Electric circuits and network theorems

BASED ON ANNA UNIVERSITY 2017 REGULATION

1. Department of Electrical and Electronics Engineering
Approved by AICTE | Affiliated to Anna University | Accredited by NAAC | Accredited NBA | Recognized by UGC under 2(f) and 12(B)
Chennai Main Road, Kumbakonam-612 501.
ARASU ENGINEERING COLLEGE
1
ELECTRIC CIRCUITS AND NETWORK THEOREMS

2. Prepared by
Mrs.K.Kalpana.,M.E.,(Ph.D).,
Assistant Professor
Department of Electrical and Electronics
Engineering
Arasu Engineering College
Kumbakonam
Department of Electrical and Electronics
Engineering
2

3. INTRODUCTION
Theinterconnectionofvariouselectricelementsina
prescribedmannercomprisesasanelectriccircuitin
ordertoperformadesiredfunction.
Anelectricalnetworkisaninterconnection
ofelectricalcomponents(e.g.,batteries,resistors,
inductors,capacitors,switches,transistors)oramodelof
suchaninterconnection,consistingofelectricalelements
(e.g.,voltagesources,currentsources,resistances,
inductances,capacitances).
Theelectricelementsincludecontrolledanduncontrolled
sourceofenergy,resistors,capacitors,inductors,etc.
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4. INTRODUCTION
Analysisofelectriccircuitsreferstocomputations
requiredtodeterminetheunknownquantitiessuchas
voltage,currentandpowerassociatedwithoneormore
elementsinthecircuit.
Manyothersystems,likemechanical,hydraulic,thermal,
magneticandpowersystemareeasytoanalyzeand
modelbyacircuit.
Tolearnhowtoanalyzethemodelsofthesesystems,
firstoneneedstolearnthetechniquesofcircuitanalysis.
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5. BASIC ELEMENTS & INTRODUCTORY CONCEPTS
•ElectricalNetwork:Acombinationofvarious
electricelements(Resistor,Inductor,Capacitor,
Voltagesource,Currentsource)connectedinany
mannerwhatsoeveriscalledanelectricalnetwork.
Wemayclassifycircuitelementsintwocategories,
passiveandactiveelements.
•PassiveElement:Theelementwhichreceives
energy(orabsorbsenergy)andtheneitherconverts
itintoheat(R)orstoreditinanelectric(C)ormagnetic
(L)fieldiscalledpassiveelement.
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6. ActiveElement:Theelementsthatsupplyenergyto
thecircuitiscalledactiveelement.
Examplesofactiveelementsincludevoltageand
currentsources,generatorsandelectronicdevicesthat
requirepowersupplies.
Atransistorisanactivecircuitelement,meaningthatit
canamplifypowerofasignal.
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7. BilateralElement:Conductionofcurrentinboth
directionsinanelement(example:Resistance;
Inductance;capacitance)withsamemagnitudeis
termedasbilateralelement.
UnilateralElement:Conductionofcurrentinone
directionistermedasunilateral(example:Diode,
Transistor)element.
Response:Anapplicationofinputsignaltothe
systemwillproduceanoutputsignal,thebehaviorof
outputsignalwithtimeisknownastheresponseofthe
system
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8. PotentialEnergyDifference:Thevoltageor
potentialenergydifferencebetweentwopointsinan
electriccircuitistheamountofenergyrequiredto
moveaunitchargebetweenthetwopoints.
Resistoropposestheflowofcurrentthroughitandit
dissipatesenergyintheformofheat.(Ohms(Ω)).
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9. Inductorstoresenergyinits
magneticfield.
UnitofinductanceisHenry(H).
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10. Capacitorstores energy in its electric field.
Unit of capacitance is Farad (F).
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11. Classification of sources( active elements)
Sources (active elements) are classified as
1.Independent sources
2.Dependent sources
1.Independentsources:
Batteriesandgeneratorsarecalledindependentsources
whichcandirectlygenerateelectricalenergy.Independent
sourcesdonotdependonotherelectricalsources.
2.Dependentsources:
TransistorsandOp-ampsarecalleddependentsources
whoseoutputenergydependsonotherindependentsources.
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EE8251-CIRCUIT THEORY

12. V-I characteristics of ideal and practical sources:
Ideal voltage sources:
Avoltagesourceisatwo-terminaldevicewhosevoltageat
anyinstantoftimeisconstantandisindependentof
thecurrentdrawnfromit.Suchavoltagesourceiscalled
anIdealVoltageSourceandhavezerointernalresistance.
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13. Practical Voltage sources:
Sourceshavingsomeamountofinternalresistancesareknown
asPracticalVoltageSource.Duetothisinternalresistance,
voltagedroptakesplace,anditcausestheterminalvoltageto
reduce.
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IdealCurrentsources:
•Anidealcurrentsourceisatwo-terminalcircuit
elementwhichsuppliesthesamecurrenttoanyload
resistanceconnectedacrossitsterminals.Herethe
currentsuppliedbythecurrentsourceisindependentof
thevoltageofsourceterminals.Ithasinfiniteresistance.

15. •Practical Current sources
•Apracticalcurrentsourceisatwoterminaldevicehaving
someresistanceconnectedacrossitsterminals.Unlike
idealcurrentsource,theoutputcurrentofpractical
sourcedependsonthevoltageofthesource.
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16. Active Network:
Anetworkthatcontainsatleastoneactive
element,suchasanindependentvoltageor
currentsource,isanactivenetwork.
Passive Network:
Anetworkthatdoesnotcontainanyactive
elementsisapassivenetwork.
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17. Lumped elements :
Physicallyseparatedelementssuchasresistor,
capacitorandinductorsarecalledaslumpedelements.
Distributedelement:
Itisnotseparableforelectricalpurposes.A
transmissionlinehasdistributedresistance,inductance
&capacitancealongitslength.
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18. Bilateral element :
 Conductionofcurrentinbothdirectionsinan
elementwithsamemagnitude.(Ex:R,L,C)
Inabilateralelement,thevoltageandcurrent
relationisthesameforcurrentflowingineither
direction.
Unilateral elements-:
Conductionofcurrentinonedirection.(Ex:
Diode,Transistor).
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19. D.Ccircuits:
Thedirectionofcurrentflow,atanypointinthecircuit,
doesnotchangewithtime.
A.Ccircuits:
Thedirectionofcurrentflow(aswellasitsmagnitude)
changeswithtime.
CHARGE
Chargeisanelectricalpropertyoftheatomicparticlesof
whichmatterconsists,measuredincoulombs(C).
Charge,positiveornegative,isdenotedbytheletterq
orQ.
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EE8251-CIRCUIT THEORY

20. CURRENT
Currentcanbedefinedastherateofflowofelectrons
inaconductiveorsemiconductivematerial,measuredin
Ampere(A).
Electriccurrent,isdenotedbytheletteriorI.
Theunitofcurrentistheampereabbreviatedas(A)and
correspondstothequantityoftotalchargethatpasses
throughanarbitrarycrosssectionofaconducting
materialperunitsecond.
Mathematically,
•Where Q is the symbol of charge measured in Coulombs
(C),
•I is the current in amperes (A) and t is the time in
second (s).
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21. The current can also be defined as the rate of charge
passing through a point in an electric circuit.
Mathematically, i= dq/dt
Two types of currents:
1)A direct current (DC) is a current that remains constant with
time.
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22. 2) An alternating current (AC) is a current that varies with time.
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23. •The difference in potential energy of the charges is
called potential difference.
•It is also known as voltage.
•It is denoted by V, unit Volts
•V=W/Q or dw/dQ
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VOLTAGE (or) POTENTIAL DIFFERENCE :

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POWER
Poweristherateofchangeofenergy,measuredin
watts(W).Power,isdenotedbytheletterporP.
Mathematically,P=dw/dt
WherePispowerinwatts(W),
wisenergyinjoules(J),and
tistimeinseconds(s).
Fromvoltageandcurrentequations,itfollowsthat;
P=dw/dt=dw/dqXdq/dt
=VXI
Thus, if the magnitude of current I and voltage are given,
then power can be evaluated as the product of the two
quantities and is measured in watts (W).

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ENERGY
Energy is the capacity for doing work, and is
measured in joules (J).
The energy absorbed or supplied by an element from
time 0 to t is given by,

26. Node:Anodeisapointinanetwork,wheretwoor
moreelementsarejoined.
Principalnode:Junctionofthreeormoreelements.
(Currentisdivided)
Simplenode:Junctionofanytwoelements.(Currentis
notdivided)
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27. Branch:Branchrepresentsasingleelementconnected
betweentwonodessuchasavoltagesourceorresistor.
Intheabovecircuit,thebranchesare:af,ab,bc,cd,be.
Loop:Loopisananyclosedpathinthecircuit.Inthe
abovecircuit,theloopsare:abefa,bcdeb,abcdfa.
Mesh:Meshisaclosedpaththatdoesnotcontainany
otherloops.Intheabovecircuit,themeshesare:abefa,
bcdeb.
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28. Ohm’s Law:
Ohm'slawstatesthatthecurrentthroughaconductor
isdirectlyproportionaltothepotentialdifferenceor
voltageacrossthetwopoints,thatthetemperatureof
theconductorremainsconstant
V∝I(Or)V=IR
Themathematicalequationthatdescribesthis
relationshipis:V/I=R
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29. whereIisthecurrentthroughtheresistanceinunitsof
amperes,
Visthepotentialdifferencemeasuredacrossthe
resistanceinunitsofvolts,and
whereRistheproportionalityconstant.
R is called Resistance of the conductor and is measured
in ‘Ohms’ (Ω).
Thepowerabsorbedbytheresistanceiscalculatedas:
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Since I=V/R

30. •The V-I relation for resistor according to Ohm’s law is
depicted in Fig.
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LimitationsofOhm’sLaw:
1.Ohm’slawisnotapplicabletonon-linearelementslike
diode,transistoretc.
2.Ohm’slawisnotapplicablefornon-metallicconductors
likesiliconcarbide.
3.Ohm’slawholdsgoodonlyforconstanttemperature.If
thetemperaturechange,thislawcannotbeapplied.

31. KIRCHOFF'S LAW
Kirchoff'sFirstLaw-TheCurrentLaw,(KCL)
Itstatesthat“thealgebraicsumofcurrentsmeetingata
junction(node)isequaltozero".
InotherwordsthealgebraicsumofALLthecurrents
enteringandleavinganodemustbeequaltozero,
I(exiting)+I(entering)=0.
IncaseofACcircuits,Kirchoff’scurrentlaw
statesthatphasorsumofincomingcurrentis
equaltothephasorsumofoutgoingcurrent.
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32. I
1+ I
2+ I
3= I
4+ I
5
Here, the 3 currents entering the node, I
1, I
2, I
3are all positive
in value and the 2 currents leaving the node, I
4and I
5are
negative in value.
Then the equation can also rewrite as; I
1+ I
2+ I
3-I
4-I
5= 0
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33. Kirchhoff’sVoltageLaw(KVL):
Thealgebraicsumofelectromotiveforcesplusthe
algebraicsumofvoltagesacrosstheimpedances
(resistances)inanyclosedelectricalcircuitisequalto
zero.
Foranyclosedpathinacircuit,thealgebraicsumof
thevoltageiszero.
Inotherwords,foranyclosedpathinacircuit,the
sumofvoltageriseisequaltosumofvoltagedrop.
Mathematically,∑emf+∑IR=0inanyclosed
electricalcircuit.
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34. ARASU
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Consider the above fig., Applying KVL: V
1-V
2 + V
3-V
4= 0
It can be also written as, V
1+V
3= V
2+ V
4

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36. ARASU
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Consider the circuit shown in figure.
Apply KVL to the circuit,
-E
1+ E
2-E
3+E
4+I
1R
1-I
2R
2-I
3R
3-I
4R
4=0

37. Problem:1
Acurrentof0.5Aisflowingthroughtheresistanceof
10Ω.Findthepotentialdifferencebetweenitsends.
Solution:
Current I= 0.5A.
Resistance R = 10Ω
Potential difference V =?
V = IR
=0.5 ×10
=5V.
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38. ARASU
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Problem:2
Asupplyvoltageof220Visappliedtoaresistor100Ω.
Findthecurrentflowingthroughit.
Solution:
VoltageV=220V
ResistanceR=100Ω
CurrentI=V/R
=220/100
=2.2A.

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Problem:3
Calculatetheresistanceoftheconductorifacurrentof2A
flowsthroughitwhenthepotentialdifferenceacrossits
endsis6V.
Solution:
CurrentI=2A.
PotentialdifferenceV=6.
ResistanceR=V/I
=6/2
R=3ohm.

40. Problem: 4
Calculatethecurrentandresistanceofa100W,200Velectric
bulb.
Solution:
Power, P = 100W
Voltage, V = 200V
Power P= VI
Current I = P/V
=100/200
=0.5A
Resistance R = V /I
=200/0.5
=400Ω
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41. ARASU
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Resistors In Series
When resistors are connected as shown in figure such that
the same current passes through all of them, they are said
to be in series.
Each resistor has a voltage drop across it given by Ohm’s
law. Thus,
V
1=I
1R
1
V
2=I
2R
2and
V
3=I
3R
3
The total drop in the three resistors put together is
V = V
1+V
2+ V
3
V =I(R
1+R
2+R
3)

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Power dissipated in R1 is given by

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ResistorsInParallel
Whenresistorsareconnectedacrossoneanothersuch
thatthesamevoltageisappliedtoeach,thentheyare
saidtobeinParallel.Suchanarrangementisshownin
fig.
Thecurrentineachresistorisdifferentandthecurrent
Itakenfromthesupplydividesamongallthethree
resistors.

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Ifwereplacethethreeresistorsinparallelbyasingle
equivalentresistorR,asshowninfigure,itwilldrawthe
samecurrent,I.
thereforeV=IR
eff(byOhm’slaw)

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If there were only two resistors in parallel, then the effective
resistance is given by

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Difference between Series and Parallel Circuit

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1.Find the voltage across the resistor R in figure.

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2.Find the current through 12 Ωresistor in figure.

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Voltage division rule:
Current division rule:

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Problem:
Findthecurrentflowingthrougheachbranchforthe
circuitshowninfigure.
Solution:
Heretworesistorareconnectedinparallel.
ThereforeR
eqis

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R
eq= 2.4Ω
Input current I= V/R = 230 / 2.4 =95.83A
Using current division rule, the current through 4Ωresistor is

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Using current division rule, the current through 6Ωresistor is
Problem:
ThreeloadsA,BandCareconnectedinparallelacrossa
250Vsource.LoadsAtakes50A.LoadBisaresistorof10
ohmsandloadCtakes6.25kW.Calculate(I)R
AandR
C(ii)
Currentsl
BandI
C(iii)PowerinloadsAadB(v)TotalPower
and(vi)Totaleffectiveresistance.

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Solution:

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Problem:
Determine the total current taken from the source.

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Solution:
Here, 50Ωand 50Ωare connected in series. So, 50+50=100Ω
100Ωand 100Ωare connected in series. So, 100+100=200Ω

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100Ωand 100Ωare connected in parallel. So,
200Ωand 50Ωare connected in parallel. So,

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40Ωand 50Ωare connected in series. So, 40+50=90Ω
90Ωand 90Ωare connected in parallel. So,
45Ωand 55Ωare connected in series. So, 45+55=100Ω

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100Ωand 100Ωare connected in parallel. So,
Total current I= V/R = 100/50= 2A
I= 2A

60. Problem:Forthecircuitshowninfigure,determine(a)the
batteryvoltageV,(b)thetotalresistanceofthecircuit,
and(c)thevaluesofresistanceofresistorsR1,R2and
R3,giventhatthep.d.’sR1,R2acrossandR3are5V,2V
and6Vrespectively.
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(a) Battery voltage V =V1 + V2 + V3 =5 + 2 + 6=13V
(b)Total circuit resistance R= V/ I
=13/4=3.25Ω
(c)Resistance R1 = V1/ I
=5/4 =1.25 Ω
Resistance R2 = V2/ I
= 2/4 =0.5 Ω
Resistance R3 = V3/ I = 6/4 =1.5 Ω
EE8251-CIRCUIT THEORY

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MeshAnalysis
Inmeshmethod,Kirchhoff'sVoltageLaw(KVL)isapplied
toanetwork.Forthisnetwork,wehavetowritemesh
equationsintermsofmeshcurrents.Here,weareusing
meshcurrentinsteadofbranchcurrents.Hereeachmesh
isassignedaseparatemeshcurrent.Assumethatthe
meshcurrentdirectionisclockwise.Then,KVLisapplied
tothenetwork,inordertowriteequationsintermsof
unknownmeshcurrents.
EE8251-CIRCUIT THEORY

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EE8251-CIRCUIT THEORY
ThebranchCurrentcanbefoundoutbytakingthe
algebraicsumofthemeshcurrentswhicharecommonto
thatbranch.
Thefollowingstepsshouldbeimplementedtofindoutthe
meshcurrentsandbranchcurrents.
Considerasimplenetworkasshowninfigure.
Step1:First,eachmeshisassignedaseparatemesh
current.Assumeallmeshcurrentsdirectionsare
clockwise.Itconsistsoftwomeshes
(PQSPandQRSQ)andtwomeshcurrents(I
1andI
2).

64. Step2:Iftwomeshcurrents(I
1andI
2)areflowing
throughanetworkelements,theactualcurrentinthe
circuitelementisthealgebraicsumofthetwomesh
currents(I
1andI
2).Inthisnetwork,meshcurrents,I
1
andI
2areflowingthroughR
2.Firstweconsiderone
directioni.e.,QtoS,currentisI
1-I
2andinforanother
direction.i.e.,StoQ,currentisI
2-I
1
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65. Step3:Writeequationforeachmeshintermsofmesh
currentsbyapplyingKVL.Whenwritingmeshequations,
weassignriseinpotentialaspositive(+)signandfallin
potentialasnegative(—)sign.
Step4:Supposeanyvalueofmeshcurrentbecomes
negativeinthesolution,theactualortruedirectionof
themeshcurrentisanticlockwise,i.e.,oppositetothe
clockwisedirection.
By applying KVL to the circuit , we get two equations,
Mesh PQSP
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EE8251-CIRCUIT THEORY

66. MESHQRSQ
ThemeshcurrentsI
1and1
2canbefoundoutbysolving
theaboveequations
Step-5:Thebranchcurrentscanbeeasilyfoundoutby
usingthemeshcurrents.
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MeshAnalysisbyInspectionMethod
Step1:Convertthecurrentsourceintoequivalentvoltage
sourcebysourcetransformation.
Step2:AlltheresistancethroughwhichtheloopcurrentI
1
flowsaresummedupanddenotedbyR
11.Itiscalledself
resistanceofloop1.
Step3:AlltheresistancethroughwhichloopcurrentsI
1
inthefirstloop,andI
2inthesecondloopfloware
summedup.ThisisdenotedbyR
12.Thesignoftheterm
R
12isnegative,ifthetwocurrentsI
1andI
2throughR
12
areinoppositedirections;otherwisethesignispositive.

68. Step4:LetV
1betheeffectivevoltageonthefirstloop
throughwhichtheloopcurrentI
1flows.ThesignofV
1is
positiveifthedirectionofV
1issameasthatofI
1(i.e.
aidingthecurrentI
1,otherwisethesignofV
1is
negative).V
1iswrittenontherighthandsideofthe
equation.
Now,zeroiswrittenontherighthandsideifthereisno
sourceinthefirstloopthroughwhichI
1flows.
Thegeneralmatrixformofmeshequationis
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70. (TheminussignindicatestheloopcurrentI
2flowsfromthe
voltageterminal+to--)
ByusingCramer’srule,wecanfindmeshorloopcurrents
I
1andI
2
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71. Nodal Analysis
Thenodalmethodisusedtoanalyzemultisource
circuits.Inthismethod,wesolvethesimultaneous
equationsusingKirchhoff’scurrentlaw(KCL)appliedat
variousnodesinanelectriccircuit.Nodedefinedas
junctionorjoiningpointoftwoormorecomponent
terminals.
Innodalmethod,weselectonenodeasareference
node,withrespecttothevoltagesatallothernodesare
measured.Thus,thereferencenodeactsasagroundor
commonforthecircuit.
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72. ARASU
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Thefirststepinnodalmethodistoconvertallvoltage
sourcestoequivalentcurrentsources.
Thesecondstepistoidentifyalltheprincipalnodesin
thecircuitandtochooseareferencenode.
Thechoiceisarbitrary,butitisusuallyconvenientto
selectthereferencenodeastheonehavingthemostof
thecomponentsconnectedtoit.
Allthenodesexceptthereferencenodearethen
numberedandtheircorrespondingvoltagesaredesignated
asV1,V2….etc.
Thereferencenodeandothernodesareindicatedas
showninfigure.

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Steps
1. Select one major node as the reference node and assign
each of the (n -1) remaining nodes with its own unknown
potential (with respect to the reference node).
2. Assign branch current to all branches.
ThearrowisdrawnfromV
1towardsV
2becausecurrent
alwaysflowsfromhigherpotentialtothelowerpotential.

74. ARASU
ENGINEERING
COLLEGE
74Department of Electrical and Electronics Engineering
3.Expressthebranchcurrentsintermsofthenode
potentials.
4.Writethecurrentequationateachofthe(n-1)unknown
nodes.
5.Substitutethecurrentexpressions(Step-3)intothe
currentequations(Step-4),whichthenbecomeasetof
simultaneousequationsofunknownnodevoltages.
6.Solvefortheunknownvoltagesandthebranch
currents.

75. ARASU
ENGINEERING
COLLEGE
75Department of Electrical and Electronics Engineering
Inmanycircuitsthereferencenodeismostconveniently
selectedasacommonterminalorthegroundterminal.The
abovecircuitdiagramconsistsofthreenodes.Itispossibleto
write(n-1)equationsforthecircuithaving‘n’nodes.Applying
KCLatnode1gives,

76. ARASU
ENGINEERING
COLLEGE
76Department of Electrical and Electronics Engineering

77. ARASU
ENGINEERING
COLLEGE
77Department of Electrical and Electronics Engineering

78. ARASU
ENGINEERING
COLLEGE
78Department of Electrical and Electronics Engineering
Thegeneralizednodeequationscanbewrittenas
[G][V]=[I]……………..3
wherethesquarematrixGiscalledthenode
conductancematrix,Visthecolumnmatrixofthenode
voltageswithrespecttothereferencenodeandIisthe
columnmatrixofinputcurrents.

79. ARASU
ENGINEERING
COLLEGE
79Department of Electrical and Electronics Engineering
NodalAnalysisbyInspectionMethod:
Step1:First,convertallthevoltagesourcestoequivalent
currentsources.
Step2:Theconductancesofallbranchesconnectedto
node1areaddedanddenotedbyG11.G11iscalledthe
selfconductanceofnode1.
Step3:Alltheconductancesconnectedtonodes1and2
areaddedanddenotedbyG12.G12iscalledmutual
conductanceofnodes1and2.ThisG12iswrittenwith
negativesign.Ifnoconductanceisconnectedbetween
nodes1and2thenG12=0,G12=G21.

80. ARASU
ENGINEERING
COLLEGE
80Department of Electrical and Electronics Engineering
Step4:I1denotesthevalueofthecurrentsourcetonode
1andiswrittenontherighthandsideoftheequation.
ThesignofI1ispositiveifitisflowingtowardsnode1,
otherwiseitisnegative.
Ifnocurrentsourceisconnectedtonode1,thenI1=0.
Forexample,considerthecircuitgivenbelow,
Firstconvertvoltagesourcesintoequivalentcurrent
sources.

81. ARASU
ENGINEERING
COLLEGE
81Department of Electrical and Electronics Engineering
The circuit consists of two nodes 1 and 2 and common (ref)
node.

82. ARASU
ENGINEERING
COLLEGE
82Department of Electrical and Electronics Engineering
By Solving this matrix, we can find out nodal voltages V
1and
V
2.

83. Problem1:Usemesh-currentanalysistodeterminethe
currentflowinginthe1Ωresistanceofthed.c.circuit
showninfig.
ARASU
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83Department of Electrical and Electronics Engineering

84. ARASU
ENGINEERING
COLLEGE
84Department of Electrical and Electronics Engineering
ThemeshcurrentsI
1,I
2andI
3areshowninFigure
UsingKirchhoff’svoltagelaw:
Forloop1,(3+5)I
1−5I
2=4……………………………… (1)
Forloop2,(4+1+6+5)I
2−(5)I
1−(1)I
3=0……..(2)
Forloop3,(1+8)I
3−(1)I
2=-5…………………………..(3)
Thus
8I
1−5I
2=4
−5I
1+16I
2−I
3=0
−I
2+9I
3=-5
BysolvingtheaboveequationsbyCramer’srule,weget
I
1=0.595A;I
2=0.152A;I
3=-0.539A

85. ARASU
ENGINEERING
COLLEGE
85Department of Electrical and Electronics Engineering
Problem2:Writethemeshequationsandsolveforthe
currentsI
1,andI
2.
Mesh 1 4I
1+ 6(I
1–I
2) = 10 –2 ................(1)
Mesh 2 6(I
2–I
1) + 2I
2+ 7I
2= 2 + 20 .............(2)

86. ARASU
ENGINEERING
COLLEGE
86Department of Electrical and Electronics Engineering
Thepreviousequationscanbewritteninmatrixformas:
SolvingtheaboveequationusingCramer’srule,weget
I
1=2.2105A
I
2=2.3509A

87. ARASU
ENGINEERING
COLLEGE
87Department of Electrical and Electronics Engineering
Problem:3
FindthecurrentinacircuitusingKirchhoff'svoltage
law.
80=20(I)+10(I)
80=30(I)
I=80/30=2.66A

88. Example 5
Write the mesh equations for the circuit shown in the
figure and solve for the current in the 12 Ωresistor.
ARASU
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88Department of Electrical and Electronics Engineering

89. ARASU
ENGINEERING
COLLEGE
89Department of Electrical and Electronics Engineering
Solution: The loop currents are assumed as shown. It is
required to find i
3. By inspection we can write,

90. ARASU
ENGINEERING
COLLEGE
90Department of Electrical and Electronics Engineering
(—) indicates that the assumed direction of I
3is not the
actual direction. The actual direction of I
3is anti-
clockwise. i. e., the current through 12 Ωis 1O A flowing
from the right to the left terminal of the resistor.

91. Example6
Writeandsolvetheequationsforthemeshcurrentsin
thenetworkshowninfigure.
ARASU
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91Department of Electrical and Electronics Engineering

92. •For solving the problems on the mesh current (loop
current) method, the preferable method is to convert the
practical current sources into practical voltage sources.
The original circuit is redrawn as in the figure below after
transforming the current sources into voltage sources.
ARASU
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92Department of Electrical and Electronics Engineering

93. ARASU
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COLLEGE
93Department of Electrical and Electronics Engineering

94. Example 7
Apply mesh current method and determine currents
through the resistors of the network shown in fig.
ARASU
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94Department of Electrical and Electronics Engineering

95. ARASU
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95Department of Electrical and Electronics Engineering

96. ARASU
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COLLEGE
96Department of Electrical and Electronics Engineering

97. ARASU
ENGINEERING
COLLEGE
97Department of Electrical and Electronics Engineering
Example8
Frame the nodal equations of the network of figure and
hence find the difference of potential between nodes 2
and 4.

98. ARASU
ENGINEERING
COLLEGE
98Department of Electrical and Electronics Engineering
Let the node 4 be the reference i.e. zero potential node. Let
V
1,V
2,V
3 be the node voltages at the nodes 1, 2 and 3 respectively.
By inspection,

99. ARASU
ENGINEERING
COLLEGE
99Department of Electrical and Electronics Engineering
It is required to find the difference of potential between
nodes 2 and 4. i.e., V
2—V
4 = V
2—0 = V
2. So, it is enough if
we know the value of V
2.

100. ARASU
ENGINEERING
COLLEGE
100Department of Electrical and Electronics Engineering
Example 9
Compute the voltage at nodes A and B in the circuit of
figure shown below.

101. ARASU
ENGINEERING
COLLEGE
101Department of Electrical and Electronics Engineering
First converting the voltage source into its equivalent current source
and re-drawing the circuit the following figure is obtained.

102. There are two nodes A and B other than the reference node.
Let V
A and V
B be the voltages of the nodes A and B respectively
with respect to reference node. By inspection,
ARASU
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102Department of Electrical and Electronics Engineering

103. ARASU
ENGINEERING
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103Department of Electrical and Electronics Engineering
EE8251&CIRCUIT THEORY

104. Example10
Use nodal voltage method and hence find the power
dissipated in the 10 ohms resistor on the circuit shown in
figure.
ARASU
ENGINEERING
COLLEGE
104Department of Electrical and Electronics Engineering

105. •Taking the node 4 as reference, and converting the voltage
source into current source, the above network is re-drawn as
below:
ARASU
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105Department of Electrical and Electronics Engineering

106. ARASU
ENGINEERING
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106Department of Electrical and Electronics Engineering

107. ARASU
ENGINEERING
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107Department of Electrical and Electronics Engineering

108. ARASU
ENGINEERING
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108Department of Electrical and Electronics Engineering

109. ARASU
ENGINEERING
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109Department of Electrical and Electronics Engineering
Example 11
In the network shown in figure, find the node voltages
V
1and V
2. Find also the current supplied by the source.

110. •The above circuit is re-drawn as below after converting the
voltage source into its equivalent current source.
ARASU
ENGINEERING
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110Department of Electrical and Electronics Engineering

111. ARASU
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111Department of Electrical and Electronics Engineering

112. ARASU
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112Department of Electrical and Electronics Engineering

113. ARASU
ENGINEERING
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113Department of Electrical and Electronics Engineering

114. ARASU
ENGINEERING
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114Department of Electrical and Electronics Engineering

115. ARASU
ENGINEERING
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115Department of Electrical and Electronics Engineering
•LOOP ANALYSIS (Dependent Sources or Controlled
Sources)
Example 1
Determine the current through 4Ωresistor of the
network shown.

116. ARASU
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116Department of Electrical and Electronics Engineering

117. ARASU
ENGINEERING
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117Department of Electrical and Electronics Engineering

118. ARASU
ENGINEERING
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118Department of Electrical and Electronics Engineering

119. Example
Determine the voltage across each conductance
of the circuit shown in the figure.
ARASU
ENGINEERING
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119Department of Electrical and Electronics Engineering
Nodal analysis (Dependent Sources or Controlled
Sources)

120. ARASU
ENGINEERING
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120Department of Electrical and Electronics Engineering

121. ARASU
ENGINEERING
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121Department of Electrical and Electronics Engineering

122. ARASU
ENGINEERING
COLLEGE
122Department of Electrical and Electronics Engineering
DCCIRCUITS:
ADCcircuit(DirectCurrentcircuit)isanelectrical
circuitthatconsistsofanycombinationofconstant
voltagesources,constantcurrentsources,andresistors.
Inthiscase,thecircuitvoltagesandcurrentsare
constant,i.e.,independentoftime.
Moretechnically,aDCcircuithasnomemory.
Thatis,aparticularcircuitvoltageorcurrentdoesnot
dependonthepastvalueofanycircuitvoltageor
current.
Thisimpliesthatthesystemofequationsthat
representaDCcircuitdonotinvolveintegralsor
derivatives.

123. ARASU
ENGINEERING
COLLEGE
123Department of Electrical and Electronics Engineering
AC CIRCUITS:
FundamentalsofAC:
Analternatingcurrent(AC)isanelectricalcurrent,wherethe
magnitudeofthecurrentvariesinacyclicalform,asopposed
todirectcurrent,wherethepolarityofthecurrentstays
constant.
TheusualwaveformofanACcircuitisgenerallythatofa
sinewave,astheseresultsinthemostefficienttransmission
ofenergy.Howeverincertainapplicationsdifferentwaveforms
areused,suchastriangularorsquarewaves.
Usedgenerically,ACreferstotheforminwhichelectricityis
deliveredtobusinessesandresidences.However,audioand
radiosignalscarriedonelectricalwirearealsoexamplesof
alternatingcurrent.

124. ARASU
ENGINEERING
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124Department of Electrical and Electronics Engineering
Sinusoidal alternating quantity
Analternatingquantitywhichvariesaccordingtothesineof
theangleθisknownassinusoidalalternatingquantity.
Allovertheworldsinusoidalvoltagesandcurrentsare
selectedforgenerationofelectricpower.
Thefollowingarethereasons:
a)Thesinusoidalvoltageandcurrentsproducelowironand
copperlossesinACrotatingmachinesandtransformers.
b)Sinusoidalvoltageandcurrentwillofferlessinterferenceto
nearbytelephonelines.
c)Theyproducelessdisturbanceintheelectricalcircuit.

125. ARASU
ENGINEERING
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125Department of Electrical and Electronics Engineering
Cycle:
Onecompletesetofpositiveandnegativevaluesofan
alternatingquantityiscalledcycle.
Timeperiod:
Thetimetakentocompleteonecompletecycle.
Frequency:
Thenumberofcyclesmadebyanalternatingquantityper
secondiscalledfrequency.TheunitoffrequencyisHertz(Hz)
AmplitudeorPeakvalue:
Themaximumpositiveornegativevalueofanalternating
quantityiscalledamplitudeorpeakvalue.Inthevoltage
waveformthepeakvalueisEm.Itoccurswhentheangleis
π/2(positivecycle)and3π/2(negativecycle).
Averagevalue:
Thisistheaverageofinstantaneousvaluesofan
alternatingquantityoveronecompletecycleofthewave.

126. Sinusoidal Voltage and current equations
A sinusoidal voltage and current can be represented as
E=E
mSinωt or V=V
mSinωt and
I=I
mSinωt
•Where E or V and I are instantaneous values and E
m or V
m
and I
mare the peak values.
•ω = angular frequency in radians per second.
V=V
mSin (2πf)t and
I=I
mSin(2πf)t
Where, ω = 2πf
f=ω / 2π= 1/ T
Where, f is the frequency in cycles per second(Hz)
ARASU
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126Department of Electrical and Electronics Engineering

127. ARASU
ENGINEERING
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127Department of Electrical and Electronics Engineering
Waveform:
Itisthegraphbetweenthealternatingquantity(only
instantaneousvalue)asordinateandtime.
Thealternatingquantitymaybeeithervoltageorcurrentor
flux.

128. ARASU
ENGINEERING
COLLEGE
128Department of Electrical and Electronics Engineering

129. ARASU
ENGINEERING
COLLEGE
129Department of Electrical and Electronics Engineering
R.M.S (Root mean square) value or effective value:
Thesteadycurrentwhichwhenflowingthrougha
givenresistorforagiventimeproducesthesameamountof
heatasproducedbythealternatingcurrentwhenflowing
throughthesameresistorforthesametimeiscalledR.M.Sor
effectivevalueofthealternatingcurrent.

130. ARASU
ENGINEERING
COLLEGE
130Department of Electrical and Electronics Engineering
To find the R.M.S. value of a Sinusoidal alternating
Quantity

131. ARASU
ENGINEERING
COLLEGE
131Department of Electrical and Electronics Engineering
For a Sinusoidal quantity, R.M.S value = 0.707 x maximum
value

132. ARASU
ENGINEERING
COLLEGE
132Department of Electrical and Electronics Engineering
Tofindtheaverage
valueofsinusoidally
varyingquantity.
i=I
mSinθ
Itisasymmetricalwave
form.
To calculatethe
averagevalue,onlyhalf
cycle must be
considered.

133. ARASU
ENGINEERING
COLLEGE
133Department of Electrical and Electronics Engineering
For a sinusoidally varying quantity, average value
= 0.637 x maximum value

134. •Form factor (K
f) : It is the ratio of rms value to average value.
Mathematically,
Form factor K
f= RMS value / Average value
•Peak factor (K
a) : It is the ratio of peak or maximum value to
r.m.s value.
•Mathematically
Peak factor K
a= Peak value / RMS Value
Peak factor is also known as amplitude factor or crest factor.
ARASU
ENGINEERING
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134Department of Electrical and Electronics Engineering

135. ARASU
ENGINEERING
COLLEGE
135Department of Electrical and Electronics Engineering
ANALYSIS OF AC CIRCUIT
It is assumed that the sinusoidally varying voltage is connected
to
(a) purely resistive
(b) purely inductive and
(c) purely capacitive circuits.
In each case it is required to find the following quantities
1. The expression for the instantaneous current.
2. The polar form of voltage and current
3. The phasor diagram.
4. Ratio of voltage to current
5. Average power
6. Phase angle between voltage and current and hence
power factor.

136. ARASU
ENGINEERING
COLLEGE
136Department of Electrical and Electronics Engineering
Purely Resistive Circuit Excited by a Sinusoidally Varying Voltage

137. ARASU
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COLLEGE
137Department of Electrical and Electronics Engineering

138. ARASU
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138Department of Electrical and Electronics Engineering

139. ARASU
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139Department of Electrical and Electronics Engineering

140. ARASU
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140Department of Electrical and Electronics Engineering

141. ARASU
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141Department of Electrical and Electronics Engineering

142. ARASU
ENGINEERING
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142Department of Electrical and Electronics Engineering

143. Circuit Consisting of Pure Inductance Excited by a
Sinusoidally Varying Voltage
ARASU
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143Department of Electrical and Electronics Engineering

144. ARASU
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144Department of Electrical and Electronics Engineering

145. ARASU
ENGINEERING
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145Department of Electrical and Electronics Engineering
i

146. ARASU
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146Department of Electrical and Electronics Engineering

147. ARASU
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147Department of Electrical and Electronics Engineering

148. ARASU
ENGINEERING
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148Department of Electrical and Electronics Engineering

149. ARASU
ENGINEERING
COLLEGE
149Department of Electrical and Electronics Engineering

150. ARASU
ENGINEERING
COLLEGE
150Department of Electrical and Electronics Engineering
Purely Capacitive Circuit Excited by Sinusoidally Varying
Voltage

151. ARASU
ENGINEERING
COLLEGE
151Department of Electrical and Electronics Engineering

152. ARASU
ENGINEERING
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152Department of Electrical and Electronics Engineering

153. ARASU
ENGINEERING
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153Department of Electrical and Electronics Engineering

154. ARASU
ENGINEERING
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154Department of Electrical and Electronics Engineering

155. ARASU
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155Department of Electrical and Electronics Engineering

156. ARASU
ENGINEERING
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156Department of Electrical and Electronics Engineering
Voltage division

157. ARASU
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157Department of Electrical and Electronics Engineering

158. CURRENT DIVISION
ARASU
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158Department of Electrical and Electronics Engineering

159. ARASU
ENGINEERING
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159Department of Electrical and Electronics Engineering

160. ARASU
ENGINEERING
COLLEGE
160Department of Electrical and Electronics Engineering

161. ARASU
ENGINEERING
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161Department of Electrical and Electronics Engineering
EE8251-CIRCUIT THEORY

162. ARASU
ENGINEERING
COLLEGE
162Department of Electrical and Electronics Engineering
Source Transformation
Voltage Source to Current Source
Ifavoltagesourcehasaresistanceconnected
series,itcanbetransformedintoanequivalentcurrent
sourcewitharesistanceconnectedinparallel

163. ARASU
ENGINEERING
COLLEGE
163Department of Electrical and Electronics Engineering
Source Transformation
Current Source to Voltage Source
If a current source has a resistance connected in parallel, it can
be transformed into an equivalent voltage source with a
resistance connected in series.

164. ARASU
ENGINEERING
COLLEGE
164Department of Electrical and Electronics Engineering
1.Convert the given current source into a voltage source

165. ARASU
ENGINEERING
COLLEGE
165Department of Electrical and Electronics Engineering
2. Convert the given voltage source into a current
source for the circuit given below.

166. Star(Y) and Delta (Δ) transformation or T and π:
ARASU
ENGINEERING
COLLEGE
166Department of Electrical and Electronics Engineering
BetweenthreeterminalsA,Band
C,wecanconnectthreeresistances
(impedances)intwofashions.
Oneiscalledstarconnection(Y),
theotherisDeltaconnection(Δ).
InstarConnection,oneendof
eachresistanceisconnectedata
pointcalledstarpointandthe
remainingthreeterminalsare
connectedtoA,BandC.
ItiscalledstarconnectionSinceit
appearslikeastar.

167. InDeltaconnection,thethree
resistances(impedances)areconnected
endtoend,soastoformDeltashape.
Insomeelectricalcircuits,calculations
becomesimplerifitisstarconnected.
Insomeotherproblems,Delta
connectionmayenableustosolveit
easily.
Henceitisnecessarytoconvertgiven
starconnectionresistancesintoits
equivalentDeltaandvice-versa.
ARASU
ENGINEERING
COLLEGE
167Department of Electrical and Electronics Engineering

168. GivenResistancesinDelta-ToFindResistancein
EquivalentStar
ARASU
ENGINEERING
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168Department of Electrical and Electronics Engineering

169. ARASU
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169Department of Electrical and Electronics Engineering

170. ARASU
ENGINEERING
COLLEGE
170Department of Electrical and Electronics Engineering
Given Resistances in Star To Find Resistance in
Equivalent Delta

171. ARASU
ENGINEERING
COLLEGE
171Department of Electrical and Electronics Engineering

172. ARASU
ENGINEERING
COLLEGE
172Department of Electrical and Electronics Engineering

173. ARASU
ENGINEERING
COLLEGE
173Department of Electrical and Electronics Engineering
EE8251-CIRCUIT THEORY

174. ARASU
ENGINEERING
COLLEGE
174Department of Electrical and Electronics Engineering
Problem:Forthepassivecircuitconsistingofresistances
(inohms)asshown,calculatetheequivalentresistance
betweentheterminalsAandD.

175. ARASU
ENGINEERING
COLLEGE
175Department of Electrical and Electronics Engineering
Solution:BetweentheterminalsAandDinthecircuit
given,thecombinationisneitherseriesnorparallel.
Hencesimplificationisnotpossibleasitis.So.wecan
convertthedeltabetweenA,BandCintoequivalentstar.
Asaresultthecircuitbecomesasdrawnbelow:

176. ARASU
ENGINEERING
COLLEGE
176Department of Electrical and Electronics Engineering
Now,itisseriesparallelcombinationofresistances.
HencethetotalresistancebetweenAandDisequalto,

177. ARASU
ENGINEERING
COLLEGE
177Department of Electrical and Electronics Engineering
Problem:Forthenetworkshowninthefigurebelow,findthe
equivalentresistancebetweentheterminalsBandC.

178. ARASU
ENGINEERING
COLLEGE
178Department of Electrical and Electronics Engineering
Solution:Thegivencombinationofresistancesbetweenthe
terminalsBandCisneitherseriescombinationnorparallel
combination.
IfthestarConnectionbetweenA,BandCisconverted
intoequivalentdelta,wewillhaveaknowncombination
whichcanbesimplifiedbySeriesparallelsimplification.

179. ARASU
ENGINEERING
COLLEGE
179Department of Electrical and Electronics Engineering
The given figure is re-drawn after replacing the star by its
equivalent Delta.
TheresistancebetweenBandCterminalsisequivalentto
theparallelcombinationof3and1.5Ohms.

180. Super position theorem
With the help of this theorem, we can find the current
through or the voltage across a given element in a linear
circuit consisting of two or more sources.
The statement is as follows:
“In a linear circuit containing more than one source, the
current that flows at any point or the voltage that exists
between any two points is the algebraic sum of the
currents or the voltages that would have been produced
by each source taken separately with all other sources
removed.”
ARASU
ENGINEERING
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180Department of Electrical and Electronics Engineering

181. Note:
1. Removal of an ideal voltage source means short
circuiting.
2. Removal of an ideal current source means replacing it
by an open circuit.
3. Removal of practical voltage source means replacing it
by an internal resistance.
ARASU
ENGINEERING
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181Department of Electrical and Electronics Engineering

182. Example
Compute the current through 23 ohm resistor of the
figure below by using superposition theorem.
ARASU
ENGINEERING
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182Department of Electrical and Electronics Engineering

183. Solution: Step 1 : Allow only the voltage source to act. The
corresponding circuit is as below:
ARASU
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183Department of Electrical and Electronics Engineering
EE8251-CIRCUIT THEORY

184. ARASU
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184Department of Electrical and Electronics Engineering

185. Step 2: Allow only the current source to act. The circuit becomes
as
ARASU
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185Department of Electrical and Electronics Engineering

186. ARASU
ENGINEERING
COLLEGE
186Department of Electrical and Electronics Engineering

187. ARASU
ENGINEERING
COLLEGE
187Department of Electrical and Electronics Engineering
Example
Using superposition theorem or otherwise, obtain the
current in EA in figure below:

188. ARASU
ENGINEERING
COLLEGE
188Department of Electrical and Electronics Engineering
EE8251-CIRCUIT THEORY
•Step 1 : First allow only 1V source to act. Let the current in EA
be I’. The relevant circuit diagram is as below:

189. ARASU
ENGINEERING
COLLEGE
189Department of Electrical and Electronics Engineering
•I’ can be found by applying mesh current method or
converting delta to star and then applying Ohm’s law.
The above circuit becomes as drawn below after
converting delta connection between A, B and C into its
equivalent star.

190. ARASU
ENGINEERING
COLLEGE
190Department of Electrical and Electronics Engineering

191. ARASU
ENGINEERING
COLLEGE
191Department of Electrical and Electronics Engineering
•Step 2: When only 2V source is acting and the
other source short circuited, the circuit
becomes as shown in fig. Now let the current
in EA be I”.

192. ARASU
ENGINEERING
COLLEGE
192Department of Electrical and Electronics Engineering

193. ARASU
ENGINEERING
COLLEGE
193Department of Electrical and Electronics Engineering

194. Thevenin’stheorem:
Alinearnetworkconsistingofanumberofvoltage
sourcesandresistancescanbereplacedbyan
equivalentnetworkhavingasinglevoltagesourcecalled
Thevenin’svoltage(V
Th)andasingleresistancecalled
Thevenin’sresistance(R
Th)
ARASU
ENGINEERING
COLLEGE
194Department of Electrical and Electronics Engineering

195. ARASU
ENGINEERING
COLLEGE
195Department of Electrical and Electronics Engineering
Explanation:

196. To find V
Th:
The load resistance R
Lis removed.
The current I in the circuit is I= E / (R
1 + R
2)
. The voltage across AB = Thevenin’svoltage V
Th.
V
Th= I R
2 V
Th= E R
2/ (R
1 + R
2)
ARASU
ENGINEERING
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196Department of Electrical and Electronics Engineering

197. ARASU
ENGINEERING
COLLEGE
197Department of Electrical and Electronics Engineering
•To find R
Th:
•TheloadresistanceR
Lisremoved.Thecellis
disconnectedandthewiresareshortasshown.
•TheeffectiveresistanceacrossAB=Thevenin’s
resistanceR
Th.
•R
1isparalleltoR
2andthiscombinationinserieswithR
3]

198. Theequivalentcircuitisgivenby
ARASU
ENGINEERING
COLLEGE
198Department of Electrical and Electronics Engineering
EE8251-CIRCUIT THEORY

199. Problem:UseThevenin’stheoremtofindthecurrent
flowinginthe10Ωresistorforthecircuitshownin
figure.
TofindVTh:
TheloadresistanceR
Lis
removed.
The10Ωresistanceisremoved
fromthecircuitasshownin
figure
ARASU
ENGINEERING
COLLEGE
199Department of Electrical and Electronics Engineering

200. •Thereisnocurrentflowinginthe5Ωresistorand
currentI
1isgivenby:
I
1=10/(R
1+R
2)
=10/(2+8)
=1A
V
Th=IR
2=1X8=8V
TofindR
Th:
TheloadresistanceR
Lisremoved.Thecellis
disconnectedandthewiresareshortasshown.
ARASU
ENGINEERING
COLLEGE
200Department of Electrical and Electronics Engineering

201. Resistance, R
Th= R
3+ (R
1R
2 / R
1+ R
2)
=5+ (2×8/2+8)
= 5 + 1.6 = 6.6 Ω
ARASU
ENGINEERING
COLLEGE
201Department of Electrical and Electronics Engineering

202. ARASU
ENGINEERING
COLLEGE
202Department of Electrical and Electronics Engineering
Thevenin’s equivalentcircuitisshowninfig.
Load Current IL= E / (R+R
Th) = 8/ (10+6.6)
= 8/16.6=0.482A

203. ARASU
ENGINEERING
COLLEGE
203Department of Electrical and Electronics Engineering
•Problem
•Find the Thevenin ‘s equivalent for the network of the
figure between a and b.

204. ARASU
ENGINEERING
COLLEGE
204Department of Electrical and Electronics Engineering
•Solution: The network given is a combination of
voltage and current sources. By converting voltage
source into current source and vice versa, wherever
necessary and simplifying we can obtain the required
network.
•Step 1: Converting or transforming the voltage source
of 1OV in series with resistance 3Ωinto equivalent
current source, the following circuit is obtained.

205. ARASU
ENGINEERING
COLLEGE
205Department of Electrical and Electronics Engineering
Step 2: Replacing the 2 current sources in parallel by its
equivalent current source, we get the following circuit.

206. ARASU
ENGINEERING
COLLEGE
206Department of Electrical and Electronics Engineering
Step 3: Transforming the 2 current sources in series by
their equivalent voltage sources we get the following
network.

207. ARASU
ENGINEERING
COLLEGE
207Department of Electrical and Electronics Engineering
Step 4: Transforming the voltage sources which are
parallel in the above circuit, into their equivalent current
sources we get the following network.

208. ARASU
ENGINEERING
COLLEGE
208Department of Electrical and Electronics Engineering
Step 5: Converting the current source into equivalent
voltage source, we get the Thevenin’s equivalent circuit
as shown below:

209. ARASU
ENGINEERING
COLLEGE
209Department of Electrical and Electronics Engineering

210. MaximumPowerTransferTheorem
Thistheoremdescribestheconditionformaximum
powertransferfromanactivenetworktoanexternal
loadresistance.
Itstatesthatinalinear,active,bilateralDCnetwork,
themaximumpowerwillbetransferredfromsourceto
theloadwhentheexternalloadresistanceequalsto
theinternalresistanceofthesource.
ARASU
ENGINEERING
COLLEGE
210Department of Electrical and Electronics Engineering

211. ARASU
ENGINEERING
COLLEGE
211Department of Electrical and Electronics Engineering
Explanation of Maximum Power Transfer Theorem:
Basically,theconditionatwhichmaximumpower
transfercanbeobtainedbyderivinganexpressionof
powerabsorbedbytheloadusingmeshornodalcurrent
techniquesandthenfindingitsderivativewithrespectto
theloadresistance.

212. ARASU
ENGINEERING
COLLEGE
212Department of Electrical and Electronics Engineering
From the above circuit, the current flowing through the
load, ‘I’ is given as

213. IntheaboveequationR
Lisavariable,thereforethe
conditionformaximumpowerdeliveredtotheloadis
determinedbydifferentiatingloadpowerwithrespect
totheloadresistanceandequatingittozero.
ARASU
ENGINEERING
COLLEGE
213Department of Electrical and Electronics Engineering

214. ARASU
ENGINEERING
COLLEGE
214Department of Electrical and Electronics Engineering
Thisistheconditionformaximumpowertransfer,which
statesthatpowerdeliveredtotheloadismaximum,
whentheloadresistanceR
LmatcheswithThevenin’s
resistanceR
THofthenetwork.
Underthiscondition,powertransfertotheloadis

215. ForagivenvaluestheThevenin’svoltageandThevenin’s
resistance,thevariationofpowerdeliveredtotheloadwith
varyingloadresistanceisshowninbelowfigure.
ARASU
ENGINEERING
COLLEGE
215Department of Electrical and Electronics Engineering

216. ARASU
ENGINEERING
COLLEGE
216Department of Electrical and Electronics Engineering
Problem:Forthecircuitshownbelowdeterminethe
valueofloadresistance,R
Lforwhichmaximumpower
willtransferfromsourcetoload.

217. Now,thegivencircuitcanbefurthersimplifiedby
convertingthecurrentsourceintoequivalentvoltage
sourceasfollows.
ARASU
ENGINEERING
COLLEGE
217Department of Electrical and Electronics Engineering

218. ARASU
ENGINEERING
COLLEGE
218Department of Electrical and Electronics Engineering
•WeneedtofindtheThevenin’sequivalentvoltageVth
andThevenin’sequivalentresistanceRthacrosstheload
terminalsinordertogettheconditionformaximum
powertransfer.Bydisconnectingtheloadresistance,the
open-circuitvoltageacrosstheloadterminalscanbe
calculatedas;

219. By applyingKirchhoff’s voltage law, we get
12 –6I –2I –16 = 0
–8I = 4
I = –0.5 A
The open-circuit voltage across the terminals A and B,
V
AB= 16 –2 ×0.5
= 15 V
Thevenin’sequivalentresistanceacrosstheterminalsAand
Bisobtainedbyshort-circuitingthevoltagesourcesas
showninthefigure.
ARASU
ENGINEERING
COLLEGE
219Department of Electrical and Electronics Engineering

220. ARASU
ENGINEERING
COLLEGE
220Department of Electrical and Electronics Engineering
Req=(6×2)/(6+2)
=1.5Ω
Sothemaximumpowerwilltransferredtotheloadwhen
R
L=1.5ohm.
Currentthroughthecircuit,I=15/(1.5+1.5)
=5A
Therefore,themaximumpower=5
2
×1.5=37.5W

221. ARASU
ENGINEERING
COLLEGE
221Department of Electrical and Electronics Engineering
Norton’s Theorem
Norton’s theorem is the dual of the Thevenin’s Theorem:
Statement: “Any linear active network with output
terminals A, B as shown in the figure can be
replaced by a single current source. I
SC(I
N) in parallel with a
single impedance Z
Th(Z
n) = R
Th (R
N)
I
scis the current through the terminals AB of the active
network when shorted.
Z
This the Thevenin’s impedance.
The current through an impedance connected to the
terminals of the Norton’s equivalent circuit must have the
same direction as the current through the same impedance
connected to the original active network.

222. ARASU
ENGINEERING
COLLEGE
222Department of Electrical and Electronics Engineering

223. ARASU
ENGINEERING
COLLEGE
223Department of Electrical and Electronics Engineering
EE8251-CIRCUIT THEORY
After evaluating the values of I
Scand R
Th, the current
through R
Lconnected across A and B can be calculated by
applying division of current formula which is expressed
below:

224. ARASU
ENGINEERING
COLLEGE
224Department of Electrical and Electronics Engineering
Example
Determine the voltage across 200Ωresistor in circuit by
Norton ‘s theorem.

225. ARASU
ENGINEERING
COLLEGE
225Department of Electrical and Electronics Engineering
Solution :
Step 1: To find the short circuit current I
SC .Replace the
200 Ωby short circuit. The current through short
circuited AB is I
SC. Refer the following figure.

226. ARASU
ENGINEERING
COLLEGE
226Department of Electrical and Electronics Engineering
The voltage source will drive a current of 10/50= 0.2 A,
which flows through short circuited AB only. Similarly the
current of 1A flows through 20 and then through short-
circuited AB only. 0.2 A will not flow through 2OΩand 1A will
not flow through 50Ω.It is because of short-circuit. So, I
sc=
0.2+1=1.2A.
Step 2: To find R
Th, : From the given circuit, disconnect R
L
= 200 Ωbetween A and B and also kill the sources. The
resultant circuit becomes as below:

227. ARASU
ENGINEERING
COLLEGE
227Department of Electrical and Electronics Engineering

228. ARASU
ENGINEERING
COLLEGE
228Department of Electrical and Electronics Engineering
Step 3: Drawing Norton’s equivalent circuit and showing
the load resistance R
L, we get the following circuit:

229. ARASU
ENGINEERING
COLLEGE
229Department of Electrical and Electronics Engineering
Reciprocity Theorem
Statement
In a linear, bilateral network a voltage source V volts in
a branch gives rise to a current I, in another branch. If
V is applied in the second branch, the current in the
first branch will be I.‘This V/I is called transfer
impedance or resistance.

230. ARASU
ENGINEERING
COLLEGE
230Department of Electrical and Electronics Engineering
On changing the voltage source from branch 1 to
branch 2, the current I in the branch 2 appears in
branch 1.

231. ARASU
ENGINEERING
COLLEGE
231Department of Electrical and Electronics Engineering
Example
For the circuit shown in the figure below, find I
3and
verify reciprocity theorem.

232. ARASU
ENGINEERING
COLLEGE
232Department of Electrical and Electronics Engineering
Solution : Step 1: By inspection putting in the matrix
form, we get

233. ARASU
ENGINEERING
COLLEGE
233Department of Electrical and Electronics Engineering
Step 2: Transferring the battery, to the branch with
1.375 Ωresistor, the following circuit is obtained. The
current through 1Ωis taken as I
3. According to
reciprocity theorem this I
3= 2A.
Let us calculate the value of I
3in figure below:

234. ARASU
ENGINEERING
COLLEGE
234Department of Electrical and Electronics Engineering
Thus reciprocity theorem is verified.

235. ARASU
ENGINEERING
COLLEGE
235Department of Electrical and Electronics Engineering
GENERATION OF 3-PHASE VOLTAGE
 The 3-phase voltage can be produced in a
stationary armature with rotating field or in a rotating
armature with a stationary field.
3-phase voltages are generated in three separate but
identical sets of windings or coils which are displaced by
120 electrical degrees in the armature.
Hence, the voltages generated in them are 120 degree
apart in time phase. This arrangement is shown in Fig.
Here, RR’ constitutes one coil (R-phase) ; YY’ another
coil (Y-phase) and BB’ constitutes the third coil (B phase).
The field magnets are assumed to be rotating in clock-
wise direction.

236. ARASU
ENGINEERING
COLLEGE
236Department of Electrical and Electronics Engineering

237. ARASU
ENGINEERING
COLLEGE
237Department of Electrical and Electronics Engineering
Three phase wave form

238. ARASU
ENGINEERING
COLLEGE
238Department of Electrical and Electronics Engineering
Representation of three phase voltage in polar and rectangular form

239. ARASU
ENGINEERING
COLLEGE
239Department of Electrical and Electronics Engineering

240. ARASU
ENGINEERING
COLLEGE
240Department of Electrical and Electronics Engineering

241. ARASU
ENGINEERING
COLLEGE
241Department of Electrical and Electronics Engineering
There are six types of wires
Vulcanized Indian rubber wire(VIR)
Tough rubber sheathed wire(TRS)
Poly vinyl chloride wire(PVC)
Lead alloy sheathed wire
Weather proof wires
Flexible wires

243. ARASU
ENGINEERING
COLLEGE
243Department of Electrical and Electronics Engineering

244. ARASU
ENGINEERING
COLLEGE
244Department of Electrical and Electronics Engineering
General construction of cable

245. ARASU
ENGINEERING
COLLEGE
245Department of Electrical and Electronics Engineering
Cable Structure

246. Distribution of Electric Energy to
Domestic Consumer

247. •Types of wiring
•Suitability of a particular wiring system for a
given installation
•Corridor and staircase lighting
•Necessity of earthing
•Different types of earthing
•Fuses
•Different types of fuses
•Imp VTU Questions
Wiring

248. Introduction
•A network of wires drawn from the meter board to
the various energy consuming points(lamps,
fans, motors etc) through control and protective
devices for efficient distribution of power is known as
electrical wiring.
•Electrical wiring done in residential and commercial
buildings to provide power for lights, fans, pumps and
other domestic appliances is known as domestic
wiring. There are several wiring systems in practice.
They can be classified into different Types

249. Types of
Wiring:
1.Cleat wiring
2.CTS wiring or TRS wiring or batten wiring
3.Metal sheathed wiring or lead sheathed
wiring
4.Casing and capping
5.Conduit wiring

250. Cleat
wiring
•: In this type of wiring, insulated conductors (usually
VIR, Vulcanized Indian Rubber) are supported on
porcelain or wooden cleats.
•The cleats have two halves one base and the other
cap.
•The cables are placed in the grooves provided in the
base and then the cap is placed. Both are fixed
securely on the walls by 40mm long screws.
•The cleats are easy to erect and are fixed 4.5 –15
cms apart.
•This wiring is suitable for temporary installations
where cost is the mPravaeeinnkumcarr.Cit, Deeprt.iaof EEbE,uSKtIT,not
the appearance.
Bangalore

251. Cleat
wiring

253. Advantages and
Disadvantages:
Advantages:
1.Easy installation
2.Materials can be retrieved for
reuse
3.Flexibility provided for
inspection, modifications and
expansion.
4.Relatively economical
5.Skilled manpower not
required.
Disadvantages:
1.Appearance is not good
2.Open system of wiring

254. CTS ( Cable Tyre Sheathed) / TRS ( Tough
Rubber Sheathed ) / Batten wiring:
•In this wiring system, wires sheathed in tough
rubber are used which are quite flexible. They are
clipped on wooden battens with brass clips (link
or joint) and fixed on to the walls or ceilings by
flat head screws.
•These cables are moisture and chemical proof.
•They are suitable for damp climate but not
suitable for outdoor use in sunlight.
•TRS wiring is suitable for lighting in low voltage
Bangaloreinstallations.
Praveen kumar .C, Dept. of EEE,SKIT,

255. CTS ( Cable Tyre Sheathed) / TRS ( Tough
Rubber Sheathed ) / Batten wiring:

256. Advantages and
Disadvantages:
Advantages:
1.Easy installation and is
durable
2.Lower risk of short circuit.
3.Cheaper than casing and
capping system of wiring
4.Gives a good appearance
if properly erected.
Disadvantages:
1.Danger of mechanical
injury.
2.Danger of fire hazard.
3.Should not be exposed to

257. Metal Sheathed or Lead Sheathedwiring
•The wiring is similar to that of CTS but the
conductors (two or three) are individually
insulated and covered with a common outer
lead-aluminum alloy sheath.
•The sheath protects the cable against
dampness, atmospheric extremities and
mechanical damages.

258. Metal Sheathed or Lead Sheathedwiring
•The sheath is earthed at every junction to
provide a path to ground for the leakage
current.
•They are fixed by means of metal clips on
wooden battens.
•The wiring system is very expensive. It is
suitable for low voltage installations.

259. Metal Sheathed or Lead Sheathedwiring
Precautions to be taken during
installation
•The clips used to fix the cables on battens
should not react with the sheath
2.Lead sheath should be properly earthed to
prevent shocks due to leakage currents.
3.Cablesshouldnotberunindampplaces
andinareaswherechemicals(mayreactwith
thelead)areused.

260. Advantages and
Disadvantages:
Advantages:
1.Easy installation and
is aesthetic in
appearance.
2.Highly durable
3.Suitable in adverse
climatic conditions
provided the joints are
not exposed
Disadvantages:
1.Requires skilled
labor
2.Very expensive

261. Casing andCapping
•It consists of insulated conductors laid inside
rectangular, teakwood or PVC boxes having
grooves inside it.
•A rectangular strip of wood called capping having
same width as that of casing is fixed over it.
•Both the casing and the capping are screwed
together at every 15 cms.
•Casing is attached to the wall. Two or more wires
of same polarity are drawn through different
grooves.
•The system is suitable for indoor and domestic
Bangaloreinstallations.
Praveen kumar .C, Dept. of EEE,SKIT,

262. Casing andCapping

263. Advantages and
Disadvantages:
Advantages:
1.Cheaper than lead sheathed
and conduit wiring.
2.Provides good isolation as
the conductors are placed
apart reducing the risk of
short circuit.
3.Easily accessible for
inspection and repairs.
4.Since the wires are not
exposed to atmosphere,
insulation is less affected by
dust, dirt and climatic
variations.
Bangalore

264. Conduitwiring
Praveen kumar .C, Dept. of EEE,SKIT,
•In this system PVC (polyvinyl chloride) or VIR cables
are run through metallic or PVC pipes providing good
protection against mechanical injury and fire due to
short circuit.
•They are either embedded inside the walls or
supported over the walls, and are known as concealed
wiring or surface conduit wiring (open conduit)
respectively.
•The conduits are buried inside the walls on wooden
gutties and the wires are drawn through them with fish
(steel) wires.
•The system is best suited for public buildings,
industries and workshop
Bs
an.
galore

266. Conduitwiring

267. Advantages and
Disadvantages:
Advantages:
1.No risk of fire and good protection
against mechanical injury.
2.The lead and return wires can be
carried in the same tube.
3.Earthing and continuity is assured.
4.Waterproof and trouble shooting is
easy.
5.Shock-proof with proper earthing
and bonding
6.Durable and maintenance free
7.Aesthetic in appearance
Disadvantages:
1.Very expensive system of wiring.
2.Requires good skilled workmanship.
3.Erection is quiet complicated and is

268. FACTORS AFFECTING THE CHOICE
OF WIRING SYSTEM
The choice of wiring system for a particular installation
depends on technical factors and economic viability.
1.Durability: Type of wiring selected should conform to
standard specifications, so that it is durable i.e. without
being affected by the weather conditions, fumes etc.
2.Safety:Thewiringmustprovidesafetyagainst
leakage,shockandfirehazardsfortheoperating
personnel.
3.Appearance:Electricalwiringshouldgivean
aestheticappealtotheinteriors.
4.Cost:Itshouldnotbeprohibitivelyexpensive.

269. Cnt.
.
5.Accessibility: The switches and plug points
provided should be easily accessible. There must be
provision for further extension of the wiring system, if
necessary.
6.Maintenance Cost: The maintenance cost should
be a minimum
7.Mechanical safety: The wiring must be protected
against any mechanical damage

270. Two-way and Three-way Control
of Lamps:
•The domestic lighting circuits are quite simple
and they are usually controlled from one point.
But in certain cases it might be necessary to
control a single lamp from more than one point
(Two or Three different points).
•For example: staircases, long corridors, large
halls etc.

271. 1 Two-way Control oflamp:
•Two-way control is usually used for staircase
lighting.
•The lamp can be controlled from two different
points: one at the top and the other at the bottom -
using two-way switches which strap wires
interconnect.
•They are also used in bedrooms, big halls and large
corridors.
•The circuit is shown in the following figure.

272. 1 Two-way Control oflamp:

273. 1 Two-way Control oflamp:
Explanation: -Switches S1 and S2 are two-way
switches with a pair of terminals 1&2, and 3&4
respectively. When the switch S1 is in position1 and
switch S2 is in position 4, the circuit does not form a
closed loop and there is no path for the current to flow
and hence the lamp will be OFF. When S1 is changed to
position 2 the circuit gets completed and hence the lamp
glows or is ON. Now if S2 is changed to position 3 with
S1 at position 2 the circuit continuity is broken and the
lamp is off. Thus the lamp can be controlled from two
different points.

274. 1 Two-way Control oflamp:

275. 1 Two-way Control oflamp:
Explanation: -Switches S1 and S2 are two-way
switches with a pair of terminals 1&2, and 3&4
respectively. When the switch S1 is in position1 and
switch S2 is in position 4, the circuit does not form a
closed loop and there is no path for the current to flow
and hence the lamp will be OFF. When S1 is changed to
position 2 the circuit gets completed and hence the lamp
glows or is ON. Now if S2 is changed to position 3 with
S1 at position 2 the circuit continuity is broken and the
lamp is off. Thus the lamp can be controlled from two
different points.

276. 2. Three-way Control of
lamp:
•In case of very long corridors it may be necessary to
control the lamp from 3 different points. In such cases,
the circuit connection requires two; two-way switches
S1and S2 and an intermediate switch S3.
•An intermediate switch is a combination of two, two
way switches coupled together. It has 4 terminals ABCD.
It can be connected in two ways.

277. a)Straight connection
b)Cross connection
•In case of straight connection, the terminals or points
AB and CD are connected as shown in figure 1(a)
while in case of cross connection, the terminals AB
and C D is connected as shown in figure 1(b). As
explained in two –way control the lamp is ON if the
circuit is complete and is OFF if the circuit does not
form a closed loop.
2. Three-way Control of
lamp:

278. 2. Three-way Control of
lamp:

279. 2. Three-way Control of
lamp:

280. Truth table:
2. Three-way Control oflamp:

281. BLOCKDIAGRAM

282. 1.To protect the operating personnel from danger
of shock in case they come in contact with the
charged frame due to defective insulation.
2.To maintain the line voltage constant under
unbalanced load condition.
3.Protection of the equipments
4.Protection of large buildings and all machines
fed from overhead lines against lightning.
Necessity ofEarthing:

283. 1ohm.
•The important methods of earthing are the plate
earthing and the pipe earthing.
•The earth resistance for copper wire is 1 ohm
and that of G I wire less than 3 ohms.
•The earth resistance should be kept as low as
possible so that the neutral of any electrical
system, which is earthed, is maintained almost at
the earth potential.
•The typical value of the earth resistance at
powerhouse is 0. 5 ohm and that at substation is
Methods ofEarthing:

284. .
•In this method a copper plate of 60cm x 60cm x 3.18cm
or a GI plate of the size 60cm x 60cm x 6.35cm is used
for earthing.
•The plate is placed vertically down inside the ground at
a depth of 3m and is embedded in alternate layers of
coal and salt for a thickness of 15 cm.
•In addition, water is poured for keeping the earth
electrode resistance value well below a maximum of 5
ohms.
•The earth wire is securely bolted to the earth plate.
•A cement masonry chamber is built with a cast iron
cover for easy regular maintenance.
1. PlateEarthing:

285. PlateEarthing:

286. •Earth electrode made of a GI (galvanized) iron pipe of
38mm in diameter and length of 2m (depending on the
current) with 12mm holes on the surface is placed
upright at a depth of 4.75m in a permanently wet ground.
•To keep the value of the earth resistance at the desired
level, the area (15 cms) surrounding the GI pipe is filled
with a mixture of salt and coal..
•the efficiency of the earthing system is improved by
pouring water through the funnel periodically. The GI
earth wires of sufficient cross-sectional area are run
through a 12.7mm diameter pipe (at 60cms below) from
the 19mm diameter pipe and secured tightly at the top
as shown in the figure.
2. PipeEarthing

287. Pipe Earthing:

288. •Protection for electrical installation must be provided in
the event of faults such as short circuit, overload and
earth faults.
•The protective circuit or device must be fast acting and
isolate the faulty part of the circuit immediately.
•It also helps in isolating only required part of the circuit
without affecting the remaining circuit during
maintenance.
•The following devices are usually used to provide the
necessary protection:
PROTECTIVEDEVICES

289. 1.Fuses
2.Relays
3.Miniature circuit breakers(MCB)
4.Earth leakage circuit breakers(ELCB)
Types

290. 1.Re-wirable or kit -kat fuses: These fuses are simple in
construction, cheap and available up-to a current
rating of 200A. They are erratic in operation and their
performance deteriorates with time.
2.Plug fuse: The fuse carrier is provided with a glass
window for visual inspection of the fuse wire.
3.Cartridge fuse: Fuse wire usually an alloy of lead is
enclosed in a strong fiber casing. The fuse element is
fastened to copper caps at the ends of the casing. They
are available up-to a voltage rating of 25kV. They are
used for protection in lighting installations and power
lines.

291. circuits.
4.Miniature Cartridge fuses: These are the miniature version of the
higher rating cartridge fuses, which are extensively used in
automobiles, TV sets, and other electronic equipment‟s.
5.Transformer fuse blocks: These porcelain housed fuses are placed
on secondary of the distribution transformers for protection
against short circuits and overloads.
6.Expulsion fuses: These consist of fuse wire placed in hollow tube
of fiber lined with asbestos. These are suited only for out door use
for example, protection of high voltage circuits.
7.Semi-enclosed re-wirable fuses: These have limited use because of
low breaking capacity.
8.Time-delay fuse: These are specially designed to withstand a
current overload for a limited time and find application in motor

292. Under abnormal conditions such as short circuit, overload or
any fault the current raises above this value, damaging the
equipment and sometimes resulting in fire hazard. Fuses are
pressed into operation under such situations.
Fuse is a safety device used in any electrical installation,
which forms the weakest link between the supply and the
load. It is a short length of wire made of lead / tin /alloy of
lead and tin/ zinc having a low melting point and low ohmic
losses. Under normal operating conditions it is designed to
carry the full load current. If the current increases beyond
this designed value due any of the reasons mentioned
above, the fuse melts (said to be blown) isolating the power
supply from the load as shown in the following figures.
FUSE

293. FUSE