ELECTRICAL ENGINEERING (Eiectrical Machines and Appliances)

THEORY- 1

VOCATIONAL EDUCATION Higher Secondary - First Year

A Publication under Govemment ofTamilnadu

Distribution ofFree Textbook Programme ( 'KOT FOR SALE)

Untouchability is a Sin U ntouchability is a Crime Untouchablity is Inhuman

TAMILNADU TEXTBOOK CORPORATION College Road, Chennai - 600 006.

Government ofTamilnadu First Edition - 201 O

CHAIRPERSON Mr. K. GOVINDASAMY

Senior Lecturer 1 Electrical Bhakthavatsalam Pol:ytechnic College,

Kanchipuram- 631 552

AUTHORS

Mr. A. RAMESH Vocational Instructor Govt. Model Hr. Sec. School Saidapet, Chennai- 15.

Mr. V.V. Shanmugadoss Vocational Teacher Govt. Hr. Sec. School Perunagar Kanchipuram- 603 404

Mr. Kasinathan Vocational Instructor General Kariappa Hr. Sec. School Saligramam Chennai- 600 093

Mr. R. Balamurugan Vocational Instructor Govt. Model Hr. Sec. School Saidapet, Chennai- 15.

Mr. P. Muthusamy Vocational Instructor Govt. Boys Hr. Sec. School Namakkal South Namakkal

This book has been prepared by The Directorate of School Education on behalf ofthe Govermnent ofTamilnadu

This book has been printed on 60 GSM Paper

Printed by Offset at:

HIGHER SECONDARY- VOCATIONAL COURSE

ELECTRICALMACHINESAND APPLIANCES

Syllabus for XI Standard

Theory I ( ELECTRICAL ENGINEERING) l. INTRODUCTION OF ELECTRICAL E:\'GIJ'\o'EERING

Introduction-methods ofpower generation -electrical safety-safetyprecautions ofelectrician - electric shock- preventive method of elecnic shock - first aid.

2. MATERIALS AND TOOIJS USED IN ELECTRICAL ENGINEERING DEPARTMENT

Electrc conductor- types of electric eonductor- properties of electric conductor- electrical insulating materials-properties-types ofinsulating materals-electrical accessories- types of switches - fuse unt- socket- ceiling rose- hand tools.

3. ELECTRICAL TERMS AND DC CIRCUITS

Curren! voltage- resistance- ampere-volt- ohms- ohm's law- capacitance- krichoff's law- electrical circuit- closed electric circuit- open electric crcuit- electric short circuit -series circuit -parallel circuit- series parallel circuit- power -energy calculation.

4. ELECTRO MAGNETISM

Magnetic materials- Electro magnet-magnetic effect dueto current- flemming's right hand rule-max-well's cork screw rule- magnetic field in the coi!- end rule- magnetic reacton when the curren! passing in a conductor in same drection and opposite direction- Faraday's electro magnetic induction- induced electro motive force statically induccd e.m.f- self and mutual induced e.m.f lenz's law- hysterisis hysterisis loop- energy stored in a magnetic filed.

5. ELECTRICAL EFFECT

Electrical energy

Electrical energy

Electrical energy

Electrical Energy

electrical energy

electrical energy

6. BATTERIES

Light energy (lamp- CFL) Smmd energy (Bell Syren) Magneti e energy (Electromagnet) Heat Energy (Iron box) Chemical energy (Electroplating-Battery charging) Mechanical energy (Electlic Motor)

Battery- types of Batteries primary cells secondary ce lis- difference between primary cells and secondary cells- Lead acid cell- recharge batteries- watch cell- UPS.

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7. A.C. CIRCUITS AND ELECTRICALMEASURING INSTRL"MENTS

Ahemating current- A.C. wave fonn - power factor- R.M.S. value- phase difference- pure resist ve circuit- inductive circuit- capacitive circuit- R.L. circuit- R. C. circuit- R. L. C. circuit- star delta connection and two watrrneter method.

Ammeter- Voltmeter- ohm meter-watt meter- multi meter- Tong tester- Tecometer-megger- single phase energy meter-Three phase energy meter.

8. TRA.~SFORiVIER

Introduction- construction- operation- types oftransforrner- uses- protective devices of transforrner - transformer oil.

9. DC GENERATOR

Basic Principie- construction-parts of generator- method of functiouing- types of genemtor series generator- shunt generator- compound generator.

lO.DCMOTOR

Basic Principie- construction- parts ofDC motor -- method of functioning- types of DC motor- series motor-- Shunt motor- compound Motor.

ll.AC GENERATOR (ALTERl~ATOR)

Construction- operation- parts of ac generator- types of AC generator- single phase AC generator- three phaseAC generator.

12.ACMOTORS

Single phase motor- construction- operation- uses.

TYPES OF SINGLE PHASE MOTOR

i) Split phase motors ) Capacitor type motors ii) Repulsion type motors iv) Shaded pole motors v) Universal motors V1) Submorsible type motOI'S

Three phase induction motor- types- constructon- operation- parts ofthree phase motors stator Rotor.

SQUIRREL CAGE INDUCTION MOTOR a) Single squirrel cage induction motor. b) Double squirrel cage nduction motor

IV

13. MOTOR STARTERS

AC motor starters : DOL starter- star 1 delta starter-Auto transfonner struier slipring motor starter (resistance type starter). DC ~OTOR STARTERS

Three point starter- four point starter.

14. ELECTRONICS

Semi conductors -electrons and boles- ntrinse semi conductor- extrinsic semi conductor-dopping of semi conductor-N. type semi conductor-P. type semi conductor- PN Junction diode-halfwave rectifier- full wave rectifrre- bridge reetifire- Zenerdode-lght emitting dode- Juncton transistor- PNP, NPN transstors SCR.

V

PREFACE

This book presents simple, explict and easy for learning at the begining leve! for the

subjeet on Electrical Engineering. Considerable emphasis is laid on the fundamentals physical coneepts, principies and functions ofvarious elements.

The Government ofTamilnadu is deciding to revamp Vocational Education in Higher

Secondary Student to make them easy to understand higher studies in engineering faculty.

The students at school final leve! and the beginers on this subject can easily able to understand the Principies and Coneepts. Much care is taken to explain all the details with

neat diagram and sketches. All the topics ofths book is sclf illustrative. The students at the

begining leve! will learn this book with much interest themselves, because such care is

taken while preparation ofthis book.

I personally thank all for giving me this best opportunity to bring out a best book for

benefit of the Vocational Students at school final leve!. All the readers of this book will

enrich knowledge on basic Electrical Engineering, which makes us feel proud and happy.

vi

Thiru. K. Govindasamy Chairperson

CONTENT

SI.No. Page No.

l. lntroduction ofElectrical Engineering 1

2. Materials and Tools U sed in Electrical Engineering Department 10

3. Electrcal Terms and DC Circuits 20

4. Electro Magnetism 71

5. Electrical Effect 84

6. Batteries 92

7. A.C. Crcuits and Electrical Measuring Instruments 109

8. Transformer 151

9. DC Generator 168

10. DCMotor 184

11. AC Generator(Alternator) 193

12. ACMotors 206

13. Motor Starters 226

14. Electronics 233

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l. ELECTRICAL ENGINEERING 1.1. INTRODUCION

The subjects is introduced in Higher Secondary leve!. In this subject students are taught how to maintain and repair electrical appliances and electrical machine howto connect elecirical circuits and repairing minor or major faults in the citeuits and motorrevmding both theoretically and practically. So by learning this subjects students are able to earn oftheir own by practicing such leaming.

We know, the universe consists of five big natural energy sources like water, land, Air, fire & space. The Sixth important energy developed by human is called "Electricity". In this modem world, for our dayto da y life the electricityplays a vital role. Simply to say, e ven man lives without tood but not lives without Electricity. Because in our lite, the electrical goods play an importan! role. Hence thc students must know about this source and how t is applied.

Electrcity is one type of energy. Al! matter whether so !id, liquid or, gaseous consists of minute partid es known asAtoms. According to modem Research electric current means flow of electron. So we need to know about the Ato m.

L2.ATOM

ft has a hard central e ore known as nucleus. It contains two type of part eles one is known as pro ton and carries positive charge. The other is neutron, which is electrically neutral, .e. it carries no charge. Around the nucleus in elliptical orbit the electrons one revolving. Electrons caJTy the negative charge. The number of electro u are number of protons in a atom are equal. So the atom is electrically neutraL The nun1ber of protons in the nucleus of ato m gives the atomic number. The total numbers of neutTon and proton are known as atomic weight. Because negligible weight ofthe electron is nottaken to calculate atomic weight

1.3. METHODS OFELECTRICITY PRODUCTS

An electricity is produced bythe extraction of electrons from an atom. The energies which are used to produce an electricity are (i) Friction (ii) Light (iii) Heat (iv) Pressure (v) Chemical Aclion (vi) magnetism.

1.3.1. Electricity dueto friction

Dueto the frction oftwo material, the electrons come out from one material to join vvith the other materiaL The material which looses the electron gets +ve charge and the material attracted the electron gets --ve charge. This type of electricity is called Static Electricity.

Ex. :Materials like Glass, Rubber, Wax, Silk. Reyon, Nylon.

1 ,3.2 Electricity dueto light

When the light falls on the material, the electrons emitted from the surface and producing the low of current Forthis purpose Photocell is used. Photo cell is used to convertthe light energy into cmTent. The rnaterials which emitted electron dueto the light tal! on the surface are "photo sensitve metaL"

Ex. Sodium, Potassium, Lithimn, Cesitm1. 1

1.3.3 Electricity due to Pressurc

Electrons in the outennost orbt of an atom is extracted dueto the pressure applied toan atom and the electricity is produced. Ths is called "Pieza Electricity". In a telephone, diapharam is pressured by the sound waves. Because ofthis, Electric Waves are produced depending upon the pressure of sound waves.

1.3.4 Electricity dueto Heat

The ends oftwo metal rods are joined together and this joined part is heating. Dueto this the part opposite to the heated place is com1ected by a Galvanometer and the Electricity is known as the deflecton ofthe pointer.

In the san1e way, two metal plates are joined together and is heating, forthe purpose of producing electricity. This type is called "Thenno Coupling Method."

For the abo ve four methods, sufficientelectricity is not produced and the energy of electricity is also less. Because, the othertwo methods are used to produce the sufficient electricitywith high energy.

1.3.5.Electricity dueto chemical action

By using the method of chemical action, electrons are extracted from an atom and producing electricity. This method is used for producing electricity in primary and secondary cells.

Prin1ary cell is used in torch light and the secondary cell is u sed in cars, motorcycles etc.

1.3.6. Electricity dueto Magnetism

In this method Electrons are extracted from an atom dueto magnetism. F or this purpose generators are used. In generator the energized electricity is produced bythe magnetic poles and armature winding.

In our countiythe requirementof electricity is produced in allthe abo ve methods.

In thiswehaveto studyaboutall thepowergeneratingstations in Tamiinaduandhowtheelcctricity is produccd.

1.4. POWER GENERATING PLANTS

Today, there are seven power gencrating stations are available n our country. By ths, approximately 7000 MW curren! is produced in our country

Types ofPower Generating Plants

l. Hydro Electric Power Plant 2. Thcnnal Power Plant

3. Atomic Power Plant

4. Gas Power Plant 5. Diesel Power Plant 6. Solar Power Plant

7. Wind-Mill Power Plant 2

1.4.1. Hydro Electric Power Plant

From the waterreservoir, the water is taken thtough the jainttubes to the waterturbne. For tbe rotation ofturbine, the knetic energy ofwater is con verted into meehanical energy and s con verted imo eleetrical energy by the using of generators.

This type of plan! is placed in TamilNadu at Mettur, Kunda, Bicara, Suruliyaru & Kadamparai.

1.4.2. Tbennal Power Plant

Chemical energy is converted into heat energy by burnng of coa! or lignire in bciler plant Water in the bciler is converted nto stearn by heat energy. This steam s flowing through the steam turbine which s connected to the generator and this energy is con verted into mechanical energy by the rotation of turbine. TI1e mechanical energy is again con verted into electrical energy by the use of generator.

This type of plan! is placed in Tami!Nadu at Ennore. (Chennai), Neyvelli, Tuticorn and Metmr. Thermal Power Plants play a major role for the reqnirements of electricity in Tamilnadu

1.4.3. Atomic Power Plant

By the diffuson of an ato m ofUranium or Thorium, to gettng more heat Based on this principie the atomic powerplant is workng. The heat energy is produced and is used toro tate the steam turbine and ths energy is converted into meehanical energy. The generator convens the mechanical energy into electrical energy.

Ths plant is placed in Kalpakkam near Chennai and Tharapur in Rajastan S tate. Leakage of gas by ths plant may cause pollution and affeet the health ofthe people.

1.4.4. Gas Power Plant

For the rotation of turbine, the underground gas is used. The generator which is connected to the turbine produces the eleetricity. Ths plant is placed in Ramanathapuram and Kuthalam.

1.4.5 Diesel Power Plant

Ths type of plant is used for the place where the contnuous requirements ofelectricity is needed. Le. in big factories and refrigeration works. The electricity is produced by the generator which is connected to the big diese! engne.

Dependng upon the requirements, different capacities of small or large diese! generators are used in hotels, hospitals, J ewelleryshops, cinema theatres, shipyards etc.

1.4.6 Solar Power Plant

For the purpose of mnimum production of electricity, this type of plant is placed on the roof ofthe buildings. In this plant, the electricity is produced by using sun-rays. This is used in houses, hotels, hospitals, traffic signallights etc.

1.4.7 Wind-Mill Power Plant

The Wind-Mill is rotated by heavy speed of wind. The electricity is produced by the generator which is operated bythe wind-mi!L This plant is placed atKayathar in Nellai District and placed at Palladam-Udurnalai Pettai Road in Coimbatore District

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1.5. ELECTRICALSAFETY AI"'D PRECAUTIONS

Aman who works in the electrical department must be carefully handled the work without any damage to the equipments and also workers. Because accdent may occur heavy loss. He must know al! ihe ope:t"atons of electrical equpments. Othe1wse wrongly handled the equipments wll cause heavy loss. Electrcal accident may occur only dueto carelessness. Dueto this, workers wll get injured, damaged equipments will cause loss, because the work was stopped. To avoid tbis, electrical workers must lbllow the rules and regulations when working.

1.5.1. Electrical Precautions

Before he use the equipments, he mustknow the operation ofthat equipments. Electrical connections are made properly according to the definition.

Only the trained and efficient person is allowed to operate, testing andreparing the machine.

A person works in the electric post and tower post must wear the safety belt and glouse.

lf the sitnation is occur, the man who works on the ladder, the other persons helps to capture the ladder for safety. Ifit is essential, then the post and the ladder must be ted wth a ro pe for safety purpose.

After earthngthe overhead lines by discharge rod, then the work \Vll continue.

Check the condition of al! the hand tools, supply wires operated in current and also to check the earth v

When the battery is charging in a room, the room must be in a condition to get free air. To avoid dangerous situation, no fire is available near the battery.

For producing electrolyte, water is not added to acid. Hydrochloric acid is added in the water by drop by drop.

Swetting hand is not used to switch ON or work on the electric supply. Ifthe person has swetting on the hand continuously, he must wear the glouse.

The switch is in OFF position. Befo reto tum ON the switch to check if anybody is working in that electrical circuit.

The abo ve points are used for electrical workers and they can be work without any damage.

1.5.2. Electrical Shock

Human body has aelectrical conducting property. Without swetting ofhurnan body the resisrance is approximately 80000Q(Ohms) and during swetting resistance ofthe human body is approximately 1 OOOO.(Ohms). Ifwe touch the current carrying condnctor, the curren! is conducted through our body to earth. So the electrical circuit is closcd and we get electric shock duc to this, nervous structure, herut, lungs and brain are affected. Ifthe curren! is heavy, death rnay occur. Therefore we rnust know, even though the current is essential,ifit is used wrongly, it \Vil! cause heavy loss. i.e. death and economical loss.

To prevent this electrical shock, we know about the methods of preventive cares and protectve methods for safety precautions.

1.5.3. P1eventive method to avoid electric shock

The Operation ofelectrcal equpmenls rnust be known.

Darnaged wire is not used for wirng works or electrical connection.

The Electrical Instruments used or connections (.e. sw'tch,plug,pushings) is not having any scratch or breakable, lf it is in such a way that it rnust be replaced by new one.

Requirement hand Tools are used in proper way

The hm1d tools are insulatcd essentially.

Proper earthing s provided.

Ifthe supply is taken frorn the socket, only the plug top is used. To avoid, the supply is taken by inserting the wire with stick in the socket.

Depending upon the load, rated arnpere fi.lse wre is used.

The electrical equipment is repared after the main switch is off.

For any reason do not operate by overcoming the safety rules.

The electlical shock may be avoided for following the abo ve methods in a proper way.

5

1.5.4. First A id

Dueto unavodable reason, aman affects from sudden accident occur or electric shock, he may be treated by first aid method to protect from death, before taken into hospitaL

When a person is affected by curren! shock, first the circuit should be disconnected. Ifthe main switch is nearer put off the switch or using any wooden stck we could disconnect the person from circuit. Then immediately send him to consulta doctor.

Ifthe affected person lose his consciousness, but breathing is normal then looser hs clothes and apply cold water on hs tace and keep him in open air.

Ifthe person does not breath then immediately arrange artificial method ofbreathing clean his mouth and keep it open.

TI1ere are three rnethods of artificial breathng.

HOLGER NELSON METHOD

In this rnethod the vietim should be kept in the bed facing the ground. F old his hands and keep it in the backside ofhis head, the helper sitting at his head should rnassage his back using both hands. Ths is done with in two seconds.

Mouth to mouth

lo ths method the helperpushes air by keeping his mouth on the victim's mouth.By closing his nose then the air filllungs. So the victim get~ artificial respiration.

Backwards

Breathin

PuULower JawForward

Lay Victim on his back and Lo osen Clothing around :'lleck

Mouth to Mouth Resuscitation Procedure- 1

6

Nos Trials

Seal Patient's Lips with yours and nflate Lungs

Blow into lungs (12 times every minute) avoid patient's exhaled air

Mouth to Mouth Resuscitation

Procedure - 2

MOUTHTO MOUTII METHOD

l. Put the victim on a hed-sheet.

2. If his tongue is folded conect it.

3. U sing both ha:nds catch !lis forhead and e hin.

4. For respiration place your mouth overthe mouth ofthe victim and send an to his body for respination.

Through N ose

In this method the helper se:nd air through victim's nosc. By closingllis mouth the air is blown in his nosetill the heart ofthe victim rises bythis waythe victim gets artificial respiration. For a child the air blo"l'llll is halfthe heart leve!, compared to adult.

7

Questions

Part-A Choose the Correct Answer

l. The smallestparticle of an element is known as,

a)Atom b) molecule e) Nucleous d) Electron. 2. TheAtom is composed of,

a) Eleetrons only b) Protons only e) Neutrons only d) Electrons, proton, Neutrons 3. In case ofElectric fire use,

a) Dry sand b) wet sand e) Corbon powder 4. The number ofElectrons in anAtom are equal to

a) equal to neutrons e) Equal to the atomic number of t:he substance

5. The Supply volltage u sed for domestic purpose is,

a)ll0-120V b) 120 -130 V c)220-230V 6. Switch always be installed on

a) neutral wire b) earth wire e) Phase wire 7. Wit:hout swetring ofhuman body oft:heresistance is approximately.

a) 80 KO b) 40 Kn e) lO KQ Part-B

Answer the following questions in one word

l. \Vhat are the main part eles in an Atom?

2. Whatisl\-ucleous?

3. Neutron have which charge?

4. Proton have which charge?

5. What charge does electrons have?

6. Should we t:hrow water incase ofthe electrc fire?

d) Water

b) Equal to protons d) None ofthis.

d) 400-440 V

d) non e of t:he abo ve

d) None of ths.

7. While a person is in contact wit:h elcctric shockshould be removed by pulling his ann?

8

Answer the following questions in briefly

l. What is calledAtom?

Part- C

2. What preventive precautions should be taken to avoid electric shock?

3. What is Electricity?

4. What is current?

5. \Vhat are the difterent method ofartifieial repiration?

6. \Vhat are the methods used for production ofEJectricity?

Part- D

Answer the following questions in one page leve!

l. Explain the structure of Atom?

2. Explain the methods of prevent electric shock~

3. Explain the different types ofFirstAid?

Part-E

Answer the following questions in two page leve!

l. Explain the power generating methods'?

2. Explain the Electrical safety and precautions?

9

2. MATERIALS USED FOR ELECTRICAL WORKS

INTRODUCTION

Generally the materials used for electrical works divided into three types. There are called Conductors, lnsulators and Semiconductors. The materials which conduct the current from one place to other place are called eonductors, the materals do not conduct curren! i.e. it resists the cmTent are called Insulator and the materals whichhave half ofthepropetties ofthese two are called sencunductors. i.e. It conducted only a very low value of current Forthis purpose conductors and insulators are widely used in Electrcal department where as serniconductors are used in Electronics departrnent.

In this chapter, we ha veto study aboutthe types and properties of conductors and insulators.

2.1. CONDUCTOR

\Vhat is called cunductor? The wire which carties (Conducts) cwTent from the supplypoint to the load is called conductors. The material is operatcd byusing the current is called load. Eg. Fan, Radio, !ron box, Mixie, Grnder, Bulb etc. Generally al! types of metals are used for conducting purpose, some metals pennit easilyto allow the curren! flow throughit. Thistype of metal is callcd "Good Conductors".

2.1.1 Properties of conductor

To conduct the current easily.

Would have low resistan ce.

Would have high tensile stress.

More flexibilty.

Itvvillnot affectcd by the corrosion dueto air ( or) not affected by rain, heat. Whena crurent is flowing through the conductor, it will get heated. Therefore it is not affected

byheat.

Easy to soldering.

Cost is low and is easily available to buy it.

2.1.2 Types ofConductor

Conductors are classified into three types depending u pon the conducting property with low resistancc there are so lid conductors, Ji quid conductors and gas conductors.

So lid Conductors

Silver, Copper, Brass,Aluminium, Tungsten, Nichrome, Zine !ron are called good conductors. There are converted into thln vvi.re and thinrod or strap for the purpose of conduction.

We have to study about the metal is used for conduction and where there s used.

10

2.1.3 Liquid Conductors

The conductors in the torm ofliquids are called as Liquid Conductors. Mercury, SulphuricAcid, Nitrate are sorne ofthe liquid conductors used in batteres. Mercury is used in high power vapour lamps and automatic circuit breakers.

2.1.4 Gas Conductors

Organ, Helium, Neon, Nitrogen are sorne ofthe gas conductors. They are used in gas discharge lan1ps athigh temperature.

2.2. INSULATORS

Insulator is non-conducting material. i.e., it resists electricity. lt has high resistance value, normally in Mega Ohms.

Properties

lt has high resistance and speciiic resistan ce.

High di-electric strength.

Good Mechanical strength.

Withstands high temperature.

May not get change in shape dueto temperature.

M ay not absorb water.

Can be made to any shape.

Can not get fire easily.

Classification oflnsulators

Generally Insulators are classified into three types:

l. Hard Insulators

Ex. : Back lite, Porcelain, Wooden Plank, Glass, Mica, Ebonite

2. Soft Insulators

Ex.: Rubber, Poly" Vnyl Chloride, Vamish coated papers, Micanite, Pressphan paper

3. Liquid lnsulators

Ex. : Mineral oil, Shellac, Vamish

2.2.1. Tools and their uses

For the bettem1ent of our electrical works a number of minor and majar tools are used. In this chapter we are going to Ieam such tools.

11

l. CLITING PLIER:

In electrical tools cutting plier is the most important too l. It is used to cut the cables and to tighten them. The handles of the plier is, wrapped by rubber evento be used in cwTent supply. lt is also used to orremove screws.

2. LONG NOSE PLIER

lt is used to fix and remove screws in narrow gaps. It is widely used while repairing radios and speakers.

3.IG~IFE

It is used to remove the insulation in electric cables. The handle of aknife is made up of wood or plastic. lts length is in four or five inches to keep easily in shirt pockets.

4. SCREW DRIVER

It is used to fix and tighten the screVv'S. The point of a screw-driver should be flat to be fixed in the gap in thehead of a screw. Itis available in different sizes from 4.5 inches to 12 inches. It is named according to its length. Its handle is made up ofwood or plastic. Wooden handles are better to be used for long period than plastic handled Screw Driver.

5. CONNECTING SCREW DRIVER

It is al so a type ofScrewDriver. Its handle is made of plastic. It is available in small sizes. It is used to fix and tighten screws injoints, and peles. It is of 4.5 or 5 in ches.

12

6. TESTER

It is the essential too! of a electrician. It is also like connecting screw Driver in size.

Its handle is also made up of slots. In its handle a visible pipe like part is fixed in ita neon bu lb is fixed with a screw metal and there s a clip in its head. All these parts are connected using a cable. It is used to check current supply in electric circuits. Ifthere is a current supply in the circuit the neon bulb glows.

7.POCKER

It has a sharp end. lt is used to malee holes to fix screws in electric boards.

8.JUMPER

Itis used tomake holes on walls. It is availablein8 SWG or 6SWG sizes. Its handle ismade up of iron. By hammering its handle, required boles are rnade by the sharp points.

7. TUBEJUMPER

lt is used to rnake boles on walls. But it is used to rnake holes between the walls to eonnect electric cables. One side ofthis jumper is like a saw. The hammer is used to make holes androtate the jumper dock wise to make hales casi! y and quickly.

8.WOODSAW

It is used to cut wooden boxes, sticks and round blocks forthe required size.

13

9.HACKSAW

lt is used to cut PVC or metal pipes and metal frames. Tbe frame ofhack saw is made up oflron and the handle is made up of wood. A clip is fixed in its other end to adjust the length.

~--

10. BALLPANE HAM:VIER

As its head is round shaped like a ball it is called so. Its head s made up ofiron and handle is made up of wood. It is uscd to fix needles and bend iron rods. It s available in different weights.

Handle

11. CLAWHAMMER

In tbis type ofhammer of end is flat, the other bent and there is a claw in the end. It is used to remo ve nails and hannnering tbe nails.

12.MALLET

It is fully made up of wood. It is mostly used for woodet works.

14

13. HAND DRILLING M ACHINE

It is used to make holes in wooden materials. In one end ofthis machine a chuck is available to fix required drilling bit. Fixing t i wood by keeping the handle tightly, holes are made by rotating the cli

14.FILES

It is used to correet the sze and smooth the upper parte metals. It is named according to the size and the rough surface fo smoothing other surface.

TRYSQUARE lt is used for measuring angles of90: (Right angle) Measure-ments in mile metre are marked in its

sea! e. lt is u sed to measure 90 right angle accurately.

~ ::.

i

-: 1:

~ :;-:: -

: -

15

WlRE GAUGE PLATE

It is of round shape. It is used to measure the vv'idth ofwires. Its unit is gauge. The \vire is put into the hole in the centre of the Gauge to measure its width. Wires are available in gauge of 8 SWG, 12 SWG; 18 SWG.

o o

2.3 TYPE OF SWITCH

S.P.T Switch: This is a mechanical divice uscd for opening or closing an electrical circuit. Single poi e switch is uscd for closing ( or) opcning one phase only most ofthe switches are turmbler type but, now a days flush type switches are used.

2.3.1 Intermediate switch: To control a light from more than two different places, the intennediate switch is used for exarnple a long hall, corriders and passage ways with many doors etc.

2.3.2. Knife switch: Knife switch is made ofCopper and is generally used in laboratories for switch boards. It has a long pece ofl copper strip hinged in one end and which can go into a copper socket at d1e od1er end. It has got an insulated handle and two terrninals. Below the main sorne times there is additional small strip held by means of springs. The small strips makes contact to perrnits any number of control points.

23.3 Main S\~itch: Main s\v:itch is the one which controla the electrical supply for whole house ( or) factory. These are also called as Iron ciad switches. There are different types, Two pole and Three polc in the Two pole switch, their ,,;be two fuse units, the neutro! one will have a link and the phase l. will ha ve the rated fi.Ise wire. Thcre is also an earth terminal. The !ron ciad switch has a metalic cover which can be screwedout for changing the blown out fuse only. Afterputting ofthe switch. From the main switch leads are taken to the distribution box.

16

2.4 Fuse unit:

Function : Afuse in meant for protectingthe circuit from damage if a short circuit developes sorne were in the wiring ( or) in the cmmected appliances, Like a switch it instantly breaks the circuit and the flow of curren! in the circuit is intem1pted at once. It does so automatically by melting offit self A fuse in made of a metallic wire (tin, lead and Zinc alloy) having a low melting point and so life al any instan! any excessive curren! passed through the circuit, ts heat melts ofthe fuse. When the fuse blows t is a clear ndicaton that somethng has gone wrong so me were in the system. Every electrical circuit mus! there-fore have a fuse ofthe correct rating as a protective device. Fuses are usally rated for 5 Amps, 1 O Amps and 15 Amps

Types of fuses: Kitkate porceilin fuse unit, HRC fuse, Cartridge fuse.

2.4.1 Cartridge fu se: This type offuse in mostlyused in T. V, Radio, Record pler, Voltage stabiliser, etc. They are in the shape of a capsule in which the fuse wire is stretched in a gas tube with metalic caps at each end, The blown of fuse wire can be seen stright away. This type of fuse is easy to replace by simplypressing it into its seat

2.4.2. Kit Kat type fuse : These are the ones mostly used in domestic nstallations. This fuse consists of a proclaim base having two fixed contacts, for connecting the incoming and outgoing cables. TI1e bottom part ofthe fuse is called the base and the top is called the fuse carrier. The line anda load wires are connected in the base terminals and the carrier is provided with a fuse. The base fixed but the canier is removable.

2.5 Wall socket: It has ready to give supply to the soldering iron, Table Fan, Radio, T V and other electrical applances. It has two pin, 3 pin and 5 pin socket for connecting plus. it is usua\Iy rated or 5 Amps and 15 Amps.

2.6 Ceiling rose : Ceiling fan ( or) Tubo lamps are get supply from this ceiling rose. It has two orThree brozz plates with connecting terminal screws.

Types of Ceiling rose : Two plate ceiling rose, 1bree plate ceiling rose.

17

Questions Part-A

I. Choose the Correct Answer

l. ............. is the best conductor.

a) gold b) Sil ver e) Coppcr 2. M!CAis better.

a) Conductor b) h1su!ator e) Semiconductor 3. Intermediate switch is used to control a lamp fiom

a) Oneplace e )individual controlling

b) more than two places d) None ofthese.

4. Ceiling rose is used to talce supply for

a) Portable equipment e) heaterof2000 watts

b) florocent tu be d) an electric ron.

5. \Vhch too! is used for pulling, tvvisting, cutting and vvrupping purpose?

d) Aluminium

d) None ofthese.

a) screw driver b) Insulated combination plier e) si de cutter d) gas plier 6) Vv1ch too! sused for hammeringthenail?

a) Plier b)Screwdriver c)mallet 7) The too! used for measuring the size of the conductor wire is a) Try square b)SWG

Part- B

11. Answer the following questions in one wotd

1) Whch Switch is used to control a bell point? 2) Ofwhichmeterial these ceilingrose are made? 3) What are the rating of single way switch? 4) How the switches are connected with load? 5) What is P. V. C. stands for? 6) Give example of safety accessories? 7) Where the two way switches are used generally"

18

e) Wooden scale

d)Hammer.

d) ~one ofthese.

Part-C

III. Answer tbe following questions in briefly

1) \\'hat is conductor? 2) What is insulator?

3} \\'hat are semi conductors?

4) Give examples for conductors?

5) Give examples or Insulators? 6) Give examples for Hand tools?

Part-D

IV. Answer tbe following questions in one page leve!

1) Explain the elassfication ofconductors?

2) Explain the propertes oflnsulators?

3) Write short notes on Tester, cutting pler?

Part-E

V. Answer the following questions in two page leve!

1) Explain the propertes of conductors and lnsulatrs?

2) Explain the types ofHand too!?

19

3. ELECTRICAL TERlVIS AND DCCIRCUITS

3.1. INTRODUCION

Generally says, the cmrent flows fonn positive ( +) tenninal to negative (-) tenninal. But electrons flow from negative tenninal to positive tenninaL The flow of electron is called current Related to this. we study about sorne of the electrical tem1s.

3.1.1. Electrical Current

The continous flow of free electrons constitues an electric current Theunit ofcurrent is amper (A) and is measured by Ammeter. It is denoted by the letter ''1".

Amphere

If one columb charge cross o ver the area of cross section ofthe conductor per one second then the value of curren! flows through the conductor is called 'One Ampere'.

OneCoulomb

2rc x lO 18 number of electrons is mentioned as one conlomb.

3.1.2. Voltage

To create the curren! flow in a conductor, i.e., the electrical pressure which is used to move !he electrons is called voltage. Itis denoted by the letter 'V'. The unitofvoltage is 'Volt' and is measured by voltmeter.

OneVolt

One volt means the force to m ove one coulomb of electrons in one second.

3.1.3. Resistance

The property of conductor which opposes the flow of cmrent through it is called resistance. lt is denoted bytheletter 'R'. The Lmit ofresistance is ohms (Q) and is measmed by Ohm meter. Ohm

When a conductor having 1 V potential between the two end points, one ampere current will flowing through the conductor and the resistance value of the conductor is 1 Ohm (Q). 3.1.4. Electro Motive Force (EMF)

In a circuit, a force is used to conduct the electrons from one point to another point is called Electro Motive Force. The unit ofEMF is vol t.

Electro Motive Force Potential difference + Voltage drop

i.e, (EMF = PD + Voltage drop)

20

3.1.5 Potential Difference

It is represented by, the potental difference between any two points in the electrical circut Shottly it is called PD and the Unit is Vol t.

3.1.6. Electric Power

Power is deiined as the product ofvoltage and current. Unit ofpower is watts. The ener;;y absorbed by an appliance in one hour is called the energy consummed by the appliance. It's unit is watt and denoted by the letter "P."

p V xiwatts

Electrc work Q P x t watt hour one kili o watt hour 1 Unit

3.1. 7. Law of Resistan ce

The resistan ce of a conductor in a circuit depends upon the following states.

It depends upon the material.

Directly proportional to the length ofthe conductor.

Inversely proportional to the area of the cross-secton of the conductor.

l t also depends upon the temperature of the conductor.

Resistance calculation

Resistance Specific resistance x length Arca of the cross-section

R= pi/a R- resistan ce- ohms

p- Specific resistance - Ohm meter

L- Length ofthe conductor- meter

E ()

e"' \

A- Arca of the cross-section of a conductor- Sq.m

21

3.1.8 Specific Resistance The resistan ce that is offered by one cubic cm material is called specific resistan ce.

The followingtable shows the specificresistance of materials

Materials

Gold

Silver

Copper

Alurnnium

Rubber

Glass

Examplel

Specific resistance is ohm - meter 2A2x10'

1.63 X JO' 1.724 X 108

2.83 X 10' 8 X 107

lO X 1011

1 Cm2 cross section, 50 m long copper conductor has specific resistance 1. 72 x 1 O' ohm-cm t1nd the resistance.

Solution

Copper conductor length .e. L Cross Section (a) Specific resistance

Resistance

Resistance R

Example2

R

=

L = 50m 50x 100 cm 1 cm2

1.72 x JO' Q cm pL a

1.72 X 10' X 50 X 100 1

= 0.00860hm

0.0086 Q

Arca of cross section of the Aluminium conductor is 0.009 sq.cm .Specific resstance s 2.69 x 1 o' ohm-meter. Potential difference between the end ponts ofAiuminium conductor (PD) is 20v. If2Acurrent is flowing through this, what is the length ofthe conductor?

Arca ofthe cross-section (a)

Specitlc Resistance (p) Potential differencc (V) Current(I) Resistance (R)

=0.009cm2

0.009 x 1 O" m 2

=2.69x 10-" ohm-meter

=20V

=2A

= V/I 20/2 = 1 O O

22

R=pL -

a

Therefore L = Ra p

=334.5m

3.2. CONDUCTANCE

= LQx 0.009x!O-" 2.69 X 10'8

Conductance is reciproca! of resistance whereas resistance of a conductormeasure the opposition which offers to the flow of current, hence the conductance measures the inducement, which offers to low of cun-ent. Its unit is Mho and denoted by the letter G.

Conductance G = 1/R Mho (U) Generally the materials are classified by its conduetance as they are

l. Conductor

2. Insulator

3. Semi conductor

l. Conductor

Conductor means the material, whichshould allow cutTent flow through it. i\11 matters are conductors. Sil ver, Copper andA!uminimn are ihvthe good conductors.

2. Insulator

Jnsulator means these substances whieh totallyresist the flow of eutTent through it. This type of substances are used in electrical appliances as Insulator.

Ex : Glass, mica, Asbestos, paper, wood, rubber, Porcelin, Plastic, dry cloth, backlite, PVC.

3. Semi Conductors

The material whose conductvity le in between conductor and Insulator is called semi conductor. Ex. Gerrnaniurn, Silcon.

3.3. TEMPERATURE CO-EFFICIENT OFRESISTANCE

The dflerence in Resistance while increasing temperature from 0 to 1 e is called temperature co-efficient of resistance.

A conduct the conductor resistance increases to R ohm

Then the difference in the resistance

t>R = R, ROOhm .R depends

23

l. directly on its ntial resistance

2. drectly on the 1ise in temperature

3. on tl1e nature ofthe material ofthe conductor

or R, - R, a R x t

where t s the rise in temperature

or R,- R., aR0t

where a ( alpha) s constant and knovm as the temperature coellicient of resistance ofconductor. from the abo ve equation

= R,-R,

R0t

HJ. t= I"cthen

R-R t

Hence the temperature coellicient of a material may be defined as the increase in resistan ce per C rise is temperanue

From ex. we find thatR, R, (1+ at) Example

Find the resistance of a copper conductor resistance at 25c where the conductor resistance at O"c is !50 Q and temperature coefficient is 0.0040 perC.

Solution:

Temperature coefficent copper= 0.0040 per C

At O'c, resstance = 150 Q

Therefore R, =Ro (1 + at) 150 (!.;. 0.004 X 25) 150(1+0.1) 150 (1.1)

= 165 Q

therefore resistance at 25'c in 165 Q

Effect oftemperaturc on resistance

The effect of rse in ternperalure is

i. The resistance in crease when temperature increases in metallike copper and iron, from this we ean understand tlmt pure metals have positive temperature co-effieient.

24

2. In alloys Iike magnesium and Eureka resstance increase is relatively small with increase in temperature.

3. In Electrolyie, Jnsulators, mica, glass and rubber resstance decreases ""ith increase is temperature. Hence they ha ve negative temperature- coe:fficient of resistancc.

3.4. OHM'S LAW

A relationshp was derived by the scientst Ohm, between the curren!, voltage and resistance of the crcuit. It says,

"Al a constan! temperamre, the current tlowing through the circuit is directly proportional to the voltage and inversely proportional to the resistance".

Curren! Volta_g

According to ohm 's law,

R =V /1 230/1 O= 23 Ohms.

3. Find out !he voltage ofthe circuit when 6 Acurrent is flo'I'

In a circuit, if there is no way to the flow of current dueto disconnection of wre or i fthe switch is in OFF state, then the circuit is sad to be open circlt.

3. Short circuit

1 1

/ Fig. 3.5. (e)

1 1

1

The wires contact each other when there are connected in supply, the short circuit will occurs e .two terminals ofthe supply is connected directly without the load the current f1ow ofthe circuit is int1nite because it has no resistance.

Leakage

\:V1len any wire in the electrical connection may contact the body of a material, current leakage will occurs. In this conducton, if we touch the electrical equipment we get shock

Classification of electric circuit

l. Series Circuit

2. Parallel circuit

3. Series Para!lel circuit

4. Mesh or Network circuit

3.6. SERIES CIRCUIT

E(or)

I

Fig. 3.6

When resistors are connected as in fig. so thatthe same current passes through all ofthem, they are said to be in series.

Here the resistors R, R,, R, are connected in series with each other.

(i.e.) R 1 is connected with R.,. R, is connected with R3 and R 3 is connected with Rl through a battery supply. The curren! flow is in same direction (i.e. one direction)

'I' ampere curren! flows in all three resistors

Each resistor has a voltage drop across itas given by Ohms law. Thus

The total drop in three resistors put together is

V

V I

WhereR

V +V +V ; 2 3

I (R, + R, + R,) ;--

V I L_

-~

R 1

__..!

-rlr-1 ____, r---~ll___j r VI V2 1 1 V3 V4

V

Fig. 3.6 (a)

\\'hen one or more batteries are conneeted in series with each other, the total poten ti al difference is !he sum of the individual ones.

In the above therearefour batteries (V, V2, V3 and V J conneeted in series with eachother. Total potential difference (V) is

V =

According to Ohm's law

I

Here V=

VI+ v, + VJ + v,

YandV=IR R

28

The voltage drop in each resistor is

v,

V 4

=

=

Where R1, R,, R, and R4 are the intemal resistance of each battery. ThereforeR =

Example

200, 400 and 60Q resistors are connected in series across a240 V supply, Find out the total resistance ofthe circuit and curren! that flows through the circuiL

R, =200 R2 =400 K=600 '

20Q 40n 60Q 1 Rl 1 1 R, 1 K 1 1 1 1 1 1 ' 1

240V

Fig.

Solution

R, = 20, R, 40 = 60

E 240V,

R ? I ? ,,

According to ohms law

1 = V R

WhereR R, + R, + R, R 20 + 40 7 60 = 120

R 120

I 240 2A 120

I = 2A 29

Example

Three resistors are connected in series. The total resistance (R) ofthe circuit is 60Q. The first two resistors are 25 Q and 15 Q find out the third one.

Supply

Solution

25Q 15Q 1 R1 lr-----11 R, 1 1.___ _ __1 1.___ _ __1

In a series total resistance (R) s R

R

R,

R,

R, 60

=

=

=

Rl + R2.,. R3

25 n,

15

?

25 + 15 + R3

25 + 15 + R3

60-40

20 Q

lmportant rules of a series ci1cuit

60!:2

1 . In the series circuit, the curren! flows in one direction

2. Total Resistance

R

3. In a series circuit, the same cu!Tent passes through al! its resistors.

1 1 1 1

4. The total drop across the series circuit is the sum ofvoltage drop across each resistor.

V

5. The total series circuit will be inactive (there is no current flow) there is a fault in any one ofits resistors.

6. This type of connection is nsed in serial sets (Decorativo !amps)

30

3.7. RESISTAN CE IN PARALLEL CIRCUIT

\Vhen resistors are connected across one another so that the same voltage is applied between the end points of each, then they are sad to be in parallel. The curren! in each resstor is dfferent and the curren\ I taken from the supply is divided among the resistors.

R, 1

:t.?-R2- 1

::t~ '1 R: 1 J

V Fig. 3.7.

In parallcl circuit total curren\(!) is equal to sorne ofthe currents ! 1, r, and !3

I = I + I + I ! 2 3 Accordng to Ohm's law, we can find thetotal resistance (R) as given below I

I,

buti

I

I

I V

=V R

=V R, V

~

=1-I+I 1 2 } +V+ V

R1 R2 ~

v(..!.. + l ~ R, R2

l._ ) R,

1 +1+ o I = R -R R 'oc V 1 ''"2 3

Where 1 R

31

1 R

R =

+ + 1 R, R,

R,~ + R,R, + R,R, R1~R3

R1R2R3

IMPORTANTRULES OFPARALLELCIRCUIT

l. In the parallel circuit current flows though two or more paths ata junction. That is, it gets divided. 2. r = r, + r, + r, .................. . 3. The voltage drop is same in all resistors

4. Ifthere are 3 restistors (R1, R2, R;) in thecircuit R ~R, + R1R3 + R1~ 5. Ifthere is a fault in one resistor the otherwto resistors will work. The curren! will be divided into

two parts and will flow through the two resistors.

Example

60. and 40. resistors are connected in parallel through 240v supply. Find out the total resistance and curren! flows in it. Solution

R =60. ~=40. V 240V

In parallel circuit

R R1~ 6x4= R1+R2 6 +4

R = 2.40.

Aecording to ohm's law

I = V = 240 = R 2.4

[ !OOAmp Example

1 '

1 1

t,.n-

R\ 1 1 4-..n--

R'2..l

240V

Fig.

!OOA

Three resistors 1 OQ, 50. and 20. are connected in para!! el.

32

The rotal current llowing in the circuit s 2A Fnd out the total resstance and supply voltage of the circuit. Solution

R, R, R, R

R

R

V

V

V

?

100 o

20

I=2A V '?

R,R3 + R,R3 + R,R, .10x5x2

(5x2) ~ (2xl O)+ (1 Ox5) lOO 100

- -

=

=

10 + 20 +50 80

1.250

IR

2 X 1.25 =2.5

2.5 V

2A

3.8. RESISTAN CE IN SERIES PARALLELCIRCUIT

100 R,

-:;.,_ 5Q .~

:;>.-; 2Q

i 1 R:;;

V Fig .

1.25 o

In this circuir one and more resistors connected in series wth one more resistors connected in parallel. It is a combination of series and parallel circuit.

1 R:aJ 1 R 1 r R J 1 R4 t 1 \ 1 1 2. 1 .,

V l Rsj. +

1

Fig.3.8

In the above series parallel circuit, there are five resistors (R,, R2, R,, R4, R5) placed in i among them R,, R, are connected in series and R,, R4, R5 are connected in parallel. The parallel resistors are connected in series wth R1 and R,.

33

Here the total resistance (R) ofthe circuit is

R = R,+R,+ R, X R4 X R,

Example

1 O Q and 8 Q resistors are connected in parallel with a 4 Q resistor is series. Find out the total resistance ofthe series parallel circuit.

4Q !

f

Fig.

Solution

Resistance ofthe parallel circuit R,R,

R,-,-R,

JO X 8 10+ 8

80 18

= 4.44 Q

Total resistance of the series parallel circuit

R

Example

=

4.44 +4

8.44 Q

1 1 10n

8Q

1 1

Three resistors 2. 4. and 6 ohm are connected in parallel. This parallel combinaton is connected in series with a resistor of 1.5 ohm. Find the curren! tlrrough in each resistor when the applied voltage is 10 V.

34

2Q 1 1 l 1

1.5 Q e B 1

4Q 1 A

1 1

6Q lO V 1 L

1 1

Fig.

Given Data

R, - 2Q

R, - 4Q

R, 6Q

R4 1.5 n

V 1 O Volts

To find ctm-ent throgh in each resistors

Solution :- Resistance betweenAB 1 l + l + l Rp R, R, R,

R,R, ~ R,_R3 + R1R3 R, x R,_x R1

Rp R, xR,_x R3 R,R,_ + R,_R1 + R1R3

2x4x6 -

(2 X 4) + ( 4 X 6) c.. ( 6 X 2) 48 48 1.o9 n

- -

8 + 24 + 12 44

Rp = L09Q

35

Total Resistance between AC

= Rp+R4

= 1.09 + 1.5

= 2.59 Q

Totalcurrent in the circuit =

Voltage drop across A.B =

V R

J.Q_ =3.86Amps 2.59

IxRp

3.89 X 1.09

4.24 Volt

(Voltage is constant in parallel circuit) i. Currentin 2 Q Resistor =

ii. Current n4 Q Resstor =

=

iii. Current in 6 Q Resistor =

3.9. KIRCHOFF'S LAW

V R,

2.12Amps

V AB R, 1.06Amps

V AB R,

=

0.706Amps

4.24

4

4.24

6

Kirchoff's Law s used to find out the current flow in the network circuts easily where ohm 's law s not applicable. It is applicable botb for D.C. andA. C. circuits.

Theyare

1. Cunent Iaw or Point law

2. Voltage law or Mesh law ( or) Tressure law ( or) Electro motive tora law. Currentlaw

The sum ofthe current flowing towards ajunction is equal to the sum oftbe currents t1owng away from t. This is called Kirchoff's Current Law.

36

Fig. 3.9 (a)

In the iig. abo ve, J a junction ( or node) fonned by five conductors. The curren! in these conductors are I" 12, 13, !4, and !5

Sorne ofthese currents are flo\\~ng towards J and others away from it.

According to Kirchoff's Law

(Flowing towards J) Otherwise

= I '-I 4 5 (Flowing away from J) =

I, + 12 + 13 -14 - 15 =O. This is known as KCL equation.

VoltageLaw

At any closed circuit the Potential Drop (IR) at each Resistance is equal to the total voltage given lo the circut

A 1 1

+ __ v

... -.

l D 1

R,

V=IR 1 1

V3=IR,

R,

Fig. 3.9 (b) 37

1 B J .

j ~ r'-

V,= IR, R,

'-r-,.

...

{ e

In a closed circuit, the sum of the potential drop is equal to the sum of the potential rises. This is called Kirchoff's Voltage Law.

V=

Example

In the circuit ofFig, find using Kirchoff's laVvo, the current in the varions elernents.

Solution

According to Kirchoff's first law mark the direction of curren! flow. According to Second law, Wrte down the KVL equation in the closed circuits.

A B e ' ::r:, 1 :T?. ~

--

CJov 251- 1 10 \( '-r-

__j ::I-,+1.?.-~if }' 6

Fig.

L ABEFAfonns a el o sed circuit

611 + 2 (1 1 + !2) 90

8!, + 21, 90

2. CBEDC fonns another closed circuit.

812 + 2 (!1 + I,) 110

2!, + 1012 110

81 + 2 [ ?. 1 90 (1)

21, 10 r, 11 o (2)

To sol ve, equation (2) is rnultiplied and subtmcted from ( 1) by 4. 81, + 21, = 90 (3)

(2) X 4 _ 811 40!2= -440 (2) -381 = -350 2 381, 350

38

I,

Substitute the I2 value in equation (1)

8!1 + 2 (9.211) 8I, + 18.422

81,

8!1

Il

r,

Fromtlris,

=

=

90

90

350 38

9.211Amps

90- 18.422

71.578

11578 8

8.947 A

8.947 Amps

Thecurrentthrough6Qresistorsis = 8.947 A lbe cun-ent through 8 Q resstor is 9.211 A The currentthrough 2 Qresstor is 8.947 9.211

= 18.158A_

Example2

By using Kirchoff's law, calculate tl1e current flo'A'ng through each resistor as shown in the

By applyng Kirchoff's second law,

In circut ABEFA, 511 + 20 (1 1 + I,) = 4 51, 201, + 20!, = 4

251, +201,=4(1) h1circuitEDCBE, 412 + 20 (I, + I,) = 6

41, + 20!, + 20!, = 6

2011 + 241, = 6

Eqn (1) x 4, 1 00!1 + 80!2 16 (3) Eqn (2) x 5, 100L + 12012 = 30 (4)

Eqn (3)- (4), -40!2 = -14 12 = 0.35 A

Tr-I _..., .. .-r,--11 mi J 1 4Q~~~'c-r,--.f 20Q ~ F 6V

i

J D

39

Substituto 12 0.35 in Eqn ( 1) 25 I, + 20 (0.35) 4 25L=-7-c4

'

I, = -0.12 A

The path of the current flow in the 5 n resistor s-ve, therefore we assume current directon is opposite to that as shov.'ll in tlg.

Currenti, in 5 n resistor = 0.12 A

Current 12 in 4 Q resistor 0.35 A

Curren! 11 + 12 in 20 Q resistor = O .23 A

3.10. CAPACITORS

INTRODUCTION

In this we are gong to study aboutan importantmaterial used in electrical circuits i.e., capacitors. Capacitor is an instrument to store electrical energy (Capacitl.mce) toa pa:tticular time and dscharge it whenneeded. It is also called as condenser.

The charge in the eapacitor is denoted by the word capacitance and is measured by the unit called Farad(F).

A capacitor can be manufactured by keeping a di-electric medium in between two electrodes. The di-electric medium can be air, wax coated paper, mica or oil etc.

3.11. CONSTRUCTION

++++

T Fig3.11

A Capaeitor can be manufaetured by keeping di-e!ectric medium in between tvvo electrodes. The capacitance ofthe capacitor differs depends upon the distance between the electrodes and the strength ofthe di-electric mediun1.

Working of a Capacitor The fig shows a capacitar is cormected across a battery. One electrode is connected to +ve

tetminal and other is cormected -ve tetminal ofthe battery.

40

1\ow, the supply is gven to the capacitor, the electrode connected in the +ve terminal gets positive charge ( +) and the electrode connected in the -ve gets negative charge(-). During this, capacitar gels chargng. i.e., The arnount of charge between the plates depends upon the delectrc material and al so the distance between the electrodes. After few seconds the curren! flow slops. Now the capactor voltage s equal to the supply voltage. [n this way the power can be stored in a capacitar. This stored power can be used again when needed.

3.12. POWER OFCAPACITOR

Power ofthe capacitor can be depends u pon the constructon e.,

directly proportional to the area ofthe electrodes.

Inversely proportional to the distan ce between the two electrodes.

Depends upon the di-eleetric strength ofthe insulating media.

3.14 CAPACITAc'IICE

The capacitance of a capacitar is defined as the ratio between the changng gven to the capacitar. Ths is denoted by the letter C and the unit is Farad.

Capacitance (C) Charge (Q) Voltage(V)

Therefore C Q_Farad V

Hence C Capactance- Parad

Q Charge given to the capacitar- Coulomb Potential difference bel\veen the plates- Volt

lfthe dielectric medhun between the two plates is stronger, then the capacitar can have high charging capacity.

Lower value of capacitance is called as micro firrad & Pi ceo farad.

3.14. ONE J

3.15.1 Fixed Capacitor

In this, three types of capacitors are mostly used. This type is ba~ed on the electrodes and dielectric material used between the two electrodes. These are used in Radio circuit. In this capacitance val u e cannot be changed. Let us study about these types of capacitors.

Paper Capacitor

Wax paper is rolled in the fonn of cylinder and dpped in wax solution in order to exhaust the air and placed in between two thin aluminium plates. This type of capacitar is used in dc-coupling cirCLts.

Mica Capacitor

In this instead of paper mica is used as the dielectric medium. Sil ver mica is coated on the swface ofthe rrca sheet and used as conductive electrodes. This type of capacitors are used in high frequency filters, coupling and tuning circuits.

Cera me Capacitor

These are the modem capacitors. In this Ceramic is used as dielectrc medium. The performance ofthis capacitar may not be affected even it get heated.

Electrolytic Capacitor

In this, aluminium Borate is used as the dielectric medium. When the paste is used in the fonn of wet it is tenned as Wet Electrolytic capacitar and it is in Dry state it is termed to be Dry Electrolytic capacitor. Otcourse wet electrolytic capacitor is not in use nowadays.

The dry electrclytic capacitar should be connected carefully. This is small in size but having high capacity.

3.15.2 Variable Capacitor

This type ofcapacitor is used in tuner circuit ofRadio receivcr: In this, air is used as dielectric medium between the two aluminium electrodes. The value ofthe capacitan ce can be changed wthin a particular limit.

Capacitance in Series

V

Fig3.15.2

Three capacitors are connected in series as show11 in fig.ln this connection ifthe total capacitan ce of the circuit is e then,

42

eapacitor in Series

V Q e

Thereforc V1 - Q e,

V = Q 2 e,

\13 Q e 2

V. + V2+V3 = Q Q + Q e, e, e,

V Q ~-l_,_ 1 e_ e,J 'e e, ' 1

V = 1 +_!_ 1 Wherc 1 - + -Q e, e, el Q e

1 + + 1 e e, e, el 1 CC +CC +Ce

..., :\ ; 7

e e, x e, X e,

e = c,c,e3 1 C2 C3 + e,e1 + e,c, e = c,c,c,

c,c3 + c,c3 + c,c2

3.15.3. Capacitors in Parallel

t V

t Fig3.15.3

43

Three capacitors (e 1, ez, e3) are cormected in parallel as shown in fig. In this circuit, fthe total capacitance ofthe circuit is e, then

Q ve

ThereforeQ - ve l

Q = ve 2 2

Q = ve J

Q +Q +Q ve +Ve +Ve 1 2 J 2 3

Q v (e +e +el 1 2 3

- e +C +e

V 1 2 J

e

Advantages

U sed in rectifier circuits to filter ac ripples.

U sed as suppressor capacitor in fluorescent lamp.

U sed in Ceiling fim, Radio and Television circuits.

Problem

\\lhere Q =e V

l. Find out the total capacitance of a crcuit when three capaetors 1 O mfd, 20 mfd and 30 mfd are connected in series and also in parallel?

Solution:

eapacitors in Series

l_=l.+l. l e e, e, e, l.= l.+ l.+ l.= "--'--"--'--"' e 10 20 30 60

ll e 60

e 60 = 5.45 MFD 11

eapacitors in Parallel

e= e!+ ez + e3 e= 10 + 20 + 30 = 60 MFD.

Fig.A

Fig. B

44

1 omta 20mta 30mfd

~---1 ---~1 1~--------~r

V

t I I l V

+

lOmtl:l 20mfd 30mfd

T T T

3.16. WORK, POWER, EFFICIENCY AND ENERGY

Work

If a force ofF moves a body through adistance S in its directon of application is cal!ed work. The unit ofwork is N ewton meter. If 1 Newton force displaces a body through a distance of 1 meter then the work done is 1 Nm (Newton-meter)

The potential difference applied across the coi! causes to flow through it. This implies that there is an electrical work done. Unit of work done in Joule. In Electric circuit f 1 volt electric poten tia! causes one coulomb ofelectric charge to pass through a circuit then the electric work done is equal to 1 Jo u! e.

1 Jouel = Volt x Coulomb

Coulomb Ampere x Time

J = Vx lxt

Power

It is the rate of doing work lts units is Watt (W) Power Work done Joules

Time Tnne

p = J

t

P = Vxlxt

t

P = VI watts

According to Ohrn's law V= IR p LR.L = 2 IR watt

2 P=VI (or) P =IR I =

p p

y R

VI

vv ....

R

P Power in Watt

V Voltagein Volt

v' _watt R

I = C'urrentinAmpere

45

R Resistan ce in Ohm.

1 kilo watt 1000 watts.

Mechanical power is mcasured in horsepower. The relatinship between mechanical power and the electric power is fmmd to be

1 Horsepower = 7 46 Watts

Watt meters are u sed to measure the power. Power is denoted by the letter "P".

Efficiency

Efficiency means the ratio between the input power to the output power. In all machineries the output pmver is les ser than input power.

Efficiency

Percentage of efficiency --

Output pov,-er

Inputpower

Outputx 100

Input

Energy

Energy means the amount of work done by aequipment during a time period of t seconds. Unit of energy is Joules.

Energy Power x time watt. sec

The energy spent for the appliance is l kilo watt hour. It is al so called as one unit.

1 unit 1000 watt hour.

Example: 1

The resistance of a lamp is 1 O Ohms, and curren! through it 2A, calculate the power.

Solution

Resistance (R) CmTent(l) Power

=

10

fR 2 X 2 X !0

40\V

So the power ofthe lamp is 40 watts

Example2

Calculate the energy mtit when a 500 watts lamp is ON for 6 hours

46

Solution

1000 watts x 1 Hour = 1 Unit

Energy consumed = 500 x 6 3000 Watt-hrs. - 3000 3 Unit

1000 3 Unit of energy is spent by using 500 watts lamp for 6 hours.

Example3

In a 100 V circuit the curren! is 4 A

Calculate

( 1) Resistance (2) Power (3) Ener,')' for 30 mn Solution

Cunent(I) Voltage(V) Time(t)

=

According to Ohm's Law

R

1) Resistance

2) Power(P) =

3) Energy (P) t

Energy =

Example4

4Ampere

100 V

30Mn.

V I

lQQ = 4

VI

lOOx 4watts

400W

25 n

30 min. 0.5 hrs

400 X 0.5 1000

0.2 Unit

watt-hrs

In a factory the following appliance are in operation

L 2 HP '\1otor 3 hours daily.

2. lOOWlamp 12hourdaily.

47

3. 1 000 w heater, 3 homs daily.

Calculate the cost for energy consumed. (1 Unit cost = Rs. 4.00) in a month consisting of30 days. Solution

l. The energy for 2 HP motor 3 homs daily

Energy 746x2x3 1000

4.476kWh 2. 100 W lamp 12 hours for daly

Energy 100 x 12

1000 1.2kWh

3. IOOOWHeater3hoursdaily

Energy = 1 000 x 3

1000

for 30 days total unit is

=(4.476+ 1.2+3)x30 = 8.676 X 30

=260.28kwh Total cost for 30 days

= 260.28 X 4.00

Rs. 1041.12

48

WORKED EXAMPLES Examplcl

Find the curren! in the 2 O resistor path CF.

Solution

Mar k the curren! in various Branches as shown fig. As thcre are two unknovvn quantities !1 and 1,

two equations ha veto be fonned by considering 1\vo el o sed circuits ofloops.

Br-~1~:2:(':":r--TC~--{:4:D:J~I~,--,D

LoopABCFA

30 - 21, - 2(11 + 12) or 411 + 212

LoopABCDEFA

30-211+412-35

21, 41,

-.;:-! 30V ~

-r-

A

=O

= 30

=O

5

1,+1,

2D

F Fig.

Multiply eq (1) by 2 and then add itto eq (2) we get

10!1 =55

r, =5.5A

Substituting the value ofl1 in eq (!) we get 22 + 21, = 30

12 4A

Curren! in 2 O resistor = 5.5 4 4 = 9 .S A

Example2

CD

-:.;;- 35V -~

E

Three resistors of 1 ohm, 2 ohm and 3 ohm are connected in series across a 12 V battery. Calculate thevoltage drop across each resistor and also determine 1he power dissipated in each resistor.

49

Given data:

R = IQ J

~=2Q

12V R3 = 3Q V= 12 volts ~--------~~1--------------~

Tofind:

a) Voltage drop across each resistor(V, V2, V3) b) Powerdissipated in each resistor (P, P2, P) Solution : In series circuit

R = R + R + R = 1 + 2 ~ 3 = 6 ohm t 1 2 3

V 12 I= = = 2A

FI o

a) Voltage drop across 1 W resistor (V 1) = 1 R1 = 2 x 1 2 volts Voltage drop across 2\V resistor (V2) = I ~ = 2 x 2 = 4 volts Voltage drop across 3\V resistor (V,)= I R3 = 2 x 3 = 6 volts

b) Powerdisspatedinthe 1Wresistor(P1) PR1 =(2)2 x 1 =4watts Power dissipated n the 2\V resistor (P,) P~ = (2)2 x 2 8 watts Powerdissipated in the 3\V resistor(P3) FR, = (2)2 x 3 12 watts

Example3

A lamp has a noted voltage 1 10V and hotresistance55 ohms. Find thevalue ofseries resistance required to operate the lamp from a 220V supply.

Currentthroughlamp =110 2A

55 Voltage across R

220-110= l!OV

Series resistance requred lo operare at 220V mans.

= 110 =55Q 2

50

R

- 10V ---*-- 11DV ----fo!

1>+-----,--- 220V -------+

Example4

Three resistors 4Q, 6Q and 8f:l are connected in parallel across 36 VDC supply lind the total resistauce and the cmTent through each resistauce.

1 1 1 1 =-

R, R, R2 R. '

1 1 1 = -+ +-

4 6 8

1 13 R, 24

24 :. R, = -- = 1.846Q

13

~= 36=9A R, 4

1 = V = 36 = 6A 2 R. 6

"

V 36 I~ = 4.5A R, 8

Example5

1 1 "

I, 60

I, 8D

361J

Fig.

Three resistor2, 4 and 12 ohms are connectedin parallel across a 12Vbattery. Find d1e cunent through each resistors and the battery. Also find the power dissipated in each resistor.

1

12 R..=4Q '

!

J3 R,=l20 .

12V

Fig.

51

Givendata: R 1 2Q

R= 12Q '

V= 12 volts

Tofind: a) Curren! through each resstor (!1, 12, !3) b) Cunent supplied by the battery (!)

Soluton:

e) Power dissipated in each resistor(P1, P2, P) In parallel circuit

1 1 1 +-+-

R R, R,

- =.! + 1 + 1 i.e .. _!_ = 0.833 R, 2 4 12 . R,

:. R, 1

= !.2ohm 0.833

12 Current supplied by the battery (I) = 1.2 = 1 Oamps

a) V 12 CuJTent through 2 ohm resstor (I) = - = --- = 6amps R, 2 V 12

Current through 4 ohm resistor (l) = - = = 3amps R, 4

V 12 Cunent through 12 ohm resistor (!,) = - = = lamps R. 12

'

b) CmTent supplied by the battery = Sum ofindvidual branch currents.

e) I=(I1+!2 + !) 6-3 + 1 = 10 amps Power dissipated in 2 ohm resistor (P 1)

V' (12)2 72watts -=-

Power dissipated in 4 ohm resistor (P ,) V2 (12)2 -=--

R, 4

V' Powerdissipated in 12 ohm resstor (P3) =-: = 2

'

52

36waits

12watls

Example6

Three resistances ofvalues 8Q, 12Q and 24Q are connected in series. Find the equivalen! resistan ce. Also find the equivalent resistan ce when they are connected in parallel.

Solution

Case (1) 12Q 24Q

+ V

Fig.

Total resistance R"' =R +R,+R. 1 '

=8 12+24=44Q

Case (2)

1

I, R,~ 12n !

I, R_,~24Q

V

1 1 1 1 Fig.

=-+-Rer R1 R2 R,

1 1 1 =-+-+

8 12 24 3+2+1 6

= ----- -24 24

R 24/6 = 4Q '"

53

Example7

A circuit consists of two resistor 3 Q and 6 Q are in parallel and the combnation is connected in series with 8 Q resistor. Calculate the equivalen! resistance and currem drawn from 100 volts supply

Solution

Parallel resistance Rp

Req=

.R1R2 R, +R2 3X6 - 2Q 3+6 R + K = 2 + 8 = 1 on

p '

r--C=:J--- B.Q H===:J~--

64------100V _____ ttS

Fig. V 100

Curren! dravm. from the supply = R = 10 = 1 O A (~fj 6

Current in 3 Q resistor = 1 Ox 6 + 3 6.67A

8D 2.\i --c==J- '--------'

! ot-----100V ___ _.~.!,

Curren! in 6Q resistor = 1 r--3 -=3.33A 6+3

Current in 8D resistor = 1 O A

Example8

Fig.

lOQ

100V ~------ ------~

A resistor R is co1mected in series with a parallel circuit comprises oftwo resistance of 12 o !un and 8 ohm. The total power dissipated in the circuit is 70 watts when the apptied voltage is 20 volts. Calculate the value ofR.

54

Given: Total power (P) 70 watts Find: Value ofR

R

R

----l

Rrsn !--

20 Volts

~----------20Vo~lt~s--------~

Solution v2

Rt= p

1

(20) 2 --= 5.714ohms

70

1 1 1 = -+--= R, R 1 R 2 12

-

1- = 0.0833 + 0.125

R,

1 :. = 0.2083

R:2

1 :. R = -- -ce 4.8ohms 12 02083

R=R -R r 12 = 5.714-4.8 = 0.914 ohrns

Example9

Fig.

1 +-

8

A cunent of 15 amp flows through two ammeters A and B joined in series. The voltage drop aeross 'A' is 0.15V and across 'B' is 0.3V. Findhow sorne cunent will divide betweenAand B. whcn they are eonnected in parallel.

55

Given:

A B

1=15A

V8 = 0.3V

Fig.

Find : When they are connected in parallel, sarne current will divide betweenA & B i.e., !A and lB. 0.15

Solution : RA ~ J

~

15 ~0.01 ohms

V 8 0.3 R 0 = -~ ce - = O . 02 Q ~ l 15

1 1 1 =

~-----.. -

-~----

R AB R A R

1 1 ~- + -- = lOO + 50 0.01 0.02

]50

R AB = -1~ = 6.6667 x!O 3 !50

V =IRAB

=15x6.6667xl03 0.1V

ExamplelO

fA V o .1 JO amps = -- = RA 0.01

JB V o. J Samps = = RH 0.02

A resistor of 1 O ohms s connected in series with two resistan ce each of 15 ohms mTanged in parallel. What resistance must be shunted across this eombination so that total current taken shall be 1.5 amps 20 V appled?

56

::r; R,=IO!l r---1 _--

:oMiA l ~

Find : The value ofR

Solution: 1 1 1

--+ R 23 R, R 3

1 1 =-+- = 0.667 + 0.0667

15 15

-

1- = 0.1334 R,,

R 23 = ---= 7.5ohms o .1334 R 23 = R 1 + R 23

= 10 + 7.5 = 17 .5ohms V 20 11 = -- = -- = 1.143 amps R 23 17.5

1 2 =1-1 1 =1.5-1.143 =0.357amps

R = ~ = 20 = 56 ohms 12 0.357

Examplell

1- Rrl5.(l.. -

p.

20Vofbl

r-

A 11 OV 60W lamp is connected in series with another lamp rated 11 OV, 1 OOW across 220V mains. Calculate the value of resistance to be shunted across the first lamp so that both lamp take their rated power.

Given: Lamp 1 (L1): P1 = 60 W; V1 = 110V Lamp 2 (L,): P2 = 100 W; V2 = 110V V=220volts

Find : The value of shunting resistance across the first lamp (R 1) "

57

i, l 60W L1 1, 00

lsh Rsh L..r--~R

110 V

(V,).

'

Solution:

I,

100 = 0.909 amps

110

P1 60 = - = 0.545 amps

v, 110

= 0.909 0.5454 = 0.3636 amps

110 = -- = 302.5 obrns

0.3636 R =

:.h 1 sh

Examplel2

1 .,

220 Volts V ( )

Fig.

1COW L,

00

110 V (V,)

Two resistors of l 000 ohm and 4000 ohm are connected in series across a constant voltage supply of250v. Fnd the p.d. across each. If a voltmeter of 12000 ohm resistance is connected across the larger resistors, what is the reading on the meter.

Given:

R,=4000l

250 Volts ( ' 1 1,

Find: i)a)V, b)V, ii) Voltmeter reading (V 2)

R,=1000l R,;::40UOO .---j G12000~ 1

250 Volts (ii)

fl'ig.

58

Solution: i) ~ R, + R,

= 1000 + 4000 = 5000 ohms

1 250 = = 0.05amps R, 5000 V

V,= I x R1 = 0.05 x 1000 =50 vo!ts

V,= 1 x R, = 0.05 x 4000 = 200 volts

ii) 1 1 1 ~=-+-R,v R, Rv

1 ~----j-~----

4000 12000

2.5XI0-4 + 8.333Xl o'

~1~ = 3.3333%10"' R,v

R , =- 1 = 3000ohms " 3.3333X1 o-4

Ry =R, + R,v

V2

= 1000+ 3000

4000ohms V 250

= 0.0625amps 4000 R,.

= 0.0625 x 3000 = 187.5 volts

Example13

Detennine the currents in different branches ofthe circuit sho\vn in Fig.

Find: Currents in the different branches ofthe circuit (11, I,, I, 1) Solution: C1osedloopABEFA

311 + 20(!1 + 12) 100 =O 3L + 201, + 20!, = 100

23!, + 20!, = l 00

59

C1osed loop CBEDC

4!2 20(1, + 1,) 110 =o 4!2 + 20I, + 2011 = 110

2012 + 2412 = 110

100V [L -------'~----~r~' 110V u20[l !

l\ 30 1, 8

Current through 20 Q resistor = I 1 - I, = 1.3 1 5 + 3.486 = 4. 801 A

CmTentthrough 3W resistor (AB) (I,) = 1.3 14A Currentthrough 4W resstor (CB) (12) = 3.486A

' Currentthrough 20W resistor (BE) (11 + 12) 4.801 A Examplel4

Using Kirchoff's laws, detemne the cmTents in the unbalanced bridge circuit shown in Fig.

A

B

D

2V Fig.

Find : Currents in the unbalanced bridge circuit

Solution: Closed loopABDA

Assume drection

r, + sr, - 4I, = o

1 4T +51 = 0 1 " 3

1

2V lQ Fig.

61

e

1Q

8

I, + 12 D

Closed loop BCDB Fig.

2(1 -I ) - 3 (I + I)- SI = O 1 3 2 l J

=O Closed loop ABCA Fig.

11 + 2(11 13) + (11 + 12) 2 11 + 21 1 - 213 + 11 + 12 411 + 12 213

[] -4 5] rl,l [o" 2 -3 -1 O I 2 = oJ 4 1 -2 _IJ 2 1 -4 5

B

e

D =O -"J .... .::...

=2

"' = 2 3 10 = 1(6+10)+4(-4+40)+5(2+12) 4 1 -2

= 16 + 144 + 70=230

o -4 5 Li., = o -3 -10 = O( 6 + 1 O) + 4 ( O + 20 ) + 5 (O + 6)

2 1 -2 ~80+30=110

1 o 5 Li., 2 o -10 = 1(0+20)+0(-4+40)+5( 4+0) = 20 +20 = 40

4 2 -2

1 -4 o

A

Ll.J = 2 ~ o = 1( 6+0)+ 4( 4+0)+0(2+ 12)=- 6+ 16=10 -.) 4 1 2

11 = llO = 0.4782 A Ll. 230

1, = L\. 1 = 40 = 0.174 A - \ 230

L\~ 1 o ! 3 = / = ,~ 0 = 0.43 A D. ~-'

62

8

e

Curren! through 1 Q resistor (AB) ! 1 0.4782A

Current through 2 Q resistor (BC) (11 - !3) 0.4782 -0.043

0.4352 A

Curren! through 3Q resistor (CD) (!2 + 13) 0.174 + 0.043

0.217A

Current through 4 Qresistor (DA)(!,) O.l74A

Curren! through the battery (I, + !2) = 0.4782 +o. 174

0.6522A

Example 15

1. Detenninethe cuiTents in different branches oflhe circuit as shown in fig.

!ocv:::- LJ r Fig.

To find: Find the currents in the different branches ofthe circuit (I, I,, 11 + 12)

Solution

Closed loop ABEFA

31, + 20 (11 "- I,)- 1 00= O

31, + 201, + 201, = 100

3I. + 20I, = 100

63

Closed loop CBEDC

412 + 20 (L + !2) --110= O 41, -;- 201, + 201, 110

= 110 201, + 241,

23 20 20 24

= i 100 1 ltl o i

. 23 201 t- = 1

20 24 ! = (23X 24) - (20 X 20) =552-400= 152

"" = 152 100

t-, i llO ~o' - l= (lOO X 24)- (110 X 20) 24

- 2400 2200 = 200

"" =200 1

23 100: t-2 20 ll O 1 = (23 X 11 O) - (20 X 1 00)

=~ I, ""

2530-200 530

= 530

200 = 1.315 Amps

!52

! 1 + I, = 1.315 + 3.486 = 4.801Amps

Current through 3n Resistor (1 1) = 1.315 Amps Current through 4Q Resistor (I,) = 3.486 Amps Currentthrough20n Resistor (11 + 1:)=4.801 Amps

64

Example16

Using Kirchoffs Jaw detemrine the currents in the tmbalanced bridge circuit shown in Fig.

~

'2-.!1--

A 5.-1

[ \ .r_

:2-V

Fig.

To Find : Detemrine the current in the unbalaneed bridge circuit.

Solution

Closed loop ABDA

r, + si,- 41, = o L - 4L + 5L =O---~ --1

l .:. '

Closed loop BCDB

2(11 - 1)- 3 (12 + I)- 51,= O 21,- 213 312 - 313 - 5I1 =O

2I.- 31, -IOL O---, ~ ,'!

Closed loop ABCA

1, + 2(1, - 1) + (1, + r,) - 2 o 11 + 21, - 21, + r, + r, 2

-----2

411 + I, - 213 = 2 --- 3

65

(.).,;,'i~)

J

1 -4 2 -3 4 1

1 4 Ll.= 2 -3

4 1

5 ~ 1 o ~-2

5 -10

2

J, o I, = o I,

2

-101 12-3 1 + 5.

-21 4 1 ' 1

-10 12 +4 -2 ' 4

= 1 (6+ 10)-'-4(-4+40)+5 (2+ 12) =16+144+70

=230

1 o -4 Ll., = 1 o ~- 3

5 ' -10 1 = o 1 - 3

' 1 1 101 :o -Jol o

+ 4 ' +5 2 : 12 -2 1 2

: 2 1 -2 =0(6+ 10)"74(0+20)+5(0 6)

=o + 80 + 30

1 o 5 2 o -10 4 2 ~2

1 -4 o t.., = 2 -3 o

4 2 1

' 3 = 1 1

= 110

-JO -2

1 (O+ 20)- O (-4 + 40) 5 (4 +O) 20 +O+ 20 =40

o 1 + 4 12 2 ' 4

o 1 1 2 + o:

2 i 4

= 1 (-6 +O)+ 4 (4 -e O)+ O ( 2 + 12) A, 110 =-6+16+0 10 1 = ... _ = ~ = 0.478Amps 1 A 230

40 ~ = 0.174 Amps 230

10 230

0.043Amps

66

3

Currentthrough 1 QResistor(AB) 1, =0.478Amps Cunent through 2 Q Resistor (BC) (I1 13) = 0.478 0.043 = 0.435 Amps Cunentthrough 3 Q Resistor(CD)(I2 + I,) = 0.174 + 0.043 = 0.217 Amps Current through 4 Q Resistor(DA) (I2) O. 1 74Amps Currentthrough the battery(I, + I2) 0.478 + 0.174

0.652Amps

67

l. Choose the Correct Answer

1) The Unit ofResstance is

a) Joule b) Ohm 2) The value ofResistance measured by,

a) Voltmeter b) Wattmeter

Questions Part-A

c)Ampere

c)Ohmmeter

3) The EME nduced in a conductors measured n,

a) Ohm b) Watts e) Volt

4) One Kilo ohm is equal to a) 103Q b) 10 60 e) 10 20

5) The unit of power is,

a) Farad b) Volt e) watts 6) ln series circut has,

a) two path b) Three path c)Onepath

7) According to ohm 's law R s equal to, a)V2/R b)FR c)V/I

Part-B

II. Answer the following questions in one word

l) Wbat is the unt of current? 2) \VhatistheuntofEMF?

3) \Vhat s the unit ofResistance?

4) Wbat s the unitofCapacitor?

5) Which letter used for specific Resistance?

6) Jfyou add more resistanee in series Vvill the current increase?

7) All the appliance connected in your home are in series ( or) in parallel?

8) How many laws are these in Krichoff's law?

68

d)Watts

d) None ofthese.

d) ampere.

d) !OQ

d)Hertz

d) none ofthese.

d) VI

Part- C 111. Answer the following questions in briefly 1) What is Voltage? 2) What is Resistance? 3) State the ohm's law. 4) Vihatisopencircuit? 5) How you can calculate the power consumed in circuit? 6) The supplyvoltage ofthe circuit240 volts and the resistance value s 600 calculate the cmTent

flowing throug the circuit~

7) Power ofheating element 1000 watts, the voltage applied is 240 volt, calculate the cu!Tent? 8) 20, 60, 80 resistors are connected in series calculate the total resistan ce?

Part-D

IV. Answer the following questions in one page leve! 1) Explain theKrichoff's Lawwithneatdiagram? 2) Write the condition of series circuit? 3) Write the condition of parellel Circuit? 4) 301 6nt 120 Resistors are connected in parallel supplyvoltage in 240 vol!, calculate the total

resistance and cU!TCnt?

Part-E V. Answer the following questions in two page leve! 1. Calculate the CU!Tent in each resistantusing Krichoff's law.

10 n 12 o

. 20

T J::

69

2. Calculate the cmrent flow through each resistance of abo ve wheat stone bridge circuit.

G

'-------tll-1 --~ :;;.v

3) The following electrical appliances are working in a faetory, a) 100 Walts, powerof40 lamps each working 8 hours inaday, b) J 500 watts heater working 6 hours per da y e) 85% efticency of 3 HP motor working 5 hours per da y. The charge ofone units 4.00 calculate the electric bll for 30 days?

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4. ELECTRO MAGNETISM

4.1. INTRODUCION

Magnetism plays an importan! role in electricity. Electrical appliances like Generator, Motor, Measuring instruments and Transformer are based on the electromagnetic principie and al so the importsnt components ofTelevison, Radio and Aeroplane are working on the same principie.

4.1.1. Magnetic Material

Magnetic materials are classified based on the property called penneability as

l. Da Magnetic Materials

2. Para Magnetic Naterials

3. Ferro Magnetic Materials

1. Dia Magnetic Materials

The materials whose permeability s belowtmity are called Da magnetic materials. They are repled bymagnet

Ex. Lead, gold, copper, glass, mercnry

2. Para Magnetic Materials

The materials v,ith penneability above unity are called paramagnetic materials. The orce of attraetion by a magnet towards these materials s low.

Ex.: Copper Sulphate, Oxygen, Platinum,Alurnnum.

3. Ferro Maguetic Materials

The materials vvith penneability thousands oftimes more than that of paramagnetic mateJials are called feuo magnetic materials. Thcy are very much attracted bythe magnet.

Ex. Iron, Cobalt, Nckel.

4.2. PERMANENT :\1AGNET

Pennanent magnet means, the magnetc materals whch will retan the magnetc property al all times pennanently. This type ofmagnets are manufactnred by alumnium, nckel, ron, cobalt steel (ALKICO).

To make a pennanent magnet acoil s wound o ver a magnetic material and DC supply is passed through the coil.

4.2.1. Electro Magnet

Insulated wire wound on a bobbin in many tnms and Jayers in which enrrent is flowing anda soft iron pece placed in the bobbn is called electromagnet.

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N

Supply

Fig. 4.2.1. This is used in a!l electrical machines, transfonners, electric bells. It is also used in a machine used

by doctors to pull out iron filing from eyes, etc. 4.3. MAGNETIC EF:FECT BY EI"ECTRIC CURRE1'

This m! e is used to know the direction of magnetic field when the current passes through the conductor.

According to this rule

Flemings righthand rule states that, ifwe spread out thumb tore finger a:nd middle tinger mutually at right angles to each others and the thurnb points to the directions of motio:n ofthe conductor and the forfinger points the direction of magnetio flux, then the middle finger gives the direction ofinduced EMF.

4.5. MAXWELLS CORK SCREW RULE

Fig. 4.5.

1bis rule al so used to know the direction of magnetic field when a current pass es through a conductor. According to this rule arighthanded screwis held with the axis ofthe conductor to advance in the direction ofcurrent when screwed. Then the rotating direction ofthe head ofthe screw indicates the direction of magnetic field. 4.6. PERMEABILITY

The perrneability of a magnetic material is defined as the ratio of flux created in that maerial to the flux created in air, provided thatmmf and dimensions ofthe magnetic circuitremain the same. Irs smbol is .t and

.t = B/.F!

where B is the flux density H is the magnetisiug force

Being a ratio it has no unit and it is expressed as a mere nurnber. The perrneability of air .t air = unity. The relativo perrneability .tr ofiron and steel ranges from 50 to 2000. The perrneability of a given material vares with its flux densty. 4.6.1. Magnetic Field

The space around a magnet in which the intluence of the magnet can be detected is called the magnetic field. 4.6.2. Magnetic Lines

Magnetic lines of force (flux) are assumed to be continous loops, the flux lines continuing on through the magne. They do not stop at the poi es.

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4.6.3. Magnetic Flux

The magnetic flux in a magnetic circuit is equal to the total number oflines existing on the cross-section ofthe magnetic core at rigbt angle to the direction of the flux. Its symbol is

4.6.6. Magneto -Motive Force

The amount of flux density set up in the core is dependent upon five factors the curren!, number of tums, material of the magnetic core, length of core and the cross-sectional arca of the core. More curren! and the more turns of wire we use, the greater will be the magnetising effect. We cal! this product ofthe turns and current the magneto motive force (mmf), similar lo the electro motive torce (emf).

MMF =NI ampere - tums

where mrnf- is the magneto motive force in ampere tums

N - is the number of turns wrapped on the core

I- is the current in the coi!, in amperes, A.

4.6.7. Magnetic Reluctance

In the magnetic circut there is something analogous to electrical resistan ce, and is called reluctance, (syrnbol S). Thetotal flux is inverselyproportional to thereluctanceand so ifwe denote mrnfby ampere tums. we can write

NI e _. where

6.

7.

8.

Currenti = EMF

Resistance

Resistivty

Curren! density

4.8.1. Residual Magnetism

Flux

Reluctivity

FltLX density

MMF

Reluctance

It is the magnetism which remains in a material when the effective rnagnetizng torce has been reduced to zero.

4.8.2. Magnetic Saturation

The limit beyond which the strength of amagnet cannot be increased is called magnetic saturation.

4.9. SOLENOID

Fig. 4.9.

Ahelically wound coi! that is made to produce a strongmagnetic feld is called a solenoid. The flttx lines in