Network Analysis (Basic Concepts)
Chapter 1 – Basic Concepts
Objectives
Upon completion of this chapter you will be able to:

Understand basic terminologies associated with Electrical Circuits

Understand the behaviour and characteristics of basic circuit elements

Solve a circuit for current and voltage using Kirchoff’s Laws

Apply Network Theorems for simplification of an Electrical Circuit.
Introduction:
Electric circuit theory and electromagnetic theory are the two fundamental theories upon
which all branches of electrical engineering are built. Many branches of electrical engineering
such as power, electric machines, control, electronics, communications, and instrumentation,
are based on electric circuit theory.
In electrical engineering, we are often interested in communicating or transferring energy
from one point to another. To do this requires an interconnection of electrical devices. Such
interconnection is referred to as an electric circuit, and each component of the circuit is
known as an element.
Basic Terminologies
Circuit
A circuit is a path between two or more points along which an electrical current can be
carried.
 The current ‘i’ is intended to flow through all the components
Network
Practically system are big inter connected network, but we do circuit analysis to some of its
points, so circuits are building blocks of network.
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Ex. Power System, Network, communication system N/W etc.
Node (n)
Node is a connection point between two or more branches.
Branch (b)
It is an element connection between two nodes.
Degree of node ( δ )
The number of branches incident or connect at any node represents its degree.
Ex.  = 2 => Simple node ( n )

 > 2 => Principle node ( n )
P
n
i  2 b

i1
Mesh (m)
A mesh is closed path which should not have further closed paths in it.
Loop (l)
Loops are all possible closed paths in an electrical network.
 For any circuit or network m  b  n  l
 The minimum number equation to setup any circuit or network is also equal to m.
 Meshes are specially called as independent loops.
 All meshes are by default loops, but all loops are not meshes
 In nodal analysis we may neglect simple nodes & one of the principle node is considered
as a reference node (V=0)
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Ex.
n=5, b=6, m=2, l=2+1=3
n=9, b=12, m=4
Planar and Non-Planar Networks
The networks which can be drawn on a two dimensional plane are termed as Planar
Networks and those which cannot be drawn in two dimensions are termed as Non-Planar
Networks.
For a network two to be planar no two branches in a network should intersect when drawn
on a two dimensional plane.
Solved Example
Problem: Find total number of node, branch and mesh.
Solution: No current will flow through 2 ohm resistor, so it is not considered as branch.
n=7, b=8, m=2
Problem: Find the network is planar or not? Find the minimum number of equations
required to solve this circuit?
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Solution: circuit can be redrawn as
It is a planar network.
Minimum number of equation= total number of meshes i.e. m=4
Problem: Find the network is planar or not?
Solution: Circuit can be redrawn as
It is a planar network.
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Types of Circuit Elements
1. Linear and Non-linear
A linear element is an electrical element with a linear relationship between current and
voltage.
Ex. capacitors, inductors, and transformers
Linearity implies that it must follow principles of Superposition and Homogeneity.
A nonlinear element is one which does not have a linear input/output relation.
Ex. diode, semiconductor devices
2. Active and Passive
The elements which generates or produces electrical energy are called active elements.
Ex. batteries, generators, transistors, operational amplifiers, vacuum tubes etc.
All elements which consume rather than produce energy are called passive elements.
Ex. resistors, inductors and capacitors.
3. Bilateral and Unilateral
In bilateral circuits, the property of circuit does not change with the change of direction of
supply voltage or current. In other words, bilateral circuit allows the current to flow in both
directions.
Ex. Transmission line
In unilateral circuits, the property of circuit changes with the change of direction of supply
voltage or current. In other words, unilateral circuit allows the current to flow only in one
direction.
Ex. Diode rectifier
4. Distributed and Lumped
An element is classified as a distributed element if the net effect of electrical phenomena
taking place within that element can’t be described in terms of only its terminal voltage and
current variables.
An element is classified as a lumped element if the net effect of electrical phenomena taking
place within that element can be described in terms of only its terminal voltage and current
variables, irrespective of its internal details and geometry.
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5. Time invariant and Time variant
An element is time invariant if the value of parameters that characterize it are independent of
time.
Ex. Resistance, capacitance, inductance
An element is time variant if the value of parameters that characteristic it are dependent of
time.
Ex. Synchronous generator
Components of electrical networks
Resistor (R)
A resistor is an electrical component that limits or regulates the flow of electrical current in
an electronic circuit. It is Linear and bilateral.
In time domain
V(t)  Ri(t) =>i(t) 
V(t)
R
In s-domain
V(s)  RI(s) =>I(s) 
V(s)
R
Using phasors for sinusoidal excitation, V  V ; I  I
V  IR => I=
V
R
Electric power (P)
P(t) 
dw
dw dq


 v i
dt
dq
dt
P(t)  i(t)2 R 
 W  =Instantaneous power
V(t)2
R
Electrical Energy (W)
dw
 dw  Pdt=> W= Pdt
dt
V(t)2
W  i(t)2 R dt 
dt J
R

P(t) 


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
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Inductors (L)
The property of an electric conductor or circuit that causes an electromotive force to be
generated by a change in the current flowing.
In time domain
t
v(t)  L
di
1
=> i(t) 
v(t). dt
dt
L0

ZL  sL , YL 
1
sL
In s-domain (assume initial condition=0)
V(s)  sLI(s) => I(s) 
V(s)
sL
Using phasors for sinusoidal excitation ( s  j;   0 )
V  jL I => I 
1
V
, ZL  jL; YL 
jL
jL
Electric power (P)
Instantaneous power = P(t)  v  i =L
di
di
 i  Li
W
dt
dt
Electrical Energy (W)


W= Pdt  Li
d 1
1 2
 dt = Li  J
2

2
 dt  2 Li
W=
di
dt  J
dt
 For DC excitation L
di
=0, VL  0 , inductor is short circuited for DC.
dt
 Inductor never allows sudden change in current through it.
 Ideal inductor is called wire with zero internal resistance so, power dissipated is zero.
Practical inductors have small resistance in series with coil.
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 Inductors are used as filters, alternators, compensators, current limiting reactors etc. in
communication and power system.
Capacitor (C)
A capacitor is an electronic component used for storing charge and energy. The usual
capacitor is a pair of plates.
In time domain
t
dv(t)
1
i(t)  C
=> v(t) 
i(t). dt
dt
C0

In s-domain or Laplace (assume zero initial condition)
I(s)  sCV(s) => V(s) 
I
1
, Y  sC
, ZC 
sC C
jC
Using phasors for sinusoidal excitation ( s  j;   0 )
I  jCV => V 
I
jC
Electric power (P)
Instantaneous power = P(t)  v  i =vC
dv
W
dt
Electrical Energy (W)


W= Pdt  Cv
d 1
 dt  2 Cv
W=
dv
dt  J
dt
1 2
 dt = Cv  J
2

2
 For DC excitation C
dv
=0, iC  0 , capacitor is Open Circuited for DC.
dt
 Capacitor never allows sudden change in voltage across it.
 Ideal capacitor is construct to have infinite dielectric resistance between the electrodes. So
power dissipated is zero. Practical capacitor are construct to have very large dielectric
resistance (MΩ) between the electrodes.
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 Inductors are used as filters, alternators, compensators, current limiting reactors etc. in
communication and power system.
s= 0
s= 
s= s
L
S.C
O.C
sL
C
O.C
S.C
1/sC
Common properties for inductor and capacitor
 Linear Element: the variation between the terminal voltage and the terminal current is
linear in the time domain or in s-domain or in both domains.
 Bilateral Element: the terminal current flows in either direction irrespective of the polarity
of the voltage applied between the two terminals of the element.
 When the excitation is non-sinusoidal the analysis of the network can be performed either
in the time domain or in the s-domain.
 When the excitation is sinusoidal then analysis of the network is performed only by using
phasors.
 When the circuit analysis is performed in the Laplace domain following advantages are
obtained
1. Integral or differential equation is transformed to a linear equation and the
manipulation becomes simpler.
2. Initial condition, if any is taken automatically.
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Network Analysis (Basic Concepts)
Transformer
Device that converts an alternating current of a certain voltage to an alternating current of
different voltage, without change of frequency, by electromagnetic induction.
V1 n1 I2


V2 n2 I1
n2  n1 ; step up
n1  n2 ; step down
If n1  1, n2  10 then
V2
V1

n2
n1

10
1
(  1);
I2
I1

n1
n2

1
10
V2 I2
.  1 ; i.e. V2 I2  V1 I1
V1 I1
 Any step up transformer can’t work as an amplifier since the power at the output is equal
to power at the input.
 For any device to work as an amplifier the power at the Output should be greater than
power at Input.
Power Sources in Electrical Circuit
Sources
Independent
Voltage
Ideal
Practical
Dependent
current
Ideal
VCVS
VCCS
CCVS
CCCS
Practical
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Independent Voltage Source
It maintains a voltage (fixed or varying with time) which is not affected by any other quantity
 If V(t)  u(t) , If t>0, then u(t) will be DC sources.
 If V(t)  2e-tu(t), 10te-4tu(t) then voltage source is
time varying non- sinusoidal source
 If in the AC source, voltage magnitude is constant
then it is sine wave or cosine, otherwise it is non-sinusoidal.
Ex. 4 cos 200t , 20600
Dependent Voltage Source
Dependent voltage source is a voltage source whose value depends on a voltage or current
of someplace else in the network.
They can be classified into two categories:

VCVS : Voltage Controlled Voltage Source
Here, the voltage of dependent source depends on
voltage of some other point inside the network.

CCVS: Current Controlled Voltage Source
Here, the voltage of dependent source depends on current through some element
inside the network.
Independent Current Source
It maintains a current (fixed or varying with time) which is not affected by any other quantity
 If I(t)  u(t) , If t>0, then u(t) will be DC sources.
-t
-4t
 If I(t)  2e u(t), 10te u(t) then current source is time
varying non- sinusoidal source
 If in the AC source, current magnitude is constant
then it is sine wave or cosine, otherwise it is non-sinusoidal.
Ex. 4 cos 200t , 20600
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Dependent current Source
Dependent current source is a current source whose value depends on a voltage or current
of someplace else in the network.
They can be classified into two categories:

VCCS : Voltage Controlled Current Source
Here, the current of dependent source depends on
voltage of some other point inside the network.

CCCS: Current Controlled Current Source
Here, the current of dependent source depends on current through some element
inside the network.
Note: While analyzing any electrical network the Independent & dependent voltage and
current sources are handled exactly in the same manners except in the following two cases 1. Analysis of the network using superposition theorem.
2. Analysis using the Thevenin’s and Norton’s theorem.
In such cases,
 All Independent voltage sources are short circuited or replaced by the internal
impedances.
 All independent current sources are open circuited or replaced by their internal
impedances.
 All dependent voltage and current sources remain as is it. These sources are neither Short
Circuited nor Open Circuited.
Kirchhoff’s Laws
1. Kirchhoff Current Law (KCL): Total current or charge entering a junction or node is
exactly equal to the charge leaving the node assuming there is no charge storage at the
node.
In other words the algebraic sum of all the currents entering and leaving a node must be
equal to zero, I(exiting) + I(entering) = 0. This idea by Kirchhoff is commonly known as
the conservation of charge. It is always defined at a node.
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By KCL ∑leaving currents + ∑ entering currents=0
i1  i3  i4  i2  i5  0  i1  i3  i4  i2  i5
i.e. sum of leaving current is equal to sum of entering current
i
dq
dq
, at a given time ‘t’
is same for all branches
dt
dt
 q1  q3  q4  q2  q5
i.e. sum of leaving charge is equal to sum of entering charge
 KCL applies to any lumped electric circuit, it does not matter whether the elements are
linear, non-linear, active, passive, time varying, time invariant, etc. KCL is independent of
the nature of the elements connected to the node.
 Since there is no accumulation of the charge at any node, KCL expresses conservation of
charge at each and every node in a lumped electric circuit.
2. Kirchhoff Voltage Law (KVL): In any closed loop network, the total voltage around the
loop is equal to the sum of all the voltage drops within the same loop which is also equal to
zero.
In other words the algebraic sum of all voltages within the loop must be equal to zero. This
idea by Kirchhoff is known as the conservation of energy. It is always defined in loop or
mesh i.e. in a closed path.
By KVL, ∑Branch voltages=0
V  VR  VL  VC  0 => V  VR  VL  VC
V
W
dq
and i=
q
dt
For series connection charge flow will be same for all elements

W WR WL WC



=> W=WR  WL  WC
q
q
q
q
 KVL applies to any lumped electric circuit, it does not matter whether the elements are
linear, non-linear, active, passive, time varying, time invariant, etc. KVL is independent of
the nature of the elements present in the loop.
 KVL expresses conservation of energy in every loop of a lumped electric circuit.
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Assumptions
For KCL
For KVL
Voltage drop = -ve
Voltage rise = +ve
Note
 Parallel combination of voltage source and current source act as constant voltage source.
 Voltage across any current source is purely arbitrary and depends upon externally
connected voltage source or externally connected elements.
 We can't write any KVL equation in a closed loop which contains a current source since
voltage across the current source is purely arbitrary and in general is unknown.
 Series connection of voltage and current source act as a constant current source.
 The current through any voltage source is purely arbitrary, and depends upon externally
connected current source or externally connected elements.
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 We can’t write a KCL equation at a node where voltage source is connected because the
current through the voltage source is unknown.
Nodal Analysis
KCL+ Ohm’s Law =Nodal Analysis
Steps:
1. Identify total number of nodes.
2. Assign one of the node to ground node i.e. voltage of that node is always zero.
3. Use KCL and Ohm’s law to write the nodal equations.
Solved Examples
Problem: Write nodal equations for the circuit given below.
Solution: Nodal analysis
Total number of nodes =4
Assign one node voltage to zero
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Apply KCL at node ‘A’
Va  Vc
V  0 Va  Vb
2 a

 3  0 ………………….(1)
5
1
2
Apply KCL at node ‘B’
3 
Vb  Va Vb Vc Vc  Va



 4  0 ……....……..(2)
2
3
4
5
And Vb  Vc  5 ………………...……………………………….(3)
From equations (1) & (3)
17Va  7Vc  25 ……………………………………………….(4)
From equations (2) & (3)
42Va  77Vc  310 ……………………………………..….(5)
From equations (4), (5) & (3)
Va  4.03, Vb =-1.226, Vc  6.226
Mesh Analysis
KVL + Ohm’s Law = Mesh analysis
Steps:
1. Identify total number of meshes.
2. Assign mesh currents.
3. Use KVL and Ohm’s law to write the mesh equations.
Solved Examples
Problem: Find the power delivered by the voltage source
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Solution: Mesh analysis
Total number if meshes = 3
i1, i2 and i3 are mesh currents
Apply KVL in lower loop
-1 i1  i3   i1  2i2  1 i2  i3   0
2i1  2i3  3i2  0...............(1)
Apply KVL in upper loop
 i1  i3   3i3  5  1 i2  i3   0
i1  5i3  i2  5....................(2)
And i1  i2  10.....................(3)
From eq. (1) & (2)
8i1  13i2  10.................(4)
From eq. (3) & (4)
i1 
20
10
5
, i2 
& i3  
3
3
3
5
25
Power delivered by the voltage source =  5 
W
3
3
Connections of Elements
Series Connection
Elements are said to be in series when elements are connected to a single node and current
through these elements are same.
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For series connection Z eq  Z1  Z 2 ,

1 
 ZR  R , ZL  jL, Z C 

jC 

R : R eq  R1  R 2
L : L eq  L1  L 2
C : Ceq 
1 1

C1 C2
 Two current sources of different values can never be exist in series because they violate
KCL. If they exist in series they must be equal.
 Voltage sources of any value can be exist in series.
Parallel Connection
Elements are said to be in parallel if they form a loop containing no other elements and
voltages across the elements are same.
For Parallel connection
R:
1
1
1


R eq R1 R 2
L:
1
1 1
 
L eq L1 L 2
1
1
1

 ,
Z eq Z1 Z 2

1 
 ZR  R , ZL  jL, Z C 

jC 

C : Ceq  C1  C2
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 Two voltage source of different value can never be exist in parallel because they violate
KVL. If they exist in parallel they must be equal in both magnitude and polarity.
 Current sources of any value can be exist in parallel
Voltage Division
V  IZ eq => I=
V
Z1  Z 2
 Z1 
V1  IZ1=V 

 Z1  Z2 
 Z2 
V2  IZ2=V 

 Z1  Z2 
 R1 
 R2 
R : V1=V 
 ; V2=V 

 R1  R2 
 R1  R 2 
 L1 
 L2 
L : V1=V 
 ; V2=V 

 L1  L2 
 L1  L2 
 C2 
 C1 
C : V1=V 
 ; V2=V 

 C1  C2 
 C1  C2 
Current Division
V  IZ eq




1   Z1 Z 2 

=> V=I
 I

 1 1   Z1  Z 2 
  
 Z1 Z 2 
I1 
V  Z2 
=I 

Z1  Z1  Z2 
I1 
V  Z1 
=I 

Z2  Z1  Z2 
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 R2 
 R1 
R : I1=I 
 ; I2=I 

 R1  R2 
 R1  R2 
 L2 
 L1 
L : I1=I 
 ; I2=I 

 L1  L2 
 L1  L2 
 C1 
 C2 
C : I1=I 
 ; I2=I 

 C1  C2 
 C1  C2 
Delta to Star
Z1 
 Y  Transformation
Z aZ c
Z a  Zb  Z c
;
Z2 
Z a Zb
Z a  Zb  Z c
;
Z3 
Zb Z c
Z a  Zb  Z c
R: R1 
R aR c
R a  Rb  R c
R2 
R aR b
R a  Rb  R c
R3 
R cRb
R a  Rb  R c
L: L1 
LaL c
L a  Lb  L c
L2 
LaLb
L a  Lb  L c
L3 
LcLb
L a  Lb  L c
1 . 1
Cc
Ca
1
C:

1
1
1
C1


Ca Cb Cc
1 . 1
Ca
Cb
1

1
1
1
C2


Ca Cb Cc
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1 . 1
Cc
Cb
1

1
1
1
C3


Ca Cb Cc
20
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Network Analysis (Basic Concepts)
Star to Delta
 Y  Transformation
Z Z  Z2 Z3  Z3Z1
Za  2 1
;
Z3
Z Z  Z2 Z3  Z3Z1
Zb  2 1
;
Z1
R 2R1  R 3R1  R 2R 3
R1
Rc 
L1L3  L3L 2  L1L 2
L1
Lc 
R: R a 
R1R 3  R 3R 2  R1R 2
R3
Rb 
L: L a 
L1L3  L3L 2  L1L 2
L3
Lb 
C:
Z Z  Z2 Z3  Z3Z1
Zc  2 1
;
Z2
1 1
1 1
1 1


C C
C3 C2 C1 C2
1
 1 3
1
Ca
C3
R 3R 2  R1R 2  R 3R1
R2
L1L3  L3L 2  L1L 2
L2
1 1
1 1
1 1


C C
C3 C2 C1 C2
1
 1 3
1
Cb
C1
1 1
1 1
1 1


C C
C3 C2 C1 C2
1
 1 3
1
Cc
C2
Solved Examples
Problem: find Zin?
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Network Analysis (Basic Concepts)
Solution: Converting delta in to star
Z1 
3j  6 j
 18 j
4 j  3j  6 j
Z2 
4 j  6 j
 24 j
4 j  3j  6 j
Z3 
3j  4 j
 12j
4 j  3j  6 j
Zin  12j ||  12j  
Source Transformation Technique
The source transformation of a circuit is the transformation of a power source from a voltage
source to a current source, or a current source to a voltage source. It is to eliminate extra
nodes present in the network. KVL and KCL is not applicable for practical sources, so source
transformation technique is not applicable for practical sources.
Current source to Voltage source
Voltage source to current source
Note: Source transformation is also applicable for dependent sources, provided that
controlled variable must be outside the branch, where the source transformation is applied.
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Network Analysis (Basic Concepts)
Open Circuit (O.C.)
A circuit is called open circuit if current in the circuit is zero. If any passive element is
connected in series with O.C. can be neglected.
R O.C. 
V

I
R can be neglected
Short Circuit (S.C.)
A circuit is called short circuit if voltage across two points is zero. If any passive element is
connected parallel with S.C. can be neglected.
R S.C. 
V
0
I
R3, R4 can be neglected
Ratings or specification
They represent maximum permissible safe values for continuous operation of electrical
device.
 Increase the cross-sectional area of conductor to increase the current.
 Increase the insulation with standing capacity to increase the voltage.
 Most of the electrical utilities are designed for constant voltage, but the current depends
up on their load level.
Finding Equivalent Resistance
 Short circuit all the voltage sources present in the circuit
 Open circuit all the current sources present in the circuit
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Network Analysis (Basic Concepts)
Solved Example
Problem: Find Req=?
Z
Z 
Solution: It is a balanced wheatstone bridge  1  2 
 Z3 Z 4 
Circuit can be redrawn as
R eq  4 || 8 
8

3
Problem: Find Req=?
Solution: convert delta in to star
R eq   2 || 3   1 
11

5
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Network Analysis (Basic Concepts)
Problem: Find Rxy=?
Solution: After Short circuit the voltage source, circuit will become
R xy   3 || 1    2 || 4  
25

12
Problem: Find Rab=?
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Network Analysis (Basic Concepts)
Solution: After Open circuit the current source circuit will become
No current will flow in 3Ω , so it can be neglected
R ab   3 || 7  
21

10
Network Theorems
Solving for currents and voltages in multi-loop electric circuits can be quite complicated,
particularly for AC circuits. The voltage law and current law always apply, but using the may
lead to long systems of equation. Network Theorems used to simplify Electrical networks:
Superposition Theorem
In any linear, active, bilateral network consisting of number of energy sources resistance etc.
the effect produced by any element when all sources act at a time is equal to the sum of
effect produced by the same element when each source is considered individually.
Consider these points when apply super position theorem: All other independent voltage sources are S.C or replaced by their internal impedances.
 All other independent current sources are O.C or replaced by their internal impedances.
 All dependent voltage and current sources remain as they are and these sources are
neither S.C nor O.C.
 The theorem is not applicable to the networks containing
(1) Non-linear elements
(2) Unilateral elements such as P-N diode.
2
 The theorem is also not applicable to non-linear parameters such as power since P  V .
 The presence of dependent sources makes the network an active and hence super
position theorem is used for both active as well as passive networks
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Network Analysis (Basic Concepts)
Homogeneity Principle
In a linear network if the excitation is multiplied with a constant ‘K’, then the response in all
other branches of the network are also multiplied with the same constant K.
Here the excitation is multiplied by 1/3 and the response of the each branch also multiplied
by 1/3.
 Homogeneity principle is applicable only for those networks which contains one source.
Solved Example
Problem: Find the value of I=?
Solution: Apply super position theorem
I due to 1A current source,
 0.5 
I1  1 
  0.25A
 1.5  0.5 
I due to 1V voltage source,
I2  0A  Balanced bridge
I  I1  I2  0.25A
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Network Analysis (Basic Concepts)
Problem: Network is nonlinear containing only resistors.
If i1  8A;i2  12A; Vx  80V
i1  8A;i2  4A; Vx  0
i1  i2  20A; Vx  ?
Solution: If the direction of current or the voltage polarity is changed then equations will not
change because condition given will be independent of polarity.
Apply super position theorem
i2  0; Vx1  i1R1
i1  0; Vx2  i2R2
Vx  Vx1  Vx2  i1R1  i2R2
Put values:
80  8R1  12R2 .........(1)
0  8R1  4R2 ..........(2)
From Eq. (1) & (2)
16R2  80 => R2  5
20 5
  2.5
8
2
5
Vx  20   20  5  150 V
2
R1 
Thevenin’s Theorem
A linear, active, RLC network which contains one or more independent or dependent voltage
and current sources can be replaced by single voltage source VOC in series with an
equivalent impedances.
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Network Analysis (Basic Concepts)
Norton’s Theorem
A linear active RLC network which contains one or more independent voltage and current
sources can be replaced by single current source in shunt with an equivalent impedance.
R eq Represented equivalent impedances between two terminals for both Thevenin’s and
Norton when: All independent voltage sources are S.C. or replaced by their internal impedances.
 All Independent current sources are O.C. or replaced by their internal impedances.
 All dependent voltage and current sources remain as they are and these sources are
neither S.C nor O.C.
 These theorems is always applicable irrespective of
(a) Type of the elements contain.
(b) Type of voltage and Current sources are present.
 These theorems is not applicable to the network containing
(a) Non-linear elements
(b) Unilateral elements such as P-N junction diode.
 Thevenin’s and Norton duals of each other because they are source transferable.
Methods for finding Rth
Category 1: Circuit with independent sources only. In this case we can find Rth or RN by
direct method. S.C. the voltage source and O.C. the current source and find Rth or RN across
the desired points.
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Network Analysis (Basic Concepts)
Solved Examples
Problem: Find current i, using both thevenin’s and Norton’s theorems.
Solution: Thevenin’s equivalent
Open circuit terminals across which Theorem is applied,
Short circuit Voltage source and open circuit the current source
Vth  Voc  10  2  12V
Rth  Req  1
i
12
 4A
3
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Network Analysis (Basic Concepts)
Norton’s equivalent
ISC  10  2  12A
Rth  Req  1
1
i  12    4A
3
Category 2: Problems with both independent and dependent sources. Since dependent
sources can’t be expressed in terms of resistance, determining Rth or RN is not possible
V
directly instead we use ohm’s law where R th or RN  oc at the targeted terminals.
Isc
Solved Examples
Problem: Find current i, using both thevenin’s and Norton’s theorems.
Solution: Thevenin’s equivalent
Apply KVL in closed loop
10  Vx  Voc and 10  2Vx  Voc
 Vx  0  Voc  10
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Network Analysis (Basic Concepts)
Norton’s equivalent
Apply KCL at node A
VX  2VX  ISC => VX

ISC
3
Apply KVL
10  VX  2ISC  0
10 
ISC
30
 2ISC  0 => ISC 
3
7
 R th or RN 
10  7 7
 
30
3
Category 3: Problem with only dependent source only. Such networks can’t function on their
own, as there is no independent active source to drive it. In such models Thevenin’s
equivalent will not have Thevenin’s voltage and Norton equivalent will not have Norton’s
current but they have only resistance and this resistance can be indirectly determine by
V
externally exciting them with the source of Voltage ‘V’ (ohm’s law) where R th or RN  .
1
Ex. H parameter equivalent circuit of an amplifier.
Solved Examples
Problem: Determine R th or RN between XY
Solution: connect 1V voltage source across XY
V0  2 2V0  I => V0   2I
3
Apply KVL


1  4I  1 2V0  I   V0  0
1  4I 
2  2I
2I
I 0
3
3
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Network Analysis (Basic Concepts)
1
A
3
1
R th   3
I
I
Problem:
Find Voc ,R th & Isc ,R th across ab.
Solution: To find Voc circuit can be redrawn as
Apply KVL in 1st loop
2  103  2  103 i1 
Voc
5000
Apply KVL in 2nd loop
20  103  50i1  Voc  0
10
Voc 
9
For Isc circuit will be
103  2  2i1  103  i1  106
Isc  50i1  50  106  50A
For R eq , independent voltage source will be short circuited. Dependent current and voltage
source are remain same.
R eq 
Voc
Isc


10
9
50  106

200
k or we can connect voltage or current source of 1V or 1A
9
across a & b.
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Network Analysis (Basic Concepts)
Apply KVL in1st loop
2  103i1  1
0
5000
i1  0.1A
Apply KVL in 2nd loop
1  (20  103 )(I 50  0.1  106 )
9
mA
200
1 200
Req  
k
I
9
I
Problem: For the given system. Find Voc ,R th & Isc ,RN
Solution: For i=0, Voc  20V and For V=0, Isc  i  4A
V
20
R th  oc 
 5
Isc
4
Thevenin’s and Norton equivalent circuits
Problem: Find R th across ab, for the circuit shown below.
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Network Analysis (Basic Concepts)
Solution: Dependent source is there, so we can’t find R th directly. To find R th connect 1V
voltage source across ab.
Apply KCL at node A,
I  100i
Apply KVL in outer loop
3
1  10 i  i=-1mA
So, I=0.1A
1
R th   10
I
Problem: Find R th across 12, for the circuit shown below.
Solution: Dependent source is there, so we can’t find R th directly. To find R th connect 1V
voltage source across 12.
1  ib  10k => ib  0.1mA
Apply KCL at 1V node
1
1

 99ib  I
10k 100
101
Put value of ib,
 99  0.1  10 3  I =>I=20mA
10k
1
R th   50
I
Problem: Find ix using Norton’s Theorem
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Network Analysis (Basic Concepts)
Solution: Apply Norton’s theorem across 2 ohm. Dependent source is there, so we can’t
V
find RN directly, we use ohm’s law where RN  oc
Isc
For Voc
Apply KVL
10  VOC  3  0 => VOC  7V
For Isc
Apply KVL
10  1   ISC  3   2ISC => ISC 
 RN 
7
A
3
Voc 7  3

 3
Isc
7
Norton’s equivalent circuit
 7  3 7
i      A
x  3 5  5
Maximum Power Transfer Theorem
In any linear, active, bilateral consisting of number of energy sources with internal resistances
driving an external load, then power transfer to the load is maximum when load resistance is
equal to the source resistance
Condition for maximum power transfer from source to load.
VS
IL 
RS  RL
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Network Analysis (Basic Concepts)
PL  IL2RL 
VS2 .RL
(RS  RL )2
For maximum power
VS2
P

W
 0 =>RL  RS and max
4RS
dRL
d PL 
 The theorem is always valid for the variable load only.
 To apply maximum power transfer theorem any given network has to be first transformed
to the thevenin’s equivalent model.
 The maximum power transferred to the load is only 50% of the total power generated by
the source.
 The theorem is always valid irrespective of(a) Nature of source and load impedances.
(b) Type of voltage or the current sources.
Solved Example
Problem: In the circuit shown, find the value of load resistance R L , so that maximum power
is transferred to it? Hence calculate maximum power transferred to the load?
Solution: Convert it into Thevenin’s equivalent circuit
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Network Analysis (Basic Concepts)
Apply KVL in loop 1
20
20  (1  2) I1  I1 
3
Apply KVL in loop 2
20
20  (3  4) I2  I2 
7
40
VOC  2I1  4I2 
V
21
To find Rth, short the voltage source
2 12 50
R th   2 || 1    3 || 4   


3 7
21
50
For Maximum power RL  Req 
21
40
VOC
2
I
 21 
50 5
(RL  R eq )
2
21
2
 2   50  8
2
Pmax  (I) RL      
W
5
21
21
  

Solved Example
Problem: In the circuit shown, find the value of the load resistance R L for maximum power
transfer? Hence calculate the maximum power transferred to the load?
Solution: Convert it into Thevenin’s equivalent circuit
Two port equations:
I1  5V1  4V2      (1)
I2  4V1  5V2      (2)
100  I1  V1      (3)
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Network Analysis (Basic Concepts)
For VOC;
VOC = V2when I2=0
From eq. (1),
I1  5V1  4VOC      (4)
From eq. (2),
0  4V1  5VOC      (5)
From eq. (4) & (5) I1  5V1 
4  4V1
5
=> 5I1  9V1
500  9V1  5V1 => V1  250 V & I1  450 A
From eq. (3)
7
7
From eq. (5)
VOC  
200
7
For ISC;
ISC =-I2 when V2=0
From eq. (1) & (3)
From eq. (2)
V1 
100
6
Isc   I2  4V1  
200
3
200
7 3
R th 

200 7
Isc

3
Voc

3
For maximum power RL  R eq 
7
2
2
 V

VOC
2
OC


Pmax  I RL 
R 
 476.19W
 Req  RL  L 4RL


Reciprocity Theorem
In a linear, passive, bilateral, single source network, the ratio of the excitation to response is
always constant when the positions of the excitation and response are interchanged.
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Network Analysis (Basic Concepts)
I
I
I
Apply Homogeniety Theorem
 Cons tant 
 1  2
V
V1 V2
 The basis of the theorem is the symmetry of impedance and admittance matrix.
 Z11

Z   Z21
 Z31
Z12
Z22
Z32
Z13 

Z23  For reciprocal network, reciprocity theorem is always applicable.
Z33 
Z12  Z21 ; Z13  Z31 ; Z23  Z32
 In verifying reciprocity theorem the network configuration remains same whereas only
the external condition of the network changed
 Theorem is not valid for the network containing:
(a) Non linear elements
(b) Unilateral (P-N junction diodes)
(c) Multiple voltage and current sources
(d) Single dependent source in the network
 While writing the reciprocal network of given network ideal voltage source is connected
to series with target branch and ideal current source is connected to parallel to the target
branch.
Ex. Telephone network, excitation is telephone exchange and response is our telephone
Example:
Solved Example
Problem: Find I2?
Solution: Apply reciprocity theorem
V'1 V'2
4 30

=> 
=> I1  52.5A
7
I2
I'2
I'1
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Network Analysis (Basic Concepts)
Tellegen’s Theorem
In any linear, time invariant network the sum of instantaneous power absorbed by various
elements in all the branches of the network is always equal to zero.
b
Vk .ik  0
k 1

Where Vk  Branch voltages; ik  Branch currents ; b =no. of branches in network
 In any network, total instantaneous power supplied by various voltage and current
sources is always equal to total power absorbed by various passive elements in different
branches of the network.
 The theorem is always applicable irrespective of:(a) Shape of the network
(b) Type of elements contained in the network.
(c) Value of each element.
 The theorem is always applicable as long as the KVL and KCL equation are applicable to
the network.
 This theorem is verification of law of conservation of energy.
Conventions:
1.
2.
3.
Power delivered =VI
Power absorbed =VI
Power absorbed =VI
Solved Example
Problem: In the circuit shown, find the missing branch voltages and branch current.
Calculate power absorbed by each element in various branches of the network. Hence find
the total power absorbed by various elements in the different branches of the network?
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i1  2A, v1  4V
i3  4A, v2  2V
i6  4A, v 4  3V
Solution: All currents are absorbed currents for all elements
Apply KVL in loop 1
V1  V2  V3  0 => 4  2  V3  0 => V3  2V
Apply KVL in loop 2
V3  V4  V5  0 => 2  3  V5  0 => V5  5V
Apply KVL in loop 3
V2  V6  V4  0 => 2  V6  3  0 => V6  1V
Apply KCL at node 2
i1  i2  i6  0 => 2  i2  4  0 => i2  6A
Apply KCL at node A
i2  i3  i4  0 => 6  4  i4  0 => i4  2A
Apply KCL at node 4
i4  i5  i6  0 => 2  i5  4  0 => i5  2A
1
4
2
2
3
-2
4
3
5
5
6
-1
ik
2
-6
4
2
2
4
Vk .ik
8
-12
-8
6
10
-4
Vk
6
 Vk .ik  8  12  8  6  10  4  0
k 1
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Network Analysis (Basic Concepts)
Milliman’s Theorem
n
 EiYi
Where E  i1
n

i1
and Z 
Yi
1i
n
 Yi
i1
 The theorem is an extension of the Thevenin’s theorem and is useful whenever there are
large number of branches in the given network.
 The theorem is applicable only when any given network is rearranged in the standard
format.
Dual Miliman’s Theorem

This theorem is an extension of the Norton’s theorem and is useful whenever there are
large number of current sources in the given network.

The theorem is applicable only when any given network is re arranged in the standard
format.
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