DC Circuits and Networks for BTech First Year in Electrical Engineering

Basics of DC Circuits and Networks

1. BASICS OF CIRCUIT

2. Electrical Safety
“Danger—High Voltage.”

3. Introduction to Electric Circuits
An electric circuit is an interconnection of electrical elements.
Functions:
To transfer energy from one point to another.
Basic concepts:
Charge
Current
Voltage
Power
Energy
Circuit elements

4. Charge
Chargeisanelectricalpropertyoftheatomicparticleswhichmatterconsistsof.
Theunitofchargeisthecoulomb(C).
ThesymbolforthechargeisQ(or)q.
��������=
�
�.??????����
−��
=??????.�??????��
��
??????�??????������
Typesofcharge
1.Positive charge
2.Negative charge
A single electron has a charge of -1.602x10-19 c.
A single proton has a charge of +1.602x10-19 c.

5. Current
•The flow of free electrons in a conductor is called electric
current.
•The electric current is defined as the time rate of charge.
•The unit of current is the ampere (A).
•The symbol for the current is I (or) i.
•1ampere=1coulomb/second

6. Voltage
•Thepotentialdifferencebetweentwopointsinanelectriccircuit
calledvoltage.
•Definedtobethechargerateofdoingwork.
•Energyrequiredtomoveaunitchargethroughanelement.
•VoltageisrepresentedbyV(or)v.
•Voltagemaybeconstant/varying.
•Theunitofvoltageisvolt.
•1volt=1joule/coulomb=1newtonmeter/coulomb
•Voltage,

7. PrincipleofoperationofelectricCells
If two electrode plates of
different conducting
material (e.g. copper and
zinc) are placed in a
solution (the electrolyte) of
salts, acids or alkaline, a
voltage will appear
between them.

8. ANALOGY
V1
V2
V1>V2
V1–V2=Potential
difference

9. Power
Therateatwhichworkdonebyelectricalenergy(or)energysuppliedperunit
timeiscalledthepower.
Poweristherateatwhichenergyisexpandedortheabsorbing.
ThepowerdenotedbyeitherPorp.P=VxI
Measuredinwatts(W).
Powercanbeabsorbedorsuppliedbycircuitelements.
Positivepower→elementabsorbspower.
Negativepower→elementsuppliespower.
‘Sign’determinedbyvoltageandcurrent.
Anidealcircuit:
∑P
supplied+∑P
absorbed=0.

10. DCCircuits and Networks

11. Network
Interconnection of two or more
simple circuit elements is called
an electric network.
Circuit
A network contains at least one
closed path, it is called electrical
circuit.

12. “An Electric Circuit is an interconnection of
electrical elements.”
There are many elements used in a circuit:
1.Sources
2.Branch & Node
3.Loop & Mesh
4.Resistor
5.Inductor
6.Capacitor
7.Switch
8.Wire
Electric Circuit

13. Circuit Elements
•An element is the basic building block of a circuit.
•Electric circuit is interconnecting of the elements.
•Types of elements:
✓Active elements →Capable of generating energy (i.e.
batteries, generators).
✓Passive elements →Absorbs energy (i.e. resistors,
capacitors and inductors).
✓Voltage and current sources →the most important active
elements.

14. ActiveElements
Thesourcesofenergyarecalledactiveelement.Theymaybe
voltagesourceorcurrentsource.
Example:
Generators,Transistors,etc.
Dividedinto:
Independentsource:Doesnotdependtootherelementstosupply
voltageorcurrent.
Dependentsource:Reverseofindependent.
Constantvoltagesource:Voltagesameforallelements.
Constantcurrentsource:Currentsamethroughoutthecircuits.

15. 1)IdealVoltageSources
Anidealvoltagesourceisatwo-terminalelementthat
maintainsthesamevoltageacrossitsterminalsregardless
ofthecurrentflowingthroughit.
V
t=constant,nomatterwhattheloadcurrentis.
V
t
I
L
V
o
L
t
+
-
V
o

16. 2)IdealCurrentSources
Anidealcurrentsourceisatwo-terminalelementthat
maintainsthesamecurrentregardlessofthevoltageacross
itsterminals.
I
S=constant,nomatterwhattheloadvoltageis.
I
O
V
O
I
S

17. DEPENDENT(CONTROLLED)SOURCES
Dependentsourcesarewhoseoutput
(currentorvoltage)isafunctionofsome
othervoltageorcurrentinacircuit.
Thesymbolstypicallyusedtorepresent
dependentsourcesareintheshapeofa
diamond.

18. PassiveElements
Theseelementsstores(intheformofelectrostatic,
electromagneticenergy)ordissipatesenergy(intheformof
heat).
Example:
Resistance(R),Inductor(L),Capacitor(C).

19. Resistance
It is the property of a substance which opposes the flow of current
through it.
The resistance of element is denoted by the symbol “R”.
It is measured in Ohms Ω.

20. Inductor
It is the property of a substance which stores energy
in the form of electromagnetic field.
The inductance of element is denoted by the symbol
“L”.
It is measured in Henry Η.

21. Capacitor
It is the property of a substance which
stores energy in the form of electrostatic
field.
The capacitance of element is denoted by
the symbol “C”
It is measured in Farads Ϝ.
21

22. BilateralAndUnilateral Element
▪The bilateral elements have sameV-I
relationship for current flowin either direction.
Eg : Any conducting wire, Resistors.
▪Unilateral Elements = don’t have same V-I
relationship for current flow in either direction
Eg: Vaccum diodes, Silicon
diodes, rectifiersetc.,

23. LinearandNon-linear Element
▪LinearelementshavelinearV-Irelationship(i.e.Straightline)passing
throughorigin.
▪Linearelementsobeyssuperpositiontheorem. Eg:Resistors
▪Non-LinearElementshavenon-linearV-Irelationship
▪Eg:SCR,Triac

24. LUMPEDANDDISTRIBUTED
▪Small in size and simultaneous action
takes place for any given cause at
same time of instant.
▪size is very small compared to
wavelength of signal applied)
▪Eg: R,L,C
▪Distributed Elements are not electrically
separable for any analytical purpose.
▪Eg: Transmission line has its distributedR,L,C
throughout its entire length.

25. Fundamental laws
•Fundament laws that govern electric circuits:
•Ohm’s Law.
•Kirchoff’s Law.
•These laws form the foundation upon which electric circuit analysis is
built.
•Common techniques in circuit analysis and design:
•Combining resistors in series and parallel.
•Voltage and current divisions.
•Wye to delta and delta to wye transformations.
•These techniques are restricted to resistive circuits.

26. Ohm’s Law
Relationship between current and voltage within a
circuit element.
The voltage across an element is directly
proportional to the current flowing through it →v α i
Thus::v=iRand R=v/i
Where:
R is called resistor.
Has the ability to resist the flow of
electric current.
Measured in Ohms (Ω)
v=iR

27. Value of R :: varies from 0 to infinity
Extreme values == 0 & infinity.
Only linear resistors obey Ohm’s Law.
Short circuit Open Circuit
Ohm’s Law

28. Conductance (G):
Unit mhoor Siemens(S).
Reciprocal of resistance R
G = 1 / R
Has the ability to conduct electric
current
Ohm’s Law
Power:
P = iv →i( iR ) = i
2
R watts
→(v/R) v = v
2
/R watts
R and G are positive quantities, thus
power is always positive.
R absorbs power from the circuit →
Passive element.

29. Example 1:
Determine voltage (v), conductance (G) and power (p) from the
figure below.
Ohm’s Law

30. Example 2:
Calculate current iin figure below when the switch is in position 1.
Find the current when the switch is in position 2.
Ohm’s Law

31. Nodes, Branches & Loops
Elements of electric circuits can be interconnected in several way.
Need to understand some basic concepts of network topology.
Branch: Represents a single element
(i.e. voltage, resistor etc.)
Node: The meeting point between two or more branches.
Loop: Any closed path in a circuit.

32. BRANCH, NODE, LOOP, MESH
Branch : any portion of a circuit with two terminals
connected to it.
A branch may consist of one or more circuit elements.

33. Node : the point of connection between two or more branches.
A node usually indicated by a dot in a circuit.

34. Loop : any closed path through the circuit in which no node
is encountered more than once.
Mesh : a loop that does not contain other loops.

35. Example 3:
Determine how many branches and nodes for the following circuit.
Nodes, Branches & Loops

36. 5 Branches
1 Voltage Source
1 Current Source
3 Resistors
3 Nodes
a
b
c
Nodes, Branches & Loops

37. Example 4:
Determine how many branches and nodes for the following circuit.
Nodes, Branches & Loops

38. Kirchoff’s Laws

39. Kirchhoff'sLaws
Kirchhoff'scircuitlawsaretwoequalitiesthatdealwiththe
conservationofchargeandenergyinelectricalcircuits.
TwoLawsofKirchhoff'sareasfollows:
1.Kirchhoff'sCurrentLaw.
2.Kirchhoff'sVoltageLaw.

40. Kirchoff’sCurrentLaw(KCL)
ThislawisalsocalledKirchhoff'sfirstlaw,Kirchhoff'spoint
rule,Kirchhoff'sjunctionrule(ornodalrule).
Theprincipleofconservationofelectricchargeimpliesthat:
“Atanynode(junction)inanelectricalcircuit,thesum
ofcurrentsflowingintothatnodeisequaltothesumof
currentsflowingoutofthatnode,
orThealgebraicsumofcurrentsinaofconductors
meetingatapointiszero.”

41. •Thecurrententeringany
junctionisequaltothecurrent
leavingthatjunction.
•i
1+i
4=i
2+i
3
I=0
•Thealgebraicsumofcurrent
entering/leavinganode(or
closedboundary)iszero.
Current enters = +ve
Current leaves = -ve
∑ current entering = ∑ current leaving
Kirchoff’sCurrentLaw(KCL)

42. Kirchoff’s Laws
Example 5:
Given the following circuit, write the equation for currents.

43. Example 6:
Current in a closed boundary
Kirchoff’s Laws

44. Example 7:
Use KCL to obtain currents i
1, i
2, and i
3in the circuit.
Kirchoff’s Laws

45. Kirchhoff'svoltagelaw(KVL)
•ThislawisalsocalledKirchhoff'ssecondlaw,Kirchhoff'sloop
(ormesh)rule.
•Theprincipleofconservationofenergyimpliesthat
“Thedirectedsumoftheelectricalpotentialdifferences
(voltage)aroundanyclosedcircuitiszero,
ortheemfsinanyclosedloopisequivalenttothesumofthe
potentialdropsinthatloop”

46. Thesumofallthevoltagesaroundthe
loopisequaltozero.
v
1+v
2+v
3-v
4=0
=IRV
emf
Kirchhoff'svoltagelaw(KVL)
The algebraic sum of the products of the resistances of the
conductors and the currents in them in a closed loop is equal to
the total emf available in that loop.
Similarly to KCL, it can be stated as:

47. Example 8:
Use KVL to obtain v
1, v
2and v
3.
Kirchhoff'svoltagelaw(KVL)

48. Example 9:
Use KVL to obtain v
1, and v
2.
Kirchhoff'svoltagelaw(KVL)

49. Example 10:
Calculate power dissipated in 5Ωresistor.
10
Kirchhoff'svoltagelaw(KVL)

50. Two or more circuit elements are said to be in series if the current
from one element exclusively flows into the next elements.
All series elements have the same current.
Series Resistors
Equivalent series resistance:
�
??????�=෍
�=�
??????
�
�=�
�+�
�+⋯+�
??????
Series Circuit

51. Example 11:
For the circuit shown,
a) Find the equivalent resistance seen by the source.
b) Find the current I .
c) Calculate the voltage drop in each resistor.
d) Calculate the power dissipated by each resistor.
e) Find the power output of the source.
Given: V= 24 V, R
1 = 1 , R
2 = 3 , and R
3= 4 .
EEE 1012 INTRODUCTION TO ELECTRICAL ENGINEERING
Series Circuit

52. EEE 1012 INTRODUCTION TO ELECTRICAL ENGINEERING
b) ??????=
??????
�
??????�
=
24??????
8Ω
=3??????
c) �
1=????????????
1=3??????1Ω=3�
�
2=????????????
2=3??????3Ω=9�
�
3=????????????
3=3??????4Ω=12�
Solution:
a)??????
??????�=??????
1+??????
2+??????
3
=1+3+4
=8
R
EQ
Note: �
1+�
2+�
3=3+9+12=24�
The total voltage drop is equal to the voltage output of the source.

53. EEE 1012 INTRODUCTION TO ELECTRICAL ENGINEERING
Note: ??????
1+??????
2+??????
3=9+27+36=72�
The total power dissipated by the resistors is the same as the power
output by the source.
e)??????=??????�=3??????24�=72�
d)??????
1=??????
2
??????
1
=3??????
2
1Ω
=9�
??????
2=??????
2
??????
2
=3??????
2
3Ω
=27�
??????
3=??????
2
??????
3
=3??????
2
4Ω
=36�

54. Two or more circuit elements are said to be in parallel if the
elements share the same terminals.
All parallel elements have the same voltage.
Parallel Resistors
Equivalent parallel resistance:
or
where
EEE 1012 INTRODUCTION TO ELECTRICAL ENGINEERING
�
�
??????�
=
�
�
�
+
�
�
�
+⋯+
�
�
??????
�
??????�=
�
ൗ
�
�
�
+ൗ
�
�
�
+⋯+ൗ
�
�
??????
�
1=�
2=�
3=�
4
Parallel Circuit

55. Various Parallel Resistors Networks
EEE 1012 INTRODUCTION TO ELECTRICAL ENGINEERING
Parallel Circuit

56. EEE 1012 INTRODUCTION TO ELECTRICAL ENGINEERING
•Example 12:
For the circuit shown,
a) Find the equivalent resistance seen by the source.
b) Find the total current I .
c) Calculate the currents in each resistor.
d) Calculate the power dissipated by each resistor.
e) Find the power output of the source.
Given: V= 24 V, R
1 = 1 , R
2 = 3 , and R
3= 4 .

57. EEE 1012 INTRODUCTION TO ELECTRICAL ENGINEERING
Solution:
a)
1
�
??????�
=
1
�
1
+
1
�
2
+
1
�
3
=
1
1
+
1
3
+
1
4
=1.583Ω
∴??????
??????�=
1
1.583
=0.632Ω
b) ??????=
??????
�
??????�
=
24??????
0.632Ω
=37.975??????≈38??????
R
EQ
I

58. EEE 1012 INTRODUCTION TO ELECTRICAL ENGINEERING
c) ??????
1=
??????
�
1
=
24??????
1Ω
=24??????
R
EQ
I
??????
2=
??????
�
2
=
24??????
3Ω
=8??????
??????
3=
??????
�
3
=
24??????
4Ω
=6??????
Note: ??????
1+??????
2+??????
3=24+8+6=38??????
The sum of the individual current is equal to the current output of the
source.

59. EEE 1012 INTRODUCTION TO ELECTRICAL ENGINEERING
Note: ??????
1+??????
2+??????
3=576+192+144=912�
The total power dissipated by the resistors is the same as the power
output by the source.
e)??????=??????�=38??????24�=912�
d)??????
1=??????
1
2
??????
1
=24??????
2
1Ω
=576�
??????
2=??????
2
2
??????
2
=8??????
2
3Ω
=192�
??????
3=??????
3
2
??????
3
=6??????
2
4Ω
=144�

60. Example 13:
TheWheatstone Bridge consists of two series circuits that are connected in
parallel with each other.
1) Find the value of the voltage
V
ab= V
ad-V
bdin terms of the four
resistances and the source voltage V
s.
2) If R
1= R
2= R
3= 1 k, V
s= 12 V,
and V
ab= 12 mV, what is the value
of R
x.
EEE 1012 INTRODUCTION TO ELECTRICAL ENGINEERING
Series and Parallel Resistor Combinations

61. EEE 1012 INTRODUCTION TO ELECTRICAL ENGINEERING
Solution:
1) �
�??????=
�
2
�1+�2
�
??????and �
�??????=
�
??????
�3+�??????
�
??????
Thus,
�
��=�
�??????−�
�??????=ቆ
�
2
�1+�2
− ൰
�
??????
�3+�??????
�
??????
2) 0.012=
1000
1000+1000
−
�
??????
1000+�
??????
12
??????
??????=996Ω
R
1
R
2
R
3
R
X
V
s
a V
ab b
c
d
R
1
R
2
R
3
R
X

62. Conductance (G)
Series conductance:
1/G
eq= 1/G
1+1/G
2+…
Parallel conductance:
G
eq= G
1+G
2+…

63. VDR is useful in determining the voltage drop across a resistance within a
series circuit.
where V
X= the voltage drop across the measured resistor,
R
X= the resistance value of the measured resistor,
R
EQ= the circuit total resistance,
V
S= the circuit applied voltage
EEE 1012 INTRODUCTION TO ELECTRICAL ENGINEERING
??????
??????=
�
??????
�
??????�
??????
�
+ V
1-
+ V
3-
+
V
2
-
S
Voltage Divider Rule (VDR)

64. Series Resistors & Voltage Division
Voltage Division:
v
1= iR1 & v2= iR2
i= v/(R1+R2 )
Thus:
v1=vR1/(R1+R2)
v2=vR2/(R1+R2)

65. Example 14:
Determine the voltage across the R
2and the R
3.
Solution:
EEE 1012 INTRODUCTION TO ELECTRICAL ENGINEERING
�
2=
??????
2
??????
??????�
�
�=
20
10+20+30
60=20V
�
3=
??????
3
??????
??????�
�
�=
30
10+20+30
60=30V

66. EEE 1012 INTRODUCTION TO ELECTRICAL ENGINEERING
CDR is useful in determining the current flow through one branch of
a parallel circuit.
where I
X= the current flow through any parallel branches,
R
X= the resistance of the branch through which the
current is to be determined,
R
EQ= the total resistance of the parallel branch,
I
S= the circuit applied current
??????
??????=
�
??????�
�
??????
??????
�
Current Divider Rule (CDR)

67. Example 15:
Find each of the branch currents in the figure shown below.
EEE 1012 INTRODUCTION TO ELECTRICAL ENGINEERING
Solution:
??????
1??????
2??????
3
1
�??????�
=
1
�1
+
1
�2
+
1
�3
=
1
3??????Ω
+
1
8??????Ω
+
1
24??????Ω
=
1
2??????Ω
??????
??????�=2??????Ω

68. Thus,
??????
�1=
??????
??????�
??????1
∙??????=
2??????Ω
3??????Ω
12????????????=8????????????
??????
�2=
??????
??????�
??????2
∙??????=
2??????Ω
8??????Ω
12????????????=3????????????
??????
�3=
??????
??????�
??????3
∙??????=
2??????Ω
24??????Ω
12????????????=1????????????
??????
�1+??????
�2+??????
�3=8+3+1=12???????????? ( KCL )
EEE 1012 INTRODUCTION TO ELECTRICAL ENGINEERING

69. For the particular case of two parallelresistors,
and, the current passing through
R
1and R
2are
??????
1=
�
??????�
�
1
∙??????=
�1�2
�1+�2
�
1
∙??????
??????
2=
�
??????�
�2
∙??????=
�1�2
�1+�2
�2
∙??????
EEE 1012 INTRODUCTION TO ELECTRICAL ENGINEERING
�
??????�=�
�ฮ�
�=
�
��
�
�
�+�
�
I
I
1 I
2
??????
�=
�
�
�
�+�
�
∙??????
??????
�=
�
�
�
�+�
�
∙??????
Note: It onlyworks for two parallel resistors.

70. Voltage and Current Division
Example 16:
Calculate v
1, i
1, v
2and i
2.

71. Voltage and Current Division
Example 17:
Determine i
1through i
4.

72. Voltage and Current Division
Example 18:
Determine vand i.
Answer v = 3v, I = 6 A.

73. Voltage and Current Division
Example 19:
Determine Iand V
s if the current through 3Ωresistor = 2A.

74. Voltage and Current Division
Example 20:
Determine R
ab.

75. Voltage and Current Division
Example 21:
Determine v
xand power absorbed by the 12Ωresistor.
Answer v = 2v, p = 1.92w.

76. Nodal Analysis
Inelectriccircuitsanalysis,nodalanalysis,node-
voltageanalysis,orthebranchcurrentmethodisa
methodofdeterminingthevoltage(potential
difference)between"nodes“)inanelectricalcircuitin
termsofthebranchcurrents.

77. Steps to Solve Nodal Analysis :
➢Noteallconnectedwiresegmentsinthecircuit.Thesearethenodesof
nodalanalysis.
➢Selectonenodeasthegroundreference.Thechoicedoesnotaffectthe
resultandisjustamatterofconvention.Choosingthenodewithmost
connectionscansimplifytheanalysis.
➢Assignavariableforeachnodewhosevoltageisunknown.Ifthevoltageis
alreadyknown,itisnotnecessarytoassignavariable.
➢Foreachunknownvoltage,formanequationbasedonKirchhoff's
currentlaw.Basically,addtogetherallcurrentsleavingfromthenode
andmarkthesumequaltozero.

78. •Iftherearevoltagesourcesbetweentwounknownvoltages,jointhetwo
nodesasasupernode.Thecurrentsofthetwonodesarecombinedina
singleequation,andanewequationforthevoltagesisformed.
•Solvethesystemofsimultaneousequationsforeachunknown
voltage.

79. StepsofNodalAnalysis
+
–
V 500W
500W
1kW
500W
500W
I
2
1.Chooseareference(ground)node.
2.Assignnodevoltagestotheothernodes.
3.ApplyKCLtoeachnodeotherthanthereferencenode;
expresscurrentsintermsofnodevoltages.

80. Currents and Node Voltages
R2
V
1 V
2
V
1−V
2
R2
R1
V
1
V
1
R1

81. KCLatNode1
R2
R1
I
1
V
1 V
2
1
V
1−V
2
+
V
1
R2R1
I=
KCLatNode2
R3
V
2 V
3V
1
=0
R2R3R4
V
2−V
1
+
V
2
+
V
2−V
3

82. KCLatNode3
2
V
3−V
2
+
V
3
R4R5
=IR5
R4
I
2
V
2 V
3

83. Network Theorems
•SuperpostionTheorem
•TheveninTheorem
•NortonTheorem
•MaximumTransferTheorem

84. •ItisusedtofindthesolutiontonetworksContainingtwoormoreVoltagesources.
•Thecurrentthrough,orvoltageacross,anelementinalinearbilateralnetworkisequalto
thealgebraicsumofthecurrentsorvoltagesproducedindependentlybyeachsource.
•Foratwo-sourcenetwork,ifthecurrentproducedbyonesourceisinonedirection,while
thatproducedbytheotherisintheoppositedirectionthroughthesameresistor,the
resultingcurrentisthedifferenceofthetwoandhasthedirectionofthelarger.
•Iftheindividualcurrentsareinthesamedirection,theresultingcurrentisthesumof
twointhedirectionofeithercurrent.
•Thetotalpowerdeliveredtoaresistiveelementmustbedeterminedusingthetotal
currentthroughorthetotalvoltageacrosstheelementandcannotbedeterminedbya
simplesumofthepowerlevelsestablishedbyeachsource
Superposition Theorem

85. How to Apply Superposition Theorem
ForapplyingSuperpositiontheorem:
•Replaceallotherindependentvoltagesourceswithashortcircuit
(therebyeliminatingdifferenceofpotential.i.e.V=0,internal
impedanceofidealvoltagesourceisZERO(shortcircuit).
•Replaceallotherindependentcurrentsourceswithanopen circuit
(therebyeliminatingcurrent.i.e.I=0,internalimpedanceofidealcurrent
sourceisinfinite(opencircuit).

86. ForExample:
Solution:
•Theapplicationofthesuperpositiontheoremis,whereitisusedtocalculate
thebranchcurrent.
•Webeginbycalculatingthebranchcurrentcausedbythevoltagesourceof
120V.
120V
3
6
12A4
2
i
1
i
2
i
3
i
4

87. •Tocalculatethebranchcurrent,thenodevoltageacrossthe3Ωresistormustbe
known.Therefore
120V
3
6
4
2
i
'
1i
'
2
i
'
3
i
'
4
v
1
6 32+4
+
v
1−120
+
v
1v
1
=0
wherev
1=30V
Theequationsforthecurrentineachbranch,
Bysubstitutingtheidealcurrentwithopencircuit,
wedeactivatethecurrentsource.

88. 120−30
=15A
i'
2=
6
30
=10A
i
'
3=i
'
4=
3
30 =5A
6
Inordertocalculatethecurrentcausebythecurrentsource,we
deactivatetheidealvoltagesourcewithashortcircuit,asshown
3
6
12A4
2
1
i
"
2
i
"
i
3
"
4
i
"
i'
1=

89. Todeterminethebranchcurrent,solvethenode voltagesacrossthe3Ωand
4Ωresistorsasshownin Figure4
Thetwonodevoltagesare
3
6
12A4
2
v
4
+
v
3
-
+
-
2
+
v
3
+
v
3v
3−v
4
36
42
++12
v
4−v
3v
4
=0
=0

90. •Bysolvingtheseequations,weobtain
v
3=-12V
v
4=-24V
Nowwecanfindthebranchescurrent,

91. Tofindtheactualcurrentofthecircuit,addthecurrentsdueto
boththecurrentandvoltagesource,

92. ApplicationofNetworkTheorems
•Networktheoremsareusefulinsimplifyinganalysisofsomecircuits.
•Butthemoreusefulaspectofnetworktheoremsistheinsightitprovides
intothepropertiesandbehaviourofcircuits
•Networktheoremalsohelpinvisualizingtheresponseofcomplex
network.
•TheSuperpositionTheoremfindsuseinthestudyofalternatingcurrent
(AC)circuits,andsemiconductor(amplifier)circuits,wheresometimes
ACisoftenmixed(superimposed)withDC

93. Real-Life Applications of Superposition Theorem
1.PowerDistributionNetworks
•Incitypowergrids,electricitycomesfrommultiplegeneratingstations.
•Superpositionhelpsanalyzehowmucheachstationcontributestothe
current/voltageatdifferentpointsinthenetwork.
•Usedinloadsharingandfaultanalysis.
2.CommunicationSystems
•Inradio,TV,andmobilecommunicationcircuits,signalsfrommultiple
sourcesoverlap.
•Superpositionprincipleisusedinmodulation,demodulation,andsignal
filtering.
•Forexample:analyzinghowAM/FMsignalscombineinareceiver.

94. Real-Life Applications of Superposition Theorem
3.ElectricalMachines(Motors&Generators)
•Motorsoftenhavemultipleinputs(e.g.,excitation+armature).
•Superpositionhelpscalculateresultingfluxes,voltages,orcurrentsby
consideringeachsourceseparately.
4.ElectronicCircuits(Amplifiers,Filters)
•CircuitsmayhavemultipleAC+DCsources(likeabiasingsource+input
signal).
•SuperpositiontheoremseparatestheDCresponseandtheACresponseto
designamplifiersproperly.
•Verycommoninop-ampcircuitsandsignalprocessinghardware.

95. Real-Life Applications of Superposition Theorem
5.HouseholdPowerSystems
•Whenmultipleappliancesareconnectedtothesamepowerline,thetotalcurrent
distributioncanbecalculatedbyconsideringeachappliance’scontributionusing
superposition.
6.RenewableEnergySystems
•Inhybridsystems(solar+wind+battery+grid),multiplevoltage/currentsources
feedintothesamebus.
•Superpositionhelpspredicthoweachsourcecontributestosupplyandensures
balancedoperation.
7.AudioEngineering
•Insoundsystems,multipleinputsignals(mic,instruments,tracks)overlap.
•Superpositionisappliedtounderstandhowdifferentsignalscombineinacircuit
(mixers,equalizers,speakers).

96. Thevenin'stheorem
Thevenin'stheoremforlinearelectricalnetworksstatesthat
“Anycombinationofvoltagesources,currentsources,andresistorswithtwo
terminalsiselectricallyequivalenttoasinglevoltagesourceVandasingle
seriesresistorR.”
Anytwo-terminal,linearbilateraldcnetworkcanbereplacedbyanequivalent
circuitconsistingofavoltagesourceandaseriesresistor.

97. Thevenin’sTheorem
TheTheveninequivalentcircuitprovidesanequivalenceat theterminalsonly.
Thistheoremachievestwoimportant objectives:
•Providesawaytofindanyparticularvoltageorcurrentinalinear
networkwithone,two,oranyothernumberofsources
•Wecanconcentrateonaspecificportionofanetwork byreplacingthe
remainingnetworkwithanequivalent circuit

98. CalculatingtheTheveninequivalent
•SequencetopropervalueofRThandETh
•Preliminary
1.Removethatportionofthenetworkacrosswhich theThévenin
equationcircuitistobefound.
Inthe figurebelow,thisrequiresthattheloadresistorRLbe temporarily
removedfromthenetwork.

99. •Marktheterminalsoftheremainingtwo-terminalnetwork.(Theimportance
ofthisstepwillbecomeobviousasweprogressthroughsomecomplex
networks)
TofindRTh:
•CalculateRThbyfirstsettingallsourcestozero
•Voltagesourcesarereplacedbyshortcircuits,andcurrentsourcesbyopen
circuits)
•Thenfindingtheresultantresistancebetweenthetwomarkedterminals.
•Iftheinternalresistanceofthevoltageand/orcurrentsourcesisincludedinthe
originalnetwork,itmustremainwhenthesourcesaresettozero.
CalculatingtheTheveninequivalent

100. TocalculateETh:
•CalculateEThbyfirstreturningallsourcestotheiroriginalpositionandfindingthe
open-circuitvoltagebetweenthemarkedterminals.
•Thisstepisinvariablytheonethatwillleadtothemostconfusionanderrors.
•Inallcases,keepinmindthatitistheopen-circuitpotentialbetweenthetwo
terminalsmarkedinstep2.
CalculatingtheTheveninequivalent

101. •Conclusion:
•DrawtheThéveninequivalentcircuit
withtheportionofthecircuit
previouslyremoved replaced
betweentheterminalsofthe
equivalentcircuit.
•Thisstepisindicatedbythe
placementoftheresistorR
Lbetween
theterminalsoftheThévenin
equivalentcircuit
InsertFigure9.26(b)
CalculatingtheTheveninequivalent

102. AnotherwayofCalculatingtheThéveninequivalent
•MeasuringVOCandISC
•TheThéveninvoltageisagaindeterminedbymeasuringtheopen-circuit
voltageacrosstheterminalsofinterest,thatis,ETh=VOC.
•TodetermineR
Th,ashort-circuitconditionisestablishedacrosstheterminals
ofinterestandthecurrentthroughtheshortcircuitI
scismeasuredwithan
ammeter
•UsingOhm’slaw:
R
Th=V
oc/I
sc

103. Example:
Solution:
•InordertofindtheTheveninequivalentcircuitforthecircuitshowninFigure
1,calculatetheopencircuitvoltage,V
ab.
•Notethatwhenthea,bterminalsareopen,thereisnocurrentflowto4Ω
resistor.
•Therefore,thevoltagev
abisthesameasthevoltageacrossthe3Acurrent
source,labeledv
1.
•Tofindthevoltagev
1,solvetheequationsforthesingularnodevoltage.By
choosingthebottomrightnodeasthereferencenode,
25V
20
+
-
v
13A
5 4
+
a
v
ab
-
b

104. •Bysolvingtheequation,v
1=32V.
•Therefore,theTheveninvoltageV
thforthecircuitis32V.
•Thenextstepistoshortcircuittheterminalsandfindthe shortcircuitcurrentfor
thecircuitshowninFigure2.
•Note thatthecurrentisinthesamedirectionasthefallingvoltage attheterminal.
−3=0
v
1−25
+
v
1
5 20
25V
20
+
-
v
23A
5
-
+
v
ab
4a
i
sc
b
Figure2

105. 4
−3+
v
2
=0
v
2−25
+
v
2
5 20
Currentisccanbefoundifv2isknown.Byusingthebottom
rightnodeasthereferencenode,theequationforv2becomes
Bysolvingtheaboveequation,v
2=16V.Therefore,theshort
circuitcurrenti
scis
TheTheveninresistanceR
This
Figure3showstheTheveninequivalentcircuitfortheFigure1.

106. Real-Life Applications of Thevenin’s Theorem
1.PowerSystemAnalysis(Grids&Distribution)
•Usedtomodelacomplexpowernetwork(generators+transmissionlines+
transformers)intoasingleTheveninequivalentsourceforanalyzingthe
effectononeload.
•Makesfaultcurrentanalysisandstabilitystudieseasier.
2.DesignofElectricalMachines(Motors&Generators)
•Theveninequivalentsareusedtorepresentcomplicatedinternalcircuitsof
synchronousmachines,inductionmotors,andtransformers.
•Simplifiesperformancecalculation(startingtorque,faultbehavior,
efficiency).

107. Real-Life Applications of Thevenin’s Theorem
3.Short-Circuit&FaultAnalysis
•Inindustriesandpowergrids,Thevenin’stheoremhelpscalculatefault
currentsataparticularbusornode.
•Veryimportantfordesigningcircuitbreakers,relays,andprotectivesystems.
4.Battery&PowerSupplySystems
•AbatterypackwithinternalresistancecanbereplacedbyitsThevenin
equivalenttostudyitseffectondifferentloads.
•Usedinbatterymanagementsystems(BMS)forEVsandsolarstorage.

108. Real-Life Applications of Thevenin’s Theorem
5.Electronics(Amplifiers,Transistors,Op-Amps)
•UsedtoreplacecomplicatedtransistorbiasingcircuitswithasimpleThevenin
source+resistance.
•HelpsinAC&DCanalysisofamplifierstages.
6.Communication&SignalProcessing
•TransmissionlinesandantennacircuitsaremodeledusingTheveninequivalents.
•Usedinimpedancematchingformaximumpowertransfer(importantinRF
circuits).
7.TestingandMeasurement
•Whenanalyzingtheeffectofameasuringdevice(likeavoltmeterorsensor)ona
circuit,Thevenin’stheoremhelpssimplifythesourcesideintoonevoltage+
resistance.

109. Nortontheorem
Norton'stheoremforlinearelectricalnetworksstatesthatanycollectionof
voltagesources,currentsources,andresistorswithtwoterminalsis
electricallyequivalenttoanidealcurrentsource,I,inparallelwithasingle
resistor.
Anytwolinearbilateraldcnetworkcanbereplacedbyanequivalent
circuitconsistingofacurrentandaparallelresistor.

110. CalculatingtheNortonequivalent
ThestepsleadingtothepropervaluesofI
NandR
N
Preliminary
•RemovethatportionofthenetworkacrosswhichtheNortonequivalentcircuitis
found
•Marktheterminalsoftheremainingtwo-terminalnetwork
TofindR
N:
•CalculateR
Nbyfirstsettingallsourcestozero(voltagesourcesarereplacedwith
shortcircuits,andcurrentsourceswithopencircuits)andthenfindingthe
resultantresistancebetweenthetwomarkedterminals.
•(Iftheinternalresistanceofthevoltageand/orcurrentsourcesisincludedinthe
originalnetwork,itmustremainwhenthesourcesaresettozero.)
•SinceR
N=R
Ththeprocedureandvalueobtainedusingtheapproachdescribedfor
Thévenin’stheoremwilldeterminethepropervalueofR
N

111. Norton’sTheorem
TodetermineIN:
•CalculateINbyfirstreturningallthesourcestotheiroriginalpositionand
thenfindingtheshort-circuitcurrentbetweenthemarkedterminals.
•Itisthesamecurrentthatwouldbemeasuredbyanammeterplaced
betweenthemarkedterminals.
Conclusion:
•DrawtheNortonequivalentcircuitwiththeportionofthecircuitpreviously
removedreplacedbetweentheterminalsoftheequivalentcircuit

112. Example:
Step1:Sourcetransformation(The25Vvoltagesourceisconvertedtoa
5Acurrentsource.)
25V
20
3A
5 4
a
b
20
3A5
4
a
b
5A

113. 4
8A
4
a
b
Step3:Sourcetransformation(combinedserialresistanceto
producetheTheveninequivalentcircuit.)
8
32V
a
b
Step2:Combinationofparallelsourceandparallelresistance

114. 8Ω
Step4:Sourcetransformation(ToproducetheNorton
equivalentcircuit.Thecurrentsourceis4A(I=V/R)
32V/8
a
b
Nortonequivalentcircuit.
4A

115. Maximumpowertransfertheorem
Themaximumpowertransfertheoremstatesthat,to
obtainmaximumexternalpowerfromasourcewithafinite
internalresistance,theresistanceoftheloadmustbeequal
totheresistanceofthesourceasviewedfromtheoutput
terminals.
Aloadwillreceivemaximumpowerfromalinearbilateraldcnetwork
whenitstotalresistivevalueisexactlyequaltotheThévenin
resistanceofthenetworkas“seen”bytheload
RL=RTh
So,R
eq=R
th+R
L
R
eq=2R
L

116. Resistancenetwork
whichcontains
dependentand
independentsources
2
4R
L
V
Th
2
V
ThR
L
P
max= =
•Maximumpowertransferhappenswhentheload resistanceR
Lisequaltothe
Theveninequivalent resistance,R
Th.TofindthemaximumpowerdeliveredtoR
L
(2R
L)
2

117. Thankyou