PLUS TWO PHYSICS

(HSE, VHSE, CBSE)

Class – XII

Prof. C K Xavier M.Sc., M.Phil.,B.Ed & Prof. P M Joy M.Sc.,M.Ed , M.Phil

Associate Professor Associate Professor

Head of the Department of Physics Department of Physics

St. Thomas’ College, Thrissur St. Thomas’ College, Thrissur

(Former member, UG Board of

Studies, University of Calicut)

ROSARY PUBLISHERS (P) LTD

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PLUS TWO PHYSICS

(HSE, CBSE, VHSE)

Class – XII

XAVIER & JOY Chiramal (H) Pellissery (H) Peacock Nagar Paradise Avenue

Pipe Line Road Palakkal, Palissery P O

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PREFACE

This book is a continuation of our previous book “PLUS ONE PHYSICS”, which was the outcome of a survey we conducted among the students of various higher secondary schools and the discussions we made with the higher secondary school teachers. Though this book is strictly based on the SCERT syllabus, it covers almost all topics of the NCERT syllabus. This book will fulfill the requirements of HSE, VHSE and CBSE students as well.

This book is presented in a most simple and lucid manner. We have taken much pain to explain the difficult points in such a manner that an average student can easily grasp even the tough topics. Small and simple sentences have been used through out the book to get rid of the hurdles of English language. We are sure that our book will cater the interest of the average, bright and the brilliant students alike.

A large number of objective questions, analysis type short questions and problems mostly asked in previous question papers have

been worked out. This is highly suitable for the students facing the

newly introduced grading system. The most important feature of this book is that we have arranged many applied and analysis type short

questions with their answers in the bottom portions of the pages where the related subject matter is dealt with. We have given sample questions

for every chapters and model question papers.

Though our treatment is simple and lucid, we have not forgotten to include new intellectual level short questions and problems at the end of each chapter, mainly based on the NCERT text book, keeping in mind the bright students. There are a lot of exercise problems with every chapter. This book contains a large number of objective questions with answer keys for the students to prepare for various entrance examinations, with much confidence

We express our sincere gratitude to the teachers and students of the Higher secondary, V.H.S.E and C.B.S.E for accepting our book PLUS ONE PHYSICS. We thankfully recognize all those who inspired and helped us in this endeavour. Suggestions or corrections for the improvement of this book from any quarter will be thankfully accepted and will be applied in the next revised edition.

Xavier & Joy

PREFACE TO THE FOURTH REVISED EDITION

We have great pleasure to introduce the fourth revised edition of the “PLUS TWO PHYSICS” for the benefit of all Higher Secondary students. We have made the necessary modifications suggested by the teachers and the students and have done the corrections wherever necessary.

This book is modified to cover all NCERT syllabus and we have made the topics as simple as possible incorporating all important points. We have added a multitude of grading level questions with their answers after every chapter based on the Source book, Edumate and Fundamentals of Physics by

Resnick and Halliday. This will enable the students to understand the principle of physics well. We have added seven question papers of the Higher Secondary examinations with their answers and three sample question papers at the end of the book. We have given answers to the questions of NCERT exercises after every chapter.

We express our deep gratitude to the teachers and students for accepting our previous editions and for making valuable suggestions.

Suggestions for improvement of the book are always welcome.

Xavier & Joy

CONTENTS

CHAPTER 1 ELECTRIC CHARGES AND FIELDS

Introduction 1.1

Electric Charges 1.1

Conductors and Insulators 1.3

Charging by Induction 1.4

Basic properties of Electric Charges 1.5

Coulomb’s Law 1.6

Forces between Multiple Charges 1.8

Electric Field 1.9

Electric Field Lines 1.10

Electric Flux 1.11

Electric Dipole 1.11

Dipole in a Uniform External Field 1.14

Continuous Charge Distribution 1.15

Gauss’ Law 1.15

Application of Gauss’ Law 1.16

CHAPTER 2 ELECTROSTATIC POTENTIAL AND CAPACITANCE

Introduction 2.1

Electrostatic Potential 2.1

Potential due to a Point Charge 2.2

Potential due to an Electric Dipole 2.3

Potential due to a System of Charges 2.4

Equipotential Surfaces 2.5

Potential Energy of a System of Charges 2.6

Potential Energy of a Dipole in an External field 2.8

Electrostatics of Conductors 2.9

Dielectrics and Polarisation 2.10

Capacitors and Capacitance 2.10

The Parallel Plate Capacitor 2.11

Effect of Dielectric on Capacitance 2.12

Combination of capacitors 2.13

Energy Stored in a Capacitor 2.15

Van de Graff Generator 2.16

CHAPTER 3 CURRENT ELECTRICITY

Introduction 3.1

Electric Current 3.1

Electric Currents in Conductors 3.1

Ohm’s law 3.2

Drift of Electrons and the Origin of Resistivity 3.2

Limitations of Ohm’s Law 3.4

Resistivity of various Materials 3.5

Temperature Dependence of Resistivity 3.6

Combination of Resistors – Series and Parallel 3.10

Cells. Emf, Internal Resistance 3.11

Cells in series and Parallel 3.12

Kirchhoff’s Laws 3.13

Wheatstone Bridge 3.15

Meter Bridge 3.16

Potentiometer 3.17

Electrical Energy, Power 3.20

CHAPTER 4 MOVING CHARGES AND MAGNETISM

Introduction 4.1

Magnetic Force 4.1

Motion in a Magnetic Field 4.3

Motion in Combined Electric and Magnetic Fields 4.4

Magnetic Field due to a Current Element, Biot-Savart Law 4.6

Magnetic Field on the Axis of a Circular Current Loop 4.7

Ampere’s Circuital Law 4.11

Magnetic Induction of Solenoid and the Toroid 4.11-4.13

Force between Two Parallel Currents, the Ampere 4.13

Torque on Moving Coil Galvanometer 4.17

CHAPTER 5 MAGANETISM AND MATTER

Introduction 5.1

The Bar Magnet 5.1

The Magnetism and Gauss’ Law 5.7

The Earth Magnetism 5.7

Magnetisation and Magnetic Intensity 5.14

Magnetic Properties of Materials 5.15

Permanent Magnets and Electromagnets 5.18

CHAPTER 6 ELECTROMAGNETIC Induction

Introduction 6.1

The Experiments of Faraday and Henry 6.1

Magnetic Flux 6.2

Faraday’s Law of Induction 6.2

Lenz’s Law and Conservation of Energy 6.4

Motional Electromotive Force 6.5

Energy Consideration: A Quantitative Study 6.5

Eddy Currents 6.6

Inductance 6.7

AC Generator 6.13

CHAPTER 7 ALTERNATING CURRENT

Introduction 7.1

AC Voltage Applied to a Resistor 7.1

Representation of AC Current and Voltage by Rotating Vectors-Phasors 7.1

AC Voltage Applied to an Inductor 7.2

AC Voltage Applied to a Capacitor 7.3

AC Voltage Applied to a Series LCR Circuit 7.4

Power in AC Circuit: The Power Factor 7.7

LC Oscillations 7.7

Transformers 7.8

CHAPTER 8 ELECTROMAGNETIC WAVES

Introduction 8.1

Displacement Current 8.1

Electromagnetic Waves 8.3

Electromagnetic Spectrum 8.5

CHAPTER 9 RAY OPTICS AND OPTICAL INSTRUMENTS

Introduction 9.1

Reflection of Light by Spherical Mirrors 9.2

Refraction 9.8

Total Internal Reflection 9.11

Refraction at Spherical Surfaces and by Lenses 9.13

Refraction through a Prism 9.20

Dispersion by a Prism 9.22

Some Natural Phenomena due to Sunlight 9.24

Optical Instruments 9.26

CHAPTER 10 WAVE OPTICS

Introduction 10.1

Huygens Principle 10.1

Refraction and Reflection of Plane Waves using Huygens Principle 10.2

Coherent and Incoherent Addition of Waves 10.4

Interference of Light Waves and Young’s Experiment 10.6

Diffraction 10.9

Polarisation 10.13

CHAPTER 11 DUAL NATURE OF RADIATION AND MATTER

INTRODUCTION 11.1

Electron Emission 11.1

Photoelectric Effect 11.2

Experimental Study of Photoelectric Effect 11.2

Photoelectric Effect and Wave Theory of Light 11.4

Einstein’s Photoelectric Equation: Energy Quantum of Radiation 11.4

Particle Nature of Light: The Photon 11.6

Wave Nature of Matter 11.7

Davisson and Germer Experiment 11.8

CHAPTER 12 ATOMS

Introduction 12.1

Alpha-Particle Scattering and Rutherford’s Nuclear Model of Atom 12.1

Atomic Spectra 12.3

Bohr Model of the Hydrogen Atom 12.4

The Line Spectra of the Hydrogen Atom 12.6

De Broglie’s Explanation of Bohr’s Second Postulate of Quantisation 12.6

CHAPTER 13 NUCLEI

Introduction 13.1

Atomic Masses and Composition of Nucleus 13.1

Size of the Nucleus 13.2

Mass-Energy and Nuclear Binding Energy 13.3

Nuclear Force 13.5

Radioactivity 13.5

Nuclear Energy 13.10

CHAPTER 14 SEMICONDUCTOR ELECTRONIC: MATERIALS, DEVICES AND SIMPLE CIRCUITS

Introduction 14.1

Classification of Metals, Conductors and Semiconductors 14.2

Intrinsic Semiconductor 14.3

Extrinsic Semiconductor 14.3

p-n Junction 14.4

Semiconductor diode 14.5

Application of Junction Diode as a Rectifier 14.7

Special Purpose p-n Junction Diodes 14.8

Junction Transistor 14.10

Digital Electronics and Logic Gates 14.19

Integrated Circuits 14.23

CHAPTER 15 COMMUNICATION SYSTEMS

Introduction 15.1

Elements of a Communication System 15.1

Basic Terminology Used in Electronic Communications Systems 15.1

Bandwidth of Signals 15.3

Bandwidth of Transmission Medium 15.3

Propagation of Electromagnetic Waves 15.3

Modulation and its Necessity 15.5

Amplitude Modulation 15.6

Production of Amplitude Modulated Wave 15.8

Detection of Amplitude Modulated Wave 15.9

APPENDIX 7 Solved and 3 Sample Question Papers 1-45

Total 592

CHAPTER 3

CURRENT ELECTRICITY

INTRODUCTION

The branch of physics which deals with the motion of electric charges is called Current Electricity. In a metallic conductor, the electrons loosely bound to the atoms are in random motion. These electrons are called free electrons. The net flow of charge at any point will be zero. The external energy required to drive the free electrons in a definite direction is called electromotive force (e.m.f). The e.m.f ‘E’

is the work done in moving a unit positive charge from one end to the other. The electric current is defined as the rate of flow of charges. The unit of e.m.f is volt or Joule/Coulomb.

ELECTRIC CURRENT

Electric current is defined as the rate of flow of charges across any cross

sectional area of a conductor. If Q is the charge flowing through the cross section

of a conductor in a time t , then the current I =t

Q

. Current at the instant„t‟ is the

ratio of Q to t in the limit 0t .

i.e, I (t) = 0t

Ltt

Q

It is a scalar quantity and its unit is ampere (A).

Ampere: If q = 1C and t =1s, then, I = t

q =

s 1

C 1= 1A. Current through a

conductor is said to be one Ampere if one coulomb charge flows through any cross section in one second. By convention, the direction of current is the direction of

flow of positive charge. In metallic conductors, the current is due to the flow of electrons. Hence the direction of current is opposite to the direction of flow of

electrons.

ELECTRIC CURRENT IN CONDUCTORS

When an electric field is applied on an electric charge it will experience a force. If it is free it will move producing an electric current. In atoms and molecule, the negatively charged electrons and positively charged nuclei are bounded to each other. Bulk matter is made up of many molecules, e.g. one gram of water contains

approximately 2210 molecule. These molecules are so closely packed that the

electrons are no longer attached to the individual nuclei. In metals there are a number of free electrons within the bulk material. When electric field is applied, these free electrons will move opposite to the direction of electric field.

SQ1) Differentiate e.m.f and potential difference.

Ans: The e.m.f is a term associated with the source of electricity and it does not depend on the resistance of the circuit. e.m.f is the difference in potentials between the terminals of the cells when the circuit is open. Potential difference is the difference in potentials between any two points of a circuit when the circuit is closed. Potential difference exists throughout the circuit and it is proportional to the resistance of the circuit. (e.m.f and potential difference are measured in volt) SQ2) When does charge move from one point to another?

Ans: Flow of charges between two points can take place only if there is a potential difference between them and a conducting path.

Current Electricity 3-2

For example consider a cylindrical conductor of radius R as shown in fig. There are two dielectric discs of same radii with a charge +Q distributed over one flat surface and –Q over the other flat surface of the cylinder. Now an electric field is created from +Q to –Q. The free electron will be accelerated towards +Q contributing a current. This will exist for a very short time and no current thereafter. For a continuous current there should be a mechanism like cells or batteries to supply fresh charges to the end of the cylinder

OHM’S LAW

George Simon Ohm established the relationship between the potential difference and current. This is known as Ohm‟s law. Ohm’s law states that at constant temperature the current flowing through a conductor is directly

proportional to the potential difference between the ends of

the conductor.i.e., I V or V I, V = IR. The proportionality constant R is known as the resistance of the conductor. The unit

of resistance is ohm .

R = I

V. When V= 1 volt and I = 1ampere, R = 1 . Hence one ohm is the

resistance of a conductor, if 1A current flows through it when a p.d of 1V is

applied.

Ohm‟s law holds good only when a steady current flows through a conductor. According to ohm‟s law the graph connecting voltage and current will be straight line passing through origin as shown in the fig.3.2

DRIFT OF ELECTRONS AND ORIGIN OF RESISTIVITY

In the absence of electric field the free electrons in a conductor will move in all possible directions. This is the thermal speed of electrons just like molecular speed of a gas. The thermal speed is the order of 105 ms-1. But the average velocity of electron is zero and hence no net current is produced

SQ3) What are the carriers of current in metals, electrolytes, semiconductors and gases? Ans: i) In metals, the current carriers are electrons. ii) In electrolytes, the current carriers are positive ions and negative ions. iii) In semiconductors, the current carriers are negatively

charged electrons and positively charged holes. iv) In gases, the current carriers are negatively and positively charged ions SQ4) What is current due to random motion of free electrons in a conductor? Ans: The current due to random motion of free electrons is zero SQ5) What is the reason for death due to electric shock? Ans: Nerve impulses are adversely affected by the electric current. So heart beat is affected leading to death. SQ6) Explain why average time between two successive collisions is called relaxation time? Ans: It is called relaxation time because, it is the time taken by the system to relax back to thermal equilibrium.

I

V

Fig.3.2

Fig.3.1

3-3 Current Electricity

When an electric field E is applied between the ends of a conductor, the electrons will experience a force F = Ee in a direction opposite to the electric field and thus they are accelerated. The accelerated electrons collide with one another and also with the positive ions in the conductor. Thus electrons lose their energy by collision. They are again accelerated by the electric field. Between two successive collisions, electrons acquire velocity in a direction opposite to the direction of electric field.

The average time between two successive collisions is called relaxation

time ( ) (the value of relaxation time is about 10-14 second)

The average velocity with which the free electrons are drifted under the

influence of electric field is called drift velocity (vd). Drift velocity is very small

and is the order of 1mm/sec.

Mobility

The acceleration produced on the electron, a = mass

electrontheonforce =

m

Ee.

The drift velocity between two successive collisions, dv = a = m

Ee =E,

where m

τe is called the mobility. Mobility is defined as the drift velocity

acquired per unit electric field. It is a positive quantity. Its unit is 112 sVm .

Relation between current and drift velocity

Consider a conductor of length l and area of cross-section A. Let n be the number of free electrons per unit volume and vd be the drift velocity. In one second the number of charges crossing any cross section will be those electrons contained in a length equal to vd of the conductor is equal to vdAn. Total charge flowing in one second or current, I = nAe vd.

Current density (j)

Current density ( j ) at a point in a conductor is defined as the amount of current flowing through unit area of the conductor around that point,

provided the area is held in a direction perpendicular to the current

direction.. i.e., Current density, j = A

I = dven

A

vnAe d . Current density is a

vector quantity. Its unit is Am-2.

SQ7) How can we get large amount of current though the free electrons have small drift speed and small charge? Ans: This is because electron number density is very large, about 1029 per m3. SQ8) Do all the free electrons of a conductor move in the same direction i.e. from lower to higher potential? Ans: No. The velocity is in a random direction for each free electron of a conductor. But all of them will have a drift from lower potential to higher potential. SQ9) Explain why the electrons acquire a steady average drift velocity even though they are accelerated by the applied field Ans: Accelerated electrons collide with the metal ions to lose their drift speed. This continues through out the motion. So only a steady average drift velocity is acquired by the free electrons

Current Electricity 3-4

LIMITATIONS OF OHM’S LAW

Conductors which do not obey Ohm‟s law are called non-ohmic conductors. Ex: vacuum tube, semiconductor diode, electrolytes and conducting gases etc.

In many cases the relation between voltage and current is different from Ohm‟s law. The important exceptions are

1) V depends on I nonlinearly. Ex: For a junction diode, the current voltage relation is non linear as shown in the fig.(a).

2) The relation between V and I depends on the sign of V. For a junction diode, the resistance depends on the sign of V as shown in the fig.(b).

3) The relation between V and I is non-unique. i.e., for the same current, there is more than one value of voltage V. For a thyristor, the V-I characteristic curve shows different current for the same potential difference V as in fig.(c). The portion BA shows that the current decreases on increasing the potential difference.

SQ10) The drift velocity is very small. How is then current established almost at the instant a circuit is closed?

Ans: The establishment of electric field takes place throughout the circuit with the speed of light. So every point of the circuit experiences an electron drift. So current is established almost instantly. SQ11) A wire is carrying current. Is it charged?

Ans: No. The current is due to the flow of charges. The number of electrons entering from one end of the wire is equal to the number of electrons leaving from the other end. SQ12) What is the nature of the graph plotted with I along X-axis and V along Y axis for a conductor? What does the slope give?

Ans: It is a straight line graph passing through the origin.

The slope of the graph gives V/I i.e. resistance

SQ13) When we switch on a heater, the bulbs get dimmer in our house. Why? Ans: The heater draws large current at the starting. So the current flowing through the bulbs becomes low. So power of bulbs decreases and bulbs get dimmer

SQ14) Explain why aluminum wires are preferred for overhead power cables?

Ans: Aluminium wire is preferred because i) It has low density (i.e. lighter) ii) It has lower resistance and hence less power loss and iii) it is cheaper, when compared with copper.

0.2

I 1.5 mA

1A V

-2

I

V

B A

2 I

V

Fig.3.3.(a) Fig.3.3.(b)

Fig.3.3.(c)

V

I

3-5 Current Electricity

Elecrtical resistance and Resistivity

The resistance of a conductor R is directly proportional to the length of the conductor l and it is inversely proportional to the area of cross- section A.

R A

l or R ρ

A

l, where ρ is called electrical resistivity or specific

resistance of the material of the conductor. The unit of is ohm-metre (m).

l

RAρ . When l = 1m and A= 1m2, R = . The resistivity of a material is

equal to the resistance offered by a conductor of unit length having unit area of cross-section. The resistivity is a constant for the material of the conductor, while the resistance depends on length, area of cross-section, material and temperature of the conductor.

Conductance and Conductivity

Conductance (G) is the reciprocal of the resistance. i.e., G =R

1. The unit of

conductance is mho or Siemens (S). Conductivity σ is the reciprocal of resistivity,

= ρ

1. The unit of conductivity is mho m-1 or Sm-1.

RESISTIVITY OF VARIOUS MATERIALS

The materials can be classified into conductors, semiconductors and insulators. Metals like Silver, Aluminium, Copper, Iron, Tungsten etc are good conductors of electricity. The resistivity of conductors is in the range of 10-8 to

m10-6. Semiconductors are partially conducting materials. Ex: Silicon, Germanium, Carbon, Graphite etc

SQ15) Deduce ohm’s law from drift velocity dv Obtain expression for resistance and

resistivity?

Ans: Drift velocity dv = a where a is the acceleration produced on the free electron on the

application of a p.d V and is the average time between two collisions.

;m

eEa But E

l

V a

lm

eV , avd

lm

eV

The relation between current and drift velocity is given by I = neAvd = nAe m

e

l

V ,

I = Vτml

nAe2

, where τ

ml

nAe2

is a constant = R

1 for the given conductor. I V

The resistance of the conductor, R = 2nAe

ml

Resistivity ρ = l

RA =

τnAe

ml2

l

A =

τne

m2

Current Electricity 3-6

The resistivity of semiconductors is in the range of 10-5 to 103 m. Insulators are bad conductors of electricity. The resistivity is in the range of 1014 to 1016 m. Ex: Glass, Mica, Paper, Amber, Quartz, Wood etc.

Material Resistivity, C0atmΩρ 0

A. Conductors

Silver 8106.1

Copper 8107.1

Aluminium 8107.2

Tungsten 8106.5

Iron 81010

Platinum 81011

Mercury 81098

Nichrome( alloy of Ni, Fe, Cr) ~ 810100

Manganin (alloy) 81048

B. Semiconductors

Carbon (Graphite) 5105.3

Germanium 46.0

Silicon 2300

C. Insulators

Glass 1410 1010

Hard Rubber 1613 1010

NaCl ~ 1410

Fused Quartz ~ 1610

TEMPERATURE DEPENDENCE OF RESISTANCE

Generally the resistance of materials (metals) will increase with the increase of temperature. The relation connecting resistance and relaxation time is

given by R=2nAe

mli.e., R

1. On heating, amplitude of vibration of lattice ions will

increase. As the relaxation time decreases, the number of collisions of the free electrons with the lattice will also increase. This is the cause of increase of resistance with temperature.

SQ16) A wire is drawn into double its original length. What will be increase in its i) resistance

and ii) resistivity?

Ans: i) When length is doubled, the area will be halved R = A

l=

2/A

2l=

A

4 l = 4R. So

increase in resistance= 4R-R = 3R ii) Resistivity will not change as it is independent of the dimensions of the wire but depends upon the material of the wire.

SQ17) Resistivity of copper, constantan and silver are 1.7 10-6 cm, 39.1 106 cm and 10-6

cm respectively. Which is the best conductor and why?

Ans: Silver is the best conductor since its resistivity is the least

3-7 Current Electricity

The resistance at temperature t0C is given by Rt = R0 (1+ 2tt ), where R0 is

resistance at 00C. and are constants. As β is very small, for small change in

temperature the term t2 can be neglected.

Rt = R0 (1+ t ), where is called the temperature coefficient of resistance

of the material and it is given by tR

RR

0

0t

Temperature coefficient of resistance

Temperature coefficient of resistance of the material of a conductor is

defined as the ratio of increase in resistance to its resistance at 0°C per

degree rise of temperature.

If R1 is the resistance at t10C and R2, the resistance at t2

0C, then

R1= R0 (1+ 1t ) and R2 =R0 (1+ 2t ) or 2

1

R

R=

2

1

t1

t1

By cross multiplying and arranging, we get R1+ R1 2t = R2+ R2 1t

or (R1 2t -R2 1t ) = R2- R1 = 1221

12

tRtR

RR

Negative temperature coefficient

For certain materials, the resistance will decrease with increase of temperature. i.e., the temperature coefficient of resistance is negative. Ex:

Silicon, Germanium, Carbon in the form of graphite etc.

Resistance of alloys

Alloys can be prepared such that temperature coefficient of resistance is almost equal to zero. These are the materials of zero temperature coefficient of resistance. Ex: Manganin, Constantan, Eureka, Nichrome etc. The resistivity of these alloys is high, and hence these are used in making resistance boxes and standard resistance coils.

Resistivity of semiconductors and insulators with temperature

The temperature dependence of resistivity of semiconductors and insulators is

given by /kTE

0geρρ where Eg is the energy gap between conduction band and

valence band, k Boltzmann constant and T absolute temperature. For semiconductors Eg1eV and for insulators Eg 5eV. Hence the resistivity is very high for the insulator compared with semiconductors.

SQ18) Which has greater resistance, metal or alloy? Ans: The resistance of alloy is generally greater than that of metal. SQ19) What is the nature of graph plotted with temperature in degree Celsius along X-axis and resistance along Y- axis? What does i) the slope and ii) y- intercept give? Ans: Rt = R0(1+ t ). i.e., Rt = R0+ R0 t

This is similar to the equation of a straight line y=mx + c. Therefore the graph between Rt and t will be a straight line.

i) Slope gives (R0) and ii) The y-intercept gives R0.

R0

Rt

t 0c

A

B

Current Electricity 3-8

SUPERCONDUCTIVITY

The electrical resistivity of many metals and alloys drops suddenly to

zero when they are cooled to a very low temperature called the critical temperature Tc (Transition temperature). This phenomenon is called superconductivity. The materials showing super conductivity are called superconductors.

The phenomenon of superconductivity was first discovered by H.Kamerlingh Onnes in 1911 in mercury. The critical temperature of mercury was found to be 4.2 K. Current once set up in a superconductor will persist for a very long time. The first theoretical explanation of super conductivity was given by Bardeen, Cooper and Schrieffer in 1957 and it is called the BCS theory.

Applications

1) Superconductors are used for making superconducting magnets.

2) Superconductors can be used as memory or storage elements in high speed computers.

3) Superconducting wires can be used in transmission lines.

4) Superconductors play an important role in material science, research and high energy particle physics.

Thermistor

Thermistors are heat sensitive resistors made from semiconductors. The resistance of thermistors changes appreciably with rise in temperature. The temperature coefficient of resistance of a thermistor may be positive or negative. Thermistors are made in the form of bead, disc or rod.

Uses: Thermistors are commonly used,

(1) In electronic industry.

2) To measure very low temperature and very low change in temperature.

3) To safeguard the filament of the picture tube of a television from current variation.

4) To control the temperature

SQ20) What is the effect of temperature on the resistivity of alloys? Ans: The resistivity of alloys increases with the rise in temperature. SQ21) What is the effect of rise in the temperature on the electrical conductivity of metals and semiconductors?

Ans: The conductivity of metal decreases and the conductivity of semiconductor increases with the rise in temperature. SQ22) Choose the correct alternative: a) Alloys usually have much lower/higher temperature coefficients of resistance than pure metals. b) The resistivity of the alloy manganin is nearly independent of/increases rapidly with, increase of temperature. Ans: a) lower b) nearly independent of

3-9 Current Electricity

Colour code of Resistance

Carbon resistors consist of ceramic core with a deposit of a thin layer of crystalline carbon. The resistance depends on the amount of carbon. The main advantages of carbon resistors are compactness and low cost.

The colour code is used to indicate the value of resistance and its tolerance or percentage of reliability. The resistors have a set of concentric rings or bands of different colours. The colour code of carbon resistance is given in the table.

The first two bands from the end

indicate the first two significant figures of the resistance. The third band indicates the number of zeroes (decimal multipliers) after the two significant figures. The fourth band indicates the tolerance in percentage. The colour code can be obtained by remembering, “B.B ROY of Great Britain has a Very Good Wife”. The capital letters of the sentence represent Black to White. First B is for black (0), Second B for brown (1), R for red (2), O for orange (3), Y for yellow (4), G for green (5), B for blue (6), V for violet (7), G for grey (8) and W for white (9).

Examples

SQ23) What is the cause of superconductivity? Ans: At and below the critical temperature the electrons move in pairs in a coherent manner. The lattice ions are not able to deflect this cooperative and coherent movement of electrons.

These pairs of electrons are called Cooper pairs.

SQ24) (a) Resistivity of copper, constantan and silver are cmcm 66 11.39,107.1 and

cm610 respectively. Which is the best conductor and why?

Ans: Silver is the best conductor since its resistivity is the least (b) What are the difference between conductance and conductivity? Ans: The reciprocal of the resistance is conductance. The reciprocal of resistivity is conductivity. Conductance is the property associated with the given conductor. But conductivity is the property associated with the material of the conductor.

Colour Number Multiplier Tolerance

(%)

Black 0 1

Brown 1 101

Red 2 102

Orange 3 103

Yellow 4 104

Green 5 105

Blue 6 106

Violet 7 107

Gray 8 108

White 9 109

Gold 10-1 5

Silver 10-2 10

No

Colour

20

Yellow Violet Orange

Silver

47 47kΩorΩ47000or103

Tolerance %10 (silver) Hence resistance is 47kΩ %10

Yellow-4 Violet-7 Orange-103

Brown Black Yellow

Gold

10 kΩ001orΩ000001or104

Tolerance %5 (Gold) Hence resistance is 100kΩ %5

brown-1 black-0 yellow-104

Fig.3.4

Current Electricity 3-10

COMBINATION OF RESISTORS – SERIES AND PARALLEL

There are two types of combinations for the resistors

1) Series combination and 2) Parallel Combination

Resistors in series

Three resistors with resistances R1, R2 and

R3 are connected in series (end to end connection) as shown in the figure.

Let V be the p.d. applied between the ends of series combination. The same

current I flows through each resistor. This is the property of series combination. If V1, V2 and V3 are the potential differences between the ends of each resistor, then applied p.d, V = V1 + V2 + V3 …….. (1)

Then p.d across each resistors are given by

V1= IR1 V2= IR2 V3= IR3 …...……….. …… (2)

If the combination is replaced by a single resistor with effective resistance R, then V = IR ………….. ……………………….(3)

Substituting (2) and (3) in (1), we get

IR = IR1 + IR2 + IR3 or R = R1 + R2 + R3

Thus the equivalent resistance of a number of resistors in series is the

sum of resistances of the individual resistors. If there are n resistors connected in series, the effective resistance R is given by

R = R1 + R2 + R3 +……………….+ Rn

The effective resistance is higher than the highest value of the resistance in the combination. Series combination is used to increase the effective resistance.

Resistors in parallel

Three resistors of resistances R1,R2 and R3 are connected in parallel as shown in the figure between two common points A and B. A p.d. V is applied between the points A and B. The p.d across each resistor is the same. This is the property of parallel combination. The main current I entering the combination is divided into I1, I2 and I3 through R1, R2 and R3 respectively.

SQ25) Which is true, in series connection current/p.d remains the same in each resistor? Ans: In series connection current remains the same through each resistor. SQ26) ‘n’ equal resistors are connected in series. The effective resistance is Rs. They are then

disconnected and connected in parallel. The effective resistance is Rp. Find the ratio sR / pR

.Ans: Rs= r n, Rp=n

r,

p

s

R

R =

2n

The effective resistance in series connection = 2n times the resistance in parallel connection.

R1 R2 R3 I I

R

V2 V3 V1

Fig.3.5

Fig.3.6

R1

R2

R3

I1

I2

I3

V

R

A B

I I

3-11 Current Electricity

I = I1 +I2 +I3 ….......................... (1)

The current through each resistor

1

1R

VI ,

2

2R

VI and

3

3R

VI ….(2)

If the combination is replaced by a single resistor of effective resistance R, then

R

VI ……………………………… (3)

Substituting (2) and (3) in (1), 321 R

V

R

V

R

V

R

V or

321 R

1

R

1

R

1

R

1

Thus the reciprocal of the effective resistance is equal to the sum of the reciprocals of the individual resistances. For only two resistances

21 RandR the effective resistance R =21

21

RR

RR

If there are n resistors connected in parallel, the effective resistance R is given

byn321 R

1.....

R

1

R

1

R

1

R

1 .

Parallel combination is used to reduce the effective resistance. The effective resistance will be less than the lowest resistance in the combination.

CELLS, EMF, INTERNAL RESISTANCE

Consider an external resistance R connected to the terminal of a cell of e.m.f E. A current I will flow through R from the positive terminal of the cell to its negative terminal. In order to maintain continuity, current has to flow through the electrolyte of the cell, from its negative terminal to its positive terminal. The materials inside the cell will offer a resistance r to the flow of current.The resistance offered by the material of the cell when an electric current flows through it, is called internal resistance of the cell.

SQ27) Which is true, in parallel connection current/p.d remains the same in each resistor? Ans: In parallel connection p.d remains the same for each resistor. But the current is different through each resistor.

SQ28) A person has ten 1/10 resistances. What are the maximum and minimum resistances he can make using all of them?

Ans: Maximum is obtained when they are connected in series. Rmax= r n= 11010

1

Minimum is obtained when they are connected in parallel. Rmin= n

r= 01.010

10

1

SQ29) A person has two resistances 1R and 2R . He can use them either individually or in

combination. The possible resistances are 2, 3, 6 and 9. What are 1R and 2R ?

Ans: The smallest resistance 2 ohm is for parallel connection and the highest resistance 9 ohm is for series combination. So individual resistance must be 3 and 6 ohms

9RR 21 …..… (1) 2RR

RR

21

21

29

RR 21 or 18RR 21 …(2)

Solving equation (1) and (2) we get R1 = 3 or R2 = 6 or vice versa.

E

R

r

I

Fig.3.7

Current Electricity 3-12

The total resistance to the flow of current is equal to r + R,

current Rr

EI

Or E = I (R + r) = IR + Ir …(1), where IR is called the

terminal potential difference V of the cell and Ir is the potential drop, v across the cell.

E = V + v .... (2) or V = E - v …… (3)

Thus the terminal p.d. is less than the e.m.f. of the cell by an amount v, the potential drop across the cell. When the current drawn from the cell increases, the potential drop across the cell increases and hence the terminal p.d, V = E–Ir decreases. When no current is drawn or when the circuit is open, the terminal p.d. is equal to e.m.f of cell i.e., V = E

We know E = Ir +IR and V=IR or I =R

V

RR

Vr

R

VE Vr

R

VE R

V

VEr

Thus the internal resistance increases with external resistance. A fresh cell has low internal resistance and it increases with long use of the cell.

Factors affecting internal resistance

The internal resistance depends on: 1) Nature of electrolyte, 2) Nature of electrode, 3) Area of the plates immersed in the electrolyte and 4) Distance between the plates.

Distinction between Potential difference and E.m.f. of a cell

E.m.f P.d.

1. E.m.f is a source of electrical

energy

1. P.d is the difference of potential

between any two points of a closed circuit

2. E.m.f. is terminal p.d. when circuit

is open

2. Potential difference is distributed

throughout the circuit

3. E.m.f. is greater than p.d. 3. P.d. is less than e.m.f

CELLS IN SERIES AND IN PARALLEL

We can replace a combination of cells by an equivalent cell just like equivalent resistance for a combination of resistances

SQ30) Why is the terminal voltage of a cell less than its e.m.f? Ans: The terminal voltage of a cell is less than its e.m.f as a part of the energy is dissipated by the internal resistance. SQ31) What happens when a battery is short circuited? Ans: The entire power delivered by the battery is dissipated inside it as heat energy. The heat

energy produced will be r

V2

, where r is the internal resistance of the battery and V the

terminal voltage.

3-13 Current Electricity

For a series combination, the negative terminal of one cell of e.m.f (E1)is connected to positive terminal of the second cell of e.m.f (E2). The effective e.m.f will be E = E1+ E2. The effective internal resistance will be r = r1 + r2. If negative terminals are joined in series connection. The effective e.m.f E = E1- E2 (if E1> E2) and the effective resistance will be the same as r1 + r2.

In the parallel combination of two cells joing the positive terminals together

and the negative terminals together. The effective resistance r is given by 21

111

rrr

and the effective e.m.f E is given by 2

2

1

1

r

E

r

E

r

E . If there are n cells in the

combination the effective resistances r is given by 21

111

rrr +….

nr

1 and the effective

e.m.f E is given by 2

2

1

1

r

E

r

E

r

E + …….

n

n

r

E If the negative terminal of the second is

connected to the positive terminal of the first, the effective resistance will be the

same. But effective e.m.f E is given by 2

2

1

1

r

E

r

E

r

E .

KIRCHHOFF’S RULES

Ohm‟s law is applicable only for simple circuit. For complicated circuits Kirchhoff‟s laws can be used to find the current or voltage. There are two Kirchhoff‟s laws. (1) Kirchhoff’s current law and (2) Kirchhoff’s voltage law.

Kirchhoff’s first law (Current law or Junction rule)

Kirchhoff‟s current law states that the algebraic sum of currents meeting at any junction in a circuit is zero.

SQ32) It is easier to start a car engine on a warm day than a chilly day. Why? Ans: When temperature increases, internal resistance decreases. So more current can be drawn from the battery on a warm day. Hence it is easy to start SQ33) We have a low voltage source and we want to draw a high current from it. What should be the value of internal resistance of the source, small or high? Why?

Ans: We have rR

EI

.When the internal resistance (r) is very small, maximum current is

drawn and it is equal to R

E.

SQ34) What is the effect of temperature on internal resistance of a cell? Ans: The internal resistance of a cell decreases with rise in temperature. SQ35) A battery giving high voltage should have high internal resistance. Why? Ans: If there is a short circuit, then the current should not exceed the safety limits. For this a battery of high internal resistance is to be used.

I1

1

I2

2

I3 I4

I5

3 4

5

O

Fig.3.8

Current Electricity 3-14

Explanation of first law

Consider a junction of five resistances as shown in the figure. The currents I1, I4, and I5 enter the junction O while the current I2 and I3 leave the junction O. By convention, the current flowing towards the junction is taken as positive and the current leaving the junction is taken as negative.

By Kirchhoff‟s first law the algebraic sum of the currents meeting the junction is zero. i.e., I1 + (-I2) + (-I3) + I4 + I5 = 0 or I1 + I4 + I5 = I2 + I3

i.e., the sum of the currents entering the junction is equal to sum of the

currents leaving the junction. This law is a consequence of conservation of charges.

Kirchhoff’s second law (Voltage law or Loop rule)

Kirchhoff‟s voltage law or loop rule states that the algebraic sum of product of resistance and current in each branch of any closed circuit is equal to the algebraic sum of the e.m.fs in that closed circuit.

Another statement

Kirchhoff‟s voltage law states that in any closed circuit, the algebraic sum of all the potential difference is zero.

Physical Idea

This law supports the conservation of energy. i.e., the net change in energy of a charge in a closed path will be zero. This follows from the conservative nature of electrostatic force. Hence the work done by the charge in any closed path is zero.

Explanation of Kirchhoff’s second law using a closed network

If the clockwise current in a loop is taken as positive, then the anticlockwise current is taken as negative or vice versa. The direction of e.m.f of a cell is from negative to positive within the cell.

Applying Kirchhoff‟s second law to the closed path ABEFA, we have

-I3R2 – I1R1 = -E1 (The direction of e.m.f and the current are in anticlockwise direction.) I3R2 + I1R1 = E1

For the closed path ABCDEFA, we have I2R3 – I1R1 = E2 – E1

SQ36) Can Kirchhoff’s laws be applied to AC circuits?

Ans: Yes. Kirchhoff’s laws can be applied to AC circuits also.

A

B

C D

E

F I1

E1

I3

I1

I3

I1

R1

I2

E2

I2

I2 R3

I2

R2

Fig.3.9

3-15 Current Electricity

WHEATSTONE BRIDGE

Wheatstone bridge consists of four resistances P, Q, R and S connected to form a closed path as shown in the figure.

A cell of e.m.f E is connected between A and C and a galvanometer G between D and B.

Bridge is said to be balanced when the

galvanometer shows null deflection. This is when the current through the galvanometer Ig is

zero. The main current I will be branched into I1 and I2 at the junction A i.e., I = I1 + I2. Now I1 passes through resistance P and Q and I2 through R and S. The currents I1 and I2 combine at C.

Consider the loop ABDA. I1 P - I2 R = 0 I1 P = I2 R ………. (1)

For the loop BCDB, I1 Q - I2 S = 0 I1Q = I2S ………..(2)

)2(

)1(gives

S

R

Q

P . This is the condition for balancing the

bridge. If the resistance S is unknown, it can be calculated by knowing the values of resistances P, Q and R.

SQ37) When the Wheatstone bridge is balanced, galvanometer and cell are interchanged. What will be the reading of the galvanometer?

Ans: The original balancing condition is S

R

Q

P We can write this

as R

P

S

Q . This is the balancing condition in the changed

position. Hence no current is flowing through the galvanometer. Hence galvanometer shows null deflection SQ38) A pattern as shown in figure is made with a uniform wire

with a resistance 4 per metre. If AB = 2 m, find the resistance between A and B?

Ans: The resistance of AB = 24= 8.

Resistance of ACB = resistance of ADB = 41 4

The effective resistance X is given by

4

1

4

1

8

1

X

1

8

4

4

8X (C and D are at same

potential they can be joined. Hence CD is neglected.) SQ39) In a Wheatstone bridge there are four resistances. In a meter bridge which are the four resistances when compared with Wheatstone bridge? Ans: There are two resistance boxes between the two gaps of a meter bridge. The other two resistances are those corresponding to the length of the wire on either side of jockey.

Q

I1 A

B

D

C G

I1

I2

I2 I

P

R S

E

Ig= 0

I

Fig.3.10

A B

C

D

Q

A

B

D

C

I2

I1

I2

P

R S

E

G Ig= 0

I

I1

Current Electricity 3-16

Application of Wheatstone bridge for

temperature measurement

The connections are made as shown in the figure. The resistance S is immersed in a water bath. Its resistance at various temperatures is calculated by using the balancing condition of Wheatstone bridge.

A graph is plotted with temperature along the x-axis

and resistance along the y-axis. It will be straight line graph as shown in figure.

To find the unknown temperature (T), the resistance S is kept in liquid of unknown temperature and resistance is measured. The unknown temperature can be found out from the temperature – resistance graph.

METER BRIDGE

Meter bridge is a practical application of Wheatstone bridge. It is used to measure the unknown resistance.

Meter bridge consists of a long wire AB of 1 meter length made of manganin or constantan stretched on a wooden frame. Its ends are soldered to two L-shaped copper strips.

SQ40) Why are the connections between resistors in a wheatstone or meter bridge made of

thick copper strips? Ans: To reduce the resistance of connections to be minimum. SQ41) What is the balancing length in the metre bridge when the resistances in the two gaps are equal? Ans: When the resistances in the two gaps are equal, the balancing length will be 0.5m. SQ42) Why is it necessary to connect positive of cells in the primary and secondary circuits to a common point? Ans: This is to oppose the e.m.f of cells in primary and secondary so that null deflection can be obtained in the galvanometer.

l m

G1 G2

A B

C

K E

P Q

G H.R J

(1-l)m Fig.3.12

Fig.3.11.(a)

S R

esis

tan

ce

Temperature T

Q

G

P

R

S

Fig.3.11.(b)

3-17 Current Electricity Another copper strip is fixed in between the L-shaped copper strips leaving two gaps G1 and G2 as shown in the figure. The unknown resistance P is connected in the gap G1 and standard (known) resistance box Q in the gap G2. A jockey J is connected to C through a galvanometer G and a high resistance box. The jockey can make contact at any point on the wire AB. A Leclanche cell is connected between the points A and B through a key K. A meter scale is provided below the wire between A and B.

Working

The circuit is closed using the key K. Take a suitable resistance R from the standard resistance box Q. Adjust the position of the jockey on the meter bridge wire so that the deflection in the galvanometer is zero. Let J represent this point. Measure the length AJ as l meters. The length JB = (1–l)meters.

If r is resistance per unit length of the wire, the resistance of wire AJ = rAJ = r l The resistance of JB = rJB = r(1–l).

By comparing with Wheatstone bridge, the resistance of AJ stands for R, resistance of JB stands for S. Applying the balance condition of Wheatstone bridge.

)r(1

r

Q

P

l

l

l

l

1

l

l

1

QP

To eliminate the error due to end resistances, another set of readings are taken by interchanging P and Q. Now the length l is measured from the end B to jockey. The average value of P is gives the unknown resistance.

POTENTIOMETER

Potentiometer is an instrument used for the measurement of potential difference and e.m.f, comparison of e.m.f and the internal resistance of a cell.

SQ43) Give a method to find the ratio of the radii

2

1

r

rof two wires using a metre bridge.

Ans: The wires R1 and R2 are connected in the left gap and right gap respectively. If L is the

balancing length, then 2

1

R

R=

L1

L

. But

2

1

R

R=

2

1

2

2

r

r

2

1

l

l (R

A

l =

2r

l).

2

1

2

2

r

r

2

1

l

l=

L1

L

2

1

l

l

L

L1

r

r

2

1 , where 1l and 2l are lengths of the wires.

Q

A

C

J

B G

I1

I

P

R S

E

Ig=0

I

Fig.3.13

Pri

mary

cir

cu

it

K

Rh

B

A L.cell(E)

Secondary circuit

G

J

Fig.3.14

Current Electricity 3-18

It consists of a uniform wire made of manganin or constantan of 10 meter length. The wire is stretched on a wooden board between the point A and B in parallel rows, each of length 1m. The ends of the wire are fixed to copper strips with binding screws. A jockey J is used for making contact with the wire. A meter scale is fixed on the board parallel to the wire.

Principle of potentiometer

An accumulator (battery) is connected between the points A and B through a rheostat (Rh) and a key K. The positive of the battery is to be connected to A. This forms the primary circuit. When this circuit is closed, a steady current (I) will flow through the potentiometer wire. The p.d. is uniformly distributed over the wire as the wire is uniform. Thus the p.d across any length is proportional to that length of the wire.

A primary cell of e.m.f (E) is connected in series with the potentiometer, a galvanometer, a high resistance and a jockey. The positive of the cell is to be connected to A. This forms the secondary circuit.

Adjust the position of jockey so that the deflection in the galvanometer is zero. The length AJ is called the balancing length (l). The potential difference of the length AJ balances the e.m.f of the cell, E.

lIrE , where r is the resistance per unit length of potentiometer wire i.e.,

lE . Thus the e.m.f of the cell in the secondary is proportional to the

balancing length.

Comparison of E.m.f’s of two cells

The primary circuit is formed between A and B using an accumulator, rheostat and a key. A secondary circuit is formed using a six terminal key, galvanometer, and a high resistance. Two cells of e.m.f E1 and E2 are connected using the six terminal key.

SQ44) What is the use of rheostat in the primary of a potentiometer?

Ans: By adjusting the rheostat, we can obtain maximum balancing length which can increase the accuracy. To repeat the experiment the rheostat has to be adjusted. SQ45) Does a current flow through the potentiometer from the cell when we find the e.m.f of a cell? Ans: No. At the balancing point no current is drawn from the cell. SQ46) A potentiometer is the best instrument to measure the e.m.f of a cell. Why? Ans: At the balance point (null point), potentiometer draws no current from the source of e.m.f. So we get the actual e.m.f of the battery.

K

Rh B

A

G

J

E1

HR 1 2

3 4 E2

Fig.3.15

3-19 Current Electricity

It is to be noted that the positive terminal of cells in the secondary and

positive terminal of the accumulator in the primary are connected to common point A.

The primary circuit is closed using the key K. The cell of e.m.f E1 is introduced in the secondary circuit by putting the keys in the gap 1 and 2. Now the jockey is moved on the wire until the deflection in the galvanometer is zero. Measure the balancing length as l1.

By the principle of potentiometer, E1 1l ………………… (1)

Now, the cell of e.m.f E2 is introduced in the secondary circuit by transferring keys to 3 and 4 from 1 and 2. Measure the balancing length l2.

22E l …………………. (2)

)2(

)1( gives

2

1

2

1

E

E

l

l . The experiment can be repeated by adjusting

the rheostat in the primary.

Determination of internal

resistance(r) of a cell

Primary circuit is made between A and B using an accumulator, a rheostat and a key K1. Secondary circuit is made between A and J using a cell, galvanometer and a high resistance.

To find the internal resistance of the cell in the secondary circuit, a parallel connection is made with the cell using a resistance box R and a key K2.

Keeping the key K2 open, find the balancing length l1 of the cell of e.m.f E.

1E l …….... (1)

SQ47) Will the balancing length change if a resistance is introduced in series with primary circuit? Ans: Now the potential gradient will decrease causing an increase in balancing length. SQ48) How can you increase the sensitivity of a potentiometer wire? Ans: Greater sensitivity means that even a small p.d. can be measured. So the potential gradient (p.d per metre) must be small. Hence, increasing the length of the potentiometer wire increases its sensitivity.

SQ49) Explain why copper is not used in potentiometer. Ans: Resistance of copper wire is very small. Hence when copper is used for the potentiometer wire, it is as if short circuited. No p.d will be available on the wire. SQ50) Which has more resistance, a 1000W bulb or 1000W heater? Give reason. Ans: Power P = V2/R. As the power P and voltage V are the same for both the bulb and the heater they have the same resistance. SQ51) We use parallel connection, in house wiring. Why? Ans: The two reasons are, i) each appliance gets the same voltage and ii) each appliance can be controlled individually by using separate switches.

Fig.3.16

Pri

mary

cir

cu

it

K1

Rh

B

G H.R.

R

A

J

K2

Current Electricity 3-20

Take a suitable resistance R from the box keeping the key K2 closed, Find the balancing length l2. This balancing length is for a potential drop across R (or terminal p.d. of the cell).

P.d. across R = rR

ER

2

rR

ERl

….....(2)

)2(

)1(gives,

R

rR

2

1

l

l or R + r = R

2

1

l

l

r=R

1

2

1

l

l or r = R

2

21

l

ll

The experiment can be repeated for different values of R. It is found that the internal resistance increases with external resistance.

ELECTRIC ENERGY AND POWER

In a source of e.m.f, chemical or some other form of energy is converted into electrical energy. To maintain the current, the source goes on spending its energy. A part of this energy is for useful work and the rest for heating the resistors in the circuit. The rate at which work is done by the source of e.m.f in maintaining the current in an electric circuit is called electric power of the circuit.

If I is the current flowing through a resistance R in a time t, then the quantity of charge Q = It. If the current is flowing between two points of potential difference V, then the work done in moving the charge = V It.

The electric power, P = t

ItV

time

workdone = VI.

i.e., Electric power is the product of potential difference and current. Since V = IR, power P = I2 R.

Unit of power

The unit of power is volt-ampere or watt (Js-1).

Electrical energy

The electrical energy consumed by an electric device can be expressed in terms of its power. If P is the power of the device and t the time of working, then electric energy W = P t. The S.I unit of electric energy is Joule or watt-second. The

practical unit of electrical energy is kilowatt hour (kWh).

Kilowatt hour: One kilowatt hour is the energy consumed in one hour at

a rate of one kilowatt.

1 kWh = 1000 Wh = 10006060 J = 3.6106 J

SQ52) You are given two bulbs of power 60W and 500W working on 220V a.c supply. Which of these has more resistance?

Ans: PowerR

VP

2

. P R

1 , or R

P

1 60W bulb has more resistance.

SQ53) When a current is passed through a conductor, heat energy is produced. What is the source of heat energy? Ans: The electrons collide with the atoms of the conductor. Thus electrical energy supplied is converted into mechanical energy and an equivalent heat energy is produced.

3-21 Current Electricity Example: Consider a bulb of 100 W working in 1 hour at a p.d of 230 volt.

Energy consumed in one second = 100 J

Energy consumed in one hour = 1006060J

=6103.6

6060100

kWh (1kWh = 3.6106 J)

= 0.1 kWh = 0.1 unit (1 kWh = 1 unit)

Thermal effect of electric current: Joule’s law

Consider a resistor with resistance R connected in series with a cell. If „I‟ is the steady current, the energy dissipated in the form of heat on the resistor, H = I2Rt, where t is the time of flow of the current. This

is known as Joule‟s law of heating.

Joule’s law states that the amount of heat generated in a conductor by the flow of a steady current is directly

proportional to (1) square of the current (2) the resistance of the conductor and (3) time of flow of current.

But according to Ohm‟s law, I = R

V H = VIt or t

R

V2

.

Practical application of Joule’s heating

Joule‟s heating has direct application in electric iron, electric heater, electric oven etc. The electric heating is used for producing light in an electric bulb. When the filament gets very hot, it emits light. Most of the power consumed by the filament appears as heat but a small part appears as light. A common application of Joule‟s heating is in the case of fuse wires.

Fuse wire

When high current flows, the fuse wire gets heated and melts making a break in the circuit. This is to protect the electrical devices from high currents. For making fuse wires, metals or alloys (Tin and Lead) of suitable melting points are used. Example: Aluminium, copper, iron etc.

For domestic purposes fuse wires are rated as 1A, 2A, 5A, 10A etc. Fuse wire

will melt when a current shown by the rating, flows through the wire. When a 5A fuse wire is cut into two parts, each part will have the same 5A rating. i.e., the

rating will not depend upon the length of the fuse wire but depends on the

radius of the wire.

SQ54) You are given two electric bulbs of 60W and 100W. Which of these will be brighter when connected i) in series and ii) in parallel?

Ans: i) In series connection, the same current flows through each bulb. So power consumed, P

= I2R when current I is constant P R and therefore the resistance is greater for 60W bulb. So

60W bulb will have more brightness. ii) In parallel connection, potential difference is same for

each bulb. So power consumed, R

VP

2

or P R

1. Therefore 100W bulb will have more

brightness as its resistance is lower SQ55) Transmission of electric power is done at high voltage. Why? Ans: When a given power is transmitted at high voltage, low current is passed through the lines. Hence power loss, I2R will be very small.

C

K

A V

B A R

I

Fig.3.17

Current Electricity 3-22

For example, consider an electric iron which consumes 1kW electric power

when operated at 220 volt. The current through the Iron will be A4.54220

1000 . To

protect this electric iron, a fuse wire of 5A is recommended.

* THERMO ELECTRICITY

Thermoelectricity is the conversion of thermal energy into electrical energy. Ex: Seebeck effect and Thomson effect. These are reversible process. The thermoelectricity is a reverse process of Joule‟s heating which is an irreversible process. Peltier effect is the conversion of electrical energy into thermal energy.

See beck Effect

In 1821, Seebeck, a German physicist, discovered the thermoelectric effect called the Seebeck effect. This is the flow of current in an electrical circuit made of two different conductors, when the two junctions in the circuit were kept at different temperatures.

Two dissimilar metals connected to form two junctions are called a thermo couple. The e.m.f. developed in the circuit is called thermoelectric e.m.f. The current due to thermo electric e.m.f is called thermoelectric current.

The figure shows a thermo couple made of Cu and Fe. When the junctions are kept at two different temperatures, current flows through the circuit from Copper to Iron through the hot junction (Hot Cu-Fe can be remembered as hot coffee). The direction of current will be reversed when hot and cold junctions are interchanged.

SQ56) Two bulbs when connected in parallel to a source take 50W each. What is total power consumed when they are connected in series with the same source?

Ans: In parallel connection, power consumed, P =R

V 2

, where V is the source voltage and R is

the resistance of bulb. Here the two bulbs have the same resistance R as their powers are same. When they are connected in series the total resistance =R +R = 2R.

So total power consumed =2R

V 2

=2

50=25W

SQ57) When do we get the maximum power output from a device? Ans: When the internal resistance of the device equals external resistance, we get the maximum power output. (Maximum power transfer theorem) SQ58) A heater coil is cut in to two parts and only one of them is used in the heater. What is the ratio of the heat produced by this half to that by the original coil?

Ans: Heat produced R and R l. When length is halved, resistance of coil is also halved.

From Ohm’s law, the current is doubled. Power, P=I2Rt = tR

I2

2 2= 2I2Rt. So heat produced will

be twice the initial value.

Fig.3.18

Fe

G Cu

Hot

junction Cold junction

2θ °C 1θ °C

(This is not in the Syllabus)

3-23 Current Electricity Seebeck made an extensive study using thermocouples of different materials. He arranged a thermo electric series. A part of Seebeck series is Bi, Ni, Co, Pd, Pt, Cu, Mn, Hg, Pb, Sn, Au, Ag, Zn, Cd, Fe, Sb, Te. The direction of the current at the hot junction is from the metal appearing earlier in the series to the metal appearing later in the series. The magnitude of thermo e.m.f is larger for metals appearing farther apart in the series.

Variation of Thermo e.m.f with temperature

One junction of the thermocouple is kept at 00C and the other junction is heated from 00C.

As the temperature of the hot junction increases, thermo e.m.f also increases. At a particular

temperature called neutral temperature ( n ),

thermo e.m.f reaches a maximum value and thereafter decreases to zero. At a particular temperature, called the temperature of

inversion ( i ) the thermo e.m.f reverses its

direction on further heating. A graph plotted with the temperature of the hot junction along X-axis and the thermo e.m.f along Y-axis is as shown in the fig.3.19.

The thermo e.m.f has a dependence on the temperature difference ( ) between

the junctions. The thermo e.m.f, V = 2

2

1 , where and are constants

depending on the materials of the thermocouple and the temperature difference.

Neutral temperature ( nθ )

The temperature of the hot junction at which the thermo e.m.f becomes maximum is called neutral temperature. It is a constant for the given pair. For

Copper – Iron thermocouple, n = 270 0C

SQ59) Explain how voltmeter and ammeter are connected in an electrical circuit, giving reason for each Ans: Voltmeter measures potential difference between two points of a circuit. It is connected in parallel to a circuit. Since the resistance of the voltmeter is very high it will not affect the current

through the circuit Ammeter measures current. It is connected in series to a circuit. Since the resistance of the ammeter is very small, it will not affect the current through the circuit SQ60) When the temperature of the two junctions of a thermocouple are the same, is there any p.d’s at the two junctions? If there exists a p.d.,why don’t the p.d. produce a current in the thermocouple?. Ans: There exist contact potential differences at the junctions. When the junctions are at the same temperature, the contact p.d. is equal and opposite. SQ61) What is the difference between thermo e.m.f.-temperature graph and thermo power- temperature graph?

Th

erm

o-e

.m.f

(m

V)

270°C

540°C

0 iθ

nθ

θeTemperatur

For Cu-Fe thermocouple

Fig.3.19

Current Electricity 3-24

Ans: First graph is parabolic and the second graph is a straight line.

Temperature of inversion ( iθ )

The temperature of the hot junction at which the thermo e.m.f becomes zero is called temperature of inversion. It is not a constant for the given pair of metals. It depends on the temperature of the cold junction. Temperature of inversion is as much above neutral temperature as the cold junction is below the neutral temperature.

i.e., i - n = n - c or 2

θθθ ci

n

i.e., the neutral temperature is the

average of the temperature of cold junction and the temperature of inversion.

When c = 00C then, 2

0 in

, i =2 n The temperature of inversion is

twice the neutral temperature when the cold junction is at 0°C. The thermo electric power (Seebeck coefficient S)

The thermo electric power, S = d

dV (thermo e.m.f, V = 2

2

1 )

Peltier Effect

In 1834, a French scientist Peltier found that when electric current is passed through a circuit of a thermocouple, heat is liberated at one junction and absorbed at the other junction. This phenomenon is called Peltier effect

In a Cu-Fe thermocouple at the junction (1), where the current flows from Cu to Fe, heat is absorbed. At the junction (2) where current flows from Fe to Cu, heat is liberated. i.e. junction (1) gets cooled and junction (2) gets heated (Fig.a). When the direction of current is reversed, junction (1) gets heated and junction (2) gets cooled (Fig.b). Thus Peltier effect is reversible.

Peltier coefficient (π )

Peltier Coefficient is the amount of heat energy absorbed or evolved in one of the junctions of a thermocouple when 1A current flows for 1sec. Its unit is volt. If a current of 1ampere flows for t seconds, the heat absorbed or evolved at one junction, H=π It

Relation between Peltier coefficient (π ) and Seebeck coefficient (S)

The relation between and S is = T d

dV = TS, where T is the temperature of

the junction in Kelvin scale.

SQ62) A graph is drawn with thermoelectric power (S) against the

temperature difference θ . What does i) the slope give and ii) Y- intercept?

Ans: The equation representing the thermoelectric power is a straight line

graph, given by, S= . Comparing with the equation of the straight line

y = mx+c, we get

Cooled Heated

1 2

Cu

Fe

Fig.3.20.(b)

Fig.3.20.(a)

Cooled Heated

1 2 I

Cu

Fe

D

O

S

3-25 Current Electricity i) The slope gives the value of β ii) Y-intercept give the value of.

Thomson effect

William Thomson (Lord Kelvin) suggested that each element of a single conductor would be a source of e.m.f whenever there is a temperature gradient along the conductor. If two points of a conductor are kept at different temperatures, an e.m.f is developed between these two points. This effect is called Thomson effect. Thomson effect is reversible.

Thomson Coefficient (σ )

Thomson coefficient is defined as the potential difference between the two points of a conductor which are maintained at a temperature difference of 1°C. Thomson coefficient σ depends on temperature and material of the

conductor. For a temperature difference T the Thomson e.m.f = T .

Origin of Thomson Effect

In a conductor with a temperature gradient, the free electrons will have higher energy in the hot region than the cold region. Hence free electrons will diffuse from high temperature region to low temperature region. This give rise to potential difference. The flow of electrons continues till the potential difference is enough to counteract the net diffusion of electrons. This is the origin of Thomson effect.

Origin of Thermo e.m.f

The number of free electrons per unit volume (electron density) will be different for different metals. When two different metals are in contact, there will be a net electron diffusion from a metal of higher electron density into the metal of lower electron density. This gives rise to a contact potential at the junction. But this contact potential is a function of temperature.

The contact potentials at the two junctions of a thermocouple are equal and opposite when they are maintained at the same temperature. Hence the net e.m.f is zero. When the junctions are at different temperatures, the contact potentials will be different. This leads to a net e.m.f called thermo e.m.f.

Applications of thermo electricity

The thermoelectric effects are used for i) measurement of temperature, ii) thermoelectric generator, and iii) thermoelectric refrigerators.

SQ63) Give the difference between Peltier effect and Joule’s heating. Ans: i) During Joule’s heating heat is produced when current passes. But in Peltier effect, heat is generated at one junction and heat is absorbed at the other junction when current is passed. ii) During Joule’s heating, heat produced is proportional to the square of the current. But in Peltier effect, heat produced or absorbed is proportional to current.

Current Electricity 3-26

iii) Joule’s heating is independent of the direction of heating. But in Peltier effect the hot and cold junctions are interchanged when the direction of current changes.

Measurement of temperature using thermocouple

a) Thermoelectric thermometer

One junction of a thermocouple is kept at any fixed temperature (Ex: melting point of ice or room temperature). The other junction is kept in the region whose temperature is to be found out. A sensitive galvanometer is connected in series with the thermocouple. The galvanometer gives the temperature of the hot junction as it is calibrated to give temperature directly. Different thermoelectric thermometers are to be used for measuring different ranges of temperatures.

Examples: 1) Copper-alloy of gold and iron (1K to 50K), 2) Copper-constanton (50K to 500K), 3) Platinum-rhodium (up to 2000K)

Advantages

1) It is very accurate and convenient for use.

2) It can be used for a wide range of temperatures (0.001 K to 2000 K).

3) Rapidly varying temperatures can be measured.

4) It can be used to measure temperature inside a small cavity.

Disadvantages

1) For measuring different ranges, different thermometers are required.

2) Difficult to maintain the temperature of cold junction constant for a long time.

b) Thermoelectric Detector (Thermopile)

Thermocouple can be used as a detector of radiation. A large number of Antimony-Bismuth thermocouples are connected in series. Alternate junctions are arranged on one side and blackened to absorb all the thermal radiations falling on it. All other junctions are concealed from the radiations and kept at a fixed temperature. Heat radiations falling on the blackened junctions will give a deflection in the galvanometer. This deflection is proportional to the intensity of radiations falling on it.

c) Thermoelectric generator

Thermocouples of high thermoelectric power is used for generating electricity. Such thermocouples should have low electrical resistivity and low thermal conductivity. Electricity can be generated in remote area by heating one junction of the thermocouple using kerosene lamp. Using semiconductor alloys, it is able to

produce a thermoelectric power of about 5kW.

d) Thermoelectric refrigerator

Thermoelectric refrigerator works on the principle of Peltier effect. When current is passed through the thermocouple, one junction will become hot and the other junction will become cold. This cold junction can be used to cool a closed region of the refrigerator.

The two main advantages of this refrigerator are

Heat

Fig.3.21

3-27 Current Electricity 1) Thermoelectric refrigerator is noiseless as there are no motors or moving parts 2) There is no threat to the ozone layer as no gas is used.

MODEL QUESTIONS WITH ANSWERS

1. Consider a metallic conductor of length l and uniform area of cross section a.

A potential difference is applied between the ends of the conductor. a) What happens to the free electrons when p.d is applied?

b) What is the motion of the free electrons before the p.d is applied?

c) Is the direction of current in the direction of the flow of electrons when p.d is

applied?

d) If a charge Q coulombs flows across the cross sectional area in one second, what is the current?

e) What is the unit of current?

Ans. a) When a p.d is applied to a conductor the free electrons are drifted in a

direction opposite to the field.

b) Before the p.d is applied, the free electrons move in all possible direction.

Hence no net current is produced. c) No. The direction of current is opposite to the direction of flow of electrons.

d) Current Q1

Q

t

QI ampere

e) ampere (A)

2. The free electrons inside a metallic conductor are in random motion.

a) Can we get a current in the conductor due to the random motion?

b) In the electric field, what happens to the motion of the electrons?

c) When an electric field (E) is applied to the conductor, what is the force experienced

by an electron? d) If „m‟ is the mass of an electron, what is its acceleration?

Ans. a) No In random motion the average velocity of electrons will be zero

b) Electrons are accelerated opposite to the applied field.

c) The force experienced by an electron=Ee

d) If E is the electric field, the acceleration producedm

Ee

3. a) What is the relaxation time? What is its order? b) What is drift velocity?

c) What is the relation between drift velocity and relaxation time?

d) The order of drift velocity is

i) mm/s ii) m/s iii) km/s iv) velocity of light

Ans. a) The average time between two successive collisions is called relaxation time. It is of order 10-14s

b) The average velocity with which the free electrons are drifted under the

influence of electric field is called drift velocity (vd).

c) The drift velocity between two successive collisions, dv = m

Ee

d) mm/s

4. a) What do you mean by current density?

b) Current is a scalar quantity while current density a vector quantity. Explain.

c) What is the unit of current density?

Ans. a) Current density j at a point in a conductor is defined as the amount of

current flowing through unit area of the conductor around that point, provided the

area is held in a direction perpendicular to the current direction.

Current Electricity 3-28

b) Current means the quantity of charge flowing per second. No direction is

required. But current density is the current flowing per unit area perpendicular to the

area. Hence this is a vector quantity

c) The unit of current density is Am2 5. a) Compare the drift velocity of electrons with thermal velocity. b) Though the drift velocity is very small, there is no time delay to complete the

circuit. Why?

c) Do all the free electrons experience the same velocity when field is applied?

Ans. a) Drift velocity of the electron is in the order of 1mm/s while the thermal speed

is in the order of 105m/s

b) The establishment of electric field takes place through out the circuit with the speed of light. So every point of the circuit experiences an electron drift. Thus

current is established almost instantly.

c) No. The free electrons are moving in all possible directions with different

velocities. When field is applied the electrons modify their random motion in such a

way that they drift slowly in the direction opposite to that of field.

6. A circuit diagram is given as shown in the figure.

a) What are the symbols used in the diagram standing for?

b) What are the instruments used to measure the current

through the circuit and voltage across the resistance?

c) Draw the circuit diagrams for the measurement of the current and voltage for the

resistance using the measuring instruments. d) What is the use of rheostat?

Ans. a) The lower symbol stands for resistance. Top left symbol stands for key and

the right symbol for a cell.

b) Ammeter is used as the instrument to measure the current through the

circuit and voltmeter to measure voltage across the resistance.

c)

d) Rheostat is used to vary the resistance in order to control the current through the

circuit.

7. a) Draw the graph between potential difference and current for a resistance.

b) What is the nature of the graph?

c) How can you state the relation between V and I as a law?

d) How can you find the resistance from the graph? Give the unit of resistance. e) What is conductance? Give its unit.

Ans. a)

b) It is a straight line graph passing through the origin

c) At constant temperature the current flowing through a conductor is directly

proportional to the potential difference between the ends of the conductor.(Ohm’s

Law) d) The reciprocal of the slope of the graph gives resistance. Unit is Ohm

e) Reciprocal of resistance is conductance i.e. R

1G . Unit is mho.

I

V

A

V

3-29 Current Electricity 8. a) What is the variation of resistance with i) length and ii) area of cross section of

conductor?

b) What is resistivity? Give its units.

c) What is conductivity? Give its units. d) Get an expression for resistivity in terms of relaxation time.

Ans. a) (i) R l (ii) R A

1

b) The resistivity of a material is equal to the resistance offered by a conductor of unit length having unit area of cross-section. Unit of resistivity is Ohm-metre

c) Reciprocal of resistivity is conductivity ie

1

.Unit is mho m-1or Sm-1

d) Resistivity

2en

m.Here n is the number of electrons per unit volume.

9. a) When the temperature increases, what happens to relaxation time? b) When relaxation time decreases, what happens to resistivity?

c) What is the variation of resistivity with temperature?

Ans. a) When the temperature increases, relaxation time decreases.

b) When relaxation time decreases, resistivity increases. c) The resistivity increases when temperature increases.

10. a) What happens to the resistivity when the temperature is absolute zero?

b) What is super conductivity?

c) Give some practical application of super conductivity.

Ans. a) At absolute zero resistivity become zero.

b) The electrical resistivity of many metals and alloys drops suddenly to zero

when they are cooled to a very low temperature called the critical temperature Tc

(Transition temperature). This phenomenon is called superconductivity.

c) Superconductors are used (i) in making superconducting magnets,(ii) in memory or storage elements in high speed computers, (iii) in transmission lines

and (iv)in material science, research and high energy particle physics.

11. Temperature variation of resistance is given by t1RR 0 .

a) What is R0 and what is ?

b) What is the unit of ? c) Define the temperature coefficient of resistance

Ans. a) R0 is the resistance at 00C and is the temperature coefficient of resistance b) 0C-1

c) Temperature coefficient of resistance of the material of a conductor is

defined as the ratio of increase in resistance to its resistance at 0°C per degree rise of temperature.

12.a) Draw a graph with resistance along y-axis and temperature along x-axis.

b) What is the nature of the graph?

c) What does the y-intercept give?

d) What does the slope of the graph give? e) Can you find the temperature at which the resistance becomes zero using the

graph?

Ans. a)

R0

Rt

t 0c

A

B

Current Electricity 3-30

b) Graph is a straight line with y intercept

c) y-intercept gives R0

d) The slope gives R0 e) Yes. The graph is produced to cut the x-axis. The temperature corresponding

to this point gives the temperature at which the resistance becomes zero. 13. The resistances of a metallic wire is R1 at temperature t1 and R2 at temperature t2

a) Give the mathematical expression to find b) Is it possible to find the resistance at 0oc?

c) What is the resistance at a absolute zero (or zero Kelvin)

d) When ohm‟s law is stated „at given temperature‟ is mentioned. Why?

Ans. a) = 1221

12

tRtR

RR

b) Yes. Where t1=0 we get 20

02

tR

RR or 0220 RRtR

220 Rt1R or 2

20

t1

RR

c) Resistance will be zero at a absolute zero. d) When temperature changes resistance will change. Therefore Ohm’s law is

not true. Hence Ohm’s law is stated at constant temperature.

14. A few coloured rings are found at one end of a carbon resistor

a) How many rings are usually found?

b) Give the significance of the rings c) If the rings are brown, black, red and silver. What is the resistance?

Ans. a) four rings are usually found.

b) The first two bands from the end indicate the first two significant figures of

the resistance. The third band indicates the number of zeroes (decimal multipliers) after the two significant figures. The fourth band indicates the tolerance in

percentage.

c) 1k 10%

15. There are two types of combinations for resistances - series and parallel.

a) In series combination, what is the current through each resistor?

b) In parallel combination, what is the current through each resistor?

c) In series combination, what is the p.d across the each resistor? d) In parallel combination, what is the p.d across the each resistor?

Ans. a) In series combination the current is the same through each resistor

b) In parallel combination the current through each resistor is inversely

proportional to each resistance.

c) In series combination the p.d across each resistor will be proportional to each resistance. It is the product of current and resistance.

d) In parallel combination the p.d is the same for each resistor

16.Consider a series combination of resistances R1 , R2 and R3.

a) What is the formula for effective resistance for the series combination?

b) Give the formula for effective resistance for the parallel combination

c) State whether true or false. Correct if false i) „Effective resistance in the parallel combination is smaller than the smallest‟

ii)„Effective resistance in the series combination is larger than the largest‟

Ans. a) The effective resistance in series combination will be the sum of the

resistances. ie 321 RRRR

3-31 Current Electricity b) The effective resistance R in parallel combination is given by

321 R

1

R

1

R

1

R

1 (Refer text page 3-10)

c) (i) True (ii) True 17.

a) What is the effective resistance between A and C? b) How much current enters the circuit through A?

c) What is the current through each resistor?

d) What is the amount of current leaving the point C?

Ans. a) Resistance of each branch R1 =2+2=4

The effective resistance R of the parallel combination is given by 21 R

1

R

1

R

1

2

8

16

44

44

RR

RRR

21

21

b) The main current through the circuit is A12

2

R

V

c) Since effective resistance of each branch is the same current is split in to

equal parts. Hence the current through each resistance will be 0.5A d) As the split up currents meet at C current leaving the point C is 1A

18. a) Apply the Kirchhoff‟s law at the junctions A and C of the above figure

b) What is the potential difference between points A and C?

c) Apply Kirchhoff‟s law for the net works A, B, C, D, A

d) Apply Kirchhoff‟s law for the net works A, D, C, E, F, A

Ans. a) At the junction A current entering the junction is equal to current leaving

the junction. ie IA=IB+ID . At the junction C current entering the junction is equal

to current leaving the junction. ie IB+ID=IC b) The potential difference between points A and C is the p.d of the cell and it

is equal to 2V

c) IB 2 +IB 2 ID 2 ID 2= 0

d) ID 2 + ID 2 = 2

19. In the above diagram B and D are connected by a galvanometer. If the potential at C is

zero

a) What is the potential at B? b) What is the potential at D?

c) What is the current through the galvanometer? Why?

d) If the resistances between A and D and A and B are both replaced by 4 what is

the current through the galvanometer?

Ans. a) Potential at B is 1V

I

2 2

2 2

E R

B R

D R

C R

F R

2 V

A R

Current Electricity 3-32

J

b) Potential at D is 1V

c) The current through the galvanometer is equal to zero as B and D are at the

same potentials.

d) The current through the galvanometer will be zero as the points B and D are at equal potentials.

20. The figure shows the Wheatstone‟s bridge

a) Give the condition for balanced Wheatstone‟s

bridge

b) What is the current through the galvanometer

when the bridge is balanced? c) If P and Q are interchanged in the balanced

condition, is there are any current through the

galvanometer.

d) If P and S are interchanged in the balanced

condition, is there are any current through the galvanometer.

Ans. a) The condition for balanced Wheatstone’s

bridge is S

R

Q

P

b) Current through the galvanometer is zero when the bridge is balanced

c) As the balancing condition is violated there will be a flow of current through

the galvanometer.

d) As the balancing condition remains the same now, there is no current

through the galvanometer.

21. Consider a circuit as shown in the figure. The cell has an internal resistance r

a) What is the current in the circuit?

b) Is terminal p.d of cell equal to its e.m.f?

c) What is the potential difference across the

terminals of the cell? d) Find the condition under which potential

difference and e.m.f are equal?

Ans. a) Current in the circuit will be, I= rR

E

b) No. The terminal p.d of the cell is not equal to its e.m.f.

c) Potential difference across the terminals of the cell )rI(E

d) If no current is drawn then the potential difference and the e.m.f are equal.

22.a) How many wires are there in a meter Bridge?

b)Meter bridge is a modified form of ………. bridge

c) What are the uses of a meter bridge?

d) If the meter bridge is balanced at the midpoint of wire. What is the ratio of the

resistances? Ans. a) There is one wire of length one meter in a meter Bridge

b) Meter bridge is a modified form of Wheatstone’s bridge

c) The meter bridge is used to find the resistance and the resistivity of the

material of a wire. It is also used to compare the resistances of two wires.

d) If the meter bridge is balanced at the

midpoint of the wire, the ratio of the resistances will be equal to one

23. If a 2.4 volt bulb is connected to a meter bridge as

shown in the figure. One terminal of the bulb is

Q

A

B

D

C G

I

P

R S

2v

E R

I

R R

2 V

B

A

2.4 V bulb

3-33 Current Electricity connected to one end of the wire, A. The other terminal is connected to the sliding contact, J.

a) What is the observation regarding the brightness as the point J is moved towards

B.?

b) To get maximum brightness where should the terminals of the bulb be connected? Ans. a) The brightness increases as the point J is sliding towards B.

b) To get maximum brightness the terminals of the bulb is connected at the

ends of the meter wire.

24. a)What is the difference between potentiometer and meter bridge

b) To find the resistance which one (potentiometer or a meter bridge) is used?.

c) To measure e.m.f of a cell which one is used?

d) To compare e.m.f of two cells which one is used?

Ans. a) A potentiometer has ten wires of one meter each while a meter bridge has a

wire of one meter length

b) To find the resistance a meter bridge is used

c) To measure e.m.f’s of a cell a potentiometer is used.

d) To compare e.m.f’s of two cells a potentiometer is used?

25. a) How can we increase the accuracy of measurements using a potentiometer?

b) Why the potentiometer method is said to be null method?

c) Explain how to find the internal resistance of a cell using the potentiometer?

Ans. a) Better accuracy of measurements is obtained by using a potentiometer when

the balancing length is a maximum. This is achieved by adjusting the rheostat in the

primary so that the balancing length is on the 10th wire.

b) During the e.m.f measurement of a cell using potentiometer no current is

drawn from the cell. Hence it is said to be null method.

c) Refer text page 3-19

26. a) A charge Q flows from a higher potential V1 to a lower potential V2 in a conductor.

What is the work done by the cell?

b) If the current through a conductor is I and the p.d is V, what is power?

c) How does the work done appear?

d) How much heat is produced in a time„t‟ when „I‟ flows through the conductor under

a p.d „V‟

Ans. a) Work done by the cell d.peargch Q 21 VV

b)Power = current p.d =I V

c) The work appears as heat( Sometimes work appears as heat and light)

d) The heat produced will be equal to V I t

27. a) What is a thermo couple ?

b) What is thermo e.m.f? What is Seebeck effect?

c) What is Peltier effect?

d) What are thermo electric series?

e) Which two metals give larger e.m.f when made into a thermo couple?

Ans. a) Two dissimilar metals connected to form two junctions are called a thermo

couple

b) When the two junctions of a thermo couple are kept at different temperatures an e.m.f is developed in the circuit. This is the thermo e.m.f. This

phenomena is called Seebeck effect.

c) When electric current is passed through a circuit of a thermo couple heat is

liberated at one junction and is absorbed at the other junction. This phenomenon is

called Peltier effect.

d) The thermo electric series or Seebeck series is Bi, Ni, Co, Pd, Pt, Cu, Mn, Hg,

Pb, Sn, Au, Ag, Zn, Cd, Fe, Sb and Te.

Current Electricity 3-34

e) The thermo e.m.f is larger for metals appearing farther apart in the series.

Hence Bi and Te give larger e.m.f when made in to a thermo couple.

28. a) Draw a graph showing thermo e.m.f (thermo electric current) with temperature for Cu-

Fe thermocouple

b) What are neutral temperature and temperature of inversion?

c) Give the relation ship between neutral temperature, temperature of inversion and

temperature of cold junction.

d) What is the reason for thermoelectricity?

Ans. a)

b) The temperature of the hot junction at which the thermo e.m.f becomes

maximum is called neutral temperature. The temperature of the hot junction at which

the thermo e.m.f becomes zero is called temperature of inversion.

c) Neutral temperature 2

icn

, where c is temperature of the cold

junction and i is the temperature of inversion. d) When the junctions are kept at different temperatures the contact potentials will be different. This leads to a net e.m.f called thermo e.m.f.

29. a) Give the expression for thermo e.m.f with explanation for the symbols

b) If the temperature of the cold junction is raised by t0 c. how much the temperature

of inversion and neutral temperature are effected?

Ans. a) The thermo e.m.f, V = 2

2

1 , where and are constants depending on

the materials of the thermocouple and the temperature difference.

b) If the temperature of the cold junction is raised by t0 c the temperature of

inversion is lowered by t0 c. The neutral temperature remains same

MULTIPLE CHOICE QUESTIONS

1) „n‟ similar resisters each of resistance r when connected in parallel has the total resistance

R. The total resistance in series connection will be (92A)

a) nR b) R/n c) n2R d) R/n2

2) The voltage generated by a thermocouple is given by V = 2

2

1 , where and are

constants. Then the neutral temperature is (92A)

a) / b) c) - /2 d)- /

3) The resistance of a wire is R . The wire is stretched to double its length. New resistance

of wire is (92S)

a) 2

R b) 4R c) 2R d)

4

R

4) At the neutral temperature, thermoelectricity is (92S) a) Maximum b) half of the maximum c) Zero d) other than zero value.

5) Ten identical resistances each having a resistance of 1 are joined together in parallel.

The combination has a resistance of (93A)

a) 10 b) 1 c) 0.1 d) 0.01

Th

erm

o-e

.m.f

(mV

)

270°C

540°

C 0 iθ

nθ

θeTemperatur

3-35 Current Electricity 6) In electrolytic conduction, the current is due to (93A)

a) electrons b) holes c) ions d) protons

7) Above the neutral temperature, thermo e.m.f (93A)

a) is constant b) increases with rise in temperature c) decreases with rise in temperature d) can not be predicted.

8) Three bulbs of ratings 40W,60W, and 100W are designed to work on 220V mains. The bulb

which burns most brightly when connected in series across the mains is (93S)

a) 100W b) 60W c) 40W d) All will burn equally bright

9) What is immaterial for an electric fuse is (94A) a) Its resistivity b) its radius c) its length d) the current through it

10) When cells are arranged in parallel (94A)

a) Current capacity decreases. b) Current capacity increases.

c) e.m.f increases. d) e.m.f decreases.

11) Which of the bulb of 25, 40, and 60 watts has the lowest resistance? (94A)

a) 25 watts b) 40 watts c) 60 watts d) insufficient data 12) Wheatstone bridge is most sensitive when (94A)

a) ratio of arms is equal to one b) the ratio of arms is equal to ten

c) Any value for the ratio of arm d) both a and b are true

13) When the temperature of a conductor is increased, its resistance (94S)

a) Decreases b) Increases c) Fluctuates d) remains constant

14) Minimum resistance obtained by combining five5

1 resistances (94S)

a) 10

1 b)

25

1 c)

5

1 d)

50

1

15) An electric bulb is rated 220volt and 110watts. Power consumed by it when operated on

110volt is a) 50 watts b) 75 watts c) 90 watts d) 27.5 watts

16) In Wheatstone bridge, P=9 , Q=11 , R=4 and S=6 . How much resistance must be

put in parallel to S to balance the bridge? (95A)

a) 24 b) 9

44 c)26.4 d) 18.7

17) The best instrument to accurately measure the e.m.f of a cell is (95A)

a) volt meter b) ammeter c) potentiometer d) Wheatstone bridge

18) In which one of the following substances the resistance decreases with increase in

temperature? (95S)

a) carbon b) constantan c) copper d) silver

19) Three resistances, 1 each are connected to form a triangle. The resistance between any

two corners will be (96A)

a) 3 b) 2

3 c)

3

2 d)

3

1

20) Of the bulbs in a house, one glows brighter than the other. Which of the two has higher

resistance? (96A)

a) The bright bulb b) The dim bulb c)Both have same resistance

d) Brightness does not depend upon the resistance

21) If the heating element of an electric heater should be made with a material which should

have (96A)

a) high specific resistance and high melting point

b) high specific resistance and low melting point

c) low specific resistance and low melting point

d) low specific resistance and high melting point

22) The equivalent resistance of two resistors in parallel is always (96S)

a) the sum of the resistances b) the difference of the resistances

Current Electricity 3-36

c) the average of the resistances d) lower than the smallest resistance value

23) A flash light cell of e.m.f 2V gives a current of 20A when connected to an ammeter of

resistance 0.03 . The internal resistance of the cell is (97A)

a) 0.03 ohm b) 0.1 ohm c) 1 ohm d) 0.07 ohm

24) A uniform resistance wire of resistance 3 is cut into three equal parts. The resistance

of a part is a) 1 b) 2 c) 3 d) 4

25) From the following, which is the most suitable connecting wires? (97A)

a) tungsten b) constantan c) copper d) nichrome

26) Potentiometer is preferred than a voltmeter to measure voltage since (97S)

a) potentiometer is handy compared to voltmeter to measure voltage

b) potentiometer is cheap

c) no current is drawn by a potentiometer during measurement

d) range of voltmeter is small compared to potentiometer

27) The value of resistance of a fuse wire should be (97S) a) very high b) very low c) moderate d) zero

28) During charging of a lead accumulator, what is formed

a) Sulphuric acid b) Water c) Lead sulphate d) Nothing

29) A wire of resistance 10 is bent to form a circle. Effective resistance between two points

on any diameter of the circle is (98M)

a) 10 b) 5 c) 15 d) 2.5

30) Two wires X and Y are connected in series. When it is heated the combination has the

same resistance at all temperatures. If X is of silicon, the Y should be (98S)

a)Germanium b) Copper c) Aluminium d) Alloy of silicon and copper

31) A potentiometer, a 2V cell and a resistance of 1950 are connected in series. Length of the potentiometer wire is 1m and potential difference per cm of the wire is 0.1mV. Resistance

of potentiometer wire is (98S)

a) 8 b) 10 c)12 d) 20

32) A wire of resistance R is doubled by stretching. Half of the stretched wire will have the resistance (98S)

a) 2R b)4R c) R d) 4

R

33) An effective resistance R is obtained by connecting n identical resistors each of resistance

r in series. When connected in parallel, the effective resistance of the combination is (99M)

a)R

n2

b)R

r2

c)2Rr d)

2Rn

34) The thermo e.m.f of a thermocouple (97S)

a) is independent of the nature of the material used.

b) depends upon the temperature of cold junction alone c) depends upon the temperature of hot junction alone

d)depends upon the temperature difference between two junctions as well as temperature of

the cold junction.

35) The time required for 1kW heater to raise the temperature of 10 litres of water through

10 0C is (99S) a) 210 sec b) 420 sec c) 42 sec d) 840 sec

36) A certain charge liberates 0.8 g of oxygen. The same charge will liberate how many grams

of silver? (99S)

a) 108g b) 10.8g c) 0.8g d) g.80

108

37) A wire has resistance 16 . It is bent in the form of a circle. The effective resistance

between the two points on any diameter of the circle is (2000M)

3-37 Current Electricity a) 16 b) 32 c) 8 d) 4

38) The four resistances of a Wheat stone‟s bridge are 1 each. When the bridge is balanced, the cell has an e.m.f of 1volt. The current drawn from the cell is a) 1A b) 5A c)

0.25A d) 4A (2000SAY)

39) Three resistors each of 3 are connected in parallel. The equivalent resistance is

a) 1 b) 9 c) 6 d) 3

1 (2001M)

40) In metallic conduction, the current is due to the movement of

a) free electrons b) holes c) ions d) protons

41) A wire of resistance R is drawn to thrice its length. Its resistance will be

a) 9R b) 7R c) 5R d) 3R 42) The heating element in an electric iron is made of (2001SAY)

a) Constantan b) Tungsten c) Nichrome d) Copper

43) The electrochemical equivalent of two substances are in the ratio 3:2. Then their

chemical equivalents are in the ratio (2001SAY)

a) 2:3 b) 3:2 c) 4:3 d) 3:4

44) In the circuit given, when the key K is not closed, the reading of the ammeter will be (2002 M)

a) A9

10 b) 5A c) 2A d) A

10

9

45) An electric iron is marked 1000W, 250V. The resistance of

the heating element is

a) 625 b) 6.25 c) 62.5 d) 250

46) When a 200W and 100W bulb filaments for operations at 220V are connected in series to

a 220V supply, the power consumed will be (2002M)

a) 33W b) 67W c) 100W d) 300W

47) When cells are arranged in series a) e.m.f increases b) e.m.f decreases

c) Current capacity increases d) current capacity decreases

48) How many electrons do pass through a lamp in one minute if the current is 300 mA?

(2002SAY)

a) 3000 b) 3910 c) 1.12

2010 d)182010

49) If the radius of a wire of constant length is doubled, its resistance becomes (2002SAY)

a) double b) half c) four times d) one-fourth

Key to Multiple Choice Questions

1.c 2.d 3.b 4.a 5.c 6.c 7.c 8.c 9.c 10.b 11.c 12.a

13.b 14.b 15.d 16.c 17.c 18.a 19.c 20.b 21.a 22.d 23.d 24.a

25.c 26.c 27.a 28.a 29.d 30.a 31.b 32.a 33.b 34.d 35.b 36.b

37.d 38.a 39.a 40.a 41.a 42.c 43.a 44.c 45.c 46.b 47.a 48.c

49.d

PROBLEMS

Problem 1: One coulomb of charge is passed through a conductor in three seconds.

Calculate the electric current.

Ans: Electric current, t

qI 0.33

3

1A

Problem 2: What will be the number of electrons passing through the filament of an

electric bulb in 10 seconds if the current of 2 ampere passes through it?

10V

5

4

K

A

Current Electricity 3-38

Ans: I=t

q

t

ne

19106.1

102

e

Itn 1.25×1020 electrons

Problem 3: What time does an electron take to drift from one end of a copper wire of

length 1m to its other end? The number density of conduction electrons in copper is 8.5×1028 m-3, cross-sectional area of wire is 10-6m2 and current is 1A. Ans: n = 8.51028m-3, A =10-6 m2, e =1.610-19 C, I =1A, l = 1m

Current, I = nAevd

15

19628dms1035.7

106.110105.8

1

nAe

Iv

Time required to cover the length dd v

1

v

l

51035.7

1

= 1.34×104 sec.

Problem 4: What is the drift velocity of free electrons in an aluminium wire having area

of cross-section 30mm2 and one ampere current passes through it? Assume that each aluminium atom contributes 3 free electrons. (Density of aluminium is 2.7×103 kgm-3

and atomic weight of aluminium is 26.98 g.)

Ans: I =1A, e =1.610-19C, A = 30mm2 = 310-7 m2, Atomic weight of aluminium =

26.98 g, Density of Al = 2.7103 kg m-3,

Drift velocity, neA

Ivd

Number of Al atoms in 26.98 g of Al = Avogadro number of atoms

In 1g of aluminium, the number of atoms = 98.26

1002.6 23atoms

In1kg of Al, the number of atoms, N 98.26

10001002.6 23 251023.2 atoms

From the given density of aluminium, the mass of 1m3 of Al = 2.7103kg

Number of atoms in 1m3 of Al = 2.23×1025 ×2.7×103 = 6.02×1028 m-3

As each atom contribute 3 free electrons, the number of free electrons in 1 m3 of Al

=3×6.02×1028 m-3 =29101.806 m-3

71929d

103106.110806.1

1

neA

Iv

= 1.15 × 10-4 ms-1

Problem 5: The number density of free electrons in copper is 8.5×1028 m-3. A p.d of 5V

is applied to the ends of a copper wire of 5m long. The conductivity of copper is

5.81×107 -1Ω m-1. Calculate the average drift velocity of electrons.

Ans: Length of the wire, l = 5m; electric field, l

VE =

1Vm15

5

We have j= nevd. But j= , where is the conductivity of the medium

= nevd Drift velocity, 2819

7

d105.8106.1

11081.

en

Ev

= 0.43 cm s-1

Problem 6: A 12 volt potential difference is applied between the ends of a conductor

having a resistance of 4Ω . Calculate the current flowing through the conductor.

Ans: V =12 volt and R = 4 V = IR or R

VI

4

12 = 3A

Problem 7: A very small current is passed through a wire of resistance 60Ω and

length 2m having an area of cross-section 210-6m2. Calculate the resistivity of the material at that temperature.

Ans: R = 60, l =2m, A = 2 10-6m2

3-39 Current Electricity

Resistivity

l

AR =

2

10260 6= 6×10-5 Ωm-1

Problem 8: Two pieces of wires having lengths 5cm and 18cm cut from a uniform wire

at an area of cross-section 310-6m2 join together in series. It is connected to a p.d. of 25V and the current is equal to 3A. Calculate the resistance of two wires.

(Ans: R of 5cm wire =1.81, R of 18cm wire = 6.52)

Problem 9: The resistance of a wire is 4Ω . What is its new resistance if it is stretched

to double its length?

Ans: Resistance of the wire 1

1A

R 1l , where l1 is the length of the wire, A1 is the area of

cross section and specific resistance of the material. When length is doubled, the new length, l2 = 2l1.

Since volumes are equal A1l1= A2 l2 = A2 (2l1) or 2

AA 1

2

The new resistance, 2

22

AR

l

2

A

2

1

1l

1

1

A4

l=4 1R 44 16 Ohms

Problem 10: If the ratio of lengths of two wires is 1:3 and ratio of their diameters is

3:1, find the ratio of their resistances. (97M, 85S)

Ans: A

Rl

, or R 2r

l (since A =

2r )

For 1st wire, 21

11

rR

l …(1) and for 2nd wire, R2

22

2

r

l …(2)

2

1

2

2

1

2

22

21

1

2

1

r

rr

rR

R

l

l

l

l=

27

1

3

1

3

12

R1:R2 = 1:27

Problem 11: Resistances of coil at 30 0C and 100 0C are 10.2Ω and 12.7Ω

respectively. Calculate its temperature co-efficient of resistance().

Ans: 1221

12

tRtR

RR

103 C103.9

307.121002.10

2.107.12

Problem 12: A heating element of nichrome (α =1.7 10-4 /0C) has a resistance of

200Ω at a temperature 25oC. What is the resistance of the same heating element at a

temperature 325oC?

Ans: R1 =200 =1.710-4 /oC t1 = 25oC t2 = 325°C R2=?

= 1221

12

tRtR

RR

i.e., 1.710-4 =

25R253200

002R

2

2

1.710-4 25R253200 2 = R2 200

11.05 4.25 23R10 = 2R 200 R2 = Ω210.16

00425.1

05.211

Problem 13: Resistance of a conductor at 0oC is 8.5Ω . At what temperature does its

resistance become 10Ω ? = 0.00351 K-1. (Ans: 50.3 0C)

Problem 14: At what temperature would resistance of a conductor be doubled its

resistance at 0°C? The temperature co-efficient of resistance is . (78S)

Current Electricity 3-40

Ans: tR

RR

0

0t But 0t R2R t

1

tR

RR2

0

00 or α

1t i.e. when temperature is

equal to

1, the resistance of the conductor will be doubled.

Problem 15: A uniform wire has a resistance of 100Ω at 25oC and = 2×10-2 oC-1. The wire is stretched so that the length becomes doubled and the resistance becomes

405Ω . Calculate the temperature of the wire just after stretching.

Ans: The resistance of the wire after stretching at 250C =4 100=400 (Ref SQ.16)

So R1=400 , R2=405 =210-2 /oC t1 = 25oC t2=?

= 1221

12

tRtR

RR

or 210-2 =

25054t004

004054

2

0.02 (400 t2 – 10125) = 5 t2 = 8

5.2025 =

8

5.207= 25.94 0C

Problem 16: Three resistors 12Ω , 7Ω and 1Ω are connected in series. Find the

resultant resistance of the combination.

Ans: Total resistance of the series combination, R = R1 + R2 +R3

=12 + 7+1 = 20Ω

Problem 17: Calculate the effective resistance(R) of a parallel combination of 2Ω ,

4Ω and 8Ω resistors.

Ans: 321 R

1

R

1

R

1

R

1 =

8

7

8

124

8

1

4

1

2

1

R Ω

7

8

Problem 18: The effective resistance of two coils is 20Ω when connected in series and

4.8 Ω when connected in parallel. Find the resistances.

Ans: Let R1 and R2 be the two resistances.

In series, R1 +R2 = 20 or R1 = 20 –R2 …….(1)

In parallel,

8.4RR

RR

21

21 i.e., 8.420

RR 21

Or R1R2 = 204.8 = 96 ……. (2)

Substituting for R1 in (2), we get (20 – R2) R2 = 96

Or 096R20R 222

i.e. (R2 – 8) (R2 – 12) = 0 R2 = 8Ω or 12Ω

So if R2 = 8, then R1 = 12Ω and if R2 = 12, then R1 = 8Ω

The two resistances are 12Ω and 8Ω

Problem 19: Three resistors are connected in parallel. Two of them are 4Ω and 12Ω . What is the value of the third resistance if the effective resistance of the combination

is11

24Ω ? (Ans: 8)

Problem 20: Calculate the effective resistance between A and B in the figure.

Ans: R2 and R3 are in series

211RRR 32a

This combination is parallel to R5

A

B

R1

R2

R3

R4 R5

1

1 1

1

1

3-41 Current Electricity

3

2

12

12

RR

RRRor

R

1

R

1

R

1

5a

5aA

Aa5

This RA is series to R1

The resultant of this combination = RA + R1 = 3

2+1=

3

5. But R4=1

This combination is parallel to R4 and the total resistance

Problem 21: Calculate the total current and current through R1,

R2 and R3 in figure. R1 = 10Ω , R2= 25Ω , R3 = 15Ω . The

terminal voltage of the source is 6V.

(Ans: Total current through the circuit =1.24A

Current through 1R 0.6A

Current through 2R 0.24A

Current through 3R = 0.4A)

Problem 22: A 5Ω resistor is connected to a battery of e.m.f 6V. The current through the circuit is 1A. Calculate the internal

resistance of the battery.

Ans: Current rR

EI

1Ω 5

1

6R

I

Er

Problem 23: A load resistance of 100Ω connected to a battery of e.m.f 24V and

internal resistance 3Ω . Calculate the terminal voltage.

Ans: Terminal voltage, V =rR

E

× R 100

3100

24

= 23.3 V

Problem 24: Calculate the potential difference and current across R1, R2, R3 and R4. if

the circuit as shown is connected to a battery of terminal voltage of 100V. R1= R2 = R3 =

R4 = 2Ω .

(Ans: Total current through the circuit = 20 A

Current through R1 and R2 = 20A

Current through 3R =10 A

Current through 4R =10 A

P.d. across R1 = 40V

P.d. across R2 = 40V

P.d. across R3 and R4 =20V)

Problem 25: Calculate the minimum voltage to charge a 6V secondary battery of

internal resistance 0.5Ω through a load resistance 10Ω . Ans: The required voltage, E = Voltage of cell + I × internal resistance

E 5.05.010

E6

or 6

5.10

5.01E

5.10

10

6E 6.3 V

Problem 26: The current in a resistor connected to a battery with a terminal voltage 4.5V is 200mA. Calculate the current through the resistor if it is connected to a p.d. of

25V. (Ans: 1.11A)

Problem 27: A 6V battery of 3Ω internal resistance charged with 12V d.c. supply

through a 12 Ω resistor connected in series. Calculate the charging current.

(Ans: 0.4 A)

R1 R2

I

R3

R4

I/2

I/2

100V

I

I

R1=10

R3=15

R2=25

6V

Current Electricity 3-42

Problem 28: The terminal p.d. of a cell is 28.5 volt when the current is 1A and 27 V

when the current is 2A. Find the e.m.f. and internal resistance of the cell.

Ans: Let E be the e.m.f. and r be the internal resistance of the cell.

But terminal p.d. = e.m.f – p.d. across internal resistance

In the first case 28.5 = E - 1r, i.e. 28.5 = E-r ………………(1)

In the second case 27 = E - 2 r, i.e. 27 = E – 2r …………..(2)

(1) – (2) r = 1.5Ω Substituting for r in (1), E = 28.5 + 1.5 = 30V

Internal resistance of the cell is 1.5Ω and e.m.f is 30V

Problem 29: Find the current I in

the given circuit.

Ans: Applying Kirchhoff‟s voltage

law to the closed loop ABCDA,

10 I + 0.5 I + 5 I+ 0.5 I + 8 I + 0.5 I + 5 I + 0.5 I + 10 I = 50 – 70 – 30 + 40

I (10 + 0.5 + 5 + 0.5 + 8 + 0.5 + 5 + 0.5 + 10) = – 10

40 I = – 10 I = A0.25

40

10

The negative sign indicates that the current flows in the anticlockwise direction.

Problem 30: A Daniel cell has an e.m.f. 1.08V. When a 8Ω resistance is connected in

series with the cell, the p.d across the resistance is found to be 0.8V. Find the internal

resistance of the cell. (Ans: 8.2 ohm)

Problem 31: A coil resistance 1.5Ω and a 2Ω resistance wire of length 1m are

connected in series with 2V cell having internal resistance 0.5Ω . Find the current in

the circuit and the fall of potential per cm of the wire.

Ans: Total resistance in the circuit, R = 1.5+2+0.5=4

Current in the circuit, I = R

V=

cetanresis

d.P=

4

2= 0.5A

Total p.d. across 2 resistance wire = 0.52 = 1volt

Fall off potential per cm=1 volt /100cm= 10-2 Vcm-1

Problem 32: Using Kirchhoff’s laws, determine the currents I1, I2 and I3 for the given

network.

Ans: Applying Kirchhoff‟s current law to the node A,

I1 = I2 + I3 ………(1) Applying Kirchhoff‟s voltage law to the closed loop ADCBA,

– 4 I2 – 2 I1 = – 20 + 10

2 I2 + I1 = + 5

i.e. 2 I2 + I2 + I3 = 5

3 I2 + I3 = 5 ………(2) Applying Kirchhoff‟s voltage law to the closed loop ADEFA,

4 I2 – 6 I3 = – 10

or 2 I2 – 3 I3 = – 5 ……(3)

A

D C

B I 10 5

5

10 8

40 V 30 V

70 V 50 V

0.5 0.5

0.5 0.5

B C I1

2 20 V

A D

4 10 V

F E

6

I2

I3

3-43 Current Electricity

Solving (2) and (3), we get A11

102I and A

11

253I

Substituting in (1), we get A11

35

11

25

11

10I1

Problem 33: In the given circuit, E, F, G and H are cells of e.m.f.

2, 1, 3 and 1V and their internal resistances are 2, 1, 3 and 1Ω

respectively. Calculate i) the p.d. between B and D ii) the p.d. across the terminals of each of the cells G and H.

Ans: The distribution of currents in the network following the

Kirchhoff‟s first law is as shown in the figure.

According to Kirchhoff‟s law, in a closed circuit DCBD,

132I1I)3I( 1

2I2II3 1 or 2I2I4 1

or 1II2 1 .............(1)

In a closed circuit DBAD,

121)II(2)II(I2 111

1)II(3I2 11 or 1I3I3I2 11

1I3I5 1 ………….(2)

Solving (1) and (2), we get A13

6IandA

13

1I1

i) The p.d between B and D = 13

2I2 1 0.154 V

ii) Terminal p.d. across G = e.m.f – voltage drop across its internal resistance =

3 13

63 1.61 V

The voltage drop across the internal resistance of H is

Terminal p.d. across H = e.m.f + voltage drop across its internal resistance

= 1 13

61 1.46 V

Problem34: Find the effective resistance between A and B.

Each resistance R is 1Ω .

Ans: The circuit can be redrawn as shown. By symmetry, we can

disconnect at O as shown.

Resistance of upper part = 1+1+21

21

=

3

8. The resistance of the

middle part = 1+1= 2. The effective resistance (X) between A and B

is given by

8

10

8

3

2

1

8

3

X

1

10

8X 0.8Ω

Problem 35: A cube is made by using 12 identical wires each of

resistance R. What is the effective resistance

between two diametrically opposite corners A

and G?

Ans: By symmetry B, D, E are at same potential. We can connect them. C, F and H are also at same

potential.

Then we get the equivalent circuit as shown.

E

H

G

F - +

+ -

- +

+ - A B

C D

2

E

H

G

F - +

+ -

- +

+ - A B

C D

2

I-I1

I-I1

I I

I

I1

I1 1V, 1

1V, 1

2V, 2

3V, 3

A B

R

R R

R

R

R R

R R R

A B

R

R R

R

R

R R

R R R

O

O

B

A

C G

F

H E

D

A G B,D,

E

C,F,

H

Current Electricity 3-44

The effective resistance between A and G

= Ω6

5R

3

R

6

R

3

R

Problem 36: If 129Ω is connected to the left gap and 200Ω is connected to the right gap of a meter bridge, then find the balancing length from left.

Ans: l

l

100

QP , where lengths are in cm. Here P =129, Q = 200

P (100 -l) = Q l or 100P – Pl= Q l, 100P = Q l+ Pl or 100P= l (P+Q)

200129

129100

QP

P100

l = 39.21 cm

Problem 37: Calculate the effective resistance between A and B in the figures.

i) ii)

Ans:i) It is a Wheatstone bridge since 3

4

2

1

R

R

R

R . There is no significance for R5.

(R1 = R2 =R3 = R4 = R5 = 1Ω )

There are only two branches of R1 +R2 and R3 +R4. These are in parallel connection.

Total resistance of (i) =4

4

22

22

= 1Ω

ii) is exactly equal to fig.(i). So resultant resistance =1Ω

Problem 38: Calculate the total current and current through each resistor shown in the

figure. The terminal voltage of the source is 6V and R1=10Ω ,

R2=30Ω , R3=6Ω , R4=2 Ω and R5 =10Ω .

Ans: It is a Wheatstone bridge where R1, R4, R2, R3 are the resistances

instead of P, Q, R, S in Wheatstone bridge respectively.

52

10

R

R

4

1 56

30

R

R

3

2 i.e., 3

2

4

1

R

R

R

R

So the bridge is balanced. Then R5 is not significant.

Total resistance of the bridge 3612

3612

RRRR

RRRR

3241

3241

= 9

Total current 9

6I = 0.667 A. Current through R2 and R3 =

32 RR

V

=

36

6= 0.167 A

Current through R1 and R4 =12

6

RR

V

41

= 0.5 A

Problem 39: An unknown resistance is connected to the left gap and 1Ω resistance is connected to the right gap of a meter bridge. The balancing length from left is 0.6m.

Calculate the value of unknown resistance.

Ans: Here Q =1 and balancing length l = 0.6m

A

B

R1

R2

R3

R4 R5

1

1 1

1

1

A

B

1

1 1 1

1

R1 R4 R5

R3 R2

6V

3-45 Current Electricity

Resistance connected at left, l

l

1

QP

6.01

6.01

= 1.5Ω

Problem 40: In a metre bridge experiment to find the resistance of a coil, the balancing

length is found to be 55cm, when 2Ω resistance is taken from the resistance box. Calculate the resistance of the coil. If the length of the coil is 90cm and radius 0.028

cm, also calculate it’s specific resistance.

Ans: In meter bridge experiment, the unknown resistance is P.

l-

l

100

QP , where lengths are in cm. Q = 2, l =55cm, 100-l =45cm

P =

45

5522.44Ω

Specific resistance or resistivity of the material of the coil, L

Pr2

r = 0.028 cm = 2.810-4 m L = 90cm = 0.9m

9.0

44.2108.214.324

= 6.674 × 10-7Ωm

Problem 41: In a potentiometer experiment, the balancing length for a Leclanche cell is

640 cm and for a Daniel cell is 560 cm. Compare the ratio of their e.m.fs.

Ans: The ratio of emf‟s, 2l

l1

2

1

E

E

560

6401.143

Problem 42: A standard cell of e.m.f. 2V is connected to a potentiometer wire of length

10m and resistance 90Ω through a 10Ω resistance. Calculate the p.d. per cm of potentiometer wire. If 5.6m of potentiometer wire is required to balance the e.m.f. of

an external cell, calculate its e.m.f.

Ans: Current through the potentiometer circuit,

resistancetotal

d.pI =

1090

2

= 0.02A

P.d. across the potentiometer wire = I resistance = 0.02 90 = 1.8V

P.d. per cm of the potentiometer wire =length

d.p

1000

8.11.8×

310Vcm-1

The e.m.f of the external cell = balancing length P.d./cm

= 32 108.1106.5 = 1.008 V

Problem 43: A 10m potentiometer wire of resistance 36Ω is connected in series with a

battery of e.m.f 5V and internal resistance 4Ω together with a resistance box also.

What resistance unplugged from the box will give a potential drop of 2V/mm? (94S) Ans: Let RP be the resistance of potentiometer wire and Rb be the resistance unplugged

from the box. Let „r‟ be the internal resistance of the cell. Here Rb is instead of the rheostat in

the primary circuit.

E =5V r = 4 Rp = 36 Rb =?

Potential drop per meter = 210-6103 = 2 10-3 Vm-1

P.D over 10 m wire of potentiometer = 2 10-3 Vm-1 10 =2 10-2 V

P.D. across potentiometer wire = p

bp

RRRr

E

36364

5102 2

bR bR40

180

40

102

1802

bR = 8960Ω

Current Electricity 3-46

Problem 44: The balancing length in a Daniel cell in an open circuit is 660cm. When a

resistance of 4 connected parallel to the cell, the new balancing length is 330 cm. Find the internal resistance of the cell.

Ans: Internal resistance r = R

2

21

l

ll. But R = 4, l1 = 660 cm l2 = 330 cm

r = 4

330

303606= 4

Problem 45: In a potentiometer arrangement to find the internal resistance of a cell a

cell is balanced by a length 76.3 cm of the potentiometer wire. The balancing point

changes by 12.5cm when the cell is shunted by 9.5. Find the internal resistance.

(Ans: 1.7) Problem 46: In a potentiometer experiment the null point is at 2.5m when the cell

whose internal resistance is to be measured is shunted by a resistance of 5Ω . When

the cell is shunted by 20Ω , the null point is at 3.75m. Calculate the internal resistance of the cell.

Ans: Let E be the e.m.f and r be the internal resistance of the cell. Then the p.d. across the

5 shunt =5r

5E

. By the principle of potentiometer, p.d. is proportional to the balancing

length.

5.25r

E5

…..........(1)

The p.d. across 20 shunt =20

20

r

E 75.3 … ..(2)

Dividing (1) by (2), we get 75.3

5.2

)5r(4

20r

75.3

10

5r

20r

r = 4Ω .

Problem 47: An electric bulb is marked as 60W, 230V. Calculate the current through

the filament of the bulb and resistance of the filament.

Ans: Power = current voltage. Current =voltage

power=

V230

W60= 0.261 A

Resistance of the filament =current

voltage =

A261.0

V230= 881.67Ω

Problem 48: 1000 watts electric heater is used in 250 volts main. Calculate the

resistance of the heater and the current drawn by it. Also calculate heat produced by it

in one minute.

Ans: Electric power P = VI V

PI I =

250

1000 = 4 A

Resistance of the heater = 4

250

I

V62.5Ω

Heat produced in one minute = VIt = P t =1000 60 J = 60 kJ Problem 49: A 100W, 230 volt bulb is connected to 100V supply. Calculate the energy

consumed in one second.

Ans: Power, R

VP

2

, Resistance of the bulb =100

230

P

V 22

=529

3-47 Current Electricity

Energy consumed per second = power =R

V2

; where V =100V

Power = 529

1002

=18.9 watt

Problem 50: Calculate the efficiency of an electric motor which works on a 24V supply

and draws a current of 75A yielding a mechanical power of 1000W.

Ans: Efficiency = %100inputpower

outputPower , V =24V, I =75A, P = VI

%1007524

W1000

= 55.56%

Problem 51: How long a 100W and a 230V electric bulb take to consume a unit of electrical energy. (Here 1unit =1kWh)

Ans: Energy in kWh or units = Power in kW × Time in hours

1 unit = 0.1× t t = 1.0

1 = 10 hrs

Problem 52: The temperature of cold junction is 15oC. Find the neutral temperature if

the thermo e.m.f reverses at 500oC.

Ans: 2

icn

2

515

2

50015257.5 0C

Problem 53: The thermo e.m.f of a copper- constantan thermo couple with cold

junction at 0oC, obeys the equation V = + 2 where V is in V and temperature

difference () in C0. =41V oC-1 and =0.041V oC-2. Find the temperature of the hot

junction when the thermo-e.m.f is 6mV.

Ans: =41VoC-1 = 4110-6 V oC-1 = 0.041 V oC-2 = 4110-9 V oC-2

V = 6mV = 610-3 V V= +2 or 2 +-V = 0

2

V42

9

39266

10412

1061041410411041

= 129.56 C0 (Positive root is taken because cannot be negative.)

Temperature of the hot junction = 129.56 C0

Problem 54: Calculate the maximum thermo e.m.f of the copper-iron thermocouple

with =17.2 V oC-1 and = -0.048V oC-2 keeping the cold junction at 0 oC.

Ans: We have2

2

1V …………….(1)

For maximum thermo e.m.f, = n and d

dV0n or

n ,

Substituting n in (1), maxV =

222

2

1

2

1

=

2

2

=

6

26

10048.02

102.17

= 3.08210-3V = 3.082 mV

PROBLEMS BASED ON NCERT

Current Electricity 3-48

Q1. The storage battery of a car has an e.m.f. of 12 V. If the internal resistance of the battery is

0.4 , what is the maximum current that can be drawn from the battery?

Ans. IRV , R

VI . Here 4.0rR and V= 12 V AI 30

4.0

12

Q2. A battery of e.m.f.10 V and internal resistance 3 , is connected to a resistor. If the current

in the circuit is 0.5 A, what is the resistance of the resistor? What is the terminal voltage of the battery when the circuit is closed?

Ans. E=V + Ir, E = IR + Ir, ie. 1735.0

10r

I

ER

Again using E = V + Ir

Terminal Voltage, V = E – Ir = 10 – 0.5 3 = 10 – 1.5 = 8.5V

Here by short cut V = IR = 0.5 17 = 8.5 V

Q3. a) Three resistors 1 , 2 , and 3 , are combined in series. What is the total resistance

of the combination?

b) If the combination is connected to a battery of e.m.f.12 V and negligible internal resistance, obtain the potential drop across each resistor.

Ans. a) 321 RRRReq , 6321eqR

b) IRV , AR

V

R

VI

eq

26

12

Then voltage drop across R1 is given by, VIRV 21211

Voltage drop across R2 is given by, V2 = IR2 = 2 2 = 4V

Voltage drop across R3 is given by, V3 = IR3 = 2 = 6V

Q4. a) Three resistors 2 , , 4 , and 5 , are combined in parallel. What is the total resistance

of the combination? b) If the combination is connected to a battery of e.m.f.20 V and negligible internal resistance, determine the current each resistor, and the total current drawn from the battery.

Ans. a)

321

1111

RRRR ,

20

19

20

4510

5

1

4

1

2

11

R or

19

20R

b) V = IR, I = R

V

i.e. Total Current AI 1919/20

20

Current through 2 resistor = 102

20

1

R

VA

Current through 4 resistor = AR

V5

4

20

2

Current through 5 resistor = AR

V4

5

20

3

Q5. At room temperature ( )0.27 0C the resistance of a heating element is 100 . What is the

temperature of the element if the resistance is found to be 117 , given that the temperature

coefficient of the material of the resistor is .1070.1 .104 C

3-49 Current Electricity

G H

5

5

5 10

10

I-I1+I2

10V

I-I1

I1-I2

I2 I1

I

A

B

C

D

10

Ans. TRR 112 ,

1

12

R

RRT

i.e.

1000

7.1

1017

100107.1

100117 2

4

T oC

i.e. T 12 TT = 1000oC 10272710001000 12 TT oC

Q6. A negligibly small current is passed through a wire of length 15 m and uniform cross-

section ,100.6 27 m and its resistance is measured to be 5.0 . What is the resistivity of the

material at the temperature of the experiment?

Ans. A

lr ,

l

AR i.e.

77

10215

1065

m

Q7. A silver wire has a resistance of 2.1 at 27.5 C0, and a resistance of 2.7 at 100 C0

.

Determine the temperature co-efficient of resistivity of silver.

Ans. TRR 112 , 1

1

12 0039.0)5.72(1.2

6.0

)5.27100(1.2

1.27.2

C

TR

RR

Q8. A heating element using nichrome connected to a 230 V supply draws an initial current of 3.2 A which settles after a few seconds to a steady value of 2.8 A. What is the steady

temperature of the heating element if the room temperature is 27.0 C0? Temperature co-

efficient of resistance of nichrome averages over the temperature range involved is

.1070.1 04 C

Ans. 88.712.3

2301 R and 14.82

8.2

2302 R

TR

RR

1

12 , 1

12

R

RRT

ie.

1

1212

R

RRTT

Or 8678402727107.188.71

88.7114.8241

1

122

T

R

RRT

oC

Q9. Determine the current in each branch of the network shown in Figure.

Ans. Applying Kirchhoff‟s second law to net work ABDA, we get

05510 121 IIII

i.e. 03 21 III …(i)

For BCDB, 22121 5105 IIIIII

0243 21 III … (ii)

For ABCHGA, 1051010 211 IIII

or 223 21 III …(iii)

Adding (ii) and (iii), we get 256 21 II …(iv)

Multiplying (i) by 2 and adding to (iii), 29 21 II

i.e. 12 92 II …(v)

Using (v) for (iv), 29256 11 II or 1251 1 I or AI17

41

Substituting this value in (v), AI17

2

17

4922

Current Electricity 3-50

)100(

B

D A

G

C

Y X

From (i), AIII17

10

7

2

17

433 21

(1) Current in AIAB17

41

(2) Current in DC = AIII17

4

17

2

17

4

17

1021

(3) Current in AD = AII17

6

17

4

17

101

(4) Current in BC = AII17

6

17

2

17

421

(5) Current in AIBD

17

22

The –ve sign shows that the current is opposite to that shown in the fig.

(6) The total Current AI17

10

Q10 a) In a meter bridge (Figure), The balance point is found to be at

39.5 cm from the end A, when the resistor Y is of 12.5 . Determine the resistance of X. Why are the connections between resistors in a Wheatstone or meter bridge made of thick copper strips? b) Determine the balance point of the bridge above if X and Y are interchanged. c) What happens if the galvanometer and cell are interchanged at the balance point of the bridge? Would the galvanometer show any current?

Ans. a (i) l

lYX

100,

2.85.60

5.395.12

5.39100

5.395.12X

(ii) Thick copper strips are used to reduce their resistance of the connection wires because

this resistance is not accounted for in the calculations.

b) Interchanging Y and X, we get l

lXY

100

i.e. lXlYY 100 i.e. XY

Yl

100cm5.60

2.85.12

1005.12

from end A.

c) In this case also the galvanometer will not show any current. Because Meter bridge is

similar to a Wheatstone‟s bridge. In a Wheatstone‟s bridge the cell and the galvanometer can

be exchanged.

Q11. A storage battery of e.m.f. 8.0V and internal resistance 0.5 is being charged by a 120V

dc supply using a series resistor of 15.5 what is the terminal voltage of the battery during charging? What is the purpose of having a series resistor in the charging circuit?

Ans. During charging, the terminal e.m.f. of the battery is, IrEV

Effective e.m.f. in the circuit is 812021 EE

Current in the circuit Rr

EEI

21

A716

112

5.155.0

8120

Terminal voltage IrEV = 8 +7 0.5 = 11.5 V 0.5

15.5

120 V

8V

3-51 Current Electricity

The series resistance will reduce the high current so that the charging battery is not

damaged

Q12. In a potentiometer arrangement, a cell of e.m.f. 1.25V gives a balance point at 35.0 cm length of the wire. If the cell is replaced by another cell and the balance point shifts to 63.0 cm, what is the e.m.f. of the second cell?

Ans. Here lE ,

2

1

2

1

l

l

E

E i.e. 1

1

22 E

l

lE i.e. 25.225.1

35

632 E V

Q13. The number density of free electrons in a copper conductor is estimated is

.105.8 328 m How long does an electron take to drift from one end of a wire 3.0m long to its

other end? The area of cross-section of the wire is 26100.2 m and it is carrying a current of

3.0A.

Ans. Current AnevI d ,

Anevd

1

61928 102106.1105.8

3

= 4101.1 ms-1

Time taken, 4

41072.2

101.1

3

dv

lt s

Q14. The earth’s surface has a negative surface charge density of .10 29 mC The potential

difference of 400 kV between the top of the atmosphere and the surface results (due to the low conductivity of the lower atmosphere) in a current of only 1800 A over the entire globe. If there were no mechanism of sustaining atmosphere electric field, how much time (roughly) would be required to neutralize the earth’s surface? (This never happens in practice because there is a mechanism to replenish electric charges, namely the continual thunderstorms and lightning in

different parts of the globe.) (Radius of earth = .)1037.6 6 m

Ans. Surface area of earth A = 12662 1064.5091037.61037.614.344 r m2

Charge, Q = (charge density, ) (Surface area, A) C3129 1064.5091064.150910

Time, 2831800

1064.509 3

I

Qt s

t

QI

Q15. a) Six lead-acid type of secondary cells each e.m.f. 2.0V and internal resistance

0.015 are joined in series to provide a supply to a resistance of 8.5 . What are the current drawn from the supply and its terminal voltage?

b) A secondary cell after long use has an e.m.f. of 0.9V and large internal resistance of 380 . What maximum current can be drawn from the cell? Could the cell drive the starting motor of a car?

Ans. a) 015.065.8

26

MrR

nEI = 1.397 A

Terminal voltage, VIRV 88.115.8397.1

b) maxI is obtained when 0R , Ar

EI 005.0

380

9.1max

This current cannot start the motor of a car because a starter motor requires a large

current for a few seconds.

Q16. Two wires of equal length, one of aluminum and the other of copper have the same resistance. Which of the two wires is lighter? Hence explain why aluminum wires are preferred

Current Electricity 3-52

for overhead power cables. ( .1072.1,1063.2 88 mm uCAI Relative density of

Al= 2.7, of Cu = 8.9.)

Ans. Two wires have same length, and resistance, so the areas are different, as resistances are equal

For Cu wire

CuCuCu

AR

for aluminium wire

111

AAA

AR

So,

11

AA

CuCu

AA

172

263

1072.1

1063.28

811

Cu

A

Cu

A

A

A

Ration of masses 46.01729.8

2637.2111

CuCu

AA

Cu

A

Ad

Ad

M

M

CuA MM 1

Thus the aluminium wire for the same resistance is very light than copper and that is

why aluminium wires are preferred for overhead power cables. Aluminium is cheaper than

copper.

Q17. What conclusion can you draw from the following observation on a resistor made of alloy manganin?

Current Voltage Current Voltage

A

0.2 0.4

0.6

0.8

1.0

2.0

V

3.94 7.87

11.8

15.7

19.7

39.4

A

3.0 4.0

5.0

6.0

7.0

8.0

V

59.2 78.8

98.6

118.5

138.2

158.0

Ans.

Current A Voltage V Resistance

0.2

0.4

0.6

0.8

1.0

2.0

3.94

7.87

11.8

15.7

19.7

39.4

19.7

19.675

19.667

19.625

19.7

19.7

Here we can conclude that ohm‟s law is valid to a high accuracy and resistance of the

alloy is nearly constant at all currents. This is the reason, why manganin is used to make

standard resistance

Q18. Answer the following questions:

a) A steady current flows in a metallic conductor of non-uniform cross-section. Which of these quantities is constant along the conductor: current, current density, electric field, drift speed? b) Is Ohm’s law universally applicable for all conducting elements? If not, give examples of elements which do not obey Ohm’s law. c) A low voltage supply from which one needs high currents must have very low internal resistance. Why? d) A high tension (HT) supply of, say, 6kv must have a very large internal resistance. Why?

3-53 Current Electricity Ans. a) Current remains the same as it is steady.

All other quantities depend inversely upon area of cross-section.

Current density A

IJ . It changes as cross sectional area changes.

Electric field .A

I

A

IIRVE

As cross sectional area changes, electric field also

changes.

Drift velocity .nAe

IVd As area of cross section changes, drift velocity will also change

b) No, The devices like vacuum tube, semiconductor diodes, transistors, thermistors,

thyristors etc do not obey Ohm‟s law.

c) Maximum current drawn = .tanint

...

r

E

ceresisernal

fme clearly, low value of r will ensure

high current.

d) If internal resistance is not large, then the heavy current drawn during accidental short

circuit, can damage the supply.

Q19. Choose the correct alternative: a) Alloys of metals usually have (greater/less) resistivity than that of their constituent metals. b) Alloys usually have much (lower/higher) temperature co-efficient of resistance than pure metals. c) The resistivity of the alloy manganin is nearly independent of/increases rapidly with increase of temperature. d) The resistivity of a typical insulator (e.g., amber) is greater than that of a metal by a factor of

the order of (322 10/10 ).

Ans. a) Greater b) Lower c) Nearly independent d) 2210 .

Q20. a) Given n resistors each of resistance R, how will you combine them to get the (i) maximum (ii) minimum effective resistance? What is the ratio of the maximum to minimum resistance?

b) Given the resistance of 1 , 2 , 3 , how will be combine them to get an equivalent

resistance of (i) (11/3) , (ii) (11/5) , (iii) 6 , (iv) (6/11) ? c) Determine the equivalent resistance of networks shown in figure.

(a)

Ans. a) (i) Maximum resistance can be obtained by combining them in series with each other.

The maximum resistance ...........max nRtimesnRRRR

(ii) Minimum effective resistance can be obtained by combining them in parallel with each

other.

Minimum resistance maxR is found as

R

ntimesn

RRRR ........

1111

min

. .2

min

maxmin n

n

R

nR

R

R

n

RR

2 2 2 2

1 1 1 1

R

R

R

R

R

Current Electricity 3-54

1 1 1 1

1 1 1 1 1

)5.0,12( V

b) Here 321 2,1 RandRR

(i) If we combine 1R and 2R in parallel and 3R in series with them as shown, then equivalent

resistance

3

113

3

23

21

213

1

21

2

RR

RRR

R

(ii) If we combine 2R and 3R in parallel with each other and this combination is connected in

series with ,1R then the equivalent resistance is

32

321

32

321

RR

RRRR .

5

11

5

61

(iii) If 1R , 2R and 3R are combined in series, then

6321321 RRRR

(iv) If 1R , 2R and 3R are combined in parallel, then

11

6,

6

11

6

236

3

1

2

1

1

11111

321

RRRRR

c) (i) The network consists of four similar units.

Resistance of one unit is given by

3

4

42

42eqR

Total resistance 3

16

3

44

(ii) All the resistors are in series RRRRRRReq 5 .

Q21. Determine the current drawn from a 12 V supply with

internal resistance 0.5 by the infinite network shown in

figure. Each resistor has 1 resistance

Ans. Let the total resistance of the circuit be Z and a set of three resistors of value R each be

connected to it as shown in figure.

33R

22R

11R

Z

R

R

R Z

R R

R

3

2

1

2

1 1

2

2

4

3-55 Current Electricity

Adding of these resistors will not change the value of Z because the network is infinite

then

RZ

ZRRR

RZ

ZRRZ

2 or 022 22 RRZZ

Or 3132

842 22

RRRRR

Z (R = 1)

31Z ohm or 31 ohm

Since resistance cannot be negative, therefore Z= 73.273.1131

Current drawn, 71.35.73.2

12

rZ

EI A

Q22. Figure shows a potentiometer with a cell of 2.0 V and resistance 0.40 maintaining a potential drop across the resistor wire AB. A standard cell which maintains a constant e.m.f. of 1.02V (for very moderate currents up to a few mA) gives a balance point at 67.3 cm length of the wire. To ensure very low currents drawn from the standard cell, a very high resistance of 600

k is put in series with it, which is shorted close to the balance point. The standard cell is then

replaced by a cell of unknown e.m.f. and the

balance point found similarly, turns out to be at

82.3 cm length of the wire.

a) What is the value ?

b) What purpose does the high resistance of 600

k have?

c) Is the balance point affected by this high

resistance?

d) Is the balance point affected by the internal

resistance of the driver cell?

e) Would the method work in the above situation if the driver cell of the potentiometer had an

e.m.f. 1.0V instead of 2.0V?

f) Would the circuit work cell for determining an extremely small e.m.f., say of the order of a few

mV (such as the typical e.m.f. of a thermo-couple) if not, how will you modify the circuit?

Ans. a) Since E l,

1

2

1

2

l

l

E

E or 25.102.1

3.67

3.821

1

22 E

l

lE V

b) Purpose of using high resistor is that it will allow very small current to flow through

galvanometer when the circuit is not balanced. Once the null point is closely known the 600

k is short circuited and correct balancing length is found out.

c) No. The balance point is not affected by the high resistance as no current flows when

balanced.

d) No. The balance point is not affected by the internal resistance of the driver cell as no

current flows when balanced.

e) No, the method will not work in this situation because the balance point is not obtained if

driver cell of e.m.f. 1V instead of 2V is used. If the driver cell of standard circuit has an emf

smaller than emf of the cell to be balanced then null point will not be achieved in the length

of potentiometer wire and galvanometer will provide only one side deflection always.

Current Electricity 3-56

f) The circuit will not work well for determining extremely small e.m.f. because in such cases

the balance point will be almost on end A. To make suitable arrangement the potential drop

in potentiometer wire is brought down to few mV, so that balance point is obtained at

suitable length and this is done by introducing a high resistance in series in the standard caliberation circuit.

Q23. Figure shows a potentiometer circuit for comparison of two resistances. The balance point with a standard resistor

0.10R is found to be 58.3 cm, while that with the

unknown resistance X is 68.5 cm. Determine the value of X. What might you do if you failed to find a balancing point with

the given cell of emf ?

Ans. a) R

X

E

E

1

2 , Rl

lR

E

EX

1

2

1

2

75.11103.58

5.68

b) To obtain balancing point, either a cell of less e.m.f. is used or a suitable resistor be put in

series with R and X so that potential drop across the wire AB is reduced. When a series

resistance is connected, take care to find balancing lengths for R and X separately

Q24. Figure shows a 2.0V potentiometer used for the determination of resistance of a 1.5V cell.

The balance point of the cell in open circuit is 76.3 cm. When a resistor of 9.5 is used in the external circuit of the cell, the balance point shifts to 64.8 cm length of the potentiometer wire. Determine the internal resistance of the cell.

Ans. 2

21l

Rllr , where 1l is the balancing length in the open circuit, when R is not

included. 2l is the balancing length when R is included in the circuit.

3.761 l cm; 8.642 l cm and R = 8.5 68.18.64

5.98.643.76 r = 1.7

-

K1

K2

IMPORTANT INSTRUCTIONS TO THE STUDENTS !!!

*** A BRILLIANT STUDENT may study the whole book inevitably including the problems based on NCERT given at the end of the every chapter

* AN AVERAGE STUDENT may study at least all the text matter in this book along with the sample questions given at the bottom of the pages of the text matter, the worked out problems and the five solved question papers given at the end of the book

** A BRIGHT STUDENT may study at least all the text matter, the sample questions given at the bottom of the pages of the text matter, the model questions with answers, the multiple choice questions, all the problems, the f five solved question papers given at the end of the book

Wish you all successes The authors