming that there is negligible power loss due to leakage of current:}\\ \text{Essential power supplied by the electric power plant}\\ \text{= 800 kW+600 kW =1400 kW}\\ \left(\text{c}\right)\text{Voltage drop on the line = IR}\\ \text{=200 A ×15}\mathrm{\Omega }\text{=3000 V}\\ \therefore \text{Voltage transmitted from the power plant}\\ \text{= 3000 V + 4000 V =7000 V}\\ \text{As the power is generated at 440 V,}\\ \therefore \text{The rating of the step-up transformer situated at the power plant is 440 V-7000V.}\end{array}$

Q.27 Do the same exercise as above with the replacement of the earlier transformer by a 40,000-220 V step-down transformer (Neglect, as before, leakage losses though this may not be a good assumption any longer because of the very high voltage transmission involved). Hence, explain why high voltage transmission is preferred?

Ans.

$\begin{array}{l}\text{Here, }\mathrm{t}\text{he rating of a step}-\text{down transformer is}\\ \text{4}0000\text{ V}-\text{22}0\text{V}.\\ \text{Input voltage},\text{}\\ {\text{V}}_{\mathrm{i}}=\text{4}0000\text{V}\\ \text{Output voltage},\text{}\\ {\text{V}}_{\mathrm{o}}=\text{22}0\text{V}\\ \text{Total requirement of the electric power},\text{P}=\text{8}00\text{kW}\\ =\text{8}00×\text{1}{0}^{\text{3}}\text{W}\\ \text{Source voltage},\text{V}=\text{22}0\text{V}\\ \text{Voltage at which the electric plant is generating power},\text{}\\ \text{V}‘=\text{44}0\text{V}\\ \text{Distance between power generating station and the}\\ \text{town},\text{d}=\text{15 km}\\ \text{The }\mathrm{r}\text{esistance of the two wire lines carrying power}\\ =0.\text{5}\mathrm{\Omega }{\text{km}}^{\text{-1}}\\ \text{Total resistance of wire lines},\text{}\\ \text{R}=\text{2}×\text{15 km}×0.\text{5}\mathrm{\Omega }{\text{km}}^{\text{-1}}\\ \mathrm{R}=\text{15}\mathrm{\Omega }\\ \text{P}={\text{V}}_{\mathrm{i}}\text{I}\\ \\ \text{Rms current in the wire line, I=}\frac{\mathrm{P}}{{\text{V}}_{\mathrm{i}}}\\ =\frac{\text{8}00×\text{1}{0}^{\text{3}}\mathrm{W}}{\text{4}0000\text{}\mathrm{V}}\\ =20\text{ }\mathrm{A}\\ \left(\text{a}\right)\text{Line power loss}={\text{I}}^{\text{2}}\text{R}\\ ={\left(\text{2}0\right)}^{\text{2}}×\text{15}\\ =\text{6 kW}\\ \left(\text{b}\right)\text{Assuming that there is negligible power loss due to}\\ \text{the leakage of current}.\\ \therefore \mathrm{P}\text{ower supplied by the plant}=\text{8}00\text{kW}+\text{6 kW}\\ =\text{8}0\text{6 kW}\\ \left(\text{c}\right)\text{Voltage drop on the line}=\text{IR}\\ =\text{2}0\text{}\mathrm{A}×\text{15}\mathrm{\Omega }\text{}\\ =\text{3}00\text{V}\\ \therefore \text{Voltage transmitted by the power plant}\\ =\text{3}00\text{}\mathrm{V}+\text{4}0000\text{}\mathrm{V}\text{}\\ =\text{4}0\text{3}00\text{V}\\ \text{As }\mathrm{t}\text{he power is generated at 44}0\text{V,}\\ \therefore \text{The rating of the step}-\text{up transformer needed}\\ \text{at the plant is 44}0\text{V}-\text{4}0\text{3}00\text{V}.\\ \therefore \text{Power loss during transmission}=\frac{6\text{}\mathrm{kW}}{806\text{kW}}×100\\ =0.744\text{}\mathrm{%}\\ \text{In the previous exercise},\text{the power loss due to the}\\ \text{same reason =}\frac{600\text{}\mathrm{kW}}{1400\text{kW}}×100\\ =42.8\text{}\mathrm{%}\\ \\ \therefore \text{High voltage transmission is preferred for this purpose.}\\ \end{array}$