INSTITUTE OF DISTANCE EDUCTION

1

THE UNIVERSITY OF ZAMBIA

INSTITUTE OF DISTANCE EDUCTION

BACHELOR OF SCIENCE

With Education (BSc.Ed)

THE UNIVERSITY OF ZAMBIA

INSTITUTE OF DISTANCE EDUCATION (IDE)

BACHELOR OF SCIENCE

With Education (BSc.Ed)

(MODULE 3)

PHY1010: INTRODUCTORY PHYSICS

2

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All rights reserved. No part of this publication may be reproduced, stored in a retrieval

system, or transmitted in any form or by any means, electronic or mechanical, including

photocopying, recording or otherwise without the permission of the University of Zambia,

Institute of Distance Education.

Inquiries concerning reproduction or rights and requests for additional training materials

should be addressed to:

The Director,

Institute of Distance Education

The University of Zambia

P.O. Box 32379

Lusaka

Zambia

Tel: +211 290719

Fax: +211 253952

E-mail: [email protected]

Website: www.unza.zm

3

TABLE OF CONTENTS

ACKNOWLEDGEMENTS ................................................................................................ 6

MODULE STRUCTURE ................................................................................................... 7

INTRODUCTION .............................................................................................................. 8

AIM ..................................................................................................................................... 9

SPECIFIC OBJECTIVES ................................................................................................... 9

ASSESSMENT ................................................................................................................... 9

PRESCRIBED READINGS ............................................................................................. 10

RECOMMENDED READINGS ...................................................................................... 10

TIME FRAME .................................................................................................................. 10

STUDY SKILLS ............................................................................................................... 10

NEED HELP? ................................................................................................................... 11

UNIT 3.1........................................................................................................................... 12

ELECTRIC FORCES AND FIELDS ................................................................................... 12

3.1.1 Introduction...................................................................................................... 12

3.1.2 Learning Objectives ......................................................................................... 12

3.1.3 Coulomb’s Law and Unit of Charge ............................................................... 12

3.1.4 The Electric Field Strength ............................................................................. 14

3.1.5 Electric Flux and Gauss’ Flux Theorem ......................................................... 17

UNIT 3.2........................................................................................................................... 21

ELECTRIC POTENTIAL ......................................................................................... 21

3.2.1 Introduction...................................................................................................... 21


3.2.3 Electric Potential Energy and Potential Difference ........................................ 21

3.2.4 Equipotentials and The Electronvolt ............................................................... 23

3.2.5 Absolute Potentials .......................................................................................... 25

3.2.6 Capacitors in Parallel and Series ................................................................... 27

3.2.7 Dielectrics ........................................................................................................ 28

3.2.8 Energy Storage in Capacitors and Energy Density ......................................... 29

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UNIT 3.3........................................................................................................................... 33

DIRECT CURRENT CIRCUITS (D.C) ................................................................... 33

3.3.1 Introduction...................................................................................................... 33


3.3.3 Conventional Current and Ohm’s Law ............................................................ 33

3.3.4 Resistivity and its Temperature Dependence ................................................... 35

3.3.5 Resistors in Series and Parallel ...................................................................... 36

3.3.6 Kirchhoff’s Laws and Applications ................................................................. 37

3.3.7 Emf and Terminal Potencial Difference ......................................................... 40

UNIT 3.4........................................................................................................................... 43

MAGNETISM ............................................................................................................ 43

3.4.1 Introduction...................................................................................................... 43


3.4.3 Force on a Conductor Carrying Current in a Magnetic Field ....................... 43

3.4.4 Magnetic Forces on Moving Charges ............................................................. 46

3.4.5 Motion of a Charge in a Magnetic Field ........................................................ 47

3.4.6 Magnetic Field of a Current in (i) a long straight wire, (ii) a circular loop,

(iii) a long solenoid ..……………………………………………………………….50

3.4.7 The Definition of an Ampere ........................................................................... 51

3.4.8 The Torque on a Current Loop ....................................................................... 53

3.4.9 Moving Coil Meters: (i) Ammeters, and (ii) Voltmeters ................................. 55

UNIT 3.5 .......................................................................................................................... 58

ELECTROMAGNETIC INDUCTION .................................................................... 58

3.5.1 Introduction...................................................................................................... 58


3.5.3 Magnetic Flux, ϕ .............................................................................................. 58

3.5.4 Faraday’s Law of Induced EMF ..................................................................... 60

3.5.5 Mutual Inductance .......................................................................................... 61

3.5.6 Self - Inductance .............................................................................................. 62

3.5.7 Inductance – Resistance Circuits .................................................................... 63

3.5.8 Energy Stored in a Magnetic Field ................................................................. 65

3.5.9 Motional Electro-Motive Force ...................................................................... 66

3.5.10 Transformers .................................................................................................. 68

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UNIT 3.6........................................................................................................................... 72

ALTERNATING CURRENT CIRCUITS ................................................................ 72

3.6.1 Introduction...................................................................................................... 72


3.6.3 Charging and Discharging a Capacitor ......................................................... 72

3.6.4 Root Mean Square (rms) Values ..................................................................... 74

3.6.5 Resistance Circuits .......................................................................................... 76

3.6.6 Capacitance Circuits ....................................................................................... 77

3.6.7 Inductance Circuits .......................................................................................... 79

3.6.8 Series LRC Circuits ......................................................................................... 80

3.6.9 Electrical Resonance ....................................................................................... 84

UNIT 3.7........................................................................................................................... 87

REFLECTION AND REFRACTION OF LIGHT ................................................... 87

3.7.1 Introduction...................................................................................................... 87


3.7.3 The Concept of light ........................................................................................ 87

3.7.4 Focal Length of Spherical Mirrors ................................................................. 88

3.7.5 Mirror Equation and Magnification ............................................................... 90

3.7.6 Refraction of Light and Snell’s Law ................................................................ 92

3.7.7 Total-Internal Reflection .................................................................................. 95

3.7.8 Image Formation by Lenses ............................................................................ 95

3.7.9 Thin Lens Formula and Magnification ............................................................ 97

3.7.10 Combination of Lenses and Lenses in Close Contact .................................... 99

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ACKNOWLEDGEMENTS

The University of Zambia (UNZA), Institute of Distance Education (IDE) wishes to

thank DR. SYLVESTER HATWAAMBO for writing this

MODULE 3: ELECTRICITY, MAGNETISM AND LIGHT as part of the

PHY 1010: INTRODUCTORY PHYSICS course.

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MODULE STRUCTURE

I. Introduction

II. The Aim of the Module

III. Module Objectives [Learning outcomes]

IV. Assessment

V. Prescribed and Recommended Readings

VI. Time frame

VII. Study skills [Learning tips]

VIII. Need help [Studying at a distance]

This module is divided into seven (7) units. Each unit addresses some of the learning

outcomes. You will be asked to complete various tasks so that you can demonstrate your

competence in achieving the learning outcomes.

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INTRODUCTION

Traditionally, PHY 1010: Introductory Physics course has been divided into three modules

namely: Module 1; Mechanics, Module 2; Mechanical and thermal properties of matter and

oscillations, and Module 3; Electricity, magnetism and Light. You have already studied

modules 1 and 2. We now welcome you to Module 3, the last part of the Introductory

Physics (PHY 1010) course offered as a foundation course for all students in the School of

Natural Sciences (BSc NQS) and those from the School of Education following a Bachelor

of Science with Education programme (BSC. Ed).

How to study physics

Studying physics may seem to be quite daunting at first. However, the whole exercise

becomes quite easy when one learns to ‘marry’ concepts from one area of the subject to

another. In other words, one must not learn the various concepts in isolation. Some

concepts such as resonance cut across all areas. It (resonance) may look quite different

when it appears in mechanics than when it appears in electric circuits. Careful observation

will reveal that the two are one and the same thing.

In this module we assume that you have already gone through modules 1 and 2 thoroughly

before embarking on this particular module. Hence, pause a little and reflect what you did

in the last two modules. Some of the concepts and terminologies will have been dealt with

already.

In particular, this module attempts to introduce you to the concepts of electric forces and

fields, electric potential, direct current circuits, magnetism, electromagnetic induction,

alternating circuits and reflection and refraction of light in that order. A number of worked

examples have been included to help your understanding on how to solve problems.

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Aim

The aim of this module is to introduce you to the lifeblood of modern technological

civilization brought about the discovery of electricity. Most household appliances such as

telephones, televisions, computers etc would not have been possible without the discovery

of electrons. Furthermore, the concepts of reflection and refraction of light are very

important in a variety of applications such as in sight, energy transmission, telescopes, etc

would not have been possible without light.

Specific Objectives

By the end of this module, you should be able to:

i. State Coulomb’s law and solve problems in electric forces and fields

ii. Define electric potential energy and electric potential difference and solve

problems associated with electrical potential

iii. Define Ohm’s law, resistivity, emf and terminal potential difference and apply

the Kirchhoff’s Voltage Law (KVL) and Kirchhoff’s Current Law (KCL) to

solve for currents and voltages in circuit networks

iv. Describe a force on a conductor carrying current and explain the principle of

moving coil meters

v. Analyze electromagnetic induction

vi. Solve problems in alternating current circuits

vii. Solve problems on reflection and refraction of light.

ASSESSMENT

Continuous Assessment (CA): 50%

2 tests – 30%

Assignments – 5%

Lab. Practical – 15%

Final Examination: 50%

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Prescribed Readings

1. Raymond A. Serway, Chris Vuille (2012) Introductory Physics, 9th edition, New

York: Harcourt, Brace, Jovanovich.

Recommended Readings

1. Frederick J. Bueche, David A. Jerde (1995), Principles of Physics, 10th edition,

McGraw-Hill, Inc.

2. Hugh D. Young and Roger A. Freedman, Sears and Zemansky (2007), University

Physics (with Modern Physics) 12th edition, Pearson/Addison Wesley

3. Halliday D. and Resnick R., Physics Parts I & II, 3rd Edition, John Wiley and Sons,

1978.

Time frame

You are expected to spend at least 60 hours of study time on this module. In addition, there

shall be arranged contacts with lecturers from the University from time to time during the

course. You are requested to spend your time judiciously so that you reap maximum benefit

from the course.

Study Skills

You may not have studied by distance education before especially in science subjects. Here

are some simple tips for you to follow which will help you do better in your learning and

keep you focused-

1. Set goals such as: I will succeed in this course. You might not have time to do a

full lesson in one night, so plan how much you can do, then stick to it until you are

done.

2. Establish a regular study/learning schedule

3. Determine what time is best for you to study

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4. Have a dedicated study place with all the supplies you might need

5. Tell people what you are doing because only then are you more likely to stick to a

course.

6. If you do not understand something ask your local learning center or your tutor,

who will be able to help you.

7. Search for the meaning of principles and concepts instead of just memorizing

Need help?

In case you have difficulties during the duration of the course, please get in touch with the

Director, Institute of Distance Education, or the resident lecturer in your province.

All enquiries in connection with the payment of fees should be directed to the Director,

Institute of Distance Education:

The Director,

Institute of Distance Education,

University of Zambia,

P. O. Box 32379,

10101 Lusaka

Coordinator, Learner Support Services (Mobile): +260 978772248

Senior Administrative Officer

(Programme Development & Production) +260 977639993

IDE Land Line: +260 211 290719

IDE Fax: +260 211 290719

IDE E-mail: [email protected]

http://www.unza.zm

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UNIT 3.1

ELECTRIC FORCES AND FIELDS

3.1.1 Introduction

We recall that an atom consists of the central nucleus and electrons around the nucleus.

The nucleus consists of protons and neutrons. The protons are bound to the nucleus and

electrons can easily be removed from the atom under certain conditions. For example, when

the negatively charged ebonite rod touches an uncharged metal ball, electrons are

conducted off the rod onto the ball. Some of the charging processes were demonstrated

during your high school days. In particular, charging by rubbing and charging by induction

were demonstrated to you. In this unit, you will be introduced to the Coulomb’s law and

the unit of charge, define and calculate the electric field experienced by point charges, and

describe Gauss’ flux theorem for calculating the electric flux.

3.1.2 Learning Objectives

By the end of this unit you should be able to;

1) Define Coulomb’s law and state the unit of charge

2) Use the superposition principle to calculate the electric force of point charges

3) Evaluate the electric field intensity of point charges

4) Describe electric flux and Gauss’s flux theorem

3.1.3 Coulomb’s Law and Unit of Charge

We are familiar that there are only two charges (positive and negative) and that like charges

repel each other and unlike charges attract each other. The magnitudes and directions of

these attractive and repulsive forces are governed by Coulomb’ law. Coulomb’ law states

that the force between any two charges is proportional to the product of the magnitudes of

the two charges and inversely proportional to the square of the distance between them.

Mathematically, the Coulomb’ law may be expressed as in figure 3.1.1.

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Figure 3.1.1: Two point charges placed at distance r from each other.

Thus, 𝑭 = 𝒌𝒒𝟏𝒒𝟐

𝒓𝟐 (𝑵). (3.1.1)

Where,

𝒌 =𝟏

𝟒𝝅𝜺𝒐≈ 𝟗 × 𝟏𝟎𝟗 𝑵.

𝒎𝟐

𝒄𝟐 . (3.1.2)

𝜀𝑜 = 8.85 × 10−12 𝒄𝟐

𝑁𝑚2 = permittivity of free space. (3.1.3)

The force also depends on the medium in which the charges are placed. The S.I unit for

electric charge is the coulomb “C”. The charge of an electron 𝒆 = −𝟏. 𝟔 × 𝟏𝟎−𝟏𝟗 𝑪.

When two charges are close to the third charge, each of the two charges exerts a force and

the total force on the third charge is simply the vector sum of the two separate forces. This

is called the superposition principle for coulomb’s law of forces.

Activity

A3.1.1: Find the force on the 4 μC charge shown below.

Solution: The force on q1 due to the presence charge of q2 is F2 and the charge on q1

due to the presence of q3 is F3 as shown below.

+q1 -q2

r

q1 = +4μC q2 = - 5μC q3 = + 3μC

2 m 4 m

14

𝐹2 =9 𝑥 109 (4 × 10−6)(5 × 10−6)

22 = 0.045 N to the right.

𝑭𝟑 =𝟗 × 𝟏𝟎𝟗 (𝟒 × 𝟏𝟎−𝟔)(𝟑 × 𝟏𝟎−𝟔)

𝟔𝟐 = -0.003 N to the left.

Hence, the net force will be

𝑭𝒏𝒆𝒕 = 𝑭𝟐 + 𝑭𝟑 = 𝟎. 𝟎𝟒𝟓 + (−𝟎. 𝟎𝟎𝟑) = 𝟒. 𝟐 × 𝟏𝟎−𝟐 𝑵 to the right.

3.1.4 The Electric Field Strength (𝑬)

The electric field is a region around a charge where the electric force can be experienced.

The strength of the field is very high near the charge and decreases as 1

𝑟2 as you will see

shortly.

We define the electric field strength E or electric field intensity due to a charge qt at a

distance r from it, as the force per unit positive charge qt.

𝑬 = 𝑭

𝑞𝑡 (N/C). (3.1.4)

The electric field strength E is a vector and is directed along the direction of the force. At

a distance r from a point charge q, the force on charge qt is 𝒌𝒒𝒒𝒕

𝒓𝟐 , then the magnitude

of the electric field strength E is

𝑬 = 𝑭

𝒒𝒕=

𝒌𝒒𝒒𝒕

𝒒𝒕𝒓𝟐= 𝒌

𝒒

𝒓𝟐 (N/C). (3.1.5)

F3 F2

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Activity

A3.1.2: Find the magnitude of E at point B shown in the figure below due to the two

point charges.

Solution: Resolving all the electric field strength at point B, we have,

𝑬𝟏 = 𝒌𝒒𝟏

𝒓𝟐=

𝟗 × 𝟏𝟎𝟗 × −𝟑.𝟔 × 𝟏𝟎−𝟔

(𝟎.𝟎𝟔)𝟐 = 9 x 106 N/C

𝑬𝟐 = 𝒌𝒒𝟐

𝒓𝟐=

𝟗 × 𝟏𝟎𝟗 × 𝟓 × 𝟏𝟎−𝟔

(𝟎.𝟏)𝟐 = 4.5 x 106 N/C

q2 =5 μC q1 = -3.6 μC

37 o

10 cm

B

6 cm

8 cm

q2 =5 μC q1 = -3.6 μC

37 o

8 cm

6 cm

10 cm

B

E1

E2

E2x

E2y

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𝑬𝟐𝒙 = 𝑬𝟐𝒄𝒐𝒔𝟑𝟕𝒐 = 3.6 x 106 N/C

𝑬𝟐𝒚 = 𝑬𝟐𝒔𝒊𝒏𝟑𝟕𝒐 = 2.7 x 106 N/C

∴ 𝐸𝑥 = 𝐸2𝑥 = 3.6 x 106 N/C

𝐸𝑦 = 𝐸1 + 𝐸2𝑦 = (9 − 2.7) × 106 = 6.3 x 106 N/C

Hence, magnitude 𝑬 = √𝑬𝒙𝟐 + 𝑬𝒚

𝟐 = √(𝟑. 𝟔 × 𝟏𝟎𝟔)𝟐 + (𝟔. 𝟑 × 𝟏𝟎𝟔)𝟐

= 7.26 x 106 N/C

Activity

A3.1.3: A charge q1 = 5 μC is placed at the origin while a charge of q2 = 10 μC is placed

at a distance of 50 cm from the origin along the positive x – axis. Find a point where the

electric field intensity E due to these two charges will be zero.

Solution: From simple inspection, the point will be located at a point x distance from the

test charge qt as shown on the x – axis below.

The resultant electric field intensity at x being zero means that,

𝑬 = 𝑬𝟏 + 𝑬𝟐 = 0

Thus, 𝑬𝟏 =𝑘𝑞1

𝑟2=

𝑘𝑞1

𝑥2 and

𝑬𝟐 =𝑘𝑞2

𝑟2=

𝑘𝑞2

(0.5−𝑥)2 .

q1 x (0.5-x)

qt

q2

x = 0 X = 50 cm E1 E2

x-axis

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∴ 𝒌𝒒𝟏

𝒙𝟐=

𝒌𝒒𝟐

(𝟎.𝟓−𝒙)𝟐 . Cross multiplying and substituting the values of 𝒒𝟏 and 𝒒𝟐, we

have (𝟓 × 𝟏𝟎−𝟔)(𝟎. 𝟓 − 𝒙)𝟐 = (𝟏𝟎 × 𝟏𝟎−𝟔 )( 𝒙𝟐)

Taking square root both sides, rearranging and solving for 𝒙 have,

𝟎. 𝟓 − 𝒙 = ±√𝟐 𝒙 → 𝒙 =𝟎.𝟓

𝟏+√𝟐 = 0.21 m. Or 𝒙 =

𝟎.𝟓

𝟏−√𝟐= -1.21 m

Since x cannot be negative, i.e it cannot be neither to the left of x = 0 nor to the right of

x = 50 cm, then x = 0.21 m or x = 21 cm.

3.1.5 Electric Flux and Gauss’s Flux Theorem:

The electric flux, ϕ is defined as the number of field lines that pass through a given area.

Consider the number of field lines passing through an area A shown in figure 3.1.2.

Figure 3.1.2: Electric flux through an area A

From the figure above, the electric flux ϕ = ∑ 𝑬. 𝑨 = ∑ 𝑬𝑨𝒄𝒐𝒔𝜽, where θ is the angle

between the area normal and the direction of the electric field intensity E. If the area normal

is perpendicular to the electric field strength, then no flux passes through the area. This was

E

A

θ

𝒏ෝ (area normal)

field lines

18

first demonstrated by Gauss and stated that the sum of all contributions of electric flux over

a closed surface is proportional to the total amount of charge enclosed by the surface.

Mathematically,

∑(𝐸˔𝛥𝐴) =𝑞(𝑒𝑛𝑐𝑙𝑜𝑠𝑒𝑑)

𝜀𝑜. (3.1.6)

This is known as the Gauss’s theorem. Note that if the surface encloses no charge, the

total flux through the surface will be zero and hence E = 0, as many field lines leave the

enclosed region as they enter it.

The Gauss’ flux theorem helps us to calculate the electric field strength for simple

geometries like spheres, lines or planes etc. We now examine the application of the

Gauss’ flux theorem to simple geometries such as a hollow sphere, and a long cylinder.

a) Spherical symmetry

Consider a hollow conducting sphere of charge q and radius r2 as shown in figure 3.1.3.

Figure 3.1.3: A hollow sphere of radius r2

It is easy to see that,

E =

𝟎 𝒓 < 𝒓𝟏𝒒

𝟒𝝅𝜺𝒐𝒓𝟐𝒓 > 𝒓𝟐

r1

q

r

r2

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Hint: Area of the sphere is 4𝜋𝑟2 .

b) Cylindrical Symmetry

Consider a straight rod of length l along which charge q is uniformly distributed as shown

in figure 3.1.4.

Figure 3.1.4: A straight line with a uniformly distributed charge.

We define a linear density λ (charge per unit length). Thus, q = λl. From symmetry

considerations;

1. ∑(𝑬˔𝜟𝑨) =0 at the ends of the cylinder since at these ends, E is not

perpendicular to the surface.

2. On the sides of the cylinder enclosing total charge q = λl we have

∑(𝐸𝛥𝐴) = E.2πrl = 𝑞

𝜀𝑜=

𝜆𝑙

ɛ𝑜 . (Note that 2πrl is the cross-sectional area of the

cylinder).

∴ 𝑬 = 𝝀

𝟐𝝅𝜺𝒐𝒓 (N/C).

E

Line of charge

Gaussian surface

r Etot

Δl

Δl

q= λΔl

q= λΔl

r

20

SAE3.1.1 Find the force on the 20 μC charge shown below.

Answers: The magnitude of force is 3.40 N and the direction is 64.9o or 65o.

SAE3.1.2 Two chares equal in magnitude q but of opposite sign are placed along the x -

axis at x = b and at x = -b. Show that the electric field due to these charges at a point on

the y – axis in a direction parallel to the x – axis and its magnitude is given by

𝑬 = 𝟐𝒌𝒒𝒃

(𝒚𝟐+𝒃𝟐)𝟑 𝟐⁄ .

SAE3.1.3 Two identical tiny metal balls have charges q1 and q2. The repulsive force one

exerts on the other when they are 20 cm apart is 1.35 x 10-4 N. After the balls are touched

together and then separated once again to 20 cm, the repulsive force is found to be 1.406

x 10 -4 N. Find the value of charges q1 and q2. (Answer: q1 = 20 x 10-9 C and q2 = 30 x 10-9

C or Vice versa).

SAE3.1.4 Sparking occurs through air when the electric field strength exceeds about 3 x

106 N/C. (This is the electric field strength of air). About how much charge can a 10.0

cm diameter metal sphere hold before sparking occurs? (Answer: 8.3 x10-7 C).

3.1 Self-Assessment Exercises

10 μC 4 μC

37 o

100 cm

20 μC

21

UNIT 3.2

ELECTRIC POTENTIAL

3.2.1 Introduction

In this unit, you will be introduced to the work done in moving a charge in an electric field.

The work done turns out to be the electrical potential energy of that charge. The concept

of equipotential lines will be explained and that the work done in carrying a test charge

from any point on the same equipotential line is zero. The concept of absolute potential

will be explained with respect to the work done in carrying a unit positive charge from

infinity to that point. The concept of capacitance of a capacitor will be explained both in

vacuum and in a dielectric material. Finally, you will also be shown how to derive an

expression for the energy stored in a capacitor.



1) Calculate the potential difference between two points

2) Describe the equipotential lines and that no work is done in moving a charge on

the same equipotential line

3) Calculate the absolute potential at a given point in the vicinity of point charges.

4) Calculate the capacitance of a parallel plate capacitor

5) Describe a dielectric material and show how it is able to increase the capacitance

of a capacitor

6) Calculate the energy stored in a capacitor

3.2.3 Electrical Potential Energy and Potential Difference

Consider a particle of positive charge q that is allowed to move in an electric field E

produced between two oppositely charged parallel plates shown in figure 3.2.1.

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Figure 3.2.1: Two oppositely charged parallel plates A and B.

The positive charge will freely move from plate B to plate A, gaining kinetic enrgy.

However, if the charged particle is moved from A to B, an external force F = - qE must

be applied to move the charge against the electric field. Hence, work W = qEd has to

done on the particle as it is moved from plate A to plate B. Thus, this work done is stored

as an electrical potential energy. The electrical potential energy of a particle at point B

relative to point A is equal to the work done against the electric forces required to move

the particle from A to B.

We define the electrical potential difference ΔV between two points A and B as the work

done in carrying a unit positive charge from A to B or simply the work per unit positive

charge.

∆𝑉 = 𝑉𝐴𝐵 = 𝑉𝐵 − 𝑉𝐴 =𝑊

𝑞=

𝐹𝑑

𝑞=

𝑞𝐸𝑑

𝑞

Therefore, 𝑉𝐴𝐵 = 𝐸𝑑 (𝑉𝑜𝑙𝑡𝑠) . (3.2.1)

Since 𝑬 = 𝒌𝒒

𝒅𝟐,

𝑽𝑨𝑩 = 𝒌𝒒𝒅

𝒅𝟐= 𝒌

𝒒

𝒅 (Volts). (3.2.2)

E

q

d

B A

+

+

+

_

_

_

23

ACTIVITY

A3.2.1: Calculate the potential difference between two points A and B if it requires

8 x 10-4 J of external work to move a charge of +4 μC from A to B. Which point is at a

higher potential?

Solution:

∆𝑽 = 𝑽𝑨𝑩 = 𝑽𝑩 − 𝑽𝑨 =𝑾

𝒒

∴ ∆𝑉𝐴𝐵 = 𝑉𝐵 − 𝑉𝐴 =𝑊

𝑞=

8𝑥10−4

4𝑥10−6= 200 𝑉

Since the potential difference between A and B is 200 V and positive, B is at a higher

potential.

A3.2.2: If a proton (q = e, mp = 1.67 x 10-27 kg) is released from point B as shown. Find

its speed just before it strikes the plate at A. Take 𝑽𝑨𝑩 = 45 V.

Solution:

From energy conservation, we have,

∆𝑲𝑬 = ∆𝑬𝑷𝑬

𝟏

𝟐𝒎𝒗𝑨

𝟐 = 𝒒𝑽𝑨𝑩

𝑽𝑨 = √𝟐𝒒𝑽𝑨𝑩

𝒎𝒑 = √

𝟐×𝟏.𝟔 × 𝟏𝟎−𝟏𝟗 × 𝟒𝟓

𝟏.𝟔𝟕 × 𝟏𝟎−𝟐𝟕 = 9.29 x 104 m/s.

3.2.4: Equi-potentials and the Electronvolt

Consider a point charge +q shown in figure 3.2.2. The radial lines represent the electric

field lines. The dashed line circles indicate the equipotential lines surrounding a charge.

The electric field lines are perpendicular to the equipotential lines.

q

A B

A B

+

- +

-

- +

24

Figure 3.2.2: A positive point charge surrounded by equipotential lines.

From figure 3.2.2, we note that the potential at points A and B are given in equation (3.2.3).

𝑽𝑨 = 𝒌𝒒

𝒓𝑨 , 𝑽𝑩 = 𝒌

𝒒

𝒓𝑩 (3.2.3)

∆𝑽 = 𝑽𝑨𝑩 = 𝑽𝑩 − 𝑽𝑨 = 𝒌𝒒(𝟏

𝒓𝑩−

𝟏

𝒓𝑨) = 0. Since 𝒓𝑨 = 𝒓𝑩.

Hence, ∆𝑊 = 𝑞∆𝑉 = 0. We conclude that no work is done in moving a test charge from

any point on the same equipotential line. Hence, static electric field is conservative field,

i.e it is independent of the path taken.

We state without proof that since charge resides on the outer surface of a conductor, the

electric field strength E inside the conductor is zero. Thus, the volume of a conductor

constitutes an equipotential volume.

The electron-volt (eV).

One electron-volt (eV) is the amount of energy acquired or lost by an electron as it traverses

(moves) through a potential difference of one volt.

𝟏𝒆𝑽 = 𝟏. 𝟔 × 𝟏𝟎−𝟏𝟗 J. (3.2.4)

Electric field line

Equipotential line

B

A

+q

rA

rB

25

3.2.5: Absolute Potentials

Consider a point B which is at radius r from the center of a charge shown in figure 3.2.3.

Figure 3.2.3: Point B at a radius r from the center of a positive charge

We define the absolute potential at B as the work done in carrying a unit positive charge

from infinity to point B. We consider the potential at infinity to be zero since

𝑉 = 𝑘𝑞

𝑟 ; if r = ∞, therefore 𝑉 ≈ 0.

The superposition principle is applicable here. Thus the potential produced by any charge

is not affected by the presence of other charges. If a charge distribution consists of N

discrete charges q1, q2, q3, ……………qN, the potential at point P, located at distances r1, r2,

r3, ……………..rN respectively, from the charges, may be obtained by the algebraic sum of

the potentials due to individual charges i.e,

𝑽𝑷 = 𝑽𝟏 + 𝑽𝟐 + 𝑽𝟑 … … … … … … … … … . . +𝑽𝑵. (3.2.3)

= ∑ 𝑉𝑖 = 𝑘𝑁𝑖=1 ∑

𝑞𝑖

𝑟𝑖

𝑁𝑖=1 . (3.2.4)

Note that potential is a scalar quantity.

B

+q r

26

Activity

A3.2.3: Calculate the absolute potential at point B in the vicinity of the three point charges

shown in the figure below.

Solution:

𝑽𝑩 = 𝑽𝟏 + 𝑽𝟐 + 𝑽𝟑

= 𝑘 [𝑞1

𝑟1+

𝑞2

𝑟2+

𝑞3

𝑟3]

= 𝟗 × 𝟏𝟎𝟗 [𝟓𝒙𝟏𝟎−𝟖

𝟎.𝟏+

𝟖𝒙𝟏𝟎−𝟖

𝟎.𝟏−

𝟒𝟎𝒙𝟏𝟎−𝟖

𝟎.𝟐] = - 6300 V

10cm

10cm

20cm

q3 = -40 x 10-8 C

q2 = +8 x 10-8 C

q1 = +5 x 10-8 C

B

27

3.2.6: Capacitors in Parallel and Series

A capacitor is a device that stores charge. It consists of two neighboring conductors that

have equal and opposite charges. Figure 3.2.4 shows three types of capacitors. The symbols

(a) and (b) are for a fixed capacitance and (c) is for variable capacitance.

Figure 3.2.4: Three types of capacitors

The capability of a capacitor to store charge is called its capacitance C. The space in the

capacitor may be filled with air or any other insulator.

We saw in chapter 3.2.1, that

𝐸 ∝ 𝑞 and 𝑉 ∝ 𝐸. Hence, 𝑞 ∝ 𝑉. Mathematically, we may write,

𝑞 = 𝐶𝑉, (3.2.5)

where C is called the capacitance C of the capacitor.

∴ 𝐂 =𝒒

𝑽 (Farads F). (3.2.6)

The capacitance depends on the geometry of the capacitor. For example, the capacitance

of a parallel plate capacitor is directly proportional to the cross-sectional area A of the

plates and inversely proportional to their separation distance.

𝑪 = 𝜺𝒐𝑨

𝒅. (3.2.7)

Where 𝜀𝑜 = 8.85 x 10-12 F/m (permittivity of free space). Smaller units of Faraday are used

like μF = 10-6 F and picofarad (pF = 10-12 F or μμF).

(c) (b) (a)

28

3.2.7: Dielectrics

Sometimes we wish to increase the capacitance of a capacitor. This is achieved by inserting

a non-conducting material (insulator) between the plates of the capacitor. These insulators

are called dielectric materials. Figure 3.2.5 shows a dielectric inserted in a capacitor.

Figure 3.2.5: Insulator inserted between two plates of a capacitor.

Let 𝑪𝒅𝒊 =𝒒

𝑽𝒅𝒊, and 𝑪𝒗𝒂𝒄 =

𝒒

𝑽𝒗𝒂𝒄 . Taking ratios, we have

𝑪𝒅𝒊

𝑪𝒗𝒂𝒄=

𝒒𝑽𝒗𝒂𝒄

𝒒𝑽𝒅𝒊= 𝒌. (3.2.8)

Where k is defined as the dielectric constant of the insulator, 𝒅𝒊 stands for dielectric and

𝒗𝒂𝒄 stands for vacuum.

Thus 𝑪𝒅𝒊 = 𝒌𝑪𝒗𝒂𝒄. (3.2.9)

For parallel plate capacitor, 𝑪𝒗𝒂𝒄 = 𝜺𝒐𝑨

𝒅 , then using equation (3.2.9) we have,

𝐶𝑑𝑖 = 𝑘𝜀𝑜𝐴

𝑑 . We define ϵ = 𝑘𝜀𝑜 as the permittivity of the material (dielectric). The

presence of a dielectric also lowers the effective voltage of a capacitor.

+

+

+

+

+

+

-

-

-

-

-

-

+

+

-

-

+ -

Dielectric material

29

Capacitors in Parallel.

From the conservation of charge, we have

q = q1 + q2 +q3

𝐶𝑉 = 𝐶1𝑉1 + 𝐶2𝑉2 + 𝐶3𝑉3

But, V = 𝑽𝟏 = 𝑽𝟐 = 𝑽𝟑

∴ 𝑪 = 𝑪𝟏 + 𝑪𝟐 + 𝑪𝟑 (3.2.10)

Capacitors in series

From the conservation of energy,

have,

𝑽 = 𝑽𝟏 + 𝑽𝟐 + 𝑽𝟑

𝒒

𝑪𝑻=

𝒒

𝑪𝟏+

𝒒

𝑪𝟐+

𝒒

𝑪𝟑

Since we are talking of the same charge flowing through the circuit, we have

1

𝐶𝑇=

1

𝐶1+

1

𝐶2+

1

𝐶3 (3.2.11)

V, q

C3

C1

C2

q1

q2

q3

C1 C2 C3

V

V1 V2 V3

30

3.2.8: Energy Stored in a Capacitor and Energy Density

The process of charging a capacitor is equivalent to transferring charge from one plate to

the other. Work must be done due to potential difference across the plates. This work is

stored as energy in the capacitor.

If ΔW is the work done in delivering a charge Δq against a potential V, then ΔW = Δq V.

If the final charge on the capacitor is q and the initial voltage is 0, then the average

potential difference during the transfer process is 1

2(0 + 𝑉) =

1

2𝑉. Hence, the total work

done is

𝑾 =𝟏

𝟐𝒒𝑽. (3.2.12)

But 𝑞 = 𝐶𝑉 and then

𝑾 =𝟏

𝟐𝑪𝑽𝟐 (𝐉) =

𝟏

𝟐

𝒒𝟐

𝑪 (J) (3.2.13)

Thus 𝑼 =𝟏

𝟐𝑪𝑽𝟐 (𝑱) (3.2.14)

Since energy is stored in the electric field, it is useful to express energy in terms of the

electric field strength E. Hence, for parallel plate capacitor, we have,

𝑼𝒆 =𝟏

𝟐𝑪𝑽𝟐 but V = Ed and then 𝑼𝒆 =

𝟏

𝟐𝑪𝑬𝟐𝒅𝟐. But 𝑪 = 𝜺𝒐

𝑨

𝒅 , for parallel plate

capacitor. Thus, 𝑼𝒆 =𝟏

𝟐𝜺𝒐𝑬𝟐𝑨𝒅, where Ad is the volume of the space between plates.

Hence, the energy density

𝒖𝒆 =𝟏

𝟐𝜺𝒐𝑬𝟐. (3.2.15)

For dielectrics,

𝑪 = 𝝐𝑨

𝒅 , where 𝝐 = 𝒌𝜺𝒐. Hence, 𝑼𝒆 =

𝟏

𝟐𝝐𝑬𝟐𝑨𝒅 and the energy density becomes

𝑢𝑒 =𝐸

𝐴𝑑=

1

2𝜖𝐸2 (3.2.16)

as the energy per unit volume.

31

Activity

A3.2.5: The parallel plates of a capacitor measure 5 cm by 4 cm and are separated by a

distance of 0.5 cm. To start with the potential difference between the plates is 2000 V, but

when filled with a dielectric, the potential drops to 1000 V. Calculate the following;

(a) the initial capacitance

(b) the initial charge on each plate

(c) the final capacitance when filled by a dielectric

(d) the dielectric constant k

(e) the initial energy stored in the capacitor

(f) the final energy stored in the capacitor.

Solution:

(a) A = 5 x 10-4 x 4 x 10-4 = 20 x 10-4 m2 ; d = 0.5 x 10-2 m

𝐶𝑣𝑎𝑐 = 𝜀𝑜𝐴

𝑑 = 8.85 × 10−12 × 20 × 10−4 ×

1

0.5×10−2

= 3.54 x 10-12 F

(b) 𝑞 = 𝐶𝑣𝑎𝑐𝑉 = 3.54 x 10-12 x 2000 = 7.08 x 10-9 C

(c) When the dielectric is placed between the plates, the charge remains

same, but the voltage drops.

𝐶𝑑𝑖 =𝑞

𝑉=

7.08 × 10−9

1000 = 7.08 x 10-12 F

(d) 𝑘 = 𝐶𝑑𝑖

𝐶𝑣𝑎𝑐=

7.08 𝑝𝐹

3.54 𝑝𝐹= 2

(e) 𝑈𝑖 =1

2𝐶𝑣𝑎𝑐𝑉2 =

1

2× 3.54 × 10−12 × (2000)2 = 7.08 x 10-6 J

(f) 𝑈𝑓 =1

2𝐶𝑑𝑖𝑉2 =

1

2× 7.08 × 10−12 × (1000)2 = 3.54 x 10-6 J

32

SAE 3.2.1: A pendulum of length L hungs from the ceiling of a room in which a downward

electric field strength E exists. The pendulum ball has a mass m and carries a positive

charge q. Find the frequency of the pendulum for small angle oscillations.

Answer: 𝒇 =𝟏

𝟐𝝅√

𝒈

𝑳+

𝒒𝑬

𝒎𝑳

SAE3.2.2: The Bohr model of the hydrogen atom, consists of an electron

(e = -1.6 x 10-19 C) moving in a circular radius r = 0.053 x 10-9 m with a proton at the

center. Calculate the potential energy, and the kinetic energy of the electron in this orbit.

Answers. EPE = -27.2 eV, and |𝑲𝑬| = +13.6 eV.

SAE 3.2.3: A capacitor is connected across a 120 V battery and stores a charge of 45 μC.

(a) What is the capacitance of the capacitor?

(b) How much energy does the capacitor store?

Answer: (a) q = 3.75 x 10-7 C

(b) U = 2.7 x 10-3 J

SAE 3.2.4: A parallel-plate capacitor of capacitance C is given the charge of Q and then

disconnected from the circuit. How much work is required to pull the plates of this

capacitor to twice their original separation?

Answer: 𝑄2

2𝐶


33

UNIT 3.3

DIRECT CURRENT CIRCUITS (D.C)

3.3.1 Introduction

An electric current consists of a flow of charge from one place to another. Currents are

involved in nearly all practical applications of electricity. In this chapter you will

understand how electric energy is transferred from a source such as battery or a generator

to a load or appliance. You will appreciate that the flow of current is associated with some

resistance governed by Ohm’s law. You will also be able to appreciate the dependence of

the resistivity on temperature. You will also be able to use Kirchhoff’s current law and

Kirchhoff’s voltage law to solve for unknown currents at junctions and unknown voltages

along closed loops. You will also be able to distinguish between electro-motive force and

the terminal potential difference.



1) Distinguish between flow of electrons and conventional current.

2) Solve simple problems on resistivity dependence on temperature

3) Decompose a system of resistors in series and parallel

4) Apply Kirchhoff’s current law and Kirchhoff’s voltage law to compute currents in

different branches of a circuit and voltage between any two points respectively.

5) Describe the charging and discharging processes of a battery

3.3.3 Conventional Current and Ohm’s Law

An electric current is the rate flow of charge. 𝐼 =∆𝑄

∆𝑡 (A). This is the conventional current

which flows in the direction of the electric field E as shown in figure 3.3.1. If the average

motion of charges is always in the same direction, then we have a direct current (D.C).

34

Figure 3.3.1: The flow of conventional I, same as direction of electric field

Ohm’s Law

Experimentally, it has been found that the potential difference V is proportional to the

current I. The constant of proportionality is called the resistance R. Hence, Ohm’s law

obeys equation (3.3.1) as

𝑽 = 𝑰𝑹 (𝐕). (3.3.1)

The figure 3.3.2 shows the I-V characteristic behaviour of ohmic and non-ohmic

materials.

Figure 3.3.2: Current-Voltage curves for Ohmic and non-ohmic behaviour.

I VB VA

+ -

+

+ +

+

-

- - -

E

Curr

ent

Voltage

Slope = 1

𝑅

Non-ohmic

Non-ohmic

Ohmic

35

3.3.4 Resistivity and its Temperature Dependence

At constant temperature, resistance R depends on the material and its dimensions. Since

𝑅 ∝ 𝑙, 𝑎𝑛𝑑 𝑅 ∝1

𝐴. Then,

𝑅 = 𝜌𝑙

𝐴 . (Ohms) (3.3.2)

Where, l is the length of the material, A is the cross-sectional area and 𝝆 (rho) is the

resistivity of the material.

Consider a conducting material across which a potential difference is applied. The total

charge on the conductor is zero, since the number of negative charges equals the number

of positive charges. The free charges (electrons) in the conductor move and collide with

the ion cores. The number of collisions are in the order of 1013 to 1014 collisions per

second. Due to these collisions, the temperature of the conductor increases. Thus,

collisions provide resistance to the motion of charges.

Thus, ∝ 𝒍 . And, experimentally,

𝑅𝑇 = 𝑅𝑜(1+∝ 𝑇 + 𝛽𝑇2 + … … . . . ) (3.3.3)

𝑹𝑻 ≈ 𝑹𝒐(𝟏+∝ 𝑻) to first approximation. (3.3.4)

A plot of 𝑅𝑇 versus T gives a straight line with slope ∝𝑅𝑜, where ∝ is the temperature

coefficient of resistance.

A3.3.1: A light bulb whose filament is made of tungsten has a resistance of 240 Ω when

white-hot (about 1800oC). Find the approximate resistance of the bulb at room

temperature (20 oC). (∝ for Tungsten is 0.0045 oC-1).

Solution:

Recall that ∆𝑹 = 𝑹𝒐 ∝ ∆𝑻 where 𝑹𝑻 = 𝟐𝟒𝟎 Ω at T = 1800 oC. To find 𝑹𝒐.

Thus, 𝑅𝑇 − 𝑅𝑜 = 𝑅𝑜 ∝ (𝑇𝑻 − 𝑇𝑜)

36

𝟐𝟒𝟎 − 𝑹𝒐 = 𝑹𝒐[𝟎. 𝟎𝟎𝟒𝟓(𝟏𝟖𝟎𝟎 − 𝟐𝟎)]

240 − 𝑅𝑜 = 8𝑅𝑜

240 = 9𝑅𝑜

⸫ 𝑹𝒐 = 𝟐𝟕 𝛀

3.3.5 Resistors in Series and Parallel

(a) Resistors in series

Figure 3.3.3: Shows resistors in series.

From figure 3.3.3, we see that,

𝑽 = 𝑽𝟏 + 𝑽𝟐 + 𝑽𝟑 , and 𝑰𝑹𝑻 = 𝑰𝑹𝟏 + 𝑰𝑹𝟐 + 𝑰𝑹𝟑. This reduces to

𝑹𝑻 = 𝑹𝟏 + 𝑹𝟐 + 𝑹𝟑. (3.3.5)

(b) Resistors in parallel

Figure 3.3.4: Resistors in parallel.

R2

I

V

V1

R1

V3 V2

R3

V

R1

I

I2

I1

R2

R3 I3

37

From figure 3.3.4, we see that

𝐼𝑇 = 𝐼1 + 𝐼2 + 𝐼3 and 𝑉

𝑅𝑇=

𝑉1

𝑅1+

𝑉2

𝑅2+

𝑉3

𝑅3. But, 𝑉 = 𝑉1 = 𝑉2 + 𝑉3, and hence,

𝟏

𝑹𝑻=

𝟏

𝑹𝟏+

𝟏

𝑹𝟐+

𝟏

𝑹𝟑 . (3.3.6)

3.3.6 Kirchhoff’s Laws and their Applications

We use these laws to compute currents in different branches of the circuit or potential

difference between any two points in a circuit.

(a) Kirchhoff’s Current Law (KCL)

This rule states that the algebraic sum of all the currents into a junction (point) must equal

the sum of all the currents leaving the junction. This is an alternative statement of charge

conservation. It states that charge can neither be destroyed nor created and also it does

not accumulate. We now illustrate Kirchhoff’s current rule in figure 3.3.5 for currents

arriving and leaving a junction.

Figure 3.3.5: Illustrating Kirchhoff’s current law.

I4 I3

I2 I1

38

From figure 3.3.5 above, we have

𝑰𝟏 + 𝑰𝟒 = 𝑰𝟐 + 𝑰𝟑 (3.3.7)

(b) Kirchhoff’s Voltage Law (KVL)

This law states that the algebraic sum of the voltage changes around a closed circuit is

equal to zero. This is an alternative statement of energy conservation. We now apply

Kirchhoff’s voltage rule to the loop shown in figure 3.3.6.

Figure 3.3.6: Shows a closed loop abcdfa.

𝒂 → 𝒃 + 𝜺𝟏 (𝐯𝐨𝐥𝐭𝐚𝐠𝐞 𝐫𝐢𝐬𝐞)

𝒃 → 𝒄 − 𝑰𝑹𝟏 (𝐯𝐨𝐥𝐭𝐚𝐠𝐞 𝐝𝐫𝐨𝐩)

𝒄 → 𝒅 − 𝑰𝑹𝟐 (𝐯𝐨𝐥𝐭𝐚𝐠𝐞 𝐝𝐫𝐨𝐩)

𝒅 → 𝒇 − 𝜺𝟐 (𝐯𝐨𝐥𝐭𝐚𝐠𝐞 𝐝𝐫𝐨𝐩)

𝒇 → 𝒂 − 𝑰𝑹𝟑 (𝐯𝐨𝐥𝐭𝐚𝐠𝐞 𝐝𝐫𝐨𝐩)

From Kirchhoff’s Voltage Law we have,

+𝜺𝟏 − 𝑰𝑹𝟏 − 𝑰𝑹𝟐 − 𝜺𝟐 − 𝑰𝑹𝟑 = 𝟎. (3.3.8)

Thus, the work done in carrying a charge around a closed circuit or loop is zero. Hence,

the static force field is conservative.

R2

R3 R1

I

ɛ2

a b

c

ɛ1

1

d f

39

Rules for solving circuit problems

1. Draw the circuit

2. Assign a current to each important wire both symbol and direction

3. Reduce series and parallel resistance systems whenever possible

4. Write junction equations and loop equations separately

5. Solve these equations for the unknowns.

Activity

A3.3.2: Find the currents I1, I2 and I3 shown below.

Solution: By applying the junction rule at point A in the figure have,

𝑰𝟑 = 𝑰𝟏 + 𝑰𝟐. (1)

Applying the loop rule to loop I starting at A, have

𝟑𝟎𝑰𝟐 − 𝟔𝟎 − 𝟓𝑰𝟏 + 𝟏𝟎 − 𝟏𝟎𝑰𝟏 = 𝟎

This reduces to

𝟑𝟎𝑰𝟐 − 𝟏𝟓𝑰𝟏 = 𝟓𝟎. (2)

Applying the loop rule to loop II starting at A, we have

40Ω

20Ω

5Ω

30Ω

10Ω

60V

50V 10V

A

I II

I1

I2

I3

40

−𝟐𝟎𝑰𝟑 − 𝟓𝟎 − 𝟒𝟎𝑰𝟑 + 𝟔𝟎 − 𝟑𝟎𝑰𝟐 = 𝟎 .

This reduces to

𝟑𝟎𝑰𝟐 − 𝟔𝟎𝑰𝟑 = 𝟏𝟎. (3)

Solving equations (1), (2) and (3) simultaneously get

𝑰𝟏 = −𝟏. 𝟑𝟑 𝑨

𝑰𝟐 = 𝟏. 𝟎 𝑨

𝑰𝟑 = −𝟎. 𝟑𝟑 𝑨

Note that the negative sign means that the actual directions are opposite to those shown.

3.3.7 Electro-Motive-Force (emf) and Terminal Potential Difference

There is always a contact potential difference between two dissimilar bodies in contact.

For example, when two bodies are rubbed, they become electrically charged, thus acquire

a potential difference.

The emf of a battery is generated by the chemical action in the battery. When connected in

the circuit, the battery has some internal resistance r due to some resistance as the charges

move in the electrolyte. The schematic diagram of a battery and its equivalent electric

circuit connected to a load is shown in figure 3.3.8 (I) and (II) respectively.

When no current is being drawn from the battery (not connected), there is no potential drop

across the internal resistance r. Hence, the potential difference between the terminals is

equal to the electro-motive force. However, if the battery is connected across an external

resistor R, figure (II), the current I and the potential difference across the terminals is,

Terminal potential 𝑉𝑇 = 𝜀 − 𝐼𝑟 . (Discharging battery) (3.3.9)

41

Figure 3.3.8: Schematic representation of a battery with its internal resistance.

For the battery that is being charged i.e current flowing through the battery from the

positive to the negative terminal, we have

Terminal potential 𝑉𝑇 = 𝜀 + 𝐼𝑟 . (Charging battery) (3.3.10)

SAE 3.3.1: A water heater draws 30 A from a 120 V power supply 10 m away. Find the

minimum cross sectional area of the copper wire that can be used if the voltage applied to

the heater is not to be lower than 115 V. Resistivity of copper is 1.7 x 10-8 Ωm.

Answer: 2.0 mm2.

SAE 3.3.2: A resistance thermometer makes use of the temperature variation of resistivity.

When a coil of platinum wire whose resistance at 20oC is 11 Ω is placed in a furnace, its

r

ɛ

r

R

ɛ

(I) (II)

battery

<=>


42

resistance triples. Find the temperature of the furnace, assuming the temperature coefficient

of resistivity of platinum (0.0036/oC) remains constant.

Answer: 576oC.

SAE 3.3.3: Find I1, I2, and I3 for the circuit shown below.

Answers: I1 = 0.67 A, I2 = 0.4 A, and I3 = - 0.27 A.

SAE 3.3.4: A “D” cell of emf 1.5 V and internal resistance 0.3 Ω is connected to a flashlight

bulb whose resistance is 3.0 Ω. Find (a) the current in the circuit and (b) the terminal

voltage of the cell.

Answers: (a) I = 0.45 A and (b) V = 1.4 V

SAE 3.3.5: Four batteries, each of emf 6.0 V and internal resistance 0.30 Ω, are

connected in parallel with a load of 2.0 Ω. Find the current in the load.

Answer: 2.9 A.

30Ω

10Ω

15Ω

60V

50V

40V

I2 I3

I1

43

UNIT 3.4

MAGNETISM

3.4.1 Introduction

We are all familiar with magnets in their ability to attract nails and other iron or steel

objects. However, not many of you appreciate how closely related electricity and

magnetism are. You will observe that a conductor carrying current in a magnetic field

experiences a force. Similarly, this concept is generalized to moving charges as well. We

will also be introduced to the concept of a magnetic field of a current in a long straight

wire, in a circular loop, and in a long solenoid. We will extend further to the definition of

the ampere and hence describe the working principle of the moving coil meters such as

ammeters and voltmeters.



1) Calculate a force experienced by a current carrying wire in a magnetic field

2) Derive the magnetic force on moving charges in a magnetic field

3) Calculate the radius of the circular path followed by a charged particle in a

magnetic field

4) Calculate the force per unit length exerted by two parallel wires a given distance

apart.

5) Explain the turning effect of a coil in a magnetic field

6) Describe how a galvanometer may be converted into an ammeter and into a

voltmeter

3.4.3 Force on a Conductor carrying Current in a Magnetic Field

A current in a wire is capable of generating a magnetic field. The direction of the magnetic

field is given by the right hand rule which states that, grasp a wire in your right hand with

44

your thumb pointing in the direction of the current, the fingers circle the wire in the

direction of the magnetic field as shown in figure 3.4.1 (a).

Figure 3.4.1: Shows the right-hand rule in (a) and the magnetic field around a wire

carrying current in (b).

A wire carrying a current passing through a region in which an external magnetic field

exists experiences a force due to that field. The magnitude of the magnetic force is

proportional to both the current in the wire and the length of the wire in the magnetic

field. The force is also proportional to the magnetic field strength B.

The force that a wire of length l carrying a current i in a magnetic field strength B

experiences is given by equation (3.4.1)

𝑭 = 𝒊𝑩˔𝒍 (𝐍). (3.4.1)

Where F, B, and l are all perpendicular to each other. The S.I unit for B is N/Am, which

is called tesla (T). 1 Tesla = 1Wb/m2. Also 1G (Gauss) = 10-4 T.

Note that the direction of the force is given by the extension of the right –hand rule. It

states as; point the fingers of your right hand along the lines of the magnetic field and the

In

I I

+

+

+

+

+

+

+

+

+

+

- -

- -

- -

- -

- -

Out

(a) (b)

45

thumb in the direction of the current. The force on the wire is in the direction in which

your palm would push.

Consider the orientation of a wire carrying a current in a magnetic field as shown in

figure 3.4.2.

Figure 3.4.2: Wire carrying current in a magnetic field

Since the wire makes an angle of θ with the magnetic field, it is the perpendicular

component of 𝑩˔ which exerts a force on the wire.

Thus, 𝑩˔ = 𝐵𝑠𝑖𝑛𝜃. (3.4.2)

Hence, |𝑭| = 𝒊𝒍𝑩𝒔𝒊𝒏𝜽 and in vector notation becomes

𝑭 = 𝒊𝒍 × 𝑩. (3.4.3)

Where × stands for cross product.

Note: If both l and B are parallel, then no force is experienced by the wire.

Activity

A3.4.1: A wire 10 cm long, carrying current 5 A, is placed in a magnetic field of 3 Wb/m2

making an angle of θ = 0o, 45o, and 90o with the magnetic field. Calculate the force on the

wire in each case.

i B

Bll 𝐵˔

θ

θ

46

Solution:

From F =iBlsin θ = 5 x 3 x 0.1 x sin θ = 1.5 sin θ

For θ = 0o ⸫ F = 1.5 x sin 0o = 0 N

For θ = 45o ⸫ F = 1.5 x sin 45o = 1.06 N

For θ = 90o ⸫ F = 1.5 x sin 90o = 1.5 N

3.4.4 Magnetic Forces on moving Charges

Consider a wire of length l, cross-sectional area A and having n charge carriers per unit

volume.

Figure 3.4.3: Wire of length l carrying current i and cross-sectional area A

Number of charge carriers in length l = nAl.

Force per charge = 𝑭𝒐𝒓𝒄𝒆 𝒐𝒏 𝒘𝒊𝒓𝒆

𝑵𝒖𝒎𝒃𝒆𝒓 𝒐𝒇 𝒄𝒂𝒓𝒓𝒊𝒆𝒓𝒔=

𝑩˔ 𝒊 𝒍

𝒏 𝑨 𝒍=

𝒊𝑩˔

𝒏 𝑨 (3.4.4)

But, 𝑖 =𝑐ℎ𝑎𝑟𝑔𝑒 𝑡ℎ𝑎𝑡 𝑝𝑎𝑠𝑠𝑒𝑠 𝑎 𝑔𝑖𝑣𝑒𝑛 𝑝𝑜𝑖𝑛𝑡

𝑠𝑚𝑎𝑙𝑙 𝑡𝑖𝑚𝑒 ∆𝑡=

𝑞𝑛𝐴∆𝑙

∆𝑡=

𝑞𝑛𝐴𝑣∆𝑡

∆𝑡 (3.4.5)

Where q is the charge of each carrier.

Therefore, 𝑭𝒎 =𝑩˔𝒒𝒏𝑨𝒗∆𝒕

𝒏𝑨∆𝒕 (3.4.6)

𝐹𝑚 = 𝑞𝑣𝐵˔ (N). (3.4.7)

Δl = vΔt

i

A

47

As we had for a wire, the force F is always perpendicular to both magnetic field B and

the velocity v of the charge q. Thus, F is always perpendicular to the plane defined by v

and B as in figure 3.4.4.

Figure 3.4.4 : The magnetic force is perpendicular to both velocity and the field

3.4.5 Motion of a Charge in a Magnetic Field

Consider a charged particle of mass m and charge q moving with a velocity v in a

direction perpendicular to the magnetic field B as shown in figure 3.4.5.

Figure 3.4.5: Motion of a charge in a magnetic field

The force on a charge q according to equation F =qvB is always perpendicular to

velocity. Thus, the effect of the force is to change the direction of the velocity without

changing its magnitude. This results in a uniform circular motion with centripetal

acceleration 𝑣2

𝑟 and hence centripetal force 𝐹𝑐 = 𝑚

𝑣2

𝑟. Thus, equating the magnetic force

and the centripetal force, and rearranging, have

B(out)

v

F

v r

v

B (into paper)

48

𝒓 =𝒎𝒗

𝒒𝑩. (3.4.8)

This is the radius of the circular path followed by a charged particle in a magnetic field.

Activity

A3.4.2: An electron is moving horizontally through a vertical field of 0.03 Wb/m2 with a

velocity of 1.5 x 108 m/s. Calculate the radius of the electron orbit and the cyclotron

frequency. (Ignore the relativistic effects).

Solution:

(a) Using equation (3.4.8), have

𝑟 =𝑚𝑣

𝑞𝐵=

9.11𝑥10−31𝑥1.5𝑥108

1.6𝑥10−19𝑥0.03 = 2.85 cm

(b) Since 𝒗 = 𝒓𝝎 𝒂𝒏𝒅 𝝎 = 𝟐𝝅𝒇. We have

𝜔 =𝑞𝐵

𝑚=

1.5𝑥10−19𝑥0.03

9.11𝑥10−31 = 5.27 x 109 rad/s

49

3.4.6 The Magnetic Field of a Current:

(a) in a long straight wire:

The magnetic field B generated by a current in a long straight wire circles the wire. The

magnitude of this field at a distance r from the wire is given by

𝑩 =𝝁𝒐𝒊

𝟐𝝅𝒓. (Tesla) (3.4.9)

Where i = current in the wire.

𝝁𝒐 = 𝟒𝝅 𝒙 𝟏𝟎−𝟕 𝐓. 𝐦/𝐀, called

the permeability of free space.

And r is the distance from the wire axis.

(b) in a circular loop

Figure 3.4.7: A circular loop of radius r showing circular magnetic field lines around a

wire carrying current I.

r i

Figure 3.4.6: Magnetic field around a

long current carrying wire

50

For a circular loop of radius a, the magnitude of the field at the loop’s center is

𝑩 =𝝁𝒐𝒊

𝟐𝒂. (3.4.10)

A coil consisting of N loops tightly packed together produces a field N times larger at its

center.

(c) In a solenoid

A long wire wound in a close-packed helix around a cylindrical form surface, forms a

solenoid.

Figure 3.4.8: The magnetic field pattern in a long solenoid (a) and the magnetic pattern

is nearly uniform as shown in (b).

51

Inside a long, hollow solenoid that carries a current i and has n loops of wire per meter of

length, the magnetic field is nearly uniform and its magnitude is given by

𝑩 = 𝜇𝑜𝑛𝑖. (3.4.11)

Activity

A3.4.3: A solenoid 40 cm long has 800 loops and carries a current of 5.0 A. Find (a) the

number of loops per meter and (b) the magnitude of the magnetic field at the center of the

solenoid.

Solution:

(a) The number of loops per meter 𝑛 =800

40×10−2 = 2000

(b) The magnitude of the magnetic field 𝐵 = 𝜇𝑜𝑛𝑖 = 4𝜋 × 10−7 × 2000 × 5

= 12.6 mT

3.4.7 The definition of an Ampere

Consider two parallel wires a distance b apart carrying currents i1 and i2 in the same

direction as shown in figure 3.4.9.

Figure 3.4.9: Two parallel wires carrying currents in same direction.

Wire 1 generates a magnetic field 𝐵1 =𝜇𝑜𝑖1

2𝜋𝑏 on wire 2 directed downward. Thus, wire 2

will experience an inward force F2 given by 𝐹2 = 𝑖2𝑙𝐵1. Substituting the expression for

B1 have,

𝑖1

b

2

1

F2

B1

𝑖2

52

𝑭𝟐 = 𝒊𝟐𝒍𝝁𝒐𝒊𝟏

𝟐𝝅𝒃. The force per unit length that current 𝒊𝟏 exerts on unit length of 𝒊𝟐 is

𝐹2

𝑙=

𝜇𝑜𝑖1𝑖2

2𝜋𝑏. (3.4.12)

From the action – reaction law, there is an equal and opposite force F1 exerted on 𝑖𝟏 by

𝑖𝟐. For antiparallel currents, there is repulsion between the wires. Hence, when two equal

parallel currents of one ampere (A) are placed one meter apart, they exert on each other

a force equal to 2 x 10-7 N per meter of their length.

Activity

A3.4.4: Two parallel conducting wires carry a current of 50 A. The force per unit length

between these wires is equal to the force of the gravitational pull on a wire 1 m long. If

the mass of 1 m of wire is 1 g, calculate the distance between the two parallel wires.

Solution:

Since 𝒊𝟏 = 𝒊𝟐 = 𝟓𝟎 𝑨 and 𝑭

𝒍= 𝒎𝒈. But,

𝑭

𝒍=

𝝁𝒐𝒊𝟐

𝟐𝝅𝒃. Equating these two have

𝒎𝒈 =𝝁𝒐𝒊𝟐

𝟐𝝅𝒃, giving 𝐛 =

𝝁𝒐𝒊𝟐

𝟐𝝅𝒎𝒈=

𝟒𝝅×𝟏𝟎−𝟕×(𝟓𝟎𝟐)

𝟐𝝅×𝟏×𝟏𝟎−𝟑×𝟗.𝟖 = 5.1 cm

𝑖1 1

b

𝑖2 2

53

3.4.8 The Torque on a Current Loop

Many practical devices such as motors and many meters , make use of the torque that a

current loop experiences when placed in a magnetic field.

Consider a current carrying coil mounted on an axle to be able to rotate in a magnetic

field as shown in figure 3.4.10.

Figure 3.4.10: A current carrying coil mounted on an axle to be able to rotate in a

magnetic field.

The loop has length a and width b. Notice that only the two force Fh cause a torque about

the axis of rotation. Maximum torque occurs when the magnetic field lines skim past the

surface of the coil. Each of the two forces Fh gives a torque

𝝉 = (𝑭𝒉)(𝒍𝒆𝒗𝒆𝒓 𝒂𝒓𝒎) (3.4.13)

Note that Fh has to be perpendicular to the lever arm a. Thus

𝑻𝒐𝒓𝒒𝒖𝒆 = 𝟐𝑭𝒉𝒔𝒊𝒏𝜽𝒙𝒂 = 𝟐𝑭𝒉𝒂𝒔𝒊𝒏𝜽. (3.4.14)

Where θ is the angle between B and perpendicular to surface area of the coil. Since Fh

acts only on the vertical sides of the coil, each contributes a force ibB to Fh.. For N loops,

have,

54

𝑻𝒐𝒓𝒒𝒖𝒆 = (𝟐𝒂𝒃)(𝑵𝒊)𝑩𝒔𝒊𝒏𝜽. (3.4.15)

Since 2ab = Area A of the coil, therefore

𝑻𝒐𝒓𝒒𝒖𝒆 = 𝑵𝒊𝑨𝑩𝒔𝒊𝒏𝜽. (3.4.16)

We define a quantity called the magnetic moment μ of a current loop as

𝝁 = 𝒊𝑨 (𝑪𝒖𝒓𝒓𝒆𝒏𝒕 𝒙 𝑨𝒓𝒆𝒂), (3.4.17)

Directed in the axis of the magnet.

Therefore 𝑻𝒐𝒓𝒒𝒖𝒆 = 𝝁𝑩𝒔𝒊𝒏𝜽, (3.4.18)

where θ is the angle between µ and B.

Activity

A3.4.5: A rectangular loop measuring 12 cm by 10 cm with 50 turns and a current of 3 A

is placed in a uniform magnetic field of 0.40 Wb/m2. Calculate

(a) The magnetic moment of the loop

(b) The torque on the loop if the plane of the coil is (i) perpendicular and (ii) parallel

to the magnetic field B.

Solution:

(a) Area A = 0.12 x 0.1 = 1.2 x 10-2m2. I = 3 A, N = 50,

µ = NiA = 1.2 x10-2 x 3 x 50 = 1.8 Am2.

(b) 𝝉 = 𝝁𝑩𝒔𝒊𝒏𝜽

(i) For plane perpendicular to B, θ =0

τ = 0.

(ii) For plane parallel to B, θ = 90o

τ = 𝝁𝑥𝐵 = 1.8 x 0.4 = 0.72 N-m.

55

3.4.9 Moving Coil Meters

These work on the principle that a current-carrying coil in magnetic field experiences a

torque. The torque is proportional to the current and its effect can be used to measure

current.

A galvanometer consists of a coil that is freely pivoted and placed in a magnetic field. A

pointer attached to the coil moves over a calibrated scale as shown in figure 3.4.11.

Figure 3.4.11: A galvanometer with a coil that is freely pivoted and a pointer attached to

the coil moves over a calibrated scale.

When a current passes through the coil, it interacts with the magnetic field, producing a

deflection of the coil. The deflection is directly proportional to the current. Thus,

deflection (rotation) can be used to measure current.

Ammeters:

An ammeter is a low resistance instrument used to measure current. To convert a

galvanometer into an ammeter, we should increase the range and reduce the effective

resistance as low as possible.

56

For example, to make an ammeter that would read Io = 10 A full scale, from a

galvanometer which reads IG = 2mA full scale and has a resistance RG = 50 Ω. This can

be achieved by putting an extra resistor Rs called the shunt, parallel to RG as shown in

figure 3.4.12.

Figure 3.4.12: Conversion of a galvanometer into an ammeter

Since Rs and RG are parallel, voltage across them is the same. Thus,

𝑽 = 𝑰𝑮𝑹𝑮 = 𝑹𝒔𝑰𝒔, but 𝑰𝒔 = 𝑰𝒐 − 𝑰𝑮

Therefore 𝑹𝒔 =𝑰𝑮

𝑰𝒔. 𝑹𝑮 =

𝑰𝑮

𝑰𝒐−𝑰𝑮. 𝑹𝑮 =

𝟐𝒎𝑨 × 𝟓𝟎

(𝟏𝟎−𝟐𝒎𝑨)= 0.010 Ω

Voltmeter:

A voltmeter is a high resistance instrument used to measure voltage. It is always put

across the circuit. To convert a galvanometer into a voltmeter, add resistance Rs in series

with resistance RG of the galvanometer as shown figure 3.4.13.

Figure 3.4.13: Conversion of a galvanometer into a voltmeter

RG

IG

Is Io

R

R

Rs

Rs

RG

VB VA

57

To find the value of Rs, we impose a condition that the potential drops across Rs and RG

must add up to Vo (the potential drop of the voltmeter).

Therefore, 𝑽𝒐 = 𝑰𝑮𝑹𝑮 + 𝑹𝒔𝑰𝑮

where IG is the current in the galvanometer.

𝑹𝒔 =𝑽𝒐

𝑰𝑮− 𝑹𝑮. (3.4.19)

SAE 3.4.1: Calculate the magnetic field in air 10 cm from a wire that carries a current of

1.0 A.

Answer: 2 x 10-5 T

SAE 3.4.2: A solenoid 20 cm long and 40 mm in diameter with an air core is wound with

a total of 200 turns of wire. The solenoid’s axis is parallel to the earth’s magnetic field at a

place where the latter is 3.0 x 10-5 T in magnitude. What should the current in the solenoid

be for its field to exactly cancel the earth’s field inside the solenoid?

Answer: 0.024 A.

SAE 3.4.3: A long, straight wire carries a current of 100 A. (a) Find the force on an electron

moving parallel to the wire, in the opposite direction to the current, at a speed of 1 x 107

m/s when it is 10 cm from the wire. (b) Calculate the force on the electron under the same

circumstances when it is moving perpendicular toward the wire.

Answers: (a) 3.2 x 10-16 N

(b) 3.2 x 10-16 N in the same direction as the current.

SAE 3.4.4: The cables that connect the starter motor of a car with its battery are 10 cm

apart for a distance of 40 cm. Find the forces between the cables when the currents in

them is 300 A.

Answer: 0.72 N


58

UNIT 3.5

ELECTROMAGNETIC INDUCTION

3.5.1 Introduction

We have seen that one way in which electricity and magnetism are related is the presence

of a magnetic field around every electric current. We now turn to the opposite of the above

phenomenon that a magnetic field produces a current. This is what is known as

electromagnetic induction established by Michael Faraday in 1831. In this unit, we shall

look at the concept of magnetic flux and Faraday’s law of induced emf. You will be

introduced to the idea of mutual inductance and self-inductance. We shall further look at

the inductance-resistance circuits and formulate an expression for the energy stored in a

magnetic field. You will also be introduced to the concept of motional emf and the working

principles of transformers.



1) Define the magnetic flux as the number of field lines through a given area

2) Calculate the induced emf in a coil

3) Describe mutual inductance

4) Calculate self-inductance for a solenoid

5) Evaluate the energy stored in an inductor

6) Calculate the total energy stored in a magnetic field

7) Describe the motional electro-motive force (emf)

8) Use the transformer equation to solve problems in transformers

3.5.3 Magnetic Flux ϕ

We define the magnetic flux ϕ as the number of magnetic field lines that pass

perpendicular through an area A.

𝜙 = 𝐵˔𝐴 (T.m2 or Wb) (3.5.1)

59

In general, we express magnetic flux as

𝜙 = (𝐵cos𝜃)𝐴 = 𝐵𝐴𝑐𝑜𝑠θ. (3.5.2)

Where θ is the angle between the magnetic field B and the area normal 𝒏ෝ.

Figure 3.5.1: The number of the magnetic field line through area A

Activity

A3.5.1: In a room, the magnetic field has a magnitude of 6 x 10-5 T and is directed at an

angle of 70o below the horizontal. Find the magnetic flux through a 400 cm x 80 cm table

top in the room.

Solution:

θ = 20o

B˔ = Bcosθ

𝜙 = 𝐵𝑐𝑜𝑠20 𝑥 𝐴

= 6 x 10-5 x cos 20 x 4 x 0.8 = 1.8 x 10-4 T.m2

B

A

θ

𝒏ෝ (area normal)

field lines

θ

B˔

Bll

B

H

60

3.5.4 Faraday’s Law of Induced EMF

Consider a coil with N loops that has flux through it changed from 𝝓𝟏 to 𝝓2 in time

interval Δt. The induced emf in the coil while flux is changing is

𝑒𝑚𝑓 = 𝑁∆𝜙

∆𝑡 (V) (3.5.3)

Equation (3.5.3) is called Faraday’s law.

Thus, changing magnetic flux produces an induced emf as shown in figure 3.5.2.

Figure 3.5.2: In (a) there is no current in a stationary wire in a magnetic field. In (b), a

current is induced when the wire in moved across the field.

Lenz’s Law: States that the direction of the induced emf or current is such that it opposes

the change that produces it. The flux through a coil can change in any of the three ways:

(i) Changes in B

(ii) Changes in area A

(iii) Changes in θ

61

Activity

A3.5.2: A circular loop of radius 5 cm is placed with its plane perpendicular to magnetic

field B. If the magnetic field B changes from 0.1 Wb/m2 to 0.5Wb/m2 in 0.025 sec,

calculate the induced emf.

Solution:

𝐸𝑚𝑓 =∆𝜙

∆𝑡

Since θ = 0o and B is parallel to the area normal, then A = πr2 = 0.008 m2.

𝐸𝑚𝑓 =∆𝜙

∆𝑡=

∆𝐵𝐴

∆𝑡=

𝐴∆𝐵

∆𝑡= 0.008 ×

(0.5−0.1

0.025 = 0.128 V.

3.5.5 Mutual Inductance

Consider two coils placed side by side as shown in figure 3.5.3.

Figure 3.5.3: Two coils placed side by side. When switch S is open, both coils have zero

flux. When switch is suddenly closed, flux is induced in secondary coil.

When the switch is open, both coils have zero flux. When the switch is suddenly closed,

the primary coil will act as an electromagnet and will generate flux some of which will go

through the secondary coil. Since the flux in the secondary coil will be proportional to

62

the current in the primary, the induced emf in the secondary will be proportional to the

rate of change of current in the primary, ∆𝑰𝒑

∆𝒕 . Thus, we may write

𝑒𝑚𝑓𝑠𝑒𝑐 = 𝑀∆𝑰𝒑

∆𝒕 . (3.5.4)

Where M = mutual inductance of the two coils in Henrys (H) or Vs/A. It is also a

property of the geometry of the two coils.

3.5.6 Self Inductance

It is not necessary to have two coils to show inductance effects. An induced emf appears

in a single coil if the current in this coil is changed. If ∆𝑰

∆𝒕 is the rate change of current

through the coil, we can write for average induced emf as

𝐸𝑚𝑓 = 𝐿∆𝐼

∆𝑡 . (3.5.5)

Where L = self-inductance in units of Henry (H).

Activity

A3.5.3: (a) Calculate the self-inductance of a solenoid of radius 3.0 cm, length 30 cm,

and 3000 total number of turns. (b) Suppose the current changes from 0 to 5 Ain 0.2

seconds, calculate the induced emf across the solenoid.

Solution:

Given: N = nl; 𝐵 = 𝜇𝑜𝑛𝑖 , 𝜙 = 𝐵𝐴.

Thus, using 𝑬𝒎𝒇 = 𝑵∆𝝓

∆𝒕= 𝑳

∆𝑰

∆𝒕, we have,

⸫ 𝐿 =(nl)∆𝜙

∆𝐼 =

(𝑛𝑙)𝜇𝑜𝑛𝑖𝐴

𝑖= 𝜇𝑜𝑛2𝑙𝐴 = 4𝜋 × 10−7 × (

3000

0.3)

3× 0.3 × 𝜋 ×

0.032 = 0.11 Henry.

(b) Using 𝑬𝒎𝒇 = 𝑳∆𝑰

∆𝒕 = 0.11 x 5/0.2 = 2.75 V.

63

3.5.7 Inductance-Resistance Circuits

An inductor L is simply loops of wire as shown in figure 3.5.4. It has an ability to store

energy.

Figure 3.5.4: Shows an inductor L connected in a circuit

The resistance R is that of an inductor L. Applying Kirchhoff’s loop rule to the circuit,

have

𝜺 = 𝒊𝑹 + 𝑳∆𝒊

∆𝒕 . (3.5.6)

Multiplying both sides by i, we get

𝑖𝜀 = 𝑖2𝑅 + 𝐿𝑖∆𝑖

∆𝑡, (3.5.7)

where 𝒊𝜺 is the rate at which the battery is supplying energy to the circuit, 𝒊𝟐𝑹 is the rate

at which energy is dissipated as heat across the resistor R and 𝑳𝒊∆𝒊

∆𝒕 is the rate at which

energy is being used across the inductor and stored in the magnetic field. Solving the

differential equation above requires calculus, but the solution is

𝑖(𝑡) = 𝐼𝑓 (1 − 𝑒−(𝑅𝐿⁄ )𝑡) (3.5.8)

where 𝑰𝒇 =𝜺

𝑹. 𝒆 ≈ 𝟐. 𝟕, (Base natural logarithm).

A plot of 𝑖(𝑡) versus t is shown in figure 3.5.5.

ℰ

64

Figure 3.5.5: Dependence of current through an inductor over time t.

𝑳

𝑹= inductive time constant; it is the time the current takes to reache 0.63 or 63% of its

maximum saturation value. If t = 𝑳 𝑹⁄ , then

𝑖(𝑡) = 𝐼𝑓(1 − 𝑒−1 ) = 𝐼𝑓 (1 −1

2.7) = 𝐼𝑓(1 − 0.37) = 0.63 𝐼𝑓

As the current rises, the flux generated in the inductor induces an emf whose direction is

such as to oppose the increasing current. The induced emf is 𝐿∆𝑖

∆𝑡.

When there is a current through the coil, the charges flow through a potential difference

𝑳∆𝒊

∆𝒕. The work the battery does as it carries a charge Δq through the inductor is

∆𝑾 = ∆𝒒𝑽 = ∆𝒒(𝑳∆𝒊

∆𝒕) (3.5.9)

∆𝑊 = 𝐿𝑖∆𝑖 (since i = Δq/Δt) (3.5.10)

When the current in the inductor starts from i =0 at time t =0 and reaches a value of i = If

in time t, the total work done by the battery is obtained by replacing i by the average

value of (0+𝐼𝑓)

2=

𝐼𝑓

2,

I(t)

Time

65

Therefore 𝑾 =𝟏

𝟐𝑳𝑰𝒇

𝟐 (J) (3.5.11)

Hence, an inductor L through which there is a current I has stored in it an energy

1

2𝐿𝐼2 (J).

3.5.8 Energy Stored in a Magnetic Field

Here, we consider a particular inductor. We compute the energy stored inside a solenoid

of length l and cross-sectional area A. The self-inductance is 𝐿 = 𝜇𝑜𝑛2𝑙𝐴 and the

magnetic field 𝐵 = 𝜇𝑜𝑛𝑖. The energy stored in a magnetic field is

𝑼𝑩 = 𝟏

𝟐𝑳𝒊𝟐 =

𝟏

𝟐𝝁𝒐𝒏𝟐𝒍𝑨 x

𝑩𝟐

𝝁𝒐𝟐𝒏𝟐

=𝟏

𝟐

𝑩𝟐𝒍𝑨

𝝁𝒐.

Where 𝑙𝐴 = volume enclosed by the solenoid.

Therefore, Energy density 𝒖𝑩 =𝑼𝑩

𝑨𝒍=

𝑩𝟐

𝟐𝝁𝒐=

𝟏

𝟐

𝑩𝟐

𝝁𝒐. (3.5.12)

Activity

A3.5.4: Calculate the total energy 𝑈𝐵 and the energy density 𝑢𝐵 of the magnetic field in

a solenoid 0.5 m long, having 6000 turns and a current of 9 A. The radius of the solenoid

is 5 cm.

Solution:

Given, 𝐵 = 𝜇𝑜𝑛𝑖 using ,

𝒖𝑩 =𝑩𝟐

𝟐𝝁𝒐=

𝝁𝒐𝟐𝒏𝟐𝒊𝟐

𝟐𝝁𝒐,

=𝟏

𝟐× 𝟒𝝅 × 𝟏𝟎−𝟕 × (

𝟔𝟎𝟎𝟎

𝟎. 𝟓)

𝟐

× 𝟗𝟐

=7.33 x 103 J/m3

66

The volume of solenoid is V = lxA = l x πr2 = 0.5 x π x (0.05)2 =0.0039 m3

𝑈𝐵 = 𝑢𝐵 𝑥 𝑉𝑜𝑙𝑢𝑚𝑒 = 7.33𝑥103 𝑥 0.0039

= 28.6 J

3.5.9 Motional Electro-Motive Force (emf)

According to Faraday’s law, changing magnetic flux produces induced emf. However,

induced emf can result by having a relative motion of the conductor in a steady (time

independent) magnetic field. This is called the motional emf.

Consider a rod of length l which rolls with velocity v along the parallel wires that form a

U-shaped loop as shown in figure 3.5.6.

Figure 3.5.6: A rod that rolls along a parallel U-shaped wire.

As the rod pq moves, the flux through the area of the loop increases because the area

increases. In time Δt, the rod rolls a distance v Δt. As a result, the area increases by

ΔA=l(vΔt). The accompanying flux change is

∆𝜙 = 𝐵˔∆𝐴 = 𝐵˔lvΔt. The resulting emf induced is 𝑒𝑚𝑓 =∆𝜙

∆𝑡= 𝐵˔𝑣𝑙

Therefore, 𝑒𝑚𝑓 = 𝐵˔𝑣𝑙. (3.5.13)

67

Equation (3.5.13) is called Motional EMF.

Activity

A3.5.5: A rod of length 500 cm is held horizontal with its axis in east-west direction. It is

allowed to fall straight down. What is the induced electromotive force when its speed is

4.5 m/s if the earth’s magnetic field is 0.8 G at an angle of 53o below the horizontal?

(Note: 1 G = 10-4 T)

Solution:

𝑩˔ = 𝑩𝒄𝒐𝒔𝜽 = 𝑩𝒄𝒐𝒔𝟓𝟑𝒐, since v is

perpendicular to the rod.

⸫ 𝒆𝒎𝒇 = 𝑩˔𝒗𝒅

= (𝐵𝑐𝑜𝑠53𝑜)𝑥𝑣𝑥𝑙

= 0.8 x 10-4 x cos53o x 4.5 x 5

= 1.08 mV

B˔

v

B

53o

Bₗₗ

East

N

68

3.5.10 Transformers

A transformer is a device that changes the effective value of the emf. If the transformer

changes the input voltage to a higher output value, it is called a step-up transformer. If

the output voltage is smaller than the input voltage, it is called a step-down transformer.

A typical step-up transformer is shown in figure 3.5.7.

Figure 3.5.7: A typical step-up transformer

Suppose that a.c power is supplied to the primary coil, a varying current in this coil

produces a varying flux confined to the iron core. Because the flux lines tend to follow the

iron core, the lines circle through the secondary coil and thus flux in the primary is same

as flux in the secondary coils. The flux in the secondary coil gives rise to an induced emf

in it given by 𝜺𝒔 = −𝑵𝒔∆𝝓

∆𝒕. The induced emf in the primary coil is equal to the voltage

of the power supply. Thus, 𝜺𝒑 = −𝑵𝒑∆𝝓

∆𝒕. Taking ratios, we have

𝜀𝑠

𝜀𝑝=

𝑁𝑠

𝑁𝑝. (3.5.14)

Equation (3.5.14) is called the transformer equation.

It is necessary to transmit power at high voltages than on high currents because the joule

heating loss (I2R) will be much smaller if I is small.

69

Activity

A3.5.6: A power station has to transmit 10,000 W. It can be done either by sending

100 A at 100 V or 10 A at 1,000 V. If the resistance of the transmission line is 0.25 Ω,

calculate the power loss in each case. What transformer is needed to convert the lower

voltage into higher voltage?

Solution:

a) When I = 100 A, Power loss = I2R = (100)2 x 0.25 = 2500 W

When I = 10 A, Power loss = I2R = (10)2 x 0.25 = 25 W

Thus, in the first case, the generator will have to supply power 10,000 + 2500 = 12,500

W, while in the latter case, we have 10,000 + 25 = 10,025 W.

b) From above, 𝑽𝒔

𝑽𝒑=

𝑵𝒔

𝑵𝒑, =>

𝑽𝒔

𝑽𝒑=

𝟏𝟎𝟎𝟎

𝟏𝟎𝟎= 𝟏𝟎.

Thus, Ns = 10Np , the secondary transformer should have 10 times as many turns as the

Primary in order to convert 100 V into 1000 V before transmitting.

SAE 3.5.1: A square wire loop 5 cm on a side is perpendicular to a magnetic field of 0.08

T. If the field drops to 0 in 0.2 seconds, calculate the average emf induced in the loop

during this time.

Answer: 1.0 mV

SAE 3.5.2: Calculate the self-induced emf in a 0.10 H coil when the current in it is

changing at the rate of 200 A/s.

Answer: 20 V


70

SAE 3.5.3: A square wire loop 10 cm on a side is oriented with its plane perpendicular to

a magnetic field. The resistance of the loop is 5.0 Ω. How rapidly should the magnetic field

change if a current of 2.0 A is to flow in the loop?

Answer: 1.0 x 103 T/s

SAE 3.5.4: Consider a square coil of sides 5.0 cm with 100 turns positioned perpendicular

to a magnetic field of 0.60 T as shown below.

If the coil is pulled slowly to a region with B = 0 in time t = 0.10 seconds, find,

(a) the induced emf and the induced current

(b) the dissipation of energy in the coil if its resistance is 100 Ω.

Answers:

(i) Induced emf = 1.5 V and induced current I = 15 mA.

(ii) Energy dissipation = 2.3 x 10-3 J.

SAE 3.5.5: A potential difference of 50 V is suddenly applied across a 12.0 mH, 8.0 Ω

coil. Find (i) the initial current and the initial rate of change of current, (b) the current when

the rate of change of current is 2000 A/s, and (c) the final current and final rate of change

of current.

Answers: (a) 4.17 x 103 A/s; (b) 3.25 A (c) 6.25 A, final rate of change of current.

B (into)

+

+

+

+

+

+

+ +

+ v

71

SAE 3.5.6: A transformer rated at a maximum power of 10 kW is used to couple a 5000

V transmission line to a 240 V circuit. (a) What is the ratio of turns in the windings of the

transformer? (b) What is the maximum current in the 240 V circuit?

Answers: (a) 21 (b) 42 A

72

UNIT 3.6

ALTERNATING CURRENT (AC) CIRCUITS

3.6.1 Introduction

Nearly all the world’s electric energy is carried by alternating current. This is achieved by

the economy of high voltage transmission which minimizes heat losses. Alternating current

is also preferred in industry because ac electric motors are, as a class, cheaper, more durable

and require less maintenance than dc motors. In this unit, we shall explore the mechanism

of charging and discharging of a capacitor. We shall describe the root mean square values

of the current and voltage and come up with effective power losses in ac circuits. We shall

also look at resistance, capacitance and inductance circuits and their combined impedance

of their circuit. The power factor and the phase angle diagrams will be dicussed. Finally,

we will study the condition for electrical resonance.



1) Describe the charging and discharging of a capacitor

2) Calculate the power loss across a resistor in an ac circuit

3) Show that the current and the voltage are in phase across a resistor in an ac circuit

4) Evaluate the capacitive reactance for a capacitor connected in an ac circuit

5) Calculate the inductive reactance in an inductive circuit

6) Define the phase angle and the power factor in an LRC circuits

7) Derive the resonance frequency in a capacitor-inductor circuits

3.6.3 Charging and Discharging a Capacitor

The figure 3.6.1 shows a circuit diagram of a capacitor being charged by a battery through

a resistor. When switch S is closed, the battery tries to send charge clockwise around the

circuit. Initially, there is no charge on the capacitor and the current is limited only by the

resistor.

73

Figure 3.6.1: Charging a capacitor through a resistor using a battery source

Thus, at time t = 0, q = 0 and 𝑖 =𝑉𝑜

𝑅. As time goes on, the current decreases as the charge

builds on the capacitor plates. When the current decreases to zero, the charge on the

capacitor reaches the maximum value of 𝑞𝑜. Mathematically, the dependence of current i

and the charge q as a function of time during the charging process is depicted as in

equations (3.6.1) and (3.6.2) respectively.

𝒊(𝒕) = 𝒊𝒐𝒆−𝒕

𝑹𝑪⁄. (3.6.1)

Where, 𝑖𝑜 =𝑉𝑜

𝑅. The product 𝑅𝐶 is called the capacitive time constant. Equation (3.6.1)

may be written in terms of charge q as

𝒒(𝒕) = 𝒒𝒐 (𝟏 − 𝒆−𝒕

𝑹𝑪⁄ ) = 𝑪𝑽𝒐 (𝟏 − 𝒆−𝒕

𝑹𝑪⁄ ). (3.6.2)

If a resistor R is connected directly across a charged capacitor C, the capacitor will

discharge through the resistor as

𝒒(𝒕) = 𝒒𝒐𝒆−𝒕

𝑹𝑪⁄ . (3.6.3)

Thus, equation (3.6.3) is for discharging a capacitor through a resistor R.

74

Activity

A3.6.1: Some television (TV) sets have a capacitor charged to a potential difference of

about 20,000 V when connected to a power supply. As a safety measure, a resistor called

a bleeder is connected across the capacitor’s plates so that the capacitor discharges after

the set has been turned off. Given that the bleeder resistor is 106 Ω and capacitor is 10µ F,

how long must you wait after turning off the set before it is safe to touch the capacitor?

How much charge would have remained after this time?

Solution:

a) The time constant for this circuit is

𝝉𝑪 = 𝑹𝑪 = 1 x 106 x 10-6 = 10 sec

b) From 𝒒(𝒕) = 𝒒𝒐𝒆−𝒕

𝑹𝑪⁄ = 𝒒𝒐𝒆−𝟏𝟎𝑹𝑪

𝑹𝑪⁄ = 𝒒𝒐𝒆−𝟏𝟎 = 4.45 x 10-5 x qo

Thus, it is safe to touch after 10 time constants.

3.6.4 Root-Mean Square Values (RMS)

In dc circuits, the voltages and currents remain constant with time while in ac circuits, the

voltages and currents keep changing with time as shown in figure 3.6.3.

Figure 3.6.3: The changing voltage or current as a function of time from an ac source.

ac source

time

-Vo

~

B

A

Vo

75

We may write for sinusoidal voltage and current as

𝒗 = 𝒗𝒐𝒔𝒊𝒏𝝎𝒕 = 𝒗𝒐𝒔𝒊𝒏𝟐𝝅𝒇𝒕. (3.6.4)

𝑖 = 𝑖𝑜𝑠𝑖𝑛𝜔𝑡 = 𝑖𝑜𝑠𝑖𝑛2𝜋𝑓𝑡. (3.6.5)

Where f is the rotation frequency of the generator coil (50Hz in Zambia), 𝑣𝑜 is the voltage

amplitude and 𝑖𝒐 is the current amplitude. The average value of 𝒗 and 𝑖 over a cycle is

zero because a sine function is negative as much as it is positive. We therefore need to

use average value of 𝑣2 and 𝑖2. Thus, squaring the above equation, have

𝒗𝟐 = 𝒗𝒐𝟐𝒔𝒊𝒏𝟐𝟐𝝅𝒇𝒕 . (3.6.6)

The average value of 𝑠𝑖𝑛22𝜋𝑓𝑡 is 1

2 .

We define the root mean square current 𝒊𝒓𝒎𝒔 as the square root of (𝒊𝟐)𝑨𝑽.

Since, (𝑖2)𝐴𝑉 = 𝑖𝑜

2

2,

⸫ 𝑖𝑟𝑚𝑠 = √(𝑖2)𝐴𝑉 = 𝑖𝑜

√2 = 0.707𝑖𝑜. (3.6.7)

Similarly, 𝒗𝒓𝒎𝒔 =𝒗𝒐

√𝟐 = 0.707𝒗𝒐. (3.6.8)

The root mean square values are also called effective values and 𝑖𝑜 and 𝑣𝑜 are peak values.

Thus, 𝐼 =𝒊𝑜

√2 and 𝑉 =

𝒗𝑜

√2 are effective values.

Most ac meters read rms voltage and current. The advantage of using rms values is that

ohm’s law can be used for devices in ac circuits provided we replace i and v by their rms

values of I and V respectively. Thus, the power loss across a resistor R in an ac circuit is

𝑷𝑨𝑽 = (𝒊𝒗)𝑨𝑽 =𝒊𝒐

√𝟐.

𝒗𝒐

√𝟐= 𝒊𝒓𝒎𝒔. 𝒗𝒓𝒎𝒔 = 𝑰𝑽 = 𝑰𝟐𝑹. (3.6.9)

Note, in a dc system, the rms, average and instantaneous current are all the same.

76

Activity

A3.6.2: If the effective values of current and voltage are 5 A and 100 V, what are their

maximum values? What power loss will take place across a 10 resistor connected to a

power source with such rms values?

Solution:

Since, 𝐼 =𝒊𝑜

√2 or 𝑖𝒐 = 𝐼𝑥√2 = 5 x 1.414 = 7.07 A.

And 𝑉 =𝒗𝑜

√2 or 𝑣𝒐 = 𝑉𝑥√2 = 100 x 1.414 = 141.4 V

Power loss = 𝑰𝟐𝑹 = 52 x 10 = 250 W

3.6.5 Resistance Circuits

Figure 3.6.4 shows a resistor connected to an ac power supply. At any particular instant,

ohm’s law tells us that 𝑉 = 𝐼𝑅 and since the voltage across the resistor is

𝑣 = 𝑣𝑜𝑠𝑖𝑛2𝜋𝑓𝑡, then the current through the resistor is 𝑖 =𝑣

𝑅=

𝑣𝑜

𝑅𝑠𝑖𝑛2𝜋𝑓𝑡. Thus

the current and the voltage are in phase (vary in unison) but differ in amplitudes. Note

also that all power losses in simple ac circuits occur in resistors.

𝒗𝑹 = 𝒗𝒐𝒔𝒊𝒏𝟐𝝅𝒇𝒕

Figure 3.6.4 shows a resistor connected across an ac power supply.

Time

77

3.6.6 Capacitance Circuits

Figure 3.6.5 shows a capacitor connected across an ac power supply.

Figure 3.6.5 shows a capacitor connected to an ac power supply and a current curve

output for a corresponding voltage waveform input.

The voltage across the capacitor 𝑣𝐶 is given in equation (3.6.10).

𝒗𝑪 =𝒒

𝑪= 𝒗𝒐𝒔𝒊𝒏𝟐𝝅𝒇𝒕. (3.6.10)

Since C is constant, the charge oscillates in value in the same way as source voltage. The

rate change of charge on the capacitor, ∆𝒒

∆𝒕, is the current in the circuit, which is the slope

of the curve obtained when q is plotted against time t. The current has its maximum value

at t = 0, but the voltage does not reach its maximum until ¼ of cycle later. Thus current

leads voltage by 90o. The instantaneous power is

𝑷 = 𝑰𝑽 = 𝒊𝒐 𝐜𝐨𝐬 𝟐𝝅𝒇𝒕 × 𝒗𝒐𝒔𝒊𝒏𝟐𝝅𝒇𝒕. (3.6.11)

But, 𝒔𝒊𝒏𝟐𝜽 = 𝟐𝒔𝒊𝒏𝜽𝒄𝒐𝒔𝜽, then the power instantaneous power may be written as

𝑃 =1

2𝑖𝑜𝑣𝑜𝑠𝑖𝑛4𝜋𝑓𝑡. (3.6.12)

Thus, the power to the capacitor varies sinusoidally at twice the frequency of the ac voltage.

Hence the average power to the capacitor is zero. From the definition of current, ∆𝒒

∆𝒕 , if the

78

frequency is very low, (one cycle per day) the current in the circuit will essentially be zero

because the charge on the capacitor will essentially be constant. Hence, the capacitor

almost blocks (impedes) the current at low frequencies. At high frequencies, (106 Hz) the

charge on the capacitor changes so many times (million) per second. Hence, at high

frequencies, the current is high.

The ability of a capacitor to block the flow of charge is called capacitive reactance 𝑿𝒄.

From ohm’s law, 𝑽𝒄 = 𝑰𝑿𝒄 and from simple calculus, it can be shown that

𝑋𝑐 =1

2𝜋𝑓𝐶 . From here, we see that the impeding effect is large when the frequency is

small and low when frequency is large.

Activity

A3.6.3: An ac voltmeter across a capacitor reads 100 V and that C = 0.65 µF. Find the

rms current in the circuit if the frequency of the ac voltage is (a) 20 Hz and (b) 2 x 106

Hz.

Solution:

Recall: 𝑉𝑐 = 𝐼𝑋𝑐 , where 𝑉𝑐 = 100𝑉, 𝐼 = 𝑖𝒓𝒎𝒔

𝑿𝒄 =𝟏

𝟐𝝅𝒇𝑪=

𝟏

𝟐𝝅 × 𝒇 × 𝟔. 𝟓 × 𝟏𝟎−𝟕=

𝟐. 𝟒𝟓 × 𝟏𝟎𝟓

𝒇

(a) at 20 Hz, ⸫ 𝑿𝒄 =1.22 x 104 Ω

(b) at 2MHz, ⸫ 𝑿𝒄 =0.122 Ω

⸫ For f = 20 Hz, 𝐼 =𝑉

𝑋𝑐=

100

1.22×104 = 8.2 x 10-3 A.

⸫ For f = 2 x 106 Hz, 𝑰 =𝑽

𝑿𝒄=

𝟏𝟎𝟎

𝟎.𝟏𝟐𝟐= 820 A

79

3.6.7 Inductance Circuits

Figure 3.6.6 shows an inductor connected over an ac power supply.

Figure 3.6.6: An inductor connected across an ac power supply and a corresponding

current wave form in the inductor.

The voltage from the voltage source is equal to the voltage induced in the inductor.

𝑣(𝑡) = 𝑣𝑜𝑠𝑖𝑛2𝜋𝑓𝑡 = 𝐿∆𝑖

∆𝑡. (3.6.13)

Since 𝑳 is constant, the source voltage is proportional to the rate change of current. The

ac voltage leads the ac current by ¼ cycle or by 90o. On average, an inductor does not

consume energy. It stored energy in one cycle and gives it back to the source in the other

portion of the cycle. The impeding effect of an inductor is large at high frequencies and

small at low frequencies. The impeding effect of an inductor called inductive reactance

𝑿𝑳 is related to the rms current and voltage by

𝑉 = 𝐼𝑋𝐿, where

𝑿𝑳 = 𝟐𝝅𝒇𝑳. (3.5.14)

Note that the impedance effect of a resistor is independent of the frequency.

80

Activity

A3.6.4: An inductor has L = 15m H. The source voltage as read by the ac meter is 40 V

and its frequency is 50 Hz. Find the current through the inductor. Repeat for a frequency

of 6 x 105 Hz.

Solution:

For 50 Hz; 𝑿𝑳 = 𝟐𝝅𝒇𝑳 = 2 x π x 50 x 15 x 10-3 = 4.7 Ω

From 𝑽 = 𝑰𝑿𝑳 => 𝑰 =𝑽

𝑿𝑳=

𝟒𝟎

𝟒.𝟕 = 8.5 A

For f = 6 x 105 Hz, 𝑋𝐿 = 2𝜋𝑓𝐿 = 2 x π x 6 x 105 x 15 x 10-3 = 5.7 x 104 Ω

⸫ 𝑽 = 𝑰𝑿𝑳 => 𝑰 =𝑽

𝑿𝑳=

𝟒𝟎

𝟓.𝟕×𝟏𝟎𝟒 = 7.1 x 10-4 A

3.6.8 Series LRC Circuits

Figure 3.6.7 shows series connection of the capacitor, inductor and resistor to the ac

power supply.

Figure 3.6.7 Series connection of the capacitor, inductor and resistor to a power supply

81

The current in the circuit is 𝒊(𝒕) = 𝒊𝒐𝒔𝒊𝒏𝟐𝝅𝒇𝒕 and is the same in all the elements since

they are series connected. We can immediately graph the voltage v across each of the

elements as a function of time as shown in figure 3.6.8.

Figure 3.6.8: Graphs of the voltage across each of the elements shown in figure 3.6.7

i

VR

VL

VC

82

Note: 𝑽𝑹 = 𝑽𝑶𝑹𝒔𝒊𝒏𝝎𝒕 = 𝒊𝑶𝑹𝒔𝒊𝒏𝝎𝒕 (3.6.15)

𝑉𝐿 = 𝑉𝑂𝐿𝑠𝑖𝑛𝜔𝑡 = 𝑖𝑂𝑋𝐿𝑐𝑜𝑠𝜔𝑡 (3.6.16)

𝑉𝐶 = −𝑉𝑂𝐶𝑐𝑜𝑠𝜔𝑡 = 𝑖𝑂𝑋𝐶𝑐𝑜𝑠𝜔𝑡 (3.6.17)

𝑿𝑳 = 𝝎𝑳 and 𝑿𝑪 = 𝟏

𝝎𝑪. (3.6.18)

We can also see from the graphs in figures 3.6.8 that 𝑉𝐿 and 𝑉𝐶 are always of opposite

sign. An ac voltmeter across the inductor reads 𝑉𝐿 and across the capacitor reads 𝑉𝐶. If

𝑉𝐿 = 𝑉𝐶, then a voltmeter connected between a and c would read zero and NOT 𝑉𝐿 + 𝑉𝐶.

The LRC circuit obeys ohm’s law relation of 𝑽 = 𝑰𝒁, where

𝑍 = √𝑅2 + (𝑋𝐿 − 𝑋𝐶)2 . (3.6.19)

Where 𝒁 is called the impedance of the circuit. Squaring and multiplying both sides of

equation (3.6.19) by I2 , we get

𝐼2𝑍2 = 𝐼2𝑅2 + (𝐼𝑋𝐿 − 𝐼𝑋𝐶)2 and since 𝐼2𝑍2 = 𝑉2 ; 𝐼2𝑅2 = 𝑉𝑅2

, 𝐼𝑋𝐿 = 𝑉𝐿 and

𝑰𝑿𝑪 = 𝑽𝑪, we have

𝑉2 = 𝑉𝑅2 + (𝑉𝐿 − 𝑉𝐶)2 . (3.6.20)

By the Pythagorean Theorem, we have the phasor diagrams for the LRC circuit given in

figure 3.6.9. The power loss in the resistor R is I2R and can be written as

Power loss = 𝑰𝟐𝑹 =𝑽

𝒁𝑰𝑹 = 𝑽𝑰 (

𝑹

𝒁). (3.6.21)

But 𝑅

𝑍 = cosϕ and then the power loss becomes

Power loss = 𝐼𝑉𝑐𝑜𝑠𝜙 . (3.6.22)

Where 𝜙 is the phase angle between R and Z, called the phase angle. Cos𝜙 is called the

power factor. Tan𝜙 = 𝑋𝐿−𝑋𝐶

𝑅, where 𝜙 is the phase difference between the current and

voltage in the LRC circuit.

83

Figure 3.6.9: Phasor diagrams for the LRC circuit shown in figure 3.6.7

Activity

A3.6.5: A power source (V = 80 V, f = 2000 Hz) is connected in series across a 500 Ω

resistor and a 0.40 µF capacitor. Find

(a) the current in the circuit

(b) the voltmeter readings across the resistor,

(c) the reading across the capacitor and the power loss in the circuit.

Solution:

𝑿𝑳 = 𝟎, since there is no inductor.

𝑋𝐶 =1

2𝜋𝑓𝐶=

1

2𝜋𝑥2000𝑥0.4𝑥10−6= 199𝛺

𝒁 = √𝑹𝟐 + (𝑿𝑪)𝟐 = √𝟓𝟎𝟎𝟐 + (𝟏𝟗𝟗)𝟐 = 538 Ω

(a) 𝑰 =𝑽

𝒁=

𝟖𝟎

𝟓𝟑𝟖 = 0.149 A

(b) 𝑉𝑅 = 𝐼𝑅 =0.149 x 500 = 74.3 V

(c) 𝑉𝐶 = 𝐼𝑋𝐶 = 0.149 𝑥 199 = 29.7 V

(d) Power loss = 𝐼2𝑅 = 0.1492𝑥 500 = 11.1 W

VL

V (VL- VC) ϕ ϕ

Z

XL

(XL- XC)

VR R

VC XC

84

3.6.9 Electrical Resonance

Alternating current circuits that contain both capacitance and inductance show resonance.

Figure 3.6.10 shows an inductor and a capacitor connected in series to an ac circuit.

Figure 3.6.10: An inductor and a capacitor connected in series to an ac circuit.

The current in this circuit is 𝑰 =𝑽

𝒁=

𝑽

(𝑿𝑳−𝑿𝑪) . When 𝑿𝑳 = 𝑿𝑪, the current in the

circuit becomes infinite. This can be achieved since 𝑿𝑳 increases with frequency whilst

𝑿𝑪 decreases with frequency.

The dependence of the capacitive reactance and the inductive reactance is shown in

figure 3.6.11.

Figure 3.6.11: Dependence of capacitive reactance and inductive reactance as a function

of frequency.

Rea

ctan

ce

200

300

100

4 8 12

2

XL

Z=|XL - XC|

Resonance (fo)

XC

6 10 14

Frequency (kHz)

C L

~

Variable frequency

85

At resonance frequency (fo),we have 𝑿𝑳 = 𝑿𝑪, and then 𝟐𝝅𝒇𝒐𝑳 =𝟏

𝟐𝝅𝒇𝒐𝑪 , and from

which the resonance frequency may be written as

𝑓𝒐 =𝟏

𝟐𝝅√

1

𝐿𝐶 . (3.6.23)

As the frequency of the source is varied, the current in the circuit behaves as shown in

figure 3.6.12. Of course, the voltage of the oscillator must be kept constant for all

frequencies. In practical circuits, the peak is finite rather than infinite because all wires

have some resistance.

Figure 3.6.12: As the frequency of the source is varied, the current behaves as shown.

c)

d)

SAE3.6.1: What is the current in a coil of negligible resistance and inductance 0.40 H when

it is connected to a 120 V, 60 Hz power supply?

Answer: I = 0.80 A.


86

SAE3.6.2: A capacitor whose capacitive reactance is 80 Ω at 50 Hz is used in a 60 Hz

circuit. What is its capacitive reactance in the latter circuit?

Answer: 𝑿𝑪 = 66 Ω

SAE3.6.3: A resistor, a capacitor, and an inductor are connected in series across an ac

power source. The effective voltages across the circuit components are 𝑽𝑹 = 𝟓 𝑽,

𝑽𝑪 = 𝟏𝟎 𝑽, and 𝑽𝑳 = 𝟏𝟐 𝑽. Find the effective voltage of the source and the phase angle

in the circuit.

Answers: V = 5.4 V and phase angle ϕ = 22o.

SAE3.6.4: The antenna circuit of a certain radio receiver consists of a 10 mH coil and a

variable capacitor; the resistance in the circuit is 50 Ω. An 880 kHz radio wave produces a

potential difference of 1.0 x 10-4 V across the circuit. Find the capacitance required for

resonance and the current at resonance.

Answers: C = 3.3 x 10-12 F and I = 2.0 x 10-6 A.

SAE3.6.5: A circuit that contains inductance and resistance has an impedance of 50 Ω at

100 Hz and an impedance of 100 Ω at 500 Hz. What are the values of the inductance and

the resistance of the circuit?

Answers: L = 28 mH and R = 47 Ω

87

UNIT 3.7

REFLECTION AND REFRACTION OF LIGHT

3.7.1 Introduction

Geometric Optics consists of two components namely Reflection and Refraction of Light

and Optical Devices. In this unit we shall concentrate on Reflection and Refraction of Light

whilst the latter will be covered in your second year of physics education. We shall look at

the concept of light, followed by the description of the focal length for curved mirrors. We

shall derive the mirror equation and verify Snell’s law. We shall look at the conditions for

total internal reflection and image formation by lenses. Finally, we shall describe the image

formation by a combinations of lenses and evaluate their power in diopters.



1) State the speed of light in vacuum as 3.0 x 108 m/s.

2) Describe the formation of images in spherical mirrors.

3) Solve problems using the mirror equation and their magnification.

4) State Snell’s law and use it to solve problems under refraction of light.

5) Describe image formation in lenses.

6) Derive the thin lens formula and its magnification.

7) Calculate the power of lenses in combination in diopters.

3.7.3 The Concept of Light

Light waves are electromagnetic in nature and consist of an electric field perpendicular to

and in phase with magnetic field. Hence, light waves are transverse since their vibrations

are perpendicular to the direction of propagation. The visible light spectrum consists of

waves of different wavelengths as shown in figure 3.7.1.

88

Figure 3.7.1: The visible spectrum as a function of the wavelength

The letters in figure 3.7.1 refer V for violet, B for blue, G for green, Y for yellow, O for

orange, and R for red. Light waves can be reflected, refracted, interfered, polarized,

diffracted etc. The speed of light in vacuum is c = 2.99792458 x 108 m/s or approximately

c = 3.0 x 108 m/s.

3.7.4 Focal Length of a Spherical Mirror

A spherical mirror is a portion of a surface of a hollow sphere as shown in figure 3.7.2.

Figure 3.7.2: A spherical mirror showing center of curvature C, radius of the mirror R,

and principal axis of the mirror PA.

If light is reflected from the inside surface of the sphere, the mirror is called a concave

mirror. If light is reflected from the outside surface of the sphere, the mirror is called a

convex mirror as shown in figure 3.7.3 respectively.

λ (nm) V

700 400 500 600

B G Y O R

C

R

P A

Spherical mirror

Hollow sphere

89

Figure 3.7.3: Concave mirror and convex mirror respectively.

The focal length of a concave spherical mirror is half the mirror radius as seen in figure

3.7.3 above. Thus,

𝑓 =𝑅

2. (3.7.1)

Rules for locating position of the image in spherical mirrors:

1) A ray parallel to the mirror axis is reflected through the focus or appear to be

reflected from the focus for the convex mirror

2) A ray through the focus is reflected parallel to the mirror axis for concave mirror

or appears to proceed in that direction for convex mirror

3) A ray through the center of curvature is reflected through the center of curvature.

Convex mirror Concave mirror

90

3.7.5 Mirror Equation and Magnification

If p is the object distance, i is the image distance, and f, is the focal length of a mirror, an

analytical method exists that relates these quantities as shown in equation 3.7.2, known as

the Mirror Equation.

1

𝑝+

1

𝑖=

1

𝑓 . (3.7.2)

Thus, when p = ∞, i = f and when p = f, i = ∞, as it should be. We define linear

(lateral) magnification as,

𝑀 =𝑖𝑚𝑎𝑔𝑒 𝑠𝑖𝑧𝑒

𝑜𝑏𝑗𝑒𝑐𝑡 𝑠𝑖𝑧𝑒=

𝑖𝑚𝑎𝑔𝑒 𝑑𝑖𝑠𝑡𝑎𝑛𝑐𝑒

𝑜𝑏𝑗𝑒𝑐𝑡 𝑑𝑖𝑠𝑡𝑎𝑛𝑐𝑒 .

Or 𝑴 =𝒊

𝒑 𝒐𝒓 𝑴 = −

𝒊

𝒑 (for an inverted image). (3.7.3)

Activity

A3.7.1: A 3 cm high object is located a distance 50 cm from a concave mirror having a

radius of curvature of 40 cm. Calculate the (i) position (ii) size and (iii) nature of the

image.

Solution:

Height of object = 3 cm = Object size, p = 50 cm, R = 40 cm

Therefore 𝑓 =𝑅

2=

40

2 = 20 cm

(i) Using equation (3.7.2) have

𝟏

𝒑+

𝟏

𝒊=

𝟏

𝒇 =>

𝟏

𝒊=

𝟏

𝒇−

𝟏

𝒑=

𝟏

𝟐𝟎−

𝟏

𝟓𝟎=

𝟑

𝟏𝟎𝟎

=> 𝑖 =100

3= 33.3 cm (Real image since positive)

(ii) 𝑀 =𝑖𝑚𝑎𝑔𝑒 𝑠𝑖𝑧𝑒

𝑜𝑏𝑗𝑒𝑐𝑡 𝑠𝑖𝑧𝑒= −

𝒊

𝒑= −

33.3

50 = - 0.667

(iii) Image size = - 0.667 x object size = - 0.667 x 3 = - 2 cm (Negative

sign shows that image is inverted.)

91

Conventional rules on mirrors:

1) Take p, i, and f as positive on the side which light is incident on a converging

mirror (concave mirror) forming a real image of a real object. Any deviation from

this, add a negative sign.

2) For a convex mirror, no matter what the location and size of the object, the image

is always smaller, virtual and erect. Used in rear-view mirrors of cars and in big

shops to provide wider view as shown in figure 3.7.4.

Figure 3.7.4: Image formation for a convex mirror

Activity

A3.7.2: Suppose you use a convex mirror of radius of curvature of 100 cm to reflect light

from an object placed 75 cm in front of the mirror, find the nature of the image.

Solution:

Given that, p = 75 cm, f = -50 cm and 1

𝑝+

1

𝑖=

1

𝑓 . Hence,

𝟏

𝒊=

𝟏

𝒇−

𝟏

𝒑=

𝟏

−𝟓𝟎−

𝟏

𝟕𝟓 . => 𝒊 = - 30 cm (Virtual image)

𝑴 = −𝒊

𝒑= −(−

𝟑𝟎

𝟕𝟓) = +4 (Erect image)

F F O I C

p i

92

3.7.6 Refraction of Light: Snell’s Law

When light goes from one optical medium to another optical medium, the speed of light

changes at the interface, and this change in speed produces bending of light and this

bending is called refraction. Denote the angle of incidence as θ1 and angle of refraction as

θ2.

Consider the motion of the wave front in a plane wave shown in figure 3.7.5. Suppose at

time t = 0, the wave front now is at AC. After time t = t, the wave front now is at BD.

Figure 3.7.5: Refraction of a wave front as it moves from one optical medium to the

other.

The distance travelled in each material is 𝑣1𝑡 and 𝑣2𝑡, where 𝑣1 is the velocity of light in

material 1 and 𝑣2 is the velocity of light in material 2. From the two right-angled

triangles;

𝒔𝒊𝒏𝜽𝟏 =𝒗𝟏𝒕

𝑨𝑩 and 𝒔𝒊𝒏𝜽𝟐 =

𝒗𝟐𝒕

𝑨𝑩 . (3.7.4)

Taking ratios of the above equations have,

𝑠𝑖𝑛𝜃1

𝑠𝑖𝑛𝜃2=

𝑣1

𝑣2. (3.7.5)

n2

n1

Material 2

Material 1 𝜃1 𝜃1

𝜃2

𝜃2

𝑣1𝑡

𝑣2𝑡

A

C

B

D

93

Using the definition of refractive index, 𝒏 =𝑪

𝒗 => 𝒗 =

𝑪

𝒏 , we have,

𝑠𝑖𝑛𝜃1

𝑠𝑖𝑛𝜃2=

𝑣1

𝑣2=

𝐶

𝑛1𝐶

𝑛2

=𝑛2

𝑛1. (3.7.6)

Rearranging equation (3.7.6) have,

𝑛1𝑠𝑖𝑛𝜃1 = 𝑛2𝑠𝑖𝑛𝜃2 as Snell’s Law. (3.7.7)

Note:

1) If light goes from less dense medium to more dense medium, it bends towards the

normal.

2) If light goes from more optically dense medium to less optically dense medium, it

bends away from the normal. When light is incident perpendicularly to a surface

between one medium and another, it suffers no bending.

94

Activity

A3.7.3: Light is incident upon the surface od water level in a flat-bottomed dish (see

figure below). At what angle does light emerges from the bottom of the dish?

Solution:

At air-water interface, 𝑛1𝑠𝑖𝑛𝜃1 = 𝑛2𝑠𝑖𝑛𝜃2, 1.0𝑠𝑖𝑛𝜃1 = 𝑛𝑤𝑠𝑖𝑛𝜃2 -----(1)

At water-glass interface, 𝑛2𝑠𝑖𝑛𝜃2 = 𝑛3𝑠𝑖𝑛𝜃3, 𝑛𝑤𝑠𝑖𝑛𝜃2 = 𝑛𝑔𝑠𝑖𝑛𝜃3 -----(2)

⸫ 𝟏. 𝟎𝒔𝒊𝒏𝜽𝟏 = 𝒏𝒈𝒔𝒊𝒏𝜽𝟑 --------(3).

This relation shows as if water was not present.

At glass-air interface,

𝑛3𝑠𝑖𝑛𝜃3 = 𝑛1𝑠𝑖𝑛𝜃4, 𝑛𝑔𝑠𝑖𝑛𝜃3 = 1.0𝑠𝑖𝑛𝜃4 ----------(4)

Comparing equations 3 and 4 have,

𝒔𝒊𝒏𝜽𝟏 = 𝒔𝒊𝒏𝜽𝟒 and therefore, 𝜽𝟏 = 𝜽𝟒

𝜃1

𝜃2

𝜃3

𝜃4

𝜃2

𝜃3

water

𝑛2

𝑛1

𝑛3

𝑛1

95

3.7.7 Total Internal Reflection

Figure 3,7.6: Shows the total internal reflection as the angle of incidence is greater than

the critical angle.

When the angle of refraction is exactly 90o, then the angle of incidence is called the

critical angle θc. At any angle of incidence greater than the critical angle, the beam is

reflected into the same medium and is called the total internal reflection. This property

has great use in optical fiber optics as shown in figure 3.7.7

Figure 3.7.7: Application of total internal reflection in light “pipes”.

3.7.8 Image Formation by lenses

Lenses are objects made of optically transparent material, usually glass of such forms as

to converge or diverge a beam of light when it passes through them. The converging and

diverging patterns of lenses are shown in figure 3.7.8.

𝜃1

𝑛2

𝑛1

𝜃2

𝜃𝐶 𝜃1 𝜃1

90o

96

Figure 3.7.8: Converging or convex lens and diverging or concave lens respectively

Rules for locating position of images for lenses

1) A ray parallel to the principal axis passes through the focal point F of a convex

lens after refraction through the lens or appear to come from the focal point F of a

concave lens

2) A ray which passes through the center of the lens does so without bending. A

small displacement is negligible because the lens is thin.

3) A ray after passing through the focal point, hits the convex lens surface and

emerges parallel to the principal axis. In the case of a concave lens, the ray

appears to proceed towards the focal point.

F F F F

97

3.7.9 The Thin Lens Formula and Magnification

We now derive the lens formula using figure 3.7.9 show below.

Figure 3.7.9: Derivation of the lenses formula diagram

OC = p = Object distance, CI = i = Image distance, f = Focal length,

OS = Object size, IT = Image size.

From similar triangles ACF and TIF, and noting that AC = OS, we have

𝑂𝑆

𝐼𝑇=

𝐴𝐶

𝐼𝑇=

𝐶𝐹

𝐼𝐹=

𝑓

𝑖−𝑓, (3.7.8)

And similarly, from similar triangles SOC and TIC, we get

𝑆𝑂

𝑇𝐼=

𝑂𝐶

𝐶𝐼=

𝑝

𝑖. (3.7.9)

Equating equations (3.7.8) and 3.7.9), we have,

𝑓

𝑖−𝑓=

𝑝

𝑖. Cross multiplying and rearranging, we get,

=> 𝑖𝑓 + 𝑝𝑓 = 𝑝𝑖 . (3.7.10)

Dividing both sides of equation (3.7.10) by 𝑖𝑝𝑓 have,

𝟏

𝒑+

𝟏

𝒊=

𝟏

𝒇 . (3.7.11)

Equation (3.7.11) is known as the thin lens formula. This equation holds for both

converging and diverging lenses under the following sign conventions.

T

2F 2F

F

F O

S

C

A

I

p i

f

98

1) f is positive for a convex lens and negative for concave lens

2) p is positive for real object and negative for virtual object.

3) i is positive for a real image and negative for a virtual image

Hence, we define linear magnification = 𝒊𝒎𝒂𝒈𝒆 𝒔𝒊𝒛𝒆

𝒐𝒃𝒋𝒆𝒄𝒕 𝒔𝒊𝒛𝒆=

𝒊

𝒑=

𝑰𝑻

𝑶𝑺 . (3.7.12)

Or Magnification M = −𝑖𝑚𝑎𝑔𝑒 𝑠𝑖𝑧𝑒

𝑜𝑏𝑗𝑒𝑐𝑡 𝑠𝑖𝑧𝑒= −

𝒊

𝒑 . (3.7.13)

Activity

A3.7.4: A diverging lens with a 20 cm focal length forms an image of a 3.0 cm high

object placed 40 cm in front of the lens. Find the nature of the image (i.e image position,

magnification, upright or inverted image).

Solution:

Given; Object size = 3.0 cm, f = - 20 cm (since diverging lens), p = 40 cm.

Rearranging the lens formula have,

𝟏

𝒊=

𝟏

𝒇−

𝟏

𝒑=

𝟏

−𝟐𝟎−

𝟏

𝟒𝟎

𝑖 =𝑝𝑓

𝑝−𝒇= −13.3 cm

M = −𝒊𝒎𝒂𝒈𝒆 𝒔𝒊𝒛𝒆

𝒐𝒃𝒋𝒆𝒄𝒕 𝒔𝒊𝒛𝒆= −

𝒊

𝒑.

⸫ image size = i/p x object size = - (-13.3/40) x 3 = +2.99 cm (upright)

99

3.7.10 Combination of Lenses and Lenses in Close Contact

Most optical instruments you will study later, use more than one lens. We take the image

formed by the first lens to be the object to the second lens and use the lens formula as

explained in section 3.7.9.

More than one lens used together almost in contact, cooperate and produce an effect

depending on their powers. The power of the lens is measured in Diopters D.

Thus,

𝑫𝑪 = 𝑫𝟏 + 𝑫𝟐 . (3.7.14)

Where 𝐷𝐶 =1

𝑓𝐶 , 𝐷1 =

1

𝑓1 , and 𝐷2 =

1

𝑓2 . And all focal lengths have to be in meters.

Thus, the power of a lens in diopters, is the reciprocal of the focal length in meters.

Activity

A3.7.5: An object 4 cm high is placed at a distance of 40 cm from a converging lens of

focal length 15 cm. A diverging lens of focal length -10 cm is placed a distance of 30 cm

from the first lens. Find the position, size and nature of the final image.

Solution:

1st Lens: 𝒑𝟏 = 𝟒𝟎 𝒄𝒎, 𝒇𝟏 = 𝟏𝟓 𝒄𝒎

𝟏

𝒊𝟏=

𝟏

𝒇𝟏−

𝟏

𝒑𝟏=

𝟏

𝟏𝟓−

𝟏

𝟒𝟎 . => 𝒊𝟏 = 𝟐𝟒 𝒄𝒎.

𝑚1 = −𝑖1

𝑝1= −

24

40= −

3

5 (Smaller and inverted)

2nd Lens: 𝒑𝟐 = 𝟑𝟎 − 𝟐𝟒 = 𝟔 𝒄𝒎, 𝒇𝟐 = −𝟏𝟎 𝒄𝒎

𝟏

𝒊𝟐=

𝟏

𝒇𝟐−

𝟏

𝒑𝟐=

𝟏

−𝟏𝟎−

𝟏

𝟔 . => 𝒊𝟐 = −𝟑. 𝟕𝟓 𝒄𝒎. (Virtual image)

𝑚2 = −𝑖2

𝑝2= − (

−3.75

6) = +

5

8 (Upright relative to its object i.e first image)

100

Thus, the total 𝑴 = 𝒎𝟏 × 𝒎𝟐 = −𝟑

𝟓×

𝟓

𝟖= −

𝟑

𝟖. (Inverted with respect to 1st object

and smaller in size.

SAE 3.7.1: A candle 5 cm high is placed 40 cm from a concave mirror whose radius of

curvature is 60 cm. Find the (a) position (b) nature of the image and (c) size of image.

Answers: (a) i = 120 cm, (b) Real (c) Size is -15 cm, or three times as large as the

candle itself.

SAE 3.7.2: A movie director holds a diverging lens 5.0 m from an actress and sees her

one-tenth of her normal size. Calculate the focal length of the lens.

Answer: -556 mm.

SAE 3.7.3: Find the critical angle for total internal refection in a diamond crystal when

the diamond is (a) in air and (b) immersed in water.

Answers: (a) 24o (b) 33o

SAE 3.7.4: An erect object is placed a distance in front of a converging lens equal to

twice the focal length 𝑓1 of the lens. On the other side of the lens is a converging mirror

of focal length 𝑓2 separated from the lens by a distance 2(𝑓1 + 𝑓2). Find the (a) position

(b) nature (real or virtual) and (c) relative size of the final image.

Answers: (a) 𝒊𝟐 = 𝟐𝒇𝟐 (b) Since 𝒊𝟐 is positive, then real image.

(c) M = 1 (Erect relative to 1st object).


101

SAE 3.7.5: A converging lens 30 cm from a candle produces a. inverted image of the

candle on a screen 30 cm from the other side of the lens. A diverging lens is placed midway

between the candle and the other lens. When the candle is moved back 10 cm from its

original position, its image reappears on the screen. Find the focal lengths of the two lenses.

Answers: (i) f (Converging lens) = +15 cm and (ii) f (Diverging lens) = - 37.5 cm.

GOOD - LUCK

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