Chapter

Circuit Concepts and

Network Simplification
1 Techniques

1.1 Introduction

Today we live in a predominantly electrical world. Electrical technology is a driving force

in the changes that are occurring in every engineering discipline. For example, surveying
is now done using lasers and electronic range finders.
Circuit analysis is the foundation for electrical technology. An indepth knowledge of
circuit analysis provides an understanding of such things as cause and effect, feedback
and control and, stability and oscillations. Moreover, the critical importance is the fact
that the concepts of electrical circuit can also be applied to economic and social systems.
Thus, the applications and ramifications of circuit analysis are immense.
In this chapter, we shall introduce some of the basic quantities that will be used
throughout the text. An electric circuit or electric network is an interconnection
of electrical elements linked together in a closed path so that an electric current
may continuously flow. Alternatively, an electric circuit is essentially a pipe-line that
facilitates the transfer of charge from one point to another.

1.2 Current, voltage, power and energy

The most elementary quantity in the analysis of electric circuits is the electric charge.
Our interest in electric charge is centered around its motion results in an energy transfer.
Charge is the intrinsic property of matter responsible for electrical phenomena. The
quantity of charge q can be expressed in terms of the charge on one electron. which is
1:602  10 19 coulombs. Thus, 1 coulomb is the charge on 6:24  1018 electrons. The
current flows through a specified area A and is defined by the electric charge passing
through that area per unit time. Thus we define q as the charge expressed in coulombs.
Charge is the quantity of electricity responsible for electric phenomena.
2 j Network Theory

The time rate of change constitutes an electric current. Mathemetically, this relation
is expressed as

dq (t)
i(t) = (1.1)
Z dt
t
or q (t) = i(x)dx (1.2)
1
The unit of current is ampere(A); an ampere is 1 coulomb per second.
Current is the time rate of flow of electric charge past a given point .
The basic variables in electric circuits are
current and voltage. If a current flows into
terminal a of the element shown in Fig. 1.1,
then a voltage or potential difference exists
between the two terminals a and b. Nor-
mally, we say that a voltage exists across Figure 1.1 Voltage across an element
the element.
The voltage across an element is the work done in moving a positive charge
of 1 coulomb from first terminal through the element to second terminal. The
unit of voltage is volt, V or Joules per coulomb.
We have defined voltage in Joules per coulomb as the energy required to move a
positive charge of 1 coulomb through an element. If we assume that we are dealing with
a differential amount of charge and energy,
dw
then v= (1.3)
dq

Multiplying both the sides of equation (1.3) by the current in the element gives
 
dw dq dw
vi =
dq dt
) dt
=p (1.4)

which is the time rate of change of energy or power measured in Joules per second or
watts (W ).
p could be either positive or negative. Hence it
is imperative to give sign convention for power.
If we use the signs as shown in Fig. 1.2., the
current flows out of the terminal indicated by x,
Figure 1.2 An element with the current
which shows the positive sign for the voltage. In
leaving from the terminal
this case, the element is said to provide energy
with a positive voltage sign
to the charge as it moves through. Power is then
provided by the element.
Conversely, power absorbed by an element is p = vi, when i is entering through the
positive voltage terminal.
 Circuit Concepts and Network Simplification Techniques j 3

Energy is the capacity to perform work. Energy and power are related to each
other by the following equation:
Z t
Energy = w = p dt
1
EXAMPLE 1.1
Consider the circuit shown in Fig. 1.3 with
v = 8e t V and i = 20e t A for t  0. Find
the power absorbed and the energy supplied
by the element over the first second of oper-
ation. we assume that v and i are zero for Figure 1.3
t < 0:
SOLUTION
The power supplied is
t
p = vi = (8e )(20e t )
2t
= 160e W

The element is providing energy to the charge flowing through it.

The energy supplied during the first seond is
Z 1 Z 1
2t
w= p dt = 160e dt
0 0
= 80(1 e 2 ) = 69.17 Joules

1.3 Linear, active and passive elements

A linear element is one that satisfies the prin-

ciple of superposition and homogeneity.
In order to understand the concept of super-
position and homogeneity, let us consider the Figure 1.4 An element with excitation
element shown in Fig. 1.4. i and response v

The excitation is the current, i and the response is the voltage, v . When the element
is subjected to a current i1 , it provides a response v1 . Furthermore, when the element is
subjected to a current i2 , it provides a response v2 . If the principle of superposition is
true, then the excitation i1 + i2 must produce a response v1 + v2 .
Also, it is necessary that the magnitude scale factor be preserved for a linear element.
If the element is subjected to an excitation i where  is a constant multiplier, then if
principle of homogencity is true, the response of the element must be v .
We may classify the elements of a circuir into categories, passive and active, depending
upon whether they absorb energy or supply energy.
4 | Network Theory

1.3.1 Passive Circuit Elements

An element is said to be passive if the total energy delivered to it from the rest of the
circuit is either zero or positive.
Then for a passive element, with the current flowing into the positive (+) terminal as
shown in Fig. 1.4 this means that
t
w= vi dt ≥ 0
−∞

Examples of passive elements are resistors, capacitors and inductors.

1.3.1.A Resistors
Resistance is the physical property of an ele-
ment or device that impedes the flow of cur-
rent; it is represented by the symbol R.
Figure 1.5 Symbol for a resistor R
Resistance of a wire element is calculated us-
ing the relation:
ρl
R= (1.5)
A
where A is the cross-sectional area, ρ the resistivity, and l the length of the wire. The
practical unit of resistance is ohm and represented by the symbol Ω.
An element is said to have a resistance of 1 ohm, if it permits 1A of
current to flow through it when 1V is impressed across its terminals.
Ohm’s law, which is related to voltage and current, was published in 1827 as
v = Ri (1.6)
v
or R=
i
where v is the potential across the resistive element, i the current through it, and R the
resistance of the element.
The power absorbed by a resistor is given by
 v  v2
p = vi = v = (1.7)
R R
Alternatively,
p = vi = (iR)i = i2 R (1.8)
Hence, the power is a nonlinear function of current i through the resistor or of the
voltage v across it.
The equation for energy absorbed
 by or delivered
 to a resistor is
t t
w= pdτ = i2 R dτ (1.9)
−∞ −∞

Since i2 is always positive, the energy is always positive and the resistor is a passive
element.
 Circuit Concepts and Network Simplification Techniques j 5

1.3.1.B Inductors

Whenever a time-changing current is passed through a coil or wire, the voltage across
it is proportional to the rate of change of current through the coil. This proportional
relationship may be expressed by the equation
di
v=L (1.10)
dt
Where L is the constant of proportionality known as induc-
tance and is measured in Henrys (H). Remember v and i are
both funtions of time.
Let us assume that the coil shown in Fig. 1.6 has N turns and
the core material has a high permeability so that the magnetic
fluk  is connected within the area A. The changing flux
creates an induced voltage in each turn equal to the derivative Figure 1.6 Model of the
of the flux , so the total voltage v across N turns is inductor
d
v=N (1.11)
dt

Since the total flux N  is proportional to current in the coil, we have

N  = Li (1.12)

Where L is the constant of proportionality. Substituting equation (1.12) into equa-

tion(1.11), we get
di
v=L
dt
The power in an inductor is
 
di
p = vi = L i
dt

The energy stored in the inductor is

Z t
w= p d
1
Z i(t)
1 2
=L i di = Li Joules (1.13)
i( 1) 2

Note that when t = 1; i( 1) = 0. Also note that w(t)  0 for all i(t); so the
inductor is a passive element. The inductor does not generate energy, but only stores
energy.
6 j Network Theory

1.3.1.C Capacitors

A capacitor is a two-terminal element that is a model of a

device consisting of two conducting plates seperated by a di-
electric material. Capacitance is a measure of the ability of C
a deivce to store energy in the form of an electric field.
Capacitance is defined as the ratio of the charge
stored to the voltage difference between the two con- 1.7 Circuit symbol for
ducting plates or wires, a capacitor
q
C=
v
The current through the capacitor is given by
dq dv
i= =C (1.14)
dt dt
The energy stored in a capacitor is
Zt
w= vi d
1
Remember that v and i are both functions of time and could be written as v (t) and
i(t).

dv
Since i=C
dt
Zt
dv
we have w= v C d
d
1
Zv(t)
1 2? ?v(t)
=C v dv = Cv ?
2 v ( 1)
v( 1)
Since the capacitor was uncharged at t = 1, v( 1) = 0.
Hence w = w (t)
1
= Cv 2 (t) Joules (1.15)
2
Since q = Cv; we may write
1 2
w (t) = q (t) Joules (1.16)
2C
Note that since w(t)  0 for all values of v(t), the element is said to be a passive
element.
 Circuit Concepts and Network Simplification Techniques | 7

1.3.2 Active Circuit Elements (Energy Sources)

An active two-terminal element that supplies energy to a circuit is a source of energy. An

ideal voltage source is a circuit element that maintains a prescribed voltage across the
terminals regardless of the current flowing in those terminals. Similarly, an ideal current
source is a circuit element that maintains a prescribed current through its terminals
regardless of the voltage across those terminals.
These circuit elements do not exist as practical devices, they are only idealized models
of actual voltage and current sources.
Ideal voltage and current sources can be further described as either independent
sources or dependent sources. An independent source establishes a voltage or current
in a circuit without relying on voltages or currents elsewhere in the circuit. The value of
the voltage or current supplied is specified by the value of the independent source alone.
In contrast, a dependent source establishes a voltage or current whose value depends on
the value of the voltage or current elsewhere in the circuit. We cannot specify the value
of a dependent source, unless you know the value of the voltage or current on which it
depends.
The circuit symbols for ideal independent sources are shown in Fig. 1.8.(a) and (b).
Note that a circle is used to represent an independent source. The circuit symbols for
dependent sources are shown in Fig. 1.8.(c), (d), (e) and (f). A diamond symbol is used
to represent a dependent source.

Figure 1.8 (a) An ideal independent voltage source

(b) An ideal independent current source
(c) voltage controlled voltage source
(d) current controlled voltage source
(e) voltage controlled current source
(f) current controlled current source
8 | Network Theory

1.4 Unilateral and bilateral networks

A Unilateral network is one whose properties or characteristics change with the direction.
An example of unilateral network is the semiconductor diode, which conducts only in one
direction.
A bilateral network is one whose properties or characteristics are same in either direc-
tion. For example, a transmission line is a bilateral network, because it can be made to
perform the function equally well in either direction.

1.5 Network simplification techniques

In this section, we shall give the formula for reducing the networks consisting of resistors
connected in series or parallel.

1.5.1 Resistors in Series

When a number of resistors are connected in series, the equivalent resistance of the com-
bination is given by
R = R1 + R2 + · · · + Rn (1.17)
Thus the total resistance is the algebraic sum of individual resistances.

Figure 1.9 Resistors in series

1.5.2 Resistors in Parallel

When a number of resistors are connected in parallel as shown in Fig. 1.10, then the
equivalent resistance of the combination is computed as follows:
1 1 1 1
= + + ....... + (1.18)
R R1 R2 Rn
Thus, the reciprocal of a equivalent resistance of a parallel combination is the sum of
the reciprocal of the individual resistances. Reciprocal of resistance is conductance and
denoted by G. Consequently the equivalent conductance,
G = G1 + G2 + · · · + Gn

Figure 1.10 Resistors in parallel

 Circuit Concepts and Network Simplification Techniques | 9

1.5.3 Division of Current in a Parallel Circuit

Consider a two branch parallel circuit as shown in Fig. 1.11. The branch currents I1 and
I2 can be evaluated in terms of total current I as follows:
IR2 IG1
I1 = = (1.19)
R1 + R2 G1 + G2
IR1 IG2
I2 = = (1.20)
R1 + R2 G1 + G2

Figure 1.11 Current division in a parallel circuit

That is, current in one branch equals the total current multiplied by the resistance of the
other branch and then divided by the sum of the resistances.

EXAMPLE 1.2
The current in the 6Ω resistor of the network shown in Fig. 1.12 is 2A. Determine the
current in all branches and the applied voltage.

Figure 1.12

SOLUTION

Voltage across 6Ω = 6 × 2
= 12 volts
10 j Network Theory

Since 6Ω and 8Ω are connected in parallel, voltage

across 8Ω = 12 volts. 
Therefore, the current through 12
= = 1:5 A
8Ω (between A and B) 8
Total current in the circuit = 2 + 1:5 = 3:5 A
Current in the 4Ω branch = 3.5 A

Current through 8Ω (betwen C and D) = 3:5 

20
20 + 8
= 2.5 A
Therefore, current through 20Ω = 3:5 2:5
= 1A
6  8 8  20
Total resistance of the circuit =4+ +
6 + 8 8 + 20
= 13:143Ω
Therefore applied voltage, V = 3:5  13:143 (∵ V = IR)
= 46 Volts

EXAMPLE 1.3
Find the value of R in the circuit shown in Fig. 1.13.

Figure 1.13

SOLUTION
Voltage across 5Ω = 2:5  5 = 12:5 volts
Hence the voltage across the parallel circuit = 25 12.5 = 12.5 volts

Current through 20Ω = I1 or I2

12:5
= = 0:625A
20
 Circuit Concepts and Network Simplification Techniques j 11

Therefore, current through R = I3 = I I1 I2

= 2:5 0:625 0:625
= 1:25 Amps
12:5
Hence; R= = 10Ω
1:25

1.6 Kirchhoff’s laws

In the preceeding section, we have seen how simple resistive networks can be solved
for current, resistance, potential etc using the concept of Ohm’s law. But as the network
becomes complex, application of Ohm’s law for
solving the networks becomes tedious and hence
time consuming. For solving such complex net-
works, we make use of Kirchhoff’s laws. Gustav
Kirchhoff (1824-1887), an eminent German physi-
cist, did a considerable amount of work on the
principles governing the behaviour of eletric cir-
cuits. He gave his findings in a set of two laws: (i)
current law and (ii) voltage law, which together Figure 1.14 A simple resistive network
are known as Kirchhoff’s laws. Before proceeding for difining various circuit
to the statement of these two laws let us familar- terminologies
ize ourselves with the following definitions encoun-
tered very often in the world of electrical circuits:

(i) Node: A node of a network is an equi-potential surface at which two or more circuit
elements are joined. Referring to Fig. 1.14, we find that A,B,C and D qualify as
nodes in respect of the above definition.

(ii) Junction: A junction is that point in a network, where three or more circuit elements
are joined. In Fig. 1.14, we find that B and D are the junctions.

(iii) Branch: A branch is that part of a network which lies between two junction points.
In Fig. 1.14, BAD,BCD and BD qualify as branches.

(iv) Loop: A loop is any closed path of a network. Thus, in Fig. 1.14, ABDA,BCDB and
ABCDA are the loops.

(v) Mesh: A mesh is the most elementary form of a loop and cannot be further divided
into other loops. In Fig. 1.14, ABDA and BCDB are the examples of mesh. Once
ABDA and BCDB are taken as meshes, the loop ABCDA does not qualify as a mesh,
because it contains loops ABDA and BCDB.
12 j Network Theory

1.6.1 Kirchhoff’s Current Law

The first law is Kirchhoff’s current law(KCL), which states that the algebraic sum of
currents entering any node is zero.
Let us consider the node shown in Fig. 1.15. The sum of the currents entering the
node is
ia + ib ic + id = 0
Note that we have ia since the current ia is leaving the node. If we multiply the
foregoing equation by 1, we obtain the expression

ia ib + ic id = 0

which simply states that the algebraic sum of currents leaving a node is zero. Alternately,
we can write the equation as
ib + id = ia + ic
which states that the sum of currents entering a node
is equal to the sum of currents leaving the node. If the
sum of the currents entering a node were not equal
to zero, then the charge would be accumulating at a
node. However, a node is a perfect conductor and
cannot accumulate or store charge. Thus, the sum of
Figure 1.15 Currents at a node
currents entering a node is equal to zero.

1.6.2 Kirchhoff’s Voltage Law

Kirchhoff’s voltage law(KVL) states that the algebraic sum of voltages around any closed
path in a circuit is zero.
In general, the mathematical representation of Kirchhoff’s voltage law is
X
N

vj (t) = 0
j =1

where vj (t) is the voltage across the j th branch (with proper reference direction) in a loop
containing N voltages.
In Kirchhoff’s voltage law, the algebraic sign
is used to keep track of the voltage polarity.
In other words, as we traverse the circuit, it is
necessary to sum the increases and decreases
in voltages to zero. Therefore, it is impor-
tant to keep track of whether the voltage is
increasing or decreasing as we go through each
element. We will adopt a policy of consider-
ing the increase in voltage as negative and a
decrease in voltage as positive. Figure 1.16 Circuit with three closed paths
 Circuit Concepts and Network Simplification Techniques j 13

Consider the circuit shown in Fig. 1.16, where the voltage for each element is identified
with its sign. The ideal wire used for connecting the components has zero resistance,
and thus the voltage across it is equal to zero. The sum of voltages around the loop
incorporating v6 ; v3 ; v4 and v5 is
v6 v3 + v4 + v5 = 0

The sum of voltages around a loop is equal to zero. A circuit loop is a conservative
system, meaning that the work required to move a unit charge around any loop is zero.
However, it is important to note that not all electrical systems are conservative. Ex-
ample of a nonconservative system is a radio wave broadcasting system.

EXAMPLE 1.4
Consider the circuit shown in Fig. 1.17. Find each branch current and voltage across
each branch when R1 = 8Ω; v2 = 10 volts i3 = 2A and R3 = 1Ω. Also find R2 .

Figure 1.17

SOLUTION

Applying KCL (Kirchhoff’s Current Law) at node A, we get

i1 = i2 + i3

and using Ohm’s law for R3 , we get

v3 = R3 i3 = 1(2) = 2V

Applying KVL (Kirchhoff’s Voltage Law) for the loop EACDE, we get

10 + v1 + v3 = 0
) v1 = 10 v3 = 8V
14 | Network Theory

Ohm’s law for R1 is

v1 = i1 R1
v1
⇒ i1 = = 1A
R1
Hence, i2 = i1 − i3
= 1 − 2 = −1A
From the circuit, v2 = R2 i2
v2 −10
⇒ R2 = = = 10Ω
i2 −1

EXAMPLE 1.5
Referring to Fig. 1.18, find the follow-
ing:
(a) ix if iy = 2A and iz = 0A
(b) iy if ix = 2A and iz = 2iy
(c) iz if ix = iy = iz

SOLUTION Figure 1.18

Applying KCL at node A, we get

5 + iy + iz = ix + 3

(a) ix = 2 + iy + iz
= 2 + 2 + 0 = 4A

(b) iy = 3 + ix − 5 − iz
= −2 + 2 − 2iy

⇒ iy = 0A

(c) This situation is not possible, since ix and iz are in opposite directions. The only
possibility is iz = 0, and this cannot be allowed, as KCL will not be satisfied (5 = 3).

EXAMPLE 1.6
Refer the Fig. 1.19.

(a) Calculate vy if iz = −3A

(b) What voltage would you need to replace 5 V source to obtain vy = −6 V if

iz = 0.5A?
 Circuit Concepts and Network Simplification Techniques j 15

Figure 1.19

SOLUTION

(a) vy = 1 (3 vx + iz )
Since vx = 5V and iz = 3A;
we get vy = 3(5) 3 = 12V
(b) vy = 1 (3 vx + iz ) = 6
= 3 vx + 0.5
) 3 vx = 6:5
Hence, vx = 2.167 volts

EXAMPLE 1.7
For the circuit shown in Fig. 1.20, find i1 and v1 , given R3 = 6Ω.

Figure 1.20

SOLUTION
Applying KCL at node A, we get
i1 i2 + 5 = 0
From Ohm’s law, 12 = i2 R3
) i2 =
12
R3
=
12
6
= 2A
Hence; i1 = 5 i2 = 3A
16 j Network Theory

Applying KVL clockwise to the loop CBAC, we get

v1 6i1 + 12 = 0
) v1 = 12 6i1
= 12 6(3) = 6volts

EXAMPLE 1.8
Use Ohm’s law and Kirchhoff’s law to evaluate (a) vx , (b) iin , (c) Is and (d) the power
provided by the dependent source in Fig 1.21.

Figure 1.21

SOLUTION
(a) Applying KVL, (Referring Fig. 1.21 (a)) we get

2 + vx + 8 = 0
) vx = 6V

Figure 1.21(a)
 Circuit Concepts and Network Simplification Techniques j 17

(b) Applying KCL at node a, we get

8vx
Is + 4vx + =
4 2
) Is + 4( 6)
6
4
=4
) Is 24 1:5 = 4
) Is = 29:5A

(c) Applying KCL at node b, we get

2 vx
iin = + Is + 6
2 4
) iin = 1 + 29:5
6
4
6 = 23A

(d) The power supplied by the dependent current source = 8 (4vx ) = 8 4 6 = 192W

EXAMPLE 1.9
Find the current i2 and voltage v for the circuit shown in Fig. 1.22.

Figure 1.22

SOLUTION
v
From the network shown in Fig. 1.22, i2 =
6
The two parallel resistors may be reduced to
36
Rp = = 2Ω
3+6
Hence, the total series resistance around the loop is

Rs = 2 + Rp + 4
= 8Ω
18 j Network Theory

Applying KVL around the loop, we have

21 + 8i 3i2 = 0 (1.21)

Using the principle of current division,

i2 =
iR2
=
i 3
R1 + R2 3+6
3i i
= =
9 3
) i = 3i2 (1.22)

Substituting equation (1.22) in equation (1.21), we get

21 + 8(3i2 ) 3i2 = 0
Hence; i2 = 1A
and v = 6i2 = 6V

EXAMPLE 1.10
Find the current i2 and voltage v for resistor R in Fig. 1.23 when R = 16Ω.

Figure 1.23

SOLUTION

Applying KCL at node x, we get

4 i1 + 3i2 i2 = 0
v v
Also; i1 = =
4+2 6
v v
i2 = =
R 16
v v v
Hence; 4 +3 =0
6 16 16
) v = 96volts
v 96
and i2 = = = 6A
6 16
 Circuit Concepts and Network Simplification Techniques j 19

EXAMPLE 1.11
A wheatstone bridge ABCD is arranged as follows: AB = 10Ω, BC = 30Ω, CD = 15Ω
and DA = 20Ω. A 2V battery of internal resistance 2Ω is connected between points A
and C with A being positive. A galvanometer of resistance 40Ω is connected between B
and D. Find the magnitude and direction of the galvanometer current.
SOLUTION

Applying KVL clockwise to the loop ABDA, we get

10ix + 40iz 20iy = 0
) 10ix 20iy + 40iz = 0 (1.23)
Applying KVL clockwise to the loop BCDB, we get
30(ix iz ) 15(iy + iz ) 40iz = 0
) 30ix 15iy 85iz = 0 (1.24)
Finally, applying KVL clockwise to the loop ADCA, we get
20iy + 15(iy + iz ) + 2(ix + iy ) 2 = 0
) 2ix + 37iy + 15iz = 2 (1.25)
Putting equations (1.23),(1.24) and (1.25) in matrix form, we get
2 32 3 2 3
10 20 40 ix 0
4 30 15 85 5 4 iy 5 = 4 0 5
2 37 15 iz 2
Using Cramer’s rule, we find that
iz = 0.01 A (Flows from B to D)
20 j Network Theory

1.7 Multiple current source networks

Let us now learn how to reduce a network having multiple current sources and a number
of resistors in parallel. Consider the circuit shown in Fig. 1.24. We have assumed that
the upper node is v (t) volts positive with respect to the lower node. Applying KCL to
upper node yields

i1 (t) i2 (t) i3 (t) + i4 (t) i5 (t) i6 (t) = 0

) i1 (t) i3 (t) + i4 (t) i6 (t) = i2 (t) + i5 (t) (1.26)
) io (t) = i2 (t) + i5 (t) (1.27)

Figure 1.24 Multiple current source network

where io (t) = i1 (t) i3 (t) + i4 (t) i6 (t) is the

algebraic sum of all current sources present
in the multiple source network shown in Fig.
1.24. As a consequence of equation (1.27), the
network of Fig. 1.24 is effectively reduced to
that shown in Fig. 1.25. Using Ohm’s law, the
currents on the right side of equation (1.27)
can be expressed in terms of the voltage and Figure 1.25 Equivalent circuit
individual resistance so that KCL equation
reduces to  
1 1
io (t) = + v (t)
R1 R2
Thus, we can reduce a multiple current source network into a network having only one
current source.

1.8 Source transformations

Source transformation is a procedure which transforms one source into another while
retaining the terminal characteristics of the original source.
Source transformation is based on the concept of equivalence. An equivalent circuit is
one whose terminal characteristics remain identical to those of the original circuit. The
term equivalence as applied to circuits means an identical effect at the terminals, but not
within the equivalent circuits themselves.
 Circuit Concepts and Network Simplification Techniques j 21

We are interested in transforming the circuit shown in Fig. 1.26 to a one shown in
Fig. 1.27.

Figure 1.26 Voltage source connected Figure 1.27 Current source connected
to an external resistance R to an external resistance R

We require both the circuits to have the equivalence or same characteristics between the
terminals x and y for all values of external resistance R. We will try for equivanlence of
the two circuits between terminals x and y for two limiting values of R namely R = 0
and R = 1. When R = 0, we have a short circuit across the terminals x and y . It is
obligatory for the short circuit to be same for each circuit. The short circuit current of
Fig. 1.26 is
vs
is = (1.28)
Rs

The short circuit current of Fig. 1.27 is is . This enforces,

vs
is = (1.29)
Rs

When R = 1, from Fig. 1.26 we have vxy = vs and from Fig. 1.27 we have vxy = is Rp .
Thus, for equivalence, we require that

vs = is Rp (1.30)
vs
Also from equation (1.29), we require is = . Therefore, we must have
Rs
 
vs
vs = Rp
Rs
) Rs = Rp (1.31)

Equations(1.29) and (1.31) must be true simulaneously for both the circuits for the two
sources to be equivalent. We have derived the conditions for equivalence of two circuits
shown in Figs. 1.26 and 1.27 only for two extreme values of R, namely R = 0 and R = 1.
However, the equality relationship holds good for all R as explained below.
22 j Network Theory

Applying KVL to Fig. 1.26, we get

vs = iRs + v

Dividing by Rs gives
vs v
=i+ (1.32)
Rs Rs

If we use KCL for Fig. 1.27, we get

v
is = i + (1.33)
Rp

Thus two circuits are equal when

vs
is = and Rs = Rp
Rs

Transformation procedure: If we have embedded within a network, a current source

i in parallel with a resistor R can be replaced with a voltage source of value v = iR in
series with the resistor R.
The reverse is also true; that is, a voltage source v in series with a resistor R can be
v
replaced with a current source of value i = in parallel with the resistor R. Parameters
R
within the circuit are unchanged under these transformation.

EXAMPLE 1.12
A circuit is shown in Fig. 1.28. Find the current i by reducing the circuit to the right of
the terminals x y to its simplest form using source transformations.

Figure 1.28
 Circuit Concepts and Network Simplification Techniques j 23

SOLUTION
The first step in the analysis is to transform 30 ohm resistor in series with a 3 V source
into a current source with a parallel resistance and we get:

Reducing the two parallel resistances, we get:

The parallel resistance of 12Ω and the current source of 0.1A can be transformed into
a voltage source in series with a 12 ohm resistor.

Applying KVL, we get

5i + 12i + 1:2 5=0

) 17i = 3:8
) i = 0:224A
24 j Network Theory

EXAMPLE 1.13
Find current i1 using source transformation for the circuit shown Fig. 1.29.

Figure 1.29

SOLUTION
Converting 1 mA current source in parallel with 47kΩ resistor and 20 mA current source
in parallel with 10kΩ resistor into equivalent voltage sources, the circuit of Fig. 1.29
becomes the circuit shown in Fig. 1.29(a).

Figure 1.29(a)

Please note that for each voltage source, “+” corresponds to its corresponding current
source’s arrow head.
Using KVL to the above circuit,

47 + 47  103 i1 4i1 + 13:3  103 i1 + 200 = 0

Solving, we find that

i1 = 4.096 mA

EXAMPLE 1.14
Use source transformation to convert the circuit in Fig. 1.30 to a single current source in
parallel with a single resistor.
 Circuit Concepts and Network Simplification Techniques j 25

Figure 1.30

SOLUTION

The 9V source across the terminals a0 and b0 will force the voltage across these two
terminals to be 9V regardless the value of the other 9V source and 8Ω resistor to its
left. Hence, these two components may be removed from the terminals, a0 and b0 without
affecting the circuit condition. Accordingly, the above circuit reduces to,

Converting the voltage source in series with 4Ω resistor into an equivalent current
source, we get,

Adding the current sources in parallel and

reducing the two 4 ohm resistors in parallel,
we get the circuit shown in Fig. 1.30 (a):

Figure 1.30 (a)

26 j Network Theory

1.8.1 Source Shift

The source transformation is possible only in the case of practical sources. ie Rs 6= 1

and Rp 6= 0, where Rs and Rp are internal resistances of voltage and current sources
respectively. Transformation is not possible for ideal sources and source shifting methods
are used for such cases.
Voltage source shift (E shift):
Consider a part of the network shown in Fig. 1.31(a) that contains an ideal voltage source.

Figure 1.31(a) Basic network

Since node b is at a potential E with respect to node a, the network can be redrawn
equivalently as in Fig. 1.31(b) or (c) depend on the requirements.

Figure 1.31(b) Networks after E-shift Figure 1.31(c) Network after the E-shift

Current source shift (I shift)

In a similar manner, current sources also can be shifted. This can be explained with an
example. Consider the network shown in Fig. 1.32(a), which contains an ideal current
source between nodes a and c. The circuit shown in Figs. 1.32(b) and (c) illustrates the
equivalent circuit after the I - shift.
 Circuit Concepts and Network Simplification Techniques j 27

Figure 1.32(a) basic network

Figure 1.32(b) and (c) Networks after I--shift

EXAMPLE 1.15
Use source shifting and transformation techiniques to find voltage across 2Ω resistor shown
in Fig. 1.33(a). All resistor values are in ohms.

Figure 1.33(a)
28 j Network Theory

SOLUTION
The circuit is redrawn by shifting 2A current source and 3V voltage source and further
simplified as shown below.

Thus the voltage across 2Ω resistor is

V =3 2 1 +4
1
1 +4 1
=3V
 Circuit Concepts and Network Simplification Techniques | 29

EXAMPLE 1.16
Use source mobility to calculate vab in the circuits shown in Fig. 1.34 (a) and (b). All
resistor values are in ohms.

Figure 1.34(a) Figure 1.34(b)

SOLUTION
(a) The circuit shown in Fig. 1.34(a) is simplified using source mobility technique, as
shown below and the voltage across the nodes a and b is calculated.

b a

Voltage across a and b is

1
Vab = =2V
3−1 + 10−1 + 15−1
30 j Network Theory

(b) The circuit shown in Fig. 1.34 (b) is reduced as follows.

Figure 1.34(c) Figure 1.34(d)

Figure 1.34(e)

From Fig. 1.34(e),

12 1  6
Vbc =
12 1 + 10 1 + 15 1
 12 = 24 V

Applying this result in Fig. 1.34(b), we get

vab = vac vbc

= 60 24 = 36 V

EXAMPLE 1.17
Use mobility and reduction techniques to solve the node voltages of the network shown
in Fig. 1.35(a). All resistors are in ohms.
 Circuit Concepts and Network Simplification Techniques j 31

Figure 1.35(a)

SOLUTION
The circuit shown in Fig. 1.35(a) can be reduced by using desired techniques as shown in
Fig. 1.35(b) to 1.35(e).

Figure 1.35(b)

From Fig. 1.35(e)

34
i= =2A
17
Using this value of i in Fig. 1.35(e),

Va = 92= 18 V
and Ve = Va 22 20 = 42V
32 j Network Theory

Figure 1.35(c)

Figure 1.35(d)

Figure 1.35(e)

From Fig 1.35(a)

Vd = Ve + 30 = 42 + 30 = 12V
 Circuit Concepts and Network Simplification Techniques j 33

Then at node b in Fig. 1.35(b),

Vb Vb Vd
45 + =0
2 8
Using the value of Vd in the above equation and rearranging, we get,
 
1 1 12
Vb + = 45
2 8 8
) Vb = 69:6 V

At node c of Fig. 1.35(b)

Vc Vc Ve
+ 45 + =0
5  10 
1 1 42
Vc + = 45
5 10 10
) Vc = 164 V

EXAMPLE 1.18
Use source mobility to reduce the network shown in Fig. 1.36(a) and find the value of Vx .
All resistors are in ohms.

Figure 1.36(a)

SOLUTION
The circuit shown in Fig. 1.36(a) can be reduced as follows and Vx is calculated.
Thus
Vx =
5
25
 18 = 3:6V
34 j Network Theory

1.9 Mesh analysis with independent voltage sources

Before starting the concept of mesh analysis, we want to reiterate that a closed path or
a loop is drawn starting at a node and tracing a path such that we return to the original
node without passing an intermediate node more than once. A mesh is a special case of
a loop. A mesh is a loop that does not contain any other loops within it. The network
shown in Fig. 1.37(a) has four meshes and they are identified as Mi , where i = 1; 2; 3; 4.
 Circuit Concepts and Network Simplification Techniques j 35

Figure 1.37(a) A circuit with four meshes. Each mesh is identified by a circuit

The current flowing in a mesh is defined as mesh

current. As a matter of convention, the mesh cur-
rents are assumed to flow in a mesh in the clock-
wise direction.
Let us consider the two mesh circuit of Fig.
1.37(b).
We cannot choose the outer loop, v ! R1 ! R2 !
v as one mesh, since it would contain the loop v !
R1 ! R3 ! v within it. Let us choose two mesh
currents i1 and i2 as shown in the figure. Figure 1.37(b) A circuit with two meshes

We may employ KVL around each mesh. We will travel around each mesh in the
clockwise direction and sum the voltage rises and drops encountered in that particular
mesh. We will adpot a convention of taking voltage drops to be positive and voltage rises
to be negative . Thus, for the network shown in Fig. 1.37(b) we have
Mesh 1 : v + i1 R1 + (i1 i2 )R3 = 0 (1.34)
Mesh 2 : R3 (i2 i1 ) + R2 i2 = 0 (1.35)

Note that when writing voltage across R3 in mesh 1, the current in R3 is taken as
i1 i2 . Note that the mesh current i1 is taken as ‘+ve’ since we traverse in clockwise
direction in mesh 1, On the other hand, the voltage across R3 in mesh 2 is written as
R3 (i2 i1 ). The current i2 is taken as +ve since we are traversing in clockwise direction
in this case too.
Solving equations (1.34) and (1.35), we can find the mesh currents i1 and i2 .
Once the mesh currents are known, the branch currents are evaluated in terms of
mesh currents and then all the branch voltages are found using Ohms’s law. If we have
N meshes with N mesh currents, we can obtain N independent mesh equations. This set
of N equations are independent, and thus guarantees a solution for the N mesh currents.
36 j Network Theory

EXAMPLE 1.19
For the electrical network shown in Fig. 1.38, determine the loop currents and all branch
currents.

Figure 1.38

SOLUTION
Applying KVL for the meshes shown in Fig. 1.38, we have

Mesh 1 : 0:2I1 + 2(I1 I3 ) + 3(I1 I2 ) 10 = 0

) 5:2I1 3I2 2I3 = 10 (1.36)
Mesh 2 : 3(I2 I1 ) + 4(I2 I3 ) + 0:2I2 + 15 = 0
) 3I1 + 7:2I2 4I3 = 15 (1.37)
Mesh 3 : 5I3 + 2(I3 I1 ) + 4(I3 I2 ) = 0
) 2I1 4I2 + 11I3 = 0 (1.38)

Putting the equations (1.36) through (1.38) in matrix form, we have

2 32 3 2 3
5:2 3 2 I1 10
4 3 7:2 4 5 4 I2 5 = 4 15 5
2 4 11 I3 0

Using Cramer’s rule, we get

I1 = 0:11A
I2 = 2:53A
and I3 = 0:9A
 Circuit Concepts and Network Simplification Techniques j 37

The various branch currents are now calculated as follows:

Current through 10V battery = I1 = 0:11A
Current through 2Ω resistor = I1 I3 = 1:01A
Current through 3Ω resistor = I1 I2 = 2:64A
Current through 4Ω resistor = I2 I3 = 1:63A
Current through 5Ω resistor = I3 = 0:9A
Current through 15V battery = I2 = 2:53A

The negative sign for I2 and I3 indicates that the actual directions of these currents
are opposite to the assumed directions.

1.10 Mesh analysis with independent current sources

Let us consider an electrical circuit source

having an independent current source as
shown Fig. 1.39(a).
We find that the second mesh current i2 = is
and thus we need only to determine the first
mesh current i1 , Applying KVL to the first
mesh, we obtain

(R1 + R2 )i1 R2 i2 = v
Since i2 = is ;
we get (R1 + R2 )i1 + is R2 = v
v is R2 Figure 1.39(a) Circuit containing both inde-
) i1 =
R1 + R2 pendent voltage and current sources

As a second example, let us take an electri-

cal circuit in which the current source is is
common to both the meshes. This situation
is shown in Fig. 1.39(b).
By applying KCL at node x, we recognize
that, i2 i1 = is
The two mesh equations (using KVL) are

M esh 1 : R1 i1 + vxy v=0

M esh 2 : (R2 + R3 )i2 vxy = 0
Figure 1.39(b) Circuit containing an independent
current source common to both meshes
38 j Network Theory

Adding the above two equations, we get

R1 i1 + (R2 + R3 )i2 = v
Substituting i2 = i1 + is in the above equation, we find that
R1 i1 + (R2 + R3 )(i1 + is ) = v
(R2 + R3 )is
v
) i1 =
R1 + R2 + R3
In this manner, we can handle independent current sources by recording the relation-
ship between the mesh currents and the current source. The equation relating the mesh
current and the current source is recorded as the constraint equation.
EXAMPLE 1.20
Find the voltage Vo in the circuit shown in Fig. 1.40.

Figure 1.40
SOLUTION

Constraint equations:
I1 = 4  10 3 A
I2 = 2  10 3 A

Applying KVL for the mesh 3, we get

4  103 [I3 I2 ] + 2  103 [I3 I1 ] + 6  103I3 3=0
Substituting the values of I1 and I2 , we obtain
I3 = 0:25 mA
Hence; Vo = 6  103I3 3
= 6  103 (0:25  10 3
) 3
= 1:5 V
 Circuit Concepts and Network Simplification Techniques j 39

1.11 Supermesh

A more general technique for mesh analysis method,

when a current source is common to two meshes,
involves the concept of a supermesh. A supermesh is
created from two meshes that have a current source
as a common element; the current source is in the
interior of a supermesh. We thus reduce the number
of meshes by one for each current source present.
Figure 1.41 shows a supermesh created from the two
meshes that have a current source in common. Figure 1.41 Circuit with a supermesh
shown by the dashed line

EXAMPLE 1.21
Find the current io in the circuit shown in Fig. 1.42(a).

Figure 1.42(a)

SOLUTION

This problem is first solved by the techique explained in Section 1.10. Three mesh currents
are specified as shown in Fig. 1.42(b). The mesh currents constrained by the current
sources are

i=2  10 3
A
i2 i3 = 4  10 3
A

The KVL equations for meshes 2 and 3 respetively are

2  103 i2 + 2  103 (i2 i1 ) vxy = 0

6 + 1  103 i3 + vxy + 1  103 (i3 i1 ) = 0
40 j Network Theory

Figure 1.42(b) Figure 1.42(c)

Adding last two equations, we get

6 + 1  103 i3 + 2  103 i2 + 2  103 (i2 i1 ) + 1  103(i3 i1 ) = 0 (1.39)

Substituting i1 = 2  10 3A and i3 = i2 4  10 3A in the above equation,

we get
   
6 + 1  103 i2 4  10 3
+ 2  103 i2 + 2  103 i2 2  10 3
 
+1  103 i2 4  10 3
2  10 3
=0

Solving we get
10
i2 = mA
3
Thus; io = i1 i2
10
=2
3
4
= mA
3

The purpose of supermesh approach is to avoid introducing the unknown voltage vxy .
The supermesh is created by mentally removing the 4 mA current source as shown in
Fig. 1.42(c). Then applying KVL equation around the dotted path, which defines the
supermesh, using the orginal mesh currents as shown in Fig. 1.42(b), we get

6 + 1  103 i3 + 2  103 i2 + 2  103 (i2 i1 ) + 1  103(i3 i1 ) = 0

Note that the supermesh equation is same as equation 1.39 obtained earlier by introduc-
ing vxy , the remaining procedure of finding io is same as before.
 Circuit Concepts and Network Simplification Techniques j 41

EXAMPLE 1.22
For the network shown in Fig. 1.43(a), find the mesh currents i1 ; i2 and i3 .

Figure 1.43(b)
Figure 1.43(a)

SOLUTION
The 5A current source is in the common
boundary of two meshes. The supermesh
is shown as dotted lines in Figs.1.43(b) and
1.43(c), the branch having the 5A current
source is removed from the circuit diagram.
Then applying KVL around the dotted path,
which defines the supermesh, using the orig-
inal mesh currents as shown in Fig. 1.43(c),
we find that

10 + 1(i1 i3 ) + 3(i2 i3 ) + 2i2 = 0

Figure 1.43(c)
For mesh 3, we have
1(i3 i1 ) + 2i3 + 3(i3 i2 ) = 0
Finally, the constraint equation is
i1 i2 = 5
Then the above three eqations may be reduced to
Supemesh: 1i1 + 5i2 4i3 = 10
Mesh 3 : 1i1 3i2 + 6i3 = 0
current source : i1 i2 = 5
Solving the above simultaneous equations, we find that,
i1 = 7.5A, i2 = 2.5A, and i3 = 2.5A
42 j Network Theory

EXAMPLE 1.23
Find the mesh currents i1 ; i2 and i3 for the network shown in Fig. 1.44.

Figure 1.44
SOLUTION
Here we note that 1A independent current source is in the common boundary of two
meshes. Mesh currents i1 ; i2 and i3 , are marked in the clockwise direction. The supermesh
is shown as dotted lines in Figs. 1.45(a) and 1.45(b). In Fig. 1.45(b), the 1A current
source is removed from the circuit diagram, then applying the KVL around the dotted
path, which defines the supermesh, using original mesh currents as shown in Fig. 1.45(b),
we find that
2 + 2(i1 i3 ) + 1(i2 i3 ) + 2i2 = 0

Figure 1.45(a) Figure 1.45(b) .

For mesh 3, the KVL equation is

2(i3 i1 ) + 1i3 + 1(i3 i2 ) = 0
 Circuit Concepts and Network Simplification Techniques j 43

Finally, the constraint equation is

i1 i2 = 1

Then the above three equations may be reduced to

Supermesh : 2i1 + 3i2 3i3 = 2
Mesh 3 : 2i1 + i2 4i3 = 0
Current source: i1 i2 = 1
Solving the above simultaneous equations, we find that
i1 = 1.55A, i2 = 0.55A, i3 = 0.91A

1.12 Mesh analysis for the circuits involving dependent sources

The persence of one or more dependent sources merely requires each of these source
quantites and the variable on which it depends to be expressed in terms of assigned mesh
currents. That is, to begin with, we treat the dependent source as though it were an
independent source while writing the KVL equations. Then we write the controlling
equation for the dependent source. The following examples illustrate the point.

EXAMPLE 1.24
(a) Use the mesh current method to solve for ia in the circuit shown in Fig. 1.46.
(b) Find the power delivered by the independent current source.
(c) Find the power delivered by the dependent voltage source.

Figure 1.46

SOLUTION
(a) We mark two mesh currents i1 and i2 as shown in Fig. 1.47. We find that i = 2:5mA.
Applying KVL to mesh 2, we find that
2400(i2 0:0025) + 1500i2 150(i2 0:0025) = 0 (∵ ia = i2 2:5 mA)
) 3750i2 = 6 0:375
= 5:625
) i2 = 1:5 mA
ia = i2 2:5 = 1:0mA
44 j Network Theory

(b) Applying KVL to mesh 1, we get

vo + 2:5(0:4) 2:4ia = 0
) vo = 2:5(0:4) 2:4( 1:0) = 3:4V
Pind.source = 3:4  2:5  10 3

= 8.5 mW(delivered)
(c) Pdep.source = 150ia (i2 )
= 150( 1:0  10 3
)(1:5  10 3)

= 0.225 mW(absorbed) Figure 1.47

EXAMPLE 1.25
Find the total power delivered in the circuit using mesh-current method.

Figure 1.48

SOLUTION
Let us mark three mesh currents i1 , i2 and i3 as shown in Fig. 1.49.
KVL equations :
Mesh 1: 17:5i1 + 2:5(i1 i3 )
+5(i1 i2 ) = 0
) 25i1 5i2 2:5i3 = 0
Mesh 2: 125 + 5(i2 i1 )
+7:5(i2 i3 ) + 50 = 0
) 5i1 + 12:5i2 7:5i3 = 75
Constraint equations :

i3 = 0:2Va
Va = 5(i2 i1 )
Thus; i3 = 0:2  5(i2 i1 ) = i2 i1 :
Figure 1.49
 Circuit Concepts and Network Simplification Techniques j 45

Making use of i3 in the mesh equations, we get

Mesh 1 : 25i1 5i2 2:5(i2 i1 ) = 0
) 27:5i1 7:5i2 = 0
Mesh 2 : 5i1 + 12:5i2 7:5(i2 i1 ) = 75
) 2:5i1 + 5i2 = 75

Solving the above two equations, we get

i1 = 3:6 A; i2 = 13:2 A
and i3 = i2 i1 = 9:6 A

Applying KVL through the path having 5Ω ! 2:5Ω ! vcs ! 125V source, we get,
5(i2 i1 ) + 2:5(i3 i1 ) + vcs 125 = 0
) vcs = 125 5(i2 i1 ) 2:5(i3 i1 )
= 125 48 2:5(9:6 3:6) = 62 V
Pvcs = 62(9:6) = 595:2W (absorbed)
P50V = 50(i2 i3 ) = 50(13:2 9:6) = 180W (absorbed)
P125V = 125i2 = 1650W (delivered)

EXAMPLE 1.26
Use the mesh-current method to find the power delivered by the dependent voltage source
in the circuit shown in Fig. 1.50.

Figure 1.50
SOLUTION

Applying KVL to the meshes 1, 2 and 3 shown in Fig 1.51, we have

Mesh 1 : 5i1 + 15(i1 i3 ) + 10(i1 i2 ) 660 = 0
) 30i1 10i2 15i3 = 660
46 j Network Theory

Mesh 2 : 20ia + 10(i2 i1 ) + 50(i2 i3 ) = 0

) 10(i2 i1 ) + 50(i2 i3 ) = 20ia
) 10i1 + 60i2 50i3 = 20ia
Mesh 3 : 15(i3 i1 ) + 25i3 + 50(i3 i2 ) = 0
) 15i1 50i2 + 90i3 = 0

Figure 1.51
Also ia = i2 i3
Solving, i1 = 42A, i2 = 27A, i3 = 22A, ia = 5A.
Power delivered by the dependent voltage source = P20ia = (20ia )i2
= 2700W (delivered)

1.13 Node voltage anlysis

In the nodal analysis, Kirchhoff’s current law is used to write the equilibrium equations.
A node is defined as a junction of two or more branches. If we define one node of the
network as a reference node (a point of zero potential or ground), the remaining nodes of
the network will have a fixed potential relative to this reference. Equations relating to all
nodes except for the reference node can be written by applying KCL.
Refering to the circuit shown
in Fig.1.52, we can arbitrarily
choose any node as the reference
node. However, it is convenient
to choose the node with most con-
nected branches. Hence, node 3 is
chosen as the reference node here.
It is seen from the network of Fig. Figure 1.52 Circuit with three nodes where the
1.52 that there are three nodes. lower node 3 is the reference node
 Circuit Concepts and Network Simplification Techniques j 47

Hence, number of equations based on KCL will be total number of nodes minus one.
That is, in the present context, we will have only two KCL equations referred to as node
equations. For applying KCL at node 1 and node 2, we assume that all the currents leave
these nodes as shown in Figs. 1.53 and 1.54.

Figure 1.53 Simplified circuit for Figure 1.54 Simplified circuit for
applying KCL at node 1 applying KCL at node 2

Applying KCL at node 1 and 2, we find that

(i) At node 1: i1 + i2 + i4 = 0

v1 va v1 v2 v1
) R1
+
R2
+
R4
0
=0
 
va
) v1
1
R1
+
1
R2
+
R4
1
v2
R2
1
=
R1
(1.40)

(ii) At node 2: i2 + i3 + i5 = 0

v2 v1 v2 vb v2
) R2
+
R3
+
R5
=0
   
vb
) v1
1
R2
+ v2
1
R2
+
1
R3
+
1
R5
=
R3
(1.41)

Putting equations (1.40) and (1.41) in matrix form, we get

2 32 3 2 v 3
1 1 1 1 v1 a

6 R1 + + 76 7 6 R1 7
6 R2 R4 R2 76 7=6 7
6 76 7 6 7
4 1 1 1 1 54 5 4 vb 5
+ + v2
R2 R2 R3 R5 R3

The above matrix equation can be solved for node voltages v1 and v2 using Cramer’s
rule of determinants. Once v1 and v2 are obtainted, then by using Ohm’s law, we can find
all the branch currents and hence the solution of the network is obtained.
48 j Network Theory

EXAMPLE 1.27
Refer the circuit shown in Fig. 1.55. Find the three node voltages va , vb and vc , when all
the conductances are equal to 1S.

Figure 1.55
SOLUTION

At node a: (G1 + G2 + G6 )va G2 v b G6 v c = 9 3

At node b: G2 va + (G4 + G2 + G3 )vb G4 v c = 3
At node c: G6 v a G4 vb + (G4 + G5 + G6 )vc = 7
Substituting the values of various conductances, we find that
3va vb vc = 6
va + 3vb vc = 3
va vb + 3vc = 7
Putting the above equations in matrix form, we see that
2 32 3 2 3
3 1 1 va 6
4 1 3 1 5 4 vb 5 4
= 3 5
1 1 3 vc 7
Solving the matrix equation using cramer’s rule, we get
va = 5:5V; vb = 4:75V; vc = 5:75V
The determinant Δ used for computing va , vb and vc in general form is given by
 P 
 
 a G Gab Gac 
 P 

G =  Gab G Gbc 
 b P 
 Gac Gbc G 

c

P
where G is the sum of the conductances at node i, and Gij is the sum of conductances
i
conecting nodes i and j .
 Circuit Concepts and Network Simplification Techniques j 49

The node voltage matrix equation for a circuit with k unknown node voltages is
Gv = is ;
2 3
va
6 vb 7
v=6 .
6 7
where; 7
4 .. 5
vk

is the vector consisting of k unknown node voltages.

2 3
is1
6 is2 7
6
ia = 6 .
7
The matrix 7
4 .. 5
isk

is the vector consisting of k current sources and isk is the sum of all the source currents
entering the node k . If the k th current source is not present, then isk = 0.

EXAMPLE 1.28
Use the node voltage method to find how much power the 2A source extracts from the
circuit shown in Fig. 1.56.

Figure 1.56

SOLUTION
Applying KCL at node a, we get
va va 55
2+ + =0
4 5
) va = 20V
P2Asource = 20(2) = 40W (absorbing)

Figure 1.57
50 j Network Theory

EXAMPLE 1.29
Refer the circuit shown in Fig. 1.58(a).
(a) Use the node voltage method to find the branch currents i1 to i6 .
(b) Test your solution for the branch currents by showing the total power dissipated equals
the power developed.

Figure 1.58(a)
SOLUTION

(a) At node v1 :
v1 110 v1 v2 v1 v3
+ + =0
2 8 16
) 11v1 2v2 v3 = 880

At node v2 :
v2 v1 v2 v2 v3
+ + =0
8 3 24
) 3v1 + 12v2 v3 = 0

At node v3 :
v3 + 110 v3 v2 v3 v1
+ + =0
2 24 16
) 3v1 2v2 + 29v3 = 2640
Figure 1.58(b)
Solving the above nodal equations,we get
v1 = 74:64V; v2 = 11:79V; v3 = 82:5V

110 v1
Hence; i1 = = 17:68A
2
v2
i2 = = 3:93A
3
 Circuit Concepts and Network Simplification Techniques j 51

v3 + 110
i3 = = 13:75A
2
v1 v2
i4 = = 7:86A
8
v2 v3
i5 = = 3:93A
24
v1 v3
i6 = = 9:82A
16
(b) Total power delivered = 110i1 + 110i3 = 3457:3W
Total power dissipated = i21  2 + i22  3 + i23  2 + i24  8 + i25  24 + i26  16
= 3457.3 W

EXAMPLE 1.30
(a)Use the node voltage method to show that the output volatage vo in the circuit of
Fig 1.59(a) is equal to the average value of the source voltages.
(b) Find vo if v1 = 150V, v2 = 200V and v3 = 50V.

Figure 1.59(a)

SOLUTION
Applying KCL at node a, we get
vo v1 vo v2 vo v3 vo vn
+ + +


+ =0
R R R R

) nvo = v1 + v2 +


+ v n

= [v1 + v2 +


+ v ]
1
Hence; vo n
n
1X
n

= vk
n
k =1

1
(b) vo = (150 + 200 50) = 100V Figure 1.59(b)
3
52 j Network Theory

EXAMPLE 1.31
Use nodal analysis to find vo in the circuit of Fig. 1.60.

Figure 1.60

Figure 1.61
SOLUTION

Referring Fig 1.61, at node v1 :

v1 + 6 v1 v1 + 3
+ + =0
6 3 2
v1 v1 v1
) 6
+
3
+
2
= 2:5
) v1 = 2:5 V
 
v1
vo =
2+1
1
2:5
=
3
1
= 0:83volts

EXAMPLE 1.32
Refer to the network shown in Fig. 1.62. Find the power delivered by 1A current source.

Figure 1.62
 Circuit Concepts and Network Simplification Techniques j 53

SOLUTION
Referring to Fig. 1.63, applying KVL
to the path va ! 4Ω ! 3Ω, we get
va = v1 v 3

v2 = 12V

v1 v1 v2
At node v1 : + 1=0
4 2
v1 v1
) 4
+
2
12
1=0
v1 = 9:33 V

v3 v3 v2
At node v2 : + +1=0
3 2 Figure 1.63
v3 v3
) 3
+
2
12
+1=0
) v3 = 6V
Hence; va = 9:33 6 = 3:33 volts
P1A source = va 1
= 3:33  1 = 3:33W (delivering)

1.14 Supernode

Inorder to understand the concept of a supernode, let us consider an electrical circuit as

shown in Fig. 1.64.
Applying KVL clockwise to the loop containing R1 , voltage source and R2 , we get
va = vs + vb
) va vb = vs (Constraint equation) (1.42)
To account for the fact that the source voltage
is known, we consider both va and vb as part
of one larger node represented by the dotted
ellipse as shown in Fig. 1.64. We need a larger
node because va and vb are dependent (see
equation 1.42). This larger node is called the
supernode.
Applying KCL at nodes a and b, we get
va
ia = 0
R1
vb
and + ia = is Figure 1.64 Circuit with a supernode
R2
incorporating va and vb .
54 j Network Theory

Adding the above two equations, we find that

va vb
+ = is
R1 R2
) va G1 + vb G2 = is (1.43)

Solving equations (1.42) and (1.43), we can find the values of va and vb .
When we apply KCL at the supernode, mentally imagine that the voltage source vs
is removed from the the circuit of Fig. 1.63, but the voltage at nodes a and b are held at
va and vb respectively. In other words, by applying KCL at supernode, we obtain

va G1 + va G2 = is

The equation is the same equation (1.43). As in supermesh, the KCL for supernode
eliminates the problem of dealing with a current through a voltage source.
Procedure for using supernode:
1. Use it when a branch between non-reference nodes is connected by an independent
or a dependent voltage source.
2. Enclose the voltage source and the two connecting nodes inside a dotted ellipse to
form the supernode.
3. Write the constraint equation that defines the voltage relationship between the two
non-reference node as a result of the presence of the voltage source.
4. Write the KCL equation at the supernode.
5. If the voltage source is dependent, then the constraint equation for the dependent
source is also needed.

EXAMPLE 1.33
Refer the electrical circuit shown in Fig. 1.65 and find va .

Figure 1.65
 Circuit Concepts and Network Simplification Techniques j 55

SOLUTION
The constraint equation is,
vb va = 8
) vb = va + 8

The KCL equation at the supernode

is then,

va + 8 (va + 8) 12 va12
+ +
500 125 250
va
+ =0
500
Therefore; va = 4V Figure 1.66

EXAMPLE 1.34
Use the nodal analysis to find vo in the network of Fig. 1.67.

Figure 1.67
SOLUTION

Figure 1.68
56 j Network Theory

The constraint equation is,

v2 v1 = 12
) v1 = v2 12

KCL at supernode:

v2 12 (v2 12) v3 v2 v2 v3
1  10 1  10 1  10 1  103
3
+ 3
+ 3
+ =0

) 4  10 3 v2 2  10 3 v3 = 24  10 3
) 4v2 2v3 = 24

At node v3 :
v3 v2 v3 (v2 12)
= 2  10 3
1  103 1  103
+

) 2  10 3 v2 + 2  10 3 v3 = 10  10 3

2v2 + 2v3 = 10

Solving we get v2 = 7V
v3 = 2V
Hence; vo = v3 = 2V

EXAMPLE 1.35
Refer the network shown in Fig. 1.69. Find the current Io .

Figure 1.69

SOLUTION

Constriant equation:
v3 = v1 12
 Circuit Concepts and Network Simplification Techniques j 57

Figure 1.70

KCL at supernode:
v1 12 v1 v1 v2
3  10 2  10 3  103
3
+ 3
+ =0

) 7
6
 10 3 v1
1
3
 10 3v2 = 4  10 3

) 7
6
v1
1
3
v2 = 4

KCL at node 2:
v2 v1 v2
+ 4  10 3
3 3  103
+ =0
103

) 1
3
 10 3
v1 +
2
3
 10 3
v2 = 4  10 3

) 1
3
v1 +
2
3
v2 = 4

Putting the above two nodal equations in matrix form, we get

2 7 1
32 3 2 3
v1 4
6 6 7
3 76 6 7 6 7
6 7=6 7
4 1 2 54 5 4 5
3 3 v2 4

Solving the above two matrix equations using Cramer’s rule, we get

v1 = 2V
v1
) Io =
2 103
=
2
2  103
= 1mA
58 j Network Theory

EXAMPLE 1.36
Refer the network shown in Fig. 1.71. Find the power delivered by the dependent voltage
source in the network.

Figure 1.71

SOLUTION
Refer Fig. 1.72, KCL at node 1:
v1 80 v1 v1 + 75ia
+ + =0
5 50 25
v1
where ia =
50 v 
1
v1 v1 v1 + 75
) 5
80
+
50
+
25
50 =0
Solving we get v1 = 50V
Figure 1.72

v1
) ia =
50
=
50
50
= 1A
v1 ( 75ia )
Also; i1 =
(10 + 15)
v1 + 75ia
=
(10 + 15)
50 + 75  1
= = 5A
(10 + 15)
P75ia = (75ia )i1
= 75  1  5
= 375W (delivered)

EXAMPLE 1.37
Use the node-voltage method to find the power developed by the 20 V source in the circuit
shown in Fig. 1.73.
 Circuit Concepts and Network Simplification Techniques j 59

Figure 1.73
SOLUTION

Figure 1.74

Constraint equations :
va = 20 v2
v1 31ib = v3
v2
ib =
40
Node equations :
(i) Supernode:
v1 v1 20 v3 v2 v3
+ + + + 3:125va = 0
20 2 4 80
v1 v1 (v1 35ib ) v2 (v1 35ib )
) 20
+
2
20
+
4
+
80
+ 3:125(20 v2 ) = 0
 v2   v2 
v1 v1 v1 35 v2 v1 35
) 20
+
2
20
+
4
40 +
80
40 + 3:125(20 v2 ) = 0
60 j Network Theory

(ii) At node v2 :
v2 v2 v3 v2 20
+ + =0
40 4 1

v2 v2 (v1 35ib ) v2
) 40
+
4
+
1
20
=0
 v2 
v2 v2 v1 35 v2
) 40
+
4
40 +
1
20
=0

Solving the above two nodal equations, we get

v1 = 20:25V; v2 = 10V

Then v3 = v1 35ib
v2
= v1 35
40
= 29V
20 v1 20 v2
Also; ig = +
2 1
20 + 20:25 (20 10)
= +
2 1
= 30.125 A
P20V = 20ig = 20(30:125)
= 602.5 W (delivered)

EXAMPLE 1.38
Refer the circuit shown in Fig. 1.75(a). Determine the current i1 .

Figure 1.75(a)
 Circuit Concepts and Network Simplification Techniques j 61

SOLUTION

Constraint equation:
Applying KVL clockwise to the loop containing 3V source, dependent voltage source,
2A current source and 4Ω resitor, we get

v1 3 0:5i1 + v2 = 0
) v1 v2 = 3 0:5i1

v2 4
Substituting i1 = , the above equation becomes
2
4v1 3v2 = 8

Figure 1.75(b)

KCL equation at supernode:

v1 v2
4
+
2
4
= 2 ) v1 + 2v2 = 0

Solving the constraint equation and the KCL equation at supernode simultaneously,
we find that,

v2 = 727:3 mV
v1 = 2v2
= 1454:6 mV
v2 4
Then; i1 =
2
= 1:636A
62 j Network Theory

EXAMPLE 1.39
Refer the network shown in Fig. 1.76(a). Find the node voltages vd and vc .

Figure 1.76(a)

SOLUTION
vb vc
From the network, shown in Fig. 1.76 (b), by inspection,vb = 8 V, i1 =
2
Constraint equation: va = 6i1 + vd
va vb va vd vc
KCL at supernode: + + = 3vc
2 2 2
 
) va
1 1
+
2 2
1
2
1
vb + [vd
2
vc ] = 3vc (1.44)

Figure 1.76(b)
 Circuit Concepts and Network Simplification Techniques j 63

Substituting vb = 8 V in the constrained equation, we get

(vb vc )
va = 6 + vd
2
= 3(vb vc ) + vd
= 3(8 vc ) + vd (1.45)
Substituting equation (1.45) into equation (1.44), we get
1 1
[3(8 vc ) + vd ] (8) + [vd vc ] = 3vc
2 2
) 24 3vc + vd 4 + 21 vd 21 vc = 3vc
) 6:5vc + 1:5vd = 20 (1.46)
vc vb vc vd
KCL at node c: + =4
2 2
vc 8 vc vd
Substituting vb = 8V, we have + =4
2 2
) vc 8 + vc vd = 8
) 2vc vd = 16
) vc 0:5vd = 8 (1.47)
Solving equations (1.46) and (1.47), we get
vc = 1.14V
vd = 18.3V
EXAMPLE 1.40
For the circuit shown in Fig. 1.77(a), determine all the node voltages.

Figure 1.77(a)
84 j Network Theory

By inspection, we find that I2 = 2 /0 A.

Applying KVL clockwise to mesh 1, we get
12 + I1 (2 j 1) + (I1 I2 )(4 + j 2) = 0
Substituting I2 = 2 /0 in the above equation yields,
12 + I1 (2 j 1 + 4 + j 2) 2(4 + j 2) = 0
= 3:35 /1:85 A
20 + j 4
) I1 =
6 + j1
Hence Vo = 4(I1 I2 )
= 5:42 /4:57 V
Wye  Delta transformation
For reducing a complex network to a single impedance between any two terminals, the
reduction formulas for impedances in series and parallel are used. However, for certain
configurations of network, we cannot reduce the interconnected impedances to a single
equivalent impedance between any two terminals by using series and parallel impedance
reduction techniques. That is the reason for this topic.
Consider the networks shown in Fig. 1.103 and 1.104.

Figure 1.103 Delta resistance network Figure 1.104 Wye resistance network

It may be noted that resistors in Fig. 1.103 form a Δ (delta), and resistors in Fig.
1.104. form a Υ (Wye). If both these configurations are connected at only the three
terminals a, b and c, it would be very advantageous if an equivalence is established be-
tween them. It is possible to relate the resistances of one network to those of the other
such that their terminal characteristics are the same. The relationship between the two
configurations is called Υ Δ transformation.
We are interested in the relationship between the resistances R1 , R2 and R3 and the
resitances Ra , Rb and Rc . For deriving the relationship, we assume that for the two
networks to be equivalent at each corresponding pair of terminals, it is necessary that
the resistance at the corresponding terminals be equal. That is, for example, resistance
at terminals b and c with a open-circuited must be same for both networks. Hence, by
equating the resistances for each corresponding set of terminals, we get the following set
of equations :
 Circuit Concepts and Network Simplification Techniques j 85

(i) Rab (Υ) = Rab (Δ)

R2 (R1 + R3 )
) Ra + Rb =
R2 + R1 + R3
(1.57)

(ii) Rbc (Υ) = Rbc (Δ)

R3 (R1 + R2 )
) Rb + Rc =
R3 + R1 + R2
(1.58)

(iii) Rca (Υ) = Rca (Δ)

R1 (R2 + R3 )
) Rc + Ra =
R1 + R2 + R3
(1.59)

Solving equations (1.57), (1.58) and (1.59) gives

R1 R2
Ra = (1.60)
R1 + R2 + R3
R2 R3
Rb = (1.61)
R1 + R2 + R3
R1 R3
Rc = (1.62)
R1 + R2 + R3

Hence, each resistor in the Υ network is the product of the resistors in the two adjacent
Δ branches, divided by the sum of the three Δ resistors.
To obtain the conversion formulas for transforming a wye network to an equivalent
delta network, we note from equations (1.60) to (1.62) that

R1 R2 R3 (R1 + R2 + R3 ) R1 R2 R3
Ra Rb + Rb Rc + Rc Ra = = (1.63)
(R1 + R2 + R3 )2 R1 + R2 + R3

Dividing equation (1.63) by each of the equations (1.60) to (1.62) leads to the following
relationships :
Ra Rb + Rb Rc + Ra Rc
R1 = (1.64)
Rb
Ra Rb + Rb Rc + Ra Rc
R2 = (1.65)
Rc
Ra Rb + Rb Rc + Ra Rc
R3 = (1.66)
Ra

Hence each resistor in the Δ network is the sum of all possible products of Υ resistors
taken two at a time, divided by the opposite Υ resistor.
Then Υ and Δ are said to be balanced when

R1= R2 = R3 = RΔ and Ra = Rb = Rc = RΥ
86 j Network Theory

Under these conditions the conversions formula become

1
RΥ = RΔ
3
and RΔ = 3RΥ

EXAMPLE 1.55
Find the value of resistance between the terminals a b of the network shown in
Fig. 1.105.

Figure 1.105

SOLUTION
Let us convert the upper Δ to Υ

(6k)(18k)
Ra 1 = = 3 kΩ
6k + 12k + 18k
(6k)(12k)
Rb1 = = 2 kΩ
6k + 12k + 18k
(12k)(18k)
R c1 = = 6 kΩ
6k + 12k + 18k Figure 1.106

The network shown in Fig. 1.106 is now reduced to that shown in Fig. 1.106(a)
 Circuit Concepts and Network Simplification Techniques j 87

Hence; Rab = 4 + 3 + 7:875 + 2

= 16.875kΩ

Figure 1.106(a)

EXAMPLE 1.56
Find the resistance Rab using Υ Δ transformation.

Figure 1.107

SOLUTION

Figure 1.108
88 j Network Theory

Let us convert the upper Δ between the points a1 , b1 and c1 into an equivalent Υ.
6  18
Ra 1 = = 3:6Ω
6 + 18 + 6
66
Rb1 = = 1:2Ω
6 + 18 + 6
6  18
R c1 = = 3:6Ω
6 + 18 + 6
Figure 1.108 now becomes

Rab = 5 + 3:6 + 7:2 27:6jj

= 8:6 +

7:2 27:6
7:2 + 27:6
= 14:31Ω
EXAMPLE 1.57
Obtain the equvivalent resistance Rab for the circuit of Fig. 1.109 and hence find i.

Figure 1.109
 Circuit Concepts and Network Simplification Techniques j 89

SOLUTION
Let us convert Υ between the terminals a, b and c into an equivalent Δ.

Ra Rb + Rb Rc + Rc Ra
Rab =
Rc
10  20 + 20  5 + 5  10
= = 70Ω
5
Ra Rb + Rb Rc + Rc Ra
Rbc =
Ra
10  20 + 20  5 + 5  10
= = 35Ω
10
Ra Rb + Rb Rc + Rc Ra
Rca =
Rb
10  20 + 20  5 + 5  10
= = 17:5Ω
20

The circuit diagram of Fig. 1.109 now becomes the circuit diagram shown in Fig.
1.109(a). Combining three pairs of resistors in parallel, we obtain the circuit diagram of
Fig. 1.109(b).

Figure 1.109(a)

70  30
70jj30 = = 21Ω
70 + 30
12:5  17:5
12:5jj17:5 = = 7:292Ω
12:5 + 17:5
15  35
15jj35 = = 10:5Ω
15 + 35
Rab = (7:292 + 10:5)jj21 = 9:632Ω
vs
Thus; i= = 12:458 A Figure 1.109(b)
Rab
Many electric circuits are complex, but it is an engineer’s goal to reduce their complexity to
analyze them easily. In the previous chapters, we have mastered the ability to solve networks
containing independent and dependent sources making use of either mesh or nodal analysis. In
this chapter, we will introduce new techniques to strengthen our armoury to solve complicated
networks. Also, these new techniques in many cases do provide insight into the circuit’s operation
that cannot be obtained from mesh or nodal analysis. Most often, we are interested only in the
detailed performance of an isolated portion of a complex circuit. If we can model the remainder
of the circuit with a simple equivalent network, then our task of analysis gets greatly reduced and
simplified. For example, the function of many circuits is to deliver maximum power to load such
as an audio speaker in a stereo system. Here, we develop the required relationship betweeen a
load resistor and a fixed series resistor which can represent the remaining portion of the circuit.
Two of the theorems that we present in this chapter will permit us to do just that.

3.1 Superposition theorem

The principle of superposition is applicable only for linear systems. The concept of superposition
can be explained mathematically by the following response and excitation principle :
i1 ! v1

i2 ! v2

then; i1 + i2 ! v1 + v2
The quantity to the left of the arrow indicates the excitation and to the right, the system
response. Thus, we can state that a device, if excited by a current i1 will produce a response
v1 . Similarly, an excitation i2 will cause a response v2 . Then if we use an excitation i1 + i2 , we
will find a response v1 + v2 .
The principle of superposition has the ability to reduce a complicated problem to several easier
problems each containing only a single independent source.
160 j Network Theory

Superposition theorem states that,

In any linear circuit containing multiple independent sources, the current or voltage at any
point in the network may be calculated as algebraic sum of the individual contributions of each
source acting alone.
When determining the contribution due to a particular independent source, we disable all
the remaining independent sources. That is, all the remaining voltage sources are made zero by
replacing them with short circuits, and all remaining current sources are made zero by replacing
them with open circuits. Also, it is important to note that if a dependent source is present, it must
remain active (unaltered) during the process of superposition.
Action Plan:

(i) In a circuit comprising of many independent sources, only one source is allowed to be active
in the circuit, the rest are deactivated (turned off).
(ii) To deactivate a voltage source, replace it with a short circuit, and to deactivate a current
source, replace it with an open circuit.
(iii) The response obtained by applying each source, one at a time, are then added algebraically
to obtain a solution.
Limitations: Superposition is a fundamental property of linear equations and, therefore, can be
applied to any effect that is linearly related to the cause. That is, we want to point out that,
superposition principle applies only to the current and voltage in a linear circuit but it cannot be
used to determine power because power is a non-linear function.

EXAMPLE 3.1
Find the current in the 6 Ω resistor using the principle of superposition for the circuit of Fig. 3.1.

Figure 3.1

SOLUTION
As a first step, set the current source to zero. That is, the current source appears as an open circuit
as shown in Fig. 3.2.
6 6
i1 = = A
3+6 9
 Circuit Theorems j 161

As a next step, set the voltage to zero by replacing it with a short circuit as shown in Fig. 3.3.

i2 =
2 3 6
= A
3+6 9

Figure 3.2 Figure 3.3

The total current i is then the sum of i1 and i2

12
i = i1 + i2 = A
9
EXAMPLE 3.2
Find io in the network shown in Fig. 3.4 using superposition.

Figure 3.4

SOLUTION
As a first step, set the current source to zero. That is, the current source appears as an open circuit
as shown in Fig. 3.5.

Figure 3.5
162 j Network Theory

6

0
io = = 0:3 mA
(8 + 12) 103
As a second step, set the voltage source to zero. This means the voltage source in Fig. 3.4 is
replaced by a short circuit as shown in Figs. 3.6 and 3.6(a). Using current division principle,
iR2
iA =
R1 + R2

where R1 jj
= (12 kΩ 12 kΩ) + 12 kΩ
= 6 kΩ + 12 kΩ
= 18 kΩ
and R2 = 12 kΩ

) 
4 10 3 12 103  
iA =
(12 + 18) 103 
= 1:6 mA Figure 3.6
Again applying the current division principle,
00
io =
iA 12= 0:8 mA
12 + 12
Thus; io = io 0 + io 00 = 0:3 + 0:8 = 0:5 mA

Figure 3.6(a)
 Circuit Theorems j 163

EXAMPLE 3.3
Use superposition to find io in the circuit shown in Fig. 3.7.

Figure 3.7

SOLUTION

As a first step, keep only the 12 V source active and rest of the sources are deactivated. That is,
2 mA current source is opened and 6 V voltage source is shorted as shown in Fig. 3.8.

12

0
io =
(2 + 2) 103
= 3 mA

Figure 3.8

As a second step, keep only 6 V source active. Deactivate rest of the sources, resulting in a
circuit diagram as shown in Fig. 3.9.
164 j Network Theory

Applying KVL clockwise to the upper loop, we get

2  10 3
io
00
2  10 3
io
00
6=0
) io
00
=
4 
6
103
= 1:5 mA

Figure 3.9

As a final step, deactivate all the independent voltage sources and keep only 2 mA current
source active as shown in Fig. 3.10.

Figure 3.10

Current of 2 mA splits equally.

000
Hence; io = 1mA

Applying the superposition principle, we find that

io = io 0 + io 00 + io 000
=3 1:5 + 1
= 2:5 mA
 Circuit Theorems j 165

EXAMPLE 3.4
Find the current i for the circuit of Fig. 3.11.

Figure 3.11

SOLUTION

We need to find the current i due to the two independent sources.

As a first step in the analysis, we will find the current resulting from the independent voltage
source. The current source is deactivated and we have the circuit as shown as Fig. 3.12.
Applying KVL clockwise around loop shown in Fig. 3.12, we find that

5i1 + 3i1 24 = 0
) i1 =
24
8
= 3A

As a second step, we set the voltage source to zero and determine the current i2 due to the
current source. For this condition, refer to Fig. 3.13 for analysis.

Figure 3.12 Figure 3.13

Applying KCL at node 1, we get

v1 3i2
i2 +7= (3.1)
2
v1 0
Noting that i2 =
3
we get, v1 = 3i2 (3.2)
166 j Network Theory

Making use of equation (3.2) in equation (3.1) leads to

3i2 3i2
i2 + 7 =
2
) i2 =
7
4
A
Thus, the total current
i = i1 + i2
7 5
=3 A= A
4 4
EXAMPLE 3.5
For the circuit shown in Fig. 3.14, find the terminal voltage Vab using superposition principle.

SOLUTION Figure 3.14

As a first step in the analysis, deactivate the in-
dependent current source. This results in a cir-
cuit diagram as shown in Fig. 3.15.
Applying KVL clockwise gives
4 + 10 0+3 Vab1 + Vab1 = 0
) 4Vab1 = 4
) Vab1 = 1V
Figure 3.15
Next step in the analysis is to deactivate the
independent voltage source, resulting in a cir-
cuit diagram as shown in Fig. 3.16.
Applying KVL gives

10 2+3 Vab2 + Vab2 = 0

) 4Vab2 = 20
) Vab2 = 5V

Figure 3.16
 Circuit Theorems j 167

According to superposition principle,

Vab = Vab1 + Vab2

= 1 + 5 = 6V

EXAMPLE 3.6
Use the principle of superposition to solve for vx in the circuit of Fig. 3.17.

Figure 3.17

SOLUTION
According to the principle of superposition,
vx = vx 1 + vx 2

where vx1 is produced by 6A source alone in the circuit and vx2 is produced solely by 4A current
source.
To find vx1 , deactivate the 4A current source. This results in a circuit diagram as shown in
Fig. 3.18.
KCL at node x1 :

vx 1 vx 1 4ix1
+ =6
2 8
vx 1
But ix1 =
2
v
vx 1 vx 1 4 x21
Hence; + =6
2 8

) vx 1
2
+
vx 1
8
2vx1
=6

) 4vx1 + vx1 2vx1 = 48 Figure 3.18

) vx 1 =
48
3
= 16V
168 j Network Theory

To find vx2 , deactivate the 6A current source, resulting in a circuit diagram as shown in Fig.
3.19.
KCL at node x2 :

vx 2 ( 4ix2 )
v x2
+ =4
8 2
) vx 2
8
vx + 4ix2
+ 2
2
=4 (3.3)

Applying KVL along dotted path, we get

vx 2 + 4ix2 2ix2 = 0
) vx 2 = 2ix2 or ix2 =
vx 2
2
(3.4)

Substituting equation (3.4) in equation (3.3), we get

 vx 2

vx 2 +4
vx 2 2
+ =4
8 2
) vx 2
8
+
vx 2
2
2vx2
=4

) vx 2 vx 2
=4
)
8 2
vx 2 4vx2 = 32
) vx 2 =
32
3
V

Hence, according to the superposition principle,

vx = vx 1 + vx 2 Figure 3.19
32
= 16 = 5:33V
2
EXAMPLE 3.7
Which of the source in Fig. 3.20 contributes most of the power dissipated in the 2 Ω resistor ?
The least ? What is the power dissipated in 2 Ω resistor ?

Figure 3.20
 Circuit Theorems j 169

SOLUTION
The Superposition theorem cannot be used to identify the individual contribution of each source
to the power dissipated in the resistor. However, the superposition theorem can be used to find the
total power dissipated in the 2 Ω resistor.

Figure 3.21

According to the superposition principle,

i1 = i01 + i02
where i01 = Contribution to i1 from 5V source alone.
and i02 = Contribution to i1 from 2A source alone.
Let us first find i01 . This needs the deactivation of 2A source. Refer to Fig. 3.22.
0
5
i1 = = 1:22A
2 + 2:1
Similarly to find i02 we have to disable the 5V source by shorting it.
Referring to Fig. 3.23, we find that
0
i2 =

2 2:1
= 1:024 A
2 + 2:1

Figure 3.22 Figure 3.23

170 j Network Theory

Total current,

i1 = i01 + i02
= 1:22 1:024
= 0:196 A
Thus; P2Ω = (0:196)2 2
= 0:0768 Watts
= 76:8 mW

EXAMPLE 3.8
Find the voltage V1 using the superposition principle. Refer the circuit shown in Fig.3.24.

Figure 3.24

SOLUTION
According to the superposition principle,

V1 = V10 + V100

where V10 is the contribution from 60V source alone and V100 is the contribution from 4A current
source alone.
To find V10 , the 4A current source is opened, resulting in a circuit as shown in Fig. 3.25.

Figure 3.25
 Circuit Theorems j 171

Applying KVL to the left mesh:

30ia 60 + 30 (ia ib ) =0 (3.5)

Also ib = 0:4iA
= 0:4 ( ia ) = 0:4ia (3.6)

Substituting equation (3.6) in equation (3.5), we get

30ia 60 + 30ia 
30 0:4ia = 0
) ia =
60
= 1:25A

48
ib = 0:4ia = 0:4 1:25
= 0:5A
Hence; V1
0
= (ia ib )  30
= 22:5 V

To find, V100 , the 60V source is shorted as shown in Fig. 3.26.

Figure 3.26

Applying KCL at node a:

00
Va Va V1
+ =4
20 10
) 30Va 20V100 = 800 (3.7)

Applying KCL at node b:

00 00
V1 V1 Va
+ = 0:4ib
30 10
Also; Va = 20ia ) 20
ib =
Va

00 00
V1 V1 0:4Va
Va
Hence; + =
30 10 20
) 7:2Va + 8V100 = 0 (3.8)
172 j Network Theory

Solving the equations (3.7) and (3.8), we find that

00
V1 = 60V
Hence V1 = V10 + V100
= 22:5 + 60 = 82:5V

EXAMPLE 3.9
(a) Refer to the circuit shown in Fig. 3.27. Before the 10 mA current source is attached to
terminals x y , the current ia is found to be 1.5 mA. Use the superposition theorem to find
the value of ia after the current source is connected.
(b) Verify your solution by finding ia , when all the three sources are acting simultaneously.

Figure 3.27

SOLUTION
According to the principle of superposition,

ia = ia1 + ia2 + ia3

where ia1 , ia2 and ia3 are the contributions to ia from 20V source, 5 mA source and 10 mA source
respectively.
As per the statement of the problem,

ia1 + ia2 = 1:5 mA

To find ia3 , deactivate 20V source and the 5 mA source. The resulting circuit diagram is


shown in Fig 3.28.
10mA 2k
ia3 = = 1 mA
18k + 2k
Hence, total current

ia = ia1 + ia2 + ia3

= 1:5 + 1 = 2:5 mA
 Circuit Theorems j 173

Figure 3.28

(b) Refer to Fig. 3.29

KCL at node y:

18
Vy
 103
+
Vy
2 
20
103

= (10+5) 10 3

Solving, we get Vy = 45V:

Vy 45
Hence; ia =
18  10 3
=

18 103
= 2:5 mA
Figure 3.29
3.2 Thevenin’s theorem

In section 3.1, we saw that the analysis of a circuit may be greatly reduced by the use of su-
perposition principle. The main objective of Thevenin’s theorem is to reduce some portion of a
circuit to an equivalent source and a single element. This reduced equivalent circuit connected to
the remaining part of the circuit will allow us to find the desired current or voltage. Thevenin’s
theorem is based on circuit equivalence. A circuit equivalent to another circuit exhibits identical
characteristics at identical terminals.

Figure 3.30 A Linear two terminal network Figure 3.31 The Thevenin’s equivalent circuit

According to Thevenin’s theorem, the linear circuit of Fig. 3.30 can be replaced by the one
shown in Fig. 3.31 (The load resistor may be a single resistor or another circuit). The circuit to
the left of the terminals x y in Fig. 3.31 is known as the Thevenin’s equivalent circuit.
174 j Network Theory

The Thevenin’s theorem may be stated as follows:

A linear two–terminal circuit can be replaced by an equivalent circuit consisting of a
voltage source Vt in series with a resistor Rt , Where Vt is the open–circuit voltage at the termi-
nals and Rt is the input or equivalent resistance at the terminals when the independent sources
are turned off or Rt is the ratio of open–circuit voltage to the short–circuit current at the
terminal pair.
Action plan for using Thevenin’s theorem :

1. Divide the original circuit into circuit A and circuit B .

In general, circuit B is the load which may be linear or non-linear. Circuit A is the balance of
the original network exclusive of load and must be linear. In general, circuit A may contain
independent sources, dependent sources and resistors or other linear elements.

2. Separate the circuit A from circuit B .

3. Replace circuit A with its Thevenin’s equivalent.
4. Reconnect circuit B and determine the variable of interest (e.g. current ‘i’ or voltage ‘v ’).

Procedure for finding Rt :

Three different types of circuits may be encountered in determining the resistance, Rt :
(i) If the circuit contains only independent sources and resistors, deactivate the sources and find
Rt by circuit reduction technique. Independent current sources, are deactivated by opening
them while independent voltage sources are deactivated by shorting them.
 Circuit Theorems j 175

(ii) If the circuit contains resistors, dependent and independent sources, follow the instructions
described below:
(a) Determine the open circuit voltage voc with the sources activated.
(b) Find the short circuit current isc when a short circuit is applied to the terminals a b
voc
(c) Rt =
isc

(iii) If the circuit contains resistors and only dependent sources, then
(a) voc = 0 (since there is no energy source)
(b) Connect 1A current source to terminals
a b and determine vab .
vab
(c) Rt =
1
Figure 3.32

For all the cases discussed above, the Thevenin’s equivalent circuit is as shown in Fig. 3.32.

EXAMPLE 3.10
Using the Thevenin’s theorem, find the current i through R = 2 Ω. Refer Fig. 3.33.

Figure 3.33

SOLUTION

Figure 3.34
176 j Network Theory

Since we are interested in the current i through R, the resistor R is identified as circuit B and
the remainder as circuit A. After removing the circuit B, circuit A is as shown in Fig. 3.35.

Figure 3.35

To find Rt , we have to deactivate the independent voltage source. Accordingly, we get the
circuit in Fig. 3.36.

Rt jj
= (5 Ω 20 Ω) + 4 Ω

=

5 20
+4=8Ω
5 + 20 Rt

Referring to Fig. 3.35,

50 + 25I = 0 ) I = 2A
Figure 3.36
Hence Vab = Voc = 20(I ) = 40V

Thus, we get the Thevenin’s equivalent circuit which is as shown in Fig.3.37.

Figure 3.37 Figure 3.38

Reconnecting the circuit B to the Thevenin’s equivalent circuit as shown in Fig. 3.38, we get
40
i = = 4A
2+8
 Circuit Theorems j 177

EXAMPLE 3.11
(a) Find the Thevenin’s equivalent circuit with respect to terminals a b for the circuit shown
in Fig. 3.39 by finding the open-circuit voltage and the short–circuit current.
(b) Solve the Thevenin resistance by removing the independent sources. Compare your result
with the Thevenin resistance found in part (a).

Figure 3.39
SOLUTION

Figure 3.40
(a) To find Voc :
Apply KCL at node 2 :
V2 V2 30
+ 1:5 = 0
)
60 + 20 40
V2 = 60 Volts

Hence; Voc =I
 V 60 0 
=
2
60 + 20
60 
= 60
60
80
= 45 V
178 j Network Theory

To find isc :

Applying KCL at node 2:

V2 V2 30
+ 1:5 = 0
)
20 40
V2 = 30V
V2
isc = = 1:5A
20
Voc 45
Therefore; Rt = =
isc 1:5
= 30 Ω
Figure 3.40 (a)

The Thevenin equivalent circuit with respect to the terminals a b is as shown in Fig. 3.40(a).
(b) Let us now find Thevenin resistance Rt by deactivating all the independent sources,

Rt Rt

Rt jj
= 60 Ω (40 + 20) Ω
60
= = 30 Ω (verified)
2

It is seen that, if only independent sources are present, it is easy to find Rt by deactivating all
the independent sources.
 Circuit Theorems j 179

EXAMPLE 3.12
Find the Thevenin equivalent for the circuit shown in Fig. 3.41 with respect to terminals a b.

Figure 3.41

SOLUTION
To find Voc = Vab :
Applying KVL around the mesh of
Fig. 3.42, we get

20 + 6i 2i + 6i = 0
) i = 2A

Since there is no current flowing in

10 Ω resistor, Voc = 6i = 12 V
To find Rt : (Refer Fig. 3.43)
Since both dependent and indepen- Figure 3.42
dent sources are present, Thevenin resis-
tance is found using the relation,
voc
Rt =
isc

Applying KVL clockwise for mesh 1 :

20 + 6i1 2i + 6 (i1 i2 ) =0
) 12i1 6i2 = 20 + 2i

Since i = i1 i2 , we get

12i1 6i2 = 20 + 2 (i1 i2 )

) 10i1 4i2 = 20

Applying KVL clockwise for mesh 2 :

10i2 + 6 (i2 i1 ) =0
) 6i1 + 16i2 = 0 Figure 3.43
180 j Network Theory

Solving the above two mesh equations, we get

i2 =
120
136
A ) isc = i2 =
120
136
A
voc 12
Rt = = = 13:6 Ω
isc 120
136
EXAMPLE 3.13
Find Vo in the circuit of Fig. 3.44 using Thevenin’s theorem.

Figure 3.44

SOLUTION
To find Voc :
Since we are interested in the voltage across 2 kΩ resistor, it is removed from the circuit of
Fig. 3.44 and so the circuit becomes as shown in Fig. 3.45.

Figure 3.45
By inspection, i1 = 4 mA
Applying KVL to mesh 2 :
12 + 6  10 3
(i2

+ 3  103 i2 = 0
i1 )
3

) 12 + 6  10 3
i2 4  10 3
+3  10 i2 =0
 Circuit Theorems j 181

Solving, we get i2 = 4 mA
Applying KVL to the path 4 kΩ ! a b ! 3 kΩ, we get
4  10 3
i1 + Voc 3  10 = 03
i2

) Voc = 4  10 + 3  10
3
i1
3
i2

= 4  10  4  10 + 3  10  4  10
3 3 3 3
= 28V

To find Rt :
Deactivating all the independent sources, we get the circuit diagram shown in Fig. 3.46.

Figure 3.46

Rt jj
= Rab = 4 kΩ + (6 kΩ 3 kΩ) = 6 kΩ
Hence, the Thevenin equivalent circuit is as shown in Fig. 3.47.

Figure 3.47 Figure 3.48

If we connect the 2 kΩ resistor to this equivalent network, we obtain the circuit of Fig. 3.48.

Vo =i 2  10
3

=
28
(6 + 2)  10
 2  10
3
3
= 7V

EXAMPLE 3.14
The wheatstone bridge in the circuit shown in Fig. 3.49 (a) is balanced when R2 = 1200 Ω. If the
galvanometer has a resistance of 30 Ω, how much current will be detected by it when the bridge
is unbalanced by setting R2 to 1204 Ω ?
182 j Network Theory

Figure 3.49(a)
SOLUTION
To find Voc :
We are interested in the galavanometer current. Hence, it is removed from the circuit of Fig.
3.49 (a) to find Voc and we get the circuit shown in Fig. 3.49 (b).
120 120
i1 = = A
900 + 600 1500
120 120
i2 = = A
1204 + 800 2004
Applying KVL clockwise along the path
1204Ω ! b a !
900 Ω, we get
1204i2 Vt 900i1 = 0
) Vt = 1204i2900i1
= 1204 
120
2004
900  1500
120

= 95:8 mV
Figure 3.49(b)
To find Rt :
Deactivate all the independent sources and look into the terminals a b to determine the
Thevenin’s resistance.

Figure 3.49(c) Figure 3.49(d)

 Circuit Theorems j 183

Rt jj
= Rab = 600 900 + 800 1204 jj
=

900 600 1204 800
+

1500 2004
= 840:64 Ω

Hence, the Thevenin equivalent circuit consists of the

95.8 mV source in series with 840.64Ω resistor. If we
connect 30Ω resistor (galvanometer resistance) to this
equivalent network, we obtain the circuit in Fig. 3.50. Figure 3.50

iG =

95:8 10 3
= 110:03 A
840:64 + 30 Ω

EXAMPLE 3.15
For the circuit shown in Fig. 3.51, find the Thevenin’s equivalent circuit between terminals a and b.

Figure 3.51

SOLUTION
With ab shorted, let Isc = I . The circuit after
transforming voltage sources into their equiv-
alent current sources is as shown in Fig 3.52.
Writing node equations for this circuit,

At a : 0:2Va 0:1 Vc + I = 3
At c : 0:1Va + 0:3 Vc 0:1 Vb = 4
At b : 0:1Vc + 0:2 Vb I =1

As the terminals a and b are shorted Va = Vb

Figure 3.52
and the above equations become
184 j Network Theory

0:2Va 0:1 Vc + I = 3
0:2Va + 0:3 Vc = 4
0:2Va 0:1 Vc 1=1

Solving the above equations, we get the short circuit current, I = Isc = 1 A.
Next let us open circuit the terminals a and b and this makes I = 0. And the node equations
written earlier are modified to

0:2Va 0:1 Vc = 3
0:1Va + 0:3 Vc 0:1 Vb = 4
0:1Vc + 0:2 Vb = 1

Solving the above equations, we get

Va = 30V and Vb = 20V

Hence, Vab = 30 20 = 10 V = Voc = Vt

Voc 10
Therefore Rt = = = 10Ω
Isc 1
The Thevenin’s equivalent is as shown in Fig 3.53
Figure 3.53

EXAMPLE 3.16
Refer to the circuit shown in Fig. 3.54. Find the Thevenin equivalent circuit at the terminals a b.

Figure 3.54

SOLUTION
To begin with let us transform 3 A current source and 10 V voltage source. This results in a
network as shown in Fig. 3.55 (a) and further reduced to Fig. 3.55 (b).
 Circuit Theorems j 185

Figure 3.55(a)
Again transform the 30 V source and following the reduction procedure step by step from
Fig. 3.55 (b) to 3.55 (d), we get the Thevenin’s equivalent circuit as shown in Fig. 3.56.

Figure 3.55(b) Figure 3.55(c)

Figure 3.55(d) Figure 3.56 Thevenin equivalent

circuit

EXAMPLE 3.17
Find the Thevenin equivalent circuit as seen from the terminals a b. Refer the circuit diagram
shown in Fig. 3.57.
186 j Network Theory

Figure 3.57
SOLUTION
Since the circuit has no independent sources, i = 0 when the terminals a b are open. There-
fore, Voc = 0.
The onus is now to find Rt . Since Voc = 0 and isc = 0, Rt cannot be determined from
Voc
Rt = . Hence, we choose to connect a source of 1 A at the terminals a b as shown in Fig.
isc
3.58. Then, after finding Vab , the Thevenin resistance is,
Vab
Rt =
1
KCL at node a : Va 2i Va
+ 1=0
5 10
Va
Also; i =
Va

10
Va 2 10 Va
Hence; + 1=0
5 10
) Va =
50
13
V
50 Va
Hence; Rt =
Ω =
1 13
Alternatively one could find Rt by connecting a 1V source at the terminals a b and then find
1
the current from b to a. Then Rt = . The concept of finding Rt by connecting a 1A source
iba
between the terminals a b may also be used for circuits containing independent sources. Then
set all the independent sources to zero and use 1A source at the terminals a b to find Vab and
Vab
hence, Rt = .
1
For the present problem, the Thevenin equivalent circuit as seen between the terminals a b
is shown in Fig. 3.58 (a).

Figure 3.58 Figure 3.58 (a)

 Circuit Theorems j 187

EXAMPLE 3.18
Determine the Thevenin equivalent circuit between the terminals a b for the circuit of Fig. 3.59.

Figure 3.59

SOLUTION
As there are no independent sources in the circuit, we get Voc = Vt = 0:
To find Rt , connect a 1V source to the terminals a b and measure the current I that flows
from b to a. (Refer Fig. 3.60 a).
1
Rt = Ω
I

Figure 3.60(a)

Applying KCL at node a:

Vx
I = 0:5Vx +
4
Since; Vx = 1V
1
we get, I = 0:5 + = 0:75 A
4 Figure 3.60(b)
1
Hence; Rt = = 1:33 Ω
0:75
The Thevenin equivalent circuit is shown in 3.60(b).
Alternatively, sticking to our strategy, let us connect 1A current source between the terminals
Vab
a b and then measure Vab (Fig. 3.60 (c)). Consequently, Rt = = Vab Ω:
1
188 j Network Theory

Applying KCL at node a:

0:5Vx +
Vx
4
=1 ) Vx = 1:33V
Vab Vx
Hence Rt = = = 1:33 Ω
1 1

The corresponding Thevenin equivalent

circuit is same as shown in Fig. 3.60(b) Figure 3.60(c)

3.3 Norton’s theorem

An American engineer, E.L. Norton at Bell Telephone Laboratories, proposed a theorem similar
to Thevenin’s theorem.
Norton’s theorem states that a linear two-terminal network can be replaced by an
equivalent circuit consisting of a current source iN in parallel with resistor RN , where iN
is the short-circuit current through the terminals and RN is the input or equivalent resistance
at the terminals when the independent sources are turned off. If one does not wish to turn off
the independent sources, then RN is the ratio of open circuit voltage to short–circuit current
at the terminal pair.

Figure 3.61(a) Original circuit Figure 3.61(b) Norton’s equivalent circuit

Figure 3.61(b) shows Norton’s equivalent circuit as seen from the terminals a b of the
original circuit shown in Fig. 3.61(a). Since this is the dual of the Thevenin circuit, it is clear that
voc
RN = Rt and iN = . In fact, source transformation of Thevenin equivalent circuit leads to
Rt
Norton’s equivalent circuit.
Procedure for finding Norton’s equivalent circuit:
(1) If the network contains resistors and independent sources, follow the instructions below:
(a) Deactivate the sources and find RN by circuit reduction techniques.
(b) Find iN with sources activated.
(2) If the network contains resistors, independent and dependent sources, follow the steps given
below:
(a) Determine the short-circuit current iN with all sources activated.
 Circuit Theorems j 189

(b) Find the open-circuit voltage voc .

voc
(c) Rt = RN =
iN

(3) If the network contains only resistors and dependent sources, follow the procedure
described below:

(a) Note that iN = 0.

(b) Connect 1A current source to the terminals a b and find vab .
vab
(c) Rt =
1
Note: Also, since vt = voc and iN = isc
voc
Rt = = RN
isc

The open–circuit and short–circuit test are sufficient to find any Thevenin or Norton equiva-
lent.

3.3.1 PROOF OF THEVENIN’S AND NORTON’S THEOREMS

The principle of superposition is employed to provide the proof of Thevenin’s and Norton’s
theorems.

Derivation of Thevenin’s theorem:

Let us consider a linear circuit having two accessible terminals x y and excited by an external
current source i. The linear circuit is made up of resistors, dependent and independent sources. For
the sake of simplified analysis, let us assume that the linear circuit contains only two independent
voltage sources v1 and v2 and two independent current sources i1 and i2 . The terminal voltage v
may be obtained, by applying the principle of superposition. That is, v is made up of contributions
due to the external source and independent sources within the linear network.

Hence; v = a0 i + a1 v1 + a2 v2 + a3 i1 + a4 i2 (3.9)
= a0 i + b0 (3.10)
where b0 = a1 v1 + a2 v2 + a3 i1 + a4 i2
= contribution to the terminal voltage v by
independent sources within the linear network.

Let us now evaluate the values of constants a0 and b0 .

(i) When the terminals x and y are open–circuited, i = 0 and v = voc = vt . Making use of
this fact in equation 3.10, we find that b0 = vt .
190 j Network Theory

(ii) When all the internal sources are deactivated, b0 = 0. This enforces equation 3.10 to
become
v = a0 i = Rt i ) a0 = Rt
Rt

Vt

Figure 3.62 Current-driven circuit Figure 3.63 Thevenin’s equivalent circuit of Fig. 3.62

where Rt is the equivalent resistance of the linear network as viewed from the terminals x y .
Also, a0 must be Rt in order to obey the ohm’s law. Substuting the values of a0 and b0 in equation
3.10, we find that
v = Rt i + v1

which expresses the voltage-current relationship at terminals x y of the circuit in Fig. 3.63.
Thus, the two circuits of Fig. 3.62 and 3.63 are equivalent.

Derivation of Norton’s theorem:

Let us now assume that the linear circuit described earlier is driven by a voltage source v as shown
in Fig. 3.64.
The current flowing into the circuit can be obtained by superposition as

i = c0 v + d 0 (3.11)

where c0 v is the contribution to i due to the external voltage source v and d0 contains the contri-
butions to i due to all independent sources within the linear circuit. The constants c0 and d0 are
determined as follows :
(i) When terminals x y are short-circuited, v =
0 and i = isc . Hence from equation (3.11),
we find that i = d0 = isc , where isc is the
short-circuit current flowing out of terminal x,
which is same as Norton current iN

Thus, d0 = iN
Figure 3.64
Voltage-driven circuit

(ii) Let all the independent sources within the linear network be turned off, that is d0 = 0. Then,
equation (3.11) becomes
i = c0 v
 Circuit Theorems j 191

For dimensional validity, c0 must have the

dimension of conductance. This enforces c0 =
1
where Rt is the equivalent resistance of the
Rt
linear network as seen from the terminals x y.
Thus, equation (3.11) becomes
1
i = v isc
Rt
Figure 3.65 Norton’s equivalent of
1
= v iN voltage driven circuit
Rt

This expresses the voltage-current relationship at the terminals x y of the circuit in Fig.
(3.65), validating that the two circuits of Figs. 3.64 and 3.65 are equivalents.

EXAMPLE 3.19
Find the Norton equivalent for the circuit of Fig. 3.66.

Figure 3.66

SOLUTION
As a first step, short the terminals a b. This
results in a circuit diagram as shown in Fig. 3.67.
Applying KCL at node a, we get
0 24
3 + isc = 0
)
4
isc = 9A

To find RN , deactivate all the independent

sources, resulting in a circuit diagram as shown
in Fig. 3.68 (a). We find RN in the same way as Figure 3.67
Rt in the Thevenin equivalent circuit.

RN =
4 12
=3Ω

4 + 12
192 j Network Theory

Figure 3.68(a) Figure 3.68(b)

Thus, we obtain Nortion equivalent circuit as shown in Fig. 3.68(b).

EXAMPLE 3.20
Refer the circuit shown in Fig. 3.69. Find the value of ib using Norton equivalent circuit. Take
R = 667 Ω.

Figure 3.69
SOLUTION
Since we want the current flowing through R, remove
R from the circuit of Fig. 3.69. The resulting circuit
diagram is shown in Fig. 3.70.
To find iac or iN referring Fig 3.70(a) :
0
ia = = 0A
1000
12
isc = A = 2 mA Figure 3.70
6000

Figure 3.70(a)
 Circuit Theorems j 193

To find RN :
The procedure for finding RN is same that of Rt
in the Thevenin equivalent circuit.
voc
Rt = RN =
isc

To find voc , make use of the circuit diagram shown

in Fig. 3.71. Do not deactivate any source.
Applying KVL clockwise, we get
Figure 3.71
12 + 6000ia + 2000ia + 1000ia = 0
) ia =
4
3000
A

) voc = ia  4
1000 = V
3
4
voc

Therefore; RN = = 3 = 667 Ω
isc 2 10 3
The Norton equivalent circuit along with resistor R is as shown below:
isc 2mA
ib = = = 1mA
2 2

Figure : Norton equivalent circuit with load R

EXAMPLE 3.21
Find Io in the network of Fig. 3.72 using Norton’s theorem.

Figure 3.72
194 j Network Theory

SOLUTION
We are interested in Io , hence the 2 kΩ resistor is removed from the circuit diagram of Fig. 3.72.
The resulting circuit diagram is shown in Fig. 3.73(a).

Figure 3.73(a) Figure 3.73(b)

To find iN or isc :
Refer Fig. 3.73(b). By inspection, V1 = 12 V
Applying KCL at node V2 :

V2 V1 V2 V2 V1
+ + =0
6 kΩ 2 kΩ 3 kΩ
Substituting V1 = 12 V and solving, we get

V2 = 6V
V1 V2 V1
isc = + = 5 mA
3 kΩ 4 kΩ
To find RN :
Deactivate all the independent sources (refer Fig. 3.73(c)).

Figure 3.73(c) Figure 3.73(d)

 Circuit Theorems j 195

Referring to Fig. 3.73 (d), we get

RN jj jj
= Rab = 4 kΩ [3 kΩ + (6 kΩ 2 kΩ)] = 2:12 kΩ

Hence, the Norton equivalent circuit

along with 2 kΩ resistor is as shown in
Fig. 3.73(e).

Io =
isc  RN
= 2:57mA
R + RN
Figure 3.73(e)

EXAMPLE 3.22
Find Vo in the circuit of Fig. 3. 74.

Figure 3.74

SOLUTION
Since we are interested in Vo , the voltage across 4 kΩ resistor, remove this resistance from the
circuit. This results in a circuit diagram as shown in Fig. 3.75.

Figure 3.75
196 j Network Theory

To find isc , short the terminals a b :

 Circuit Theorems j 197

Constraint equation :
i1 i2 = 4mA (3.12)
KVL around supermesh :
4+2  10 3
i1 +4  10 3
i2 =0 (3.13)
KVL around mesh 3 :
8  10 ( 3
i3 i2 ) +2  10 ( 3
i3 i1 ) =0
Since i3 = isc , the above equation becomes,
8  10 ( 3
isc i2 ) +2  10 ( 3
isc i1 ) =0 (3.14)
Solving equations (3.12), (3.13) and (3.14) simultaneously, we get isc = 0:1333 mA.
To find RN :
Deactivate all the sources in Fig. 3.75. This yields a circuit diagram as shown in Fig. 3.76.

Figure 3.76

RN = 6 kΩ 10 kΩjj
=

6 10
= 3:75 kΩ
6 + 10
Hence, the Norton equivalent circuit is as shown
in Fig 3.76 (a).
To the Norton equivalent circuit, now connect the
4 kΩ resistor that was removed earlier to get the Figure 3.76(a)
network shown in Fig. 3.76(b).
198 j Network Theory

Vo = isc (RN jj R)

RN R
= isc
RN +R
= 258 mV

Figure 3.76(b) Norton equivalent circuit with R = 4 kΩ

EXAMPLE 3.23
Find the Norton equivalent to the left of the terminals a b for the circuit of Fig. 3.77.

Figure 3.77

SOLUTION
To find isc :

Note that vab = 0 when the terminals a b are short-circuited.

5
Then i = = 10 mA
500
Therefore, for the right–hand portion of the circuit, isc = 10i = 100 mA.
 Circuit Theorems j 199

To find RN or Rt :

Writing the KVL equations for the left-hand mesh, we get

5 + 500i + vab = 0 (3.15)
Also for the right-hand mesh, we get
vab = 25(10i) = 250i
vab
Therefore i =
250
Substituting i into the mesh equation (3.15), we get
 vab

5 + 500 + vab = 0
250
) vab= 5V
voc vab 5
RN = Rt  = = = 50 Ω
isc isc 0:1
The Norton equivalent circuit is shown in
Fig 3.77 (a).

Figure 3.77 (a)

EXAMPLE 3.24
Find the Norton equivalent of the network shown in Fig. 3.78.

Figure 3.78
200 j Network Theory

SOLUTION
Since there are no independent sources present in the network of Fig. 3.78, iN = isc = 0.
To find RN , we inject a current of 1A between the terminals a b. This is illustrated in
Fig. 3.79.

Figure 3.79 Figure 3.79(a) Norton

equivalent circuit

KCL at node 1:
v1 v1 v2
1= +
)
100 50
0:03v1 0:02v2 = 1
KCL at node 2: v2 v2 v1
+ + 0:1v1 = 0
)
200 50
0:08v1 + 0:025v2 = 0
Solving the above two nodal equations, we get
v1 = 10:64 volts )
voc = 10:64 volts
10:64
voc
Hence; R N = Rt = = = 10:64 Ω
1 1
Norton equivalent circuit for the network shown in Fig. 3.78 is as shown in Fig. 3.79(a).

EXAMPLE 3.25
Find the Thevenin and Norton equivalent circuits for the network shown in Fig. 3.80 (a).

Figure 3.80(a)
 Circuit Theorems j 201

SOLUTION
To find Voc :
Performing source transformation on 5A current source, we get the circuit shown in
Fig. 3.80 (b).
Applying KVL around Left mesh :
50 + 2ia 20 + 4ia = 0
) ia =
70
6
A

Applying KVL around right mesh:

20 + 10ia + Voc 4ia = 0

) Voc = 90 V
Figure 3.80(b)

To find isc (referring Fig 3.80 (c)) :

KVL around Left mesh :
50 + 2ia 20 + 4 (ia isc ) =0
) 6ia 4isc = 70
KVL around right mesh :

4 (isc ia ) + 20 + 10ia = 0
) 6ia + 4isc = 20

Figure 3.80(c)

Solving the two mesh equations simultaneously, we get isc = 11:25 A

voc 90
Hence, Rt = RN = = =8Ω
isc 11:25
Performing source transformation on Thevenin equivalent circuit, we get the norton equivalent
circuit (both are shown below).

Thevenin equivalent circuit Norton equivalent circuit

202 j Network Theory

EXAMPLE 3.26
If an 8 kΩ load is connected to the terminals of the
network in Fig. 3.81, VAB = 16 V. If a 2 kΩ load is
connected to the terminals, VAB = 8V. Find VAB if a
20 kΩ load is connected across the terminals.

SOLUTION
Figure 3.81

Applying KVL around the mesh, we get (Rt + RL ) I = Voc

If RL = 2 kΩ; I = 10 mA ) Voc = 20 + 0:01Rt

If RL = 10 kΩ; I = 6 mA ) Voc = 60 + 0:006Rt

Solving, we get Voc = 120 V, Rt = 10 kΩ.

Voc 120
If RL = 20 kΩ; I =
(RL + Rt )
=
(20  103 + 10  10 ) = 4 mA
3

3.4 Maximum Power Transfer Theorem

In circuit analysis, we are some times interested

in determining the maximum power that a circuit
can supply to the load. Consider the linear circuit
A as shown in Fig. 3.82.
Circuit A is replaced by its Thevenin equivalent
circuit as seen from a and b (Fig 3.83).
We wish to find the value of the load RL such that Figure 3.82 Circuit A with load RL
the maximum power is delivered to it.
The power that is delivered to the load is given by

 Vt
2
2
p =i RL = RL (3.16)
Rt + RL
 Circuit Theorems j 203

Assuming that Vt and Rt are fixed for a given source, the maximum power is a function of
RL .In order to determine the value of RL that maximizes p, we differentiate p with respect to
RL and equate the derivative to zero.
" #
dp (Rt + RL )2 2 (Rt + RL )
= Vt2 =0
dRL (RL + Rt )2
which yields RL = Rt (3.17)

To confirm that equation (3.17) is a maximum,

2
d p
it should be shown that 2 < 0. Hence, maxi-
dRL
mum power is transferred to the load when RL is
equal to the Thevenin equivalent resistance Rt .
The maximum power transferred to the load is
obtained by substituting RL = Rt in equation
3.16.
Accordingly, Figure 3.83 Thevenin equivalent circuit
is substituted for circuit A
2
Vt R L Vt
2
Pmax = =
(2RL )2 4RL

The maximum power transfer theorem states that the maximum power delivered by a source
represented by its Thevenin equivalent circuit is attained when the load RL is equal to the
Thevenin resistance Rt .

EXAMPLE 3.27
Find the load RL that will result in maximum power delivered to the load for the circuit of Fig.
3.84. Also determine the maximum power Pmax .

Figure 3.84

SOLUTION
Disconnect the load resistor RL . This results in a circuit diagram as shown in Fig. 3.85(a).
Next let us determine the Thevenin equivalent circuit as seen from a b.
204 j Network Theory

180
i = = 1A

150 + 30
Voc = Vt = 150 i = 150 V

To find Rt , deactivate the 180 V source. This results in the

circuit diagram of Fig. 3.85(b).

Rt = Rab = 30 Ω 150 Ωjj

=
30 150  = 25 Ω
Figure 3.85(a)

30 + 150
The Thevenin equivalent circuit connected to the
load resistor is shown in Fig. 3.86.
Maximum power transfer is obtained when
RL = Rt = 25 Ω:
Then the maximum power is

(150)2
2
Vt
Pmax
4RL
=
4 25
=
 Figure 3.85(b)
= 2:25 Watts
The Thevenin source Vt actually provides a total
power of

Pt = 150  i

= 150 
150
25 + 25
= 450 Watts
Thus, we note that one-half the power is dissipated in RL .
Figure 3.86
EXAMPLE 3.28
Refer to the circuit shown in Fig. 3.87. Find the value of RL for maximum power transfer. Also
find the maximum power transferred to RL .

Figure 3.87
 Circuit Theorems j 205

SOLUTION
Disconnecting RL , results in a circuit diagram as shown in Fig. 3.88(a).

Figure 3.88(a)

To find Rt , deactivate all the independent voltage sources as in Fig. 3.88(b).

Figure 3.88(b) Figure 3.88(c)

Rt jj
= Rab = 6 kΩ 6 kΩ 6 kΩ jj
= 2 kΩ

To find Vt :
Refer the Fig. 3.88(d).
Constraint equation :

V3 V1 = 12 V

By inspection, V2 =3V
KCL at supernode :
V3 V2 V1 V1 V2
+ + =0
6k 6k 6k

) V3
6k
3
+
V3
6k
12
+
V3 12 3
6k
=0
Figure 3.88(d)
206 j Network Theory

) V3 3 + V3 12 + V3 15 = 0
) 3V3 = 30
) V3 = 10
) Vt = Vab = V3 = 10 V

Figure 3.88(e)

The Thevenin equivalent circuit connected to the load resistor RL is shown in Fig. 3.88(e).

Pmax = i2 RL
 Vt
2
= RL
2RL
= 12:5 mW

Alternate method :
It is possible to find Pmax , without finding the Thevenin equivalent circuit. However, we have to
find Rt . For maximum power transfer, RL = Rt = 2 kΩ. Insert the value of RL in the original
circuit given in Fig. 3.87. Then use any circuit reduction technique of your choice to find power
dissipated in RL .
Refer Fig. 3.88(f). By inspection we find that, V2 = 3 V.
Constraint equation :

V3 V1 = 12
) V1 = V3 12

KCL at supernode :
V3 V2 V1 V2 V3 V1
+ + + =0
6k 6k 2k 6k

) V3
6k
3
+
V3 12
6k
3
+
2k
+
V3
6k
=0
V3 12

) V3 3 + V3 15 + 3V3 + V3 12 = 0
) 6V3 = 30
) V3 =5 V Figure 3.88(f)

V3
2 25
Hence; Pmax = = = 12:5 mW
RL 2k
 Circuit Theorems j 207

EXAMPLE 3.29
Find RL for maximum power transfer and the maximum power that can be transferred in the
network shown in Fig. 3.89.

Figure 3.89

SOLUTION
Disconnect the load resistor RL . This results in a circuit as shown in Fig. 3.89(a).

Figure 3.89(a)

To find Rt , let us deactivate all the independent sources, which results the circuit as shown in
Fig. 3.89(b).
Rt = Rab = 2 kΩ + 3 kΩ + 5 kΩ = 10 kΩ

For maximum power transfer RL = Rt = 10 kΩ.

Let us next find Voc or Vt .
Refer Fig. 3.89 (c). By inspection, i1 = 2 mA & i2 = 1 mA.
208 j Network Theory

Figure 3.89(b)

Applying KVL clockwise to the loop 5 kΩ ! 3 kΩ ! 2 kΩ ! , we get a b

5k  + 3k (
i2 i1 ) + 2k  + = 0
i2 i1 Vt




) 5  10 1  10
3 3 +3  1032  10 1  10 3 +2  10 2  10 3 3 3 + Vt = 0
) 5 9 4 + Vt = 0
) Vt = 18 V:
The Thevenin equivalent circuit with load resistor RL is as shown in Fig. 3.89 (d).
18
i =
(10 + 10) 103 
= 0:9 mA

Then,
Pmax = PL = (0:9 mA)2  10 kΩ
= 8:1 mW

Figure 3.89(c) Figure 3.89(d)

EXAMPLE 3.30
Find the maximum power dissipated in RL . Refer the circuit shown in Fig. 3.90.

Figure 3.90
 Circuit Theorems j 209

SOLUTION
Disconnecting the load resistor RL from the original circuit results in a circuit diagram as shown
in Fig. 3.91.

Figure 3.91
As a first step in the analysis, let us find Rt . While finding Rt , we have to deactivate all the
independent sources. This results in a network as shown in Fig 3.91 (a) :

Figure 3.91(a)

Rt jj
= Rab = [140 Ω 60 Ω] + 8 Ω
=

140 60
+ 8 = 50 Ω:
140 + 60
For maximum power transfer, RL = Rt = 50 Ω. Next step in the analysis is to find Vt .
Refer Fig 3.91(b), using the principle of
current division,

i1 =

i R2
+
R1 R2

20  170
= = 17 A
170 + 30

i2 =
 = 20  30
i R1
R1 + R2 170 + 30
600
= = 3A
200
Figure 3.91(a)
210 j Network Theory

Applying KVL clockwise to the loop comprising of 50 Ω ! 10 Ω ! 8 Ω ! a b, we get

50i2 10i1 + 8 0+ Vt =0

) 50(3) 10 (17) + Vt = 0
) Vt = 20 V

The Thevenin equivalent circuit with load resistor RL is

as shown in Fig. 3.91(c).
Figure 3.91(c)
20
iT = = 0:2A
50 + 50
2
Pmax = iT  
50 = 0:04 50 = 2 W

EXAMPLE 3.31
Find the value of RL for maximum power transfer in the circuit shown in Fig. 3.92. Also
find Pmax .

Figure 3.92

SOLUTION
Disconnecting RL from the original circuit, we get the network shown in Fig. 3.93.

Figure 3.93
 Circuit Theorems j 211

Let us draw the Thevenin equivalent circuit as seen from the terminals a b and then insert
the value of RL = Rt between the terminals a b. To find Rt , let us deactivate all independent
sources which results in the circuit as shown in Fig. 3.94.

Figure 3.94

Rt = Rab
jj
=8Ω 2Ω
=

8 2
= 1:6 Ω
8+2
Next step is to find Voc or Vt .
By performing source transformation on the circuit shown in Fig. 3.93, we obtain the circuit
shown in Fig. 3.95.

Figure 3.95

Applying KVL to the loop made up of 20 V ! 3 Ω ! 2 Ω ! 10 V ! 5 Ω ! 30 V, we get

20 + 10i 10 30 = 0
) i =
60
10
= 6A
212 j Network Theory

Again applying KVL clockwise to the path 2 Ω ! 10 V ! a b, we get

2i 10 Vt = 0
) Vt = 2i 10
= 2(6) 10 = 2 V

The Thevenin equivalent circuit with load resistor

RL is as shown in Fig. 3.95 (a).

Pmax = i2T RL
Vt
2
Figure 3.95(a) Thevenin equivalent
= = 625 mW
4Rt circuit

EXAMPLE 3.32
Find the value of RL for maximum power transfer. Hence find Pmax .

Figure 3.96
SOLUTION
Removing RL from the original circuit gives us the circuit diagram shown in Fig. 3.97.

Figure 3.97

To find Voc :
KCL at node A :
0
ia 0:9 + 10i0a = 0
) 0
ia = 0:1 A


Hence; Voc = 3 10i0a
=3  10  0 1 = 3 V
:
 Circuit Theorems j 213

To find Rt , we need to compute isc with all independent sources activated.

KCL at node A:
00
ia 0:9 + 10ia 00 = 0
) ia
00
= 0:1 A
Hence isc = 10ia = 10 00
01=1A
:
Voc 3
Rt = = =3Ω
isc 1
Hence, for maximum power transfer RL = Rt = 3 Ω.
The Thevenin equivalent circuit with RL = 3 Ω
inserted between the terminals a b gives the net-
work shown in Fig. 3.97(a).

3
iT = = 0:5 A
3+3
Pmax = i2T RL
= (0:5)2 3
= 0.75 W
Figure 3.97(a)
EXAMPLE 3.33
Find the value of RL in the network shown that will achieve maximum power transfer, and deter-
mine the value of the maximum power.

Figure 3.98(a)
SOLUTION
Removing RL from the circuit of Fig. 3.98(a), we
get the circuit of Fig 3.98(b).
Applying KVL clockwise we get

12 + 2 103 i + 2Vx0 = 0
Also 0
Vx =1  10 3
i

Hence; 12 + 2  10 3i +2 1  10 3i =0
12
4  10
i = = 3 mA Figure 3.98(b)
3
214 j Network Theory

Applying KVL to loop 1 kΩ !2 !Vx


b a, we get

1  10 3
i + 2Vx0 Vt =0


) Vt =1  10 + 2 1  10

3
i
3
i

= 1  10 + 2  10 3 3

i

= 3  10 3  10 3 3

=9V

To find Rt , we need to find isc . While finding isc ,

none of the independent sources must be deacti-
vated.
Applying KVL to mesh 1:

12 + Vx 00 + 0 = 0
) Vx
00
= 12
) 1  10 3
i1 = 12 ) i1 = 12 mA

Applying KVL to mesh 2:

1  10
3
i2 + 2Vx 00 = 0
) 1  10 3
i2 = 24
i2 = 24 mA
Applying KCL at node a:

isc = i1 i2

= 12 + 24 = 36 mA
Vt Voc
Hence; Rt = =
isc isc
9
=
36 10  3
= 250 Ω

For maximum power transfer, RL = Rt = 250 Ω.

Thus, the Thevenin equivalent circuit with RL is
as shown in Fig 3.98 (c) :
9 9
iT = = A
250 + 250 500
2
Pmax = iT 250


=
9 2
500
 250 Figure 3.98 (c) Thevenin equivalent circuit
= 81 mW
 Circuit Theorems j 215

EXAMPLE 3.34
The variable resistor RL in the circuit of Fig. 3.99 is adjusted untill it absorbs maximum power
from the circuit.
(a) Find the value of RL .
(b) Find the maximum power.

Figure 3.99

SOLUTION
Disconnecting the load resistor RL from the original circuit, we get the circuit shown in
Fig. 3.99(a).

Figure 3.99(a)
KCL at node v1 :

v1 100 v1 13i0a v1 v2
+ + =0 (3.18)
2 5 4
Constraint equations :
0
100 v1
ia = (3.19)
2
v2 v1
= va0 (applying K C L at v2 ) (3.20)
4
0
va = v1 v2 (potential across 4 Ω) (3.21)
216 j Network Theory

From equations (3.20) and (3.21), we have

v2 v1
= v1 v2

)
4
v2 v1 = 4v1 4v2
) 5v1 5v2 = 0
) v1 = v2 (3.22)

Making use of equations (3.19) and (3.22) in (3.18), we get

(100 v1 )
v1 100 v2 13 v1 v1
+ 2 + =0
2 5 4
 
) 5 (v1 100) + 2 v1 13
(100
2
v1 )
=0
) 5v1 500 + 2v1 13  100 + 13v1 = 0
) 20v1 = 1800
) v1 = 90 Volts
Hence; vt = v2 = v1 = 90 Volts
voc vt
We know that, Rt = =
isc isc

The short circuit current is calculated using the circuit shown below:

00
100 v1
Here ia =
2
Applying KCL at node v1 :

v1 100 v1 13ia v1 0
+ + =0
2 5 4
(100 v1 )
) v1 100 v1 13 v1
+ 2 + =0
2 5 4
 Circuit Theorems j 217

Solving we get v1 = 80 volts = va00

Applying KCL at node a :
0 v1
+ isc = va00
4
) isc
4
=
v1
+ va00
80
= + 80 = 100 A
4
voc vt
Hence; Rt = =
isc isc
90
= = 0:9 Ω
100
Hence for maximum power transfer,

RL = Rt = 0:9 Ω

The Thevenin equivalent circuit with RL = 0:9 Ω

is as shown.
90 90
it = =
0:9 + 0:9 1:8
2
Pmax = it 0:9 
 90 2
=
1:8
 0 9 = 2250 W
:

EXAMPLE 3.35
Refer to the circuit shown in Fig. 3.100 :
(a) Find the value of RL for maximum power transfer.
(b) Find the maximum power that can be delivered to RL .

Figure 3.100
218 j Network Theory

SOLUTION
Removing the load resistor RL , we get the circuit diagram shown in Fig. 3.100(a). Let us proceed
to find Vt .

Figure 3.100(a)

Constraint equation :
0
ia = i1 i3

KVL clockwise to mesh 1 :

200 + 1 (i1 i2 ) + 20 (i1 i3 ) + 4i1 = 0
) 25i1 i2 20i3 = 200
KVL clockwise to mesh 2 :
14i0a + 2 (i2 i3 ) + 1 (i2 i1 ) =0
) 14 (i1 i3 ) + 2 (i2 i3 ) + 1 (i2 i1 ) =0
) 13i1 + 3i2 16i3 = 0
KVL clockwise to mesh 3 :
2 (i3 i2 ) 100 + 3i3 + 20 (i3 i1 ) =0
) 20i1 2i2 + 25i3 = 100
Solving the mesh equations, we get
i1 = 2:5A; i3 = 5A
Applying KVL clockwise to the path comprising of a b ! 20 Ω, we get
Vt 20i0a = 0
) Vt = 20i0a
= 20 (i1 i3 )

= 20 ( 2:5 5)
= 150 V
 Circuit Theorems j 219

Next step is to find Rt .

Voc Vt
Rt = =
isc isc

When terminals a b are shorted, i00a = 0. Hence, 14 i00a is also zero.

KVL clockwise to mesh 1 :

200 + 1 (i1 i2 ) + 4i1 = 0

) 5i1 i2 = 200

KVL clockwise to mesh 2 :

2 (i2 i3 ) + 1 (i2 i1 ) =0
) i1 + 3i2 2i3 = 0

KVL clockwise to mesh 3 :

100 + 3i3 + 2 (i3 i2 ) =0

) 2i2 + 5i3 = 100
220 j Network Theory

Solving the mesh equations, we find that

i1 = 40A; i3 = 20A;
) i3 = 60A
isc = i1
150 Vt
Rt = = = 2:5 Ω
isc 60
For maximum power transfer, RL = Rt = 2:5 Ω. The Thevenin equivalent circuit with RL is
as shown below :

Pmax = i21 RL
 2
=
150
2:5 + 2:5
25:

= 2250 W
EXAMPLE 3.36
A practical current source provides 10 W to a 250 Ω load and 20 W to an 80 Ω load. A resistance
RL , with voltage vL and current iL , is connected to it. Find the values of RL , vL and iL if
(a) vL iL is a maximum, (b) vL is a maximum and (c) iL is a maximum.
SOLUTION
Load current calculation:
r
10
10W to 250 Ω corresponds to iL =
250
=r 200 mA
20
20W to 80 Ω corresponds to iL =
80
= 500 mA
Using the formula for division of current between two parallel branches :

i2 =
i  R1
R1 + R2
IN R N
In the present context, 0:2 = (3.23)
RN + 250
IN R N
and 0:5 = (3.24)
RN + 80
 Circuit Theorems j 221

Solving equations (3.23) and (3.24), we get

IN = 1:7 A
RN = 33:33 Ω

(a) If vL iL is maximum,

RL = RN = 33:33 Ω
iL = 1:7  33:33
33:33 + 33:33
= 850 mA
vL = iL RL = 850  10  33 33
3
:

= 28:33 V
(b) vL = IN (RN jj RL ) is a maximum when RN RL jj is a maximum, which occurs when
RL = . 1
Then, iL = 0 and
vL = 1:7  RN

= 1 7  33 33
: :

= 56:66 V
IN R N
(c) iL = is maxmimum when RL = 0 Ω
RN + RL
) iL = 1:7A and vL = 0 V

3.5 Sinusoidal steady state analysis using superposition, Thevenin and

Norton equivalents

Circuits in the frequency domain with phasor currents and voltages and impedances are analogous
to resistive circuits.
To begin with, let us consider the principle of superposition, which may be restated as follows :
For a linear circuit containing two or more independent sources, any circuit voltage or
current may be calculated as the algebraic sum of all the individual currents or voltages caused
by each independent source acting alone.

Figure 3.101 Thevenin equivalent circuit Figure 3.102 Norton equivalent circuit
240 j Network Theory

10 /0
It =
(3000 + 3535:53) + j (4000 3535:53)
= 1:526 / 4:07 mA
Pmax jj
= It 2
RL


= 1:526  10 3 2
 3535 53
:

= 8:23 mW

This power is the maximum average power that can be delivered by this circuit to a load
impedance whose angle is constant at 45 . Again this quantity is less than the maximum
power that could have been delivered if there is no restriction on ZL . In example 3.46 part (a),
we have shown that the maximum power that can be delivered without any restrictions on ZL
is 8.33 mW.

3.7 Reciprocity theorem

The reciprocity theorem states that in a linear bilateral single source circuit, the ratio of exci-
tation to response is constant when the positions of excitation and response are interchanged.

Conditions to be met for the application of reciprocity theorem :

(i) The circuit must have a single source.

(ii) Initial conditions are assumed to be absent in the circuit.
(iii) Dependent sources are excluded even if they are linear.
(iv) When the positions of source and response are interchanged, their directions should be marked
same as in the original circuit.

EXAMPLE 3.48
Find the current in 2 Ω resistor and hence verify reciprocity theorem.

Figure 3.122
 Circuit Theorems j 241

SOLUTION
The circuit is redrawn with markings as shown in Fig 3.123 (a).

Figure 3.123 (a)

1 1 1
Then; R1 = (8 +2 ) = 1:6Ω
R2 = 1:6 + 4 = 5:6Ω
R3 = (5:6 1 + 4 1 ) 1 = 2:3333Ω
20
Current supplied by the source = = 3:16 A
4 + 2:3333
Current in branch ab = Iab = 3:16  4
4 + 4 + 1: 6
= 1:32 A

Current in 2Ω; I1 = 1:32  8

10
= 1:05 A

Verification using reciprocity theorem

The circuit is redrawn by interchanging the position of excitation and response as shown in
Fig 3.123 (b).

Figure 3.123 (b)

Solving the equivalent resistances,
R4 = 2Ω; R5 = 6Ω; R6 = 3:4286Ω
Now the current supplied by the source
20
= = 3:6842A
3:4286 + 2
242 j Network Theory

Therefore,
Icd = 3:6842  8 +8 6 = 2 1053A
:

2:1053
I2 = = 1:05A
2
As I1 = I2 = 1:05 A, reciprocity theorem is verified.

EXAMPLE 3.49
In the circuit shown in Fig. 3.124, find the current through 1:375 Ω resistor and hence verify
reciprocity theorem.

Figure 3.124

SOLUTION

Figure 3.125
KVL clockwise for mesh 1 :
6:375I1 2I2 3I3 = 0
KVL clockwise for mesh 2 :
2I1 + 14I2 10I3 = 0
KVL clockwise for mesh 3 :
3I1 10I2 + 14I3 = 10
 Circuit Theorems j 243

Putting the above three mesh equations in matrix form, we get

2 32 3 2 3
6:375 2 3 I1 0
4 2 14 10 54 I2 5=4 0 5
3 10 14 I3 10
Using Cramer’s rule, we get
I1 = 2A
Negative sign indicates that the assumed direction of current flow should have been the other way.
Verification using reciprocity theorem :
The circuit is redrawn by interchanging the positions of excitation and response. The new circuit
is shown in Fig. 3.126.

Figure 3.126
The mesh equations in matrix form for the circuit shown in Fig. 3.126 is
2 32 0
3 2 3
6:375 2 3 I1 10
4 2 14 10 54 0
I2 5=4 0 5
0
3 10 14 I3 0
Using Cramer’s rule, we get
0
I3 = 2A
Since I1 = I30 = 2 A, the reciprocity theorem is verified.

EXAMPLE 3.50
Find the current Ix in the j 2 Ω impedance and hence verify reciprocity theorem.

Figure 3.127
244 j Network Theory

SOLUTION
With reference to the Fig. 3.127, the current through j 2 Ω impepance is found using series parallel
reduction techniques.
Total impedance of the circuit is
jj
ZT = (2 + j 3) + ( j 5) (3 + j 2)
( j 5)(3 + j 2)
= 2 + j3 +
j5 + 3 + j2
= 6:537 /19:36 Ω
The total current in the network is
36 /0
IT =
6:537 /19:36
= 5:507 / 19:36 A
Using the principle of current division, we find that
IT ( j 5)
Ix =
j5 + 3 + j2
= 6:49 / 64:36 A
Verification of reciprocity theorem :
The circuit is redrawn by changing the positions of excitation and response. This circuit is shown
in Fig. 3.128.
Total impedance of the circuit shown in
Fig. 3.128 is

jj
Z0T = (3 + j 2) + (2 + j 3) ( j 5)
(2 + j 3) ( j 5)
= (3 + j 2) +
2 + j3 j5
= 9:804 /19:36 Ω

The total current in the circuit is

36 /0 Figure 3.128
I0T = 0
= 3:672 / 19:36 A
ZT

Using the principle of current division,

I0T ( j 5)
Iy = = 6:49 / 64:36 A
j5 + 2 + j3

It is found that Ix = Iy , thus verifying the reciprocity theorem.

EXAMPLE 3.51
Refer the circuit shown in Fig. 3.129. Find current through the ammeter, and hence verify reci-
procity theorem.
 Circuit Theorems j 245

Figure 3.129

SOLUTION

To find the current through the ammeter :

By inspection the loop equations for the circuit in Fig. 3.130 can
be written in the matrix form as
2 32 3 2 3
16 1 10 I1 0
4 1 26 20 54 I2 5=4 0 5
10 20 30 I3 50

Using Cramer’s rule, we get

I1 = 4:6 A
I2 = 5:4 A

Hence current through the ammeter = I2 I1 = 5:4 4:6 = 0:8A. Figure 3.130

Verification of reciprocity theorem:

The circuit is redrawn by interchanging the positions of
excitation and response as shown in Fig. 3.131.
By inspection the loop equations for the circuit can be
written in matrix form as
2 32 0
3 2 3
15 0 10 I1 50
4 0 25 20 54 0
I2 5=4 50 5
0
10 20 31 I3 0

Using Cramer’s rule we get

0
I3 = 0:8 A
Figure 3.131
246 j Network Theory

Hence, current through the Ammeter = 0.8 A.

It is found from both the cases that the response is same. Hence the reciprocity theorem is
verified.
EXAMPLE 3.52
Find current through 5 ohm resistor shown in Fig. 3.132 and hence verify reciprocity theorem.

Figure 3.132

SOLUTION
By inspection, we can write
2 32 3 2 3
12 0 2 I1 20
4 0 2 + j 10 2 54 I2 5=4 20 5
2 2 9 I3 0
Using Cramer’s rule, we get
I3 = 0:5376 / 126:25 A
Hence, current through 5 ohm resistor = 0:5376 / 126:25 A
Verification of reciprocity theorem:
The original circuit is redrawn by interchanging the excitation and response as shown in Fig.
3.133.

Figure 3.133
 Circuit Theorems j 247

Putting the three equations in matrix form, we get

2 32 3 2 3
12 0 2 I01 0
4 0 2 + j 10 2 5 64 I02
75 = 4 0 5
2 2 9 I03 20

Using Cramer’s rule, we get

I01 = 0:3876 / 2:35 A

I02 = 0:456 / 78:9 A
Hence; I02 I01 = 0:3179 j 0:4335

= 0:5376 / 126:25 A

The response in both cases remains the same. Thus verifying reciprocity theorem.

3.8 Millman’s theorem

It is possible to combine number of voltage sources or current sources into a single equiva-
lent voltage or current source using Millman’s theorem. Hence, this theorem is quite useful in
calculating the total current supplied to the load in a generating station by a number of generators
connected in parallel across a busbar.
Millman’s theorem states that if n number of generators having generated emfs E1 , E2 ; En



and internal impedances Z1 ; Z2 ;



Zn are connected in parallel, then the emfs and impedances
can be combined to give a single equivalent emf of E with an internal impedance of equivalent
value Z.
E1 Y1 + E2 Y2 + : : : + En Yn
where E=
Y1 + Y2 + : : : + Yn
1
and Z=
Y1 + Y2 + : : : + Yn
where Y1 ; Y2



Yn are the admittances corresponding to the internal impedances Z1 ; Z2


Z n
and are given by
1
Y1 =
Z1
1
Y2 =
Z2
..
.
1
Yn =
Zn
Fig. 3.134 shows a number of generators having emfs E1 ; E2



En connected in parallel
across the terminals x and y . Also, Z1 ; Z2



Zn are the respective internal impedances of the
generators.
248 j Network Theory

Figure 3.134

The Thevenin equivalent circuit of Fig. 3.134 using Millman’s theorem is shown in Fig. 3.135.
The nodal equation at x gives

E1 E E2 E
Z1
+
Z2
+



+
En E
Zn
=0
E E  1 1 
) Z1
1
+
Z2
2
+


+
En
Zn
=E + + +
1




 Z1 1 Z2 Zn

) E1 Y1 + E2 Y2 +


+ En Yn = E
Z
Figure 3.135
where Z = Equivalent internal impedance.
or


+ E Y ] = EY
[E1 Y1 + E2 Y2 + n n
E Y + E Y +


+ E Y
) E=
1 1
Y
2 2 n n

where Y = Y + Y +


+ Y
1 2 n
1 1
Y + Y +


+ Y
and Z= =
Y 1 2 n

EXAMPLE 3.53
Refer the circuit shown in Fig. 3.136. Find the current through 10 Ω resistor using Millman’s
theorem.

Figure 3.136
 Circuit Theorems j 249

SOLUTION
Using Millman’s theorem, the circuit shown in Fig. 3.136 is replaced by its Thevenin equivalent
circuit across the terminals P Q as shown in Fig. 3.137.

E1 Y1 + E2 Y2 E3 Y3
E=
Y1 + Y2 + Y3
1 1 1
22 + 48 12
5 12 4
=
1 1 1
+ +
5 12 4
= 10:13 Volts
1
R=
Y1 + Y2 + Y3
1 Figure 3.137
=
0:2 + 0:083 + 0:25
= 1:88 Ω
E
Hence; IL = = 0:853 A
R + 10

EXAMPLE 3.54
Find the current through (10 j 3)Ω using Millman’s theorem. Refer Fig. 3.138.

Figure 3.138

SOLUTION
The circuit shown in Fig. 3.138 is replaced by its Thevenin equivalent circuit as seen from the
terminals, A and B using Millman’s theorem. Fig. 3.139 shows the Thevenin equivalent circuit
along with ZL = 10 j 3 Ω: