Fundamental elements of-electrical-engineering circuit theory basic stardelta

IVC
FIRST YEAR
ElectricalWiringandServicingof
ElectricalAppliances/Electrical
EngineeringTechnician
ELEMENTS
OF
ELECTRICAL
ENGINEERING
STATEINSTITUTEOFVOCATIONALEDUCATION
DIRECTOR OF INTERMEDIATE EDUCATION
GOVT. OF ANDHRA PRADESH

Telugu Academy Publication : Vocational Course - fyctec
INTERMEDIATE
VOCATIONAL COURSE
FIRST YEAR
ELEMENTS OF ELECTRICAL
ENGINEERING
FOR THE COURSE OF
ELELCTRICAL WIRING AND SERVICING OF
ELECTRICAL APPLIANCES / ELECTRICAL
ENGINEERING TECHNICIAN
STATE INSTITUTE OF VOCATIONAL EDUCATION
DIRECTOR OF INTERMEDIATE EDUCATION
GOVT. OF ANDHRA PRADESH
2005

Intermediate Vocatioinal Course, 1st
Year
ELEMENTS OF ELECTRICAL
ENGINEERING
(For the Course of E.W. & S.E.A. / E.E.T.)
Author
Sri. Lt. K. PRASAD PEM&M
Junior Lecturer, EWSEA,
Govt. Jr. College, Falaknuma, Hyd - 53.
Editor
Sri. M. PAUL PRASAD B.E. (Ele),
Junior Lecturer, EWSEA,
Govt. Jr. College, Falaknuma, Hyd - 53.
Price : Rs. /-
PrintedinIndia
Laser Typeset Reformatted by SINDOOR GRAPHICS, Dilsukhnagar, Hyderabad-60.
Phone : 24047464, 9393009995

CONTENTS
S.No. Chapter Name Page No.s
1. Magnetism 1-20
2. ElectricalCurrent-Ohm’sLawKirchoff’sLaw 21-53
3. Units of work, power and energy 54-68
4. Effectsofelectriccurrent 69-103
5. Electromagneticinduction 104-115
6. FundamentalsofAlternatingcurrents 116-124

1Magnetism
1MAGNETISM
Magnets :
1.1 Introduction :
Magnet is the substance which attracts magnetic mterial such as iron,
nikel,coblatsteel,manganeseetc.
Themagneticpropertiesofmaterialswereknownfromancienttimes. A
mineraldiscoverdaround800B.C.inthetownofMagnesiawasfoundtohave
a wondrous property. It could attract pieces of iron towards it. This mineral is
calledMagnetiteaftertheplacewhereitwasdiscovered. Further,itwasfound
thatthinstripsofmagnetitiealwaysalignthemselvesinparticulardirectionwhen
suspended freely in air. For this property, it was given the name ‘ Leading
Stone’or‘Leadstone’. Later,itwasfoundthatmagnetiteismainlycomposed
of oxides of iron( Fe3
o4
). These are now known as magnets and the study of
their property is called MAGNETISM.
WilliamGilbertdidthefirstdetailedstudyofmagnetsandtheirproperties
in1600. Magnetsarenowwidelyusedforvarietyofpurposes. Magnetsform
an essential component of all generators used for the production of electricity,
transmission and utilization of electric power. They are also used in electric
motors that are an essential component of many machines and gadgets that
operate on electricity. Modern electronic gadgets, like television, radio, tape
recorder. electric door bells also make use of magnets. Working of may of
these devices also depends on the magnetic effect of electric current.
Herewearegoingtolearnsomefundamentaldefinationsandprinciples
aboutmagnets.
1.2 Magnetic Pole - Magnetic Axis - Pole strength
Magnetic Pole :
Whenabarmagnetisdippedintoironfillingsorirondust, therearetwo
regions where fillings mainly get attracted. These two regions are calledpole
of a magnet or the strongest part of the magnet near the ends are called poles.

2 ElementsofElectricalEngineering
Thepoleswillnotbeattheends. buttheyarenearertotheends. When
suspendedfreelyinair, theendpointingNorthiscalledNorthPoleandtheend
pointing South is called SouthPole.
Magnetic poles do not exist Separately. It means we can never separate
or isolate a north pole of amagnet from its South Pole. Magnetic poles always
existinoppositepairs.
Magnetic Axis
AnimaginarylinepassingthroughmagneticnorthandSouthpoleofabar
magnet is called Magnetic axis
Betweenthetwopolesthereisaregionshowingnoattraction. Thisregion
is called Magnetic equator. This is also called Neutral Line. Magnetic axis
andtheneutrallinewillbemutuallyat900
andtheneutrallinebisectsthemagnetic
axis.
Axis
N SX
Y
Magnetic Equator
Magnet showing Axis and Equator
N S N SPole
Magnet showing poles Bar magnet showing N & S
direction
Pole

3Magnetism
Pole Strength
Thepowerofthemagnettoattractorrepelliscalledpolestrengthofthe
magnet. Thegreaterthepolestrength,thehigherthepowerofthemagnet. The
pole strenght doesn’t depends on size of magnet. The magnet may be biggest
in size but may be less powerful, and vice versa. The pole strength is ex-
pressedintermsofunitpolesorwebers. Oneunitpoleemanatesoneweberof
flux
1.3 Properties of Magnet :
1) The magnet always attracts magnetic substances.( iron, steel, cobalt,
nickel etc)
2) The magnet has two poles and when it is freely suspended, it comes to
rest pointing North and South directions. This is called directive
propertyof a magnet.
3) Like poles repel and unlike poles attract each other.
4) If a magnet is broken into pieces, each piece becomes an independent
magnet
5) A magnet losses its properties when it is heated, hammered or dropped
fromheight.
6) Amagnetcanimpartitspropertiestoanymagneticmaterial. Thismeans
when a bar magnet is rubbed over an un magnetised piece of iron or
steel,itchangesintoatemporarymagnet.
7) Repulsionisthesuresttestofmagnetism.
8) Magnetic force can easily pass through non - magnetic substances.
1.4 Shapes of Magnets :
Magnets are made in different shapes according to use and application.
The common shapes are given below.

1.5 Uses of Magnets :
The magnets are widely used in many ways for example 1) To findout
N-S direction at any place on earth 2)To find out the direction at any point on
sea ( navigation) 3) To detect magnetic materials. 4) For electrical machines
5)InmeasuringInstrumentsetc.
1.6 Classification of Magnet :
Permanent Temporary
Magnets Magnets
Bar U- shaped Horse Shoe Compass Electromagnet
Magnet Magnet Magnet Magnet
Bar Magnet
‘U’ Shape
ArtificalMagnets
ClassificationofMagnets
NaturalMagnets
lead stones

5Magnetism
Natural Magnets:
The magnets found in nature Such as lead stone which can be used in
navigation is known as Natural Magnets. The natural magnet has a chemical
composition of Fe3
O4
.
Artificial Magnets :
The magnet prepared by Artificial methods or by man made methods is
knownasArtificialmagnet. Itisfurtherclassifiedasi)PermanentMagnetand
ii)TemporaryorElectromagnet.
Permanent Magnet :
The magnet which retains the magnetic properties for a long period
(indefinitely) is known as Permanent Magnetin many applications we need
permenet magnets. Most permanent magnets are made of ALNICO, an alloy
of Aluminium, Nickel and Cobalt. Permanent magnets of different shapes and
sizes are being made form ferrite. These being light, strong and permanent.
Most of the electrical measuring instruments, Such as ammeters, voltmeters
galvnometersetccontainapermanentmagnet.
1.7 Electromagnet
Magnetismandelectricitywereconsideredtobetwoseparatephenomena
foralongtime. Howeverin1820,theDanishphysicistHansChristianOerested
(1777-1851)madeanimportantdiscoverythatestablishedarelationbetween
electricityandmagnetism.
Manual method is used to prepare magnets of small strength only like
compassneedle. Topreparestrongmagnets,electricalmethodistobeused. If
a coil of insulated copper wire be wrapped round on a cylinder of card board
( forms a solenoid ) the bar to be magnetised inserted in the cylinder and a
strong electric current passes through the coil, the bar will be found to be
magnetised.orElectromagnetsaremadebywindingthecoilsofinsulatedcopper
wire over a softiron or steel pieces. The core becomes a powerful magnet as
longascurrentispassing.

If the piece is of steel, when the current is stopped and the bar be
removed,itbecomesapermanentmagnet. Ifthepieceisofsoftiron,itwillbe
astrongmagnet.
1.8 Applications of Electromagnets :
ElectromagnetsarewidelyusedinIndustriesandalsoinmanysituations
in dialy life. They are used in cranes to lift heavy loads of scrap iron and iron
sheets.
1) One of the most importent uses of electromangnets is in generators and
motors, Where they are used to create the intense magnetic fields which are
necessary for the conversion of machanical energy into electrical energy and
vice versa. The coils wound on the field poles are to create the magnetic field.
2) The fact that certain materials get struck, to magnets is used to make i)
magnetic door closers ex. in refrigerators in which a weak magnetic strip all
roundthedoorensuresthatthedoorremainsfirmlyshut2)Magneticlatchesor
catches, used in windows , cupboard doors 3) magnetic strickers 4) magnetic
clasps in handbags 5) magnetic pin, paper, clip holders and so on.
3) Electromagnetsareusedtoseparatemagneticsubstances,likeironnickel
and cobalt, from non - magnetic substance, like copper. Zinc, brass, Plastic
andpaper. Theyarealsousedtoremove‘foreignbodies’likeironfillingsfrom
the eyes of a patient. They are used in all electrical machines, transformers,
electricbells,telegraphs,telephones,speakers,audioandvideotaperecorders
and players, relays etc.
4. Data(incomputerharddisks,floppiesandtapes)andaudiovisualsignals
(videotapes)canbestoredbycoatingspecialsurfaceswithmagneticmaterial.
Inallthese, theparticlesofthemagneticcoatinggetallignedinaparticularway
by a magnetic field produced by a recording head, much the same way as
domainsgetallignedinthepresenceofmagneticfield. Thedifferentlyalligned
particlesthenrepresentdata,soundoraudiovisualsignals. Thesameprinciple
isusedtostoreinformationinthemagneticstripesfoundoncreditcards,ATM
cards, some air line and train tickets, telephone cards etc.

7Magnetism
1.9 Comparison of Magnetic Properties of soft iron & Steel :
Soft Iron Steel
1)Itcanbehighlymagnetised 1)Itcannotbemagnetisedveryhighly
2) It loses its magnetism as 2) It does not lose its magnetism as the
inducing magnet is removed inducing magnet is removed. The
i.e. its magnetism is temporary magnetism is permanent in nature.
3) It is used for making temporary 3) It is used for making permanent bar
magnets (electromagnets) magnet, horse shoe magnet etc.
1.10 Comparison Between Electromagnet and Parmanent Magnet :
Electromagnet Permanent magnet
1. Polarity can be changed easily. 1. Polarity can not be changed easily.
2. Strength can be varied . 2. Strength cannot be varied.
3. cost is more. 3. cost is less.
4. Suitable in case of motors and 4. not suitable for larger
generators of large size. size motor and generators.
5.Electricbells,signals,Indicators 5. not possible.
can be made.
6. Can be used in lifiting work, 6. Not possible.
holding the job.
7. Can not be used in Navigation. 7. Mostly used in Navigation as a
magnetic needle.
8. Can not be used in cycle dynamo 8. can be used in cycle dynamo and
and motor cycle magnetos small toys.

1.11 Rules to findout polarity :
Endrule:
1. Looking at the end of the bar, if the current in the coil is counter - clock
wise in direction that end will be a north pole. If it is clock wise it will be
South pole. This rule is used to findout the polarity of the electromagnet.
Hand Rule: (or) Helix Rule :
Holdthethumboftherighthandatrightanglestothefingers. Placethe
hand on the wires withthe palm facing the bar and the fingers pointing in the
direction of the current. The thumb will point the N - pole of the bar. This
ruleisusedtofindthepolarityofthepolesofanelectromagnet. Righthand
thumb rule, or right hand palm, rule is used for determining the direction of
magneticlinesofforcearoundstraightconductor.
Ampere Rule :
Imagineamanswimminginthecircuitinthedirectionofthecurrentwith
his face to the bar, his left hand points towards north pole of the bar. It is used
for finding the direction of lines of force around a wire carrying current. Am-
pererulecanalsobeusedforfindingdirectionofmagneticneedle.
CoilCarryingCurrents
Hand Rule

9Magnetism
1.12 Magnetic Field :
Theregionaroundamagnet,inwhichtheforceofattractionandrepulsion
canbedetectediscalledamagneticfield. Themagneticfieldisfilledwiththe
magneticlinesofforce.
Magneticlinesofforceisnothingbutthepathalongwhichironfillingswill
re - adjust in a magnetic field.
1.13 Magnetic Lines of force:
Itisaclosedcontiniouscurveinamagneticfiledalongwhichanisolated
North pole could travel is called lines of force. or the paths along which iron
fillingswillre-adjustinamagneticfieldwillbethemagneticlinesofforceor
flux or It is a closed curve starts from North pole and ends at south pole of a
magnet.
Aline of force has no real existence, it is only imaginary. The lines of
forcearealsocalledmagneticlinesofforceormagneticflux.Themagneticflux
is expresses in webers. One weber is equal to 108
lines of force.
Tocarrytheelectriccurrentwegenerallyusecopperoraluminiumcables
because the resistance of these metals is low compared with the reisstance of
other materials. Similarly, to carry a magnetic flux. We generally use iron or
soft steel material because the reluctance of these materials is low compared
withthereluctanceofothermaterials.
Ampere’s Swimming Rule

1.14 Flux Density :
The number of lines of magnetic flux per unit area, represented by letter B.
Flex density )(
)(
AArea
Flux
B
φ
= webers / metre 2
1.15 Properties of Magnetic Lines of Force :
Force:
1. Theyareclosedcontiniouscurves.
2. They travel from north to south outside the magnet and from south to
northinsidethemagnet.
3. Theycontractlaterally,thatistheybendalongthelengthofamagnet.
4. Theymutuallyrepeleachother.
5. Theyneverintersectwitheachother.
6. Theyareimaginaryandhavenorealexistence.
Typical field pattern of different Magnets :
1.16 Field Pattern of An Isolated N - Pole:
The flux emanates from N - pole and reaches an isolated south pole
atafarofplace. Thisisonlyimaginary.

11Magnetism
1.17 Field Pattern of Bar magnet :
The flux emanates from ‘N’ - pole and reaches ‘S’ - pole outside the
magnetandinsidefromsouthtonorthofthemagnetandtheydonotcrosseach
other. Eachlineformsacompleteloop. Themagneticfieldisalwaysthoughtof
afluxorcurrentofmagnetismwhichgoesarounditscircuit. Themagneticflux
going through a magnetic circuit is not as real as an electric current flowing
through an electric circuit, but can be treated in a similar way. The magnetic
lines can be thought of like a bundle of stretched rubber bands.
N
South and North poles showing field pattern
Bar Magnet showing how iron fillings arrange them-
selves in line along the lines of force.

1.18 Field pattern of ‘U’ Shaped Magnet :
Note that there is no magnetism at the bends.
1.19 Magnetic Field Strength :
Fieldintesityormagnetisingforce(H)atanypointistheforce
exerted over a unit N - Pole of 1 weber placed at that point.
Suppose ‘N’ is the pole and it is required to find the field strength at
point ‘p’ at a distance of dm from N. Let the pole strength of the pole be ‘m’
webers. Imagine a unit N pole placed at point ‘P’. By applying inverse square
law, the force ( replulsion here) between the two poles is give by.
Thefieldstrengthisavectorquantity.
)/(/
4
/
4
1
2
0
2
mATorwbN
d
m
H
orHF
wbN
d
m
F
πµ
µπ
=
=
=
dm
m/wb n/wb
N P
Fig shows how iron fillings allign themselve with the magnetic field of
horse shoe magnet

13Magnetism
1.20 Inverse Square Law :
( Law of Magnetic force - Attraction or Repulsion )
The force between two magnetic poles are :
1) Directly proportional to the product of the pole strengths.
2) Inversely proportional to the square of the distance between the two pole
( inverse square law)
3)Inverselyproportionaltotheabsolutepermeabilityofthesurrounding
medium
Let A and B are the two poles placed at a distance of dm. Let m1
and m2
be
the pole strengths of A and B poles in webers.
Force between two poles according to coulomb’s law of magnetism can be
expressed by
newtons
d
mm
kF
or
d
mm
F
2
21
2
21
=
=
Where K is a constant and depends on the medium. If medium is air or
vaccum the value of K is equal to
04
1
µπ and in any other medium with rµ
relativepermeabilityitsvalueequalto
rµµπ 04
1
. Therefore
dm
m1
m2
A B

mediumotherinNw
d
mm
F
mediumairinnewtons
d
mm
F
r
2
21
0
2
21
0
.
4
1
.
4
1
µµπ
µπ
=
=
Thephenomenonofmagnetismdependsuponacertainpropertyofthemedium
calleditspermeability.
1.21 Unit Magnetic pole or pole strength :
Let m1
= 1, m2
= 1 and d =1 m the force
04
1
µπ
=F Newtons. Hence
a unit magnetic pole may be defined as the pole which when placed in air at a
distance of one meter from a similar and equal pole strength repel with a force
of
04
1
µπ newtons - 63340 Nw. or One Unit magnetic pole may be
defined as the strength of the magnet that it exerts a force of a dyne when
placed at a distance of one cm from another unit pole in a medium of unit
permeability.
1.22 Some Importent Fundamentals:
Magnetic Reluctance :
It is the opposition offered by a magnetic path to the establishment of a
magneticflux,itisjustlikearesistanceinelectricalcircuit. ItsunitisAT/Wb.
or Reluctance is the resistance offered by a substance to the magnetic flux.
ThereluctanceofironandsteelislowApieceofironinamagneticfield
offerslessreluctancethantheairtothemagneticlines. Thelinesthusbecomes
denserintheiron, makingitamagnet. Thisexplainstheattractionofironand
steel by a magnet.
Residual Magnetism :
Whenamagnetismovedawayfromcertainmagneticmaterialslikesoft
iron,itsmagneticfieldnolongerinfluencesthedomains.(Themagnetisedregions
arecalleddomains.Actaullydomainsarefoundinsideallmagneticmaterials)

15Magnetism
insidethem. Theirdomains,therforerearrangethemselvesandpointindiffer-
ent directions once again. As a result, these materials lose their temporary
magnetisation. In some cases, however, some of the domains retain the
allignment. When this happens, the materials remains weakly magnetic even
after the magnet ( i.e. the magnetic field is removed) is removed. You will
noticethiswhenyouremoveanailorpinfromamagnet. Thenailorthepinwill
show weak magnetism and pick up small magnetic objects. The magnetism
left in a magnetic material after being temporarily under the influence
of a magnetic field is called Residual Magnetism.
Demagnetisation :
When we strike a steel needle with a magnet, we allign the domains
inside the needle and thus magnetise it. If the allignment of the domains is
distrubed, theneedlewillloseitsmagnetismoritwillbecome demagnetised.
A permanent magnet loses its magnetism (or demagnetised) if it is heated or
hammered.
Retentivity :
Thepowerofretainingmagnetismwhentheinducingforce(mainmagnet)
isremovediscalledretentivityandthemagnetismretainediscalled‘Permanent’
orresidualMagnetism’. Theretentivityofsteelisgreaterthanthatofsoftiron.
Thisretainingabilityisalsocalled‘Coercivity’. Undertheinfluenceofmain
magnet the soft iron pieces are more powerful than the steel pieces.
Susceptibility :
Thepowerofacquiringthemagnetisminthepresenceofinducingforce
(main magnet) is called the susceptibility. The Susceptibility of soft iron is
greater then that of steel.
Magnetic Leakage :
Itisthatpartofmagneticfluxwhichfollowsapathwhichisineffectivefor
the purpose desired, Just like leakage current in the electric circuits.

Permeability :
It is the ratio of magnetic flux produced , by a magnetic force of a
materialormedium,tothemagneticfluxwhichwouldbeproducedbythesame
magneticforceinaperfectvaccumoritistheratiobetweenfluxdensitytoflux
intensity. It is denoted by µ and µ = B / H. it is only a ratio and no units.
1.23 Problems :
1. Two unlike magnetic poles are placed in air at a distance of 20 cm from
each other and their pole strengths are 5 m wb and 3 mwb. Determine the
force of attraction between them.
m1
= 0.005 wb ; m2
= 0.003 wb ; d = 0.2 m
force
)104(
77.23
2.044
10003.0005.0
4
7
0
2
7
2
0
21
−
×=∴
=
××
××
==
πµ
ππµπ d
mm
F
2. Twopolesofwhichoneis6timesstrongerthantheotherexertsoneach
other a force of 8N when placed 100 cm a part in air. Find the pole strength ?
Let the strength of one pole be ‘m’ wb. and the strength of other
pole = 6m . wb
Distance = 100 cm = 1 mtr.
N S
20Cm
m1
= 5 mwb m2
= 3 mwb
Newtons

17Magnetism
wborwbmm
x
m
m
mm
d
mm
F
r
00458.0.586.4
105.2
106
30.1263
106
168
1448106
144
106
8
4
77
2
2
72
2
7
2
0
21
=∴
==
×
×
=
×××=×
××
××
==
π
ππ
ππµµπ
3. If two similar poles are separated by a distance of 1 cm repel each other
by a force of 10-5
Newtons. Find the pole strength of each of them.
( )
( ) wbm
m
d
mm
f
mcmdmmmNWf
8
7
522
22
72
5
2
0
21
22
21
5
105.3
10
10104
104
10
10;
4
101;;10
−
−−
−
−
−
−−
×=
××
=∴
×
×
==
=====
π
πµπ
4. A magnet in the shape of square cross sectional area has a pole strength
of 0.5 x 10 -3
and cross sectional area of 2 cm x 3 cm. calculate the strength
at a distance of 10 cm from pole in air.
pole strength = m = 0.5 x 10 -3
wb
distance = d = 10 cm = 0.1 m
Fieldstrength wbN
d
m
H
r
/
4 2
0 µµπ
=
( )
wbNwH /
1.044
10105.0
2
3 7
××
××
=
−
ππ
∴ H = 3166.28 Nw / wb

5. Two identical poles having pole strengths 1 x 10-3
wb repel each
other with a force of 6.33 Nw when placed in air. Determine the distance
betweenthem.
( )
cmd
cmd
dd
mm
F
dAir
NwFwbmm
r
r
162.3
10
133.616
10101101
144
10101101
33.6;
4
?;1
33.6;101
2
733
2
2
733
2
0
21
3
21
=
=
××
××××
=∴
×××
××××
==
==
=×==
−−
−−
−
π
ππµµπ
µ
6. A pole of strength 0.005 wb placed in a magnetic field experiences a
force of 2.37 Nw. Find the intensity of magnetic field
m = 0.005 wb ; F = 2.37 NW ; H = ?
Field strength H = F / M Farads / meters
WbNwH /474
005.0
37.2
==

19Magnetism
1.24 Assignment :
1. What is a Magnet ?
2. Definepole,Magneticaxisandpolestrength
3. Explainthepropertiesofmagnet.
4. How do magnets classfied ?
5. How do you prepare a magnet by electrical methods and explain its
applications.
6. Comparepermanentmagnetwithelectromagnet.
7. Define end rule, where do you apply this rule.
8. Definehandruleandexplainitsapplication.
9. DefineAmpererule
10. Describemagneticfield.
11. Whatismagneticcurrent(orflux)
12. Definefluxdensity.
13. Draw the magnetic field pattern of a bar magnet.
14. Draw the magnetic field pattern of a a horse show magnet.
15. Mentionvariousapplicationsofamagnet.
16. Whatarethelawsofmagnetism.
17. Whatis‘Fieldintensity’?
18. State the Inverse square law of magnetism.
19. State four properties of magnetic lines of force.
20. Definepolestrength.
21. Define Magnetic Resistance ( Not - for syllabus )
22. ExplainResidualMagnetism(NotforSyllabus)
23. What is ‘demangnetism’ ? ( Not for syllabus )
24. Definepermeability.(Notforsyllabus)
25. DefineRelentivity.(Notforsyllabus)

1.25 Solve the problems :
26. Two poles of strengths 2 wb and 3 wb are placed in air at a distance
of 50 cm. Calculate the magnitude of the force. ?
( F = 1.5 X 10 6
N )
27. Two poles one of which is 8 times as strong as the other with a force
of 9 Nw, when placed 110 cm a part in air. Find the strength of each
pole ? ( 4. 634 m. wb )
28. A pole of strength 1 m is placed in a magnetic field experiences a
force of 6.33N. Find the Intensity of the magnetic field ?

21Electric Current - OHM’s Law Kirchoff’s Law
2.0ELECTRICCURRENT-OHM’SLAW-KIRCHOFF’SLAW
2.1 Electric Current :
Allelectricitycomesfromthechargescarriedbytheelectronsandprotons
of the atom. Electricity is actually the effect of either the imbalance of these
charges on a body or the movement of charges.
Theflowofelectricchargesconstitutesanelectriccurrentorsimplycurrent.
The flow of charge is due to transfer of negatively charged particles called
electrons. The current in the metal wire, is due to the flow of electrons.
Study of electricity may be classified into two parts 1) The static
electricity,whichdealswiththephysicalphenomenanproducedbychargesat
rest, and 2)The current electricity, which describes the physical affects due
tochargesinmotionoritiselectricityinactionorelectriccurrents,thatcarryout
the tasks of electrical science in a broad sense.
The flow of electric current along a conductor resembles to the flow of
water. Weather we are considering the flow of water or electricity we always
havetodealwiththreethings.
a) Current(flowofelectricityusuallyalongaconductor)
b) Pressure ( that which causes the current to flow)
c) Resistance (that which opposes or regulates the flow of current)
Like wise, an inter connected system of water pipes corresponds to an inter
connectedsystemofelectricalconductorsandequipment,knownastheelectric
circuit.
2.2 Conductors , Semiconductors, Insulators :
Conductors :
Materials that conduct electricity are called good conductors or simply
conductors. good conductors offer very low resistance to electric current or
conductrosarethosematerialwhichreadilyallowstheflowofelectronsthrough
itwithleastresistance.

Conductors are widely used for wiring circuits in domestic and com-
mercialapplications,forwindingofmotorisedappliances, forGeneration, trans-
missionanddistributionequipments.Metalsaregoodconductorsofelectricity.
But not all metals conduct electricity equally well. Silver conducts electricity
better than any other metal, But being expensive copper and alumium are
widely used as conductors. Some non - metals like graphite are also conduc-
torsofelectricity.
Semiconductor:
Semiconductoristhatmaterialwhichbehavebothasconductorandalso
insulators at different temparature, which are generally used in the field of
electronicslikeT.V.sTaperecorders,mobilephones,invariousapplications.
Eg. Silicon,Germanium.
Insulator :
Materialswhichconductalmostnoelectricityarecalledbadconductors
orinsulators. InsulatorshavehighresistanceorInsulatoristhatmaterialwhich
donotallowtheflowofelectronsthroughthem. Thesearealsocalleddieelectric
material Eg. Mica, Paper,Wood , Dry air, Dry cloth, Procelein etc.
Insulatorsplayanimportantroleinelectriccircuitsandequipments. The
insulation on wires ensure that a low - resistance circuit is not created in case
the wires touch i.e. it prevents a short circuit. The insulation on wires also
protects us from electric shocks. That is why the tools used by electricians.
Such as line tester, screw drivers, and pliers have insulated handle.
2.3 Conventional Electric Current flow :
Conventional current is that which flows from +ve to - ve through the
external resistance when connected to a source of supply. Electron current
flowisthatwhichflowsfrom-veto+vethroughexternalresistance. theunitof
electriccurrentis‘Ampere’.
Flow of Electric Current

2.4 Idea of electric potential :
If we wish to cause a current of electricity to flow from one point to
another, we must rise the potential of the first point above that of the second.
then the pressure is setup proportional to the difference in potential. This
difference of potential tends to send a current from the higher potential to the
lower as in the case of water.
A battery or generator may be thought of as a pump which pumps the
electricityupfromthelowerleveltothehigher,andkeepsonesideofthelineat
a higher potential than the other, thus setting up pressure between these two
points. Accordingly,theelectricpowercompanyrunstwowirestoaconsumer
house and simply agrees to keep the difference in the potential between them.
Even the battery acts as a pump to keep left hand side continually at a higher
potentialthantherighthandside.
The difference of potential (Referring to the fig) between A and B is
spoken as the fall of potential from A to B, or as the drop in potential from A
to B. The same applies to B and C or any other two points.
The difference of potential between two points in an electric circuit is
called the drop in potential for that part of the circuit contained between those
two points and is the cause of any current flowing between these two points.
A B C
X
Resistance Lamp
A B
Fig. showing difference of potentials
Potential difference

withreferencetotheabovefigtheterminalA isattachedtothehigh-potential
side of the generator, since it is marked ( + ) , and B is attached to the low
potential side of the generator, Since it is marked ( -). The current therefore
flowsfromAthroughR,thenthroughlamptoB,andbacktolowpotentialside
ofthegenerator. Thegeneratormustcontinualyraiseelectricityuptothehigh
potentialsidetomakeupforwhatflowsawayfromthatsidetothelowpoten-
tialside.IfnoelectricityisallowedtoflowawayfromAtoB,thegeneratorhas
to raise no more electricity up to the potential of A. It merely has to keep up
the pressure.
The term potential is sometimes used instead of voltage to designate
electrical pressure. When two points in an electric circuit have a different
voltage ( pressure), the difference between these points is called the differ-
ence in potential or the voltage drop. In the same way, the pressure
between two points in a pipe carrying water is spoken of as the drop in pres-
sure or the difference in head.
2.5 Electrical Resistance :
It is defined as the property of a substance due to which it opposes the
flow of electrons through it. Its unit is ohm ( Ω )
Let us, try to understand Resistance. In the case of both water and
electricity there may be a great pressure and yet no current. If the path of
the water is blocked by a valve turned off, these will be no flow ( current).Yet
there may be a high presure. If the path of electricity is blocked by an open
switch,therewillbenocurrent(amperes),thoughthepressure(volts)maybe
high. Thereistherefore,somethinginadditiontothepressure,thatdetermines
theamountofthecurrent,bothofwaterandelectricity. Thisistheresistanceof
the pipes and valves in the case of water and the resistance of the wires and
various devices in the case of electricity. The greater then resistance, the less
the current under the same pressure.
We say a wire has one ohm resistance when a pressure of one volt
forces a currnt of one ampere through it.
1V
1A
R = 1Ω
Unit of resistance

2.6 Law of resistance :
TheResistance‘R’offeredbyaconductordependsuponthefallowingfactors
1) Itvariesdirectlyasitslength(L)
2) It varies inversely as the cross section (a) of the conductor.
3) It depends upon the nature of the material
4) It also depends upon the temparature of the material.
Let us sum up
a
l
R
δ
= Where ‘δ ’ (Rho) is a constant represents the
nature of the material and is known as specific resistance or resistivity of a
material.
2.7 Specific Resistance :
Specificresistanceofamaterialistheresistancebetweentheopposite
faces of 1 - cm cube of the material. Good conductors will have low vlaues of
‘δ ’ and vice versa. The reciprocal of resistivity is conductivity. The higher
the conductivity the better is the conductor.
Specific resistance is measured in ohm - cm or ohm - inch or micro -
ohm - cm
2.8 Specific Resistance of some metals :
Althoughcopper,onaccountofitslowresistivity,isthemetalmostwidely
used for electrical conductros, aluminium, and even galvanized iron are some
timesused. Thereistivityofaluminiumis Ω170 /mil-footat200
C,about1.6
times that of copper. But its low specific gravity more than counter balances
this, So that for equal lengths and weights aluminium wire has less resistance
than copper and for this reason is coming into more general use.

The resistivity of iron and steel is about 7 times that of copper. These
materials, therefore, can be used only where a conductor of large cross
sectioncanbeinstalled,asinthecaseofathirdrailorwhereaverylittlecurrent
is to be transmitted as in the case of a telegraph.
2.9 Effect of temparature on resistance temparature coefficient of
resistance :
Temparature Coeffecient :
Temparaturecoefficientofamaterialmaybedefinedastheincreasein
resistance per ohm of original resistance per degree contigrade rise in the
temparture. Metals increase in resistance when their temparature is raised and
decrease in resistance when cooled As the temparture is decreased resistance
also decreases
)0(resistance
1resistance
0
0
CatOriginal
tempinrisecforinincreas
( )
ot
t
R
RR
andtRRt 0
00 1
−
=+= αα
Where R0
= Conductor resistance at 00
c .
Rt = Conductor resistance at t0
c.
t = rise in temparature.
α 0
= Temparature co - efficient and resistance at 00
C
forallpuremetalsthisco-efficientisnearly,thesamelyingbetweenthevalues
0.003 and 0.006 and depends upon the intial temparature.
Coefficient of Resistance :
‘α ‘ is called positive, if the resistance increases with the temparature,
and it is negative if the resistance decreases with the temperature. For metals
andalloys α ispositiveandforcarbon,electrolytesandInsulatorsitisnegative.
Forexample:Theresistanceofcopperwireandeurekawireisdirectlypropor-
tional to the temparature i.e. as temparature increases resistance increases and
vice versa, this is called positive temparature co - efficient of resistance.

The internal resistance of a battery and carbon is inversely propor-
tionaltothetemparaturei.e.astemparatureincreasestheresistancedecreases.
This may be known as negative temparature co - efficient.
2.10 Ohm’s Law :
ThislawisnamedaftertheGermanMathematicianGeorgeSimonohm
who first enunciated it in 1827. Ohm’s law states that the ratio of the potential
difference (v) between any two points of a circuit to the current (I) flowing
throughitisconstantprovidedthetemparatureremainsconstant. Theconstant
isusuallydenotedbyresistance(R)ofthecircuit.
Hence
I
V
R =
Ampers
volts
ohms =
R
V
I = OR V = IR
ohms
volts
Amper = Volts = Amperes x ohms
ohm’slawcanbeappliedtoanelectriccircuitasawholeoritcanbeappliedto
any part of it. This is an importent law in electrical engineering. This law is
applicable for d.c. circuits only.
Basic Circuit components :
Resistor, inductor and capacitor are the three basic components of a
network. A resistor is an element that dissipates energy as heat when current
passes through it. An inductor stores energy by virtue of a current through it.
Acapacitorstoresenergybyvirtueofavoltageexistingacrossit. Thebehaviour
of an electric device may be approximated to any desired degree of accuracy
ofacircuitformedbyinterconnectionofthesebasiccircuitelements.
Resistor :
A Resistor is a device that provides resistance in an electric circuit.
Resistance is the property of circuit element which offers oppositin or hin-
drance to the flow of current and in the process electrical energy is converted
V= voltage between two points
I. = Current flowing ,
R = Resistance of the conductor.

2.11 Resistance in series :
Three bulbs glowing very dimResistances connected in series
in to heat energy. APhysical device whose principle electrical characterstic is
resistanceiscalledresistor.
Inductors:
The electrical element that stores energy in association with flow of
current is called inductor. The basic circuit model for the inductor is called
inductance. Practicalinductorsaremadeofmanyturnsofthinwirewoundona
magnetic core or an air core. A unique feature of the inductance is that its
presence in a circuit is felt only when there is a change in current.
Capacitors :
A capacitor is a device that can store energy in the form of a charge
separationwhenitissuitablypolarizedbyanelectricfieldbyapplyingvoltage
across it. In a simplest form a capacitor consists of two parallel conducting
plates separated by air or any insulating material such as mica. It has the
charactersticofstoringelectricalenergy(charge)whichcabefullyretrivedinan
electricfield. Asignificantfeatureofthecapacitoristhatitspresenceisfeltinan
electriccircuitwhenachangingpotentialdifferenceexistsacrossthecapacitor.
The presence of an insulating material between the conducting plates does not
allow the flow of d.c. current, thus a capacitor acts as an open circuit in the
presence of d.c. current.
The ability of the capacitor to store charge is measured in terms of
capacitence.

Let R1
R2
R3
be the resistances connected in series.
‘V’ be the applied voltage.
‘I’bethecurrentpassingthroughthecircuit.
In series circuit.
i) Current remains same in each branch of resistance and line.
i.e I = I1
= I2
= I3
ii) Appliedvoltageisthesumofthebranchvoltages.i.e.V=V1
+ V2
+V3
iii) Hence total resistance is the combined resistance of all
i.e. R = R1
+ R2
+ R3
Applications of series circuits :
Seriescircuitsarecommonlyseeninapplications,suchasstreetlamps,
and airport run way lamps, Another example of an every day occurence is
lightingattemplesandhousesduringfestivalanddecorationofchristmastrees.
Whenoneofthelampinthestringburnsout, allthebulbsdonotglowbecause
thecircuitisnolongercompleteforthecurrent flow. Thefusedbulbcausean
opencircuitforcurrentflow. Ifastringcontains10bulbsandifitisconnected
to a 200 volts source, 20 v will appear across each bulb. If one bulb burns out,
then200vwillappearacrosstheremaining9bulbs,and22.2voltswillappear
across each bulb. This increased voltage can burn out another bulb and so on.
Inthecaseofairportrunwaylampandstreetlamps,normallyconstantcurrent
variable voltage sources are used to avoid burning of bulbs and to maintain
continiousillumination. Whenoneofthebulbburnsout, adeviceatthelamp
automaticallyshortcircutsthedefectivelamp,thusallowingotherbulbstoglow
continiouslyThevariablevoltagesourcewillautomaticallyreducethevoltage
across the circuit reducing the current flow through the lamps ( normal rated
currentismaintained)thuspreventingfurthurburningoutoflamps.

2.12 Resistance in parallel :
Let R1
R2
R3
are the resistances connected in parallel.
‘V’ be the voltage applied across all resistances.
‘I’ be the current among all branches.
InParallelCircuit:
i) Voltage remains same in each branch i.e. V = V1
= V2
= V3
ii) Current is devided into separate branches i.e. I = I1
+ I2
+ I3
iii) TotalResistanceinallbranches
321321
321321
321
1111111
111
..
1111
RRRR
or
RRRV
I
or
RRR
VIor
R
V
R
V
R
V
I
ei
R
V
Ilawohmsperas
RRRR
++=++=






++=++=
=
++=
Allbulbs,fansetcwillbeconnectedinparalleltothesupplyvoltage.
I1
I2
I3
R2
R3
I I
R1
Resistances connnected
in parallel Three bulbs glowing
very bright

Applications of parallel circuits :
Parallelcircuitsarewidelyusedinthelightdistributioncircuitsinhomes
andfactories. Thesecircuitsaresuppliedfromconstantvoltage-variablecur-
rentsources. Parallelcircuitsarealsousedonshipsfortheirservicedistribution
systems,wheremanybranchcircuitsareconnectedinparallelacrossthebusbars.
In home and factory distribution circuits, all parallel circuits are connected to
themaincircuitandeachparallelcircuitwillhaveafuseinit. Inactualpractice,
almostalldistributionelectricalcircuitsareparallelcircuits.
2.13 Resistance in series parallel combination :
Inthistypeofconnectionbothseriesandparallelconnectionsareused.
Hence both the rules are applicable to this circuit. In this figure Resistance R2
andR3
are in parallel and R1
is in series to this parallel combination.
Practical applications of series parallel circuits :
Seriesparallelcircuitsarecommonfeaturesofmanyelectroniccircuits.
Theyareusedinavarietyofsituationswheredifferentvoltagesandcurrentsare
required.
Voltage dividers: In many electronic devices, like radio receivers and
transmitters, television sets, the circuit requires different voltages at different
points. Thesedifferentvoltageshavetobeobtainedfromasinglevoltagesource.
Themostcommonmethodofmeetingtheserequirementsisgivenbytheuseof
voltagedividers.
R1
R1
R2
R
3
Resistance of series parallel combination

2.14 Problems :
Amanganinwireofresistance1000ohms,lengthof200mhasacrosssectional
areaof0.1mm2
. Calculate its resistivity.
Given R = Ω1000 L = 200 m ; A = 0.1 mm2
=
2
6
10
1.0
m
Required : ?=δ
solution : =
m
L
Ra
or
a
l
R
−Ω×=
×
×
===
−8
6
1050
20010
1.01000
;
δ
δδ
δ
2)Theresistance-temparaturecoefficient ofphosphorbronzeis39.4x10-4
at
O0
c. Find the temparature co - efficient at 1000
c.
Given : α 0
= 39 . 4 x 10 -4
; t = 1000
c
Required : α 100
= ?
4
0
4
4
0
0
100
1026.28
100104.391
104.39
1
:
−
−
−
×=
××+
×
=
+
=
α
α
α
α
t
Solution
3)Determinetheresistanceof91.4mtsannealedcopperwire,havingacross-
section of 1.071 cm2
, resistance of copper having 1.724 micro - ohm - cm at
200
C ?
Given:L = 91.4 mts, a = 1.071 cm 2
: δ = 0.00001724 Cm−Ω
= 91400 cm
Required: R = ?

Solution:
Ω=
×
=
∂
=
147.0
071.1
91400000001724.01
R
a
R
4) Find the resistance at 200
of annealed copper wire of 1 mm2
cross section
and 100m long with a resistivity of 1.73 x 10 -6
cm.
Ω=∴
×
×
==
====
73.1
1010
1073.1
:
.20:Required
01.01;10100:
16
4
0
224
R
a
l
RSolution
CatR
cmmmacmmlGiven
δ
5)Findtheresistanceofacoilofcopperwireof150mlengthof3sqmmcross
section. The resistivity of copper is 1.724 x 10-8
Ω -m
ohmsR
R
a
RSolution
m
mmsqmmsqamlGive
86.0
10724.150
103
15010724.11
:
?Resistance:Required
10724.1
.103.3,150:
2
6
8
8
6
=∴
××=
×
××
==
=
−Ω×=
×===
−
−
−
−
−
δ
l
6) Calculate the length of copper wire of 1.25 mm dia has a resistance of 4 Ω,
If the specific resistance of the material is 1.73 x 10 -8
Ω -m
mtsl
Ra
Lor
a
l
RSolution
LGiven
54.283
1073.14
1025.114.34
?:
8
62
=
××
×××
=∴
=
=
−
−
l
δ

7) The field winding of a motor has a resistance of 45 ohms at 00
C. What is
its resistance at 500
C ? The temparature co - efficient of resistance for copper
is 0.00428 per 0
C at 00
C ?
( )
( )
Ω==
×=
×+==
+=
=
=Ω=
5698.55
244.145
5000428.014550Resistance
1:
?50:Required
00428.0.045:
0
00
0
0
Ror
R
Rcat
tRRknowWeSolution
CatR
CatRGiven
t
copper
α
α
8) A coil of wire has a resistance of 40 Ω at 250
C. What will be its resistance
at 550
C. The temparature co - efficient of the material is 0.0043 per 0
C at 00
c.
( )tRRknowWe 001 1 α+=
Let R25
is the resistance at 250
C = 40 Ω
R55
is the resistance at 550
C - ?
α 0
is the temparature coefficient at 00
C.
( )
( )
Ω=∴
×=×=∴
=
×+
×+
=
68.44
1075.1
2365.1
40
1075.1
2365.1
1075.1
2365.1
250043.01
550043.01
25
55
55
2555
0
0
R
RR
R
R
R
R

9) If the temparature co - efficient of copper at 200
C is 0.00393 , find its
resistance at 800
C. If the resistance of electromagnet at 200
C is 30 Ω
Given: R20
= 0.00393 and R20
= 30 Ω
Solution The relation between R20
, R80
and R20
is R80
= ( )( )20801 2020 −+RR
R80
= 30 ( 1 + 0.00393 x 60 )
R80
= 37.08 ohms
10) The current passing through a lamps is 0.5 amp and the supply voltage is
250volts. Calculatetheresistanceoffilamentlamp.
Given: I = 0.5 amps; V = 250 volts
Required: R = ?
Ω==∴
=
500
5.0
250
:
R
lawohmsperas
I
V
RSolution
11) A230 volts tester has a resistance of 23 Ω . What would be the minimum
ratingofthefuseintheelectriccircuitforusingthetester?
Given: V = 230 volts Required : I = ?
R = 23
Solution : As per ohms law amps
R
V
I 10
23
230
===
Ω

Note:Theratingofthefusemeansthemaximumcurrentthefuseallowstopass
throughit,beyondwhichthefusemelts,thusdisconnectingthecircuit.
12)An electric Iron takes a current of 2.2 amp form 220 volts supply. What is
itsresistance.
Given : I = 2.2 amps Required = R = ?
Solution: Ω=== 100
2.2
220
I
V
R
13) A battery of neglegable resistance is connected to a coil of 20 Ω resis-
tance. What must be the battery emf in order that a current of 1.5 amp may
flowthecircuit.
Given: R = 20 Ω I = 1.5 amp
Required: V = ?
Solution: V = IR = 20 x 1.5 = 30 volts
14) Thepotentialdifferencebetweentheterminalsofanincandescentlamp
is 220 volts and the current is 0.22amp. What is the resistance of the lamp?
Given : P.d = 220 volts ; I = 0.22 amp
Required : R = ?
Solution : Ω=== 1000
22.0
220
I
V
R
1.5amp
V
R=20Ω
I
V = 220 Volts
Circuit

15) Three resistances of 2, 10 and 20 Ω are connected in series. Find the
equavalentvalueofresistance.
R1
= Ω2 , R2
= Ω10 , R3
= Ω20
Rt = R1
+ R2
+ R3
= 2 + 10 + 20 = 32 ohms.
16) Three resistances of 2, 10 and Ω20 are connected in parallel. Find the
equavalent resistance.
R1
= 2 Ω , R2
= 10 Ω , R3
= 20 Ω
Ω=
==
++
++=++=
61.1
13
20
20
13
20
1210
20
1
10
1
2
11
;
1111
321
R
Ror
RRRRR
R2 R3
R1
Ω2 Ω10 Ω20
R2
= 10Ω
R3
= 20Ω
R1
= 2 Ω
Series Circuit
ParallelCircuit

17) Findthetotalresistanceofthefallowingcircuit.
Resistance between R2
and R3
. i.e. B1
and C1
=
3
4
6
8
42
42
==
+
×
Resistance between BB1
=
( )
( ) 7
15
66
3
4
66
3
4
4532
4532
=
++
×+
=
+





+
×





+
RRRR
RRRR
Resistance between AD i.e. R1
+ ( R betn B1
C1
)+ R betn D.
Ω==++= 5.117786
7
15
3
18)Calculatetheeffectiveresistanceofthefallowingcombinationoftheresis-
tance and the voltage drop across each resistance when P.D. of 60 volts is
applied between A and B.
R1
= 3
R2
= 2
A
Ω
Ω
R6
= 6Ω
R4
= 6Ω
R3
= 4Ω
B
B1
C
C1
R5
= 6 Ω
D
Series parallel combination
Ω
Ω
D C
60 V
5
6
3
Ω18
A P B
Ω
Series parallel combination

Resistance between D and C.
Ω=∴==
+
=+= 2
2
1
6
3
6
12
6
1
3
1
R
Resistance between D and P = 2 + 18 = 20 Ω
Resistance A and B
Ω===
+
++=+== 4
5
20
20
5
20
41
5
1
20
1111
21 RRR
ampscurrent
ampincurrent
amps
R
V
I
31215Remaining
12
5
60
resistance5
15
4
60
=−=
==Ω
===
i) Voltage drop across 5 amps resistance will be same = 60 volts (V = IR )
ii) Voltage across Ω18 resistance = 18 x 3 = 54 volts
iii) Voltage across parallel circuit DC = 60 - 54 = 6 volts.
2.15 Kirchhoff’s laws :
The current in various branches of large network can not be found out
by ordinary methods. By applications kirchoff’s laws, complicated networks
can be solved.
Gustav Robert Kirchoff( 1824 - 1887) a German physicist, published the first
systematicdescriptionofthelawsofcircuitanalysis. Theselawsareknownas
Kirchoff’s current law (KCL) and Kirchoff’s voltage law ( KVL). His contri-
butionformsthebasisofallcircuitanalysisproblems.

1)Currentlaw: or pointlaw ItStatesthat,inanynetworkofwirescarrying
currentsthealgebraicsumofthecurrentsmeetingatJunction(orpoint)iszero.
It is also called as point law.
I1
+ I4
= I2
+ I3
+ I5
or
I1
+ I4
- I2
- I3
- I5
= 0
ThiscanalsobedefinedasthetotalcurrentsflowingtowardsaJunctionisequal
tothattotalcurrentsflowingawayfromthejunction.
2) Mesh law or voltage law : In any closed electric circuit, the sum of poten-
tial drops ( I.R) is equal to the sum of the impressed e.m.f.s.
0332211 =−+ iRiRiR
FromthecircuitABCDE
ERIRI =+ 3242
Five conductors carrying current and meeting at a point
I1
I2
I3
I4
I5
R3
R1
I1
I3
R2
I2
R1
R2
R3
R4
I2
I1
I1
I2
A
C
DB3
A
Three resistors connected in delta wheat stone bridge

2.16 Explanation of Elements of D.C. Network :
CIRCUIT: A circuit is that which allows a current to pass through it. It
consists of a number of branches.
JUNCTION: Junction is that point where different paths of current meet,
O,A,B,C are junctons.
BRANCH : Branch is a part of a circuit or network.
AB is one branch. BC is another branch or network.
LOOP : Any closed circuit is called loop. OABC is a loop.
NETWORK:Theconnectionofparameter(R-L-C)indifferentwaysiscalled
anElectricalnetwork.
CURRENT DIRECTION: In comming currents will be towards the Junc-
tionpointandoutgoingcurrentswillbeawaythejunctionpoint.
ACTIVENETWORK:Thecircuitwhichconsistsofparameters(i.e.R,L,C)
with source of e.m.f is called an Active network.
PASSIVENETWORK:Anycircuitwhichconsistsonlyparameters(R,L,C)
and no.source of emf is called passsive network.
PARAMETERS: The Resistance, Inductance, capacitence are called the
parametersofthecircuit.
LINEARCIRCUIT: Alinearcircuitisthatinwhichthevaluesofitsparam-
eters are constant.
A C
BO
Network

Non - Linear Circuit : Non - Linear circuit is that in which the values of its
parameters change with the voltage or current.
2.17Application of Kirchoff’s laws to the wheat stone bridge :
Wheatstonebridgeisusedtomeasuretheunknownresistanceinagiven
network. Itconsistsoffourarms. Ifthecurrentinthegalvanometeriszeroitis
called a balanced. Circuit, then the products of the resistance of opposite
arms are equal. Suppose the value of current through the galvanometer is not
zero then the kirchoffs laws are applied to find the currents and values of un-
knownresistancethenitiscalledan‘unbalanced’bridgecircuit.
Problems : 19) AB = Ω3 , BC = Ω6 , CD = Ω12 and DA = Ω10
2 volt cell is connected between B and D and a galvanometer of resistance
Ω20 between A and C. Find the current through the galvanometer ?
R1
R2
R3
R4
I2
I1
I3
I2
+I1
-I3
B
D
CA
E
G
12
I-I2
I-I1
+I2
A
C
BD
2V
G
3Ω
6ΩΩ
I1
I-I1
10Ω
Wheat stone bridge
Wheat stone bridge
I1
-I3
I2

LetIbethecurrentpassingthroughthecell, andvariouscurrentsareshownin
the sketch.
Consider the closed circuit DACB. We get
- 10 I1
- 20 I2
+ 12 ( I - I1
) = 0
or 12 I - 22I1
- 20 I2
= 0 ---------- (1)
Consider the closed circuit ABCA
-3 ( I - I2
) + 6 ( I - I1
+ I2
) + 20 I2
= 0
6I - 9 I1
+ 29 I2
= 0 ------------ ( 2)
AgainconsiderclosedcircuitDABED.
-10 I1
- 3 ( I 1
- I2
) = 2
or -13I1
+ 3I2
= 2 --------------- ( 3)
Solving equation (1) and (2) we get
21
2
39
II
−
=
Andsubstitutingin(3)gives
mAI 78.0
13
1
2 ==
∴ The current passing through the galvnometer is 0.78 m. amp.
20) A Battery of emf 10 volts and internal resistance 0.5 Ω is connected in
parallel with another battery of 12 volts and internal resistanced 0.8 ohm. The
terminalsareconnectedbyanexternalresistanceof20 Ω . Find the current in
each battery and the external resistance.?

Let X and y be the currents flowing from sourcesAand B respectively.
Assumingtheclockwisedirectionofcurrentasnegative. InmeshABCFG
- 0.8 x - 20 ( x + y ) = 12 volts
or - 5.2 x - 5y = 3 ---------------------- (1)
InmeshBDEG
-0.8x + 0.5 y = 12 - 10
- 0.8x + 0.5y = 2 -------------------(2)
Solving (1) and (2) we get
x = -1.74 amp and y = 1. 216
Current in source = 1.74 amp.
Current in source = 1.216 amp
current in 20 Ω resistor = 0.524 amp.
AB C D
EFG
12V
0.8
10V
0.5Ω Ω
x y
Network

2.18 Star delta transformation :
Concept of transformation :
By the application of kirchoffs laws some problems cannot be solved
and finds great difficulty due to number of equations. Such problems can be
simplified by using star - delta or delta - star transformations.
TRANSFORMATION FROM STAR TO DELTA
The figure show two systems of connections of resistances. In star or
‘Y’connectionthereisacommonpointforallthethreeresistors, andindelta
or mesh connection the three are connected in series to form the loop and the
junctions are takenout to form three supply points. The common point in star
connectioniscalled‘Neutral’. Thedeltaconnectionwillhavenoneutralpoint
Assuming that the star connection is to be converted into delta connection. If
the two networks are to be identical the resistance between any pair of lines
will be the same when the third loop is opened.
Tranformationform(Yto∆ )
B
CA
CA
B
ACCBBA
CA
A
CB
CB
A
ACCBBA
BC
C
BA
BA
C
ACCBBA
AB
R
RR
RR
R
RRRRRR
R
R
RR
RR
R
RRRRRR
R
R
RR
RR
R
RRRRRR
R
++=
++
=
++=
++
=
++=
++
=
A
RA
RCRB
RAB
A
RCA
RBC
C
B
C
B
Star
[ Y ] Delta
[ ]
Three resistor connected in star Three resistor connected in delta

How to remember : The equivalent delta resistance between any two points
isgivenbythesumofstarresistancesbetweenthoseterminalsplustheproduct
of these two star resistances devided by the third resistor.
Transformation from ( ∆ toY)
CABCAB
BCCA
C
CABCAB
ABBC
B
CABCAB
CAAB
A
RRR
RR
Rand
RRR
RR
R
RRR
RR
R
++
=
++
=
++
= ;
How to remember : As seen from the above expression it should be remem-
bered that resistance of each line of the star is given by the product of resis-
tances of the two delta sides that meet at its end divided by the sum of three
deltaresistance.
2.19 Network theorems :
A Network consists of a Three resistor connected in starnumber of
branchesorcircuitelements,consideredasasingleunit. Anetworkisconsid-
eredaspassivenetworkifitcontainsnosourceofemf. Theequaivalentresis-
tanceofsuchanetworkcanbecalculatedbydividingthevoltageappliedtothe
network by the current flowing in it. When the network contains a source of
emfitiscalledanactivenetwork.
There are certain theorems that have been developed for solving the
networkproblems. Thesetheoremseithersimplyfythenetworkifselforrender
their solution very easy and can be applied to a.c. circuits also. When applied
to a.c. circuits, the ohmic resistance of d.c. circuits are replaced by Phasor
impedence.
Someimportanttheoremswhichareveryusefulinsolvingsomecom-
plicated circuits are discussed below.

2.20 Thevenin’s and norton’s theorems :
Theveninstheoremisstatedasfallows:
The current through a load resistor R connected across any two points
A and B of an active network( containing resistors and one or more source of
emf is obtained by dividing the potential difference between A and B, with R
disconnected by ( R + r) where ‘r’is the resistance of the network measured
between points A and B with R disconnected and sources of emf replaced by
theirinternalresistances”.
Consider a network shown in fig(a). A & B are the two points of the
network which consists of resistors having resistances of R2
and R3
and a volt-
age source of emf v and internal resistance R1
. The current through the load
resistance R connected across AB has to be calculated. Suppose the load
resistance R is disconnected as shown in Fig.(b). Then current through
There is no current through R2
, therefore the voltage drop across AB
( )
( )31
31
3
31
1
3
RR
RV
RacrosspVoltagedro
and
RR
V
R
+
=
+
=
( )31
311
RR
RV
V
+
==
Fig(a) Fig(b)

The network with load disconnected and the voltage source V replaced
by its internal resistance R1
is shonw in fig(c). Now, the resistance of the
network between A and B = r = R2
+ (R1
R3
/ R1
+ R3
). As per the definition
oftheveninstheorem, theactivenetworkenclosedbythedottedlineinfig(d).
Consisting of a source having an emf equal to v1
( the open circuit potential
differencebetweenAandB)andaninternalresistancergivenbythefallowing
relationship.
( )Rr
V
IRthroughcurrent
+
=
1
2.21 Norton’s Theorems : Norton’s theorem is Similar to Thevinins theo-
rem. While thevinin’s theorem is based on the idea of an equivalent source of
emf. norton’s theorem is based on the idea of an equivalent current source.
Nortons theorem can be stated as fallows.
Any arrangement of the sources of the emf and the resistances can be
replaced by an equivalent current source in parallel with a resistance r. The
currentfromthesourceistheshortcircuitcurrentintheorginalsystemandris
the equivalent resistance of the network between its two terminals A and B.
When all sources of emf are replaced by their internal resistances.
Fig Fig

Consider the network shown in Fig. Let v1
be the potential across AB
when load resistance R is disconnected, as shown in Fig(a).
321
3
3
2
2
1
1
1
1111
RRRr
and
R
V
R
V
R
v
r
v
IcircuitshortIn Sc
++=
++==
consider the load resistance R connected as shonw in Fig(c) then
( )Rr
RrI
Rr
I
V scsc
+
××
=






+
=
11
1
Now, the network shown in fig(a) can be replaced by a current source driving
a current I through the load R as shown in fig(d). Then we have.
Fig Fig
Fig(c) Fig(d)

The current through the load,
( )Rr
rI
I sc
+
×
=
2.22 Superposition theorem :
The superposition theorem is applied to simplify complicated networks when
two or more sources of emfs are present. The theorem is started as fallows.
In any network containing more than one source of emf, the resultant
current in any branch is the algebraic sum of the currents which should be
producedbyeachemfactingalone,allothersourcesofemfbeingreplacedby
their respective internal resistances ( or impedences in case of a.c. circuits)
2.23 Problems :
Transformation from ( ∆ toY) :
Transformthegivencircuitintostartypeconnection.
1Ω 2Ω
3Ω
Network containing
more than once
source of emf
Three resistors connected in delta

convertingABCtriangle ∆ toequivalentstar.
Ω=
++
×
=
Ω=
++
×
=Ω=
++
×
=
1
132
32
2
1
132
31
3
1
132
21
c
ba
R
RR
Transformationfrom )( ∆toY
22)
Convertingstarvaluesintoequivalent ∆
=1/3Ω
= 1/2Ω= 1Ω
Ra
=1Ω
Rb
= 2 ΩRc
= 3Ω
A
BC
A
BC
Rcb
Rbc
Rca

( ) ( ) ( )
( ) ( ) ( )
( ) ( ) ( ) Ω=
×+×+×
=
Ω=
×+×+×
=
Ω=
×+×+×
=
2
11
2
133221
11
1
133221
3
11
3
133221
ca
bc
ab
R
R
R
2.24 Assignment :
1) What is current ?
2) Defineconductor,semiconductorandInsulatorswithexamples.
3) What do you understand by electric potential.
4) DefineResistance?Explainthelawsofresistance.
5) What is specific resistance ?
6) Explaintemparatureco-efficientofresistance.
7) Defineohmslaw
8) How are the resistances connected. Explain series and parallel
combination of resistances w.r.t.V,I and R.
9) Definekirchoff’slaws.
10) What is a circuit ?
11) What is a Junction ?
12) What is a loop ?
13) MentiontheequationforthetransformationofDeltacircuittostar
circuit?
14) Mentiontheequationforthetranformationofstarcircuittodeltacircuit
15) The resistance of a conductor 1mm2
in cross - section and 10 m in
length is 0.173 Ω . Determine the specific resistance of the material.
16) The temparature co -efficient α of phospher bronze is 39.4 x 10-4 0
c,
Find the coefficient α for a temparature of (a) 200
C and (b) 1000
C.

17) Findthecurrentineachbranchofthenetworkshowninfigusingkirchoffs
laws.
18) What will be the current drawn by a lamp of 250 volts, 25 watts when
connected to 230 volts supply?
19) Findtheequivalentresistanceofthecircuitgivenabove. Findthe
circuitthrough Ω6 resistor.
20) Three resistors ΩΩΩ 3020,10 and are connected in star.
Determinethevalueofequivalentresistanceindeltaconnection.
I=5A
2.3
2
3
10
4
20
6
Ω
Ω
Ω
Ω
Ω
Ω

3.0 UNITS OF WORK, POWER AND ENERGY
Introduction:
EngineeringisaAppliedSciencewithaverylargenumberofphysical
quantitieslikedistance,time,speed,temparature,force,voltage,resistanceetc.
In order to cover the entire subject of Engineering six Fundamental quantities
i.e.mass,length,time,current,temparatureandluminiousintensityhavebeen
selected, which need to be assigned proper and standard units. In this chapter,
weshallfocusourattentionsthemechaical,electricalandthermalunitsofwork,
power and energy.
3.1. WORK :
Work is said to be done when force acting on a body causes the body
to change its state. Work is done (mechanical work) when a body changes its
stateofrestoruniformmotioninastraightline.Thisworkdoneisstoredinthe
body in the form of energy.
Work done = Force x distance moved
Work = Fs
Themechanicalunitofworkisjoule.Whenaforceofonenewton acts
for a distance of 1 metre, then work done is equal to 1 joule.
1 Joule = 1 Newton x 1 meter or
1 J = 1 Nw – M
The electrical unit of work is watt-see or kwh one watt-sec of work is
said to be done in a circuit with one ampere current for one second (coloumb)
and one volt P.D across the circuit.
Work done = VIT= VQ Watt-see
It maybe noted that work done or energy possessed in an electrical
circuit or mechanical system or thermal system is measured in the same same
unitsi.e.Joules.

55Units of work, Power and Energy
3.2 Power :
Instatingtheratingofelectricalapparatusitiscustomarytogivenotonly
the voltage at which it operates, but also the rate at which it produces or con-
sumes electrical energy.The rate of producing electrical energy is called the
power,and is measured in watts and kilowatts. Thus a lamp may be rated 100
watts at 230 volts. Rate of doing work is called power or power is the rate at
which energy is expanded or the rate at which work is performed. It is the work
done per second of time .
TakenTime
donework
Power =
.watts
t
W
Power =
Where ‘W’ is the total number of joules of work performed or total
joules of energy expanded in ‘t’ seconds.
R
V
RIVIPor
R
tV
t
RtI
t
VIt
t
W
P
2
2
22
===
====
The mechanical unit is H.P.(Horse power) . It is also termed as metric
H.P. the electrical unit is watt. One metric H.P. is equal to 735.5 watts.One watt
is also equal to one joule/second. The practical unit in electrical engineering is
K.W.(kilo-watt).
A young and energetic horse can do a work of 75 kg-M/see or 4500
kg-m/minute.
Thepowerrequiredtokeepacontinuouscurrentofelectricityflowingis
the product of the current in amperes by the pressure in volts. this gives the
power in watts.
Watts = amperes x volts.
TheBiggerunitiskilowalt
1kw = 1000 watts

To determine the power in an electric circuit :
If we wish to know the power that is being consumed in a certain part
of an electric circuit, we have to insert an ammeter to measure the current in
thatpartofthecircuitandmultiplytheammeterreadingbythevoltmeterread-
ing.Thisgivespowerdirectlyinwatts.
Watts = volts x amperes
Use of the watt meter :
Instead of using two separate instruments, an ammeter and a
voltmeter, to measure the power consumed in a certain part of a curcuit, we
mayuseasingleinstrumentcalledawattmeter. Thisinstrumentisacombina-
tion of an ammeter and a voltmeter. Ammeter has a low resistance to carry
current and a voltmeter has a high resistance to carry the voltage.
3.3 Energy :
Energyiscapacityfordoingwork.Energymayexistsinseveralforms
and may be changed from one form to another. For example a lead acid bat-
terychangeschemicalenergyintoelectricalenergyondischargeandviceversa
oncharge.Ageneratorchangesmechanicalenergyintoelectricalenergyetc.
Energy of a person or an engine or an electric motor is known to us
only when it does some work.
Electric Energy :
If an electric lamp of 100-watts, gives light continuously for , say
16 hours, the electrical energy consumed is 100 x 16 = 1600 watts hours.
The Joule or watt second is very small unit of electrical energy, so for
commercialpurposeenergyismeasuredinwatthours(wh)andkilowatthours
(kwh).The kilo watthour is called board of trade unit (B.O.T.)
1. B.O.T. unit = 1kwh = 1000 wh = 36,00,000 joules

3.4 Conversion of units : Thermal to electrical :
Thermal Energy :
Heatisaparticularlyimportantformofenergyinthestudyofelectricity,
not only because it effects the electrical properties of the materials but also
becauseitisliberatedwheneverelectriccurrentflows.Thisliberationofheat is
theconversionofelectricalenergytoheatenergy.
Thethermalenergywasorginallyassingnedtheunit‘calorie’.Onecalorie
is the amount of heat required to raise the temparature of one gram of water
through 1o
C. If ‘S’is the specific heat of a body, then amount of heat required
to raise the temparature of m gm of body through Oo
C is given by
Heat gained = (msθ )calories
It has been found experimentally that 1 calories = 4.186 joules so that heat
energy in calories can be expressed in joules. Infact, the thermal unit calorie is
obsoleteandunitjouleisprefferedthesedays.Forheatingequipments,theterm
thermalefficiencyisused.Itistheratioofusefulheattothetotalheatproduced.
heatTotal
lossesHeatTotal
HeatTotal
heatUseful
Thermal
−
==η
Total heat produced electrically = CalK
VIt
.
4200
1 Joule = 4.2 calories
JoulesJoulesCalories ×=×=∴ 42.0
2.4
1
Calorie is the bigger unit, compared to joule. Heat produced in calories =
H = mst
m = mass of the substance in grams.
S = specific heat of the substance

S = 1 for water and
t = change in temperature i e . (t2
– t 1
) in o
c.
H = 0.24 VIt Calories
= 0.24 I2
Rt Calories
= 0.24 t
R
V 2
calories
3.5 Problems :
1)A force of 10 Newtons is pulling a weight, through a distance of 4 meters
calculate the work done.
Work done = force x distance
= 10 Nw X 4mts
= 40 Nw-m or
joule.
2) If a table on the floor is pulled with a force of 4 Nw through a distance of
3mtsin6seconds,calculatetherateatwhichworkdoneorcalculatethepower
of the person who pulled the table.
SecmwN
TakenTime
doneWork
Power /2
6
34
−=
×
==
Note :- (1) Newton-meter=joule
(2) Watts
Seconds
Joules
mNw
Seconds
MetresNewton
==−=
×
sec/

3) If an engine does a work of 150kg-m in 4 seconds. Calculate the H.P. of the
engine.
H.P. of the engine =
75
sec/mkgindonework −
..5.0
2
1
75
4/150
pHHp ===
4500minute ×
−
=
inTime
mkgindonework
Newtons = kgs x 9.8
4) A 220 volts lamp takes a current of 0.3 amps and gives light for 100hours.
Calculate its power in watts and in H.P.
Watts = volts x amperer
= 220 x 0.3 = 66 watts
..0897.0
5.735
66
5.735
.. PH
Watts
PH ===
5)A220voltselectriclamptakesacurrentof0.3ampsandgiveslightcontiniously
for 16 hours. Calculate the energy consumed in (a) Joules (b) watt hours (c)
Kilo-watthoursorB.O.T.units orSimplyunits
(a) Energy=VItJoules
Joules = Volts x amperes x Seconds
= 220 x 0.3 x (16 x 60 x 60)= 3801600 Joules
Joules = watts x seconds
(b) Watt-hour = watts x hours
= (220 x 0.3) x 16 = 1056 wh
(c) kilo-watt-hours=

kwh
Joules
Jouleswh
kwh
056.1
1036
3801600
1036
606010001000
55
=
×
=
×
=
××
==
The fundamental unit of electrical energy is joule, but, as it is very small, the
kwh or Board of trade unit (B.O.T.unit) orunit and has become the practical
unit.
3.6 Billing for the electrical energy consumption or the calculations for
the electrical energy consumed :
General procedure
An electric heater takes a current of 1.5 amps at 240 volts and works
for3hoursperday.Calculatethemonthlyelectricbillattherateof75paiseper
unit.Add Rs 2/- as monthly sent for the energy meter.
Kilo-watt-hours = kwh = units = perday
1000
35.1240 ××
= 1.08 units/day
units/month = 1.08 x 30 days = 32.4 units /month
cost of the electrical energy consumed in a month =
paiseRsRs −=
×
/30.24.
100
754.32
.
meter sent = Rs 2/- p.m
Total monthly electrical bill = Rs 24.30 + Rs 2/- = Rs 26.30/-Ps

7). In a house there are 4 lamps of 60 watts each working for 6 hour/day and 2
tubelightsworkingfor8hours/dayand4Fansofcapacity60wattseachworking
for 14 hours/day, and two eletric irons of ½ kw capacity each used 1½ hours/
day. The house is closed every Sunday, calculate the monthly electrical bill
based on a tariff of 75 paise per B.O.T. unit.You may take 4 Sundays/month
and add Rs 2/- Per monthly rent for the energy meter.
Energyconsumedbylamps 4 x 60 x 6 = 1440 Wh/day
EnergyconsumedbyTubeLights 2 x40 x 8 = 640 wh / day
EnergyconsumedbyFans 4 x 60 x 14 = 3360 wh / day
EnergyconsumedbyElect.Irons 2 x 500 x 3
/2
= 1500 wh / day
6940wh/ day
Workingdaysinamonths = 30 – 4 Sundays
= 26 days
monthunitsmonthKwh /44.180
1000
266940
/ =
×
=∴
Cost of electrical energy consumed = 180.44 x
100
75
Rs. 135.33 Ps
Meter Rent = Rs 2/- per month
..
. Monthly Elect. Bill = Rs 135.33 + Rs 2/- = 137.33/-
8) An Electrical installation consists of 10 light points of 60 watts each, 12
lamps of 40 watts each ,6 fans of 60 watts capacity each, and a pumpmoter of
½ H.P.Assuming that 50% of lights and fans are used for 6 hours per day and
the water pump works for 4 hours dialy, calculate the kwh consumed/month
and the monthly electrical bill based on a tariff of 70 paise/unit. Add Rs 2/- as

monthlyrentfortheenergymeter.
Totalpoweroflightpoints 10 x 60 = 600 watts
Total power of lamps 12 x 40 = 480 watts
Total power of fans 6 x 60 = 360 watts
Total = 1440 Watts
..
. 50% of 1440 watts = 720 watts
watt-hours/day = 720 x 6 = 4320 wh . taken by lights and fans
power taken by pump moter = ½ x 735.5 x 4 = 1471 wh/day
∴ Total Wh consumed /day = 4320 + 1471 = 5791 Wh
..
. Kwh/month =
1000
305791×
= 173.73 units
cost of energy consumed/month = 173.73 x
100
70
= Rs 121.611/-
meter rent = Rs 2/- per month
..
. Total monthly Elect.Bill = Rs 121.611 /- + Rs 2 /- = Rs 123.611/-
9) A man lifts a head of 100kg through a height of 2 mts in 3 seconds. Calcu-
late the energy spent or workdone.Also find out his power.
Work done or energy spent =
Force x distance moved = 100 x 2 = 200 kg mtr
Newton-meter = 98 x kg-mtr = 9.8 x200 = 1960 Nw-m
Horse power = ..
9
8
753
200
75
sec/
PH
mtsKg
=
×
=
−
(metric)

Energy spent in joule = energy spent in Nw – mtr
= 1960 joule
electric power = watts = watts
Seconds
Joules
33.653
3
1960
==
10)A 1000wimmersionheaterisusedtoheatthewater.Iftheheaterisinthe
waterfor15mtsonconnectingtothemains,thetemperatureofwaterraisedby
500
c . calculate the mass of water in kg. Assume no losses
Heat absorbed by water = ( )12 θθ −ms
= m x 1 x 50 = 50 m
heat given by the heater = calK
RtI
.
4200
2
4200
60151000 ××
or
42
9000
50=
m = 4.3 kg ..
. weight of water = 4.3 kg.
11)Ahouse has to be wired comprising the fallowing points.
a) Light points 20 no‘s of 60 watts each.
b) Fan points 8 No‘s of 100 watts each
c) Bell point 2 No‘s of 40 watts each
d) Wall plug points 4 No‘s of 500 watts each
Supplyvoltageis400voltsA.C.,calculatethefallowing:
i) Totalloadforlightandfanconnections.
ii) Total load for power connections.
iii) Sizeofmainswitchforlightandmainconnections.
iv) Sizeofmainswichforpowerconnection.

v) No of circuits proposed for light connections.
vi) No of circuit proposed for power connections
Lightpointload= 20 x 60 = 1200 watts
Fan point load = 8 x 100 = 800 watts
Wall plug load = 4 x 60 = 240 watts
Bell point load = 2 x 40 = 80 watts
Total load = 2,320 watts
Power plug point load = 4 x 500= 2000 watts
= 2 kw
Currentforlightload =
400
2320
= 5.8 Amps
Current for power load =
400
2000
= 5 Amps
As the total load will be distributed on three, phase. So on each phase
current become 6amps approximately. So for mains a I.C.T.P. 15 Amps
500voltsswitchwillbeusedandforseparatelightandpower.SeparateI.C.D.P.
switch of 15 amp will be used.
For no. of circuits total no. of light points is 34. So, as per I.E. rules
which state that each circuit should have 8 points. total circuits proposed are
4. And power as per I.E. rules. There should be only two points on each
circuit. So proposed circuits for power will be 2.
Hence i)Total light load = 2320W
ii) Total power load = 2 Kw

iii)Sizeofmainswitchforlight=I.C.D.D.15amp.
iv) Size of main switch for power = I.C.D.P. 15 amp
v) No. of circuits for light load = 4
vii) no. of circuits for power load = 2
12) Calculate the bill of electricity charges for the following load fitted in an
electricalinstallation.
1) 20 lamps 100 watts each working 6 hours/ day.
2) 10 ceiling fans 120 watts each working 12 hours/day.
3) 2 kw heater working 3 hours/day
Rate of charges for light and fans is 20 paise per unit and heater and motor 15
paise/unit.
Light load =
1000
6201000 ××
= 12 kwh
10 ceiling fans 120 watts each 12 hrs/day =
1000
1210120 ××
Total light load = 12 + 14.4 = 26.4 kwh
Power load :
2 kw heater 3 hrs/day = 2 x 3 = 6 kwh / day
2B.H.P.motorshaving85%efficiency
Motor input = watts
85
1007462 ××

Load for 4 hrs/day = 1
4
100013
1007462
×
×
××
= 7.021 Kwh
Total power load = 7.021 + 6 = 13.021 kwh
Cost light = −=
×
/28.5.
100
204.26
Rs
Power = −=
−
=
×
/95.1.
20
/063.39.
100
15021.13
Rs
Rs
Total cost = 5.28 + 1.95 = Rs. 7.23 /Ps.
13) Calculate the power of a pump which can lift 100 kg of water to
store it in a water tank at a height of 19m in 25 S? (Take the value of
g = 10 m/s2
)
Inliftingwater,thepumpworksagainstgravity.
Work done =mgh
= 100 kg x 10 m/s2
x 19m
= 19000 J
Power = w/t = 19000 J / 25 S = 760 Watts.

3.7 Assignment :
1) Define work and write its units.
2) Define power and write its units.
3)DefineEnergyandwriteitsunits.
4) What is the meaning of B.O.T?
5) What is J ?
6) How to convert H.P. into watts?
7)Whatistherelationbetweenthemechanicalunitsandelectricalunitsof
power?
8) Give the name of 10 domestic appliances.
9) A force of 10 Newtons is required to push a cycle through a distance of
4 meters. Calculate the work done in 1) Newton Metres and
2) Kg metres. Ans. 1) 40 Nw
-m 2) 4.077 kg-mts
10)Adomesticinstallationconsistsofthefollowing
a) Six 60 watts lamps working for 8 hours/day
b)fourtubelightworkingfor10hours/day.
c) three fans of 60 watts each working for 12 hours/day
d) Two electric irons of ½ kw each working for 2 hours/day.
Calculate the monthly electric bill at the rate of 75 paise/kwh. Meter
rent is Rs. 2 /- p.m. Ans :196.40 /-

11) Anofficeofelectricalinstallationcomprisesthefollowingloads.Calcu-
latetheenergychargespaidtothesupplyauthorityforthemonthofNovember.
Energyconsumption10paise/unitforpowerloadand20paiseunitforlighting
load,Air conditioner, Pump, heater. Calculators are connected to the power
circuit. The office worked for 25 days of that month except security section
which worked on all days.
60 No.’s 4 ft. 40 watt tube light works 8 hours / day
4 No.s 60 watt lamps, 12 hours/day for security purpose
16 No.’s , 60 watts ceiling fans 6 hours/day,
1 No. 1000 watt Air conditioner for 5 hours/day
1 No 750 watt lamp 2 hours/day
2 No.’s Electric calculator 500 watt 1 hrs/day
One electric heater 1500 watts for 2 hrs/day.
Metre rent at Rs. 2/- and 10/- surcharge on charges also made by
Authorities. Ans. Rs. 189.16
Ans. For lighting 710.4 kwh Rs. 142.08 /-
For power 262.5 kwh Rs. 26.25 /-

69EffectsofElectricCurrent
4.0 EFFECTS OF ELECTRIC CURRENT
4.1 Heating effect of electric currrent :
Whenelectriccurrent(i.e.flowoffreeelectrons)passesthroughacon-
ductor, there is a considerable ‘friction’between the moving electrons and the
moleculesoftheconductor. Theelectricalenergysuppliedtotheconductorto
overcomethis‘electricalfriction’(whichwereferitasresistance)isconverted
into heat. This is known as heating effect of electric current.
The heating effect of electric current is utilised in the manufacture of
many heating appliances such as electric heater, electric kettle, electric toster,
solderingironetc.Thebasicprincipleoftheseappliancesisthesame. Electric
current is passed through a high resistance(called heating element), thus pro-
ducingtherequiredheat. Theheatingelementmaybeeithernichromewireor
ribbonwoundonsomeinsulatingmaterialthatisabletowithstandheat.
4.2 Joule’s law :
According to Joules law, the heat produced in a current carrying con-
ductor is directly proportional to the square of the current and to the resistance
of the conductor and to the time of flow of current.
calories
J
RtI
H
RtIHoducedHeat
2
2
)(Pr
=∴
∴ α
∴ H = Heat produced in calories.
I = Current in Amperes
R = Resistance of the conductor in ohms.
T= Time for flow of current in seconds and
J=Joulesmechanicalequivalentofheatwhichisaconstant.
= 4.2 Joules / calories.
i.e. 1 calories of heat = 4.2 Joules.
1 Joules = 0.24 calories
Joules = 4.2 x calories
Calories = 0.24 x Joules

Caloriesistheamountofheatrequiredtorisethetemporatureof1gmofwater
through 1 0
c.
4.3 Practical applications of heating effects of electric current :
Theheatproducedinahighresistancewireisutilizedinthefollowing
appliances:
1) Soldering Iron 2) Electric Iron 3) Electric Kettle 4) Electric stove 5)
Immersion water heater 6) Geyser 7) Drying machine 8) Domestic oven 9)
Cooking set 10) Industrial furnaces 11) Electric welding and Brazing 12)
Electric lamps 13) Radiator or Room Heater 14) Incumbator etc.
4.4 Problems :
1)Awireofresistance Ω10 iskeptinwaterinatinandacurrentof2amperes
is passed through this wire for half an hour. Calculate the heat received by
water.
Heat produced H = 0.24 I 2
Rt calories.
= 0.24 x 22
x 10 x 30 60 Calories.
= 1728 Calories.
= 1.728 Kilo - Calories.
2) The mass of water in a calorimeter was 250 gms. A coil of resistance 15
ohmswasimmersedinwaterandacurrentof2amperesispassedthroughthis
wirefor15minites. Calculatethefinaltemparatureofwater,ifitsinitialtemparature
was 250
C. neglect all losses.
Heat produced in Calories = H = 0.24 I 2
R t
= 0.24 x ( 2) 2
x 15 x 15 x 60
= 12, 960 Calories.
But in physics heat = Mass x Specific heat x change of temp in 0
C.
Mst calories and where t = (t2
- t1
) 0
C
12,960 = 250 x 1 x ( t2
- t1
)

∴ t = C0
84.51
1250
960,12
=
×
( t2
- t1
) = 51.84 0
C.
3) An immersion water heater of resistance 25 ohms is kept in awater bucket
and connected to 230 volts supply mains. Mass of water is 10 kg. Initial
temparature is 250
C. Find the time taken for the water to reach boiling point.
neglectallthelosses.
Heat produced H = 0.24
R
V 2
x t = M x s x ( t2
- t1
)
( )
( )
( )
utes
hourstor
ondst
V
Rttsm
t
min6.246041.0
.41.0
6060
84.1466
sec84.1476
23024.0
25251001000,10
24.0
2
2
12
=×
=
×
=
=
×
×−××
=∴
×−××
=∴
4)On the name plate of electric kettle, it is written as 750 watts and 220 volts,
Determinethethermalefficiencyofthekettle, Ifittakes20minutestoraiseth
temparature of one kg. of water from 200
C to boiling point.
Thermalefficiencyofkettle = CaloriesInput
caloriesinoutput 100×
( ) ( )
%037.37
27
1000
602075024.0
2010011000
24.0
100
100
2
12
==
×××
−××
=
××
×−××
=
×
=
tRI
ttsm
kettlethebycaloriestakeIn
waterbyutilizedactuallyCalories

4.5 Magnetic effects of electric current :
H.C. Oersted demonstrated that whenever current passes through a conduc-
tor,amagneticfieldiscreatedarouindtheconductorthroughoutitslength.orA
currentcarryingconductorhasamagneticfieldassociatedwithit.Thisacciden-
taldiscoverywasthefirstevidenceofalongsuspectedlinkbetweenelectricity
andmagnetism. Theproductionofmagnetismfromelectricity(whichwecall
electro magnetism) has opened a new era. The operation of all electrical ma-
chinery is due to the applications of the magnetic effects of electric current in
one form or the other.
To detect the presence of such a field you can carryout the fallowing
activity
Makeasimpleelectriccircuitconsistingofalongstraightwire, abat-
tery and a plug key. Arrange the circuit so that the straight wire is placed
parallel to and over the compass needle. Now switch on the circuit.
Asthecurrentpassesthroughthewire,theneedlegetsdeflected. Ifthe
currentflowsfromNorthtoSouth,thenorthpoleoftheneedlemovestowards
east. If the current is reversed then the needle moves from South to North.
A compass needle placed under a long straight line pointing towards
northinwhichcurrentflowsfromnorthtoSouth. thecompassneedleisshown
deflecting-itsnorthpolemovingtowardseast.
Deflection of compass needle magnet

The current in this wire flows from South to north. The north pole of
the needle is seen deflecting towards west.
Conclusionsfromoersted’sexperiment:
1) Whenever current is passed through a straight conductor, it behaves
likeamagnet.
2) Themagnitudeofmagneticeffectincreaseswiththestrengthofcurrent.
3) Themagneticfieldsetupbyconductorisatrightanglestothedirection
offlowofcurrent. Thereasonforthisstatementisthatmagneticneedle
setsitselfatrightangletotheconductorcarryingcurrent.
4) Thedirectioninwhichthenorthpoleofmagneticneedlewillmove
depends upon
i) the direction of current in conductor
ii)therelatingpositionofconductorwithrespecttomagneticneedle. It
meansititwilldependuponwheathertheconductorisaboveneedleor
belowneedle.
Ampere Rule :(Forfindingthedirectionofmovementofmagneticneedle)
Imagineaswimmerswimminginthedirectionofcurrentandalwayslookingat
the magnetic needle such that current enters from his feet and leaves from his
head. Then the direction in which the left hand of swimmer points gives the
directionofmovementofnorthpoleoffreelysuspendedmagneticneedle.
Ampere’sSwimmingrule

FromFigure,theswimmerisswimminginthedirectionofcurrentandlooking
at needle. His left hand is pointing towards west. Hence, the north pole of
magnetic needle will deflect towards west. If a current carrying conductor is
placedatrightanglestothelinesofforceofamagneticfieldamechanicleforce
willbeexertedontheconductor. Themagnitudeoftehmechanicalforcecabe
calculatedbyusingAmpere’slaw.
4.6 Magnetic field around a straight conductor :
Takeaflatcardboardandoveritfixawhitesheetofpape. Inthemiddleof the
card board make a hole and through it pass a thick wire as shown in fig. con-
nect the ends of a wire to a battery through a connecting wire.
Plot the magnetic lines of force around the conductor with the help of
plotting compass. It is observed that the lines of force are in the form of con-
centric circles. The direction of lines of force will be clock wise.
If the experiment is repeated but the current is passed in opposite di-
rection,thelinesofforcewillbeinanticlockwisedirection.
Furthermore,itisfoundthatonincreasingthestrengthofcurrent,the
number of magnetic lines of force around conductor increases. This inturn,
increasesthemagneticstrengthofconductor.
Magnetic field arround a straight conductor

4.7RuleforDeterminingthedirectionofMagneticLinesofforcearound
straight conductor :
Right hand thumb rule :
Imagine you are holding the conductor with the palm of your righ hand, such
thatthumbpointsinthedirectionofflowofcurent. Thenthedirectioninwhich
fingerscurlaroundconductorgivesthedirectionofmagneticlinesofforce.
In. fig. The fingers are curling in anti - clock wise direction when thumb is
pointinginthedirectionofcurrent. Thereforethedirectionofmagneticlinesof
force is anti - clock wise.
4.8 Properties of Magnetic Lines of force around straight conductor :
1)Themagneticlinesofforceareintheformofconcentriccircles.
2) The plane of magnetic lines of force and hence, magnetic field is at right
angle to the plane of conductor carrying current.
3)The direction of magnetic lines of force reverses withthe changes in the
directionofflowofcurrent.
4)Onincreasingthemagnitudeofcurrentinconductor,thenumberofmagnetic
linesofforceincreases.
5)Magnetising force at ‘p’ due to a long straight current carrying conductor
at a distance of ‘r’meters is
Finding the direction of magnetic lines of force

mAT
R
H /
2
1
π
=
4.9 Magnetic field due to a current in a circular coil :
Takeadrawingboardandfixoveritawhitesheetofpaper. MaketwoholesA
and B in drawing board and pass through it a thick copper coil. Connect the
ends of coper coil to a dry cell through a Switch and variable resistance close
thecircuitandplotmagneticlinesofforcewiththehelpofplottingneedle.
It is seen that magnetic lines of force around A are in anti - clock wise
direction, Whereas that around B are in clock wise direction. However the
magneticlinesofforcenearthecentreofcoilbecomealmostparallel.Asthese
linesofforceseemtoenterthecoilfromthesideoftheexperiment,wecansay
that face of coil towards the experiment acts as south pole. Conversely, the
face opposite to experiment acts as north pole.
Magnetic lines around a straight conductors
P
r meters
Magnetic field around a circular coil

If we relate the above observation to the flow of current in the coil,
thenwecansaythatifthecurrentflowsinthecoilinclockwisedirectionfacing
the experimenter, then that face of the coil will act as south pole. In the same
way,ifthecurrentincoilfacingexperimenterisinanti-clockwisedirectin,then
thatfaceofthecoilwillbeleavelikenorthpole.
4.10 Properties of magnetic lines of force around circular coil :
1)Magnetic lines of force are circular around the points where the current en-
ters or leaves circular coil.
2)Withinthespaceenclosedbythecoilthemagneticlinesofforceareinsame
direction.
3)Near the centre of coil the magnetic lines of force are parallel. When the
magneticlinesofforceareparallel. Themagneticfieldissaidtobeuniform.
4) The magnetic lines of force are at right angles to the plane of coil. It means
ifcoilisinverticalplane,themagneticlinesofforceareinhorzontalplane.
5)With the increase in strength of current in coil, the magnetic lines of force
increase. Thisinturn,increasesthestrengthofmagneticfield.
6) Magnetising force at the centre of a circular coil of radius ‘r’metres.
H = I / 2 r AT / m forasingleturncoil
and coilturnNaformAT
r
NI
H ''/
2
=
coilcarrying
current
experimenter coil
carryingcurrent
experimentar
Fig .1
Fig .2

4.11 Magnetic field in a solenoid :
An insulated copper coil wound around some cylindrical cardboard or plastic
tube,Suchthatitslengthisgreaterthanitsdiameterandbehaveslikeamagnet
whenelectriccurrentflows throughitiscalledsolenoid.
When an electric current is made to flow through it, then each turn of the coil
behaveslikeamagnetic.Infigthecurrentisflowingintothecoilinclockwise
direction,thereforethelefthandsideofeachturnofthecoilactsassouthpole,
where as right hand side of each turn of the coil acts as north pole. Thus the
situation becomes similar to small bar magnets placed end to end with their
opposite poles facing each other, Such that they collectively act as a bar mag-
net. Thus, solenoid beheaves like a bar magnet. From this it is clear that if the
numberofturnsinsolenoidincreaes,thenthemagneticeffectalsoincreases.
Thestrengthofmagneticfieldofthesolenoiddependsuponthefallow-
ingfactors:
1)It is directly proportional to the number of turns in solenoid. It means
the more the number of turns, the more is the magnetic strength of the
solenoid.
2)It is directly proprotional to the magnitude of current flowing through sole-
noid. It means the more the magnitude of current, the more is the magnetic
strengthofthesolenoid.
3) It is directly proportinal to the diameter of coil. It means the wider the coil,
themoreisthemagneticstrength.
S
N
S
N
S
N
S
N
S
N
S
N
S
N
Insulated copper
Magnetic field in a solenoid

4)It depends upon the nature of material on which the coil is wound.
It has been found that if solenoid is wound on soft iron, then due to
magneticinduction,itgetshighlymagnetised. Thusstrengthofmagnetic field
stronglyincreases,Insuchasituationsolenoidiscalledelectromagnet.
It beheavs like magnet as long as the current flow through it.
Magnetising force at the centre of a long solenoid mAT
l
NI
H /
2
=
If it is a short sole noid mAT
l
NI
H /
2
=
4.12Force on a current carrying conductor in a magnetic field :
Immediatelyafteroersted’sdiscoveryofelectriccurrentsproducingmagnetic
fieldsandexistingforcesonmagnets,Amperesuggestedthatmagnetmustalso
exert equal and opposite force on a current - carrying conductor. The force
due to a magnetic field acting on a conductor can be demonstrated by the
followingactivity.
A current carrying rod, AB . experiences a force perpendicualr to the
length and the magnetic field. Take a small aluminium rodAB. Suspend it
horizontally by means of two connecting wires from a stand, as shown in fig.
Now, place a strong horse shoe magnet in such a way that the rod is between
A current carrying rod, AB . experiences a force
perpendicualr to the length and the magnetic field

the two poles with the field directed upwards. If a current is now passed in the
rod from B to A, you will observe that the rod gets displaced. This displace-
ment is caused by the force acting on the current - carrying rod. The magnet
exerts a force on the rod directed towards the left, with the result that the rod
willgetdeflectedtotheleft. Ifyoureversethecurrentorinterchangethepoles
ofthemagnet,. thedeflectionoftherodwillreverse,Indicatingtherebythatthe
direction of the force acting on it get reversed. This shows that there is a
relationshipamongthedirectionsofthecurrent,thefieldandthemotionofthe
conductor.
In the above , activity, you considered the direction of the current and
that of the field perpendicualr to each other and found that the force is perpen-
diculartobothofthem. thethreedirectionscanbeillustratedthroughflemings
lefthandrule.
4.13 Fleming’s left hand rule :
Stretchtheforefinger,thecentralfinger, andthethumbofyourlefthandmutu-
ally perpendicular to each other. If the fore finger shows the direction of field
and the central finger that of the current, then the thumb will point towards the
directionofmotionoftheconductor.
We have studied that current is simply a flow of charges. This means
that moving charges in a magnetic field would also experience a force. The
directionoftheforceonamovingpositivechargeisexactlythesameasthaton
acurrentandisgivenbyflemingslefthandrule.
Flemings left hand rule

4.14 Electrical Motors :
An electric motor is a device that converts electrical energy to mechanical en-
ergy. Electricmotorisusedasanimportentcomponentinelectricfans,wash-
ingmachines,refrigerators,mixersandblenders.
A coil of wire wrapped around an axle is placed between the two
polesofthemagneticfieldasshowninfig. Whenacurrentpassesthroughthe
coil,entersatthepointXandleavingatYthetwoarmswhichareperpendicu-
lar
to the direction of the magnetic field, experience force according to
flemingslefthandrule. Sincethedirectionsofthecurrentsinthetwosections
are oppsite to each other, the forces acting on them will also be opposite to
each other. These force push one section ( the arm CD) up and the other ( the
arm AB) down. Mounted free to turn about an axis, the coil rotates anti clock
wise. Athalf rotation,thecurrentintheloopisreversedindirectionbymeans
of sliding contacts and a split ring commutator. As a result, it is Q which now
contacts, the brush X and PcontactsY. The reversal of the current reverses
the forces so that, the side of the coil which was previously pushed up is now
pushed down, and the side previoiusly pushed down in now pushed up. The
coil, therefore rotates half a turn more where the current is again reversed. In
thiswayareversingprocessisrepeatedateachhalfturn,givingrisetoacontinious
rotation.
Brushes sliding
contents
N
B C
A
D
N
X Y
P Q
Fig. showing the working of electric motor

4.15 Field due to two parallel conductors :
Whenthecurrentsareinthesamedirectionthemagneticlinesbetween
the currents are neutralised and the conductrs try to come closer due to attrac-
tion.
In the second case when the currents are in opposite direction the
magnetic lines outside the conductors neutralize and the conductors try to go
furtherduetorepulsion.
Ampere : If two long straight parallel conductors carry some unknown but
equalcurentsandiftheircommonlengthisonemetreandthedistancebetween
them (centre to centre) is also one metre and if that mutual force acting be-
tweenthemis2 x10-7
newtons, thenthevalueofthatunknowncurrentisone
ampere.
4.16 Magnetic Circuit :
Magneticcircuitisthepathfallowedbymagneticflux. Magneticflux
fallowsacompletelooporcircuitcommingbacktoitsstartingpoint.
Itispossibletoestablishmagneticfluxinadefinitelimitedpathbyusing
magneticmaterialofhighpermeability. Inthismanner,themagneticfluxforms
a closed circuit exactly as an electric current does in an electric circuit.
The amount of flux produced in a magnetic circuit depends upon the
propertyofmagneticmaterialopposingtheproductionoffluxandthisproperty
iscalledreluctanceofthematerial.
Magneticcircuitsarefoundinallelectricalmachinesintransformers,in
motors and in many other devices.
Current in the same direction Current in the opposite direction

4.17 Flux : Group of magnetic lines of force is called flux. 1 wb = 10 8
lines.
4.18 Magneto motive force ( M.M.F) :
Thisissimilartoemf.inelectriccircuit. Itistheforcewhichdrivesthe
flux in magnetic paths. Its unit isAmpere - turn (AT)
MMF = Amperes x No. of turn
= Magnetising force x length of circuit
= HL - AT
4.19 Reluctance (OR Magnetic Resistance) :
Whenever we wish to setup an electric current in a circuit, we always have to
overcome the resistance of the electric circuit, which opposes the flow of
current. Similarlywehaveseenthatthereisalwaysmagneticresistance, which
we call reluctance, and which always opposes the setting up of the magnetic
flux in the circuit just as the resistance of an electric circuit depends upon the
material and dimensions of the circuit. So the reluctance of a magnetic circuit
dependsuponthematerialanddimensionsofthemagneticcircuit.
4.20 Relation between MMF, Flux and Reluctance OR Ohm’s Law for
magneticcircuit:
Inthemagneticcircuitwehavemagneticpressuresettingupamagnetic
fluxagainsttheresistanceofferedbythemagneticreluctance.
reluctancelinesmagneticsAmpereturn
reluctance
turnsampere
linesMagnetic
Flux
MMF
Reluctance
reluctancemagnetic
forceivemagnetomot
fluxMagnetic
×=
−
=
=
=

4.21Comparison between :
ElectricCircuit Magnetic Circuit
2)E.M.F is the source 2) MMF is the source to pass flux ( MMF
to pass current is caused by flow of current)
3) Current in Amperes; 3) φ is in webbers;
current density in A/m2
fluxdensitywb/m2
4) current
Resistance
EMF
4) Flux =
Reluctance
MMF
5) Resistance = R =
a
lδ
5) Reluctance = A
L
rµµ0
and is constant It veries as rµ is variable
6) Conductance = 1/ R 6) Permeanance = 1 / Reluctance
7) Energy is wasted as 7)Energyisrequiredtoestablishthefluxonly
long as the current lasts and not for maintaning it.
8) No leakage of current 8)Thereisleakageofflux
9) Current can be insulated 9)Thereisnomagneticinsulator. Theflux
i.e. it cannot pass through passes through all the mediums.
all the mediums
10) Current flow is true flow 10) There is no actual flow of flux. It is only
the effect. Hence the word “ Flow of flux”
is misleading.
R
I
I
N

11) Equivalentcircuit 11)Equivalentcircuit.
4.22 Flux density and magnetising force :
Flux Density :
It is the flux lines per cross sectional area normal to the flux lines. It is men-
tioned in webbers per square metre. It is denoted by letter ‘B’.
2
meter
webbers
Area
Flux
densityFlux ==
Magnetising force :
The magnetising force ( H) is really the magnetic pressure required to
sendagivennumberoflinesthrough1inchofthemagneticcircuit.
Itissimilartothevoltagerequiredtosendagivenelectriccurrentthrough
amileofwireofgivendimension.
H is the no. of ampere-turns required to send a given number of lines
throughoneinchofagivencircuit.
L
NI
H = Ampere turns per meter length.
4.23 Magnetisation of magnetic meterial B - H curve :
Consider a solenoid with iron core. If current is passed through the solenoid a
magneticfieldwillsetupanditsvalueinsidethecoilwillbegivenbytheequa-
tion mAT
L
NI
H /= .AndFluxdensityintheironcorewillbe HB rµµ0= .
thefluxdensityincreaseswiththeincreaseinH,themagnetisingforce,whichis
proportionaltothecurrentthroughthecoil. Agraphcanbedrawntoshowthe
relationbetweenBandHorcurrent. Suchagraphforagivenmaterialiscalled
a magnetisation curve. It is also defined as the graph drawn to show the
R
I V
L
AT

relationbetweenthefluxdensityandmagnetisingforceofanygivenmagnetic
circuit. Such a graph is called B - H curve.
As shown in the above figure, wind an insulated wire around a given iron bar
and pass alternting current through it. Already you know that H =AT/ L
= NI/LAT per meter and flux density B = φ / a webbers / sq.m keeping this
in view, let us plot - B - H curve as shown above.
Atthetimeofstarting,whencurrentiszerofluxproducedalsoiszero.
Asthecurrentisincreasingi.e.Hisincreasing,thefluxorfluxdensityBisalso
increasing. Therefore the curve rises slowly upwards. But when the iron get
saturatedatBmax
,itcannotproduceanymorefluxhoweverwemayincreasethe
current. Thereforethecurvehasbecomeslightlyhorizontal. Wecallthispoint
as the saturation point. Now, Suppose the direction of current - reverses.As
the current has reversed, it is slowly decreasing , when the current is decreas-
ing, the flux or flux density ‘B’ is also decreasing, but it is not retracing its
original path. This is due to the rate of decreasing of flux density is lesser than
previousrateofitsincrease. Thisiscalledfluxdensity. ‘B’islaggingbehindthe
magnetising force H. This property of lagging of B behind H is called hyster-
esis. Thereforeweseethatthoughthecurrentiszerointhereverseddirection,
thefluxdensityisnotzeroanditisequaltoOA. Thisiscalledresidualmagne-
tism or Residual flux. Now, to bring this residual flux back to zero we have to
passthecurrentinthesamereversedirectionupto‘D’. At‘D’weseethat“B”
iszero. SothiscurrentODiscalledco-erciveforce. Thispropertyofretaining
magnetismagainstsomeexternalforceiscalledco-ercivity,oftheiron. Now,If
we still increase the current, the flux density goes on increasing in the reverse
A coil carrying an electric current used to magnetised
steel rod inside the coil

directiontillsaturationnowaftersaturation;againthecurrentdirectionreverses.
SointhiswayHandBwillbegoiningonchanging. ThisgraphisknownasB
- H curve or magnetic reversal or Hysteresis loss. As the frequency of supply
is 50 cycles / sec. There will 50 magnetic reversals / seconds these 50 cycles
of magnetisation per second will produce heat inside the iron bar. This heat
cannotbeutilizedforpracticalpurpose. Thisproductionofheatiscalledaloss
ofenergy. Thislossofenergyisdirectlyproportionaltheareaofthehysteresis
loop. More area more loss and less area less loss.
4.24 Cycles of Magnetisation :
Make a winding of N turns on a steel bar specimen and make the connections
as shown in fig. V is the battery from which a current is drawn into the coil
through an Ammeter A, and a variable field Rheostat R connected in series.
FordifferentvaluesofH(I)thefluxdensitiesarecalculateddependinguponthe
demensions of the bar and a magnetisation curve ‘oa’ is drawn. Now the
current is reduced from maximum to zero and the curve ab is drawn. Here
when H = 0 i.e. I = 0. B has some value. Change the direction of current and
increase the current from zero the maximum ( i.e. the reverse direction) duly
calculating the value of B. Draw the curve bed. At this stage reduce the
current to zero again and plot the curve de. again change the direction of
current to original direction and increase the current to its original maximum
value. The entire curve a b c d e f a is called cycle of magnetisation.
A
I
R
V
d
B
e
c o
b
a
H(I)
Bmax
-H
Retentivity
Coercivity
B
Fig- V
Cycle of magnetisation or B - H curve

Fundamental elements of-electrical-engineering circuit theory basic stardelta

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