Download Circuit Concepts and Network Simplification Techniques

Chapter
1
1.1
Circuit Concepts and
Network Simplification
Techniques
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.
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Network Theory
The time rate of change constitutes an electric current. Mathemetically, this relation
is expressed as
i(t) =
dq (t)
q (t) =
or
(1.1)
Z dt
t
1
i(x)dx
(1.2)
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. Normally, 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,
v=
then
dw
(1.3)
dq
Multiplying both the sides of equation (1.3) by the current in the element gives
vi =
dw
dq
dq
dt
)
dw
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
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3
Energy is the capacity to perform work. Energy and power are related to each
other by the following equation:
Z
Energy = w =
EXAMPLE
t
1
p dt
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 operation. we assume that v and i are zero for
t < 0:
Figure 1.3
SOLUTION
The power supplied is
p = vi = (8e
= 160e
2t
t
)(20e t )
W
The element is providing energy to the charge flowing through it.
The energy supplied during the first seond is
Z
1
w=
Z
0
= 80(1
1.3
1
p dt =
160e
2t
dt
0
e 2 ) = 69.17 Joules
Linear, active and passive elements
A linear element is one that satisfies the principle of superposition and homogeneity.
In order to understand the concept of superposition and homogeneity, let us consider the
element shown in Fig. 1.4.
Figure 1.4 An element with excitation
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.
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1.3.1
Network Theory
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
vi dt ≥ 0
w=
−∞
Examples of passive elements are resistors, capacitors and inductors.
1.3.1.A Resistors
Resistance is the physical property of an element or device that impedes the flow of current; it is represented by the symbol R.
Figure 1.5 Symbol for a resistor R
Resistance of a wire element is calculated using the relation:
ρl
(1.5)
R=
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
=
(1.7)
p = vi = v
R
R
Alternatively,
(1.8)
p = vi = (iR)i = i2 R
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
w=
−∞
t
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
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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
v=L
di
(1.10)
dt
Where L is the constant of proportionality known as inductance 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
of the flux , so the total voltage v across N turns is
v=N
Figure 1.6 Model of the
d
inductor
(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 equation(1.11), we get
v=L
di
dt
The power in an inductor is
p = vi = L
di
dt
i
The energy stored in the inductor is
Z
w=
t
1
Z
=L
p d
i(t)
i(
1)
i di =
1 2
Li Joules
2
(1.13)
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.
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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 dielectric material. Capacitance is a measure of the ability of
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 conducting plates or wires,
C=
C
1.7 Circuit symbol for
a capacitor
q
v
The current through the capacitor is given by
i=
dq
dt
=C
dv
(1.14)
dt
The energy stored in a capacitor is
Zt
vi d
w=
1
Remember that v and i are both functions of time and could be written as v (t) and
i(t).
i=C
Since
dv
dt
Zt
w=
we have
v C
1
dv
d
d
Zv(t)
v dv =
=C
v(
Since the capacitor was uncharged at t =
1)
1 2?
?v(t)
Cv ?
2
v ( 1)
1, v( 1) = 0.
w = w (t)
Hence
1
= Cv 2 (t) Joules
2
(1.15)
1 2
q (t) Joules
2C
(1.16)
Since q = Cv; we may write
w (t) =
Note that since w(t)
element.
0 for all values of v(t), the element is said to be a passive
Circuit Concepts and Network Simplification Techniques
1.3.2
|
7
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
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1.4
Network Theory
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 direction. 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 combination 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
1.5.3
|
9
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
=
R1 + R2
G1 + G2
IR1
IG2
I2 =
=
R1 + R2
G1 + G2
I1 =
(1.19)
(1.20)
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
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Network Theory
Since 6Ω and 8Ω are connected in parallel, voltage
across 8Ω = 12 volts.
12
Therefore, the current through
= 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
Total resistance of the circuit
Therefore applied voltage,
6 8 8 20
+
6 + 8 8 + 20
= 13:143Ω
=4+
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
Current through
12.5 = 12.5 volts
20Ω = I1 or I2
12:5
= 0:625A
=
20
Circuit Concepts and Network Simplification Techniques
Therefore, current through
R = I3 = I
= 2:5
Hence;
1.6
I1
0:625
j
11
I2
0:625
= 1:25 Amps
12:5
R=
= 10Ω
1:25
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 networks, we make use of Kirchhoff’s laws. Gustav
Kirchhoff (1824-1887), an eminent German physicist, did a considerable amount of work on the
principles governing the behaviour of eletric circuits. He gave his findings in a set of two laws: (i)
Figure 1.14 A simple resistive network
current law and (ii) voltage law, which together
for difining various circuit
are known as Kirchhoff’s laws. Before proceeding
terminologies
to the statement of these two laws let us familarize ourselves with the following definitions encountered 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.
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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
currents entering a node is equal to zero.
Figure 1.15 Currents at a node
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
N
X
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 important to keep track of whether the voltage is
increasing or decreasing as we go through each
element. We will adopt a policy of considering the increase in voltage as negative and a
Figure 1.16 Circuit with three closed paths
decrease in voltage as positive.
Circuit Concepts and Network Simplification Techniques
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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. Example 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
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| Network Theory
Ohm’s law for R1 is
v1 = i1 R1
v1
i1 =
= 1A
R1
i2 = i1 − i3
⇒
Hence,
= 1 − 2 = −1A
From the circuit,
v2 = R2 i2
−10
v2
=
= 10Ω
R2 =
i2
−1
⇒
EXAMPLE
1.5
Referring to Fig. 1.18, find the following:
(a) ix if iy = 2A and iz = 0A
(b) iy if ix = 2A and iz = 2iy
(c) iz if ix = iy = iz
Figure 1.18
SOLUTION
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
Figure 1.19
SOLUTION
(a) vy = 1 (3 vx + iz )
vx = 5V and iz =
Since
vy = 3(5)
we get
3 = 12V
vy = 1 (3 vx + iz ) =
(b)
6
= 3 vx + 0.5
)
3 vx =
vx =
Hence,
EXAMPLE
3A;
6:5
2.167 volts
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
From Ohm’s law,
)
Hence;
i2 + 5 = 0
12 = i2 R3
12
12
i2 =
=
= 2A
R3
6
i1 = 5 i2 = 3A
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15
16
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Network Theory
Applying KVL clockwise to the loop CBAC, we get
)
v1
6i1 + 12 = 0
v1 = 12
= 12
EXAMPLE
6i1
6(3) =
6volts
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
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(b) Applying KCL at node a, we get
vx
8
=
4
2
6
Is + 4( 6)
=4
4
Is 24 1:5 = 4
Is + 4vx +
)
)
)
Is = 29:5A
(c) Applying KCL at node b, we get
2
vx
+ Is +
2
4
6
iin = 1 + 29:5
4
iin =
)
6
6 = 23A
(d) The power supplied by the dependent current source = 8 (4vx ) = 8 4 6 =
EXAMPLE
1.9
Find the current i2 and voltage v for the circuit shown in Fig. 1.22.
Figure 1.22
SOLUTION
From the network shown in Fig. 1.22, i2 =
v
6
The two parallel resistors may be reduced to
Rp =
36
= 2Ω
3+6
Hence, the total series resistance around the loop is
Rs = 2 + Rp + 4
= 8Ω
192W
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Network Theory
Applying KVL around the loop, we have
21 + 8i
3i2 = 0
(1.21)
Using the principle of current division,
i2 =
=
iR2
R1 + R2
3i
i
9
i = 3i2
)
=
=
i
3
3+6
3
(1.22)
Substituting equation (1.22) in equation (1.21), we get
21 + 8(3i2 )
Hence;
3i2 = 0
i2 = 1A
v = 6i2 = 6V
and
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
Also;
i1 + 3i2
i1 =
v
4+2
i2 =
Hence;
and
4
)
i2 = 0
v
v
6
v
R
=
=
+3
6
v
16
v
v
=0
16 16
v = 96volts
v
96
i2 = =
= 6A
6
16
Circuit Concepts and Network Simplification Techniques
EXAMPLE
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19
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
)
10ix
20iy = 0
20iy + 40iz = 0
(1.23)
Applying KVL clockwise to the loop BCDB, we get
)
30(ix
iz )
15(iy + iz )
40iz = 0
30ix
85iz = 0
15iy
(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
Putting equations (1.23),(1.24) and
2
10
20
4 30
15
2
37
(1.25) in matrix form, we get
3 2 3
32
0
40
ix
85 5 4 iy 5 = 4 0 5
iz
2
15
Using Cramer’s rule, we find that
iz = 0.01 A (Flows from B to D)
(1.25)
20
1.7
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Network Theory
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
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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
is =
vs
(1.28)
Rs
The short circuit current of Fig. 1.27 is is . This enforces,
is =
vs
(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
Also from equation (1.29), we require is =
vs
Rs
vs =
)
(1.30)
. Therefore, we must have
vs
Rs
Rs = Rp
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
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Network Theory
Applying KVL to Fig. 1.26, we get
vs = iRs + v
Dividing by Rs gives
vs
Rs
=i+
v
Rs
(1.32)
If we use KCL for Fig. 1.27, we get
is = i +
v
Rp
(1.33)
Thus two circuits are equal when
is =
vs
Rs
and Rs = Rp
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
in parallel with the resistor R. Parameters
replaced with a current source of value i =
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
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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
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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 =
EXAMPLE
4.096 mA
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
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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
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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
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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
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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
1
+4
1
+4
1
=3V
Circuit Concepts and Network Simplification Techniques
EXAMPLE
| 29
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
Voltage across a and b is
Vab =
1
3−1
+
10−1
+ 15−1
=2V
a
30
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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),
Vbc =
12 1 6
12 1 + 10 1 + 15
1
12 = 24 V
Applying this result in Fig. 1.34(b), we get
vab = vac
= 60
EXAMPLE
vbc
24 = 36 V
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
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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)
i=
34
=2A
17
Using this value of i in Fig. 1.35(e),
Va =
and
92=
Ve = Va
22
18 V
20 =
42V
32
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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
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Then at node b in Fig. 1.35(b),
Vb
45 +
2
Vd
Vb
8
=0
Using the value of Vd in the above equation and rearranging, we get,
Vb
)
1 1
+
2 8
12
8
Vb = 69:6 V
= 45
At node c of Fig. 1.35(b)
Vc
5
)
EXAMPLE
+ 45 +
Vc
Ve
=0
10 42
1
1
= 45
+
Vc
5 10
10
Vc = 164 V
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
5
18 = 3:6V
Vx =
25
34
1.9
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Network Theory
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
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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 currents are assumed to flow in a mesh in the clockwise 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
i2 . Note that the mesh current i1 is taken as ‘+ve’ since we traverse in clockwise
i1
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
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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 :
Mesh 2 :
Mesh 3 :
)
)
0:2I1 + 2(I1
I3 ) + 3(I1
I2 )
5:2I1
3(I2
I1 ) + 4(I2
3I2
)
2I3 = 10
I1 ) + 4(I3
2I1
4I3 =
15
5:2
3
4 3 7:2
2
4
32
4I2 + 11I3 = 0
3
2
I1 = 0:11A
and
3
2
10
I1
4 5 4 I2 5 = 4 15 5
I3
11
0
Using Cramer’s rule, we get
I2 =
2:53A
I3 =
0:9A
(1.37)
I2 ) = 0
Putting the equations (1.36) through (1.38) in matrix form, we have
2
(1.36)
I3 ) + 0:2I2 + 15 = 0
3I1 + 7:2I2
5I3 + 2(I3
10 = 0
(1.38)
Circuit Concepts and Network Simplification Techniques
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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
i2 =
Since
we get
R2 i2 = v
is ;
(R1 + R2 )i1 + is R2 = v
)
i1 =
v
is R2
Figure 1.39(a) Circuit containing both independent voltage and current sources
R1 + R2
As a second example, let us take an electrical 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 :
M esh 2 :
R1 i1 + vxy
(R2 + R3 )i2
v=0
vxy = 0
Figure 1.39(b) Circuit containing an independent
current source common to both meshes
38
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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
R1 + R2 + R3
In this manner, we can handle independent current sources by recording the relationship 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
i1 =
v
1.20
Find the voltage Vo in the circuit shown in Fig. 1.40.
Figure 1.40
SOLUTION
Constraint equations:
I1 = 4
I2 =
10 3 A
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;
103I3 3
= 6 103 (0:25 10
Vo = 6
=
1:5 V
3
)
3
Circuit Concepts and Network Simplification Techniques
1.11
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39
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
10
i3 = 4 10
i=2
i2
3
A
3
A
The KVL equations for meshes 2 and 3 respetively are
2 103 i2 + 2 103 (i2
i1 )
6 + 1 103 i3 + vxy + 1 103 (i3
vxy = 0
i1 ) = 0
40
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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
Substituting i1 = 2 10
we get
6 + 1 103 i2
3A
4 10
and i3 = i2
4 10
3
i1 ) + 1
3A
103(i3
4 10
(1.39)
in the above equation,
+ 2 103 i2 + 2 103 i2
+1 103 i2
i1 ) = 0
3
2 10
2 10
3
3
=0
Solving we get
i2 =
Thus;
10
mA
3
io = i1
i2
=2
10
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 introducing vxy , the remaining procedure of finding io is same as before.
Circuit Concepts and Network Simplification Techniques
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 original 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
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41
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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
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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
Network Theory
j
(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
=
EXAMPLE
3
)(1:5 10
3)
0.225 mW(absorbed)
Figure 1.47
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
Thus;
i3 = 0:2
i1 )
5(i2
i1 ) = i2
i1 :
Figure 1.49
Circuit Concepts and Network Simplification Techniques
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Making use of i3 in the mesh equations, we get
Mesh 1 :
25i1
)
Mesh 2 :
5i2
2:5(i2
i1 ) = 0
27:5i1
5i1 + 12:5i2
)
7:5i2 = 0
7:5(i2
i1 ) = 75
2:5i1 + 5i2 = 75
Solving the above two equations, we get
i1 = 3:6 A; i2 = 13:2 A
i3 = i2
and
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
= 125
5(i2
48
i1 )
2:5(i3
2:5(9:6
i1 )
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
30i1
i2 )
10i2
660 = 0
15i3 = 660
46
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Network Theory
Mesh 2 :
Mesh 3 :
)
)
)
20ia + 10(i2
i1 ) + 50(i2
i3 ) = 0
10(i2
i1 ) + 50(i2
i3 ) = 20ia
10i1 + 60i2
15(i3
i1 ) + 25i3 + 50(i3
15i1
50i3 = 20ia
i2 ) = 0
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 connected 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
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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
i1 + i2 + i4 = 0
(i) At node 1:
va
v1
)
R1
)
v1
1
R1
+
+
v2
v1
R2
1
R2
+
v1
+
R4
1
0
v2
R4
1
R2
=0
=
va
(1.40)
R1
i2 + i3 + i5 = 0
(ii) At node 2:
v1
v2
)
)
v1
1
R2
R2
+
+ v2
1
R2
vb
v2
R3
+
1
R3
+
+
v2
R5
1
R5
=0
=
vb
(1.41)
R3
Putting equations (1.40) and (1.41) in matrix form, we get
2
1
6 R1
6
6
4
+
1
R2
+
32
1
1
R4
R2
1
1
R2
R2
+
1
R3
+
76
76
76
1 54
R5
v1
v2
3
2 v
a
7 6 R1
7=6
7 6
5 4 vb
3
7
7
7
5
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
At node b:
G2 va + (G4 + G2 + G3 )vb
G4 v c = 3
At node c:
G6 v a
3
G4 vb + (G4 + G5 + G6 )vc = 7
Substituting the values of various conductances, we find that
3va vb vc = 6
va + 3vb
va
Putting the above equations
2
3
4 1
1
vc = 3
vb + 3vc = 7
in matrix form, we see that
3 2 3
32
1
1
6
va
5
4
5
4
vb
3
1
= 3 5
vc
1 3
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
G = Gab
Gac
where
P
i
Gab
P
b
G
Gbc
Gbc P G Gac
c
G is the sum of the conductances at node i, and Gij is the sum of conductances
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
6
6
va
vb
v=6 .
4 ..
where;
3
7
7
7
5
vk
is the vector consisting of k unknown node voltages.
2
6
6
is1
is2
ia = 6 .
4 ..
The matrix
3
7
7
7
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
2+
va
4
+
va
)
55
5
=0
va = 20V
P2Asource = 20(2) = 40W (absorbing)
Figure 1.57
50
Network Theory
j
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
2
)
+
v2
v1
8
11v1
+
v3
v1
2v2
16
=0
v3 = 880
At node v2 :
v2
)
v1
8
+
v2
+
v2
v3
=0
3
24
3v1 + 12v2 v3 = 0
At node v3 :
v3 + 110
v3 v2
v3 v1
+
+
=0
2
24
16
)
3v1 2v2 + 29v3 = 2640
Solving the above nodal equations,we get
v1 = 74:64V; v2 = 11:79V; v3 =
Hence;
i1 =
110
v1
= 17:68A
2
v2
= 3:93A
i2 =
3
Figure 1.58(b)
82:5V
Circuit Concepts and Network Simplification Techniques
v3 + 110
i3 =
v1
i4 =
v2
i5 =
v1
i6 =
2
8
v2
v3
24
16
v3
j
51
= 13:75A
= 7:86A
= 3:93A
= 9:82A
(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
R
+
v2
vo
R
)
Hence; vo
+
v3
vo
R
+ +
vo =
R
=0
+ v
= [v1 + v2 + + v ]
n
nvo = v1 + v2 +
n
1
n
=
(b)
vn
vo
1
(150 + 200
3
n
1X
n
vk
k =1
50) = 100V
Figure 1.59(b)
52
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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
)
)
6
+
3
+
v1 + 3
2
v1
v1
v1
+
+
=
6
3
2
v1 = 2:5 V
v1
=0
2:5
1
2+1
2:5
=
1
3
= 0:83volts
vo =
EXAMPLE
v1
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
At node v1 :
)
At node v2 :
)
)
v1
4
v1
4
v3
3
+
v2
v1
+
v1
2
12
2
1=0
1=0
v1 = 9:33 V
+
v2
v3
v3
3
2
+
+1=0
Figure 1.63
v3
12
2
Hence;
+1=0
v3 = 6V
va = 9:33
6 = 3:33 volts
1
= 3:33 1 = 3:33W (delivering)
P1A source = va
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
and
R1
vb
R2
ia = 0
+ ia = is
Figure 1.64 Circuit with a supernode
incorporating va and vb .
54
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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
SOLUTION
The constraint equation is,
vb
)
va = 8
vb = va + 8
The KCL equation at the supernode
is then,
va + 8
500
+
(va + 8)
125
Therefore;
EXAMPLE
12
va
12
250
va
+
=0
500
va = 4V
+
Figure 1.66
1.34
Use the nodal analysis to find vo in the network of Fig. 1.67.
Figure 1.67
SOLUTION
Figure 1.68
j
55
56
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Network Theory
The constraint equation is,
v2
)
v1 = 12
v1 = v2
12
KCL at supernode:
v2
)
)
12 (v2 12) v3
v2
v2 v3
+
+
+
=0
3
3
3
1 10
1 10
1 10
1 103
4 10 3 v2 2 10 3 v3 = 24 10 3
4v2
2v3 = 24
At node v3 :
)
v3 (v2 12)
v3 v2
+
= 2 10 3
1 103
1 103
2 10 3 v2 + 2 10 3 v3 = 10 10
2v2 + 2v3 =
Solving we get
10
v2 = 7V
v3 = 2V
Hence;
EXAMPLE
vo = v3 = 2V
1.35
Refer the network shown in Fig. 1.69. Find the current Io .
Figure 1.69
SOLUTION
Constriant equation:
v3 = v1
12
3
Circuit Concepts and Network Simplification Techniques
Figure 1.70
KCL at supernode:
v1
)
)
12
v1
v1 v2
+
+
=0
3
3
3 10
2 10
3 103
7
1
10 3 v1
10 3v2 = 4 10
6
3
7
1
v1
v2 = 4
6
3
3
KCL at node 2:
v2
3
)
v1
103
1
3
)
+
v2
3 103
10
3
v1 +
+ 4 10
2
10
3
1
v1 +
3
3
3
=0
v2 =
2
v2 =
3
4 10
3
4
Putting the above two nodal equations in matrix form, we get
2 7
6 6
6
4 1
3
32
3
2
3
1
v1
4
7
7
6
6
7
3 76
7=6
7
5 4
5
2 54
v2
4
3
Solving the above two matrix equations using Cramer’s rule, we get
)
Io =
v1
2
v1 = 2V
103
=
2
= 1mA
2 103
j
57
58
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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
5
where ia =
)
v1
+
v1
v1
50
+
v1 + 75ia
25
50
80
v1
+
+
5
50
Solving we get
v1 + 75
)
Also;
=0
v 25
1
50
=0
v1 = 50V
v1
Figure 1.72
50
= 1A
50
50
v1 ( 75ia )
i1 =
(10 + 15)
v1 + 75ia
=
(10 + 15)
50 + 75 1
=
= 5A
(10 + 15)
P75ia = (75ia )i1
ia =
=
= 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
Figure 1.73
SOLUTION
Figure 1.74
Constraint equations :
v2
va = 20
v1
31ib = v3
v2
ib =
Node equations :
(i) Supernode:
v1
20
v1
)
)
20
v1
20
+
+
v1
v1
2
20
2
20
+
+
+
v1
v1
20
2
+
40
v2
v3
4
35ib )
4
(v1
35
v2 40
4
v2
v2
+
+
+
v3
80
(v1
v1
+ 3:125va = 0
35ib )
+ 3:125(20
80
35
80
v2 ) = 0
v2 40
+ 3:125(20
v2 ) = 0
j
59
60
j
Network Theory
(ii) At node v2 :
v2
+
40
v2
)
40
v2
)
40
+
v2
v1
+
4
(v1
4
v2
+
v3
v2
35ib )
35
+
v2 40
4
+
v2
20
1
v2
20
1
v2
20
1
=0
=0
=0
Solving the above two nodal equations, we get
v1 =
20:25V;
v3 = v1
Then
35ib
= v1
=
Also;
ig =
v2 = 10V
35
v2
40
29V
v1
20
2
+
v2
20
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
)
Substituting i1 =
4v1
v2
0:5i1 + v2 = 0
3
v1
v2 =
0:5i1
3
4
, the above equation becomes
2
3v2 = 8
Figure 1.75(b)
KCL equation at supernode:
v1
4
+
v2
4
2
=
2
)
v1 + 2v2 = 0
Solving the constraint equation and the KCL equation at supernode simultaneously,
we find that,
v2 = 727:3 mV
v1 =
=
Then;
i1 =
=
2v2
1454:6 mV
4
2
1:636A
v2
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
From the network, shown in Fig. 1.76 (b), by inspection,vb = 8 V, i1 =
Constraint equation:
va = 6i1 + vd
)
vb
va
KCL at supernode:
va
2
1 1
+
2 2
+
va
2
+
1
1
vb + [vd
2
2
Figure 1.76(b)
vc
vd
2
vc
vb
2
= 3vc
vc ] = 3vc
(1.44)
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
vc ) + vd
= 3(8
(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
vb
vc
KCL at node c:
2
vc
Substituting vb = 8V, we have
8
2
)
)
)
vc
+
+
vd
vc
2
vc
2
8 + vc
2vc
vc
vd
=4
=4
vd = 8
vd = 16
0:5vd = 8
Solving equations (1.46) and (1.47), we get
EXAMPLE
vc =
1.14V
vd =
18.3V
1.40
For the circuit shown in Fig. 1.77(a), determine all the node voltages.
Figure 1.77(a)
(1.46)
(1.47)
64
j
Network Theory
SOLUTION
Refer Fig 1.77(b), by inspection, v2 = 5V
Nodes 1 and 3 form a supernode.
Constraint equation:
v3 = 6
v1
KCL at super node:
v1
v2
10
+
v3
1
+2=0
Substituting v2 = 5V, we get
v1
)
)
v1
5
+
v3
=
10
1
5 + 10v3 =
2
20
v1 + 10v3 =
15
Figure 1.77(b)
Solving the constraint and the KCL equations at supernode simultaneously, we get
v1 = 4.091V
v3 =
1.909V
KCL at node 4 :
v4
2
+
v2
v4
4
2=0
Substituting v2 = 5V, we get
v4
2
Solving we get,
1.15
+
v4
5
4
2=0
v4 = 4:333V:
Brief review of impedance and admittance
Let us consider a general circuit with two accessible terminals, as shown in Fig. 1.78. If
the time domain voltage and current at the terminals are given by
v = vm sin(!t + v )
i = im sin(!t + i )
then the phasor quantities at the terminals are
Figure 1.78 General phasor
circuit
Circuit Concepts and Network Simplification Techniques
j
65
V = Vm v
I = Im i
We define the ratio of V to I as the impedence of the circuit, which is denoted as Z.
That is,
V
Z=
I
It is very important to note that impedance Z is a complex quantity, being the ratio
of two complex quantities, but it is not a phasor. That is, it has no corresponding
sinusoidal time-domain function, as current and voltage phasors do. Impedence is a
complex constant that scales one phasor to produce another.
The impedence Z is written in rectangular form as
Z = R + jX
where R = Real[Z] is the resistance and X = Im[Z] is the reactance. Both R and X , like
Z, are measured in ohms.
p
2
The magnitude of Z is written as |Z| = R2+ X
and the angle of Z is denoted as Z = tan
1
X
R
.
The relationships are shown graphically in Fig. 1.79.
The table below gives the various forms of Z for
different combinations of R; L and C .
Figure 1.79 Graphical representation
of impedance
Type of the circuit
Impedance Z
1.
Purely resistive
Z=R
2.
Purely inductive
Z = j!L = jXL
3.
Purely capactive
Z=
4.
RL
Z = R + j!L = R + jXL
5.
RC
Z=R
6.
RLC
Z = R + j!L
j
!C
The reciprocal of impendance is denoted by
Y=
1
Z
=
jXC
j
!C
=R
j
!C
‘
jXC
= R + j (XL
XC )
66
j
Network Theory
is called admittance and is analogous to conductance in resistive circuits. Evidently, since
Z is a complex number, so is Y. The standard representation of admittance is
Y = G + jB
The quantities G = Re[Y] and B = Im[Y] are respectively called conductance and suspectence. The units of Y, G and B are all siemens.
1.16
Kirchhoff’s Laws: Applied to alternating circuits
If a complex excitation, say vm ej (!t+) , is applied to a circuit, then complex voltages, such
as v1 ej (!t+1 ) ; v2 ej (!t+2 ) and so on, appear across the elements in the circuit. Kirchhoff’s
voltage law applied around a typical loop results in an equation such as
v1 e
j(ωt+θ1 )
+ v2 e
j(ωt+θ2 )
+ : : : + vN e
j (ωt+θN )
=0
Dividing by ej!t , we get
)
where
v1 ej1 + v2 ej2 + : : : + vN ejN = 0
V1 + V2 + : : : + VN = 0
Vi = Vi i ; i = 1; 2; N
are the phasor voltage around the loop.
Thus KVL holds good for phasors also. A similar approach will establish KCL also.
At any node having N connected branches,
I 1 + I2 + + IN = 0
Ii = Ii i ; i = 1; 2 N
where
Thus, KCL holds good for phasors also.
EXAMPLE
1.41
Determine V1 and V2 , the node voltage phasors using nodal technique for the circuit
shown in Fig. 1.80.
Figure 1.80
Circuit Concepts and Network Simplification Techniques
j
67
SOLUTION
First step in the analysis is to convert the circuit of Fig. 1.80 into its phasor version
(frequency domain representation).
1
F
2
)
j
!C
5 / 0 ; ! = 2 rad=s
1
j!L = j Ω
2
)
)
5 cos 2t
1
H
4
=
j 1Ω;
)
1F
j
!C
=
1
2
j Ω
Figure 1.80(a)
Figure 1.80(b)
Fig. 1.80(a) and (b) are the two versions of the phasor circuit of Fig. 1.80.
Z1 = j 1Ωjj
j1
=
j1
1
j Ω
2
1
j
2
1
j
2
=
j 1Ω
68
j
Network Theory
1
Z2 = j Ωjj1Ω
2 1
(1)
j
1 + j2
2
=
=
Ω
1
5
j +1
2
KCL at node V1 :
5 /0 ) +
V1
V1 V2
+
=0
j1
j1
(2 + j 2)V1 j 1V2 = 10
2 (V1
)
KCL at node V2 :
V1
V2
= 5∠0
+
1
+ j2
j1
5
j V1 + V2 2j V2 = 5
V2
)
)
j V2
j 1V1+ (1
j 1)V2 = 5
Putting the above equations in a matrix form, we get
2
4
2 + j2
j1
j1
1
32
54
j1
V1
3
5=4
V2
Solving V1 and V2 by Cramer’s rule, we get
V1 = 2
j1 V
V2 = 2 + j 4 V
In polar form,
V1 =
p
p5 /
26:6 V
V2 = 2 5 /63:4 V
In time domain,
v1 =
F
F5 cos(2t
2
26.6 ) V
v2 = 2 5 cos(2t + 63.4 ) V
10
5
3
5
Circuit Concepts and Network Simplification Techniques
EXAMPLE
j
69
1.42
Find the source voltage Vs shown in Fig. 1.81 using nodal technique. Take I = 3 /45 A:
Figure 1.81
SOLUTION
Refer to Fig. 1.81(a).
KCL at node 1:
V1
)
Vs
V1
V1 V2
+
+
=0
10
j5
5 + j2
(11 + j 12)V1 (5 + j 2)Vs = 10V2
(1.48)
Figure 1.81(a)
KCL at node 2:
V2 V1
V2
+I+
=0
5 + j2
8 + j3
)
Also;
(8 + j 3)V1 = (13 + j 5)V2 + (34 + j 31)I
V2 = 4I = 4(3 /45 ) = 12 /45
p
p
= 6 2 + j6 2
(1.49)
(1.50)
70
j
Network Theory
Substituting equations (1.48) and (1.50) in equation (1.49), we get
)
(8 + j 3)V1 = 74:24 + j 290:62
300 /75:7
V1 =
8:54 /20:6
= 35:1 /55:1
= 20:1 + j 28:8 V
Substituting V1 and V2 in equation (1.48) yields
(5 + j 2)Vs =
Therefore;
EXAMPLE
209:4 + j 473:1
517:4 /113:9
Vs =
= 96.1 /92.1 V
5:38 /21:8
1.43
Find the voltage v (t) in the network shown in Fig. 1.82 using nodal technique.
Figure 1.82
SOLUTION
Converting the circuit diagram shown in Fig. 1.82 into a phasor circuit diagram, we get
Figure 1.83
Circuit Concepts and Network Simplification Techniques
j
71
At node V1 :
V1
)
( 1 + j ) V1 V1 V2
+
+
=0
j2
2
j2
V1 j V2 = 1 + j
V1
V2
At nodeV2 :
j2
Also
+
Ic = 2Ix =
V2
Hence;
)
V2
j2
(1.51)
Ic = 0
2( 1 + j )
=
j2
V1
V2
+
=
j2
j2
j V1 + j 2V2 =
1
j
1
j
2
j2
Solving equations (1.51) and (1.52) using Cramer’s rule we get
V2 =
2 /135 V
v(t) = v2 (t) =
Therefore;
EXAMPLE
p
F
2 cos(4t + 135 ) V
1.44
Refer to the circuit of Fig. 1.84. Using nodal technique, find the current i.
Figure 1.84
SOLUTION
1
1
F capacitor =
=
5
j!C
1
1
j 5000 10
5
The parallel combinations of 2kΩ and j 1kΩ is
Reactance of
Zp =
2
2 103 ( j 103 )
= (1
3
3
2 10
j 10
5
=
6
j 1kΩ
j 2)kΩ
(1.52)
72
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Network Theory
Figure 1.85
The phasor circuit of Fig. 1.84 is as shown in Fig. 1.85.
Constraint equation :
V2 = V1 + 3000I
KCL at supernode :
V1
4 /0
+
2
500
(1
5
V1
j 2)
103
+
V2
=0
j 1) 103
(2
Substituting V2 = V1 + 3000I in the above equation, we get
V1
4 /0
+
2
500
(1
5
V1
j 2)
103
+
V1 + 3000I
=0
(2 j 1) 103
Also,
I=
V1
4 /0
500
(1.53)
Hence,
V1
4∠0
+
2
500
(1
5
V1
j 2)
103
4
V1
V1 + 3000
500
+
(2 j 1) 103
=0
Solving for V1 and substituting the same in equation (1.53), we get I = 24 /53:1 mA
Hence, in time-domain, we have
i = 24 cos(5000t + 53.1 )mA
Circuit Concepts and Network Simplification Techniques
EXAMPLE
j
73
1.45
Use nodal analysis to find Vo in the circuit shown in Fig. 1.86.
Figure 1.86
SOLUTION
The voltage source and its two connecting nodes form the supernode as shown in
Fig. 1.87.
Figure 1.87
Constraint equation:
Applying KVL clockwise to the loop formed by 12 /0 source, j 2Ω and
12 /0 + Vo
)
V1 = Vo
V1 = 0
12 /0
KCL at supernode:
V1 V1 V2 Vo V2
Vo
+
+
+
=0
j2
1
1
j4
j 4Ω we get
74
j
Network Theory
Substituting V1 = Vo
we get,
j
2
(Vo
12 in the above equation
12) + (Vo
)
Vo
12
j
2
V2 ) + Vo
+1+1+
j
+ V2 ( 1
4
)
j
V2 + Vo = 0
4
1
j
4
Vo 2
1) = 12
j6
2V2 = 12
j6
V1 V2 V2 Vo
+
+
=0
1
2
1
Substituting V1 = Vo 12 /0 in the above equation
1
we get,
V2 (Vo 12 /0 ) + V2 + V2 Vo = 0
2
5
)
2Vo + V2 = 12 /0
2
Solving the two nodal equations,we get
V2
KCL at V2 :
Vo = 11.056
EXAMPLE
j8.09 = 13.7 / 36.2 V
1.46
Find i1 in the circuit of Fig. 1.88 using nodal analysis.
Figure 1.88
SOLUTION
The phasor equivalent circuit is as shown in Fig. 1.88(a).
KCL at node V1 :
V1
)
KCL at node V2 :
20 /0
V1
V1 V2
+
+
=0
10
j 2:5
j4
(1 + j 1:5)V1 + j 2:5V2 = 20
V1
V2
j4
But
I1 =
+
V2
= 2I1
j2
V1
j 2:5
Circuit Concepts and Network Simplification Techniques
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75
Figure 1.88(a)
V1
V2
V2
2V1
=
j4
j2
j 2:5
j 0:55V1 j 0:75V2 = 0
Hence;
)
+
Multiplying throughout by j 20, we get
11V1 + 15V2 = 0
Putting the two nodal equations in matrix form, we get
2
4
1 + j 1:5 j 2:5
11
15
32
54
V1
3
2
5=4
V2
20
3
5
0
Solving the matrix equation, we get
V1 = 18:97 /18:43 V
V2 = 13:91 / 161:56 V
The current
I1 =
V1
= 7:59 /108:4 A
j 2:5
Transforming this to the time-domain, we get
i1 = 7.59 cos(4t + 108.4 )A
EXAMPLE
1.47
Use the node-voltage method to find the steady-state expression for vo (t) in the circuit
shown in Fig. 1.89 if
vg 1 = 10 cos(5000t + 53:13 )V
vg 2 = 8 sin 5000t V
76
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Network Theory
Figure 1.89
SOLUTION
The first step is to convert the circuit of Fig. 1.89 into a phasor circuit.
10 cos(5000t + 53:13 )V; ! = 5000rad=sec
8 sin 5000t = 8 cos(5000t
90 )V
L = 0:4 mH
C = 50F
)
)
)
)
10 /53:13 = 6 + j 8V
8 / 90 =
j 8V
j!L = j 2Ω
1
j!C
=
j 4Ω
The phasor circuit is shown in
Fig. 1.89(a).
KCL at node 1:
Vo (6 + j 8) Vo
+
j2
6
+
Vo
( j 8)
j4
=0
Solving we get Vo = 12 /0 V
Figure 1.89(a)
Hence, the steady-state expression is
vo (t) = 12 cos 5000t
EXAMPLE
1.48
Solve the example (1.47) using mesh-current method.
SOLUTION
Refer Fig. 1.90.
KVL to mesh 1 :
[6 + j 2]I1
KVL to mesh 2 :
6I1 + (6
6I2 = 10 /53:13
j 4)I2 = 8 / 90
Circuit Concepts and Network Simplification Techniques
j
77
Figure 1.90
Putting the above equations in matrix form, we get
2
4
6 + j2
6
32
6
6
j4
54
I1
I2
3
"
5=
10 /53:13
#
8 / 90
Solving for I1 and I2 , we get
I1 = 4 + j 3
I2 = 2 + j 3
Now;
Vo = (I1
Hence in time domain,
vo = 12 cos 5000t Volts
EXAMPLE
I2 )6 = 12
= 12 /0 V
1.49
Determine the current Io in the circuit of Fig. 1.91 using mesh analysis.
Figure 1.91
SOLUTION
Refer Fig 1.92
KVL for mesh 1 :
)
(8 + j 10
j 2)I1
( j 2)I2
j 10I3 = 0
(8 + j 8)I1 + j 2I2 = j 10I3
(1.54)
78
j
Network Theory
KVL for mesh 2 :
j2
(4
)
j 2)I2
( j 2)I3 + 20 /90 = 0
( j 2)I1
j 2I1 + (4
j 4)I2 + j 2I3 =
For mesh 3;
j 20
I3 = 5
(1.56)
Sustituting the value of I3 in the equations (1.54) and (1.55), we get
(8 + j 8)I1 + j 2I2 = j 50
j 2I1 + (4
j 4)I2 =
j 20
=
j 30
j 10
Putting the above equations in matrix
form,we get
2
4
8 + j8
j2
j2
4
j4
32
54
I1
3
2
j 50
5=4
3
5
j 30
I2
Figure 1.92
Using Cramer’s rule,we get
I2 = 6:12 / 35:22 A
The required current:
Io =
I2
= 6.12 /144.78 A
EXAMPLE
1.50
Find Voc using mesh technique.
Figure 1.93
SOLUTION
Applying KVL clockwise for mesh 1 :
)
600I1
(600
j 300(I1
I2 )
(1.55)
9=0
j 300)I1 + j 300I2 = 9
Circuit Concepts and Network Simplification Techniques
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79
Figure 1.94
Applying KVL clockwise for mesh 2 :
j 300(I2
2Va + 300I2
Also;
Va =
Hence;
)
2( j 300(I1
I1 ) = 0
j 300(I1
I2)
j 300 (I2
I2 )) + 300I2
j 3I1 + (1
I1 ) = 0
j 3)I2 = 0
Putting the above two mesh equations in matrix form, we get
2
4
600
j 300
j 300
j3
1
j3
32
54
I1
I2
3
"
5=
9
#
0
Using Cramer’s rule, we find that
I2 = 0:0124 / 16 A
Voc = 300I2 = 3.72 / 16 V
Hence;
EXAMPLE
p
1.51
Find the steady current i1 when the source voltage is vs = 10 2 cos(!t + 45 ) V and
the current source is is = 3 cos !t A for the circuit of Fig. 1.95. The circuit provides the
impedence in ohms for each element at the specified ! .
Figure 1.95
80
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Network Theory
SOLUTION
Figure 1.96
The first step is to convert the circuit of Fig. 1.95 into a phasor circuit. The phasor
circuit is shown in Fig. 1.96.
p
vs = 10 2 cos(!t + 45 )
is = 3 cos !t
)
)
p
Vs = 10 2 /45 = 10(1 + j )
Is = 3 /0
Figure 1.96(a)
Constraint equation:
I2
I1 = Is = 3 /0
Applying KVL clockwise around the supermesh we get
I1 Z1 + I2 (Z2 + Z3 )
I2 = I1 + Is
Substituting
we get,
)
)
Vs = 0
(from the constraint equation)
I1 Z1 + (I1 + Is )(Z2 + Z3 ) = Vs
(Z1 + Z2 + Z3 )I1 = Vs (Z2 + Z3 )Is
(10 + j 10) (2
Vs (Z2 + Z3 )Is
=
I1 =
Z1 + Z2 + Z3
2
= 2 + j 8 = 8:25 /76 A
Hence in time domain,
i1 = 8.25 cos(ωt + 76 ) A
j 2)3
Circuit Concepts and Network Simplification Techniques
EXAMPLE
j
81
p
1.52
Find the steady-state sinusoidal current i1 for the circuit of Fig. 1.97, when vs = 10 2 cos
(100t + 45 ) V:
Figure 1.97
SOLUTION
The first step is to convert the circuit of Fig. 1.97 int to a phasor circuit. The phasor
circuit is shown in Fig. 1.98.
p
vs = 10 2 cos(100t + 45 )
L = 30 mH
C = 5 mF
p
Vs = 10 2 /45 ;
)
)
! = 100 rad= sec
XL = j!L
= j 100 30 10
1
XC =
)
3
= j 3Ω
j!C
=
1
j 100 5 10
3
=
j 2Ω
KVL for mesh 1 :
(3 + j 3)I1
j 3I2 = 10 + j 10
KVL for mesh 2 :
(3
j 3)I1 + (j 3
j 2)I2 = 0
Putting the above two mesh equations in matrix form, we get
2
4
3 + j3
3
j3
j3
j1
32
54
I1
I2
3
2
5=4
10 + j 10
0
3
5
82
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Network Theory
Using Cramer’s rule,we get
I1 = 1:05 /71:6 A
Thus the steady state time response is,
i1 = 1.05 cos(100t + 71.6 )A
Figure 1.98
EXAMPLE
1.53
Determine Vo using mesh analysis.
Figure 1.99
SOLUTION
Figure 1.100
Circuit Concepts and Network Simplification Techniques
| 83
From Fig. 1.100, we find by inspection that,
I1 = 2Ia = 2(I2 − I3 )
I2 = 4 mA
Applying KVL clockwise to mesh 3, we get
1 × 103 (I3 − I2 ) + 1 × 103 (I3 − I1 ) + 2 × 103 I3 = 0
Substituting I1 = 2(I2 − I3 ) and I2 = 4 mA in the above equation and solving for I3 ,
I3 = 2 mA
we get,
Vo = 2 × 103 I3
Then,
= 4V
EXAMPLE
1.54
Find Vo in the network shown in Fig. 1.101 using mesh analysis.
Figure 1.101
SOLUTION
Figure 1.102
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
)
Hence
j 1 + 4 + j 2)
20 + j 4
I1 =
2(4 + j 2) = 0
= 3:35 /1:85 A
6 + j1
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 between 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
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85
Rab (Υ) = Rab (Δ)
(i)
)
Ra + Rb =
R2 (R1 + R3 )
(1.57)
R2 + R1 + R3
Rbc (Υ) = Rbc (Δ)
(ii)
)
Rb + Rc =
R3 (R1 + R2 )
(1.58)
R3 + R1 + R2
Rca (Υ) = Rca (Δ)
(iii)
)
Rc + Ra =
R1 (R2 + R3 )
(1.59)
R1 + R2 + R3
Solving equations (1.57), (1.58) and (1.59) gives
Ra =
Rb =
Rc =
R1 R2
(1.60)
R1 + R2 + R3
R2 R3
(1.61)
R1 + R2 + R3
R1 R3
(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
Ra Rb + Rb Rc + Rc Ra =
R1 R2 R3 (R1 + R2 + R3 )
(R1 + R2 + R3
)2
=
R1 R2 R3
R1 + R2 + R3
(1.63)
Dividing equation (1.63) by each of the equations (1.60) to (1.62) leads to the following
relationships :
R1 =
R2 =
R3 =
Ra Rb + Rb Rc + Ra Rc
Rb
Ra Rb + Rb Rc + Ra Rc
Rc
Ra Rb + Rb Rc + Ra Rc
Ra
(1.64)
(1.65)
(1.66)
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
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Network Theory
Under these conditions the conversions formula become
1
RΔ
3
RΔ = 3RΥ
RΥ =
and
EXAMPLE
1.55
Find the value of resistance between the terminals a
Fig. 1.105.
b of the network shown in
Figure 1.105
SOLUTION
Let us convert the upper Δ to Υ
(6k)(18k)
= 3 kΩ
6k + 12k + 18k
(6k)(12k)
=
= 2 kΩ
6k + 12k + 18k
(12k)(18k)
= 6 kΩ
=
6k + 12k + 18k
Ra 1 =
Rb1
R c1
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
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
j
87
88
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Network Theory
Let us convert the upper Δ between the points a1 , b1 and c1 into an equivalent Υ.
6 18
= 3:6Ω
6 + 18 + 6
66
=
= 1:2Ω
6 + 18 + 6
6 18
= 3:6Ω
=
6 + 18 + 6
Ra 1 =
Rb1
R c1
Figure 1.108 now becomes
jj
Rab = 5 + 3:6 + 7:2 27:6
7:2 27:6
= 8:6 +
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
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89
SOLUTION
Let us convert Υ between the terminals a, b and c into an equivalent Δ.
Rab =
Ra Rb + Rb Rc + Rc Ra
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
= 21Ω
70 + 30
12:5 17:5
= 7:292Ω
12:5jj17:5 =
12:5 + 17:5
15 35
= 10:5Ω
15jj35 =
15 + 35
Rab = (7:292 + 10:5)jj21 = 9:632Ω
70jj30 =
Thus;
i=
vs
Rab
= 12:458 A
Figure 1.109(b)
90
j
Network Theory
Nodal versus mesh analysis
The analysis of a complex circuit can usually be accomplished by either the node voltage
or mesh current method. One may ask : Given a network to be analyzed, how do we
know which method is better or more efficient? The choice is dictated by two factors.
When a circuit contains only voltage sources, it is probably easier to use the mesh
current method. Conversely, when the circuit contains only current sources, it will be
easier to use the node voltage method. Also, a circuit with fewer nodes than meshes
is better analyzed using nodal analysis, while a circuit with fewer meshes than nodes is
better analyzed using mesh analysis. In other words, the best technique is one which gives
smaller number of equations.
Another point to consider while choosing between the two methods is, what information is required. If node voltages are required, it may be advantageous to apply nodal
analysis. On the other hand, if you need to know several currents, it may be wise to
proceed directly with mesh current analysis.
It is often advantageous if we know both the techniques. The first advantage lies in
the fact that the second method can verify the results of the first method. Also, both the
methods have limitations. For example, while analysing a transistor circuit, only mesh
method is suited and while analysing an Op-amp circuit, nodal method is only applicable.
Mesh technique is applicable for planar1 networks. However, nodal method suits to both
planar and nonplanar 2 networks.
Reinforcement Problems
R.P
1.1
Find the power dissipated in the 80Ω resistor using mesh analysis.
Figure R.P.1.1
1
A planar network can be drawn on a plane without branches crossing each other.
A nonplanar network is one in which crossover is identified and cannot be eliminated by redrawing
the branches.
2
Circuit Concepts and Network Simplification Techniques
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91
SOLUTION
KVL clockwise to mesh 1 :
14I1
4I2
8I3 = 230
KVL clockwise to mesh 2 :
4I1 + 22I2
16I3 = 260
KVL clockwise to mesh 3 :
8I1
16I2 + 104I3 = 0
Putting the above mesh equations in matrix form, we get
2
6
6
6
4
14
4
4
22
8
16
8
32
76
76
54
I1
3
2
7
7
5
6
6
4
230
3
7
7
5
16 7 6 I2 7 = 6 260 7
104
I3
0
The current I3 is found from the above matrix equation by using Cramer’s rule.
Thus;
R.P
P80 =
I3
2
I3 R80
= 5A
= 52 80 = 2000W(dissipated)
1.2
Refer the circuit shown in Fig. R.P. 1.2. The current io = 4A. Find the power dissipated
in the 70 Ω resistor.
Figure R.P.1.2
SOLUTION
By inspection, we find that the mesh current i3 = io = 4A
KVL clockwise to mesh 1 :
75i1 70i2 5i3 = 180
92
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Network Theory
Substituting i3 = 4A; we get 75i1 70i2 = 200
KVL clockwise to mesh 2 :
70i1 + 88i2 10i3 = 0
70i1 + 88i2 = 40
Substituting the value i3 = 4A; we get
Puting the two mesh equations in matrix from, we get
75
70
70
88
i1
i2
=
200
40
Using Cramer’s rule, we get
i1 = 12A; i2 = 10A
i2 )2 70 = 4
P70 = (i1
70
= 280 W (dissipated)
R.P
1.3
Solve for current I in the circuit of Fig. R.P. 1.3 using nodal analysis.
Figure R.P.1.3
SOLUTION
KCL at node V1 :
V1
)
20 / 90
V1
V1 V2
+
+
+ 5 /0 = 0
2
j2
j1
(0:5 j 0:5)V1 + j V2 = 5 j 10
KCL at node V2 :
V2
V1
j1
Also,
Hence,
+
V2
4
I=
V1
V2
j1
+
2I
V1
j2
V2
2
+ V1
4
j2
5 /0 = 0
5 /0 = 0
Circuit Concepts and Network Simplification Techniques
)
)
(0:25
j )V2 = 5
5
0:25 j
Making use of V2 in the nodal equation at node V1 ; we get
j5
= 0:5(1 j )V1
5 j 10
0:25 j
j 40
)
(1 j )V1 = 10 j 20
1 j4
)
V1 = 15:81 / 46:5 V
15:81 / 46:5
V1
=
Hence;
I=
j2
2 / 90
= 7:906 /43:5 A
R.P
V2 =
1.4
Find Vo shown in the Fig. R.P. 1.4 using Nodal technique.
Figure R.P.1.4
Figure R.P.1.4(a) .
j
93
94
j
Network Theory
SOLUTION
We find from Fig RP 1.4(a) that,
V1 = Vo
Constraint equation:
Applying KVL clockwise along the path consisting of voltage source, capacitor, and 2Ω
resistor, we find that
12 /0 + V2
V1 = 0
V1 = V2 + 12 /0
)
or
V2 = V1
12
KCL at Supernode :
V1
)
(2
V3
V1
V2
V2 V3
+
+
=0
j2
2
j4
4
j 2)V1 + (1 + j )V2 + ( 1 + j 2)V3 = 0
+
KCL at node 3 :
V1
V3
V2
V3
0:2Vo = 0
(1.67)
j 2)V1 + V2 + ( 1 + j 2)V3 = 0
(1.68)
j2
+
4
Substituting Vo = V1 , we get
(0:8
Subtracting equation (1.68) from (1.67), we get
1:2V1 + j V2 = 0
Substituting V2 = V1
12 (from the constraint equation), we get
)
Hence
R.P
1:2V1 + j (V1 12) = 0
j 12
V1 =
= Vo
1:2 + j
Vo = 7.68 /50.2 V
1.5
Solve for i using mesh analysis.
Figure R.P. 1.5
(1.69)
Circuit Concepts and Network Simplification Techniques
j
SOLUTION
The first step in the analysis is to draw the phasor circuit equivalent of Fig. R.P.1.5.
Figure R.P. 1.5(a)
! = 2
6 sin 2t = 6 cos(2t
10 /0 V
)
)
)
)
10 cos 2t
90)
L = 2H
C = 0:25F
6 / 90 =
j 6V
XL = j!L = j 4Ω
XC =
1
j!C
1
= j 2Ω
=
j2
1
4
Applying KVL clockwise to mesh 1 :
)
j 2)I1 + j 2I2 = 0
10 + (4
(2
j 1)I1 + j I2 = 5
Applying KVL clockwise to mesh 2 :
j 2I1 + (j 4
j 2)I2 + ( j 6) = 0
I1 + I2 = 3
Putting the above mesh equations in a matrix form, we get
"
j
2
1
j
#"
I1
I2
1
#
"
=
5
#
3
Using Cramer’s rule, we get
I1 = 2 + j 0:5;
I2 = 1
Io = I1
Hence
j 0:5;
I2 = 1 + j = 1:414 /45
io (t) = 1:414 cos (2t + 45 ) A
95
96
j
Network Theory
R.P
1.6
Refer the circuit shown in Fig. R.P. 1.6. Find I using mesh analysis.
Figure R.P.1.6
SOLUTION
Figure R.P.1.6(a)
Constraint equation:
)
)
I3
I2 = 2I
I3
I2 = 2(I1
I3 = 2I1
Also, for mesh 4,
I2 )
I2
I4 = 5 A
Applying KVL clockwise for mesh 1 :
)
j 2)I1 + j 2I2 = 0
( j 20) + (2
(1
j )I1 + j I2 =
j 10
(1.70)
Circuit Concepts and Network Simplification Techniques
j
97
Applying KVL clockwise for the supermesh :
j 2)I2 + j 2I1 + 4I3
(j
Substituting
I3 = 2I1
I2
j I4 = 0
and I4 = 5A
(8 + j 2)I1
we get
(4 + j )I2 = j 5
(1.71)
Putting equations (1.70) and (1.71) in matrix form, we get
1 j
8 + j2
j
(4 + j )
I1
I2
=
j 10
j5
Solving for I1 and I2 , we get
I1 =
(5:44 + j 4:26) A
I2 =
(11:18 + j 9:7) A
I = I1
I2
= 5:735 + j 5:44
= 7.9 /43.49 A
R.P
1.7
Calculate Vo in the circuit of Fig. R.P. 1.7 using the method of source transformation.
Figure R.P. 1.7
SOLUTION
Transform the voltage source to a
current source and obtain the circuit
shown in Fig. R.P.1.7(a).
Is =
20 / 90
= 4 / 90 A
5
Figure R.P.1.7(a)
Zp = 5Ωjj3 + j 4 =
5 (3 + j 4)
= 2:5 + j 1:25Ω
5 + (3 + j 4)
98
j
Network Theory
Converting the current source in Fig. R.P. 1.7(b) to a voltage source gives the circuit
as shown in Fig. R.P. 1.7(c).
Figure R.P.1.7(b)
Figure R.P.1.7(c)
4j (2:5 + j 1:25)
Vs = Is Zp =
j 10V
=5
Vo = 10I
Vs
10
=
Zp + Z2 + 10
5 j 10
=
[2:5 + j 1:25 + 4 j 13 + 10]
= 5.519 / 28 V
R.P
10
1.8
Find vx and ix in the circuit shown in Fig. R.P. 1.8.
Figure R.P. 1.8
SOLUTION
Constraint equation:
)
i2
i1 = 3 +
vx
4
i2 = i1 + 3 +
v3
4
The above equation becomes very clear if one writes KCL equation at node B of Fig.
R.P. 1.8(a).
Circuit Concepts and Network Simplification Techniques
j
99
Figure R.P. 1.8(b)
Figure R.P. 1.8(a)
Applying KVL clockwise to the supermesh in Fig. R.P. 1.8(b), we get
50 + 10i1 + 5i2 + 4ix = 0
But ix = i1 . Hence,
)
50 + 10i1 + 5i2 + 4i1 = 0
Making use of vx = (i1
14i1 + 5i2 = 50
i2 )
2 in the constraint equation, we get
(i1 i2 ) 2
i =i +3+
2
)
)
)
)
(1.72)
1
i2 = i1 + 3 +
2i2 = 2i1 + 6 + i1
3i1
4
i2
i1
2
i2
3i2 + 6 = 0
i1
i2 =
2
Solving equations (1.72) and (1.73) gives i1 = 2:105 A; i2 = 4:105 A
Thus;
vx = 2(i1 i2 ) = 4 V
ix = i1 = 2:105 A
and
R.P
1.9
Obtain the node voltages v1 , v2 and v3 for the following circuit.
(1.73)
100
j
Network Theory
SOLUTION
We have a supernode as shown in Fig. R.P. 1.9(a). By inspection, we find that V2 = 12V.
Refer Fig. R.P. 1.9(b) for further analysis.
Figure R.P.1.9(a)
Figure R.P.1.9(b) .
KVL clockwise to mesh 1 :
v1
10 + 12 = 0
)
v1 = 2
KVL clockwise to mesh 2 :
)
v3 =
8V
v1 = 2 V, v2 = 12 V, v3 =
Hence;
R.P
12 + 20 + v3 = 0
8V
1.10
Find the equivalent resistance Rab for the circuit shown in Fig. R.P.1.10.
Figure R.P. 1.10
Circuit Concepts and Network Simplification Techniques
j
101
SOLUTION
The circuit is redrawn marking the nodes c to j in Fig. R.P. 1.10(a). It can be seen that
the network consists of four identical stars :
(i) ae; ef; cb
(ii) ac; cf; cd
(iii) dg; gf; gj
(iv) bh; f h; hj
Converting each stars in to its equivalent delta, the network is redrawn as shown in
Fig. R.P. 1.10(b), noting that each resistance in delta is 100 3 = 300Ω, eliminating
nodes c, e, g , h.
Figure R.P.1.10(a)
Figure R.P.1.10(b) .
Reducing the parallel resistors, we get the circuit as in Fig. R.P. 1.10(c).
Figure R.P.1.10(c)
Hence, there are two identical deltas af d and bf j . Converting them to their equivalent
stars, we get the circuit as shown in Fig. R.P.1.10(d).
102
j
Network Theory
300 150
= 75 Ω
600
1502
= Rf l =
= 37:5 Ω
600
Rak = Rbl = Rkd = Rlj =
Rkf
Figure R.P.1.10(d)
Figure R.P.1.10(e)
The circuit is further reduced to Fig. R.P. 1.10(e) and then to Fig. R.P. 1.10(f) and
(g). Then the equivalent resistance is
Rab =
Figure R.P.1.10(f)
R.P
214:286 300
= 125 Ω
514:286
Figure R.P.1.10(g)
1.11
Obtain the equivalent resistance Rad for the circuit shown in Fig. R.P.1.11.
a
d
Figure R.P.1.11
Figure R.P.1.11(a)
Circuit Concepts and Network Simplification Techniques
j
103
SOLUTION
The circuit is redrawn as shown Fig. 1.11(a), marking the nodes a to f to identify the
deltas in it. It contains 3 deltas abc, bde and def with 3 equal resistors of 30 Ω each. For
30
= 10Ω. Then the
each delta, their equivalent star contains 3 resistors each of value
3
circuit becomes as shown in Fig. R.P. 1.11(b) where f is isolated.
On simplification, we get the circuit as shown in Fig. R.P.1.11(c) and further reduced
to Fig. R.P.1.11(d).
Figure R.P.1.11(b)
Figure R.P.1.11(c)
Figure R.P.1.11(d)
Then the equivalent ressitance,
Rad = 10 + 13:33 + 10 = 33:33 Ω
R.P
1.12
Draw a network for the following mesh equations in matrix form :
2
5 + j5
j5
6
j5
8 + j8
4
0
6
32
3
2
3
0
I1
30 / 0
7
7
6
7
6
0
6 5 4 I2 5 = 4
5
I3
20 / 0
10
104
j
Network Theory
SOLUTION
The general form of the mesh equations in matrix form for a network having three mashes
is given by
3
2
32
3 2
Z12
Z13
Z11
I1
V1 1
7
6 Z
6
7 6
Z22
Z23 7
21
4
5 4 I2 5 = 4 V2 2 5
Z31
Z32 Z33
V3 3
I3
Z11 = Z10 + Z12 + Z13
and,
where
Z10 = Sum of the impedances confined to mesh 1 alone
Z12 = Sum of the impedances common to meshes 1 and 2
Z13 = Sum of the impedances common to meshes 1 and 3
Similiar difenitions hold good for Z22 and Z33 : Also, Zij = Zji
For the present problem,
Z11 = 5 + j 5Ω
Z12 = Z21 = j 5Ω
Z13 = Z31 = 0Ω
Z23 = Z32 = 6Ω
We know that,
)
)
Similarly;
Finally;
)
)
)
)
Z11 = Z10 + Z12 + Z13
5 + j 5 = Z10 + j 5 + 0
Z10 = 5Ω
Z22 = Z20 + Z21 + Z23
8 + j 8 = Z20 + j 5 + 6
Z20 = 2 + j 3Ω
Z33 = Z30 + Z31 + Z32
10 = Z30 + 0 + 6
Z30 = 4Ω
Making use of the above impedances, we can configure a network as shown below :
Circuit Concepts and Network Simplification Techniques
R.P
j
105
1.13
Draw a network for the following nodal equations in matrix form.
2 6
6
4
1
1
+
j 10
10
1
10
3
3
2
3 2
10 /0
7 Va
74
5
5 4
5 Vb =
1
0
1
10
1
(1
5
j) +
10
SOLUTION
The general form of the nodal equations in matrix form for a network having two nodes
is given by
Y11
Y21
Y12
Y22
V1
V2
=
I1 1
I2 2
Y11 = Y10 + Y12 and Y22 = Y20 + Y21 :
where
Y10 = sum of admittances connected at node 1 alone.
Y12 = Y21 = sum of admittances common to nodes 1 and 2.
Y20 = sum of admittances connected at node 2 alone:
For the present problem,
1
1
+
S
j 10
10
1
S
Y12 = Y21 =
10
1
Y22 = (1 j ) + 10 S
5
Y11 =
We know that, Y11 = Y10 + Y12
)
)
Similarly;
)
)
Y22
1
1
1
+
= Y10 +
j 10
10
10
1
Y10 =
S
j 10
= Y20 + Y21
1
1
1
(1 j ) +
= Y20 +
5
10
10
1
Y20 = (1 j ) S
5
106
j
Network Theory
Making use of the above admittances, we can configure a network as shown below :
Exercise problems
E.P
1.1
Refer the circuit shown in Fig. E.P.1.1. Using mesh analysis, find the current delivered
by the source. Verify the result using nodal technique.
Figure E.P. 1.1
Ans : 5A
E.P
1.2
For the resistive circuit shown in Fig. E.P. 1.2. by using source transformation and mesh
analysis, find the current supplied by the 20 V source.
Figure E.P. 1.2
Ans :
2.125A
Circuit Concepts and Network Simplification Techniques
E.P
j
107
1.3
Find the voltage v using nodal technique for the circuit shown in Fig. E.P. 1.3.
Figure E.P. 1.3
Ans : v = 5V
E.P
1.4
Refer the network shown in Fig. E.P. 1.4. Find the currents i1 and i2 using nodal analysis.
Figure E.P. 1.4
Ans :
E.P
i1 = 1 A, i2 =
1A
1.5
For the network shown in Fig. E.P. 1.5, find the currents through the resistors R1 and
R2 using nodal technique.
Figure E.P. 1.5
Ans :
3.33A, 6.67A
108
j
E.P
Network Theory
1.6
Use the mesh-current method to find the branch currents i1 ; i2 and i3 in the circuit of
Fig. E.P. 1.6.
Figure E.P. 1.6
Ans :
i1 =
E.P
1.7
1.72A, i2 = 1.08A , i3 = 2.8A
Refer the network shown in Fig. E.P. 1.7. Find the power delivered by the dependent
voltage source in the network.
Figure E.P. 1.7
Ans :
E.P
375 Watts
1.8
Find the current Ix using (i) nodal analysis and (ii) mesh analysis.
Figure E.P. 1.8
Ans :
Ix =
150(3 + j4)
A
95 + j30
Circuit Concepts and Network Simplification Techniques
E.P
j
109
1.9
Determine the current ix in the circuit shown in Fig. E.P. 1.9
Figure E.P. 1.9
Ans : ix = 3A
E.P
1.10
Determine the resistance between the terminals a
1.10.
b of the network shown in Fig. E.P.
Figure E.P. 1.10
Ans : 23.6 Ω
E.P
1.11
Determine the resistance between the points A and B in the network shown in Fig. E.P.
1.11.
Figure E.P. 1.11
Ans :
4.23 Ω
110
j
E.P
Network Theory
1.12
Determine the current in the galvanometer branch of the bridge network shown in Fig.
E.P. 1.12.
Figure E.P. 1.12
Ans :
10.62μA
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
i2
then;
i1
+ i2
!
!
!
v1
v2
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
= A
i1 =
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.
2 3
6
i2 =
= A
3+6
9
Figure 3.2
Figure 3.3
The total current i is then the sum of i1 and i2
i
EXAMPLE
= i1 + i2 =
12
A
9
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
= 0:3 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,
0
=
iA
=
io
where
R1
iR2
R1
+ R2
jj
= (12 kΩ 12 kΩ) + 12 kΩ
= 6 kΩ + 12 kΩ
= 18 kΩ
and
= 12 kΩ
4 10 3 12 103
iA =
(12 + 18) 103
= 1:6 mA
Again applying the current division principle,
12
iA
00
io =
= 0:8 mA
12 + 12
)
R2
Thus;
io
= io 0 + io 00 =
0:3 + 0:8 = 0:5 mA
Figure 3.6(a)
Figure 3.6
Circuit Theorems
EXAMPLE
j
163
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.
0
io
12
(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
io
3
00
=
io
4
00
6=0
6
= 1:5 mA
103
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.
Hence;
io
000
= 1mA
Applying the superposition principle, we find that
io
= io 0 + io 00 + io 000
=3
1:5 + 1
= 2:5 mA
Circuit Theorems
EXAMPLE
j
165
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
24
i1 =
= 3A
8
As a second step, we set the voltage source to zero and determine the current
current source. For this condition, refer to Fig. 3.13 for analysis.
Figure 3.12
i2
due to the
Figure 3.13
Applying KCL at node 1, we get
i2
+7=
Noting that
i2
=
we get,
v1
=
3i2
v1
2
v1
3
3i2
(3.1)
0
(3.2)
166
j
Network Theory
Making use of equation (3.2) in equation (3.1) leads to
3i2 3i2
i2 + 7 =
2
7
i2 =
A
4
Thus, the total current
)
i
EXAMPLE
= i1 + i2
7
5
=3
A= A
4
4
3.5
For the circuit shown in Fig. 3.14, find the terminal voltage Vab using superposition principle.
Figure 3.14
SOLUTION
As a first step in the analysis, deactivate the independent current source. This results in a circuit diagram as shown in Fig. 3.15.
Applying KVL clockwise gives
4 + 10
0+3
)
)
Vab1
+ Vab1 = 0
4Vab1 = 4
Vab1
= 1V
Next step in the analysis is to deactivate the
independent voltage source, resulting in a circuit diagram as shown in Fig. 3.16.
Applying KVL gives
10
2+3
)
)
Vab2
Figure 3.15
+ 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
2
+
4ix1
vx 1
8
But
Hence;
)
)
)
ix1
vx 1
2
vx 1
2
+
+
v
4 x21
vx 1
8
2vx1
vx 1
8
4vx1 + vx1
vx 1
=6
=
vx 1
2
=6
=6
2vx1 = 48
=
48
= 16V
3
Figure 3.18
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 :
( 4ix2 )
=4
8
2
vx 2
vx + 4ix2
+ 2
=4
8
2
vx 2
)
+
v x2
(3.3)
Applying KVL along dotted path, we get
vx 2
)
vx 2
=
+ 4ix2
2ix2
or
2ix2 = 0
ix2
=
vx 2
(3.4)
2
Substituting equation (3.4) in equation (3.3), we get
vx 2
)
)
)
)
8
vx 2
+4
+
vx 2
2
=4
2
vx 2
8
+
2vx2
vx 2
vx 2
8
vx 2
2
vx 2
2
=4
=4
4vx2 = 32
32
vx 2 =
V
3
Hence, according to the superposition principle,
vx
EXAMPLE
= vx 1 + vx 2
32
= 16
= 5:33V
2
Figure 3.19
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.
5
= 1:22A
2 + 2:1
Similarly to find i02 we have to disable the 5V source by shorting it.
0
i1
Referring to Fig. 3.23, we find that
0
i2
Figure 3.22
=
=
2 2:1
=
2 + 2:1
1:024 A
Figure 3.23
170
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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
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171
Applying KVL to the left mesh:
30ia
60 + 30 (ia
ib )
Also
ib
=0
(3.5)
=
0:4iA
=
0:4 (
ia )
= 0:4ia
(3.6)
Substituting equation (3.6) in equation (3.5), we get
)
60 + 30ia
30 0:4ia = 0
60
ia =
= 1:25A
48
1:25
ib = 0:4ia = 0:4
30ia
= 0:5A
0
Hence;
V1
= (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:
Va
+
Va
20
30Va
)
00
V1
=4
10
20V100 = 800
(3.7)
Applying KCL at node b:
00
V1
30
Also;
Va
= 20ia
V1
)
Va
V1
10
)
ib
= 0:4ib
=
Va
20
0:4Va
+
=
30
10
20
7:2Va + 8V100 = 0
00
Hence;
00
+
00
V1
Va
(3.8)
172
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Network Theory
Solving the equations (3.7) and (3.8), we find that
00
V1
Hence
V1
= 60V
= 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
= 1 mA
ia3 =
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:
Vy
18
103
+
Vy
2
20
= (10+5) 10
103
Solving, we get
Hence;
ia
=
3
= 45V:
45
=
3
10
18 103
= 2:5 mA
Vy
Vy
18
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 superposition 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
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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 terminals 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
175
j
(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
(c)
Rt
=
b
voc
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 .
(c)
Rt
=
vab
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
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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
Hence
Vab
)
I
= 2A
Figure 3.36
= 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
i
=
40
= 4A
2+8
Circuit Theorems
EXAMPLE
j
177
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
)
Hence;
30
40
V2 = 60 Volts
60 + 20
Voc
+
V2
V 60 0 =I
2
1:5 = 0
60
60 + 20
60
= 60
= 45 V
80
=
178
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Network Theory
To find isc :
_
Applying KCL at node 2:
)
Therefore;
V2
20
+
30
V2
1:5 = 0
40
V2 = 30V
isc
=
Rt
=
V2
20
Voc
isc
= 1:5A
=
45
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
EXAMPLE
3.12
Find the Thevenin equivalent for the circuit shown in Fig. 3.41 with respect to terminals a
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 independent sources are present, Thevenin resistance is found using the relation,
Figure 3.42
Rt
=
voc
isc
Applying KVL clockwise for mesh 1 :
20 + 6i1
)
2i + 6 (i1
12i1
Since i = i1
)
i2 ,
i2 )
=0
6i2 = 20 + 2i
we get
12i1
6i2 = 20 + 2 (i1
10i1
4i2 = 20
i2 )
Applying KVL clockwise for mesh 2 :
)
j
10i2 + 6 (i2
i1 )
=0
6i1 + 16i2 = 0
Figure 3.43
b.
179
180
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Network Theory
Solving the above two mesh equations, we get
EXAMPLE
i2
=
Rt
=
)
120
A
136
voc
isc
=
isc
= i2 =
120
A
136
12
= 13:6 Ω
120
136
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,
Applying KVL to mesh 2 :
)
i1
= 4 mA
12 + 6
12 + 6
10
3
10
3
i2
4
(i2
10
+ 3 103 i2 = 0
3
3
i1 )
+3
10
i2
=0
Circuit Theorems
Solving, we get
i2
j
181
= 4 mA
! a b ! 3 kΩ, we get
+
3 10 = 0
= 4 10 + 3 10
= 4 10 4 10 + 3 10 4 10
Applying KVL to the path 4 kΩ
4
)
10
3
i1
Voc
Voc
3
3
i2
3
i1
3
i2
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
jj
= Rab = 4 kΩ + (6 kΩ 3 kΩ) = 6 kΩ
Rt
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
10 28
2 10
=
(6 + 2) 10
=i 2
3
3
EXAMPLE
3
= 7V
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
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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
=
A
i1 =
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
900i1
120
900
= 1204
2004
= 95:8 mV
Vt
= 1204i2
120
1500
Figure 3.49(b)
To find Rt :
Deactivate all the independent sources and look into the terminals
Thevenin’s resistance.
Figure 3.49(c)
a
b
Figure 3.49(d)
to determine the
Circuit Theorems
Rt
jj
iG
jj
=
95:8 10 3
840:64 + 30 Ω
Figure 3.50
= 110:03 A
3.15
For the circuit shown in Fig. 3.51, find the Thevenin’s equivalent circuit between terminals
Figure 3.51
SOLUTION
With ab shorted, let Isc = I . The circuit after
transforming voltage sources into their equivalent current sources is as shown in Fig 3.52.
Writing node equations for this circuit,
0:1 Vc + I = 3
At a :
0:2Va
At c :
0:1Va + 0:3 Vc
At b :
183
= Rab = 600 900 + 800 1204
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.
EXAMPLE
j
0:1 Vb = 4
0:1Vc + 0:2 Vb
I
=1
As the terminals a and b are shorted Va =
and the above equations become
Vb
Figure 3.52
a and b.
184
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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
10
Therefore Rt =
=
= 10Ω
Isc
1
The Thevenin’s equivalent is as shown in Fig 3.53
Voc
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(d)
Figure 3.55(c)
Figure 3.56 Thevenin equivalent
circuit
EXAMPLE
3.17
Find the Thevenin equivalent circuit as seen from the terminals a
shown in Fig. 3.57.
b.
Refer the circuit diagram
186
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Network Theory
Figure 3.57
SOLUTION
Since the circuit has no independent sources, i = 0 when the terminals a b are open. Therefore, Voc = 0.
The onus is now to find Rt . Since Voc = 0 and isc = 0, Rt cannot be determined from
Rt
=
Voc
isc
. Hence, we choose to connect a source of 1 A at the terminals a
b
as shown in Fig.
3.58. Then, after finding Vab , the Thevenin resistance is,
Rt
KCL at node a :
2i
Va
5
Also;
i
Va
Hence;
)
2
5
Va
10
=
+
=
+
Vab
1
Va
1=0
10
Va
10
Va
10
50
Va =
V
13
1=0
50
Ω
1
13
Alternatively one could find Rt by connecting a 1V source at the terminals a b and then find
1
. The concept of finding Rt by connecting a 1A source
the current from b to a. Then Rt =
Hence;
Rt
=
Va
=
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
is shown in Fig. 3.58 (a).
Figure 3.58
Figure 3.58 (a)
b
Circuit Theorems
EXAMPLE
j
187
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
Ω
Rt =
I
Figure 3.60(a)
Applying KCL at node a:
I
Since;
Vx
we get,
I
Hence;
= 0:5Vx +
Vx
4
= 1V
= 0:5 +
1
= 0:75 A
4
Figure 3.60(b)
1
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
a
b
and then measure Vab (Fig. 3.60 (c)). Consequently, Rt =
Vab
1
= Vab Ω:
188
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Network Theory
Applying KCL at node a:
0:5Vx +
Hence
Rt
=
Vx
4
Vab
1
=1
=
)
Vx
1
Vx
= 1:33V
= 1:33 Ω
The corresponding Thevenin equivalent
circuit is same as shown in Fig. 3.60(b)
3.3
Figure 3.60(c)
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
. In fact, source transformation of Thevenin equivalent circuit leads to
RN = Rt and iN =
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 .
(c)
Rt
= RN =
voc
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
(c)
Rt
=
b
vab
and find vab .
1
Note: Also, since vt = voc and iN = isc
Rt
=
voc
isc
= RN
The open–circuit and short–circuit test are sufficient to find any Thevenin or Norton equivalent.
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
where
b0
(3.10)
= 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,
this fact in equation 3.10, we find that b0 = vt .
i
= 0 and
v
=
voc
=
vt .
Making use of
190
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Network Theory
(ii) When all the internal sources are deactivated,
become
v = a0 i = Rt i
b0
)
= 0. This enforces equation 3.10 to
a0
= Rt
Rt
Vt
Figure 3.62
Figure 3.63 Thevenin’s equivalent circuit of Fig. 3.62
Current-driven circuit
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
Thus, the two circuits of Fig. 3.62 and 3.63 are equivalent.
x
y
of the circuit in Fig. 3.63.
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 contributions 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
Thus, equation (3.11) becomes
i
=
=
1
Rt
1
Rt
v
isc
v
iN
y.
Figure 3.65 Norton’s equivalent of
voltage driven circuit
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
4
3 + isc = 0
)
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
Rt in the Thevenin equivalent circuit.
4 12
RN =
=3Ω
4 + 12
Figure 3.67
192
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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
= 0A
1000
12
isc =
A = 2 mA
6000
ia
=
Figure 3.70
Figure 3.70(a)
Circuit Theorems
To find RN :
The procedure for finding RN is same that of Rt
in the Thevenin equivalent circuit.
Rt
= RN =
voc
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
)
)
Therefore;
12 + 6000ia + 2000ia + 1000ia = 0
4
ia =
A
3000
4
voc = ia
1000 = V
3
4
voc
3
RN =
=
= 667 Ω
isc
2 10 3
Figure 3.71
The Norton equivalent circuit along with resistor R is as shown below:
ib
=
isc
2
=
2mA
= 1mA
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
j
193
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
6 kΩ
+
V2
2 kΩ
+
V2
V1
3 kΩ
=0
Substituting V1 = 12 V and solving, we get
V2
= 6V
isc
=
V1
V2
3 kΩ
+
V1
4 kΩ
= 5 mA
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
=
EXAMPLE
isc
R
RN
+ RN
= 2:57mA
Figure 3.73(e)
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
Constraint equation :
i2
i1
KVL around supermesh :
4+2
KVL around mesh 3 :
8
10 (
3
i3
= 4mA
10
i2 )
3
+2
i1
j
197
(3.12)
10
3
+4
i2
=0
i3
i1 )
=0
isc
i1 )
=0
10 (
3
(3.13)
Since i3 = isc , the above equation becomes,
8
10 (
3
isc
i2 )
+2
10 (
3
(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
jj
= 6 kΩ 10 kΩ
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
network shown in Fig. 3.76(b).
Figure 3.76(a)
198
j
Network Theory
Vo
= isc (RN
= isc
jj
R)
RN R
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
= 10 mA
500
Therefore, for the right–hand portion of the circuit, isc =
Then
i
=
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
Therefore
i
=
=
25(10i) =
250i
vab
250
Substituting i into the mesh equation (3.15), we get
)
RN
5 + 500
vab
+ vab = 0
250
= 5V
5
= 50 Ω
0:1
vab
= Rt voc
isc
=
vab
isc
=
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:
1=
)
v1
100
0:03v1
KCL at node 2:
)
v2
200
v2
+
v1
v2
50
0:02v2 = 1
v1
+ 0:1v1 = 0
50
0:08v1 + 0:025v2 = 0
+
Solving the above two nodal equations, we get
= 10:64 volts
)
voc = 10:64 volts
10:64
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).
v1
voc
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
70
ia =
A
6
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
=
=8Ω
Hence, Rt = RN =
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
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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;
If
RL
= 10 kΩ;
I
)
= 6 mA )
= 10 mA
I
Voc
= 20 + 0:01Rt
Voc
= 60 + 0:006Rt
Solving, we get Voc = 120 V, Rt = 10 kΩ.
If
3.4
RL
= 20 kΩ;
I
=
Voc
(RL + Rt )
=
(20
120
+ 10
103
10 ) = 4 mA
3
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
the maximum power is delivered to it.
The power that is delivered to the load is given by
p
2
=i
RL
=
Vt
Rt
+ RL
Figure 3.82 Circuit A with load RL
2
RL
(3.16)
Circuit Theorems
j
203
Assuming that Vt and Rt are fixed for a given source, the maximum power is a function of
In order to determine the value of RL that maximizes p, we differentiate p with respect to
RL and equate the derivative to zero.
RL .
"
dp
= Vt2
dRL
which yields
RL
#
(Rt + RL )2 2 (Rt + RL )
=0
(RL + Rt )2
= Rt
(3.17)
To confirm that equation (3.17) is a maximum,
2
it should be shown that
d p
2
dRL
<
0. Hence, maxi-
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,
2
2
Pmax
=
Vt R L
(2RL )2
=
Figure 3.83 Thevenin equivalent circuit
is substituted for circuit A
Vt
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
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Network Theory
180
= 1A
150 + 30
i = 150 V
Voc = Vt = 150
i
=
To find Rt , deactivate the 180 V source. This results in the
circuit diagram of Fig. 3.85(b).
Rt
jj
= Rab = 30 Ω 150 Ω
30 150
=
= 25 Ω
30 + 150
Figure 3.85(a)
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
4RL
4 25
= 2:25 Watts
The Thevenin source Vt actually provides a total
power of
2
Pmax
Pt
=
Vt
=
Figure 3.85(b)
150
= 150 25 + 25
= 150
i
= 450 Watts
Thus, we note that one-half the power is dissipated in RL .
EXAMPLE
Figure 3.86
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
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)
Rt
Figure 3.88(c)
jj
jj
= Rab = 6 kΩ 6 kΩ 6 kΩ
= 2 kΩ
To find Vt :
Refer the Fig. 3.88(d).
Constraint equation :
V3
V1
By inspection,
KCL at supernode :
= 12 V
V2
=3V
V2
V3
6k
)
V3
6k
3
+
+
12
V3
6k
V1
6k
+
+
V3
V1
V2
=0
6k
12 3
=0
6k
Figure 3.88(d)
j
205
206
Network Theory
j
)
)
)
)
V3
3 + V3
12 + V3
15 = 0
3V3 = 30
= 10
V3
= Vab = V3 = 10 V
Vt
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
V1
V2
12
KCL at supernode :
V2
V3
6k
)
)
)
)
V3
3
6k
V3
+
V3
+
12
6k
3 + V3
6k
3
+
+
V3
V3
2k
+
V3
6k
12
=0
+
=0
2k
6k
15 + 3V3 + V3 12 = 0
6V3 = 30
V3
2
Hence;
V1
Pmax
=
V3
RL
=
=5 V
25
= 12:5 mW
2k
Figure 3.88(f)
Circuit Theorems
EXAMPLE
j
207
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)
! 3 kΩ ! 2 kΩ ! , we get
) + 2k + = 0
5k + 3k (
1 10
2 10
+3 10
2 10
+2 10
Applying KVL clockwise to the loop 5 kΩ
) 5 10 1 10
)
)
3
i1
i2
3
a
i2
i1
3
3
5
9
b
Vt
3
3
3
4 + Vt = 0
Vt
= 18 V:
The Thevenin equivalent circuit with load resistor RL is as shown in Fig. 3.89 (d).
18
= 0:9 mA
i =
(10 + 10) 103
Then,
Pmax
= PL = (0:9 mA)2
10 kΩ
= 8:1 mW
Figure 3.89(c)
EXAMPLE
Figure 3.89(d)
3.30
Find the maximum power dissipated in RL . Refer the circuit shown in Fig. 3.90.
Figure 3.90
+ Vt = 0
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
i2
+
20 170
=
= 17 A
170 + 30
= 20 30
=
=
=
i
R1
R2
R2
i
R1
R1
+ R2
600
= 3A
200
170 + 30
Figure 3.91(a)
210
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Network Theory
Applying KVL clockwise to the loop comprising of 50 Ω
50i2
10i1 + 8
) 50(3)
)
0+
Vt
EXAMPLE
b,
we get
10 (17) + Vt = 0
Vt
= 20 V
20
= 0:2A
50 + 50
2
Pmax = iT
50 = 0:04 50 = 2 W
=
a
=0
The Thevenin equivalent circuit with load resistor
as shown in Fig. 3.91(c).
iT
! 10 Ω ! 8 Ω !
RL
is
Figure 3.91(c)
3.31
Find the value of
find Pmax .
RL
for maximum power transfer in the circuit shown in Fig. 3.92. Also
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
)
20 + 10i
! 3 Ω ! 2 Ω ! 10 V ! 5 Ω ! 30 V, we get
10
30 = 0
60
i =
= 6A
10
212
j
Network Theory
Again applying KVL clockwise to the path 2 Ω
2i 10 Vt = 0
)
Vt
= 2(6)
= 2i
! 10 V !
a
b,
we get
10
10 = 2 V
The Thevenin equivalent circuit with load resistor
RL is as shown in Fig. 3.95 (a).
Pmax
= i2T RL
2
=
EXAMPLE
Vt
4Rt
= 625 mW
Figure 3.95(a) Thevenin equivalent
circuit
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 :
)
Hence;
0
ia
0:9 + 10i0a = 0
0
ia
Voc
= 0:1 A
= 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:
)
Hence
isc
ia
00
0:9 + 10ia 00 = 0
ia
= 10ia = 10
=
= 0:1 A
01=1A
00
Rt
00
:
Voc
=
isc
3
=3Ω
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 network shown in Fig. 3.97(a).
iT
Pmax
=
3
= 0:5 A
3+3
= i2T RL
= (0:5)2
3
= 0.75 W
EXAMPLE
Figure 3.97(a)
3.33
Find the value of RL in the network shown that will achieve maximum power transfer, and determine 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
0
Also
Hence;
Vx
10
10
12
=
4 10
12 + 2
i
=1
3i
3
3
i
+2 1
10
= 3 mA
3i
=0
Figure 3.98(b)
214
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Network Theory
!2 !
1 10
Applying KVL to loop 1 kΩ
Vx
3
)
Vt
a,
b
i
we get
+ 2Vx0
Vt
=0
10 + 2 1 10
= 1 10 + 2 10
= 3 10 3 10
3
=1
3
i
3
3
3
i
i
3
=9V
To find Rt , we need to find isc . While finding isc ,
none of the independent sources must be deactivated.
Applying KVL to mesh 1:
12 + Vx 00 + 0 = 0
)
)
00
= 12
i1
= 12
Vx
1
10
3
)
i1
= 12 mA
Applying KVL to mesh 2:
1
)
10
3
i2
1
+ 2Vx 00 = 0
10
3
i2
=
24
i2
=
24 mA
Applying KCL at node a:
isc
= i1
i2
= 12 + 24 = 36 mA
Hence;
Rt
=
Vt
isc
=
Voc
isc
9
=
36 10
= 250 Ω
3
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
=
A
250 + 250
500
2
Pmax = iT
250
iT
=
9 2
500
= 81 mW
=
250
Figure 3.98 (c) Thevenin equivalent circuit
Circuit Theorems
EXAMPLE
j
215
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
Fig. 3.99(a).
RL
from the original circuit, we get the circuit shown in
Figure 3.99(a)
KCL at node v1 :
100
v1
2
+
v1
13i0a
+
5
v2
v1
4
=0
(3.18)
Constraint equations :
0
ia
v1
v2
4
0
va
=
100
v1
= va0
= v1
(3.19)
2
v2
(applying K C L at v2 )
(3.20)
(potential across 4 Ω)
(3.21)
216
j
Network Theory
From equations (3.20) and (3.21), we have
v2
)
)
)
v1
4
v2
v1
5v1
= v1
v2
= 4v1
4v2
5v2 = 0
v1
(3.22)
= v2
Making use of equations (3.19) and (3.22) in (3.18), we get
100
v1
2
)
)
)
)
13
v2
+
v1 )
(100
2
+
5
100) + 2
5 (v1
13
v1
500 + 2v1
5v1
13
(100
v1
v1
4
v1 )
=0
=0
2
100 + 13v1 = 0
20v1 = 1800
Hence;
We know that,
v1
= 90 Volts
vt
= v2 = v1 = 90 Volts
Rt
=
voc
isc
=
vt
isc
The short circuit current is calculated using the circuit shown below:
00
Here
Applying KCL at node v1 :
ia
100
v1
2
)
100
v1
2
+
v1
=
+
100
v1
2
13ia
0
v1
+
=0
5
4
(100 v1 )
13
v1
2
+
=0
5
4
v1
Circuit Theorems
Solving we get v1 = 80 volts = va00
Applying KCL at node a :
0
v1
4
)
+ isc = va00
v1
+ va00
4
80
=
+ 80 = 100 A
4
voc
vt
=
Rt =
isc
Hence;
=
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
=
0:9 + 0:9
1:8
2
Pmax = it
0:9
it
=
=
EXAMPLE
90 2
1:8
0 9 = 2250 W
:
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
j
217
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
25i1
i3 )
+ 4i1 = 0
i2
20i3 =
200
KVL clockwise to mesh 2 :
)
)
14i0a + 2 (i2
14 (i1
i3 )
+ 2 (i2
i3 )
+ 1 (i2
i1 )
=0
i3 )
+ 1 (i2
i1 )
=0
13i1 + 3i2
16i3 = 0
KVL clockwise to mesh 3 :
)
2 (i3
i2 )
100 + 3i3 + 20 (i3
20i1
i1 )
=0
2i2 + 25i3 = 100
Solving the mesh equations, we get
i1
=
2:5A; i3 = 5A
Applying KVL clockwise to the path comprising of a
)
Vt
20i0a = 0
Vt
= 20i0a
= 20 (i1
i3 )
= 20 ( 2:5
=
150 V
5)
b
! 20 Ω, we get
Circuit Theorems
Next step is to find Rt .
Rt
When terminals a
b
Voc
=
isc
=
Vt
isc
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
+ 3i2
i1 )
=0
2i3 = 0
KVL clockwise to mesh 3 :
)
100 + 3i3 + 2 (i3
i2 )
=0
2i2 + 5i3 = 100
j
219
220
j
Network Theory
Solving the mesh equations, we find that
i1
)
=
40A;
i3
= 20A;
= i1
i3 = 60A
150
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 :
isc
Vt
Pmax
= i21 RL
150
=
2:5 + 2:5
= 2250 W
EXAMPLE
2
25
:
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
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 :
10W to 250 Ω corresponds to iL =
i2
=
In the present context,
0:2 =
and
0:5 =
i
R1
R1
+ R2
IN R N
RN
+ 250
IN R N
RN
+ 80
(3.23)
(3.24)
Circuit Theorems
j
221
Solving equations (3.23) and (3.24), we get
IN
RN
= 1:7 A
= 33:33 Ω
(a) If vL iL is maximum,
= RN = 33:33 Ω
33:33
iL = 1:7
33:33 + 33:33
= 850 mA
RL
vL
(b) vL = IN (RN
.
RL =
Then, iL = 0 and
1
jj
= iL RL = 850
10 33 33
3
:
= 28:33 V
RL )
is a maximum when
vL
jj
RN RL
is a maximum, which occurs when
= 1 7 33 33
= 1:7
RN
:
:
= 56:66 V
(c)
3.5
iL
=
IN R N
RN
)
+ RL
is maxmimum when RL = 0 Ω
iL
= 1:7A and vL = 0 V
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
222
j
Network Theory
The superposition principle is particularly useful if a circuit has two or more sources acting
at different frequencies. The circuit will have one set of impedance values at one frequency and a
different set of impedance values at another frequency. Phasor responses corresponding to different frequencies cannot be superposed; only their corresponding sinusoids can be superposed. That
is, when frequencies differ, the principle of superposition applies to the summing of time domain
components, not phasors. Within a component, problem corresponding to a single frequency,
however phasors may be superposed.
Thevenin and Norton equivalents in phasor circuits are found exactly in the same manner
as described earlier for resistive circuits, except for the subtitution of impedance Z in place of
resistance R and subsequent use of complex arithmetic. The Thevenin and Norton equivalent
circuits are shown in Fig. 3.101 and 3.102.
The Thevenin and Norton forms are equivalent if the relations
(a) Zt = ZN
(b) Vt = ZN IN
hold between the circuits.
A step by step procedure for finding the Thevenin equivalent circuit is as follows:
1. Identify a seperate circuit portion of a total circuit.
2. Find Vt = Voc at the terminals.
3.
(a) If the circuit contains only impedances and independent sources, then deactivate all the
independent sources and then find Zt by using circuit reduction techniques.
(b) If the circuit contains impedances, independent sources and dependent sources, then
either short–circuit the terminals and determine Isc from which
Zt =
Voc
Isc
or deactivate the independent sources, connect a voltage or current source at the terminals, and
determine both V and I at the terminals from which
Zt =
V
I
A step by step procedure for finding Norton equivalent circuit is as follows:
(i) Identify a seperate circuit portion of the original circuit.
(ii) Short the terminals after seperating a portion of the original circuit and find the current
through the short circuit at the terminals, so that IN = Isc .
(iii)
(a) If the circuit contains only impedances and independent sources, then deactivate all the
independent sources and then find ZN = Zt by using circuit reduction techniques.
(b) If the circuit contains impedances, independent sources and one or more dependent
Voc
sources, find the open–circuit voltage at the terminals, Voc , so that ZN = Zt =
:
Isc
Circuit Theorems
EXAMPLE
j
223
3.37
Find the Thevenin and Norton equivalent circuits at the terminals
Fig. 3.103.
a
b
for the circuit in
Figure 3.103
SOLUTION
As a first step in the analysis, let us find Vt :
Using the principle of current division,
8 (4 /0 )
32
=
8 + j 10 j 5
8 + j5
j 320
Vt = Io (j 10) =
= 33:92 /58 V
8 + j5
Io =
To find Zt , deactivate all the independent sources. This results in a circuit diagram as shown
in Fig. 3.103 (a).
Figure 3.103(a)
Figure 3.103(b) Thevenin equivalent circuit
224
j
Network Theory
jj
Zt = j 10 (8 j 5) Ω
(j 10)(8 j 5)
=
j 10 + 8
j5
= 10 /26 Ω
The Thevenin equivalent circuit as
viewed from the terminals a
b is
as shown in Fig 3.103(b). Performing
source transformation on the Thevenin
equivalent circuit, we get the Norton
equivalent circuit.
Figure : Norton equivalent circuit
Vt
33:92 /58
=
Zt
10 /26
= 3:392 /32 A
IN =
ZN = Zt = 10 /26 Ω
EXAMPLE
3.38
Find vo using Thevenin’s theorem. Refer to the circuit shown in Fig. 3.104.
Figure 3.104
SOLUTION
Let us convert the circuit given in Fig. 3.104 into a frequency domain equiavalent or phasor circuit
(shown in Fig. 3.105(a)). ! = 1
10 cos (t
45 )
5 sin (t + 30 ) = 5 cos (t
60 )
L
C
!
= 1F !
= 1H
j !L
1
j !C
!
!
=j
=
10 / 45 V
5 / 60 V
11= 1Ω
1
11 = 1Ω
j
j
j
Circuit Theorems
j
225
Figure 3.105(a)
Disconnecting the capicator from the original circuit, we get the circuit shown in
Fig. 3.105(b). This circuit is used for finding Vt .
Figure 3.105(b)
KCL at node a :
Vt
10 / 45
Vt 5 / 60
+
=0
3
j1
Vt = 4:97 / 40:54
Solving;
To find Zt deactivate all the independent sources
in Fig. 3.105(b). This results in a network as
shown in Fig. 3.105(c) :
Zt = Zab = 3Ω j 1 Ω
j3
3
=
=
(1 + j 3) Ω
3+j
10
Figure 3.105(c)
jj
The Thevenin equivalent circuit along with the capicator
is as shown in Fig 3.105(d).
Vt
( j 1)
Zt j 1
4:97 / 40:54
(
=
0:3(1 + j 3) j 1
= 15:73 /247:9 V
Vo =
Hence;
vo
j 1)
= 15:73 cos (t + 247:9 ) V
Figure 3.105(d) Thevenin equivalent circuit
226
j
Network Theory
EXAMPLE
3.39
Find the Thevenin equivalent circuit of the circuit shown in Fig. 3.106.
Figure 3.106
SOLUTION
Since terminals a
b
are open,
Va = Is
10
= 20 /0 V
Applying KVL clockwise for the mesh on the right hand side of the circuit, we get
3Va + 0 (j 10) + Voc
Va = 0
Voc = 4Va
= 80 /0 V
Let us transform the current source with 10 Ω parallel resistance to a voltage source with 10 Ω
series resistance as shown in figure below :
To find Zt , the independent voltage source is deactivated and a current source of I A is
connected at the terminals as shown below :
Circuit Theorems
Applying KVL clockwise we get,
V0a
)
3V0a
0
4Va
j 10I
+ Vo = 0
j 10I
+ Vo = 0
V0a = 10I
Since
= Vo
Vo
= 40 + j 10Ω
Zt =
I
we get
40I
Hence;
j 10I
Hence the Thevenin equivalent circuit is as shown
in Fig 3.106(a) :
EXAMPLE
Figure 3.106(a)
3.40
Find the Thevenin and Norton equivalent circuits for the circuit shown in Fig. 3.107.
Figure 3.107
SOLUTION
The phasor equivalent circuit of Fig. 3.107 is shown in Fig. 3.108.
KCL at node a :
Voc
)
2Voc
j 10
Voc =
Voc
=0
j5
100 100
j
=
/ 90 V
3
3
10 +
j
227
228
j
Network Theory
Figure 3.108
To find Isc , short the terminals a
b
of Fig. 3.108 as in Fig. 3.108(a).
Figure 3.108 (a)
Figure 3.108 (b)
Since Voc = 0, the above circuit takes the form shown in Fig 3.108 (b).
Isc = 10 /0 A
100
/ 90
Voc
10
3
Hence;
Zt =
=
=
/ 90 Ω
Isc
10 /0
3
The Thevenin equivalent and the Norton equivalent circuits are as shown below.
Figure Thevenin equivalent
EXAMPLE
Figure
Norton equivalent
3.41
Find the Thevenin and Norton equivalent circuits in frequency domain for the network shown in
Fig. 3.109.
Circuit Theorems
Figure 3.109
SOLUTION
Let us find Vt = Vab using superpostion theorem.
(i) Vab due to 100 /0
100 /0
100
=
A
j 300 + j 100
j 200
= I1 (j 100)
100
(j 100) = 50 /0 Volts
=
j 200
I1 =
Vab1
(ii) Vab due to 100 /90
j
229
230
j
Network Theory
100 /90
j 100
j 300
Vab2 = I2 ( j 300)
100 /90
=
(
j 100
j 300
Vt = Vab1 + Vab2
I2 =
Hence;
=
j 300)
= j 150 V
50 + j 150
= 158:11 /108:43 V
To find Zt , deactivate all the independent sources.
jj
Zt = j 100 Ω
j 300 Ω
j 100( j 300)
=
= j 150 Ω
j 100
j 300
Hence the Thevenin equivalent circuit is as shown in Fig. 3.109(a). Performing source transformation on the Thevenin equivalent circuit, we get the Norton equivalent circuit.
158:11 /108:43
Vt
=
= 1:054 /18:43 A
Zt
150 /90
ZN = Zt = j 150 Ω
IN =
The Norton equivalent circuit is as shown in Fig. 3.109(b).
Figure 3.109(a)
Figure 3.109(b)
Circuit Theorems
3.6
j
231
Maximum power transfer theorem
We have earlier shown that for a resistive network, maximum power is transferred from a source to
the load, when the load resistance is set equal to the Thevenin resistance with Thevenin equivalent
source. Now we extend this result to the ac circuits.
Figure 3.110 Linear circuit
Figure 3.111
Thevenin equivalent circuit
In Fig. 3.110, the linear circuit is made up of impedances, independent and dependent sources.
This linear circuit is replaced by its Thevenin equivalent circuit as shown in Fig. 3.111. The load
impedance could be a model of an antenna, a TV, and so forth. In rectangular form, the Thevenin
impedance Zt and the load impedance ZL are
Zt = Rt + jXt
ZL = RL + jXL
and
The current through the load is
I=
Vt
Vt
=
Zt + ZL
(Rt + jXt ) + (RL + jXL )
p
The phasors I and Vt are the maximum values. The corresponding RM S values are obtained
by dividing the maximum values by 2. Also, the RM S value of phasor current flowing in the
load must be taken for computing the average power delivered to the load. The average power
delivered to the load is given by
P
=
=
jj
1 2
I RL
2
jV j
t
2
2 RL
2
(Rt + RL ) (Xt + XL )2
(3.25)
Our idea is to adjust the load parameters RL and XL so that P is maximum. To do this, we
get
@P
@RL
and
@P
@XL
equal to zero.
232
j
Network Theory
@P
@XL
@P
@RL
=
=
h
j j
Vt
2
RL (Xt
+ XL )
(Rt + RL )2 + (Xt + XL )2
j j
Vt
2
h
i2
(Rt + RL )2 + (Xt + XL )2
h
2 (Rt + RL )2 + (Xt + XL )2
@P
Setting
@XL
XL
@P
and Setting
@RL
RL
i
2RL (Rt + RL )
i2
= 0 gives
=
(3.26)
Xt
= 0 gives
q
=
Rt2
+ (Xt + XL )2
(3.27)
Combining equations (3.26) and (3.27), we can conclude that for maximum average power
transfer, ZL must be selected such that XL = Xt and RL = Rt . That is the maximum average power of a circuit with an impedance Zt that is obtained when ZL is set equal to complex
conjugate of Zt .
Setting RL = Rt and XL = Xt in equation (3.25), we get the maximum average power as
P
j j
=
Vt
2
8Rt
In a situation where the load is purely real, the condition for maximum power transfer is
obtained by putting XL = 0 in equation (3.27). That is,
RL
=
q
Rt2
j j
+ Xt2 = Zt
Hence for maximum average power transfer to a purely resistive load, the load resistance is
equal to the magnitude of Thevenin impedance.
3.6.1 Maximum Power Transfer When Z is Restricted
Maximum average power can be delivered to ZL only if ZL = Zt . There are few situations in
which this is not possible. These situations are described below :
(i)
RL and XL may be restricted to a limited range of values. With this restriction,
choose
q XL as close as possible to Xt and then adjust RL as close as possible to
Rt2
+ (XL + Xt )2 :
(ii) Magnitude of ZL can be varied but its phase angle cannot be. Under this restriction,
greatest amount of power is transferred to the load when [ZL ] = Zt .
j j
Zt is the complex conjugate of Zt .
Circuit Theorems
EXAMPLE
j
233
3.42
Find the load impedance that transfers the maximum power to the load and determine the maximum power quantity obtained for the circuit shown in Fig. 3.112.
Figure 3.112
SOLUTION
We select, ZL = Zt for maximum power transfer.
ZL = 5 + j 6
10 /0
= 1 /0
I=
5+5
Hence
Hence, the maximum average power transfered to the
load is
P
EXAMPLE
jj
1 2
I RL
2
1
= (1)2 5 = 2:5 W
2
=
3.43
Find the load impedance that transfers the maximum average power to the load and determine the
maximum average power transferred to the load ZL shown in Fig. 3.113.
Figure 3.113
234
j
Network Theory
SOLUTION
The first step in the analysis is to find the Thevenin equivalent circuit by disconnecting the load
ZL . This leads to a circuit diagram as shown in Fig. 3.114.
Figure 3.114
Vt = Voc = 4 /0
Hence
3
= 12 /0 Volts(RMS)
To find Zt , let us deactivate all the independent sources of Fig. 3.114. This leads to a circuit
diagram as shown in Fig 3.114 (a):
Zt = 3 + j 4 Ω
Figure 3.114 (a)
Figure 3.115
The Thevenin equivalent circuit with ZL is as shown in Fig. 3.115.
For maximum average power transfer to the load, ZL = Zt = 3 j 4.
It =
12 /0
3 + j4 + 3
j4
= 2 /0 A(RMS)
Hence, maximum average power delivered to the load is
P
=
jj
It
2
RL
= 4(3) = 12 W
1
It may be noted that the scaling factor
is not taken since the phase current is already
2
expressed by its RM S value.
Circuit Theorems
EXAMPLE
j
235
3.44
Refer the circuit given in Fig. 3.116. Find the value of RL that will absorb the maximum average
power.
Figure 3.116
SOLUTION
Disconnecting the load resistor RL from the original circuit diagram leads to a circuit diagram as
shown in Fig. 3.117.
Figure 3.117
Vt = Voc = I1 (j 20)
150 /30
j 20
=
(40 j 30 + j 20)
= 72:76 /134 Volts:
To find Zt , let us deactivate all the independent sources present in Fig. 3.117 as shown in
Fig 3.117 (a).
jj
Zt = (40 j 30) j 20
j 20 (40
j 30)
=
= (9:412 + j 22:35) Ω
j 20 + 40
j 30
236
Network Theory
j
The Value of RL that will absorb the maximum
average power is
RL
j j
= Zt =
p
(9:412)2 + (22:35)2
= 24:25 Ω
The Thevenin equivalent circuit with RL inserted
is as shown in Fig 3.117 (b).
Maximum average power absorbed by RL is
Pmax
where
)
=
1
2
jj
It
2
Figure 3.117 (a)
RL
72:76 /134
(9:412 + j 22:35 + 24:25)
= 1:8 /100:2 A
1
2
Pmax = (1:8)
24:25
2
= 39:29 W
It =
Figure 3.117 (b) Thevenin equivalent circuit
EXAMPLE
3.45
For the circuit of Fig. 3.118: (a) what is the value of ZL that will absorb the maximum average
power? (b) what is the value of maximum power?
Figure 3.118
SOLUTION
Disconnecting ZL from the original circuit we get the circuit as shown in Fig. 3.119. The first
step is to find Vt .
Circuit Theorems
Vt = Voc = I1 (
=
j 10)
120 /0
(
10 + j 15 j 10
j 10)
= 107:33 / 116:57 V
The next step is to find Zt . This requires deactivating the independent
voltage source of Fig. 3.119.
jj
Zt = (10 + j 15) (
=
=8
Figure 3.119
j 10)
j 10 (10
j 10
+ j 15)
+ 10 + j 15
j 14
Ω
The value of ZL for maximum average power absorbed is
Zt = 8 + j 14 Ω
The Thevenin equivalent circuit along with ZL = 8 + j 14 Ω is as shown below:
107:33 / 116:57
8 j 14 + 8 + j 14
107:33
=
/ 116:57 A
16
1
2
Pmax =
It R L
2
1 107:33 2
8
=
2
16
= 180 Walts
It =
Hence;
jj
j
237
238
j
Network Theory
EXAMPLE
3.46
(a) For the circuit shown in Fig. 3.120, what is the value of ZL that results in maximum average
power that will be transferred to ZL ? What is the maximum power ?
(b) Assume that the load resistance can be varied between 0 and 4000 Ω and the capacitive
reactance of the load can be varied between 0 and 2000 Ω. What settings of RL and XC
transfer the most average power to the load ? What is the maximum average power that can
be transferred under these conditions?
Figure 3.120
SOLUTION
(a) If there are no constraints on RL and XL , the load indepedance ZL = Zt = (3000 j 4000) Ω.
Since the voltage source is given in terms of its RM S value, the average maximum power
delivered to the load is
Pmax
RL
10 /0
3000 + j 4000 + 3000
10
=
A
2 3000
2
Pmax = It RL
100
=
3000
4 (3000)2
= 8:33 mW
jj
)
(b) Since
=
2
It =
where
XC
jj
= It
RL
and
XC
Thus,
are restricted, we firstqset
2000 Ω. Next we set RL as close to
q
RL
=
j 4000
Rt2
XC
as close to
4000 Ω as possible; hence
2
+ (XC + XL ) as possible.
30002 + ( 2000 + 4000)2 = 3605:55 Ω
Since RL can be varied between 0 to 4000 Ω, we can set
adjusted to a value
ZL = 3605:55 j 2000 Ω:
RL
to 3605:55 Ω. Hence ZL is
Circuit Theorems
10 /0
3000 + j 4000 + 3605:55
= 1:4489 / 16:85 mA
It =
j
239
j 2000
The maximum average power delivered
to the load is
Pmax
jj
= It
2
RL
= 1:4489
10
3 2
= 7:57 mW
3605 55
:
Note that this is less than the power that can be delivered if there are no constraints on
and XL .
EXAMPLE
RL
3.47
A load impedance having a constant phase angle of 45 is connected across the load terminals
a and b in the circuit shown in Fig. 3.121. The magnitude of ZL is varied until the average power
delivered, which is the maximum possible under the given restriction.
(a) Specify ZL in rectangular form.
(b) Calculate the maximum average power delivered under this condition.
Figure 3.121
SOLUTION
Since the phase angle of ZL is fixed at
that
jZ j = pjZ j
t
L
=
(3000)2 + (4000)2
= 5000 Ω:
Hence;
j j
p
ZL = ZL / 45
5000
5000
j
=
2
2
p
45 , for maximum power transfer to ZL it is mandatory
240
j
Network Theory
It =
10 /0
(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 excitation 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)
Then;
1
+2
1
)
1
R1
= (8
R2
= 1:6 + 4 = 5:6Ω
= 1:6Ω
= (5:6 1 + 4 1 ) 1 = 2:3333Ω
20
= 3:16 A
Current supplied by the source =
4 + 2:3333
4
= 1:32 A
Current in branch ab = Iab = 3:16
4 + 4 + 1: 6
8
= 1:05 A
Current in 2Ω; I1 = 1:32
10
R3
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
:
I2
=
2:1053
= 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
Putting the above three mesh equations in matrix form, we get
2
4
6:375
2
3
2
14
10
32
54
3
10
14
I1
I2
3 2
5=4
I3
0
0
10
j
243
3
5
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
4
6:375
2
3
2 3
14 10
10 14
32
54
0
I1
0
I2
3 2
5=4
0
I3
10
0
0
3
5
Using Cramer’s rule, we get
0
I3
Since I1 = I30 =
EXAMPLE
=
2A
2 A, the reciprocity theorem is verified.
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
6:537 /19:36
= 5:507 / 19:36 A
IT =
Using the principle of current division, we find that
IT ( j 5)
j5 + 3 + j2
= 6:49 / 64:36 A
Ix =
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
I0T =
36 /0
0
ZT
= 3:672 / 19:36 A
Figure 3.128
Using the principle of current division,
I0T ( j 5)
= 6:49 / 64:36 A
j5 + 2 + j3
It is found that Ix = Iy , thus verifying the reciprocity theorem.
Iy =
EXAMPLE
3.51
Refer the circuit shown in Fig. 3.129. Find current through the ammeter, and hence verify reciprocity theorem.
Circuit Theorems
j
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
4
16
1
10
1
26
20
10
20
30
32
54
I1
I2
3 2
5=4
I3
0
0
50
3
5
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
4
15
0
10
0
25
20
10
20
31
32
54
0
I1
0
I2
0
I3
3 2
5=4
50
50
0
3
5
Using Cramer’s rule we get
0
I3
= 0:8 A
Figure 3.131
245
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
4
12
0
0 2 + j 10
2
2
2
2
9
32
54
I1
I2
I3
3 2
5=4
20
20
0
3
5
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
Putting the three equations in matrix form, we get
2
4
12
0
0 2 + j 10
2
2
2
2
9
32
5 64
I01
I02
I03
3 2
75 = 4
0
0
20
j
247
3
5
Using Cramer’s rule, we get
I01 = 0:3876 / 2:35 A
I02 = 0:456 / 78:9 A
I02
Hence;
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 equivalent 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.
En
Millman’s theorem states that if n number of generators having generated emfs E1 , E2 ;
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
Y1 + Y2 + : : : + Yn
1
Z=
Y1 + Y2 + : : : + Yn
E=
where
and
where Y1 ; Y2
Yn are the admittances corresponding to the internal impedances Z1 ; Z2
and are given by
1
Z1
1
Y2 =
Z2
..
.
1
Yn =
Zn
Z
n
Y1 =
Fig. 3.134 shows a number of generators having emfs E1 ; E2
En connected in parallel
Zn are the respective internal impedances of the
across the terminals x and y . Also, Z1 ; Z2
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
En E
+
+
+
=0
Z1
Z2
Zn
1 1
E E
En
1
1
2
=E
+
+
+
+
+
+
Z1
Z2
Zn
Zn
Z1 1 Z2
+ En Yn = E
E1 Y1 + E2 Y2 +
Z
Figure 3.135
where Z = Equivalent internal impedance.
+ E Y ] = EY
E Y + E Y + + E Y
E=
Y
Y = Y + Y + + Y
1
1
Z=
=
Y
Y + Y + + Y
[E1 Y1 + E2 Y2 +
or
)
1
where
and
EXAMPLE
1
1
n
2
n
2
n
n
2
1
n
2
n
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.
E=
E1 Y1 + E2 Y2 E3 Y3
Y1 + Y2 + Y3
22
1
1
12
1
1
1
+
+
5 12 4
= 10:13 Volts
1
R=
Y1 + Y2 + Y3
1
=
0:2 + 0:083 + 0:25
= 1:88 Ω
=
5
+ 48
12
Hence;
EXAMPLE
1
4
Figure 3.137
IL
=
E
R
+ 10
= 0:853 A
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:
250
j
Network Theory
Figure 3.139
E=
E1 Y1 + E2 Y2 E3 Y3
Y1 + Y2 + Y3
1
100 /0
5
=
+ 90 /45
1
10
+ 80 /30
1
20
1
1
1
+
+
5 10 20
= 88:49 /15:66 V
Z=R=
1
=
Y1 + Y2 + Y3
1
1
5
+
1
10
+
1
20
= 2:86 Ω
I=
88:49 /15:66
E
=
= 6:7 /28:79 A
Z + ZL
2:86 + 10 j 3
E=
E1 Y1 + E2 Y2 + E3 Y3 + E4 Y4
Y1 + Y2 + Y3 + Y4
Alternately,
=
100
5
1
+ 90 45 10 1 + 80 30
5 1 + 10 1 + 20 1 + (10 j 3)
20
1
1
= 70 /12 V
Therefore;
I
=
70 /12
10 j 3
= 6:7 /28:8 A
EXAMPLE
3.55
Refer the circuit shown in Fig. 3.140. Use Millman’s theorem to find the current through (5+j 5) Ω
impedance.
Circuit Theorems
j
251
Figure 3.140
SOLUTION
The original circuit is redrawn after performing source transformation of 5 A in parallel with 4 Ω
resistor into an equivalent voltage source and is shown in Fig. 3.141.
Figure 3.141
Treating the branch 5 + j 5Ω as a branch with Es = 0V ,
E1 Y1 + E2 Y2 + E3 Y3 + E4 Y4
EP Q =
Y1 + Y2 + Y3 + Y4
4 2 1 + 8 3 1 + 20 4 1
= 1
2 + 3 1 + 4 1 + (5 j 5) 1
= 8:14 /4:83 V
Therefore current in (5 + j 5)Ω is
I=
8:14 /4:83
= 1:15 / 40:2 A
5 + j5
Alternately
EP Q with (5 + j 5) open
E1 Y1 + E2 Y2 + E3 Y3
Y1 + Y2 + Y3
4 2 1 + 8 3 1 + 20
=
2 1+3 1+4 1
= 8:9231V
EP Q =
4
1
252
j
Network Theory
Equivalent resistance R = (2 1 + 3
Therefore current in (5 + j 5)Ω is
I
EXAMPLE
=
1
+4
1) 1
= 0:9231Ω
8:9231
= 1:15 / 40:2 A
0:9231 + 5 + j 5
3.56
Find the current through 2 Ω resistor using Millman’s theorem.
in Fig. 3.142.
Refer the circuit shown
Figure 3.142
SOLUTION
The Thevenin equivalent circuit using Millman’s theorem for the given problem is as shown in
Fig. 3.142(a).
E=
where
E1 Y1 + E2 Y2
Y1 + Y2
10 /10
=
1
+ 25 /90
3 + j4
1
1
+
3 + j4 5
1
5
= 10:06 /97:12 V
1
1
Z=
=
1
1
Y1 + Y2
+
3 + j4 5
= 2:8 /26:56 Ω
Hence;
IL =
E
10:06 /97:12
=
Z + 2 2:8 /26:56 + 2
= 2:15 /81:63 A
Figure 3.142(a)
Circuit Theorems
Reinforcement problems
R.P
3.1
Find the current in 2 Ω resistor connected between A and B by using superposition theorem.
Figure R.P. 3.1
SOLUTION
Fig. R.P. 3.1(a), shows the circuit with 2V-source acting alone (4V-source is shorted).
Resistance as viewed from 2V-source is 2 + R1 Ω,
where
Hence;
Then;
∴
R1
=
3 2
+1 12
5
(1:2 + 1) 12
= 1:8592 Ω
=
14:2
2
Ia =
= 0:5182 A
2 + 1:8592
12
Ib = Ia
= 0:438 A
12 + 1 + 1:2
3
I1 = 0:438
= 0:2628 A
5
Figure R.P. 3.1(a)
With 4V-source acting alone, the circuit is as shown in Fig. R.P. 3.1(b).
Figure R.P.3.1(b)
j
253
254
j
Network Theory
The resistance as seen by 4V-source is 3 + R2 where
R2
=
2 12
+1 2
14
2:7143 2
= 1:1551 Ω
=
4:7143
4
Hence;
Ib =
= 0:9635 A
3 + 1:1551
Ib 2:7143
= 0:555 A
Thus;
I2 =
4:7143
Finally, applying the principle of superposition,
we get,
IAB = I1 + I2
= 0:2628 + 0:555
= 0:818 A
R.P
3.2
For the network shown in Fig. R.P. 3.2, apply superposition theorem and find the current I.
Figure R.P. 3.2
SOLUTION
Open the 5A-current source and retain the voltage source. The resulting network is as shown in
Fig. R.P. 3.2(a).
Figure R.P. 3.2(a)
Circuit Theorems
j
255
The impedance as seen from the voltage source is
Z = (4
j 2)
+
(8 + j 10) (
8 + j8
j 2)
Ia =
= 6:01 / 45
Ω
j 20
= 3:328 /135 A
Z
Next, short the voltage source and retain the current source. The resulting network is as shown
in Fig. R.P. 3.2 (b).
Here, I3 = 5A. Applying KVL for mesh 1 and mesh 2, we
get
Hence;
8I1 + (I1
and
(I2
I1 ) (
I2 ) (
5) j 10 + (I1
j 2)
+ (I2
5) (
j 2)
j 2)
=0
+ 4I2 = 0
Simplifying, we get
(8 + j 8)I1 + j 2I2 = j 50
and
j 2I1
+ (4
j 4)I2
=
j 10
Solving, we get
8 + j 8
j2
Ib = I2 = 8 + j 8
j 50
j 10
j2
4 j4
= 2:897 / 23:96 A
j2
Figure R.P. 3.2(b)
Since, Ia and Ib are flowing in opposite directions, we
have
I = Ia Ib = 6:1121 /144:78 A
R.P
3.3
Apply superposition theorem and find the voltage across 1 Ω resistor. Refer the circuit shown in
Fig. R.P. 3.3. Take v1 (t) = 5 cos (t + 10 ) and i2 (t) = 3 sin 2t A.
Figure R.P. 3.3
256
j
Network Theory
SOLUTION
To begin with let us assume v1 (t) alone is acting. Accordingly, short 10V - source and open i2 (t).
The resulting phasor network is shown in Fig. R.P. 3.3(a).
!
= 1rad=sec
! 5 /10
= 1H !
1
= 1F !
1
= H!
2
1
1
= F!
2
5 cos (t + 10 )
L1
C1
L2
C2
∴
V
= j1 Ω
j !L1
j !C1
j !L2
j !C2
=
j1
=j
1
Ω
2
=
j2
va (t)
Let us next assume that
in Fig. R.P. 3.3(b).
!
i2 (t)
C2
L2
)
= 5 cos [t + 10 ]
alone is acting.
The resulting network is shown
= 2 rad=sec
! 3 /0
1
= 1F !
= 1H !
1
1
= F!
2
1
= H!
2
3 sin 2t
L1
Figure R.P. 3.3(a)
Ω
Va = 5 /10 V
)
C1
Ω
A
1
Ω
j !C1
2
j !L1 = j 2 Ω
j !C2
j !L2
=
j
=
j1
Figure R.P. 3.3(b)
= j1 Ω
Vb = 3 /0
vb (t)
Ω
1 +1 51 5 = 2 5 /33 7
j :
j :
:
:
A
= 2:5 sin [2t + 33:7 ] A
Finally with 10V-source acting alone, the network is as shown in Fig. R.P. 3.3(c). Since
= 0, inductors are shorted and capacitors are opened.
Hence, Vc = 10 V
Applying principle of superposition, we
get.
!
v2 (t)
= va (t) = vb (t) + Vc
= 5 cos (t + 10 ) + 2:5 sin (2t + 33:7 ) + 10Volts
Figure R.P. 3.3(c)
Circuit Theorems
R.P
j
257
3.4
Calculate the current through the galvanometer for the Kelvin double bridge shown in Fig. R.P.
3.4. Use Thevenin’s theorem. Take the resistance of the galvanometer as 30 Ω.
Figure R.P. 3.4
SOLUTION
With G being open, the resulting network is as shown in Fig. R.P. 3.4(a).
Figure 3.4(a)
VA
I2
10
100 = 450
100 = 209 V
5 =01
10
=
=
= 1 66
45 5
45 + 5
15+
50
= 05+
10
= I1
:
;
IB
I2
:
Hence;
VB
I2
:
IB
= 2:5 V
Thus;
VAB
= Vt = VA
VB
=
20
9
2:5 =
5
Volts
18
: I2
258
j
Network Theory
To find Rt , short circuit the voltage source. The resulting network is as shown in Fig. R.P. 3.4(b).
Figure R.P. 3.4 (b)
Transforming the Δ between B , E and F into an equivalent Y , we get
RB
=
35
10 = 7 Ω
50
;
RE
=
35 5
= 3:5 Ω;
50
RF
=
5
10 = 1 Ω
50
The reduced network after transformation is as shown in Fig. R.P. 3.4(c).
Figure R.P. 3.4(c)
Hence;
RAB
350 100 4:5 1:5
+
+7
450
6
= 85:903 Ω
= Rt =
The Thevenin’s equivalent circuit as seen from A
and B with 30 Ω connected between A and B is
as shown in Fig. R.P. 3.4(d).
5
18
IG =
= 2:4mA
85:903 + 30
Negative sign implies that the current flows from
B to A.
R.P
Figure R.P. 3.4(d)
3.5
Find Is and R so that the networks N1 and N2 shown in Fig. R.P. 3.5 are equivalent.
Circuit Theorems
j
259
Figure R.P. 3.5
SOLUTION
Transforming the current source in N1 into an equivalent voltage source, we get N3 as shown in
Fig. R.P. 3.5(a).
V
I R = IS R
(3.28)
From N3 , we can write,
From N2 we can write,
Also from N2 ,
I
V
)
)
)
V
V
V
=
10Ia
3=
2Ia
3=
3=
I
5
2
I
10
I
5
=3
(3.29)
For equivalence of N1 and N2 , it is requirred that equations
(3.28) and (3.29) must be same. Comparing these equations, we
get
IR
R
R.P
=
I
5
= 0:2 Ω
and
and
IS R
IS
=3
=
3
= 15A
0:2
3.6
Obtain the Norton’s equivalent of the network shown in Fig. R.P. 3.6.
Figure R.P. 3.6
Figure R.P. 3.5(a)
260
j
Network Theory
SOLUTION
Terminals a and b are shorted. This results in a network as shown in Fig. R.P. 3.6(a)
Figure R.P. 3.6(a)
The mesh equations are
9I1 + 0I2
6I3 = 30
(3.30)
(ii)
0I1 + 25I2 + 15I3 = 30
(3.31)
(iii)
6I1 + 15I2 + 23I3 = 4VX = 4 (10I2 )
(i)
)
6I1
25I2 + 23I3 = 0
Solving equations (3.30), (3.31) and (3.32), we get
IN
= Isc = I3 = 1:4706A
With terminals ab open, I3 = 0. The corresponding equations are
Hence;
Then;
Hence;
9I1 = 30
30
A
I1 =
9
25I2 = 50
30
and I2 =
A
25
30
VX = 10I2 = 10
= 12 V
25
6I1 4VX
Vt = Voc = 15I2
=
Thus;
Rt
=
and
50 V
Voc
Isc
=
50
=
1:4706
34 Ω
Hence, Norton’s equivalent circuit is as shown in Fig. R.P. 3.6(b).
Figure R.P. 3.6(b)
(3.32)
Circuit Theorems
R.P
j
261
3.7
For the network shown in Fig. R.P. 3.7, find the Thevenin’s equivalent to show that
and
Vt
=
Zt
=
V1
2
3
(1 + a + b
ab)
b
2
Figure R.P. 3.7
SOLUTION
With xy open, I1 =
Hence,
Voc
aV1
V1
2
= Vt = aV1 + I1 + bI1
= aV1 +
=
V1
2
aV1
V1
2
[1 + a + b
+b
V
aV1
1
2
ab]
With xy shorted, the resulting network is
as shown in Fig. R.P. 3.7(a).
Figure R.P. 3.7(a)
Applying KVL equations, we get
(i)
(ii)
)
)
I1
+ (I1
2I1
(I2
I1 )
I2 )
= V1
aV1
I2
= V1
aV1
(3.33)
+ I2 = aV1 + bI1
(3.34)
(1 + b) I1 + 2I2 = aV1
Solving equations (3.33) and (3.34), we get
Isc
= I2 =
V1
(1 + a + b
3 b
ab)
262
j
Network Theory
Hence;
Zt
=
=
R.P
Voc
Isc
3
=
(1 + a + b
2 V1 (1 + a + b
V1
ab)
ab)
(3
b)
b
2
3.8
Use Norton’s theorem to determine
are in ohms.
I
in the network shown in Fig. R.P. 3.8. Resistance Values
Figure R.P. 3.8
SOLUTION
Let IAE = x and IEF = y . Then by applying KCL at various junctions, the branch currents are
marked as shown in Fig. R.P. 3.8(a). Isc = 125 x = IAB on shorting A and B .
Applying KVL to the loop ABC F EA, we get
0:04x + 0:01y + 0:02 (y
)
20) + 0:03 (x
105) = 0
0:07x + 0:03y = 3:55
(3.35)
Applying KVL to the loop EDC EF , we get
(x
y
30) 0:03 + (x
)
y
55) 0:02
(y
20) 0:02
0:05x
0:01y = 0
0:08y = 1:6
(3.36)
Circuit Theorems
j
263
Figure R.P. 3.8(a)
Solving equations (3.35) and (3.36), we get
x
Hence;
Isc
= 46:76 A
= IN = 120
x
= 78:24 A
The circuit to calculate
opened.
Rt
is as shown in Fig. R.P. 3.8(b). All injected currents have been
Rt
= 0:03 + 0:04 +
0:03 0:05
0:08
= 0:08875 Ω
Figure R.P. 3.8(b)
Figure R.P. 3.8(c)
264
j
Network Theory
The Norton’s equivalent network is as shown in Fig. R.P. 3.8(c).
I
= 78:24
0 08875
0 08875
+ 0 04
:
:
:
= 53:9A
R.P
3.9
For the circuit shown in Fig. R.P. 3.9, find R such that the maximum power delivered to the load
is 3 mW.
Figure R.P. 3.9
SOLUTION
For a resistive network, the maximum power delivered to the load is
2
Pmax
=
Vt
4Rt
The network with RL removed is as shown in Fig.
R.P. 3.9(a).
Let the opent circuit voltage between the terminals a and b be Vt .
Then, applying KCL at node a, we get
1
Vt
R
+
2
Vt
R
+
Simplifying we get
Vt = 2 Volts
With all voltage sources shorted, the resistance,
found as follows:
1
)
Rt
Rt
=
=
1
R
R
3
+
Ω
1
R
3
Vt
R
Rt
+
Figure R.P. 3.9(a)
=0
as viewed from the terminals,
1
R
=
3
R
a
and b is
Circuit Theorems
22
3
=
= 3 × 10−3
R
R
4× 3
R = 1 kΩ
Hence,
Pmax =
⇒
R.P
| 265
3.10
Refer Fig. R.P. 3.10, find X1 and X2 interms of R1 and R2 to give maximum power dissipation
in R2 .
Figure R.P. 3.10
SOLUTION
The circuit for finding Zt is as shown in Figure R.P. 3.10(a).
Zt =
=
R1 (jX1 )
R1 + jX1
R1 X12 + jR12 X1
R12 + X12
Figure R.P. 3.10(a)
For maximum power transfer,
ZL = Z∗t
⇒
Hence,
R2 + jX2 =
R2 =
⇒
R1 X12
R12 X1
−
j
R12 + X12
R12 + X12
R1 X12
R12 + X12
X1 = ±R1
X2 = −
R2
R1 − R2
R12 X1
R12 + X12
Substituting equation (3.37) in equation (3.38) and simplifying, we get
X2 = R2 (R1 − R2 )
(3.37)
(3.38)
266
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Network Theory
Exercise Problems
E.P
3.1
Find ix for the circuit shown in Fig. E.P. 3.1 by using principle of superposition.
Ans : ix =
E.P
1
A
4
Figure E.P. 3.1
3.2
Find the current through branch P Q using superposition theorem.
Figure E.P. 3.2
Ans :
E.P
1.0625 A
3.3
Find the current through 15 ohm resistor using superposition theorem.
Figure E.P. 3.3
Ans :
0.3826 A
Circuit Theorems
E.P
3.4
Find the current through 3 + j 4 Ω using superposition theorem.
Figure E.P. 3.4
Ans :
E.P
8.3 /85.3 A
3.5
Find the current through Ix using superposition theorem.
Figure E.P. 3.5
Ans :
E.P
3.07 / 163.12 A
3.6
Determine the current through 1 Ω resistor using superposition theorem.
Figure E.P. 3.6
Ans :
0.406 A
j
267
268
j
E.P
Network Theory
3.7
Obtain the Thevenin equivalent circuit at terminals a
b
of the network shown in Fig. E.P. 3.7.
Figure E.P. 3.7
Ans :
E.P
Vt = 6.29 V, Rt = 9.43 Ω
3.8
Find the Thevenin equivalent circuit at terminals x
y
of the circuit shown in Fig. E.P. 3.8.
Figure E.P. 3.8
Ans :
E.P
Vt = 0.192 / 43.4 V, Zt = 88.7 /11.55 Ω
3.9
Find the Thevenin equivalent of the network shown in Fig. E.P. 3.9.
Figure E.P. 3.9
Ans :
Vt = 17.14 volts, Rt = 4 Ω
Circuit Theorems
E.P
3.10
Find the Thevenin equivalent circuit across a
b.
Refer Fig. E.P. 3.10.
Figure E.P. 3.10
Ans :
E.P
Vt =
30 V, Rt = 10 kΩ
3.11
Find the Thevenin equivalent circuit across a
b
for the network shown in Fig. E.P. 3.11.
Figure E.P. 3.11
Ans :
E.P
Verify your result with other methods.
3.12
Find the current through 20 ohm resistor using Norton equivalent.
Figure E.P. 3.12
Ans :
IN = 4.36 A, RN = Rt = 8.8 Ω, IL = 1.33 A
j
269
270
j
E.P
Network Theory
3.13
Find the current in 10 ohm resistor using Norton’s theorem.
Figure E.P. 3.13
Ans :
E.P
IN =
4 A, Rt = RN
100
=
Ω, IL =
7
0.5 A
3.14
Find the Norton equivalent circuit between the terminals a
b for the network shown in
Fig. E.P. 3.14.
Figure E.P. 3.14
Ans :
E.P
IN
= 4.98310 / 5.71 A, Zt = ZN = 3.6 /23.1 Ω
3.15
Determine the Norton equivalent circuit across the terminals
Fig. E.P. 3.15.
Figure E.P. 3.15
Ans :
IN = 5 A, RN = Rt = 6 Ω
P
Q
for the network shown in
Circuit Theorems
E.P
j
271
3.16
Find the Norton equivalent of the network shown in Fig. E.P. 3.16.
Figure E.P. 3.16
Ans :
E.P
IN = 8.87 A, RN = Rt = 43.89 Ω
3.17
Determine the value of RL for maximum power transfer and also find the maximum power transferred.
Figure E.P. 3.17
Ans :
E.P
RL = 1.92 Ω, Pmax = 4.67 W
3.18
Calculate the value of ZL for maximum power transfer and also calculate the maximum power.
Figure E.P. 3.18
Ans :
ZL = (7.97 + j2.16)Ω, Pmax = 0.36 W
272
j
E.P
Network Theory
3.19
Determine the value of RL for maximum power transfer and also calculate the value of maximum
power.
Figure E.P. 3.19
Ans :
E.P
RL = 5.44 Ω, Pmax = 2.94 W
3.20
Determine the value of ZL for maximum power transfer. What is the value of maximum power?
Figure E.P. 3.20
Ans :
E.P
ZL = 4.23 + j1.15 Ω, Pmax = 5.68 Watts
3.21
Obtain the Norton equivalent across x
y.
Figure E.P. 3.21
Ans :
E.P
IN = ISC = 7.35A, Rt = RN = 1.52 Ω
3.22
Find the Norton equivalent circuit at terminals a
b
of the network shown in Fig. E.P. 3.22.
Circuit Theorems
j
273
Figure E.P. 3.22
Ans :
E.P
IN = 1.05 /251.6 A, Zt = ZN = 10.6 /45 Ω
3.23
Find the Norton equivalent across the terminals X
Y
of the network shown in Fig. E.P. 3.23.
Figure E.P. 3.23
Ans :
E.P
IN = 7A, Zt = 8.19 / 55 Ω
3.24
Determine the current through 10 ohm resistor using Norton’s theorem.
Figure E.P. 3.24
Ans :
0.15A
274
j
E.P
Network Theory
3.25
Determine the current I using Norton’s theorem.
Figure E.P. 3.25
Ans :
E.P
Verify your result with other methods.
3.26
Find Vx in the circuit shown in Fig. E.P. 3.26 and hence verify reciprocity theorem.
Figure E.P. 3.26
Ans :
E.P
Vx = 9.28 /21.81 V
3.27
Find Vx in the circuit shown in Fig. E.P. 3.27 and hence verify reciprocity theorem.
Figure E.P. 3.27
Ans :
Vx = 10.23 Volts
Circuit Theorems
E.P
3.28
Find the current ix in the bridge circuit and hence verify reciprocity theorem.
Figure E.P. 3.28
Ans :
E.P
ix = 0.031 A
3.29
Find the current through 4 ohm resistor using Millman’s theorem.
Figure E.P. 3.29
Ans :
E.P
I = 2.05 A
3.30
Find the current through the impedance of (10 + j 10) Ω using Millman’s theorem.
Figure E.P. 3.30
Ans :
3.384 /12.6 A
j
275
276
j
E.P
Network Theory
3.31
Using Millman’s theorem, find the current flowing through the impedance of (4 + j 3) Ω.
Figure E.P. 3.31
Ans :
3.64 / 15.23 A
4.1
Introduction
There are many reasons for studying initial and final conditions. The most important reason is that
the initial and final conditions evaluate the arbitrary constants that appear in the general solution
of a differential equation.
In this chapter, we concentrate on finding the change in selected variables in a circuit when
a switch is thrown open from closed position or vice versa. The time of throwing the switch is
considered to be t = 0, and we want to determine the value of the variable at t = 0 and at
+
t = 0 , immediately before and after throwing the switch. Thus a switched circuit is an electrical
circuit with one or more switches that open or close at time t = 0. We are very much interested
in the change in currents and voltages of energy storing elements after the switch is thrown since
these variables along with the sources will dictate the circuit behaviour for t > 0.
Initial conditions in a network depend on the past history of the circuit (before t = 0 ) and
structure of the network at t = 0+ , (after switching). Past history will show up in the form of
capacitor voltages and inductor currents. The computation of all voltages and currents and their
derivatives at t = 0+ is the main aim of this chapter.
4.2
Initial and final conditions in elements
4.2.1 The inductor
The switch is closed at t = 0. Hence t = 0 corresponds
to the instant when the switch is just open and t = 0+
corresponds to the instant when the switch is just closed.
The expression for current through the
inductor is given by
i
=
1
L
Zt
Figure 4.1 Circuit for explaining
vd
switching action of an inductor
278
j
Network Theory
i
i(t)
=
1
L
Z0
vd
= i(0 ) +
1
L
+
1
L
Zt
Zt
vd
0
vd
0
Putting t = 0+ on both sides, we get
i(0
+
i(0
+
) = i(0 ) +
1
L
Z0+
vd
0
) = i(0 )
The above equation means that the current in an inductor cannot change instantaneously.
Consequently, if i(0 ) = 0, we get i(0+ ) = 0. This means that at t = 0+ , inductor will act
as an open circuit, irrespective of the voltage across the terminals. If i(0 ) = Io , then i(0+ ) = Io .
In this case at t = 0+ , the inductor can be thought of as a current source of Io A. The equivalent
circuits of an inductor at t = 0+ is shown in Fig. 4.2.
t = 0+
Figure 4.2 The initial-condition equivalent circuits of an inductor
The final-condition equivalent circuit of an inductor is derived from the basic relationship
v
=L
di
dt
di
Under steady condition,
= 0. This means, v = 0 and hence L acts as short at t =
(final
dt
or steady state). The final-condition equivalent circuits of an inductor is shown in Fig.4.3.
t=¥
Io
Io
Figure 4.3 The final-condition equivalent circuit of an inductor
Initial Conditions in Network
j
279
4.2.2 The capacitor
The switch is closed at t = 0. Hence, t = 0
corresponds to the instant when the switch is
just open and t = 0+ corresponds to the instant
when the switch is just closed. The expression
for voltage across the capacitor is given by
=
v
v (t)
v (t)
=
1
C
1
C
Zt
Z0
id
= v (0 ) +
+
switching action of a Capacitor
Evaluating the expression at t =
v (0
Figure 4.4 Circuit for explaining
id
+
1
C
0+ ,
) = v (0 ) +
1
C
Z
Zt
id
0
t
id
0
we get
1
C
Z0+
id
v (0
+
) = v (0 )
0
Thus the voltage across a capacitor cannot change instantaneously.
If v (0 ) = 0, then v (0+ ) = 0. This means that at t = 0+ , capacitor C acts as short circuit.
q0
q0
Conversely, if v (0 ) =
then v (0+ ) = . These conclusions are summarized in Fig. 4.5.
C
C
Figure 4.5 Initial-condition equivalent circuits of a capacitor
The final–condition equivalent network is derived from the basic relationship
i
dv
=C
dv
dt
Under steady state condition,
= 0. This is, at t = , i = 0. This means that t =
dt
or in steady state, capacitor C acts as an open circuit. The final condition equivalent circuits of a
capacitor is shown in Fig. 4.6.
280
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Network Theory
Figure 4.6 Final-condition equivalent circuits of a capacitor
4.2.3 The resistor
The cause–effect relation for an ideal resistor is given by v = Ri. From this equation, we find that
the current through a resistor will change instantaneously if the voltage changes instantaneously.
Similarly, voltage will change instantaneously if current changes instantaneously.
4.3
Procedure for evaluating initial conditions
There is no unique procedure that must be followed in solving for initial conditions. We usually
solve for initial values of currents and voltages and then solve for the derivatives. For finding
initial values of currents and voltages, an equivalent network of the original network at t = 0+ is
constructed according to the following rules:
(1) Replace all inductors with open circuit or with current sources having the value of current
flowing at t = 0+ .
(2) Replace all capacitors with short circuits or with a voltage source of value vo =
is an initial charge.
q0
C
if there
(3) Resistors are left in the network without any changes.
EXAMPLE
4.1
Refer the circuit shown in Fig. 4.7(a). Find
for t < 0.
i1 (0
+)
and
iL (0
+ ).
The circuit is in steady state
=
=
t=0
Figure 4.7(a)
1W
1W
Initial Conditions in Network
j
281
SOLUTION
The symbol for the switch implies
that it is open at t = 0 and then
closed at t = 0+ . The circuit is
in steady state with the switch open.
This means that at t = 0 , inductor L is short. Fig.4.7(b) shows the
original circuit at t = 0 .
Using the current division principle,
–
–
1W
1W
Figure 4.7(b)
2 1
= 1A
iL (0 ) =
1+1
Since the current in an inductor cannot change instantaneously, we have
iL (0
+
) = iL (0 ) = 1A
At t = 0 , i1 (0 ) = 2 1 = 1A. Please note that the current in a resistor can change
instantaneously. Since at t = 0+ , the switch is just closed, the voltage across R1 will be equal to
zero because of the switch being short circuited and hence,
i1 (0
+
) = 0A
Thus, the current in the resistor changes abruptly form 1A to 0A.
EXAMPLE
4.2
Refer the circuit shown in Fig. 4.8. Find vC (0+ ). Assume that the switch was in closed state for
a long time.
Figure 4.8
SOLUTION
The symbol for the switch implies that it is closed at t = 0 and then opens at t = 0+ . Since the
circuit is in steady state with the switch closed, the capacitor is represented as an open circuit at
t = 0 . The equivalent circuit at t = 0 is as shown in Fig. 4.9.
282
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Network Theory
vC (0
) = i(0 )R2
Using the principle of voltage divider,
vC (0
)=
VS
R1
+ R2
R2
=
5 1
= 2:5 V
1+1
Since the voltage across a capacitor cannot
change instaneously, we have
vC (0
Figure 4.9
+
) = vC (0 ) = 2:5V
That is, when the switch is opened at t = 0, and if the source is removed from the circuit, still
+
vC (0 ) remains at 2.5 V.
EXAMPLE
4.3
Refer the circuit shown in Fig 4.10. Find iL (0+ ) and vC (0+ ). The circuit is in steady state with
the switch in closed condition.
Figure 4.10
SOLUTION
The symbol for the switch implies, it is closed
at t = 0 and then opens at t = 0+ . In order
to find vC (0 ) and iL (0 ) we replace the capacitor by an open circuit and the inductor by
a short circuit, as shown in Fig.4.11, because
in the steady state L acts as a short circuit and
C as an open circuit.
Figure 4.11
5
=1A
2+3
Using the voltage divider principle, we note that
iL (0
)=
vC (0
Then we note that:
vC (0
)=
+
iL (0
+
5 3
=3V
3+2
) = vC (0 ) = 3 V
) = iL (0 ) = 2 A
Initial Conditions in Network
EXAMPLE
j
283
4.4
In the given network, K is closed at t = 0 with
zero current in the inductor. Find the values
of i;
di
dt
,
2
d i
dt2
at t = 0+ if R = 8Ω and L = 0:2H.
Refer the Fig. 4.12(a).
SOLUTION
Figure 4.12(a)
The symbol for the switch implies that it is open
at t = 0 and then closes at t = 0+ . Since the
current i through the inductor at t = 0 is zero, it
implies that i(0+ ) = i(0 ) = 0.
To find
+
di(0 )
dt
and
2
+
d i(0 )
dt2
R = 8W
+
–
:
Applying KVL clockwise to the circuit shown in
Fig. 4.12(b), we get
+L
Ri
8i + 0:2
di
dt
di
dt
Figure 4.12(b)
= 12
= 12
At t = 0+ , the equation (4.1) becomes
8i(0+ ) + 0:2
8
0+02
:
di(0
+)
dt
+
di(0
dt
+
di(0
)
)
dt
= 12
= 12
12
0:2
= 60 A=sec
=
Differentiating equation (4.1) with respect to t, we get
8
di
dt
2
d i
dt2
=0
2
+
d i(0 )
dt2
=0
+ 0:2
At t = 0+ , the above equation becomes
8
Hence
di(0
+)
dt
8
+ 0:2
60 + 0 2
:
2
+
d i(0 )
dt2
2
+
d i(0 )
dt2
L = 0.2H
=0
= – 2400 A=sec2
(4.1)
284
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Network Theory
EXAMPLE
4.5
In the network shown in Fig. 4.13, the switch is closed at t = 0. Determine i;
di
dt
,
2
d i
dt2
at t = 0+ .
Figure 4.13
SOLUTION
The symbol for the switch implies that it is open at t = 0 and then closes at t = 0+ . Since there
is no current through the inductor at t = 0 , it implies that i(0+ ) = i(0 ) = 0.
R = 10W
L = 1H
C = 1mF
Figure 4.14
Writing KVL clockwise for the circuit shown in Fig. 4.14, we get
Ri
+L
di
dt
Ri
+
1
Zt
= 10
(4.2)
+ vC (t) = 10
(4.2a)
i( )d
C
+L
0
di
dt
Putting t = 0+ in equation (4.2a), we get
Ri
+
0
+L
R
0+
di
+ vC 0+ = 10
dt
0+
L
0+
di
+ 0 = 10
dt
+
0
di
dt
=
10
L
= 10 A/sec
Differentiating equation (4.2) with respect to t, we get
R
di
dt
+L
2
d i
dt2
+
i(t)
C
=0
Initial Conditions in Network
At t = 0+ , the above equation becomes
R
EXAMPLE
t
dt
Hence at
0+
di
R
+L
2
d i
10 +
100 +
0+
+
i
0+
2
d i
0+
dt2
=0
dt2
=0
+
0
285
dt2
C
2
+
d i 0
0
+
L
2
dt
C
2
d i
= 0+ ;
j
=0
=
100 A/sec2
4.6
Refer the circuit shown in Fig. 4.15. The switch
K is changed from position 1 to position 2 at
t = 0. Steady-state condition having been
reached at position 1. Find the values of i,
and
2
d i
dt2
di
dt
,
at t = 0+ .
SOLUTION
Figure 4.15
The symbol for switch K implies that it is in position 1 at t = 0 and in position 2 at t = 0+ .
Under steady-state condition, inductor acts as a short circuit. Hence at t = 0 , the circuit diagram
is as shown in Fig. 4.16.
20
= 2A
10
Since the current through
an inductor cannot change
instantaneously, i 0+ = i (0 ) = 2A. Since there is
no initial charge on the capacitor, vC (0 ) = 0. Since
the voltage across
a capacitor cannot change instanta+
neously, vC 0 = vC (0 ) = 0. Hence at t = 0+
the circuit diagram is as shown in Fig. 4.17(a).
For t 0+ , the circuit diagram is as shown in Fig. 4.17(b).
i
0
=
Figure 4.17(a)
Figure 4.16
Figure 4.17(b)
286
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Network Theory
Applying KVL clockwise to the circuit shown in Fig. 4.17(b), we get
Ri(t)
+L
di(t)
+
dt
Ri(t)
+L
Zt
1
C
i( )d
=0
(4.3)
+ vC (t) = 0
(4.3a)
0+
di(t)
dt
At t = 0+ equation (4.3a) becomes
+
0
Ri
+L
R
0+
di
dt
2+
L
+ vC 0+ = 0
0+
di
+0=0
dt
20 +
0+
di
dt
+
0
di
=0
=
dt
20 A/sec
Differentiating equation (4.3) with respect to t, we get
R
At t = 0+ , we get
R
Hence;
EXAMPLE
0+
di
dt
R
(
+L
di
dt
2
d i
20) +
+L
0+
2
d i
dt2
+
+
i
i
C
0+
dt2
C
2
+
d i 0
2
+
L
2
dt
C
2
+
d i 0
dt2
=0
=0
=0
2 106 A=sec2
4.7
In the network shown in Fig. 4.18, the switch is moved from position 1 to position 2 at t = 0. The
steady-state has been reached before switching. Calculate i,
Figure 4.18
di
dt
, and
2
d i
dt2
at t = 0+ .
Initial Conditions in Network
j
287
SOLUTION
The symbol for switch K implies that it is in position 1 at t = 0 and in position 2 at t = 0+ . Under
steady-state condition, a capacitor acts as an open circuit. Hence at t = 0 , the circuit diagram is as shown
in Fig. 4.18(a).
We know that the voltage across a capacitor
cannot change instantaneously. This means that
Figure 4.18(a)
+ = v (0 ) = 40 V.
vC 0
C
At t = 0 , inductor is not energized.
This means that i (0 ) = 0. Since current in an inductor
cannot change instantaneously, i 0+ = i (0 ) = 0. Hence, the circuit diagram at t = 0+ is as
shown in Fig. 4.18(b).
The circuit diagram for t 0+ is as shown in Fig.4.18(c).
Figure 4.18(b)
Figure 4.18(c)
Applying KVL clockwise, we get
Ri
+L
di
dt
Ri
+
1
C
+L
Zt
i( )d
=0
(4.4)
0+
di
+ vC (t) = 0
dt
At t = 0+ , we get
Ri(0
+
)+L
20
di(0
+)
dt
0+1
+ vC (0+ ) = 0
di(0
+)
dt
+ 40 = 0
+
di(0 )
dt
=
40A/ sec
Diferentiating equation (4.4) with respect to t, we get
R
di
dt
+L
2
d i
dt2
+
i
C
=0
288
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Network Theory
Putting t = 0+ in the above equation, we get
R
+
di(0 )
dt
R
(
+L
2
+
d i(0 )
dt2
40) + L
+
i(0
+)
C
2
+
d i(0 )
dt2
+
0
C
+
2
d i(0
dt2
Hence
EXAMPLE
)
=0
=0
= 800A/ sec2
4.8
t
In the network shown in Fig. 4.19, v1 (t) = e
is initially uncharged, determine the value of
0 and is zero for all
for t
2
d v
2
dt2
and
3
d v
2
dt3
t<
0. If the capacitor
at t = 0+ .
Figure 4.19
SOLUTION
Since the capacitor is initially uncharged,
+
v2 (0 ) = 0
Referring to Fig. 4.19(a) and applying KCL
at node v2 (t):
v2 (t)
v1 (t)
R1
1
R1
+
+C
dv2 (t)
dt
1
R2
v2 (t)
+
+C
v2 (t)
R2
dv2 (t)
dt
=0
=
v1 (t)
R1
0:15v2 + 0:05
Figure 4.19(a)
dv2
dt
= 0:1e
t
Putting t = 0+ , we get
0:15v2 (0+ ) + 0:05
0:15
0 + 0 05
:
dv2 (0
dt
dv2 (0
+)
+)
dt
dv2 (0
dt
+)
= 0:1
= 0:1
=
0:1
= 2 Volts= sec
0:05
(4.5)
Initial Conditions in Network
j
289
Differentiating equation (4.5) with respect to t, we get
0:15
dv2
dt
+ 0:05
2
d v2
dt2
=
t
0:1e
(4.6)
Putting t = 0+ in equation (4.6), we find that
2
+
d v2 (0 )
dt2
0:1 0:3
= 8 Volts/ sec2
0:05
Again differentiating equation (4.6) with respect to t, we get
0:15
=
2
d v2
dt2
+ 0:05
3
d v2
dt3
= 0:1e
Putting t = 0+ in equation (4.7) and solving for
3
+
d v2 (0 )
dt3
EXAMPLE
=
t
3
d v2 +
(0 ),
dt3
(4.7)
we find that
0:1 + 1:2
= 26 Volts/ sec3
0:05
4.9
Refer the circuit shown in Fig. 4.20. The circuit is in steady state with switch K closed. At t = 0,
the switch is opened. Determine the voltage across the switch, vK and
dvK
dt
at t = 0+ .
Figure 4.20
SOLUTION
The switch remains closed at t = 0 and open at t = 0+ . Under steady condition, inductor acts
as a short circuit and hence the circuit diagram at t = 0 is as shown in Fig. 4.21(a).
Therefore;
For t
0
+
vK (0
+
) = vK (0 )
=0V
the circuit diagram is as shown in Fig. 4.21(b).
Figure 4.21(a)
Figure 4.21(b)
290
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Network Theory
i(t)
At (t) =
0+ ,
=C
dvK
dt
we get
i(0
+
)=C
dvK (0
+)
dt
Since the current through an inductor cannot change instantaneously, we get
i(0
Hence;
dvK (0
+
) = i(0 ) = 2A
+
dvK (0 )
2=C
+)
dt
EXAMPLE
dt
=
2
C
=
2
1
2
= 4V/ sec
4.10
In the given network, the switch K is opened at t = 0. At t = 0+ , solve for the values of v;
and
2
d v
dt2
dv
dt
if I = 2 A; R = 200 Ω and L = 1 H
Figure 4.22
SOLUTION
The switch is opened at t = 0. This means that at t = 0 , it is closed and at t = 0+ , it is open.
Since iL (0 ) = 0, we get iL (0+ ) = 0. The circuit at t = 0+ is as shown in Fig. 4.23(a).
Figure 4.23(a)
Figure 4.23(b)
v (0
+
) = IR
=2
200
= 400 Volts
Initial Conditions in Network
Refer to the circuit shown in Fig. 4.23(b).
For t 0+ , the KCL at node v (t) gives
I
=
v (t)
+
R
(4.8)
v ( )d
0+
Differentiating both sides of equation (4.8) with respect to t, we get
1 dv (t)
1
0=
+ v (t)
R
At t = 0+ , we get
1
R
dv (0
+)
dt
+
dt
+
1
L
1 dv (0 ) 1
+
200 dt
1
L
v (0
291
Zt
1
L
j
+
(4.8a)
)=0
400 = 0
dv (0
+)
dt
=
8 104 V/ sec
Again differentiating equation (4.8a), we get
2
d v (t)
R dt2
1
At t = 0+ , we get
EXAMPLE
+
1
dv (t)
L
dt
=0
1 d2 v (0+ ) 1 dv (0+ )
+
=0
200 dt2
1 dt
2
+
d v (0 )
= 200
2
dt
8 10
4
= 16 106 V/ sec2
4.11
In the circuit shown in Fig. 4.24, a steady state is reached with switch K open. At t = 0, the
switch is closed. For element values given, determine the values of va (0 ) and va (0+ ).
Figure 4.24
292
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Network Theory
SOLUTION
At t = 0 , the switch is open and at t = 0+ , the switch is closed. Under steady conditions,
inductor L acts as a short circuit. Also the steady state is reached with switch K open. Hence, the
circuit diagram at t = 0 is as shown in Fig.4.25(a).
iL (0
)=
5
5
2
+
= A
30 10
3
Using the voltage divider principle:
va (0
)=
5 20
10
=
V
10 + 20
3
Since the current in an inductor cannot change instantaneously,
iL (0
+
) = iL (0 ) =
2
A:
3
At t = 0+ , the circuit diagram is as shown in Fig. 4.25(b).
Figure 4.25(a)
Figure 4.25(b)
Refer the circuit in Fig. 4.25(b).
KCL at node a:
va (0
+)
5
+
va (0
+)
+
va (0
10 10
1
1
1
+
+
+
va (0 )
10 10 20
+)
vb (0
+)
=0
20 1
5
+
vb (0 )
=
20
10
KCL at node b:
vb (0
+)
va (0
+)
vb (0
+)
5 2
+ =0
10
3
1
5
1
1
+
+ vb (0+ )
=
+
va (0 )
20
20 10
10
20
+
2
3
Initial Conditions in Network
j
293
Solving the above two nodal equations, we get,
va (0
EXAMPLE
+
)=
40
V
21
4.12
Find iL (0+ ); vC (0+ );
dvC (0
dt
+)
and
diL (0
dt
+)
for the circuit shown in Fig. 4.26.
Assume that switch 1 has been opened and switch 2 has been closed for a long time and
steady–state conditions prevail at t = 0 .
SOLUTION
Figure 4.26
At t = 0 , switch 1 is open and
switch 2 is closed, whereas at t = 0+ ,
switch 1 is closed and switch 2 is
open.
First, let us redraw the circuit at
= 0 by replacing the inductor
with a short circuit and the capacitor
with an open circuit as shown in Fig.
4.27(a).
From Fig. 4.27(b), we find that
iL (0 ) = 0
t
Figure 4.27(a)
Figure 4.27(b)
294
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Network Theory
Applying KVL clockwise to the loop on the
right, we get
vC (0 )
2+1 0=0
Hence, at
t
+
=0 :
vC (0
iL (0
vC (0
The circuit diagram for
Fig. 4.27(c).
+
+
t
)=
2V
) = iL (0 ) = 0A
2V
) = vC (0 ) =
0
+
is shown in
Figure 4.27(c)
Applying KVL for right–hand mesh, we get
+ iL = 0
vC
vL
At t = 0+ , we get
vL (0
+
) = vC (0+ )
=
We know that
2
=L
vL
iL (0
0=
+
)
2V
diL
dt
At t = 0+ , we get
diL (0
+)
dt
=
Applying KCL at node X ,
2
Consequently, at t = 0+
iC (0
Since
+
)=
iC
dvC (0
EXAMPLE
+)
dt
4.13
For the circuit shown in Fig. 4.28, find:
(a) i(0+ ) and v (0+ )
(b)
di(0
+)
dt
and
dv (0
dt
(c) i( ) and v ( )
+)
=
2
=
1
2A/ sec
+ iC + iL = 0
vC (0
10
+)
2
=C
=
+)
L
10
vC
We get,
vL (0
iL (0
+
)=6
dvC
dt
+
iC (0
C
)
=
6
1
2
= 12V/ sec
0=6A
Initial Conditions in Network
j
295
Figure 4.28
SOLUTION
(a) From the symbol of switch, we find that at t = 0 , the switch is closed and t = 0+ , it is
open. At t = 0 , the circuit has reached steady state so that the equivalent circuit is as shown in
Fig.4.29(a).
12
i(0 ) =
= 2A
6
v (0 ) = 12 V
Therefore, we have
i(0
+
) = i(0 )
= 2A
(b) For t
0
+,
v (0
+
) = v (0 ) = 12V
we have the equivalent circuit as shown in Fig.4.29(b).
Figure 4.29(a)
Figure 4.29(b)
Applying KVL anticlockwise to the mesh on the right, we get
vL (t)
v (t)
+ 10i(t) = 0
Putting t = 0+ , we get
vL (0
+
)
vL (0
v (0
+
)
+
) + 10i(0+ ) = 0
12 + 10
2=0
vL (0
+
)=
8V
296
Network Theory
j
The voltage across the inductor is given by
vL
vL (0
di(0
+
=L
)=L
+)
dt
=
1
L
di
dt
di(0
+)
dt
vL (0
+
)
1
( 8) =
10
Similarly, the current through the capacitor is
=
iC
dv (0
or
+)
dt
=
iC (0
=
10
+)
C
2
10
6
=C
=
=
0.8A/ sec
dv
dt
i(0
+)
C
0.2 106 V/ sec
(c) As t approaches infinity, the switch is open and the circuit
has attained steady state. The equivalent circuit at t =
is
shown in Fig.4.29(c).
) = 0
() = 0
Figure 4.29(c)
i(
v
EXAMPLE
4.14
Refer the circuit shown in Fig.4.30. Find the following:
(a)
(b)
(c)
v (0
+)
dv (0
and i(0+ )
+)
dt
and
di(0
+)
dt
) and ()
v(
i
Figure 4.30
SOLUTION
From the definition of step function,
u(t)
=
1;
0;
0
t<0
t>
Initial Conditions in Network
j
297
From Fig.4.31(a), u(t) = 0 at t = 0 .
Similarly;
u(
t)
=
or
From Fig.4.31(b), we find that u(
u(
t)
t)
=
1;
0;
1;
0;
0
t<0
t>
0
t>0
t<
= 1, at t = 0 .
t = 0+
t = 0–
Figure 4.31(a)
Figure 4.31(b)
Due to the presence of u( t) and u(t) in the circuit of Fig.4.30, the circuit is an implicit
switching circuit. We use the word implicit since there are no conventional switches in the circuit
of Fig.4.30.
The equivalent circuit at t = 0 is shown in Fig.4.31(c). Please note that at t = 0 , the
independent current source is open because u(t) = 0 at t = 0 and the circuit is in steady state.
Figure 4.31(c)
40
= 5A
3+5
v (0 ) = 5i(0 ) = 25V
i(0
Therefore
i(0
v (0
)=
+
) = i(0 ) = 5A
+
) = v (0 ) = 25V
(b) For t 0+ ; u( t) = 0. This implies that the independent voltage source is zero and hence
is represented by a short circuit in the circuit shown in Fig.4.31(d).
298
j
Network Theory
Figure 4.31(d)
Applying KVL at node a, we get
dv
4+i=C
At t = 0+ , We get
4 + i(0+ ) = C
+)
+)
dt
4 + 5 = 0:1
dv (0
dt
dv (0
v
+
dv (0
5
+
+)
dt
v (0
+)
5
25
+
5
= 40V/ sec
dt
Applying KVL to the left–mesh, we get
3i + 0:25
di
dt
+v =0
Evaluating at t = 0+ , we get
3i(0+ ) + 0:25
3
di(0
+)
dt
5 + 0 25
:
+ v (0+ ) = 0
di(0
di(0
+)
dt
+
dt
)
=
+ 25 = 0
40
1
4
=
160A/ sec
(c) As t approaches infinity, again the circuit is in steady state. The equivalent circuit at t =
is shown in Fig.4.31(e).
Figure 4.31(e)
Initial Conditions in Network
Using the principle of current divider, we get
j
299
4 5
=
i( ) =
3+5
v ( ) = (i( ) + 4) 5
2.5A
= ( 2:5 + 4)5
= 7.5V
EXAMPLE
4.15
Refer the circuit shown in Fig.4.32. Find the following:
(a) i(0+ ) and v (0+ )
(b)
di(0
+)
dt
and
dv (0
+)
dt
(c) i( ) and v ( )
Figure 4.32
SOLUTION
Here the function u(t) behaves like a switch. Mathematically,
u(t)
=
1;
0;
t>
t<
0
0
The above expression means that the switch represented by u(t) is open for t < 0 and remains
closed for t > 0. Hence, the circuit diagram of Fig.4.32 may be redrawn as shown in Fig.4.33(a).
Figure 4.33(a)
For t < 0, the circuit is not active because switch is in open state, This implies that all the
initial conditions are zero.
That is,
iL (0 ) = 0 and vC (0 ) = 0
for t
0
+,
the equivalent circuit is as shown in Fig.4.33(b).
300
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Network Theory
Figure 4.33(b)
From the circuit diagram of Fig.4.33(b), we find that
vC
i=
5
+
At t = 0 , we get
+
vC (0 )
vC (0 )
0
+
i(0 ) =
=
= = 0A
5
5
5
Also
v = 15iL
Evaluating at t = 0+ , we get
+
+
v (0 ) = 15iL (0 )
= 15iL (0 ) = 15
(b) The equivalent circuit at t =
We find from Fig.4.33(c) that
0+
iC (0
+
0 = 0V
is shown in Fig.4.33(c).
) = 5A
Figure 4.33(c)
From Fig.4.33(b), we can write
vC
dvC
dt
= 5i
=5
di
dt
Multiplying both sides by C , we get
C
dvC
dt
= 5C
di
dt
Initial Conditions in Network
di(0
= 5C
+)
1
+
iC (0 )
5C
1
5
=
5 14
= 4A/ sec
dt
Also
dv
dt
dv
dt
dv
dt
At t = 0+ , we find that
dv (0
+)
dt
dt
=
v
301
di
iC
Putting t = 0+ , we get
j
= 15iL
= 15
diL
dt
= 15 1
diL
dt
= 15vL
= 15vL (0+ )
From Fig.4.33(b), we find that vL (0+ ) = 0
+
dv (0 )
= 15 0
Hence;
dt
EXAMPLE
= 0V/ sec
4.16
In the circuit shown in Fig. 4.34, steady state is reached with switch K open. The switch is closed
at t = 0.
di1
di2
Determine: i1 ; i2 ;
and
at t = 0+
dt
dt
Figure 4.34
302
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Network Theory
SOLUTION
At t = 0 , switch K is open and at t = 0+ , it is closed. At t = 0 , the circuit is in steady state
and appears as shown in Fig.4.35(a).
20
= 1:33A
10 + 5
1:33 = 13:3V
vC (0 ) = 10i2 (0 ) = 10
i2 (0
Hence;
)=
Since current through an inductor cannot change instantaneously, i2 (0+ ) = i2 (0 ) = 1.33 A.
Also, vC (0+ ) = vC (0 ) = 13:3V.
The equivalent circuit at t = 0+ is as shown in Fig.4.35(b).
i1 (0
+
)=
20
6:7
13:3
=
= 0.67A
10
10
Figure 4.35(a)
For t
0
+,
Figure 4.35(b)
the circuit is as shown in Fig.4.35(c).
Writing KVL clockwise for the left–mesh,
we get
10i1 +
1
C
Zt
i1 ( )d
= 20
0+
Differentiating with respect to t, we get
10
di1
dt
+
1
C
i1
=0
Putting t = 0+ , we get
10
di1 (0
dt
+)
=
di1 (0
+)
dt
1
10
+
1 10
1
C
Figure 4.35(c)
i1 (0
+
i (0
6 1
+
)=0
)=
0.67 105 A/ sec
Initial Conditions in Network
Writing KVL equation to the path made of 20V
K
10Ω
2di2
10i2 +
= 20
j
303
2H, we get
dt
0+ ,
At t =
the above equation becomes
10i2 (0+ ) +
EXAMPLE
10
2di2 (0+ )
1 33 + 2
:
dt
di2 (0
+)
dt
di2 (0
dt
+)
= 20
= 20
= 3.35A/ sec
4.17
Refer the citcuit shown in Fig.4.36. The switch K is closed at t = 0. Find:
(a)
v1
and v2 at t = 0+
(b)
v1
and v2 at t =
(c)
(d)
dv1
and
dt
2
d v1
dt2
dv2
dt
at t = 0+
at t = 0+
Figure 4.36
SOLUTION
(a) The circuit symbol for switch conveys that at t = 0 , the switch is open and t = 0+ , it is
closed. At t = 0 , since the switch is open, the circuit is not activated. This implies that
all initial conditions are zero. Hence, at t = 0+ , inductor is open and capactor is short.
Fig 4.37(a) shows the equivalent circuit at t = 0+ .
Figure 4.37(a)
304
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Network Theory
10
= 1A
10
+
+
v1 (0 ) = 0; i2 (0 ) = 0
i1 (0
+
Applying KVL to the path, 10V source
(b) At = , switch
conditions, capacitor
circuit at = .
t
)=
10Ω 10Ω 2mH, we get
K
10 + 10i1 (0+ ) + v1 (0+ ) + v2 (0+ ) = 0
10 + 10 + 0 + v2 (0+ ) = 0
v2 (0
K
C
+
)=0
remains closed and circuit is in steady state. Under steady state
is open and inductor L is short. Fig. 4.37(b) shows the equivalent
t
) = 10 10+ 10 = 0.5A
() = 0
() = 0 5 10 = 5V
() = 0
i2 (
i1
v1
v2
(c) For t
0
+,
:
Figure 4.37(b)
the circuit is as shown in Fig. 4.37(c).
Figure 4.37(c)
Initial Conditions in Network
i2
Zt
1
=
L
v2 ( )d
=
0+
j
v1 (t)
R2
Differentiating with respect to t, we get
v2
1
=
L
dv1
R2 dt
Evaluating at t = 0+ we get
dv1 (0
+)
dv1 (0
R2
=
dt
+)
L2
v2 (0
+
)
= 0V/ sec
dt
Applying KVL clockwise to the path 10 V source
C
F,
K
we get
Zt
1
10 + 10i +
10Ω 4
[i( )
i2 ( )]d
=0
0+
Differentiating with respect to t, we get
10
di
dt
1
+
C
[i
i2 ]
=0
Evaluating at t = 0+ , we get
di(0
+)
dt
=
=
=
i2 (0
+)
C
i(0
10
0 1
10 4 10
+)
2
3
∵ i(0+ ) = i1 (0+ ) + i2 (0+ )
4
5
=1+0
= 1A
6
25000A/ sec
Applying KVL clockwise to the path 10 V source
we get
10Ω 10Ω 2 mH,
K
10 + 10i + 10i2 + v2 = 0
10i + v1 + v2 = 10
Differentiating with respect to t, we get
10
di
dt
+
dv1
dt
+
dv2
dt
=0
305
306
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Network Theory
At t = 0+ , we get
10
di(0
+)
dt
+
dv1 (0
+)
dt
+
10( 25000) + 0 +
dv2 (0
dt
dv2 (0
dt
dv2 (0
+)
=0
+)
=0
+)
= 25 104 V/ sec
dt
(d) From part (c), we have
Zt
1
L
v2 ( )d
0+
=
v1
10
Differentiating with respect to t twice, we get
1
dv2
L dt
=
1 d2 v1
10 dt2
At t = 0+ , we get
1
dv2 (0
L
dt
2
d v
1 (0
dt2
Hence;
EXAMPLE
+)
+)
=
1 d2 v1 (0+ )
10 dt2
= 125 107 V/ sec2
4.18
Refer the network shown in Fig. 4.38. Switch K is changed from a to b at t = 0 (a steady state
having been established at position a).
Figure 4.38
Show that at t = 0+ .
i1
= i2 =
V
R1
+ R2 + R3
;
i3
=0
Initial Conditions in Network
j
307
SOLUTION
The symbol for switch indicates that at t = 0 , it is in position a and at t = 0+ , it is in position
b. The circuit is in steady state at t = 0 . Fig 4.39(a) refers to the equivalent circuit at t = 0 .
Please remember that at steady state C is open and L is short.
iL1 (0
) = 0;
iL2 (0
) = 0;
vC2 (0
) = 0;
vC1 (0
)=0
Applying KVL clockwise to the left-mesh, we get
V
+ vC3 (0 ) + 0
R2
+0=0
vC3 (0
) = V volts:
Figure 4.39(a)
Since current in an inductor and voltage across a capacitor cannot change instantaneously, the
equivalent circuit at t = 0+ is as shown in Fig. 4.39(b).
Figure 4.39(b)
i1 (0
i3 (0
+
+
) = i2 (0+ ) since
) = 0 since
Applying KVL to the path v C3 (0+ )
V
iL1 (0
iL2 (0
+
+
)=0
)=0
R2
R3
R1
K
we get,
+ R2 i1 (0+ ) + R3 i2 (0+ ) + R1 i1 (0+ ) = 0
308
j
Network Theory
Since i1 (0+ ) = i2 (0+ ), the above equation becomes
= [R1 + R2 + R3 ] i1 (0+ )
V
+
+
i1 (0 ) = i2 (0 ) =
A
R1 + R2 + R3
V
Hence;
EXAMPLE
4.19
Refer the circuit shown in Fig. 4.40. The switch K is closed at t = 0.
Find (a)
di1 (0
dt
+)
and (b)
di2 (0
+)
dt
Figure 4.40
SOLUTION
The circuit symbol for the switch shows that at
+
t = 0 , it is open and at t = 0 , it is closed.
Hence, at t = 0 , the circuit is not activated.
This implies that all initial conditions are zero.
That is, vC (0 ) = 0 and iL (0 ) = i2 (0 ) = 0.
The equivalent circuit at t = 0+ keeping in mind
that vC (0+ ) = vC (0 ) and iL (0+ ) = iL (0 ) is
as shown in Fig. 4.41 (a).
i1 (0
+
Figure 4.41(a)
) = 0 and i2 (0+ ) = 0:
0
+.
Figure. 4.41(b) shows the circuit diagram for t
Vo sin !t
= i1 R +
1
C
Zt
i1 ( )d
0+
Differentiating with respect to t, we get
Vo ! cos !t
=R
di1
dt
+
i1
C
Initial Conditions in Network
j
309
At t = 0+ , we get
=R
Vo !
+
di1 (0 )
Also;
Vo sin !t
=
dt
di1 (0
+)
dt
+
i1 (0
+)
C
V0 ω
A/ sec
R
= i2 R + L
di2
dt
Evaluating at t = 0+ , we get
0 = i2 (0+ )R + L
di2 (0
+)
dt
EXAMPLE
di2 (0
+)
dt
Figure 4.41(b)
= 0A/ sec
4.20
In the network of the Fig. 4.42, the switch K is opened at
steady state with the switch closed.
(a) Find the expression for vK at t = 0+ .
t
(b) If the parameters are adjusted such that i(0+ ) = 1, and
the derivative of the voltage across the switch at t =
= 0 after the network has attained
di(0
0+ ,
+)
dt
dvK
dt
=
1, what is the value of
+
(0 ) ?
Figure 4.42
SOLUTION
At t = 0 , switch is in the closed state and at
+
t = 0 , it is open. Also at t = 0 , the circuit
is in steady state. The equivalent circuit at
t = 0 is as shown in Fig. 4.43(a).
i(0
)=
V
R2
and vC (0 ) = 0
Figure 4.43(a)
310
j
Network Theory
For t 0+ , the equivalent circuit is as shown in Fig. 4.43(b).
From Fig. 4.43 (b),
vK
At t = 0+ ,
vK
vK (0
+)
= R1 i +
Zt
1
C
i( )d
0+
= R1 i + vC (t)
= R1 i(0+ ) + vC (0+ )
vK (0
+
) = R1
V
+ vC (0 )
R2
= R1
V
volts
R2
Figure 4.43(b)
(b)
vK
dvK
dt
= R1 i +
= R1
di
dt
Zt
1
C
+
i( )d
0+
i
C
Evaluating at t = 0+ , we get
dvK (0
dt
+)
= R1
= R1
1
=
C
di(0
+)
dt
(
+
1) +
i(0
+)
C
1
C
R1 volts/ sec
Initial Conditions in Network
j
311
Reinforcement Problems
R.P
4.1
Refer the circuit shown in Fig RP.4.1(a). If the switch is closed at
2
+
d iL (0 )
dt2
t
= 0, find the value of
at t = 0+ .
Figure R.P.4.1(a)
SOLUTION
The circuit at t = 0 is as shown in Fig RP 4.1(b).
Since current through an inductor and voltage across a capacitor cannot change
instantaneously, it implies that iL (0+ ) = 18A and vC (0+ ) = 180 V.
The circuit for t 0+ is as shown in Fig. RP 4.1 (c).
Figure R.P.4.1(b)
Figure R.P.4.1(c)
Referring Fig RP 4.1 (c), we can write
2
10
3 diL
dt
+ 60iL + 288
10
3
Zt
iL (t)dt
0+
=0
(4.9)
312
j
Network Theory
At t = 0+ , we get
diL (0
+)
=
dt
=
60 18 + 180
2 10 3
450 103 A= sec
Differentiating equation (4.9) with respect to t, we get
2
10
2
3 d iL
dt2
diL
+ 60
dt
+ 288
10
3
At t = 0+ , we get
2
+
d iL (0 )
2
dt
R.P
iL
=0
60(450)103 288 103 (18)
2 10 3
= 1:0908 1010 A= sec2
=
4.2
For the circuit shown in Fig. RP 4.2, determine
2
+
d vC (0 )
dt2
and
3
+
d vC (0 )
:
dt3
Figure R.P.4.2
SOLUTION
Given
i(t)
= 2u(t) =
2;
0;
t
t
0
0
+
Hence, at t = 0 , vC (0 ) = 0 and iL (0 ) = 0.
For t 0+ , the circuit equations are
1 dvC (t) 1
+
64 dt
2
Zt
=
2
(4.10)
1 dvC (t)
+ iL (t) =
64 dt
2
(4.11)
vL (t)dt
0+
Initial Conditions in Network
[Note :
At t =
iC
0+ ,
+ iL =
j
313
2 because of the capacitor polarity]
equation (4.10) gives
1 dvC (0+ )
+ iL (0+ ) =
64
dt
2
Since, iL (0+ ) = iL (0 ) = 0, we get
1 dvC (0+ )
+0=
64
dt
+
dvC (0 )
=
2
128 volts= sec
dt
Differentiating equation (4.10) with respect to t we get
1 d2 vC (t) 1
+ vL (t) = 0
64 dt
2
vL
vC
Also;
24
At t = 0+ , we get
vC (0
1
=
2
+)
(4.12)
Zt
vL dt
= iL
(4.13)
0+
vL (0
+)
= iL (0+ )
24
Since vC (0+ ) = 0 and iL (0+ ) = 0, we get vL (0+ ) = 0.
At t = 0+ , equation (4.12) becomes
1 d2 vC (0+ ) 1
+ vL (0+ ) = 0
64
dt2
2
1 d2 vC (0+ ) 1
+
0=0
64
dt2
2
2
+
d vC (0 )
=0
2
dt
Differentiating equation (4.12) with respect to t we get
1 d3 vC
1 dvL
+
=0
3
64 dt
2 dt
Differentiating equation (4.13) with respect to t, we get
dvC
dvL
dt
dt
24
=
1
vL
2
(4.14)
314
j
Network Theory
At t = 0+ , we get
dvC (0
+)
dvL (0
dt
dt
24
+)
dvL (0
128
=
+)
dt
24
dvL (0
dt
1
+
vL (0 )
2
=0
+)
=
128 volts= sec
At t = 0+ , equation (4.14) becomes
1 d3 vC (0+ ) 1 dvL (0+ )
+
=0
64
dt3
2
dt
3
+
d vC (0 )
= 4096 volts= sec3
3
R.P
dt
4.3
In the network of Fig RP 4.3 (a), switch K is closed at t = 0. At t = 0 all the capacitor voltages
and all the inductor currents are zero. Three node-to-datum voltages are identified as v1 , v2 and
+
v3 . Find at t = 0 :
(i)
(ii)
v1 , v2
dv1
dt
,
and v3
dv2
dt
and
dv3
dt
Figure R.P.4.3(a)
SOLUTION
The network at t = 0+ is as shown in Fig RP-4.3 (b).
Since vC and iL cannot change instantaneously, we have from the network shown in
Fig. RP-4.3 (b),
+
v1 (0 ) = 0
v2 (0
v3 (0
+
+
)=0
)=0
Initial Conditions in Network
j
315
Figure R.P.4.3(b)
For t
0
+,
the circuit equations are
=
v C1
=
v C2
=
v C3
Zt
1
C1
i1 dt
0+
Zt
1
C2
i2 dt
0+
Zt
1
C3
i3 dt
0+
9
>
>
>
>
>
>
>
>
>
>
>
>
>
=
(4.15)
>
>
>
>
>
>
>
>
>
>
>
>
>
;
From Fig. RP-4.3 (b), we can write
i1 (0
i2 (0
and
i3 (0
+
+
+
)=
)=
v (0
+)
R1
v1 (0
;
+)
v2 (0
+)
R2
)=0
Differentiating equation (4.15) with respect to t, we get
dvC1
dt
=
i1
C1
;
dvC2
dt
=
i2
and
C2
dvC3
dt
=
i3
C3
At t = 0+ , the above equations give
dv1 (0
+)
dt
dv2 (0
+)
dt
and
dv3 (0
dt
+)
=
=
=
i1 (0
+)
C1
+
i2 (0
C2
+
i3 (0
C3
)
)
=
=
v (0
+)
R 1 C1
+
v1 (0
=0
)
v2 (0
R 2 C2
+)
=0
316
j
Network Theory
R.P
4.4
For the network shown in Fig RP 4.4 (a) with switch K open, a steady-state is reached. The circuit
paprameters are R1 = 10Ω, R2 = 20Ω, R3 = 20Ω, L = 1 H and C = 1F. Take V = 100
volts. The switch is closed at t = 0.
(a) Write the integro-differential
equation after the switch is
closed.
(b) Find the voltage Vo across C before the switch is closed and give
its polarity.
(c) Find i1 and i2 at t = 0+ .
(d) Find
di1
dt
and
di2
dt
at t = 0+ .
(e) What is the value of
t=
?
di1
dt
at
Figure R.P.4.4(a)
SOLUTION
The switch is in open state at t = 0 . The network at t = 0 is as shown in Fig RP 4.4 (b).
Figure R.P.4.4(b)
100
10
=
A
R1 + R2
30
3
10
200
VC (0 ) = i1 (0 )R2 =
20 =
volts
3
3
Note that L is short and C is open under steady-state condition.
For t 0+ (switch in closed state),
i1 (0
)=
V
=
we have
and
20i1 +
6
Zt
20i2 + 10
di1
dt
i2 dt
0+
= 100
(4.16)
= 100
(4.17)
Initial Conditions in Network
j
317
10
A
3
200
+
and
VC (0 ) = VC (0 ) =
Volts
3
From equation (4.16) at t = 0+ ,
Also
i1 (0
+
) = i1 (0 ) =
di1 (0
we have
+)
dt
= 100
=
1
100
i2 (0 ) =
20
103
100
A=sec
3
From equation (4.17), at t = 0+ , we have
+
20
200
5
= A
3
3
Differentiating equation (4.17), we get
20
di2
dt
+ 106 i2 = 0
(4.18)
From equation (4.18) at t = 0+ , we get
20di2 (0+ )
At t =
dt
+ 106 i2 (0+ ) = 0
di2 (0
dt
,
+)
) = 100
=5A
20
() = 0
i1 (
di1
dt
R.P
106 53
20
106
=
A= sec
12
=
4.5
For the network shown in Fig RP 4.5 (a), find
2
+
d i1 (0 )
.
dt2
The switch K is closed at t = 0.
Figure R.P.4.5(a)
318
j
Network Theory
SOLUTION
At t = 0 , we have vC (0 ) = 0 and i2 (0 ) = iL (0 ) = 0. Because of the switching property
of L and C , we have vC (0+ ) = 0 and i2 (0+ ) = 0. The network at t = 0+ is as shown in
Fig RP 4.5 (b).
Figure R.P.4.5(b)
Referring Fig RP 4.5 (b), we find that
i1 (0
The circuit equations for t
0
+
R2 i2
+
v (0
)=
+)
R1
are
R1 i1
and
+
1
C
|
+
Zt
1
C
Zt
(i1
= v (t)
(4.19)
=0
(4.20)
0+
(i2
0+
i2 )dt
i1 )dt +L
{z
di2
dt
}
vC (t)
At t = 0+ , equation (4.20) becomes
R2 i2 (0
+
) + vC (0+ ) + L
di2 (0
dt
di2 (0
+)
+)
dt
=0
(4.21)
=0
Differentiating equation (4.19), we get
R1
di1
dt
+
1
C
i2 )
(i1
=
dv (t)
(4.22)
dt
Letting t = 0+ in equation (4.22), we get
R1
di1 (0
dt
+)
+
1 C
i1 (0
+
)
i2 (0
+
di1 (0
dt
) =
+)
=
dv (0
+)
dt
1
R1
dv (0
dt
+)
v (0
+)
R1 C
(4.23)
Initial Conditions in Network
Differentiating equation (4.22) gives
R1
2
d i1
dt2
+
1
R.P
+
1
di1 (0
C
+)
di2 (0
dt
di1
di2
dt
dt
C
Letting t = 0+ , we get
2
+
d i1 (0 )
R1
dt2
+)
dt
2
+
d i1 (0 )
R1
dt2
2
+
d i1 (0 )
dt2
=
=
j
319
2
d v (t)
dt2
2
+
d v (0 )
dt2
=
=
1
di1 (0
C
+)
dt
1
R1 C
1
R1
+
2
+
d v (0 )
dt2
dv (0
dt
+)
1
2C
R1
v (0
+
) +
2
+
d v (0 )
dt2
4.6
Determine va (0 ) and va (0+ ) for the network shown in Fig RP 4.6 (a). Assume that the switch
is closed at t = 0.
Figure R.P.4.6(a)
SOLUTION
Since L is short for DC at steady state, the network at t = 0 is as shown in Fig. RP 4.6 (b).
Applying KCL at junction a, we get
va (0
)
10
5
+
va (0
)
vb (0
20
)
=0
Since vb (0 ) = 0, we get
va (0
)
10
5
) 0
=0
20
0:5
10
va (0 ) =
=
volts
0:1 + 0:05
3
+
va (0
Figure R.P.4.6(b)
320
j
Network Theory
Also;
For t
) = iL (0+ ) =
iL (0
0
+,
va (0
20
2
= A
3
)
+
5
10
we can write
5
va
10
va
vb
and
+
20
va
10
+
+
vb
va
5
vb
10
20
=0
+ iL = 0
Simplifying at t = 0+ , we get
1
+
va (0 )
4
1
+
va (0 ) +
20
and
Solving we get,
va (0
+)
=
1
1
+
vb (0 ) =
20
2
3
1
+
vb (0 ) =
20
6
40
= 1:905 volts
21
Exercise problems
E.P
4.1
Refer the circuit shown in Fig. E.P. 4.1 Switch K is closed at t = 0.
Find i(0+ ),
di(0
+)
dt
and
2
+
d i(0 )
.
dt2
Figure E.P.4.1
Ans: i(0+ ) = 0.2A,
di(0+ )
=
dt
2 103 A/ sec,
d2 i(0+ )
= 20 106 A/ sec2
dt2
Initial Conditions in Network
E.P
dt2
321
4.2
Refer the circuit shown in Fig. E.P. 4.2. Switch K is closed at t = 0. Find the values of i;
2
d i
j
di
dt
and
at t = 0+ .
Figure E.P.4.2
Ans: i(0+ ) = 0,
E.P
di(0+ )
d2 i(0+ )
=
= 10 A/ sec,
dt
dt2
1000 A/ sec2
4.3
Refering to the circuit shown in Fig. E.P. 4.3, switch is changed from position 1 to position 2 at
t
= 0. The circuit has attained steady state before switching. Determine i,
di
dt
and
Figure E.P.4.3
Ans: i(0+ ) = 0,
di(0+ )
=
dt
40 A/ sec,
d2 i(0+ )
= 800 A/ sec2
dt2
2
d i
dt2
at t = 0+ .
322
Network Theory
j
E.P
4.4
In the network shown in Fig. E.P.4.4, the initial voltage on
v1 (0
) = Va and v2 (0 ) = Vb . Find the values of
dv1
dt
and
C1
dv2
dt
is
Va
and on
C2
is
Vb
such that
at t = 0+ .
Figure E.P.4.4
Ans:
E.P
dv1 (0+ )
Vb Va
=
V/ sec,
dt
C1 R
dv2 (0+ )
Va Vb
=
V/ sec
dt
C2 R
4.5
In the network shown in Fig E.P. 4.5, switch K is closed at t = 0 with zero capacitor voltage and
zero inductor current. Find
2
d v2
dt2
at t = 0+ .
Figure E.P.4.5
Ans:
d2 v2 (0+ )
R2 Va
=
V/ sec2
dt2
R1 L1 C1
Initial Conditions in Network
E.P
j
323
4.6
In the network shown in Fig. E.P. 4.6, switch K is closed at t = 0. Find
2
d v1
dt2
at t = 0+ .
Figure E.P.4.6
Ans:
E.P
d2 v1 (0+ )
= 0 V/ sec2
dt2
4.7
The switch in Fig. E.P. 4.7 has been closed for a long time. It is open at
+
dv (0 )
dt
t
= 0. Find
+
di(0 )
, i( ) and v ( ).
i
Figure E.P.4.7
Ans:
di(0+ )
= 0A/ sec,
dt
dv(0+ )
= 20A/ sec, i() = 0A, v() = 12V
dt
dt
,
324
j
E.P
Network Theory
4.8
In the circuit of Fig E.P. 4.8, calculate iL (0+ ),
diL (0
+ ) dv (0+ )
C
dt
,
dt
, vR ( ), vC ( ) and iL ( ).
Figure E.P.4.8
Ans: iL (0+ ) = 0 A,
diL (0+ )
= 0 A/ sec
dt
dvC (0+ )
= 2 V/ sec,
dt
E.P
vR () = 4V, vC () =
20V, iL () = 1A
4.9
Refer the circuit shown in Fig. E.P. 4.9. Assume that the switch was closed for a long time for
+
diL (0 )
and iL (0+ ). Take v (0+ ) = 8 V.
t < 0. Find
dt
Figure E.P.4.9
Ans: iL (0+ ) = 4 A,
E.P
diL (0+ )
= 0 A/ sec
dt
4.10
Refer the network shown in Fig. E.P. 4.10. A steady state is reached with the switch K closed and
+
dv2 (0 )
.
with i = 10A. At t = 0, switch K is opened. Find v2 (0+ ) and
dt
Initial Conditions in Network
j
325
Figure E.P.4.10
Ans: v2 (0+ ) = 0,
E.P
10Ra Rc
dv2 (0+ )
=
V= sec.
dt
Ca (Ra + Rb )(Ra + Rc )
4.11
Refer the network shown in Fig. E.P. 4.11. The network is in steady state with switch K closed.
+
dvk (0 )
.
The switch is opened at t = 0. Find vk (0+ ) and
dt
Figure E.P.4.11
Ans: vk (0+ ) =
Va Rc
Volts,
Ra + Rb + Rc
dvk (0+ )
Va (Ca + Cb )
=
V/ sec
dt
(Ra + Rb + Rc )(Ca Cd + Cb Ca + Cb Cd )
E.P
4.12
Refer the network shown in Fig. E.P. 4.12. Find
2
+
d i1 (0 )
.
dt2
326
j
Network Theory
d2 i1 (0+ )
1
Ans:
=
2
dt
Ra
E.P
Figure E.P.4.12
10
10 + 2 2 A/ sec2
Ra Ca
4.13
Refer the circuit shown in Fig. E.P. 4.13. Find
steady state at t = 0 .
di1 (0
dt
+)
. Assume that the circuit has attained
Figure E.P.4.13
di1 (0+ )
10
Ans:
A/ sec
=
dt
RA
E.P
4.14
Refer the network shown in Fig. E.P.4.14. The circuit reaches steady state with switch K closed.
+
2
+
dv1 (0 )
d v2 (0 )
At a new reference time, t = 0, the switch K is opened. Find
.
and
2
dt
Figure E.P.4.14
dt
Initial Conditions in Network
Ans:
E.P
dv1 (0+ )
10
=
V/ sec,
dt
Ca (Ra + Rb )
j
327
10Rb
d2 v2 (0+ )
=
V/ sec2
dt2
La Ca (Ra + Rb )
4.15
The switch shown in Fig. E.P. 4.15 has been open for a long time before closing at t = 0. Find:
+
+
i0 (0 ), iL (0 ) i0 (0 ), iL (0 ), i0 ( ), iL ( ) and vL ( ).
Figure E.P.4.15
Ans: i(0 ) = 0, iL (0 ) = 160mA, i0 (0+ ) = 65mA, iL (0+ ) = 160mA,
i0 () = 225mA, iL () = 0, vL () = 0
E.P
4.16
The switch shown in Fig. E.P. 4.16 has been closed for a long time before opeing at t = 0.
Find: i1 (0 ), i2 (0 ), i1 (0+ ), i2 (0+ ). Explain why i2 (0 ) = i2 (0+ ).
$
Figure E.P.4.16
Ans: i1 (0 ) = i2 (0 ) = 0.2mA, i2 (0+ ) =
i1 (0+ ) =
0.2mA
328
j
E.P
Network Theory
4.17
The switch in the circuit of Fig E.P.4.17 is closed at t = 0 after being open for a long time. Find:
(a) i1 (0 ) and i2 (0 )
(b) i1 (0+ ) and i2 (0+ )
(c) Explain why i1 (0 ) = i1 (0+ )
$
(d) Explain why i2 (0 ) = i2 (0+ )
Figure E.P.4.17
Ans: i1 (0 ) = i2 (0 ) = 0.2 mA, i1 (0+ ) = 0.2 mA, i2 (0+ ) =
0.2mA
Chapter
5
5.1
Laplace Transform
Introduction
In this chapter, we will introduce Laplace transform. This is an extremely important technique.
For a given set of initial conditions, it will give the total response of the circuit comprising of both
natural and forced responses in one operation. The idea of Laplace transform is analogous to any
familiar transform. For example, Logarithms are used to change a multiplication or division problem into a simpler addition or subtraction problem and Antilogs are used to carry out the inverse
process. This example points out the essential feature of a transform: They are designed to create
a new domain to make mathematical manipulations easier. After evaluating the unknown in the
new domain, we use inverse transform to get the evaluated unknown in the original domain. The
Laplace transform enables the circuit analyst to convert the set of integrodifferential equations
describing a circuit to the complex frequency domain, where thay become a set of linear algebraic equations. Then using algebraic manipulations, one may solve for the variables of interest.
Finally, one uses the inverse transform to get the variable of interest in time domain. Also, in
this chapter, we express the impedance in s domain or complex frequency domain. Hence, we
may analyze a circuit using one of the reduction techniques such as Thevenin theorem or source
transformation discussed in earlier chapters.
5.2
Definition of Laplace transorm
A transform is a change in the mathematical description of a physical variable to facilitate computation. Keeping this definition in mind, Laplace transform of a function f (t) is defined as
Lff (t)g = F (s) =
Z1
f (t)e
st
dt
(5.1)
0
Here the complex frequency is s = + j! . Since the argument of the exponent e in equation
(5.1) must be dimensionless, it follows that s has the dimensions of frequency and units of inverse
seconds (sec 1 ).
330
j
Network Theory
The notation implies that once the integral has been evaluated, f (t), a time domain function
is transformed to F (s), a frequency domain function.
If the lower limit of integration in equation (5.1) is 1, then it is called the bilateral Laplace
transform. However for circuit applications, the lower limit is taken as zero and accordingly the
transform is unilateral in nature.
The lower limit of integration is sometimes chosen to be 0 to permit f (t) to include (t) or
its derivatives. Thus we should note immediately that the integration from 0 to 0+ is zero except
when an impulse function or its derivatives are present at the origin.
Region of convergence
The Laplace transform of a signal f (t) as seen from equation (5.1) is an integral operation. It
exists if f (t)e
t
Z1
f (t)e
is absolutely integrable. That is
t
dt < 1. Cleary, only typical
0
choices of will make the integral converge. The range of that ensures the existence of X (s)
defines the region of convergence (ROC) of the Laplace transform. As an example, let us take
x(t) = e3t , t 0. Then
Z1
X (s) =
x(t)e
( +j! )t
dt
0
Z1
=
e(
+3)t
e
j!t
dt
0
The above integral converges if and only if + 3 < 0 or > 3. Thus, > 3 defines the
ROC of X (s). Since, we shall deal only with causal signals(t 0) we avoid explicit mention of
ROC.
Due to the convergence factor, e t , a number of important functions have Laplace transforms, even though Fourier transforms for these functions do not exist. But this does not mean
that every mathematical function has Laplace transform. The reader should be aware that, for
2
example, a function of the form et does not have Laplace transform.
The inverse Laplace transform is defined by the relationship:
L
1
fF (s)g = f (t) =
1
2j
Z
+j
j
1
1
F (s)est ds
(5.2)
where is real. The evaluation of integral in equation (5.2) is based on complex variable theory,
and hence we will avoid its use by developing a set of Laplace transform pairs.
Laplace Transform
5.3
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331
Three important singularity functions
The three important singularity functions employed in circuit analysis are:
(i) unit step function, u(t)
(ii) delta function, (t)
(iii) ramp function, r (t).
They are called singularity functions because they are either not finite or they do not possess
finite derivatives everywhere.
The mathematical definition of unit step function is
u(t) =
0;
1;
t<0
t>0
(5.3)
The step function is not defined at t = 0. Thus,
the unit step function u(t) is 0 for negative values
of t, and 1 for positive values of t. Often it is
advantageous to define the unit step function as
follows:
u(t) =
1;
0;
t 0+
t0
Figure 5.1 The unit step function
A discontinuity may occur at time other than
t = 0; for example, in sequential switching, the
unit step function that occurs at t = a is expressed
as u(t a).
Figure 5.2 The step function occuring at t
= a Figure 5.3
Thus;
u(t
a) =
0;
1;
t
t
Similarly, the unit step function that occurs at t =
Thus;
u(t + a) =
0;
1;
The step function occuring at t
a < 0 or t < a
a > 0 or t > a
a is expressed as u(t + a).
t + a < 0 or t <
t + a > 0 or t >
a
a
=
a
332
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Network Theory
We use step function to represent an abrupt change in voltage or current, like the changes that
occur in the circuits of control engineering and digital systems. For example, the voltage
v (t) =
0;
K;
t<a
t>a
may be expressed in terms of the unit step function as
v (t) = Ku(t
a)
(5.4)
The derivative of the unit step function u(t) is the unit impulse function (t).
That is;
(t) =
8
<
0;
t<0
d
undefined; t = 0
u(t) =
:
dt
0;
t>0
(5.5)
The unit impulse function also known as dirac delta fucntion is shown in Fig. 5.4.
The unit impulse may be visualized as very short duration pulse of unit area. This may be
expressed mathematically as:
Z0+
(t)dt = 1
(5.6)
0
where t = 0 denotes the time just before t = 0 and t = 0+ denotes
the time just after t = 0. Since the area under the unit impulse is unity,
it is a practice to write ‘1’ beside the arrow that is used to symbolize
the unit impulse function as shown in Fig. 5.4. When the impulse has
a strength other than unity, the area of the impulse function is equal
to its strength. For example, an impulse function 5 (t) has an area
of 5 units. Figure 5.5 shows impulse functions, 2 (t + 2), 5 (t) and
2 (t 3).
Figure 5.5 Three impulse functions
Figure 5.4 The circuit
impulse function
Laplace Transform
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333
An important property of the unit impulse function is what is often called the sifting property;
which is exhibited by the following integral:
Zt2
f (t) (t
t0 )dt =
t1
f (t0 ); t1 < t0 < t2
0;
t 1 > t0 > t2
for a fintie t0 and any f (t) continuous at t0 .
Integrating the unit step function results in the unit ramp function r (t).
Zt
r (t) =
1
or
r (t) =
u( )d = tu(t)
0;
t;
(5.7)
t0
t0
Figure 5.6 shows the ramp function.
Figure 5.6 The unit ramp function
In general, a ramp is a function that changes at a constant rate.
Figure 5.7 The unit ramp function delayed by t0
Figure 5.8 The unit ramp function advanced by t0
A delayed ramp function is shown in Fig. 5.7. Mathematically, it is described as follows:
r (t
t0 ) =
t t0
t0 ; t t0
0;
t
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Network Theory
An advanced ramp function is shown in Fig. 5.8. Mathematically, it is described as follows:
r (t + t0 ) =
0;
t
t + t0 ; t t0
t0
It is very important to note that the three sigularity functions are related by differentiation as
(t) =
or by integration as
u(t) =
Zt
u(t) =
5.4
du(t)
;
dt
1
dr (t)
dt
Zt
(t)dt;
r (t) =
1
u( )d
Functional transforms
A functional transform is simply the Laplace transform of a specified function of t. Here we make
an assumption that f (t) is zero for t < 0.
5.4.1 Decaying exponential function
f (t) = e
at
u(t), where a > 0 and u(t) is the unit step function.
L
e
at
Z1
u(t) = F (s) =
f (t)dt
0
Z1
e
=
at
0
e
st
dt
1
e (s+a)t (s + a) =
t=0
1
=
s+a
5.4.2
Unit step function
f (t) = u(t)
Lfu(t)g = F (s) =
Z1
e
0
st
dt =
1
s
Laplace Transform
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335
5.4.3 Impulse function
f (t) = (t)
Lf (t)g = F (s) =
Z1
(t)e
st
dt = e
st t=0
=1
0
Please note that we have used the sifting property of an impulse function.
5.4.4 Sinusoidal function
t0
f (t) = sin !t;
sin !t =
Since
ej!t
2j
Lfe at g = s +1 a
and
we have
1 Lfsin !tg = F (s) = 21j
Z1
e
ej!t
0
1
1
=
2j s j!
=
!
s2
j!t
e
j!t
e
1
s + j!
st
dt
+ !2
Table 5.1 gives a list of important Laplace transform pairs. It includes the functions of most
interest in an introductory course on circuit applications.
Table 5.1 Important transform pairs
f (t)(t 0)
F (s)
(t)
1
1
u(t)
t
e
at
sin !t
cos !t
s
1
s2
1
s+a
!
s2 + ! 2
s
2
s + !2
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Network Theory
j
f (t)(t 0)
F (s)
n!
tn
te
sn+1
1
(s + a)2
at
e
at
sin !t
e
at
cos !t
!
(s + a)2 + ! 2
s+a
(s + a)2 + ! 2
All functions in the above table are represented without multiplied by u(t), since we have explicity declared that t 0.
5.5
Operational transforms (properties of Laplace transform)
Operational transforms indicate how mathematical operations performed on either f (t) or F (s)
are converted into the opposite domain. Following operations are of primary interest.
Note: The symbol means “by the definition”.
5.5.1 Linearity
Lff1 (t)g = F1 (s) and Lff2 (t)g = F2 (s)
If
Lfa1 f1 (t) + a2 f2 (t)g = a1 F1 (s) + a2 F2 (s)
then
Proof :
Lfa1 f1 (t) + a2 f2 (t)g Z1
0
Z1
= a1
f1 (t)e
st
dt + a2
0
5.1
bt
u(t)).
SOLUTION
We have the transform pair
Lfu(t)g = 1s
and
Lfe
bt
f2 (t)e
0
Find the Laplace transform of f (t) = (A + Be
u(t)g =
dt
Z1
= a1 F1 (s) + a2 F2 (s)
EXAMPLE
st
[a1 f1 (t) + a2 f2 (t)]e
1
s+b
st
dt
Laplace Transform
Thus, using linearity property,
Lff (t)g = F (s) = LfAu(t)g + LfBe
=
=
bt
u(t)g
B
A
+
s
s+b
(A + B)s + Ab
s(s + b)
5.5.2 Time shifting
If Lfx(t)g = X (s), then for any real number t0 ,
Lfx(t
t0 )u(t
Proof :
Lfx(t
t0 )u(t
u(t
Lfx(t
we get;
X (s)
Z1
t0 )g x(t
0
Since;
t0 s
t0 )g = e
t0 )u(t
t0 ) =
Z1
t0 )g =
1;
0;
t0 )u(t
t
t
t0 )e
x(t
t0 )e
st
dt
Using the transformation of variable,
t = + t0
Lfx(t
Z1
t0 )u(t
t0 )g =
x( )e
0
EXAMPLE
=e
t0 s
=e
t0 s
5.2
Find the Laplace transform of x(t), shown in Fig. 5.9.
Figure 5.9
dt
t0 > 0 or t > t0
t0 < 0 or t < t0
t0
we get;
st
s( +t0 )
Z1
x( )e
0
X (s)
d
s
d
j
337
338
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Network Theory
SOLUTION
Figure 5.10(a)
Figure 5.10(b)
Using Figs. 5.10(a) and 5.10(b), we can write
x(t) = x1 (t) + x2 (t) = u(t
We know that, Lfu(t)g =
1
s
u(t
2)
and using time shifting property, we have
Lfx(t)g = X (s) = 1s e
X(s) =
1
(e
s
1
2s
2s
s
e
4s
e
4s
)
5.5.3 Shifting in s domain (Frequency-domain shifting)
If Lfx(t)g = X (s), then
4)
Lfes t x(t)g = X (s
s0 )
0
Proof :
Lfe
s0 t
Z1
x(t)g es0 t x(t)e
st
dt
0
Z1
x(t)e
=
(s s0 )t
0
= X (s
s0 )
dt
Laplace Transform
EXAMPLE
j
5.3
Find the Laplace transform of x(t) = Ae
at
cos(!0 t + )u(t).
SOLUTION
Given
x(t) = Ae
at
cos(!0 t + )u(t)
= Ae
at
[cos !0 t cos = A cos e
at
sin !0 t sin ]u(t)
cos !0 tu(t)
A sin e
at
sin !0 tu(t)
We know the transform pairs,
Lfcos !0 tu(t)g =
Lfsin !0 tu(t)g =
and
s
s2
s2
+ !02
!0
+ !02
Applying frequency shifting property, we get
L
L
and
e
e
at
cos !0 tu(t) = 2
s + !2 s
0 s
at
sin !0 tu(t) = 2
s + !2 !s +a
s+a
=
(s + a)2 + !02
!0
0 s
!s +a
!0
=
(s + a)2 + !02
Finally, applying linearity property, we get
LfAe
at
cos(!0 t + )u(t)g = A cos Lfe at cos !0 tu(t)g A sin Lfe
A cos (s + a)
!0
=
A sin 2
2
(s + a) + !0
(s + a)2 + !02
A[(s + a) cos θ ω0 sin θ]
=
(s + a)2 + ω02
5.5.4 Time scaling
If Lfx(t)g = X (s), then
Lfx(at)g = a1 X
s
a
at
sin !0 tu(t)g
339
340
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Network Theory
Proof :
Lfx(at)g Z1
st
x(at)e
dt
0
at = put
)
adt = d
Lfx(at)g =
Hence
Z1
0
=
EXAMPLE
s τa
x( )e
1
Z1
x( )e
a
1
a
d
s
a
d =
0
1
a
5.4
Find the Laplace transform of x(t) = sin(2!0 t)u(t).
SOLUTION
We know the transform pair,
Lfsin !0 tu(t)g =
s2
Applying scaling property,
!0
+ !02
3
2
Lfsin 2!0 tu(t)g = 12 64 s !2 0
+ !02
2
2ω0
= 2
s + 4ω02
7
5
5.5.5 Time differentiation
If Lfx(t)g = X (s), then
L
dx(t)
dt
= sX (s)
x(0)
Proof :
y (t) =
Let
Then
dx(t)
dt
Lfy(t)g = Y (s) Z1
=
0
Z1
y (t)e
0
dx(t)
e
dt
st
dt
st
dt
X
s
a
Laplace Transform
j
341
Integrating by parts yields
Z1
1
Y (s) = e st x(t)0
0
= lim [e
t
L
Hence;
dx(t)
dt
!1
=0
st
st
x(t)( se
x(t)]
)dt
Z1
x(0) + s
x(t)e
st
dt
0
x(0) + sX (s)
= Y (s) = sX (s)
x(0)
Therefore, differentiation in time domain is equivalent to multiplication by s in the s domain.
Whenever x(t) is discontinuous at t = 0 (like a step function), then x(0) should be read as
x(0 ).
The differentiation property can be extended to yield
L
dn x(t)
dtn
= sn X (s)
sn
1
x(0) xn
1
(0)
When x(t) is discontinuous at the origin, the argument 0 on the right side of the above equation
should be read as 0 . Accordingly for a discontinuous function x(t) at the origin, we get
L
EXAMPLE
dn x(t)
dtn
= sn X (s)
sn
1
x(0 ) xn
1
(0 )
5.5
Find the Laplace transform of x(t) = sin2 !0 tu(t).
SOLUTION
We find that, x(0) = 0
We know that,
dx(t)
= 2!0 sin !0 t cos !0 tu(t)
dt
= !0 sin 2!0 tu(t)
Lfsin !0 tu(t)g = 2 !0 2
s + !0
(5.8)
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Network Theory
Applying time scaling property,
3
2
Lfsin 2!0 tu(t)g = 12 64 s !2 0
+ !02
2
=
s2
7
5
2!0
+ 2(!0 )2
Taking Laplace transform on both the sides of equation (5.8), we get
L
)
sX (s)
EXAMPLE
dx(t)
dt
= !0 Lfsin !0 tu(t)g
2!02
s2 + (2!0 )2
2ω02
X(s) =
s[s2 + (2ω0 )2 ]
x(0) =
5.6
Solve the second order linear differential equation
0
y (t) + 5y (t) + 6y (t) = x(t)
with the initial conditions, y (0) = 2, y 0 (0) = 1 and x(t) = e t u(t).
SOLUTION
Taking Laplace transform on both the sides of the given differential equation, we get
2
s Y (s)
sy (0)
0
y (0) + 5 jsY (s)
y (0)j + 6Y (s) = X (s)
X (s) = Lfe t u(t)g =
where
Substituting the initial conditions, we get
1
+ 2s + 11
s+1
2s2 + 13s + 12
Y (s) =
(s + 1)(s + 2)(s + 3)
(s2 + 5s + 6)Y (s) =
)
Using partial fraction expansion, we get
1
1
1
+6
Y (s) =
2 s+1
s+2
9
1
2 s+3
Taking inverse Laplace transform, we get
y(t) =
1 t
e + 6e
2
2t
9
e
2
3t
,
t0
1
s+1
Laplace Transform
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343
5.5.6 Integration in time domain
For a causal signal x(t),
Zt
x( )d;
y (t) =
If
0
Lfy(t)g = Y (s) = Xs(s)
then
Proof :
LBx(t)C = X(s) Z1
x(t)e st dt
0
Dividing both sides by s yields
X (s)
=
s
Z1
x(t)
st
e
s
0
dt
Integrating the right-hand side by parts, we get
Z1
1
X (s)
e ts
=
y (t)
s
s
t=0
y (t)
0
e
1
Z1
X (s)
e st = y (t)
+ y (t)e
s
s t=0
ts
s
st
( s)dt
dt
0
The first term on the right-hand side evaluates to zero at both limits, because
e
1=0
Z0
and y (0) =
x( )d = 0
0
Hence;
Y (s) =
X (s)
s
Thus, integration in time domain is equivalent to division by s in the s domain.
EXAMPLE
5.7
Consider the RC circuit shown in Fig. 5.11. The input is the rectangular pulse shown in Fig. 5.12.
Find i(t) by assuming circuit is initially relaxed.
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Figure 5.11
Figure 5.12
SOLUTION
Applying KVL to the circuit represented by Fig. 5.11, we get
Ri(t) +
)
Ri(t) +
Zt
1
i( )d = v (t)
C
0
Zt
1
i( )d = Vo [u(t
C
a)
u(t
b)]
0
Taking Laplace transforms on both the sides, we get
RI(s) +
)
Vo
(e as e bs )
Cs
s
Vo
R
I(s) =
(e as e
1
s+
RC
1
I(s) =
bs
)
We know the transform pair,
Lfe
at
u(t)g =
1
s+a
and then using the time-shift property, we can find inverse of I(s).
Vo
i(t) =
e
R
That is;
)
i(t) =
Vo h
R
t
RC
e
u(t)
(t
a)
RC
t
!t
u(t
vo
e
R
a
a)
5.5.7 Differentiation in the s domain
For a signal x(t), t 0, we have
Lf
tx(t)g =
dX (s)
ds
e
t
RC
(t
u(t)
b)
RC
t
!t
u(t
b
b)
i
Laplace Transform
Proof :
For a causal signal, x(t), the Laplace transform is given by
Lfx(t)g = X (s) =
Z1
x(t)e
st
dt
0
Differentiating both the sides with respect to s, we get
Z1
dX (s)
=
ds
In general;
EXAMPLE
)dt
[ tx(t)]e
st
dt
0
0
dX (s)
or Lftx(t)g =
ds
n
X (s)
Lftn x(t)g = ( 1)n d ds
n
Lf
Hence;
st
Z1
dX (s)
=
ds
)
x(t)( te
tx(t)g =
5.8
Find the Laplace transform of x1 (t) = te
3t u(t).
SOLUTION
We know that,
Hence
Lfe
at
u(t)g =
Lfe
3t
u(t)g =
Using the differentiation in s domain property,
1
s+a
1
s+3
Lfx1 (t)g = X1 (s) = dsd s +1 3
=
1
(s + 3)2
5.5.8 Convolution
If
and
then
Lfx(t)g = X (s)
Lfh(t)g = H (s)
Lfx(t) h(t)g = X (s)H (s)
where indicates the convolution operator.
dX (s)
ds
j
345
346
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Network Theory
Proof :
Z1
x(t) h(t) =
x( )h(t
1
)d
Since x(t) and h(t) are causal signals, the convolution in this case reduces to
Z1
x(t) h(t) =
x( )h(t
)d
0
2
Z1 Z1
Lfx(t) h(t)g = 4 x( )h(t
Hence;
3
)d 5 e
st
dt
0
0
Interchanging the order of integrals, we get
Lfx(t) h(t)g =
Z1
2
x( ) 4
0
Using the change of variable = t
3
Z1
h(t
)e
st
dt5 d
0
in the inner integral, we get
Lfx(t) h(t)g =
2
Z1
x( )e
s
0
4
Z1
h()e
s
3
d5 d
0
= X (s)H (s)
Please note that this theorem reduces the complexity of evaluating the convolution integral to
a simple multiplication.
EXAMPLE
5.9
Find the convolution of h(t) = e
t
and f (t) = e
2t .
SOLUTION
h(t) f (t) = L
1
fH (s)F (s)g
1
1
1
=L
s+1
s+2
1
1
1
=L
+
s+1
s+2
t
2
t
e
, t0
=e
1
Laplace Transform
EXAMPLE
j
347
5.10
Find the convolution of two indentical rectangular pulses. Each rectangular pulse has unit amplitude and duration equal to 2T seconds. Also, the pulse is centered at t = T .
SOLUTION
Let the pulse be as shown in Fig. 5.13.
From the Fig. 5.13, we can write
x(t) = u(t)
u(t
2T )
Taking Laplace transform, we get
X (s) =
1
1
s
s
1
= (1
Figure 5.13
e 2T s )
s
y (t) = x(t) x(t)
Let
Then;
Y (s) = X 2 (s)
=
)
2T s
e
T
T
Y (s) =
1
2T s 2
e
s
1
2
s2
s2
e
2T s
+
1
s2
e
4T s
Taking inverse Laplace transform, we get
y (t) = tu(t)
= r (t)
2(t
2T )u(t
2r (t
y(t) = x(t) x(t)
5.5.9
2T ) + (t
2T ) + r (t
4T )u(t
4T )
4T )
Figure 5.14
Initial-value theorem
The initial-value theorem allows us to find the initial value x(0)
directly from its Laplace transform X (s).
If x(t) is a causal signal,
x(0) = lim sX (s)
then;
s
!1
(5.9)
Proof :
To prove this theorem, we use the time differentiation property.
L
dx(t)
dt
Z1
= sX (s)
x(0) =
0
dx
e
dt
st
dt
The problems with are better understood after the inverse Laplace transforms are studied.
(5.10)
348
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If we let s ! 1, then the integral on the right side of equation (5.10) vanishes due to damping
factor, e st .
Thus;
lim [sX (s)
s
)
EXAMPLE
!1
x(0)] = 0
x(0) = lim sX (s)
s
!1
5.11
Find the initial value of
F (s) =
s+1
(s + 1)2 + 32
SOLUTION
f (0) = lim sX (s) = lim s
s
!1
s
s2 + s
s2 1 +
!1 s2
= lim
s
(s + 1)2 + 32
!1 (s + 1) 2 + 32
= lim
s
!1
s+1
1+
2
s
1
s
+
10
=1
s2
We know the transform pair:
Lfe
bt
cos atg =
s+b
(s + b)2 + a2
Hence, inverse Laplace transform of F (s) yields
f (t) = e t cos 3t
At t = 0, we get f (0) = 1.
This verifies the theorem.
5.5.10
Final-value theorem
The final-value theorem allows us to find the final value x(1) directly from its Laplace transform
X (s).
If x(t) is a causal signal,
then
Proof :
The Laplace transform of
lim x(t) = lim sX (s)
t
!1
s
dx(t)
is given by
dt
sX (s)
!0
Z1
x(0) =
0
dx(t)
e
dt
st
dt
Laplace Transform
Taking the limit s ! 0 on both the sides, we get
lim [sX (s)
s
!0
Z1
x(0)] = lim
!0
0
Z1
s
=
0
Z1
=
dx(t)
e
dt
st
dx(t) h
lim e
s 0
dt
!
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349
dt
st
i
dt
dx(t)
dt
dt
= x(t)j1
0
0
= x(1)
Since;
lim [sX (s)
s
we get;
!0
x(1)
Hence;
x(0)
x(0)] = lim [sX (s)]
!0
x(0) = lim [sX (s)
s !0
x(1) = lim [sX (s)]
s !0
s
x(0)
x(0)]
This proves the final value theorem.
The final value theorem may be applied if, and only if, all the poles of X (s) have a real part
that is negative.
The final value theorem is very useful since we can find x(1) from X (s). However, one
must be careful in using final value theorem since the function x(t) may not have a final value
a
as t ! 1. For example, consider x(t) = sin at having X (s) = 2
. Now we know
s + a2
lim sin at does not exit. However, if we uncarefully use the final value theorem in this case, we
t!1
would obtain:
a
=0
lim sX (s) = lim s 2
s !0
s !0 s + a 2
Note that the actual function x(t) does not have a limiting value as t ! 1. The final value
theorem has failed because the poles of X (s) lie on the j! axis. Therefore, we conclude that for
final value theorem to give a valid result, poles of X (s) should not lie to right side of the s-plane
or on the j! axis.
EXAMPLE
5.12
Find the final value of
X (s) =
Consider a function, X(s)
10
(s + 1)2 + 102
P (s)
. The roots of the denomoniator polymial, Q(s) are called poles () and the
Q(s)
roots of the numerator polynomial, P (s) are called zeros (O).
=
350
j
Network Theory
SOLUTION
lim x(t) = x(1)
t
!1
s
!0
s
s10
!0 (s + 1)2 + 102 = 0
= lim [sX (s)] = lim
We know the Laplace transform pair
L
Hence;
e
at
sin bt =
b
(s + a)2 + b2
x(t) = L
1
=L
1
fX (s)g
10
(s + 1)2 + 102
=e
t
sin 10t
x(1) = 0
Thus;
This verifies the result obtained from final-value theorem.
5.5.11
Time periodicity
Let us consider a function x(t) that is periodic as
shown in Fig. 5.15. The function x(t) can be
represented as the sum of time-shifted functions
as shown in Fig. 5.16.
Figure 5.15 A periodic function
Hence;
Figure 5.16 Decomposition of periodic function
x(t) = x1 (t) + x2 (t) + x3 (t) + = x1 (t) + x1 (t
T )u(t
T ) + x1 (t
2T )u(t
2T ) + (5.11)
where x1 (t) is the waveform described over the first period of x(t). That is, x1 (t) is the same as
the function x(t) gated over the interval 0 < t < T .
gating means the function x(t) is multiplied by 1 over the interval 0 t T and elsewhere by 0.
Laplace Transform
j
351
Taking the Laplace transform on both sides of equation (5.11) with the time-shift property
applied, we get
Ts
X (s) = X1 (s) + X1 (s)e
)
X (s) = X1 (s)(1 + e
1 + a + a2 + =
But
Hence, we get
1
1
a
+e
2T s
2T s
+ + )
jaj < 1
;
X (s) = X1 (s)
Ts
+ X1 (s)e
1
1
e
(5.12)
Ts
In equation (5.12), X1 (s) is the Laplace transform of x(t) defined over first period only.
Hence, we have shown that the Laplace transform of a periodic function is the Laplace transform
evaluated over its first period divided by 1 e T s .
EXAMPLE
5.13
Find the Laplace transform of the periodic signal x(t) shown in Fig. 5.17.
Figure 5.17
SOLUTION
From Fig. 5.17, we find that T = 2 Seconds.
The signal x(t) considered over one period is donoted as x1 (t) and shown in Fig. 5.18(a).
Figure 5.18(a)
Figure 5.18(b)
Figure 5.18(c)
352
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Network Theory
The signal x1 (t) may be viewed as the multiplication of xA (t) and g (t).
That is;
x1 (t) = xA (t)g (t)
= [ t + 1][u(t)
)
x1 (t) =
u(t
tu(t) + tu(t
1)]
1) + u(t)
=
tu(t) + (t
1 + 1)u(t
=
tu(t) + (t
1)u(t
u(t
1) + u(t)
1) + u(t
= u(t)
tu(t) + (t
1)u(t
= u(t)
r (t) + r (t
1)
1)
u(t
1) + u(t)
1)
u(t
1)
1)
Taking Laplace Transform, we get
X1 (s) =
=
Hence;
5.6
X(s) =
1
1
s
s
s2
+
1+e
1
s2
e
s
s
s2
X1 (s)
(s 1 + e s )
= 2
sT
1 e
s (1 e 2s )
Inverse Laplace transform
The inverse Laplace transform of X (s) is defined by an integral operation with respect to variable
s as follows:
1
x(t) =
2
Z
+j
j
1
1
X (s)est ds
(5.13)
Since s is complex, the solution requries a knowledge of complex variables. In otherwords,
the evaluation of integral in equation (5.13) requires the use of contour integration in the complex
plane, which is very difficult. Hence, we will avoid using equation (5.13) to compute inverse
Laplace transform.
In many situations, the Laplace transform can be expressed in the form
X (s) =
where
P (s)
Q(s)
P (s) = bm sm + bm
n
Q(s) = an s + an
(5.14)
m 1
1s
n 1
1s
+ + b1 s + b0
+ + a1 s + a0 ;
an 6= 0
The function X (s) as defined by equation (5.14) is said to be rational function of s, since
it is a ratio of two polynomials. The denominator Q(s) can be factored into linear factors.
Laplace Transform
A partial fraction expansion allows a strictly proper rational function
j
353
P (s)
to be expressed as a
Q(s)
factor of terms whose numerators are constants and whose denominator corresponds to linear or
a combination of linear and repeated factors. This in turn allows us to relate such terms to their
corresponding inverse transform.
For performing partial fraction technique on X (s), the function X (s) has to meet the following conditions:
(i) X (s) must be a proper fraction. That is, m < n. When X (s) is improper, we can use long
division to reduce it to proper fraction.
(ii) Q(s) should be in the factored form.
EXAMPLE
5.14
Find the inverse Laplace transform of
X (s) =
2s + 4
s2 + 4s + 3
SOLUTION
2s + 4
+ 4s + 3
2(s + 2)
K1
K2
=
=
+
(s + 1)(s + 3)
s+1
s+3
K1 = (s + 1)X (s)js= 1
X (s) =
where;
s2
2(s + 2) =
(s + 3) s=
1
=1
K2 = (s + 3)X (s)js=
Hence;
We know that:
Therefore;
EXAMPLE
3
2(s + 2) =
=1
(s + 1) s= 3
1
1
+
X (s) =
s+1
s+3
1
t
Lfe u(t)g = s + x(t) = [e t + e 3t ]u(t)
5.15
Find the inverse Laplace transform of
X (s) =
s2 + 2s + 5
(s + 3)(s + 5)2
354
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Network Theory
SOLUTION
K1
K2
K3
+
+
s+3
s+5
(s + 5)2
K1 = (s + 3)X (s)js= 3
X (s) =
Let
where
s2 + 2s + 5 =
(s + 5)2 s=
3
=2
1 d
[(s + 5)2 X (s)]
K2 =
1! ds
s=
5
+ 2s + 5 d
=
ds
s+3
s= 5
2
s + 6s + 1 =
= 1
(s + 3)2 s= 5
K3 = (s + 5 )2 X (s)s= 5
s2 + 2s + 5 =
= 10
(s + 3) s2
s= 5
1
s+5
2
X (s) =
s+3
Then
10
(s + 5)2
Taking inverse Laplace transform, we get
x(t) = 2e
3t
x(t) = (2e
or
e
3t
5t
e
10te
5t
5t
t0
;
10te
5t )u(t)
Reinforcement problems
R.P
5.1
Find the Laplace transform of: (a) cosh(at) (b) sinh(at)
SOLUTION
1
(a) cosh(at) = [eat + e at ]
2
We know the Laplace transform pair:
Lfe
and
at
g=
1
s+a
1
Lfeat g = s
a
Laplace Transform
Applying linearity property, we get,
Lfcosh(at)g = 12 Lfeat g + 12 Lfe
=
=
1
1
1
+
2 s a s+a
g
s
s2
1
(b) sinh at = [eat e at ]
2
Applying linearity property,
a2
Lfsinh(at)g = 12 s 1 a
a
= 2
s
a2
R.P
at
1
s+a
5.2
Find the Laplace transform of f (t) = cos(!t + ).
SOLUTION
f (t) = cos(!t + )
Given
= cos cos !t
sin sin !t
Applying linearity property, we get,
Lff (t)g = F (s)
= cos Lfcos !tg
= cos =
R.P
s
s2
+
!2
sin Lfsin !tg
sin s cos θ ω sin θ
s2 + ω 2
!
s2
+ !2
5.3
Find the Laplace transform of each of the following functions:
(a) x(t) = t2 cos(2t + 30 )u(t)
(b) x(t) = 2tu(t)
4
t
(c) x(t) = 5u
(d) x(t) = 5e
3
t
2
u(t)
d
(t)
dt
j
355
356
j
Network Theory
SOLUTION
(a) Let us first find the Laplace transform of cos(2t + 30 )u(t)
L fcos(2t + 30 )u(t)g = L fcos 30 cos 2tu(t)
= cos 30 Lfcos 2tu(t)g
= cos 30
=
sin 30 sin 2tu(t)g
sin 30 Lfsin 2tu(t)g
2
sin 30
s2 + 4
s
s2
+4
2 sin 30
s2 + 4
s cos 30
The Laplace transform of x(t) is now found by using differentiation in s domain property.
L
d2 L
fcos(2t + 30 )u(t)g
2
ds d2 s cos 30
2 sin 30
= 2
ds
s2 + 4
t2 cos(2t + 30 ) =
3
2p
3
s 17
d2 6
2
7
= 26
ds 4 s2 + 4 5
2p
3
3
s 17
d d 6
6 2
7
=
ds ds 4 s2 + 4 5
d d
=
ds ds
d
=
ds
" p
" p
3
( 2s)
= 2
(s2 + 4)2
(b) x(t) = 2tu(t)
4
8
F
12 3s
s2 + 4
1
3 2
s +4
2
p
=
!
3
s
2
!
p
3
s 1
2 2
(s2 + 4)2
6s2 +
3
s
2
2s
F
!
p
1
p
(s2 + 4)3
#
1
s2 + 4
1
3
2
+
(s2 + 4)2
2s
4L
3
s
2
(s2 + 4)3
d
(t)
dt
Lfx(t)g = X (s) = 2Lftu(t)g
#
2
p
8s2
3s2
d
(t)
dt
!
1
Laplace Transform
We know that whenever a function f (t) is discontinuous at the origin, we have L
357
j
d
f (t)
dt
= sF (s) f (0 ). Applying this relation to the second term on the right side of the above
equation, we get
X (s) = 2
=
=
1
4[s 1
s2
2
4[s
s2
2
s2
(0 )]
0]
4s
t
(c) x(t) = 5u
3
Using scaling property,
s
Lff (at)g = a1 F
a
Lfx(t)g = X (s) = 5 11=3 Lfu(t)g
we get;
1
1
=5
1=3
s
=
s
!
5
s
t
(d) x(t) = 5e 2 u(t)
We know the Laplace transform pair:
Lfe
at
u(t)g =
1
s+a
Lfx(t)g = X (s) = 5L
Hence;
=5
R.P
n
e
1
s+
1
2
1
t
2
=
5.4
Find the Laplace transform of the following functions:
(a) x(t) = t cos at
(b) x(t) =
(c) x(t) =
1
sin at sinh(at)
2a2
sin2 !t
t
o
u(t)
10
2s + 1
s
1
3
s
!
s
1
3
358
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Network Theory
SOLUTION
(a) x(t) = t cos at
d
F (s)
ds
f (t) = cos at
s
F (s) = 2
s + a2
Lftf (t)g =
We know that
Let
)
s
d
ds s2 + a2
Lft cos atg = Lftf (t)g =
Hence
=
x(t) =
(b)
s2
a2
(s2 + a2 )2
1
sin at sinh at
2a2
1
2a2
1
= 2
4a
=
1 at
e sin at
2
eat sin at
1
e
2
e
at
at
sin at
sin at
We know the shifting in s domain property:
Lfes t f (t)g = F (s)js!(s
0
s0 )
Applying this property along with linearity property, we get
Lfx(t)g = X (s)
1 Lfeat sin atg
4a2 1
a = 2
4a s2 + a2 s!s
Lfe
=
1
= 2
4a (s
=
(c) x(t) =
1
t
[(s
at
sin atg
2
2
s +a a
a
a
a)2 + a2
s
!s+a
a
(s + a)2 + a2
s
a)2 + a2 ] [(s + a)2 + a2 ]
sin2 !t
We know that
Lff (t)g = F (s) =
Z1
f (t)e
0
st
dt
Laplace Transform
Z1
Hence;
Z1
F (s)ds =
s
Z1
f (t)
s
0
Z1
f (t)
=
e
0
Z1
=
0
=L
st
e
f (t)
e
t
f (t)
t
1
t
s
1 j!t 2
e
j2
ej 2!t 2 + e j 2!t
=
4 1
1 1
1
+
F (s) =
4 s j 2!
2 s
1 j!t
=
e
j2
Hence;
Hence,
X (s) = L
=L
1
2
1
f (t)
!1
= lim
x
1
ln
4
=
R.P
Zx
x
s
F (s)ds = lim
=
1
1
4 s + j 2!
sin !t
t
t
Z1
=
dt
dt
f (t) = sin2 !t
In the present case;
359
dsdt
st
st
j
ln(x
!1
f (x)dx
s
j 2! )
x2 + 4! 2
x2
s2 + 4ω
1
ln
4
s2
ln(s
2
x
!1
+
j 2! )
1
ln
4
2 ln x + 2 ln s + ln(x + j 2! )
4
2
2
s + 4!
ln(s + j 2! )
s2
5.5
Consider the pulse shown in Fig. R.P. 5.5, where f (t) = e2t for 0 < t < T . Find F (s) for the
pulse.
360
j
Network Theory
Figure R.P. 5.5
SOLUTION
The discrete pulse f (t) could be imagined as the product of signal x(t) and g (t) as shown in Fig. R.P. 5.5(a)
and (b) respectively.
That is;
f (t) = x(t)g (t)
= e2t [u(t)
2t
= e u(t)
2t
= e u(t)
2t
= e u(t)
Hence;
u(t
T )]
2t
e u(t
e
e
2T 2(t T )
=
1
s
1
e
u(t
e
T (s 2)
2
s
e T (s
(s 2)
T)
u(t
T)
e2T e T s
Lff (t)g = F (s) = s 1 2
=
T)
2(t T +T )
2)
s
2
2
Figure R.P. 5.5(a)
Alternate method:
Lff (t)g = F (s) ZT
=
0
=
1
Z1
f (t)e
0
e2t e
st
dt
e T (s
(s 2)
2)
st
dt
Laplace Transform
R.P
5.6
Find the Laplace transform of f (t) shown in Fig. R.P. 5.6.
Figure R.P. 5.6
SOLUTION
f (t) is a discrete pulse and can be expressed mathematically as:
f (t) = x(t)g (t)
)
= sin t[u(t)
u(t
= sin tu(t)
sin tu(t
= sin tu(t)
sin[ (t
1 + 1)]u(t
1)
f (t) = sin tu(t)
sin[ (t
1)] cos (t
1)
cos[ (t
1)]
1)] sin u(t
= sin tu(t) + sin[ (t
Hence;
F (s) = Lff (t)g =
=
R.P
1)
1)
1)]u(t
1)
1s e
+ 2
s2 + 2
s + 2
π
[1 + e s ]
s2 + π 2
Figure R.P. 5.6(a)
5.7
Determine the Laplace transform of f (t) shown in Fig. R.P. 5.7.
Figure R.P. 5.7
j
361
362
j
Network Theory
SOLUTION
We can write,
f (t) = x(t)g (t)
5
t [u(t) u(t 2)]
2
5
5
= tu(t)
tu(t 2)
2
2
5
5
= tu(t)
(t 2 + 2)u(t 2)
2
2
5
5
= tu(t)
(t 2)u(t 2) 5u(t 2)
2
2 5 1
5 1
5 2s
e 2s
e
Hence; Lff (t)g = F (s) =
2
2
2 s
2 s
s
5
= 2 [1 e 2s 2se 2s ]
2s
=
R.P
Figure R.P. 5.7(a)
5.8
Find the Laplace transform of f (t) shown in Fig. R.P. 5.8.
Figure R.P. 5.8
SOLUTION
The equation of a straight line is y = mx + c, where m = slope of the line and c = intercept on
y -axis.
Hence, f (t) =
When f (t) =
5
t+5
3
2, let us find t.
Laplace Transform
j
363
5
t+5
3
)
t = 4:2 Seconds
Mathematically,
That is;
2=
f (t) = x(t)g (t)
5
=
t + 5 [u(t) u(t 4:2)]
3
5
5
=
tu(t) + tu(t 4:2) + 5u(t) 5u(t 4:2)
3
3
5
5
=
tu(t) + (t 4:2 + 4:2)u(t 4:2)
3
3
+5u(t) 5u(t 4:2)
5
5
=
tu(t) + (t 4:2)u(t 4:2) + 7u(t 4:2)
3
3
+5u(t) 5u(t 4:2)
5
5
=
tu(t) + (t 4:2)u(t 4:2) + 2u(t 4:2) + 5u(t)
3
3
Hence;
F (s) = Lff (t)g
5
5
2
= 2 + 2 e 4:2s + e
3s
3s
s
4
:
2
s
+ 6se
5 + 5e
=
3s2
R.P
Figure R.P. 5.8(a)
4:2s
+
4:2s
5.9
5
s
+ 15s
If f (0 ) = 3 and 15u(t) 4 (t) = 8f (t) + 6f 0 (t), find f (t) (hint: by taking the Laplace
transform of the differential equation, solving for F (s) and by inverting, find f (t)).
SOLUTION
Given,
15u(t) 4 (t) = 8f (t) + 6f 0 (t)
Taking Laplace transform on both the sides, we get
15
s
)
Therefore;
15
s
4 = 8F (s) + 6[sF (s)
4 = 8F (s) + 6sF (s) + 18
F (s)(6s + 8) =
)
f (0 )]
18 +
4s
15
s
18
15 4s
+
(6s + 8) s(6s + 8)
K
K2
22s + 15
= 1+
= 4
4
s
s+
6s s +
3
3
F (s) =
364
j
Network Theory
The constants K1 and K2 are found using the theory of partial fractions.
22s + 15 K1 = = 1:875
4 6 s+
3 s=0
22s + 15 K2 =
4 = 5:542
6s
s=
1:875
5:542
4
s+
3
Hence;
F (s) =
Taking the inverse, we get
f (t) = 1.875
R.P
s
h
3
5.542e
4
t
3
i
u(t)
5.10
Find the inverse Laplace transform of the following functions:
s+1
(a) F (s) = 2
s + 4s + 13
(b) F (s) =
s2
3e s
+ 2s + 17
SOLUTION
F (s) =
(a)
s+1
(s + 2)2 + 9
(s + 2) 1
(s + 2)2 + 9
s+2
=
(s + 2)2 + 32
s+2
=
(s + 2)2 + 32
=
1
(s + 2)2 + 32
3
1
3 (s + 2)2 + 32
The determination of the Laplace inverse makes use of the following two Laplace transform
pairs:
Lfe
bt
a
sin atg =
(s + b)2 + a2
Lfe bt cos atg = (s +sb+)2 b+ a2
Hence;
f (t) = L
=e
(b)
F (s) =
1
fF (s)g
2t
cos 3t
3e s
s2 + 2s + 17
1
e
3
2t
sin 3t
Laplace Transform
Let
F (s) = e
where
X (s) =
)
Since
we get;
s
j
365
X (s)
3
3
=
+ 2s + 17
(s + 1)2 + 42
3
4
=
4 (s + 1)2 + 42
3
x(t) = e t sin 4t
4
F (s) = e s X (s)
s2
f (t) = x(t
1)
3 (t 1)
Therefore;
f (t) = e
sin[4(t 1)];
t>1
4
3
f (t) = e (t 1) sin[4(t 1)]u(t 1)
4
Laplace transform method for solving a set of differential equations:
1. Identify the circuit variables such as inductor currents and capacitor voltages.
2. Obtain the differential equations describing the circuit and keep a watch on the initial conditions of the circuit variables.
3. Obtain the Laplace transform of the various differential equations.
4. Using Cramer’s rule or a similar technique, solve for one or more of the unknown variables,
obtaining the solution in s domain.
5. Find the inverse transform of the unknown variables and thus obtain the solution in the time
domain.
R.P
5.11
Referring to the RL circuit of Fig. R.P. 5.11, (a) write a differential equation for the inductor
current iL (t). (b) Find IL (s), the Laplace transform of iL (t). (c) Solve for iL (t).
Figure R.P. 5.11
366
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Network Theory
SOLUTION
(a) Applying KVL clockwise, we get
10iL (t) + 5
diL
dt
5u(t
2) = 0
(b) Taking Laplace transform of the above equation, we get
5
10IL (s) + 5[sIL (s) iL (0 )] = e 2s
s
5
)
IL (s) = s
e
2s
e
2s
=e
2s
=
+ 5iL (0 )
5s + 10
+ 5 10
s(s + 2)
1
IL (s) = e
2
Hence;
5 10 3 s
K1
K2
+
+
s
s+2
s(s + 2)
1 1
K1 =
=
s + 2 s=0
2
1
1
K2 = =
s s= 2
2
where
3s
2s
1
s
5 10 3
1
+
s+2
(s + 2)
(c) Taking Inverse Laplace transform, we get
1
u(t) e 2t u(t) t!t 2 + 5 10 3 e 2t u(t)
iL (t) =
2
1
=
u(t 2) e 2t u(t 2) + 5 10 3 e 2t u(t)
2
R.P
5.12
Obtain a single integrodifferential equation in terms of iC for the circuit of Fig. R.P. 5.12. Take
the Laplace transform, solve for IC (s), and then find iC (t) by making use of inverse transform.
Figure R.P. 5.12
Laplace Transform
j
367
SOLUTION
Applying KVL clockwise to the right-mesh, we get
Z1
iC dt + 4[iC
4u(t) + iC + 10
0:5 (t)] = 0
0
Taking Laplace transform, we get
1
10IC (s)
4 + IC (s) +
+ 4IC (s)
s
s
2=0
2s 4
5s + 10
1.6
= 0.4
s+2
)
IC (s) =
Taking inverse Laplace transform, we get
iC (t) = 0.4δ(t)
R.P
1.6e
2t u(t)
Amps.
5.13
Refer the circuit shown in Fig. R.P. 5.13. Find i(0) and i(1) using initial and final value theorems.
Figure R.P. 5.13
SOLUTION
Applying KVL we get
i+2
di
= 10
dt
Taking Laplace transform, on both the sides, we get
I (s) + 2[sI (s)
)
)
i(0 )] =
I (s) + 2[sI (s)
1] =
I (s)[1 + 2s] =
10
s
10
s
10
s
+2
368
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Network Theory
)
10
2
+
s(1 + 2s)
1 + 2s
10 + 2s
=
s(1 + 2s)
5+s
= 1
s s+
2
I (s) =
According to initial value theorem,
i(0) = lim sI (s)
s
!1
(s + 5)
= lim s s!1
1
s s+
2
5
1+
s =1
= lim
1
s!1
1+
2s
We know from fundamentals for an inductor, i(0+ ) = i(0 ) = i(0). Hence, i(0) found using
initial value theorem verifies the initial value of i(t) given in the problem.
From final value theorem,
i(1) = lim sI (s)
s
!0
5
s(s + 5)
=
= 10 A
= lim s !0
1
1=2
s s+
2
R.P
5.14
Find i(t) and vC (t) for the circuit shown in Fig. R.P. 5.14 when vC (0) = 10 V and i(0) = 0 A. The
input source is vi = 15u(t) V. Choose R so that the roots of the characteristic equation are real.
Figure R.P. 5.14
Laplace Transform
j
369
SOLUTION
Applying KVL clockwise, we get
L
di
+ vC + Ri = vi (t)
dt
(5.15)
The differential equation describing the variable vC is
C
dvC
=i
dt
(5.16)
The Laplace transform of equation (5.15) is
L[sI (s)
i(0) + VC (s) + RI (s) = Vi (s)]
(5.17)
The Laplace transform of equation (5.16) is
C [sVC (s)
vC (0) = I (s)]
(5.18)
Noting that i(0) = 0, substituting for C and L and rearranging equation (5.17) and (5.18), we
get,
[R + s]I (s) + VC (s) = Vi (s) =
I (s) +
1
sVC (s) = 5
2
15
s
(5.19)
(5.20)
Putting equations (5.19) and (5.20) in matrix form, we get
2
4
R+s
1
3"
# 2 15 3
1
I (s)
=4 s 5
1 5
VC (s)
s
5
2
(5.21)
Solving for I (s) using Cramer’s rule, we get
I (s) =
5
s2 + Rs + 2
The inverse Laplace transform of I (s) will depend on the value of R. The equation
s2 + Rs + 2 = 0 is defined as the characteristic equation. For the roots of this equation to
be real, it is essential that b2 4ac 0 .
R2
This means that;
)
The condition b2
4120
p
R2 2
4ac 0 is with respect to algebraic equaion ax2 + bx + c = 0.
370
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Network Theory
Let us choose the value of R as 3
I (s) =
Then
)
where
Hence;
I (s) =
K1 =
5
5
=
+ 3s + 2
(s + 1)(s + 2)
s2
K2
K1
+
s+1 s+2
5 s + 2 s =
1
=5
5 =
K2 =
s + 1 s = 2
5
5
I (s) =
s+1
s+2
5
Taking inverse Laplace transform, we get
i(t) = 5e t u(t)
5e
2t u(t)
Please note that t = 0 gives i(0) = 0 and t = 1 gives i(1) = 0.
Solving the matrix equation (5.21) for VC (s), using Cramer’s rule, we get
VC (s) =
10s2 + 10Rs + 30
s(s2 + Rs + 2)
Substituting the value of R, we get
VC (s) =
10(s2 + 3s + 3)
s(s + 1)(s + 2)
Using partial fraction expansion, we can write,
VC (s) =
where, K1 = 15, K2 =
Hence;
K1
K2
K3
+
+
s
s+1
s+2
10, K3 = 5
VC (s) =
15
s
10
5
+
s+1
s+2
Taking inverse Laplace transform, we get
vC (t) = 15u(t)
10e t u(t) + 5e
Verification:
Putting t = 0, we get
vC (0) = 15
10 + 5 = 10 V
vC (1) = 15
0 + 0 = 15 V
This checks the validity of results obtained.
2t u(t)
Laplace Transform
R.P
j
371
5.15
For the circuit shown in Fig. R.P. 5.15, the steady state is reached with the 100 V source. At t = 0,
1
switch K is opened. What is the current through the inductor at t = seconds ?
2
Figure R.P. 5.15
SOLUTION
At t = 0 , the circuit is as shown in Fig. 5.15(a).
i2 (0+ ) = i2 (0 ) = 2:5 A
Figure R.P. 5.15(a)
Figure R.P. 5.15(b)
For t 0+ , the circuit diagram is as shown in Fig. 5.15(b). Applying KVL clockwise to the
circuit, we get
80i(t) + 4
di
=0
dt
372
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Network Theory
Taking Laplace transform, we get
80I (s) + 4[sI (s)
)
i(0 )] = 0
80I (s) + 4sI (s) = 4 2:5
)
[20 + s]I (s) = 2:5
2:5
I (s) =
s + 20
Taking inverse Laplace transform, we get,
i(t) = 2:5e
At t = 0:5 sec, we get
i(0.5) = 2.5e
R.P
10
20t
= 1.135 10
4
A
5.16
Refer the circuit shown in Fig. R.P. 5.16. Find:
(a) vo (t) for t 0
(b) io (t) for t 0
(c) Does your solution for io (t) make sense when t = 0? Explain.
Figure R.P. 5.16
SOLUTION
Figure R.P. 5.16(a)
Laplace Transform
j
373
KCL at node A: (for t 0)
Idc =
1
L
Zt
vo dt +
0
vo
dvo
+C
R
dt
Taking Laplace transform,
Idc
Vo (s)
Vo (s)
=
+
+ sCVo (s)
s
sL
R
Idc
C
Vo (s) =
1
1
s+
s2 +
RC
LC
Hence;
Substituting the values of Idc , R, L, and C , we find that
Vo (s) =
120; 000
120; 000
=
s2 + 10; 000s + 16 106
(s + 2000)(s + 8000)
Using partial fractions, we get
Vo (s) =
where K1 = 20, and K2 =
Hence;
K1
K2
+
s + 2000
s + 8000
20
Vo (s) =
20
20
s + 2000
s + 8000
Taking inverse Laplace transform, we get
vo (t) = 20e
(b)
io (t) = C
Hence
2000t u(t)
20e
8000t u(t)
dvo
dt
Io (s) = C [sVo (s)
vo (0)]
For t 0 , since the switch was in closed state, the circuit was not activated by the source.
This means that vo (0) = vo (0 ) = vo (0+ ) = 0 and iL (0+ ) = iL (0 ) = 0:
Then;
Io (s) = CsVo (s)
25 10 9 s 120; 000
s2 + 10; 000s + 16 106
3 10 3 s
=
(s + 2000)(s + 8000)
=
=
K1
s + 2000
+
K2
s + 8000
374
Network Theory
j
We find that, K1 =
10
3,
Therefore;
and K2 = 4 10
Io (s) =
3
10 3
4 10 3
+
s + 2000
s + 8000
Taking inverse Laplace transform, we get
io (t) = 4e
(c)
io (0+ ) = 4
8000t u(t)
e
2000t u(t)mA
1 = 3mA
Yes. The initial inductor current is zero by hypothesis iL (0+ ) = IL (0 ) = 0 . Also, the initial
resistor current is zero because vo (0+ ) = vo (0 ) = 0. Thus at t = 0+ , the source current
appears in the capacitor.
R.P
5.17
Refer the circuit shown in Fig. R.P. 5.17. The circuit parameters are R = 10kΩ, L = 800 mH
and C = 100nF, if Vdc = 70V, find:
(a) vo (t) for t 0
(b) io (t) for t 0
(c) Use initial and final value theorems to check the inital and final values of current and
voltage.
Figure R.P. 5.17
SOLUTION
At t = 0 , switch is open and at t = 0+ , the switch is closed. Since at t = 0 , the circuit is not
energized by dc source, io (0 ) = 0 and vo (0 ) = 0. Then by the hypothesis, that the current in
an inductor and voltage across a capacitor cannot change instantaneously,
io (0+ ) = io (0 ) = 0 and
vo (0+ ) = vo (0 ) = 0
Laplace Transform
The KCL equation when the switch is closed (for t 0+ ) is given by
vo
1
dvo
+
+
C
dt
R
L
Zt
(vo
Vdc )d = 0
0
Zt
dvo
vo
1
C
+
+
dt
R
L
)
)
C
vo d =
0
Zt
dvo
vo
1
+
+
dt
R
L
vo d =
0
Zt
1
Vdc d
L
1
L
0
Vdc t
Laplace transform of the above equation gives
C [sVo (s)
vo (0)] +
Vo (s)
1 Vo (s)
1 Vdc
+
=
R
L s
L s2
Since vo (0) is same as vo (0 ), we get
CsVo (s) +
Vo (s)
1 Vo (s)
Vdc
+
=
R
L s
Ls2
)
Vdc
LC
Vo (s) =
s
s2
+
1
RC
s+
1
LC
Substituting the values of Vdc , R, L, and C , we get
875 106
s [s2 + 1000s + 1250 104 ]
875 106
=
s(s s1 )(s s2 )
Vo (s) =
where
s1 ; s 2 =
=
Hence;
Vo (s) =
500 p
25 104
1250 104
500 j 3500
s(s + 500)
875 106
j 3500)(s + 500 + j 3500)
Using partial fractions, we get
Vo (s) =
We find that
875 106
= 70
125 105
875 106
K2 =
( 500 + j 3500)(j 7000)
K1 =
K1
K2
K2
+
+
s
s + 500 j 3500
s + 500 + j 3500
j
375
376
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Network Theory
p
= 5 50 /171:87
p
p
70
5 50 /171:87
5 50 / 171:87
Vo (s) =
+
+
s
s + 500 j 3500
s + 500 + j 3500
Taking inverse Laplace transform, we get
h
p
vo (t) = 70 + 5 50 /171:87 e
(500 j 3500)t
p
+ 5 50 / 171:87 e
(500+j 3500)t
i
u(t)
The inverse of Vo (s) can be expressed in a better form by following the technique described
below:
Let us consider a transformed function
C + jd
C jd
+
s + a j!
s + a + j!
m/ m /
+
=
s + a j!
s + a + j!
F (s) =
m=
where
p
c2 + d2 and = tan
1
d
c
The inverse transform of F (s) is given by
f (t) = 2me
at
cos(!t + )u(t)
(For the proof see R.P. 5.19)
In the present context,
p
m = 5 50; = 171:87
! = 3500 and a = 500
L
This means that,
1
)
p
5 50 /171:87
5 50 / 171:87
+
s + 500 j 3500
s + 500 + j 3500
( p
p
= 2 5 50e 500t cos (3500t + 171:87 )
p
= 10 50e 500t cos (3500t + 171:87 )
Hence;
F
vo (t) = 70 + 10 50e
500t cos (3500t
+ 171.87 ) u(t)
Laplace Transform
(b)
vo
dvo
+C
R
dt
io (t) =
Taking Laplace transforms on both the sides, we get
Vo (s)
+ C sVo (s)
R
Vo (s)
Io (s) =
+ CsVo (s)
R
Io (s) =
)
)
Io (s) = CVo (s) s +
=
Vdc
L
2
vo (0 )
1
RC
s+
6
6 4
s s2 +
1
3
1
RC
RC
s+
1
7
7
5
LC
Substituting the values of Vdc , R, L, and C , we get
Io (s) =
=
87:5(s + 1000)
s(s + 500 j 3500)(s + 500 + j 3500)
K1
K2
K2
+
+
s
s + 500 j 3500
s + 500 + j 3500
We find that,
87:5 1000
= 7mA
1250 104
87:5(500 + j 3500)
K2 =
( 500 + j 3500)(j 7000)
= 12:5 / 106:26 mA
7
12:5 / 106:26
12:5 /106:26
+
Io (s) = +
s
s + 500 j 3500
s + 500 + j 3500
K1 =
The inverse Laplace transform yields,
h
io (t) = 7 + 12:5 / 106:26 e
= 7 + 25e
(c)
s
s2
+
1
RC
s+
1
+ 12:5 /106:26 e
106.26 ) u(t)mA
500t cos(3500t
Vdc
LC
Vo (s) =
(500 j 3500)t
(500+j 3500)t
i
u(t)
LC
Vdc LC
= 70V
LC
The same result may be obtained by putting t = 1 in the expression for vo (t).
From Final Value theorem:
vo (1) = lim vo (t) = lim sVo (s) =
t
!1
s
!0
j
377
378
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Network Theory
From initial value theorem :
vo (0) = lim sVo (s)
s
!1
!1 6 ss2 +
6s
= lim
s
Vdc
LC
1
RC
s+
1
LC
=0
This verifies our beginning analysis that vo (0+ ) = vo (0 ) = 0. The same result may be
obtained by putting t = 0 in the expression for vo (t). 1
s+
Vdc
RC
We know that,
Io (s) =
1
1
L
2
s s +
s+
RC
From final value theorem :
Io (1) = lim sIo (s)
s
!0
= lim 6 s
s
!0
Vdc
L
s2
+
1
s+
6s
LC
RC
1
RC
s+
1
LC
1
Vdc RC
L 1
LC
Vdc
70
=
= 7 mA
=
R
10 103
=
The same result may be obtained by putting t = 1 in the expression for io (t).
From initial value theorem :
io (0) = lim sIo (s)
s
!1
2
6 Vdc
= lim s 6
s!1 4 L
s+
s s2 +
=0
1
3
RC
1
RC
s+
1
7
7
5
LC
This agrees with our initial analysis that the initial current through the inductor is zero. The
same result can be obtained by putting t = 0 in the expression for io (t).
Laplace Transform
R.P
j
379
5.18
Apply the initial and final value theorems to each of the functions given below:
s2 + 5s + 10
s2 + 5s + 10
(a) F (s) =
(b) F (s) =
s+6
5(s2 + 6s + 8)
SOLUTION
Since in F (s) referred in (a) and (b) are improper 1 fractions, the corresponding time domain
counterparts, f (t) contain impulses.
Thus, neither the initial value theorem nor the final value theorems may be applied to these
transformed functions.
R.P
5.19
c + jd
c jd
+
s + a j!
s + a + j!
Find the inverse Laplace transform of F (s) =
SOLUTION
Expressing c + jd and c
jd in the exponenetial from, we get,
F (s) =
m=
where
mej
me j
+
s + a j!
s + a + j!
p
f (t) = L
Hence;
c2
1
+
d2
(a j! )t
= me e
= 2me
d
c
fF (s)g
j
= me
and = tan
1
at j ( +!t)
e
"
at
e
(a+j! )t
e
j ( +!t)
u(t) + me
j
u(t) + me
at
ej (+!t) + e
j ( +!t)
2
#
u(t)
u(t)
u(t)
= 2me at cos(θ + ωt)u(t)
R.P
5.20
Find the initial and final values of f (t) when F (s) =
s2
60
2s + 1
SOLUTION
Initial value theorem
f (0) = lim sF (s)
s
!1
= lim s
s
!1
s2
60
=0
2s + 1
1
If the degree of the numerator polynomial is greater than or equal to the degree of the denominator polynomial,
the fraction is said to be improper.
380
Network Theory
j
Final value theorem:
The poles of F (s) are given by finding the roots of the denominator polynomial. That is,
s2
2s + 1 = 0
)
(s
1)2 = 0
)
s = 1; 1
Since both the poles of F (s) lie to the right of the s plane, final value theorem cannot be used
to find f (1).
R.P
5.21
Find i(t) for the circuit of Fig.R.P. 5.21, when i1 (t) = 7e
i(1).
6t
A for t 0 and i(0) = 0. Also find
Figure R.P. 5.21
SOLUTION
Refer Fig. R.P. 5.21(a).
KCL at node v1 :
v1
5
+ i = 7e
6t
di
dt di
1
3i + 4
+ i = 7e 6t
Hence;
5
dt
4 di 8
)
+ i = 7e 6t
5 dt 5
35
di
+ 2i = e 6t
)
dt
4
Also;
v1 = 3i + 4
Figure R.P. 5.21(a)
Taking Laplace transform of the differential equation, we get
[sI (s)
)
35 1
4 s+6
35
1
I (s) =
4 (s + 2)(s + 6)
i(0)] + 2I (s) =
Laplace Transform
j
381
Using partial fraction expansion, we get
I (s) =
35
35
and K2 =
16 16
1
35
1
35
I (s) =
16 s + 2
16 s + 6
35 2t
i(t) =
e 6t u(t)
e
16
K1 =
and find that
Hence;
5.7
K1
K2
+
s+2
s+6
Circuit element models and initial conditions
In the analysis of a circuit, the Laplace transform can be carried one step further by transforming
the circuit itself rather than the differential equation. Earlier we have seen how to represent a
circuit in time domain by differential equations and then use Laplace transform to transform the
differential equations into algebraic equations. In this section, we will see how we can represent
a circuit in s domain using the Laplace transform and then analyze it using algebraic equations.
5.7.1
Resistor
The voltage-current relationship for a resistor R is given by Ohm’s law:
v (t) = i(t)R
(5.22)
Taking Laplace transform on both the sides, we get
V (s) = I (s)R
(5.23)
Fig. 5.19 (a) shows the representation of a resistor in time domain and Fig. 5.19(b) in frequency domain using Laplace transform.
Figure 5.19(a) Resistor represented in
Figure 5.19(b) Resistor represented in the
time domain
frequency domain using Laplace transform
The impedance of an element is defined as
Z (s) =
V (s)
I (s)
382
j
Network Theory
provided all initial conditions are zero. Please note that the impedance is a concept defined only
in frequency domain and not in time domain. In the case of a resistor, there is no initial condition
to be set to zero. Comparision of equations (5.22) and (5.23) reveals that, resistor R has same
representation in both time and frequency domains.
5.7.2 Capacitor
For a capacitor with capacitance C , the time-domain voltage-current relationship is
v (t) =
1
C
Zt
i( )d + v (0)
(5.24a)
0
The s domain characterization is obtained by taking the Laplace transform of the above equation. That is,
V (s) =
1
Cs
I (s) +
v (0)
s
(5.24b)
To find the impedance of a capacitor, set the initial condition v (0) to zero. Then from equation
V (s)
1
(5.24b), we get Z (s) =
=
as the impedance of the capacitor. With the help of equation
I (s)
Cs
(5.24b), we can draw the frequency domain representation of a Capacitor and the same is shown
in Fig. 5.20(b). This equivalent circuit is drawn so that the KVL equation represented by equation
(5.24 b) is satisfied. Performing source transformation on the equivalent s domain circuit for a
capacitor which is shown in Fig. 5.20(b), we get an alternate frequency domain representation as
shown in Fig. 5.20(c).
Figure 5.20(a) A capacitor represented in time domain
(b) A capacitor represented in the frequency domain
(c) Alternate frequency domain representation for a capacitor
Laplace Transform
j
383
5.7.3 Inductor
For an inductor with inductance L, the time domain voltage-current relation is
v (t) = L
di(t)
dt
(5.25)
The Laplace transform of equation (5.25) yields,
V (s) = LsI (s)
Li(0)
(5.26)
To find impedance of an inductor, set the initial condition i(0) to zero. Then from equation
(5.26), we get
Z (s) =
V (s)
= Ls
I (s)
(5.27)
which represents the impedance of the inductor. Equation (5.26) is used to get the frequency
domain representation of an inductor and the same is shown in Fig. 5.21(b). The series connection
of elements corresponds to sum of the voltages in equation (5.26). Converting the voltage source
in Fig.5.21(b) into an equivalent current source, we get an alternate representation for the inductor
in frequency domain which is as shown in Fig. 5.21(c).
To find the frequency domain representation of a circuit, we replace the time domain representation of each element in the circuit by its frequency domain representation.
Figure 5.21(a) An inductor represented in time domain
(b) An inductor represented in the frequency domain
(c) An alternate frequency domain representation
To find the complete response of a circuit, we first get its frequency domain representation.
Next, using KV L or KCL, we find the variables of interest in s doamin. Finally, we use the
inverse Laplace transform to represent the variables of interest in time domain.
384
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Network Theory
EXAMPLE
5.16
Determine the voltage vC (t) and the current iC (t) for t 0 for the circuit shown in Fig. 5.22.
C
C
Figure 5.22
SOLUTION
We shall analyze this circuit using nodal technique. Hence we represent the capacitor in frequency
domain by a parallel circuit since it is easier to account for current sources than voltage sources
while handling nodal equations.
The symbol for switch indicates that at t = 0 it is closed and at t = 0+ , it is open. The
circuit at t = 0 is shown in Fig. 5.23(a). Let us assume that at t = 0 , the circuit is in steady
state. Under steady state condition, capacitor acts as on open circuit as shown in Fig. 5.23(a).
26
12
4
=
=
6+3
9
3
4
vC (0 ) = 3 = 4V
3
vC (0) = vC (0+ ) = vC (0 ) = 4V
i1 (0 ) =
Hence;
Fig. 5.23(b) represents the frequency
domain representation of the circuit
shown in Fig. 5.22.
Figure 5.23(a)
KCL at top node:
VC (s)
2
s
+ VC (s) = 2 +
2
s
6
2
)
VC (s) =
2
s
s+
3
3
Figure 5.23(b)
Laplace Transform
Inverse Laplace transform yields
h
vC (t) = 6
Also;
IC (s) =
VC (s)
iC (t) =
2
e
3
2
t
3
2
t
3
i
2=
2
s
EXAMPLE
2e
j
385
u(t)V
2
3
s+
2
3
u(t)A
5.17
Determine the current iL (t) for t 0 for the circuit shown in Fig. 5.24.
Figure 5.24
SOLUTION
At t = 0 , switch is closed and at t = 0+ , it is open. Let
us assume that at t = 0 , the circuit is in steady state. In
steady state, capacitor is open and inductor is short. The
equivalent circuit at t = 0 is as shown in Fig. 5.25(a).
12
= 1A
8+4
vC (0 ) = 1 8 = 8V
iL (0 ) =
Therefore;
iL (0) = iL (0+ ) = iL (0 ) = 1A
vC (0) = vC (0+ ) = vC (0 ) = 8V
Figure 5.25(a)
For t 0+ , the circuit in frequency domain is as shown in Fig. 5.25(b). We will use KVL
to find iL (t). Hence, we use series circuits to represent both the capacitor and inductor in the
frequency domain. These series circuits contain voltage sources rather than current sources. It
is easier to account for voltage sources than current sources when writing mesh equations. This
justifies the selection of series representation for both the capacitor and inductor.
386
Network Theory
j
Applying KVL clockwise to the right
mesh, we get
8
s
+
20
s
IL (s) + 4sIL (s) 4+8IL (s) = 0
)
8
)
IL (s) =
)
s
+4=
20
s
+ 8 + 4s IL (s)
2+s
(s + 1) + 1
=
s2 + 2s + 5
(s + 1)2 + 4
2
s+1
1
+
IL (s) =
(s + 1)2 + 22 2 (s + 1)2 22
Figure 5.25(b)
We know the Laplace transform pairs:
s+a
(s + a)2 + b2
L e at sin bt = (s + ab)2 + b2
L
and
e
at
cos bt =
iL (t) = e
Hence;
EXAMPLE
t cos 2t +
1 t
e sin 2t u(t)A
2
5.18
Find vo (t) of the circuit shown in Fig. 5.26.
Figure 5.26
SOLUTION
The unit step function u(t) is defined as
follows:
u(t) =
1; t 0+
0; t 0
Figure 5.27(a)
Laplace Transform
j
387
Since the circuit has two independent sources with u(t) associated with them, the circuit is not
energized for t 0 . Hence the initial current through the inductor is zero. That is, iL (0 ) = 0.
Since current through an inductor cannot change instantaneously,
iL (0) = iL (0+ ) = iL (0 ) = 0
vC (0) = vC (0+ ) = vC (0 ) = 0
Also;
The equivalent circuit for t 0+ in frequency domain is as shown in Fig. 5.27(b).
Figure 5.27(b)
KCL at supernode:
V1 (s)
1+
2
)
1
+
s3
V2 (s)
2
V2 (s)
+
=
s
2
s
2
1 7
1 1
6
=
+
V1 (s) 4
+ V2 (s)
15
s
1+
)
s s
s
2
2+s
2
+ V2 (s)
=
V1 (s)
s+1
2s
s
The constraint equation:
Applying KVL to the path comprising of current source ! voltage source ! inductor,
V1 (s)
we get,
1
+ V2 (s) = 0
s+2
V2 (s)
)
V1 (s)
1
s+2
1
V2 (s) =
s+2
V1 (s) =
388
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Network Theory
Putting the above two equations in matrix form, we get
2
2
3
3
2+s 2
V1 (s)
6
2s 7
5=6
54
4
V2 (s)
1
s
6 s+1
4
1
2
3
7
7
1 5
s
s+2
Solving for V2 (s) and then applying the principle of voltage division, we get
2(3s2 + 6s + 4)
1
V2 (s) =
2
2(s + 2)(3s2 + 3s + 2)
4
s2 + 2s +
3
Vo (s) =
(s + 2)(s + 0:5 j 0:646)(s + 0:5 + j 0:646)
Vo (s) =
)
Using partial fractions, we can write
K1
K2
K2
+
+
s+2
s + 0:5 j 0:646
s + 0:5 + j 0:646
K1 = 0:5
Vo (s) =
We find that
Hence;
We know that,
L
1
K2 = 0:316 / 37:76
0:5
0:316 / 37:76
0:316 /37:76
Vo (s) =
+
+
s+2
s + 0:5 j 0:646
s + 0:5 + j 0:646
1
=e
s+a
L
at
u(t)
1
= 2me
Hence,
EXAMPLE
vo (t) = 0.5e
2t u(t)
m /
m/ +
s + a j!
s + a + j!
at
cos(!t + )u(t)
+ 0.632e
0:5t cos [0.646t
37.76 ] u(t)
5.19
For the network shown in Fig. 5.28, find vo (t), t > 0, using mesh equations.
Figure 5.28
Laplace Transform
SOLUTION
The step function u(t) is defined as follows.
u(t) =
1; t 0+
0; t 0
Since the circuit is not energized for t 0 , there
are no initial conditions in the circuit. For t 0+ ,
the frequency domain equivalent circuit is shown
in Fig. 5.29(b).
Figure 5.29(a)
Figure 5.29(b)
2
By inspection, we find that I1 (s) =
s
KVL clockwise for mesh 2:
4
s
+ 1 [I2 (s)
4
)
s
I1 (s)] + 2I2 (s) + 1 [I2 (s)
I3 (s)] = 0
I1 (s) + I2 (s) [1 + 2 + 1]
I3 (s) = 0
Substituting the value of I1 (s), we get
4
s
)
+ 4I2 (s)
4I2 (s)
I3 (s) =
I3 (s) =
2
s
6
s
KVL clockwise for mesh 3:
1 [I3 (s)
I2 (s)] + sI3 (s) + 1I3 (s) = 0
)
I2 (s) + I3 (s) [s + 2] = 0
Putting the KVL equations for mesh 2 and mesh 3 in matrix form, we get
2
4
4
1
1 s+2
32
54
I2 (s)
I3 (s)
2 6 3
5=4 s 5
3
0
j
389
390
j
Network Theory
Solving for I3 (s), using Cramer’s rule, we get
I3 (s) =
1:5
7
s s+
4
Vo (s) = I3 (s) 1 =
)
1:5
s s+
7
4
Using partial fractions, we can write
Vo (s) =
4
6
6
K1 = ; and K2 =
72
37
We find that,
Hence;
EXAMPLE
K1
K2
+
7
s
s+
Vo (s) =
6 61
4
7 s
vo (t) =
6h
1
7
7
1
75
s+
4
e
7
t
4
i
u(t)
5.20
Use mesh analysis to find vo (t), t > 0 in the network shown in Fig. 5.30.
Figure 5.30
SOLUTION
The circuit is not energized for t 0 because the independent current source is associated with
u(t). This means that there are no initial conditions in the circuit. The frequency domain circuit
for t 0+ is shown in Fig. 5.31.
By inspection we find that:
I1 (s) =
4
s
;
I2 (s) =
Ix (s) = I3 (s)
4
s
Ix (s)
2
)
2I2 (s) = I3 (s)
4
s
)
1
I3 (s)
I2 (s) =
2
4
s
Laplace Transform
j
391
KVL for mesh 3 gives
s [I3 (s)
I2 (s)] + 1 I3 (s)
4
s
+1 I3 (s) = 0
1
4
s I3 (s)
I3 (s)
2
s
4
+ I3 (s) = 0
+ I3 (s)
)
s
4(s 2)
s(s + 4)
Vo (s) = 1[I3 (s)]
4(s 2)
=
s(s + 4)
)
I3 (s) =
and
Figure 5.31
By partial fractions, we can write
K1
K2
+
s
s+4
K1 = 2; K2 = 6
Vo (s) =
We find that
Hence;
Vo (s) =
2
6
s
s+4
6e
4t u(t)
Taking inverse Laplace transform, we get
vo (t) = 2u(t)
EXAMPLE
5.21
Using the principle of superposition, find vo (t) for t > 0. Refer the circuit shown in Fig. 5.32.
Figure 5.32
SOLUTION
Since both the independent sources are associated with u(t), which is zero for
t 0 , the circuit will not have any initial conditions. The frequency domain circuit for t 0+
is shown in Fig. 5.33(a).
392
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Network Theory
Figure 5.33(a)
As a first step, let us find the contribution to Vo (s) due to voltage source alone. This needs the
deactivation of the current source.
Referring to Fig. 5.33(b), we find that
4
I (s) =
)
s
s+1+
2
+1
s
Vo1 (s) = I (s)[1] =
s2
4
+ 2s + 2
Figure 5.33(b)
Next let us find the contribution to the output due to
current source alone.
Refer to Fig. 5.33(c). Using the principle of current
division,
2
I1 (s) =
)
s
s
s+1+
2
s
Vo2 (s) = 1 [I1 (s)] =
+1
s2
2s
+ 2s + 2
Figure 5.33(c)
Finally adding the two contributions, we get
Vo (s) = Vo1 (s) + Vo2 (s)
=
=
We find that,
Hence;
s2
4
2s
2s + 4
+ 2
= 2
+ 2s + 2 s + 2s + 2
s + 2s + 2
K1
+
K1
s + 1 j1
s + 1 + j1
p
K1 = 2 / 45
p
p
2 / 45
2 /+45
Vo (s) =
+
s + 1 j1
s + 1 + j1
Laplace Transform
We know that:
m /
m/ +
s + a jb
s + a + jb
= 2me
at
393
cos(bt + )u(t)
F
vo (t) = 2 2e t cos(t + 45 )u(t)
Hence;
EXAMPLE
L
1
j
5.22
(a) Convert the circuit in Fig. 5.34 to an appropriate s domain representation.
(b) Find the Thevein equivalent seen by 1Ω resistor.
(c) Analyze the simplified circuit to find an expression for i(t).
Figure 5.34
SOLUTION
(a) Since the independent current source has u(t) in it, the circuit is not activated for t 0 . In
otherwords, all the initial conditions are zero. Fig.5.35 (a) shows the s domain equivalent
circuit for t 0+ .
Figure 5.35(a)
(b) Sine we are interested in the current in 1Ω using the Thevenin theorem, remove the 1Ω
resistor from the circuit shown in Fig. 5.35(a). The resulting circuit thus obtained is shown
in Fig. 5.35(b).
394
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Network Theory
Zt (s) is found by deactivating the independent
current source.
Zt (s) = (5 + 0:001s)jj
500
s
2500 + 0:5s
=
Ω
0:001s2 + 5s + 500
Referring to Fig. 5.35 (b),
Vt (s) =
=
3
s
[Zt (s)]
Figure 5.35(b)
106
+ 1500s
7:5 Volts
2
s(s + 5000s + 5 105 )
The Thevenin equivalent circuit along with 1Ω
resistor is shown in Fig. 5.35 (c).
I (s) =
Vt (s)
Zt (s) + 1
7:5 106 + 1500s
s(s2 + 5500s + 3 106 )
7:5 106 + 1500s
=
s(s + 4886)(s + 614)
=
Figure 5.35(c)
Using partial fractions, we get
I (s) =
2:5
s
+
0:008
s + 4886
2:508
s + 614
Taking inverse Laplace transforms, we get
i(t) = 2.5 + 0.008e
4886t
2.508e
614t
u(t)A
Check:
i(0) = 2:5 + 0:008
2:508 = 0
i(1) = 2:5:
and
These could be verified by evaluating i(t) at t = 0 and t = 1 using the concepts explained in
Chapter 4.
EXAMPLE
5.23
Refer the RLC circuit shown in Fig. 5.36. Find the complete response for v (t) if t 0+ . Take
v (0) = 2V.
Laplace Transform
j
395
Figure 5.36
SOLUTION
Since we wish to analyze the circuit given in Fig. 5.36 using KVL, we shall represent L and C in
frequency domain using series circuits to accomodate the initial conditions. Accordingly, we get
the frequency domain circuit shown in Fig. 5.36 (a).
Applying KVL
clockwiseto the circuit shown in Fig. 5.36 (a), we get
2s
2
9
I (s) + = 0
+ 6+s+
2
s + 16
s
s
32
) I (s) = 2
(s + 6s + 9)(s2 + 16)
2
9
+
Hence; V (s) = I (s)
s
=
2
s
+
s
288
s(s +
3)2 (s2
+ 16)
Using partial fraction, we get
K4
K2
K3
K4
K1
+
+
+
+
V (s) = +
2
s
s
s+3
(s + 3)
s j4
s + j4
2
Solving for K1 , K2 , K3 , and K4 , we get
Figure 5.36(a)
288
K1 =
= 2
2
2
(s + 3) (s + 16) s=0
288
d
= 2:2
K2 =
ds s(s2 + 16) s= 3
288 K3 =
s(s2 + 16) s=
3
= 3:84
288
= 0:36 / 106:2
K4 =
2
s(s + 3) (s + j 4) s=j 4
396
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Network Theory
Therefore
V (s) =
2
2
s
s
+
0:36 / 106:2
2:2
3:84
0:36 /106:2
+
+
+
2
s+3
(s + 3)
s j4
s + j4
Taking inverse Laplace Transform we get,
v(t) = 2.2e
3t
+ 3.84te
3t
106.2 )
+ 0.72 cos(4t
Verification:
Putting t = 0 in the above equation
v (0) = 2:2 + 0 + 0:72 cos( 106:2 )
= 2:2
0:2 = 2V
(The same quantity is given in the problem)
5.8
Waveform synthesis
The three important singularity functions explained in section 5.3 are very useful as building
blocks in constructing other waveforms. In this section, we illustrate the concept of waveform
synthesis with a number of exmaples, and also determine expressions for these waveforms.
EXAMPLE
5.24
Express the voltage pulse shown in Fig.5.37 in terms of unit step function and then find V (s).
dv (t)
.
Also find L
dt
Figure 5.37
SOLUTION
The pulse shown in Fig. 5.37 is the gate function. This function may be regarded as a step function
that switches on at t = 2 secs and switches off at t = 4 secs.
Laplace Transform
j
397
Figure 5.37(a)
Referring to Figs. 5.37 and 5.37 (a), we can write
v (t) = v1 (t) + v2 (t)
)
Hence;
v (t) = 5u(t
V (s) =
=
5
s
e
2) 5u(t
5 4s
2s
5
e
s
s
2s
4)
e
e
4s
Figure 5.37(b)
Taking the derivative of v (t), we get
dv (t)
= 5 [ (t
dt
Fig. 5.37(b) shows the graph of
(t
2)
4)]
dv (t)
.
dt
We can obtain Fig. 5.37(b) directly from Fig. 5.36 by observing that at t = 2 seconds, there is
a sudden rise of 5V leading to 5 (t 2). Similarly, at t = 4 seconds, a sudden fall of 5V leading
to 5 (t 4).
We know the Laplace trasnform pair
Lf (t
Hence;
L
a)g = e
dv(t)
dt
=e
as
as
=5 e
Lf (t)g
2s
e
4s
398
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Network Theory
EXAMPLE
5.25
Express the current pulse in Fig.5.38 in terms of the unit step.
Find: (i) L fi(t)g
(ii) L
R
i(t)dt .
Figure 5.38
SOLUTION
Figure 5.39(a)
Figure 5.39(b)
Referring to Figs. 5.39 (a) and (b), using the principle of synthesis, we can write
i(t) = i1 (t) + i2 (t) + i3 (t)
= 5u(t)
10u(t
2) + 5u(t
4)
Laplace Transform
j
399
The Laplace transform of the above equation yields
I (s) =
=
5
10
s
s
5
1
s
5
1
=
Zs
Let
f (t) =
then;
f (t) =
Z
2s
e
2e
e
5
+ e
s
2s
+e
4s
4s
2s 2
i(t)dt
[5u(t)
= 5r (t)
10u(t
10r (t
2) + 5u(t
2) + 5r (t
4)]dt
4)
= f1 (t) + f2 (t) + f3 (t)
The function f1 (t) is a ramp of slope = 5 as shown in Fig. 5.39 (c). To this, if we add a ramp
of slope = 10, the effect of this addition is, we get a ramp of slope = 5 10 = 5 for t 2 secs
till we encounter the next ramp. At t = 4 seconds, if we add a ramp with a slope of +5, the net
slope beyond t = 4 seconds is 5 + 5 = 0. Thus figure f (t) is drawn as shown in Fig. 5.39 (d).
Figure 5.39(c)
L ff (t)g = F (s)
= L f5r (t)
=
=
5
10
s2
s2
5 1
s2
10r (t
e
2s
2e
+
2s
2) + 5r (t
5
e
4s
+e
4s
s2
4)g
400
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Network Theory
Figure 5.39(d)
EXAMPLE
5.26
Express the sawtooth function in terms of singularity functions. Then find Lfv (t)g.
Figure 5.40
SOLUTION
There are three methods to solve this problem.
Method 1:
The function v1 (t) is a ramp function of slope = +5. This slope +5 should continue till t = 1
second. Hence at t = 1 second, a ramp of slope t = 5 is added to v1 (t). The graph of
v1 (t) + v2 (t) is shown in Fig. 5.41(a). Next, to v1 (t) + v2 (t), a step of 5V is added at t = 1
second.
Hence;
v (t) = v1 (t) + v2 (t) + v3 (t)
= 5r (t)
5r (t
V (s) = Lff (t)g =
=
5 1
2
s
e
s
5u(t
1)
5
5
s2
s2
se
s
e
s
1)
5
s
e
s
Laplace Transform
Figure 5.41(a)
Figure 5.41(b)
Method 2:
This method involves graphical manipulation.
Figure 5.41(c)
j
401
402
j
Network Theory
The equation of a straight line passing through the origin is y = mx, where m = slope of the
line. This allows us to write v1 (t) = 5t. From Fig. 5.41(c), we can write
v (t) = v1 (t)v2 (t)
Hence;
= 5t [u(t)
u(t
1)]
= 5tu(t)
5tu(t
1)
= 5tu(t)
5(t
1 + 1)u(t
= 5tu(t)
5(t
1)u(t
= 5r (t) 5r (t
5 V (s) = 2 1 e s
s
1)
se
1)
5u(t
s
1)
5u(t
1)
1)
Method 3:
Figure 5.41(d)
This method also involves graphical manipulation. We observe from Fig. 5.41(d) that v (t) is a
multiplication of a ramp function and a unit step function.
Thus;
v (t) = v1 (t)v2 (t)
= 5r (t) [u( t + 1)]
Figure 5.41(e)
Laplace Transform
From Fig. 5.41(e), we can write
v2 (t) = v3 (t) + v4 (t)
)
v2 (t) = 1
u(t
1)
)
u( t + 1) = 1
u(t
1)
Hence;
v (t) = 5r (t) [1
= 5r (t)
We know that,
r (t) = tu(t)
Hence;
v (t) = 5r (t)
Please note that, u(t)u(t
u(t
1)]
5r (t)u(t
1)
5tu(t)u(t
1)
= 5r (t)
5(t
1 + 1)u(t)u(t
= 5r (t)
5(t
1)u(t)u(t
1) = u(t
1)
1)
5u(t)u(t
1)
1) [Refer Fig. 5.41(f)]
Figure 5.41(f)
Thus;
v (t) = 5r (t)
= 5r (t) 5r (t
5 V (s) = 2 1 e s
s
Hence;
EXAMPLE
5(t
1)u(t
1)
1)
5u(t
se s
5u(t
1)
1)
5.27
Given the signal
8
<
3;
2;
x(t) =
:
2t 4;
t<0
0<t<1
t>1
Express x(t) in terms of singularity functions. Also find Lfx(t)g.
SOLUTION
The signal x(t) may be viewed as follows:
(i) in the interval, t < 0, x(t) may be regarded as 3u( t)
(ii) in the interval, 0 < t < 1, x(t) may be viewed as 2[u(t)
(iii) for t > 1, x(t) may be viewed as (2t 4)u(t 1)
u(t
1)] and
j
403
404
j
Network Theory
Thus;
x(t) = 3u( t)
)
x(t) = 3 [1
2 [u(t)
u(t)]
u(t
1)] + (2t
2u(t) + 2u(t
4)u(t
1)
1) + 2tu(t
1)
1)
=3
5u(t)
2u(t
1) + 2(t
1 + 1)u(t
=3
5u(t)
2u(t
1) + 2(t
1)u(t
=3
5u(t) + 2r(t
4u(t
1) + 2u(t
1)
1)
1)
Lfx(t)g cannot be found because x(t) contains a constant 3 for
1 < t < 0 (a noncausal
signal).
EXAMPLE 5.28
Express f (t) in terms of singularity functions and then find F (s).
Figure 5.42
SOLUTION
To find f (t) for 0 < t < 2 :
Equation of the straight line 1 is
y
x
y1
y2
=
x1
x2
y1
x1
Here, y is f (t) and x is t.
Hence,
)
f (t) 3
3 3
=
t 0
2 0
2f (t) 6 = 6t
)
f (t) = 3
3t
Figure 5.43
Laplace Transform
To find f (t) for 2 < t < 3:
Here,
f (t) + 3
0+3
=
t 2
3 2
)
f (t) + 3 = 3t
6
)
f (t) = 3t
9
8
< 3
Hence
3t; 0 < t < 2
3t 9; 2 < t < 3
f (t) =
:
0;
otherwise
The above equation may also be written as :
f (t) = [3
3t] [u (t)
= 3u (t)
3u (t
3tu (t
)
3)
u (t
2)
9] [u (t
9u (t
2) + 9u (t
2)
3tu (t) + 6tu (t
= 3u (t)
12u (t
2)
3tu (t) + 6 (t
= 3u (t)
12u (t
+12u (t
f (t) = 3u (t)
2)
3) + 9u (t
3 (t
3) u (t
3tu (t) + 6 (t
=
3
s
3) + 9u (t
3)
2)
3)
3)
2) u (t
2) u (t
9u (t
2)
F (s) = L ff (t)g
Hence;
2)
3tu (t
2)
2 + 2) u (t
3tu (t) + 6 (t
2)
3)]
3)
12u (t
3 + 3) u (t
u (t
2)
2) + 3tu (t
3tu (t) + 3tu (t
f (t) = 3u (t)
3 (t
EXAMPLE
2)] + [3t
6
3
+ 2e
2
s
s
2s
2)
3) + 9u (t
3 (t
3
e
s2
3) u (t
3)
3)
3s
5.29
Express the function f (t) shown in Fig. 5.44 using singularity functions and then find F (s).
Figure 5.44
j
405
406
j
Network Theory
SOLUTION
Equation of the straight line shown in Fig. 5.45(a) is
f1 (t) + 1
2+1
=
t 1
2 1
f1 (t) + 1 = t + 1
)
)
f1 (t) =
t
The above equation is for the values t lying between 1 and 2.
This could be expressed, by writing
f (t) = f1 (t) g (t)
Figure 5.45(a)
Figure 5.45(b)
) f (t) =
Hence;
t [u (t
u (t
1)
=
(t
1 + 1) u (t
=
(t
1) u (t
=
r (t
1)
1)
u (t
2)]
1) + (t
u (t
1) + (t
1) + r (t
F (s) = L ff (t)g
=
2 + 2) u (t
1
e s
s2
2)
2) u (t
2) + 2u (t
1 s
1
e + 2e
s
s
2) + 2u (t
2)
2s
+
2
e
s
2s
2)
Laplace Transform
EXAMPLE
5.30
Find the Laplace transform of the function f (t) shown in Fig. 5.46.
Figure 5.46
SOLUTION
Method 1:
Figure 5.47(a)
We can write;
f (t) = fA (t) + fB (t)
= sin tu (t) + sin (t
Hence;
F (s) = L ff (t)g =
=
s2
s2 + 2
π
1+e s
2
+π
1) u (t
+
s2 + 2
1)
e
s
j
407
408
j
Network Theory
Method 2 :
Figure 5.47(b)
Graphically, we can manipulate f (t) as
f (t) = fC (t) g (t)
= sin t [u (t)
u (t
1)]
= sin tu (t)
sin tu (t
= sin tu (t)
sin t (t
1 + 1) [u (t
1)]
= sin tu (t)
sin ( (t
1) + ) u (t
1)
= sin tu (t) + sin (t
F (s) = L ff (t)g =
Hence;
=
EXAMPLE
s2
s2
+
2
π
1+e s
2
+π
1)
1) u (t
+
s
s2
+ 2
5.31
Find the Laplace transform of the signal x(t) shown in Fig. 5.48.
Figure 5.48
1)
e
s
Laplace Transform
j
409
SOLUTION
Figure 5.49
Mathematically, we can write x(t) as
x (t) = xA (t)
= sin (t
1) u (t
=
s2
π
e s
+ π2
sin (t
1)
L fx (t)g = X (s) = s2 + 2 e
Hence;
EXAMPLE
xB (t)
e
s
s2 + 2
3) u (t
e
3)
3s
3s
5.32
Refer the waveform shown in Fig. 5.50. The equation for the waveform is sin t from 0 to ; sin t
s 1
coth
:
from to 2 . Show that the Lapalce transform of this waveform is F (s) = 2
s +1
2
Figure 5.50
410
j
Network Theory
SOLUTION
f (t) is a periodic waveform with a period T = seconds. Let f1 (t) be the waveform f (t)
described over only one period. The Laplace transform of f (t) and f1 (t) are related as
F1 (s)
1 e sT
Let us now proceed to find F1 (s). From Fig. 5.51 (b), we can write
F (s) =
f1 (t) = fA (t) + fB (t)
= sin tu (t) + sin (t ) u (t
1
1
) F1 (s) = 2
+ 2
e s
s +1
s +1
(1 + e s )
=
s2 + 1
F1 (s)
F1 (s)
Hence; F (s) =
=
sT
1 e
1 e s
s
(1 + e )
) F (s) = 2
(s + 1) (1 e s )
, es=2 + e
s=2
=
1 e,
2
s2 + 1
, es=2
s=2
e,
s )
)
Figure 5.51(a)
s=2
e
s=2
2
1 cosh 2
s2 + 1 sinh s
2
πs 1
= 2
coth
s +1
2
F (s) =
EXAMPLE
Figure 5.51(b)
5.33
Find the Laplace transform of the pulse shown in Fig. 5.52.
Figure 5.52
Laplace Transform
j
411
SOLUTION
We can describe Fig. 5.52 mathematically as
0<t<2
Vo ;
Vo t + 3Vo ; 2 < t < 3
f (t) =
The expression for f (t) for 2 < t < 3 is obtianed as follows :
Equation of a straight line between two points is given by
y1
x1
y2
y1
=
x1
x2
y
x
In the present context, y = f (t); x = t; (x1 ; y1 ) = (2; Vo ) and (x2 ; y2 ) = (3; 0)
0 Vo
f (t) Vo
=
t 2
3 2
f (t) = Vo t + 3Vo
Hence;
)
The time domain expression for f (t) between t = 0 and 3 could be written using graphical
manipulation as
f (t) = Vo [u (t)
u (t
2)] + [ Vo t + 3Vo ] [u (t
u (t
2)
3)]
The first term on the right-side of the above equation defines f (t) for 0 < t < 2 and the
second term on the right-side defines f (t) for 2 < t < 3.
f (t) = Vo u (t)
Vo u (t
2)
Vo tu (t
= Vo u (t)
Vo u (t
2)
Vo (t
+Vo (t
= Vo u (t)
3 + 3) u (t
Vo u (t
+3Vo u (t
= Vo u (t)
)
f (t) = Vo u (t)
2) u (t
2)
2)
3Vo u (t
3)
2)
3Vo u (t
2Vo u (t
2)
3Vo u (t
2) + Vo (t
2) + Vo r (t
2)
3) + 3Vo u (t
3)
2) + Vo (t
3) u (t
3)
3)
3) u (t
3)
3)
F (s) = L ff (t)g
Hence;
=
EXAMPLE
Vo (t
2)
2) u (t
Vo r (t
2 + 2) u (t
3) + 3Vo u (t
3) + 3Vo u (t
Vo (t
2) + Vo tu (t
Vo
s
Vo
e
s2
2s
+
Vo
e
s2
3s
5.34
Consider a staircase waveform which extends to infinity and at t = nt0 jumps to the value n + 1,
being a superposition of unit step functions. Determine the Laplace transform of this waveform.
412
Network Theory
j
SOLUTION
We can write,
F (s) = L ff (t)g =
1
1+e
=
Let
e
t0 s
s
=x
then F (s) =
t0 s
2t0 ) + 1
+ e t0 s + e 2t0 s + t0 ) + u (t
f (t) = u (t) + u (t
1
s
+e
1
s
s
2t0 s
+ 1
1 + x + x2 + s
From Binomial theorem, we have
(1
Hence;
EXAMPLE
1
= 1 + x + x2 + 1
F (s) =
s (1 x)
1
=
s (1 e t0 s )
x)
Figure 5.53
5.35
(a) Find the Laplace transform of the staircase waveform shown in Fig. 5.54. (b) If this voltage
were applied to an RL series circuit with R = 1Ω and L = 1H , find the current i(t).
Figure 5.54
SOLUTION
(a) We can express mathematically, the voltage waveform shown in Fig. 5.54 as,
v (t) =
8
1; 1 < t < 2
>
>
>
>
< 2; 2 < t < 3
3; 3 < t < 4
>
>
4; 4 < t < 5
>
>
:
0; elsewhere
Laplace Transform
v (t) = [u (t
or
1)
+3 [u (t
= u (t
u (t
2)] + 2 [u (t
u (t
3)
1) + u (t
u (t
2)
4)] + 4 [u (t
2) + u (t
3) + u (t
4)
413
3)]
u (t
4)
j
5)]
4u (t
5)
Taking the Laplace transform, we get
V (s) =
1 s
e +e
s
2s
3s
+e
+e
4s
4e
5s
(b) Assuming all initial conditions to be zero, the time domian circuit shown in Fig. 5.55 gets
transformed to a circuit as shown in Fig. 5.56.
Figure 5.55 Time Domain Circuit
Figure 5.56 Frequency
. Domain Circuit
From Fig. 5.56, we can write
I (s) =
V (s)
s+1
1
1
1
1
e s+
e 2s +
e 3s +
e
s (s + 1)
s (s + 1)
s (s + 1)
s(s + 1)
1
1
1
1
1
1
s
2s
+
e +
e
e
) I (s) =
s
s+1
s
s+1
s
s+1
1
1
1
1
+
e 4s 4
e 5s
s
s+1
s
s+1
) I (s) =
4s
4
e
s (s + 1)
5s
3s
Taking the inverse Laplace transform, we get
i (t) = u (t)
i (t) =
e
+ 1
t
!t 1 + u (t) e u (t) t!t 2 + u (t)
t
e u (t) t!t 4 4 u (t) e t u (t) t!t 5
+ u (t)
1
e t u (t)
(t 1)
e
(t
t
u (t
4)
1) + 1
u (t
4)
e
4 1
(t 2)
e
u (t
(t
5)
2) + 1
u (t
5)
e t u (t)
e
(t 3)
t
!t
3
u (t
3)
414
j
Network Theory
EXAMPLE
5.36
A voltage pulse of 10 V magnitude and 5 sec duration is applied to the RC network shown in
Fig. 5.57. Find the current i(t) if R = 10Ω and C = 0:05F .
Figure 5.57
SOLUTION
Figure 5.58(a)
Mathematically, we can express v (t) as follows :
v (t) = v1 (t) v2 (t)
= 10u(t)
Hence;
V (s) = L fv (t)g
=
10 1
10u(t
e
s
t0 s
t0 )
Assuming all initial conditions to be zero, the Laplace
transformed network is as shown in Fig. 5.58(b).
I (s) =
=
V (s)
R+
1
Cs
e
10 1
s R+
t0 s
1
Cs
Figure 5.58(b)
Laplace Transform
10Cs
1 e t0 s
s (RCs + 1)
1
10
=
1 e t0 s
1
R
s+
I (s) =
2 RC
=
10 6
4
R
1
s+
Taking inverse Laplace transform yields
i (t) =
=
10
R
10
R
e
t
RC
e
t
RC
t
u (t)
u (t)
i (t) = e 0:510
EXAMPLE
6
10
e
R
10
R
t
RC
u (t)
t
!t
10
0:5
t0 s 7
5
e
RC
t0
u (t
(t 510
e
1
s+
RC
(t t0 )
RC
e
u (t)
1
1
3
t0 )
6
6
)
u t
5 10
6
5.37
Find the Laplace transform of the waveform shown in Fig. 5.59.
Figure 5.59
SOLUTION
v (t) =
or
v (t) = 3t [u (t)
3t; 0 < t < 1
2; 1 < t < 2
u (t
1)] + 2 [u (t
= 3tu (t)
3tu (t
1) + 2u (t
= 3tu (t)
3 (t
1 + 1) u (t
= 3tu (t)
3 (t
1) u (t
1)
1)
1)
u (t
2)]
2u (t
2)
1) + 2u (t
3u (t
1)
2u (t
1) + 2u (t
1)
2)
2u (t
2)
j
415
416
Network Theory
j
v (t) = 3tu (t)
= 3r (t)
3r (t
V (s) = L fv (t)g
Hence;
=
5.9
3 (t
3
s2
3
e s
s2
1) u (t
1)
u (t
1)
u (t
1 s
e
s
2u (t
1)
2
e
s
2u (t
1)
2)
2)
2s
The System function
The system function or transfer function of a linear time-invariant system is defined as the ratio
of Laplace transform of the output to Laplace transform of the input under the assumption that all
initial conditions are zero.
Hence, for relaxed LTI system, the response Y (s) to an input X (s) is H (s) X (s), where H (s)
is the system function. The system function H (s) may be found in several ways:
1. For a system defined by a linear differential equation, by taking Laplace transform of the
Y (s)
.
differential equation and then finding the ratio
X (s)
2. From the Laplace transform of impulse response h(t).
3. From the s domain model of a physical system like an electrical system.
EXAMPLE
5.38
The output y (t) of an LTI system is found to be e
3t u(t)
when the input x(t) is 0:5u(t).
(a) Find the impulse response h(t) of the system.
(b) Find the output when the input is x(t) = e t u(t).
SOLUTION
(a) Taking Laplace transforms of x(t) and y (t), we get
1
0:5
; X (s) =
s+3
s
Y (s)
2s
H (s) =
=
X (s)
s+3
2 (s + 3) 6
6
H (s) =
=2
(s + 3)
s+3
Y (s) =
Hence
)
Taking inverse Laplace transform, we get
h (t) = 2δ (t)
6e
3t u (t)
Laplace Transform
j
417
x (t) = e t u (t)
(b)
)
Thus;
1
s+1
Y (s) = X (s) H (s)
2s
=
(s + 1) (s + 3)
X (s) =
=
K2
K1
+
s+1
s+3
2s K1 =
s + 3 s =
where
Therefore;
1
=
1
2s =3
K2 =
s + 1 s = 3
3
1
+
Y (s) =
s+1
s+3
Taking inverse Laplace transform of Y (s), we get
y (t) =
y (t) =
or
EXAMPLE
e
e
t
+ 3e
t
+ 3e
3t
; t0
3t
; u(t)
5.39
Determine the output v (t) for the circuit shown in Fig. 5.60.
Figure 5.60
SOLUTION
The transformed network of Fig. 5.60 with the assumption that all initial conditions are zero is
shown in Fig. 5.61(a).
418
j
Network Theory
V (s) =
=
1
I (s)
s2
3
1 6 Vs (s) 7
4
15
s
1+
s
1
V (s)
=
H (s) =
Vs (s)
s+1
)
Figure 5.61(a)
The inverse Laplace transform of H (s) is called the impulse response of the circuit and is
denoted by h(t).
h (t) = e t u (t)
I method :
From Convolution theorem, we have,
v (t) = h (t) vs (t)
Z1
h ( ) vs (t
=
) d
0
Z1
e
=
0
= 2e
t
u ( ) 2e
(t )
u (t
) d
Z1
u ( ) u (t
) d
0
Let us compute the product u ( ) u (t
u ( ) =
u (t
) =
Hence;
u ( ) u (t
) =
) for different values of 1; < 0
0; > 0
1; t
0; t
> 0 or < t
< 0 or > t
1; 0 < < t; t > 0
0;
otherwise
Figure 5.61(b)
Laplace Transform
Therefore;
v (t) = 2e
t
Zt
t
d = 2te
j
419
; t0
0
= 2te t u (t)
II method :
In the frequency domain, convolution operation is transformed into a multiplicative operation.
That is;
V (s) = H (s) Vs (s)
1
2
(s + 1) (s + 1)
2
=
(s + 1)2
=
Inverse Laplace transform yields,
v(t) = 2te t u(t)Volts
Reinforcement problems
R.P
5.22
(a) Find H (s) =
Vo (s)
for the circuit shown in Fig. R.P. 5.22. (b) Determine vo (t) when the
Vi (s)
intital current in the inductor is zero.
Figure R.P.5.22
SOLUTION
The Laplace transformed network with all initial conditions set to zero is shown in Fig. R.P.
5.22(a).
Vo (s) = I (s) 150 + 2 10
=
)
H (s) =
3
s
Vi (s) 150 + 2 10
100 + 3 10
Vo (s)
1.5
=
Vi (s)
2.5
3s
3s
+ 150 + 2 10
105 + 2s
105 + 5s
3s
420
j
Network Theory
(b) Vo (s) = H (s) Vi (s)
=
1:5 105 + 2s 100
2:5 105 + 5s
s
40 s + 0:75 105
=
s [s + 0:5 105 ]
=
K1
K2
+
s
s + 0:5 105
Figure R.P. 5.22(a)
= 60
s=0
5
40 s + 0:75 10 K2 =
= 20
s
5
40 s + 0:75 105
K1 =
[s + 0:5 105 ]
where
s= 0:5 10
Hence;
Vo (s) =
60
s
20
s + 0:5 105
Taking inverse Laplace transform, we get
vo (t) = 60
R.P
20e
0:5 105t
u (t) Volts
5.23
Refer the circuit shown in Fig. R.P. 5.23. The switch closes at t = 0. Determine the voltage v (t)
after the switch closes.
Figure R.P. 5.23
SOLUTION
The switch is open at t = 0 and closed at t = 0+ . Let us assume that at t = 0 , the circuit is in
steady state. The circuit at t = 0 is shown in Fig. R.P. 5.23(a).
Laplace Transform
j
421
Figure R.P. 5.23(a)
Referring to Fig. R.P. 5.23(a), we get
8
= 2A
2+2
v 0 =0
i 0
=
From switching principles, we know that the current through an inductor and the voltage
across a capacitor cannot change instantaneously. Therefore,
+
i (0) = i 0+ = i 0
v (0) = v 0
and
=v 0
= 2A
= 0V
We shall solve this probelm using nodal technique. Hence, in the frequency domain, we will
use the parallel models for the capacitor and inductor because the parallel models contain current
sources rather than voltage sources. The frequency domain circuit is shown in Fig. R.P. 5.23(b).
Figure R.P.5.23(b)
KCL at node V (s) :
V (s)
2
)
)
8
s + V (s) + V (s) + 2 = 0
1
s
s
s
1 1
+ +s =
2 s
s + 2 + 2s2
=
V (s)
2s
V (s)
4
2
s
s
2
s
422
j
Network Theory
)
V (s) =
=
=
2s2
4
+s+2
s2
2
+ 0:5s + 1
s2
2
+ 0:5s + (0:25)2
=
2
(s + 0:25) + (0:96824)2
=
2
0:96824
0:96824 (s + 0:25)2 + (0:96824)2
2
= 2:066 We know that,
L
1
0:96824
(s + 0:25)2 + (0:96824)2
a
2
(s + b) +
Hence,
v (t) = 2.066e
R.P
(0:25)2 + 1
a2
=e
bt
sin at u (t)
0:25t sin (0.96824t) u (t) Volts
5.24
Find the impulse response of the circuit shown in Fig. R.P. 5.24.
Figure R.P. 5.24
SOLUTION
The frequency domain representation of the circuit is shown in Fig. R.P. 5.24(a) by assuming that
all initial conditions to be zero.
Laplace Transform
Figure R.P. 5.24(a)
KCL at node a:
Va (s)
)
)
)
Vb (s)
Va (s)
1
+ Vg (s) +
=0
1
2
2
2s
Va (s) Vb (s)
Va (s) Vb (s)
+
+ 2sVa (s) = 0
2
2
1 1
1 1
=0
+ + 2s
Vb (s)
+
Va (s)
2 2
2 2
Va (s) [1 + 2s] Vb (s) = 0
KCL at node b:
Vb (s)
Vi (s)
s
)
Va (s)
)
)
+
Vb (s)
Va (s)
2
=0
1 1
Vi (s)
1
+ Vb (s)
=
+
2
s
2
s
Va (s)
Vi (s)
(2 + s)
+
Vb (s) =
2
2s
s
sVa (s) + (2 + s) Vb (s) = 2Vi (s)
Putting the above nodal equations in matrix form, we get
1 + 2s
s
1
2+s
Va (s)
Vb (s)
=
0
2Vi (s)
Solving, we get
2Vi (s)
2 + s + 4s + 2s2 s
Va (s)
1
=
Vi (s)
(s + 1)2
vi (t) = (t) ) Vi (s) = 1
Va (s) =
)
Given
j
423
424
j
Network Theory
Va (s)
Hence;
1
(s + 1)2
1
Va (s) =
(s + 1)2
1
)
=
Taking inverse Laplace transform, we get
va (t) = h (t) = te t u (t)
R.P
5.25
Find the convolution of h (t) = t and f (t) = e
H (s) F (s).
SOLUTION
h (t) f (t) = L
/t for t > 0, using the inverse transform of
1
fH (s) F (s)g
1
H (s) = L fh (t)g = 2
where
s
F (s) = L ff (t)g =
Hence;
H (s) F (s) =
=
1
s+
1
+ )
s2 (s
K3
K2
K1
+ 2 +
s
s
s+
Solving the partial fractions yields
1
K1 =
Hence;
H (s) F (s) =
)
2
1
=
R.P
1
s
1
+
1
1
;
1
s2
1
K3 =
+
2
1
2
1
s+
fH (s) F (s)g
1
=
K2 =
2
h (t) f (t) = L
;
2
u (t) +
1
tu (t) +
1
2
e
t
u (t)
1
t
1
+ + 2 e t u (t)
α2
α
α
5.26
Consider a pulse of amplitude 5V for a duration of 4 seconds with its starting point t = 0. Find
the convolution of this pulse with itself and draw the convolution x (t) x (t) versus time.
Laplace Transform
j
Figure R.P. 5.26(a)
SOLUTION
x (t) = 5u (t)
)
Let
5
5u (t
4)
5
e 4s
s
s
y (t) = x (t) x (t)
X (s) =
Taking Laplace transform, we get
Figure R.P. 5.26(b)
Y (s) = X (s) X (s)
=
25
50
s2
s2
4s
e
+
25
s2
e
8s
Taking inverse Laplace transform, we get
Hence;
R.P
y (t) = 25tu (t)
50 (t
y(t) = 25r(t)
50r(t
5.27
Show that
L
Ktr
(r
1 e at
1)!
=
4) u (t
4) + 25 (t
4) + 25r(t
K
(s + a)r
SOLUTION
Let
f (t) = 1
then
F (s) =
Thus;
We know that;
1
s
dn F (s)
( 1)n n!
=
dsn
sn 1
n
F (s)
L ftn f (t)g = ( 1)n d ds
n
With f (t) = 1, we get
L ft
n
n
g = ( 1)
=
n!
sn+1
( 1)n n!
sn+1
8)
8) u (t
8)
425
426
j
Network Theory
Putting n = r
1, we get
L
K
R.P
(r
tr
1
L
1
e
at
L
1)!
1
e
at
and
Therefore;
tr
tr
=
(r
1)!
sr
=
(r 1)!
(s + a)r
=
K
(s + a)r
5.28
Tests conducted on a certain network revealed that the current was i(t) = 2e t + 4e 3t when a
unit step voltage was suddenly applied to the input terminals of the network at t = 0. What voltage
must be applied to get an output current of i(t) = 2e t if the network remains unchanged?
SOLUTION
Given,
Hence;
i(t) =
2e
t
+ 4e
3t ;
I (s) =
V (s) =
and
t 0 when v (t) = u(t)
2
4
+
s+1
s+3
1
s
Laplace transform of the output
System function = H (s) =
Laplace transform of the input
I (s)
)
H (s) =
V (s)
2s (s 1)
=
(s + 1) (s + 3)
We have to find v (t) when i(t) = 2e t .
First we will find V (s) when I (s) =
Hence;
2
I (s)
using the relation H (s) =
.
s+1
V (s)
V (s) =
I (s)
H (s)
2
s+1
=
2s (s 1)
(s + 1) (s + 3)
(s + 3)
=
s (s 1)
=
K2
K1
+
s
s 1
Laplace Transform
Using partial fractions, we find that K1 =
Hence;
V (s) =
v (t) =
)
R.P
3 and K2 = 4
3
s
4
+
s 1
3u (t) + 4et u (t) Volts
5.29
Find the Laplace transform of the periodic waveform shown in Fig. R.P. 5.29.
Figure R.P. 5.29
SOLUTION
The Laplace transform of a periodic waveform is found
using the relation
F (s) =
1
F1 (s)
e sT
where F1 (s) = L ff1 (t)g = Laplace transform of f (t)
over 0 < t < T . Where T = fundamental period of
f (t).
Figure R.P. 5.29(a)
Referring to Fig. R.P. 5.29(a) we can write:
f1 (t) =
)
f1 (t) =
1
a
t [u (t)
+
1
a
0<t<a
1;
a < t < 3a
1
a
tu (t
3a < t < 4a
a)] + [u (t
t + 4 [u (t
1
+ tu (t
a
t
;
a
>
>
1
>
>
>
: a t + 4;
u (t
1
= tu (t)
a
8
>
>
>
>
>
<
3a)
a)
u (t
u (t
4a)]
a)
u (t
a) + u (t
4a) + 4u (t
3a)
4u (t
3a)]
3a)
4a)
1
a
tu (t
3a)
j
427
428
j
Network Theory
)
f1 (t) =
1
a
1
tu (t)
1
a
a
(t
+ 4u (t
3a)
1
a
a
1
a + a) u (t
3a + 3a) u (t
1
= tu (t)
(t
3a) +
4u (t
(t
a) + u (t
4a)
a) u (t
a)
1
a
a
1
= r (t)
Hence;
a
1
a
1
(t
a) u (t
r (t
1
a)
a
=
1
1
as2
as2
1
as2
1
e
as2
e
e
3as
3as
+e
4a)
u (t
4a) u (t
3a) u (t
1
as
3a)
a)
(t
3a) +
1
a
3a)
4a) + 4u (t
(t
4a) u (t
4a)
a
as
e
(t
1
a
1
3a) + r (t
r (t
a
a
F1 (s) = L ff1 (t)g
=
a) + u (t
3a) +
1
a)
u (t
4a + 4a) u (t
u (t
(t 3a) u (t 3a) 3u (t
a
+4u (t 3a) 4u (t 4a)
1
= tu (t)
(t
a)
+
4as
1
as2
e
4as
Alternate method for finding F1 (s):
From Figs. R.P. 5.29(b), (c), (d), we can write
f1 (t) = fA (t) + fB (t) + fC (t) + fD (t)
1
= tu (t)
a
+
1
a
(t
1
a
(t
a) u (t
4a) u (t
Figure R.P. 5.29(b)
4a)
a)
1
a
(t
3a) u (t
3a)
4a)
4a)
Laplace Transform
Figure R.P. 5.29(c)
Figure R.P. 5.29(d)
Hence;
F1 (s) = L ff1 (t)g
=
Finally;
1
as2
e
as
F (s) = L ff (t)g
=
where T = 4a
1
as2
1 1
F (s) =
as2
1
1
as2
F1 (s)
e sT
e as e 3as + e
(1 e 4as )
4as
e
3as
+
1
as2
e
4as
j
429
430
j
Network Theory
R.P
5.30
Find the Laplace transform of the function f (t) shown in Fig. R.P. 5.30.
Figure R.P. 5.30
SOLUTION
Let f (t) = x(t) + u(t), where x(t) is a periodic triangular wave and is as shown in Fig. R.P.
5.30(a).
Figure R.P.5.30(a)
Figure R.P.5.30(b)
Let x1 (t) be x(t) within its first period as shown in Fig. R.P.5.30(b).
Referring to Fig. R.P. 5.30(b), we can write
x1 (t) =
)
x1 (t) = 2t [u (t)
u (t
= 2tu (t)
2tu (t
= 2tu (t)
2 (t
2 (t
= 2tu (t)
2 (t
4
1)] + (4
1) u (t
x1 (t) = 2tu (t)
4 (t
)
x1 (t) = 2r (t)
4r (t
1)
1) u (t
1) + 4u (t
1) + 2 (t
2u (t
2)]
2)
2tu (t
1)
4u (t
2) u (t
1) + 2tu (t
2)
2)
1) + 4u (t
1) + 2 (t
1) + 2 (t
2)
u (t
2 + 2) u (t
2u (t
1)
1) + 2r (t
1)
4u (t
1)
1 + 1) u (t
1) u (t
)
2t) [u (t
1) + 4u (t
1 + 1) u (t
2 (t
2t;
0<t<1
2t; 1 < t < 2
2) u (t
2)
1)
4u (t
2) + 4u (t
2)
2)
2)
Laplace Transform
Hence;
X1 (s) = L fx1 (t)g
=
=
=
2
4
s2
s2
2
s
e
s2
1
2e
s2
1
e
2
+
s
2
s2
e
+e
2s
2s
s 2
Since x(t) is periodic,
X (s) = L fx (t)g =
X1 (s)
1 e sT
where T = 2 seconds
2
Hence;
We know that;
2 (1 e s )
s2 (1 e 2s )
f (t) = x(t) + u(t)
X (s) =
Applying linearity property,
F (s) = X (s) + U (s)
2 1
=
s (1
R.P
2
e s
1
+
2
s
e
)
s
5.31
Find f (t) using convolution integral for the function,
F (s) =
4s
(s + 1) (s2 + 4)
SOLUTION
Let
where
F (s) = F1 (s) F2 (s)
4
F1 (s) =
s+1
s
F2 (s) = 2
s +4
f (t) = L
Z1
1
)
f1 (t) = 4e t u (t)
)
f2 (t) = cos 2tu (t)
[F1 (s) F2 (s)]
f1 () f2 (t
=
0
) d
j
431
432
j
Network Theory
u ( ) u (t
We know that
1; 0 < < t; t > 0
0; otherwise
) =
Zt
Hence;
(t )
cos 24e
f (t) =
0
t
= 4e
Zt
d
e cos 2 d
0
Using the standard integral formula
Z
eax cos bx dx =
eax
a2 + b2
f (t) = 4e
we get
t
(a cos bx + b sin bx)
t
e
1+4
(cos 2 + 2 sin 2)
=0
4
= e t et (cos 2t + 2 sin 2t 1)
5
4
8
4 t
= cos 2t + sin 2t
e ; t0
5
5
5
8
4
cos 2t + sin 2t
f (t) =
5
5
R.P
4
e
5
t
u (t)
5.32
If h(t) = 2e
3t u(t)
(t). Find y (t) = h(t) x(t) by (a) using convolution in
and x(t) = u(t)
the time-domain (b) Finding H (s) and X (s) and then obtaining L
1
[H (s)X (s)]
SOLUTION
3t
Given
h(t) = 2e
and
x(t) = u(t)
(a)
u(t)
(t)
y (t) = x (t) h (t)
Z1
x () h (t
=
) d
0
Z1
ju ()
=
()j 2e
3(t )
u (t
) d
0
Z1
2e
=
0
3(t )
Z1
u (t
) u () d
e
2
0
3(t )
u (t
) () d
Laplace Transform
j
433
1; 0 < < t; t > 0
0; otherwise
The second integral on the right-hand side is evaluated using the sifting property for an impulse function.
We know that,
u (t
) u () =
Zt
Hence;
2e
y (t) =
0
)
y (t) = 2e
=
3t
2
1
3
3t 3
e d
e3
t
3
e
2e
0
3t
3(t )
2e
3t
2e
3t
u (t
)j=0
u (t)
u (t)
Since t > 0, we associate u(t) in the first component on the right hand side of y (t).
2
1 e 3t u (t) 2e
3
2 8 3t
u (t)
e
=
3 3
Then;
y (t) =
3t
u (t)
(b) Verification :
1
2
; X (s) =
s+3
s
Y (s) = X (s) H (s)
2 (1 s)
=
s (s + 3)
H (s) =
)
=
1
K2
K1
+
s
s+3
Using partial fractions, we find that
K1 =
Hence;
)
R.P
2
8
; K2 =
3
3 8
1
2 1
Y (s) =
3 s
3 s+3
2
8 3t
y (t) = u (t)
e u (t)
3 3
2 8 3t
u (t)
e
=
3 3
5.33
When an impulse (t) V is applied to a certain network, the ouput voltage is vo (t) = 4u(t)
4u(t 2) V. Find and sketch vo (t) if the imput voltage is 2u (t 1) V.
434
j
Network Theory
SOLUTION
When vi (t) = (t), it is given that vo (t) = 4u(t) 4u(t 2).
From this data, we can find the transfer function H (s) as follows:
L fvo (t)g
L fvi (t)g
L f4u (t) 4u (t
=
L f (t)g
H (s) =
=
4
1
e
s
2s
2)g
The transfer function H (s) can be used to find vo (t) when vi (t) = 2u(t 1) V. This procedure
is as follows:
Vo (s)
Vi (s)
Vo (s) = Vi (s) H (s)
H (s) )
2
= e
s
=
8
s2
s
e
s
4
4
s
s
8
s2
e
e
2s
3s
Taking inverse Laplace transform, we get
vo (t) = 8 (t
= 8r (t
1) u (t
1)
1)
8r (t
8 (t
3) u (t
3)
3)
The corresponding wave form for vo (t) is sketched in Fig. R.P. 5.33
Figure R.P. 5.33
R.P
5.34
Refer the two circuits shown in Fig. R.P. 5.34(a) and (b). Given that v1 (t) = sin 103 t and
v2 (t) = e 1000t for t 0 and c = 1 F.
(a) Show that it is possible to have i1 (t) = i2 (t) for all t 0.
(b) Determine the required values of R and L for the condition in part (a) to hold good.
Laplace Transform
Figure R.P. 5. 34 (a)
j
435
Figure R.P.5.34 (b)
SOLUTION
Referring Fig. R.P. 5.34(a) we can write in Laplace domain
V1 (s)
I1 (s) =
R+
1
Cs
Similarly, referring Fig. R.P. 5.34(b), we can write in Laplace domain
I2 (s) =
V2 (s)
sL +
1
Cs
i1 (t) = i2 (t) means that I1 (s) = I2 (s)
103
V1 (s) = L sin 103 t =
Also;
V2 (s) = L e
1000t
=
2
s2 + (103 )
1
s + 103
Hence, the condition I1 (s) = I2 (s) gives,
103
s2 + 106
)
R s+
103
106
R
1
R+
106
=
1
s + 103
s
1
sL +
+
106 )
s
1
=
(s2
106
L (s +
103 )
s2
+
106
L
If the above equation is satisfied, then it is possible to have i1 (t) = i2 (t). For this to happen,
it is required that
R
103
= L;
106
R
= 103
The above conditions give L = 1H and R = 103 Ω
and
106 =
106
L
436
j
Network Theory
R.P
5.35
For the circuit shown in Fig. R.P. 5.35 has zero initial conditions. At t = 0, the switch K is
p
closed. Find p
the value of R such that the response v (t) = 0:5 sin 2t volts. Take the excitation
as i(t) = te 2t A.
Figure R.P.5.35
SOLUTION
p
Given i(t) = te 2t
Taking Laplace transform of i(t) gives
I (s) =
(s +
1
p
2)2
p
Laplace transform of the response v (t) = 0:5 sin 2t is
" p #
2
1
V (s) =
2
2 s +2
Hence;
Z (s) =
V (s)
I (s)
p
1 (s + 2)2
=p
2 s2 + 2
(5.29)
For the circuit shown in Fig. R.P. 5.35 we can write
Z (s) = R +
s
2
)
1
+
1
s
2s
Z (s) = R + 2
s +2
(5.30)
Laplace Transform
j
437
Equating equations 5.29 and 5.30, we get
p 2
1 s+ 2
2s
p
=R+ 2
2
s +2
2 s +2
p 2
1
s2 + 2 R 2s = p s + 2
2
2
2
s
= p + 2s + p
2
2
)
Equating the like powers of s, we get
1
2
R= p Ω
Exercise Problems
E.P
5.1
Find the Laplace transform of the following functions :
(a) f1 (t) = sin(!t + )
(b) f2 (t) = sin2 t
(c) f3 (t) =
1
[sinh(at)
2a3
Ans: F1 (s) =
E.P
sin(at)]
s sin θ + ω cos θ
2
, F2 (s) =
, F3 (s) = 2
2
2
2
s +ω
s (s + 4)
(s
1
a2 ) (s2
+ a2 )
5.2
In the network shown in Fig. E.P. 5.2, the switch K is moved from position a to position b at
t = 0, a steady state having previously been established at position a. Solve for i(t), using the
Laplace transformation method.
Figure E.P. 5.2
Ans: i (t) =
Va
e
RA
RA +RB
L
u (t)
438
j
E.P
Network Theory
5.3
Find i1 (t) and i2 (t) for t > 0 for the circuit shown in Fig. E.P. 5.3 using Laplace transform.
Figure E.P. 5.3
h
Ans : i1 (t) = 2.4e
h
i2 (t) = 1.2e
E.P
i
+ 0.6e 6510 t + 3 u (t) mA
i
6510 t
1.2e 510 t u (t) mA
5 105 t
5
5
5
5.4
Using Laplace transform technique, find i(t) when i1 = 0:1e
5.4 when b = 105 . Assume steady state conditions at t = 0 .
A for the circuit shown in E.P.
Figure E.P. 5.4
1
27
Ans: i (t) =
e bt +
e
30
40
E.P
bt
17
e
24
6bt
10bt
u (t) A
5.5
The current source shown in Fig. E.P. 5.5 is i(t) = tu(t) A. Find vo (t) when the initial value of
vo is zero.
Ans: vo (t) = t
10
3
Figure E.P. 5.5
1
e
103 t
mV, t 0
Laplace Transform
E.P
j
439
5.6
Find the inverse Laplace transform of the following F (s):
8s 3
4s2
(a) F (s) = 2
(b) F (s) =
s + 4s + 13
(s + 3)2
2
t
Ans: (a) f (t) = 10.2e
cos (3t + 38.3 ) u (t)
(b) f (t) = 4e
E.P
3t
24te
3t
3t
+ 18t2 e
u (t)
5.7
Using convolution integral, find f (t) if
F (s) =
Ans: f (t) = 2 1
E.P
e
5t
10
s (s + 5)
u (t)
5.8
Refer the network shown in Fig. E.P. 5.8. Assume the network is in steady state for t < 0.
Determine the current i(t) for t > 0.
Figure E.P. 5.8
Ans:
E.P
i (t) = 4.22e t cos (3t
18.4 ) u (t) A
5.9
Find vo (t) in the circuit shown in Fig. E.P. 5.9.
Figure E.P. 5.9
Ans:
vo (t) = 4
8.93e
3:73t
+ 4.93e
0:27t
u (t) V
440
j
Network Theory
E.P
5.10
Find vo (t) for t > 0. Refer the circuit shown in Fig. E.P. 5.10.
Figure E.P. 5.10
4
Ans: vo (t) =
+ 2.55e
3
E.P
1
t
3
cos
F
17t + 10.1
u (t)
5.11
For the circuit shown in E.P. 5.11.
Vo (s)
Find: (a) H (s) =
Vi (s)
(c) Step response
(b) h (t)
(d) The response when vi (t) = 8 cos 2t V
Figure E.P. 5.11
2
s+4
(b) h (t) = 2e 4t u (t)
Ans: (a)
(c)
H (s) =
vo (t) = 0.5 1
(d) vo (t) = 1.5 e
E.P
e
4t
4t
u (t) V
+ cos 2t + 0.5 sin 2t u (t) V
5.12
Refer the circuit shown in Fig. E.P. 5.12. The switch is closed at t = 0
Find : (a) i1 (t) and (b) i2 (t)
Laplace Transform
Figure E.P. 5.12
Ans: (a) i1 (t) = 3.33
E.P
1.67e
6:34t
1.67e
(b) i2 (t) = 3.33 + 1.22e
6:34t
4.55e
u (t)
23:66t
23:66t
u (t)
5.13
Find the Laplace transform of the waveform shown in Fig. E.P. 5.13.
Ans:
E.P
F (s) =
A 1
e
s
s2
Figure E.P. 5.13
A
e
s
2s
5.14
Find the Laplace transform of the periodic waveform shown in Fig. E.P. 5.14.
Ans:
1
F (s) = 2
s
Figure E.P. 5.14
1
e
s2
2s
2
e
s
2s
1
1
e
2s
j
441
442
j
E.P
Network Theory
5.15
Find the Laplace transform of the waveform shown in Fig. E.P. 5.15
Ans: F (s) =
E.P
Figure E.P. 5.15
1
2
2
2
+ + 2e s
2
s
s
s
1 e s
5.16
Obtain the Laplace transform of the f (t) shown in Fig. E.P. 5.16.
Ans: F (s) =
E.P
1
5
s
Figure E.P. 5.16
3e s + 3e
3s
5e
4s
5.17
Obtain the Laplace transform of the unit impulses shown in Fig. E.P. 5.17
Figure E.P. 5.17
Ans: X(s) =
1
1 e s
Laplace Transform
E.P
j
443
5.18
Refer the circuit shown in Fig. E.P. 5.18. Let i(0) = 1A; vo (0) = 2V and vs (t) = 4e
Find vo (t) for t > 0.
2t u(t)V.
Figure E.P. 5.18
Ans:
E.P
vo (t) =
2 + 4.33e
0:5t
+ 1.33e
2t
u (t) volts
5.19
Find i(t) in the circuit shown in Fig. E.P. 5.19. Assume that the circuit is initially relaxed.
Figure E.P. 5.19
Ans:
E.P
i (t) = 0.5
0.5e
4t
te
4t
u (t)
5.20
Refer the circuit shown in Fig. E.P. 5.20. Assume zero initial conditions. Use convolution theorem
to find i(t).
Figure E.P. 5.20
Ans:
i (t) =
t
e
2
5t
(t
2)
2
e
5(t 2)
444
j
E.P
Network Theory
5.21
There is no energy stored in the circuit shown in Fig. E.P. 5.21 at the time when the switch is
opened. Show that
sI (s)
g
V2 (s) =
R1
1
2
C1 s +
s+
L1
L 1 C1
Figure E.P. 5.21
E.P
5.22
Refer the circuit shown in Fig. E.P. 5.22. If is (t) = 6u(t)mA, find v2 (t).
Figure E.P. 5.22
Ans: v2 (t) = 10e
E.P
4000t cos (3000t
90 ) u (t) V
5.23
Find Vo (s) and vo (t) in the circuit shown in Fig. E.P. 5.23 if the initial energy is zero and the
switch is closed at t = 0
Figure E.P. 5.23
Ans: vo (t) = 30
60e
5000t
+ 30e
10000t
u (t)
Laplace Transform
E.P
5.24
The initial energy in the circuit in Fig. E.P. 5.24 is zero.
(a) Find Vo (s).
(b) Use the initial and final value theorems to find vo (0+ ) and vo (1).
(c) Do the values obtained in part (b) agree with known circuit behaviour? Explain.
(d) Find vo (t).
Figure E.P. 5.24
Ans:
(a) Vo (s) =
+ 4200
21 2
s (s + 8s + 25)
103 s
(b) vo 0+ = 0, vo () = 168V
(c) YES
(d) vo (t) = 168 + 7225.95e 4t cos (3t + 91.33 ) u (t) V
E.P
5.25
Find the initial and final value of H (s) =
Ans:
1, 2
E.P
s3 + 25 + 6
s(s + 1)2 (s + 3)
5.26
Verify final value theorem and initial value theorem for the function,
f (t) = 2 + e 3t cos 2t
E.P
5.27
Using the convolution theorem, find the Laplace inverse of the following functions:
1
s
1
(i) F (s) =
(iii) F (s) =
(ii) F (s) =
2
s(s + 1)
(s a)
(s + 1)(s + 2)
Ans:
(i) f (t) = 1
e t
(ii) f (t) = teat
(iii) f (t) = e t + 2e
2t
2e t
j
445
446
j
E.P
Network Theory
5.28
In the circuit shown in Fig. E.P. 5.28, find the voltage across the resistance vR (t) using convolution
integral. Given that vg (t) = e 2t and RC = 1 second.
Figure E.P. 5.28
Ans: vR (t) = 2e
E.P
2t
e t,
t0
5.29
Find the inverse Laplace transform of the following functions:
3s
s2 + 3
1
(i) 2
(iii)
(ii)
(s + 1)(s2 + 4)
(s + 1)(s + 2)2
(s2 + 2s + 2)(s + 2)
Ans:
(i) cos t
cos 2t
(ii) e t
e
(iii)
E.P
7
e
2
2t
2t (1
+ t)
2.5e t cos t + 0.5e t sin t
5.30
In the circuit shown in Fig. E.P. 5.30, switch K is open for a long time so that steady state is
reached and at t = 0, switch is closed. Determine the current i(t) in 10 ohm resistor.
Figure E.P.5.30
Ans:
Current in each 10 Ω resistor = 2u(t)
e
5t
Laplace Transform
E.P
j
447
5.31
Synthesize the wave form shown in Fig. E.P. 5.31 using ramp function and obtain the Laplace
transform of f (t).
Figure E.P.5.31
Ans:
E.P
F (s) =
5
[1
s2
2e s + e
2s
]
5.32
Find the Laplace transform of the voltage wave form as shown in Fig. E.P. 5.32.
Figure E.P.5.32
Ans:
E.P
V (s) =
2
[1
s2
3e s + 5e
1:5s
6e
2s
+ 6e
3s
]
5.33
Find the Laplace transform of the perodic wave forms shown in Figs. E.P. 5.33(a) and (b).
Figure E.P.5.33(a)
448
j
Ans:
Network Theory
(i) F (s) =
(ii) F (s) =
E.P
1
1
1
e
1
e
1
4s s2
2
2s s
Figure E.P.5.33(b)
2
1
e s + 2 e 2s
2
s
s 4e s
2 2s
+ e
s
s
1
e
s
2s
2
+ 2e
s
3s
1
e
s2
4s
5.34
For the circuit shown in Fig. E.P. 5.34, find the current transients in both the loops using Laplace
transformation method.
Figure E.P.5.34
12
7
2
i2 (t) = +
7
Ans: i1 (t) =
E.P
5 2t
e 5t Ampere; t 0
e
7
5 7t
e 5t Ampere; t 0
e
7
5.35
Find the Laplace transform of the saw tooth wave as shown in Fig. E.P. 5.35.
Figure E.P. 5.35
Laplace Transform
Ans:
F (s) =
E.P
5.36
V (1
j
449
e T s T se T s )
T s2 (1 e T s )
For the circuit shown in Fig. E.P. 5.36 switch K is closed at t = 0. Determine the current i(t) for
t 0.
Figure E.P. 5.36
Ans:
E.P
i(t) = 0.357e
2t
5
e j 25t
25 + j2
5
25
j2
e j 25t
5.37
For the circuit shown in Fig. E.P. 5.37, determine the source current when the switch K is closed
at t = 0. Assume zero initial conditions.
Figure E.P. 5.37
Ans:
i(t) = 2.57e t
0.57e
0:3t
Amperes; t 0
Chapter
6
6.1
Resonance
Introduction
A.C Circuits made up of resistors, inductors and capacitors are said to be resonant circuits when
the current drawn from the supply is in phase with the impressed sinusoidal voltage. Then
1. the resultant reactance or susceptance is zero.
2. the circuit behaves as a resistive circuit.
3. the power factor is unity.
A second order series resonant circuit consists of R, L and C in series. At resonance, voltages
across C and L are equal and opposite and these voltages are many times greater than the applied
voltage. They may present a dangerous shock hazard.
A second order parallel resonant circuit consists of R; L and C in parallel. At resonance,
currents in L and C are circulating currents and they are considerably larger than the input current.
Unless proper consideration is given to the magnitude of these currents, they may become very
large enough to destroy the circuit elements.
Resonance is the phenomenon which finds its applications in communication circuits: The
ability of a radio or Television receiver (1) to select a particular frequency or a narrow band of
frequencies transmitted by broad casting stations or (2) to suppress a band of frequencies from
other broad casting stations, is based on resonance.
Thus resonance is desired in tuned circuits, design of filters, signal processing and control
engineering. But it is to be avoided in other circuits. It is to be noted that if R = 0 in a series
RLC circuit, the circuits acts as a short circuit at resonance and if R = in parallel RLC circuit,
the circuit acts as an open circuit at resonance.
1
452
6.2
| Network Analysis
Transfer Functions
As ω is varied to achieve resonance, electrical quantities are expressed as functions of ω, normally
denoted by F (jω) and are called transfer functions. Accordingly the following notations are used.
V (jω)
= Impedance function
I(jω)
I(jω)
Y (jω) =
= Admittance function
V (jω)
V2 (jω)
G(jω) =
= Voltage ratio transfer function
V1 (jω)
I2 (jω)
α(jω) =
= Current ratio transfer function
I1 (jω)
Z(jω) =
If we put jω = s then the above quantities will be Z(s), Y (s), G(s), α(s) respectively. These
are treated later in this book.
6.3
Series Resonance
Fig. 6.1 represents a series resonant circuit.
Resonance can be achieved by
1. varying frequency ω
2. varying the inductance L
3. varying the capacitance C
Figure 6.1 Series Resonant Circuit
The current in the circuit is
I=
E
E
=
R + j(XL − XC )
R + jX
At resonance, X is zero. If ω0 is the frequency at which resonance occurs, then
1
1
= resonant frequency.
or ω0 = √
ω0 L =
ω0 C
LC
V
The current at resonance is Im =
= maximum current.
R
The phasor diagram for this condition is shown in Fig. 6.2.
The variation of current with frequency is shown in Fig. 6.3.
2
Figure 6.2
Figure 6.3
Resonance
| 453
It is observed that there are two frequencies, one above and the other below the resonant
frequency, ω0 at which current is same.
1
and |Z| with ω.
Fig. 6.4 represents the variations of XL = ωL; XC =
ωC
1
we see that any constant product of L and C give a particular
From the equation ω0 = √
LC
L
resonant frequency even if the ratio
is different. The frequency of a constant frequency source
C
can also be a resonant frequency for a number of L and C combinations. Fig. 6.5 shows how the
L
sharpness of tuning is affected by different
ratios, but the product LC remaining constant.
C
0
0
Figure 6.4
Figure 6.5
L
For larger ratio, current varies more abruptly in the region of ω0 . Many applications call for
C
narrow band that pass the signal at one frequency and tend to reject signals at other frequencies.
6.4
Bandwidth, Quality Factor and Half Power Frequencies
At resonance I = Im and the power dissipated is
2
Pm = Im
R watts.
Im
When the current is I = √ power dissipated is
2
Pm
I2 R
= m watts.
2
2
From ω − I characteristic shown in Fig. 6.3, it is observed that there are two frequencies
Im
ω1 and ω2 at which the current is I = √ . As at these frequencies the power is only one half of
2
that at ω0 , these are called half power frequencies or cut off frequencies.
The ratio,
current at half power frequencies
1
Im
=√
=√
Maximum current
2Im
2
1
When expressed in dB it is 20 log √ = −3dB.
2
454
Network Analysis
j
Therefore !1 and !2 are also called
pIm2 = pE2R , the
p2R = jR + j(XL XC )j
As
3 dB frequencies.
magnitude of the impedance at half-power frequencies is
Therefore, the resultant reactance, X = XL XC = R.
The frequency range between half - power frequencies is
passband or band width.
BW = !2 !1 = B:
!2 !1 , and it is referred to as
R
The sharpness of tuning depends on the ratio , a small ratio indicating a high degree of
L
selectivity. The quality factor of a circuit can be expressed in terms of R and L of the inductor.
Quality factor = Q =
!0 L
R
Writing !0 = 2f0 and multiplying numerator and denominator by
2
2
1
1
Q = 2f0 12 ILIm2 R = 2 12ILI2mRT
2
m
= 2
Selectivity is the reciprocal of Q.
m
Maximum energy stored
total energy lost in a period
Q = !R0 L and !0 L = ! 1C ;
As
0
and since !0 =
6.5
2
1
pLC
, we have
Q = ! 1CR
0
rL
Q= R C
1
Expressions for ω1 and ω2 , and Bandwidth
At half power frequencies !1 and !2 ,
E
I = pE = 2
2R
fR + (XL XC )2g
jXL XC j = R i:e:; !L !C1 = R
1
2
∴
At ! = !2 ,
Simplifying,
!22 LC !2 CR
R = !2 L ! 1C
2
1=0
1 2
I , we get,
2 m
Resonance
Solving, we get
!2 =
p
RC + R C
2
2LC
2
+ 4LC
R+
=
2L
s R
2
2L
+
1
LC
j
455
(6.1)
Note that only + sign is taken before the square root. This is done to ensure that !2 is always
positive.
At ! = !1 ,
R = 1 !1 L
)
Solving;
!1 C
! LC + !1 CR p1 = 0
2 2
!1 = RC + 2RLCC + 4LC
s 2
R + 1
R
+
=
2L
2L
LC
2
1
(6.2)
While determining !1 , only positive value is considered.
Subtracting equation(6.1) from equation (6.2), we get
Since Q =
and therefore
!2 !1 = RL = Band width.
!0 L , Band width is expressed as
R
B = !2 !1 = RL = !Q0 :
Q = ! !0 ! = !B0
2
1
Multiplying equations (6.1) and (6.2), we get
2
1
!1 !2 = 4RL2 + LC
!0 = p !1 !2
or
R2 = 1
4L2
LC
= !02
The resonance frequency is the geometric mean of half power frequencies.
R << 1 ; in which case Q 5
Normally
2L
LC
1
R+ 1
Then;
!1 2RL + LC
and !2
2L
LC
p
'
=
∴
r
R
2L
+ !0 and
'
!2 = 2RL + !0
p
!0 = !1 +2 !2 = Arithmetic mean of !1 and !2
R = !0 , Equations for ! and ! as given by equations (6.1) and (6.2) can be expressed
1
2
L Q
in terms of Q as
s 2
!
0
!2 = 2Q + 2!Q0 + !02
Since
456
j
Network Analysis
= !0
Similarly
!1 = !0
2 s 3
4 21Q + 1 + 21Q 5
2
s 3
4 21Q + 1 + 21Q 5
2
2
R << p 1 and then Q > 5.
2L
LC
Consequently !1 and !2 can be approximated as
r1 R
R
!1 ' 2L + LC = 2L + !0 = B2 + !0
r1 R
R
!2 ' 2L + LC = + 2L + !0 = B2 + !0
so that
! = !1 + !2 :
Normally,
0
6.6
2
Frequency Response of Voltage across L and C
As frequency is varied, both the voltages across L and C increase with frequency upto !0 and
they are equal at !0 : But their maximum values do not occur at !0 : Vc reaches its maximum at
! < !0 and VL reaches its maximum at ! > !0 . This can be verified by calculating the frequency
at which each occurs.
6.7
Expression for ω at which VL is Maximum
Current in the circuit shown in Figure 6.1 is
I=q
Voltage across L is
VL = !LI = q
Squaring
VL2 =
This is maximum when
R2 +
E
!L
dVL 2 = 0
d!
1
!C
2
E!L
R2 + !L
E 2 ! 2 L2
R2 + !L !C1 2
1
!C
2
Resonance
"(
That is,
EL
2 2
or
)
2
+
1
2
=
LC
1
1
R2 C
2L
1
Let this frequency be !L .
Then;
!L2 = !02 1
!L = !0
That is, !L
6.8
2! ! 2 !L
L + !2C
!C !C
!C
1 2
1
1
2
R + !L !C = !L !C !L + !C
1
L 2 L#2
1
2
R2 + !2#L#
+ 2 2 2 = !#
2
!C
C
! C2
2 2 2
2
R ! C + 1 2! LC = 1
!2 (2LC R2 C 2 ) = 2
!2 = 2LC 2 R2 C 2
R
2
1
#
=0
1
1
2Q2
s
1
1
> !0 .
1
2Q2
:
Expression for ω at which VC is Maximum
E
!C R2 + !2 L
2
VC2 = 2 2 2 E
! C R + !L
q
VC =
Now
This is maximum when
That is,
" (
E 2 !2
C2
2
R
2
+
d 2
d! (VC ) = 0:
!L !C
1
!L !C
1
L !2C
1
2
=
+ 2!
(
!L !C
1
1
1
!C
!C
2
2
R2 + !L !C
1
!L + !C
1
)#
2
=0
j
457
458
j
Network Analysis
1#
L
1#
2 2
R 2 + ! 2 L2 + ! #
2 = #
2C 2
2
C ! C2 ! L
L
2! 2 L2 + R2 = 2
C
L
2
1
R2
!2 = 2 C2L2R = LC
2L2 2
1
R C = !2 1
=
1
0
LC
2 L
Let this frequency be !C
1
2Q2
r
!C = !0 1 2Q1 2
i:e:;
! C < !0
Variations of VC and VL as functions of !
are shown in Fig. 6.6.
We know that
Figure 6.6
VC = r
E
n
!C R
2
2
Consider ! 2 C 2 R2 + (! 2 LC
!C C R
2
2
2
+
(! 2 LC 1)2
!2 C 2
2 2
2
E
+ (! 2 LC
Q2
1
2
LC = !0
=
2
2
0
and
2Q2
2
2
+
!0 CR = Q1
2
g
(6.3)
2
4Q4
=
Q2
1
1
2
0
2Q4
+
4Q4
LC
2Q2
2
0
2
2
0
2
1
1)2
1)2 and at ! = !C . Then equation(6.3) becomes
1
+ (!C LC 1) = ! 1
CR + ! 1
2Q
1 11 1
1
! 1 2Q =
Q 1 2Q + !
1 1 1
1
1
1
2
2
since
2
o = p fR ! C
=
Q2
1
2
1
1
4Q2
Substituting the above expression in the denominator of equation (6.3), we get
Vcm = q EQ 1
1
6.9
4Q2
Selectivity with Variable L
In a series resonant circuit connected to a constant voltage, with a constant frequency, when L is
varied to achieve resonance, the following conditions prevail:
Resonance
j
459
XC is constant and I = q E
when L = 0.
R2 + XC2
V at X = X
2. With increase in L; XL increases and Im =
R L C
3. With further increase in L; I proceeds to fall.
All these conditions are depicted in Fig. 6.7 VC max occurs at
!0 but VL max occurs at a point beyond !0 .
L at which VL becomes a maximum is obtained in terms of
1.
other constants.
EXL
fR2 + (XL XC )2g
2 2
VL2 = R2 + (EX XL X )2
L
C
2
dVL = 0.
This is maximum when
dXL
VL =
R
Therefore;
2
+ (XL
1
2
Figure 6.7
XC )2 2XL = XL2 f2(XL XC )g
R2 + XL2 + XC2
2XL XC = XL2
XL = R X+ XC
C
2
Therefore;
Let the corresponding value of L is Lm .
Then;
and L0 = value of L at !0 such that
XL XC
2
Lm = C (R2 + XC2 )
!0 L = ! 1C :
0
6.10
Selectivity with Variable C
In a series resonant circuit connected to a constant voltage, constant frequency supply, if
varied to achive resonance, the following conditions prevail:
1. XL is constant.
2. XC varies as inversely as C
when C = 0,
3.
C is
I = 0.
V
, I = Im = .
when !C =
!L
R
with further increase in C; I starts decreasing as shown in Fig. 6.8, where Cm is the value of
capacitance at maximum voltage across C and C0 is the value of the capacitance at !0 .
1
460
j
Network Analysis
C at which VC becomes maximum can be determined in
terms of other circuit constants as follows.
EXC
R2 + (XL XC )2
2 2
VC2 = R2 + (EX XC X )2
L
C
VC = p
Figure 6.8
dVC2 = 0
dXC
R2 + (XL XC )2 2XC X 2 f2(XL XC )(
Then;
C
R2 + XL2 +XC2 2XL XC = XL XC +XC2
2
2
XC = R X+ XL
For maximim VC ;
g
1) = 0
L
Let the correrponding value of C be Cm .
Cm = R2 +L X 2 :
Then;
L
6.11
Transfer Functions
6.11.1
Voltage ratio transfer function of a series resonant circuit and frequency response
For the circuit shown in Fig. 6.9, we can
write
j!) =
R
H (j!) = VV0((j!
)
R + j !L
s
!CR
n! L 1 ! o
1
=
1 + j !L
R
=
1+j
=
=
1
1
! R!
h
0
0
1 + jQ !!0
!!0 CR
!0
!
i
1
1+Q
2
h!
!0
Figure 6.9
0
1
!C
!0
!
i
2
,
1
2
tan
1
Q ! !!0
0
!
Resonance
Let be a measure of the deviation in ! from !0 . It is defined as
= ! ! !0 = !!
0
0
1
! !0 = ( + 1) 1 = ( + 1)2
!0 !
+1
+1
For small deviations from !0 ; << 1: Then,
! !0
!0 ! ' 2
1
Then
Then;
H (j!) = 1 + j12Q = p
1
1 + 4Q2 2
=
tan
1
2 + 2
+1
2Q
The amplitfude and phase response curves are as shown in Fig. 6.10.
Figure 6.10 (a) and (b): Amplitude and Phase response of a series resonance circuit
6.11.2 Impedance function
The Impedance as a function of j! is given by
Z (j!) = R + j !L !C
!L 1 =R 1+j
R! !CR
!
0
= R 1 + jQ
!0 !
s
! !0 ! ! 0 2 ,
1
2
tan Q
=R 1+Q
!0 !
!0 !
1
For small deviations from !0 , we can write
Z (j!) ' R[1 + j 2Q ] = R
p1 + 4Q tan
2 2
1
2Q
j
461
462
j
6.12
Network Analysis
Parallel Resonance
The dual of a series resonant circuit is often considered as a parallel resonant circuit and it is as
shown in Fig. 6.11.
The phasor diagram for resonance is shown in Fig. 6.12.
The admittance as seen by the current source is
Y (j!) = YR + YL + YC 1
1
= + j !C
R
!L = G + jB
Figure 6.11 Parallel Resonance Circuit
Figure 6.12 Phasor Diagram
If the resonance occurs at !0 ; then the susceptance B is zero. That is,
!0 C = !1L
0
or
!0 = p 1 rad= sec :
LC
At resonance,
and
IC 0 = IL0 = j!0 CRI
ILC = IC 0 + IL0 = 0
The quality factor, as in the case of series resonant circuit is defined as
Maximum energy stored
Q = 2 Energy
dissipated in a period
CV 2
1
2 R T
= 2f0 CR = !0 CR:
!0 C = !1L ;
0
R
Q = ! L:
0
1
= 2 2 V 2 m
m
Since
Resonance
On either side of !0 there are two frequencies at which
the voltage is same. At resonance, the voltage is maximum and is given by Vm = IR and is evident from the
response curve as shown in Fig. 6.13. At this frequency,
p
pm
V2
= m watts. The frequencies at which the
R
1
times the maximum voltage are called half
voltage is
2
power frequencies or cut off frequencies, since at these
frequencies,
=
p
Vp m
2
p = R2
At any ! ,
V
= m = half of the maximum power.
2R
Y = 1 + j !C
2
At !1 and !2 ;
R
jY j = p
1
=
2R
s 1
2
R
Squaring,
Therefore;
At ! = !2 ,
Hence;
Figure 6.13
1
2R2
1
!C !L
=
=
1
R2
+
+
1
!L
!C !L
1
!C !L
1
2
2
1
R
!2 C !1L = R1
2
2
!2 LC 1 = !R2 L
!22 LCR R !2 L = 0 p
2 + 4LCR2
!2 = L + L2LCR
Note that only positive sign is used before the square root to ensure that !2 is positive.
Thus;
Similarly;
So that, bandwidth
!2 = 2RC +
1
!1 =
1
2RC
s
+
1
2
s2RC 1
2RC
1
B = !2 !1 = RC
+
1
LC
2
+
1
LC
j
463
464
j
Network Analysis
!1 !2 =
1
2
2RC
1
=
= !02
and
Thus;
!0 = p 1
LC
As
LC
+
1
2RC
=
and
2
rC
p
R
Q = LC = R
0
2
L
!2 = B2 +
s B +!
s2 !1 = B2 +
B = !Q0 ,
!2 = !0
!1 = !0
and
6.13
2RC
Q = !0 RC = !RL
B
and
Using
1
LC
! 0 = p !1 !2
L
Since
1
2
B
2
2
2
0
+ !02
2 s 3
4 21Q + 1 + 21Q 5
2
s 3
4 21Q + 1 + 21Q 5
2
2
Transfer Function and Frequency Response
The transfer function for a parallel RLC circuit shown in
Fig. 6.14. is H (j! ), the current ratio transfer function.
)
1
=
H (j!) = II0 ((j!
j!) RY (j!)
1
=
=
1
1
R R + j !C
1
1+j
1
!! CR
1
= 1 + jR !C
!L
!L
1! ! =
!R
1
1
0
0
1 + jQ !0
!0
!0 !L
As in the case of series resonance, here also let
0
!
= ! ! !0 = !!
0
0
1
Figure 6.14 Parallel RLC Circuit
Resonance
j
465
then,
! !0 = 2 + 2
!0 ! + 1
For << 1, for small deviations from !0
! !0 ' 2
!0 !
Therefore,
H (j!) = 1 + j12Q
6.14
Resonance in a Two Branch RL
RC Parallel Circuit
Consider the two branch parallel circuit shown in Fig. 6.15. Let E be the voltage across each of
the parallel circuit shown in the figure. The vector diagram at resonance is shown in Figure 6.1.
Figure 6.15 Two branch Parallel Circuit
Figure 6.16
The admittance of the circuit is Y (j! ) = GL
For resonance,
jBL + GC + jBC
BL = BC
If this occurs at ! = !0 ,
then
1
!0 L
!C
=
RL2 + !02 L2 RC2 + ! 1C
!0 C
= 2 2 2
RC !0 C + 1
L(1 + !02 C 2 RC2 ) = C (RL2 + !02 L2 )
!02 (LC 2 RC2 L2 C ) = RL2 C L
0
2
0
2
466
| Network Analysis
ω02 =
2 C −1
RL
2 − L2 C
LC 2 RC
2 − L
2 C −L
1 RL
1 RL
C
=
2 C −L
2 − L
LC (RC
LC (RC
C
2 − L
RL
1
C
ω0 = √
2 − L
LC RC
C
=
Therefore,
This is the expression for resonant frequency. It is to be noted that
1. resonance is not possible for certain combination of circuit elements unlike in a series
circuit where resonance is always possible.
2. resonance is also possible by varying of RL or RC .
Consider the case where
2
RC
<
or
L
2
< RL
C
L
2
< RC
C
In both these cases, the quantity under radical is negative and therefore resonance is not possible.
The admittance at resonance of the above parallel circuit is
"
!
RC
RL
S
Y0 =
2 + X 2 + R2 + X 2
RL
L0
C
C0
2
RL
<
where XL0 and XC0 are the inductive and capacitive reactances respectively at resonance.
L
If
RL = RC =
C
1
then
ω0 = √
LC
as in R, L, C series circuit.
If
RL = RC =
L
C
which means
2
2
RL
= RC
= R2 =
Then,
L
= XL XC .
C
XC
XL
− 2
2
+ XL RC + XC2
1
1
=
−
=0
XL + XC
XL + XC
BL − BC =
2
RL
Resonance
j
467
In this case, the circuit acts as a pure resistive circuit irrespective of frequency. That is, the
circuit is resonant for all frequencies.
In this case the circuit admittance is
Y
RL + RC
RL + XL2 RC2 + XC2
R2 + X 2 + R2 + X 2 L
C
=R
R4 + R2 (XL2 + XC2 ) + XL2 XC2
2R2 + XL2 + XC2
=R 4
2R + R2 (XL2 + XC2 )
=
2
2R2 + X 2 + X 2 R
L
C
= 2
R 2R2 + XL2 + XC2
rC
1
=
R= L
rL
Z =R= C
or
6.14.1 Resonance by varying inductance
If resonance is achieved by varying only L in the circuit shown in Figure 6.15 but with constant
current constant frequency source, then the condition for resonance is
BL = BC
) R2 X+LX 2 = R2 X+CX 2 = XZ 2C where ZC2 = RC2 + XC2
L
L
C
C
C
2
2
2
Then;
XL XC XL ZCq+ XC RL = 0
ZC2 ZC4 4XC2 RL2
Solving, forXL we get XL =
2XC
q 4 2 2 L
C
2
Therefore;
L = 2 ZC ZC 4XC RL since XL XC = C
The following conditions arise:
1. If ZC4
> 4XC2 RL2 ; L has two values for the circuit to resonate.
2. For ZC4 = 4XC2
3.
RL2 , L = 12 CZC2 for reasonance.
2 , No value of L makes the circuit to resonate.
For ZC4 < 4XC2 RL
468
j
6.14.2
Network Analysis
Resonance by varying capacitance
As in the previous case, we have at resonance„
BL = BC
XC
XL
2
2
2
RC2 + XC2 = ZL2 ; where ZL = RL + XL
)
Simplifying we get,
XC2 XL XC ZL2q+ RC2 XL = 0
ZL2 ZL4 4XL2 RC2
XC =
2XL
2L
q
C=
ZL2 ZL4 4XL2 RC2
Therefore;
The following conditions arise:
1. For ZL4
2.
3.
6.14.3
> 4XL2 RC2 , there are two values for C to resonante.
2 , resonance occurs at C = 2L .
For Z44 = 4XL2 RC
ZL2
2 , no value of C makes the circuit to resonate.
For ZL4 < 4XL2 RC
Resonance by varying RL or RC
It is often possible to adjust a two branch parallel combination to resonate by varying either RL
or RC . This is because, when the supply is of constant current and, constant frequency, these
resistors control inphase and quadratare components of the currents in the two parallel paths.
From the condition BL = BC , we get
XL = XC
RL + XL2 RC2 + XC2
XL R 2 + X X X 2
RL2 = X
L C
C
L
rCXL
RL = X RC2 + XL XC XL2
(6.4)
C
This equation gives the value of RL for resonance when all other quantities are constant and
2
the term under radical is positive.
Similarly if only RC is variable, keeping all other quantities constant, the value of
resonanace is given by
r XC
RC = X RL2 + XL XC XC2
L
provided the term under radical is positive.
RC for
Resonance
6.15
| 469
Practical Parallel and Series Resonant Circuits
A practical resonant parallel circuit contains an inductive coil of resistance R and inductance L in
parallel with a capacitor C as shown in Fig. 6.17. It is called a tank circuit because it stores energy
in the magnetic field of the coil and in the electric field of the capacitor. Note that resistance RC
of the capacitor is negligibly small.
Condition for parallel resonance is shown by the phasor diagram of Fig. 6.18.
IC = IL sin φ
That is,
⇒
BC = BL
ωL
= ωC.
R2 + ω 2 L2
Let the value of ω which satisfy this condition be ω0 .
Then,
R2 + ω02 L2 =
L
C
1
L
1
=
− R2
C
L2
LC
1
R2 C
ω0 = √
1−
L
LC
ω02 =
Figure 6.17
1−
R2 C
L
(6.5)
(6.6)
Figure 6.18
Admittance of the circuit shown in figure 6.17 is
1
+ jωC
R + jωL
R
jωL
= 2
−
+ jωC
R + ω 2 L2 R2 + ω 2 L2
Y (jω) =
At ω = ω0 , Y (jω) is purely real.
Hence,
Y (jω0 ) =
R2
R
+ ω02 L2
(6.7)
470
| Network Analysis
Substituting for ω0 in equation (6.7),
Y (jω0 ) =
R2
R
R
RC
= L =
2
2
L
+ ω0 L
C
(6.8)
L
, which is called the dynamic resistance of
CR
R2 C
> 1, there is
the circuit. This is greater than R if there is resonance. However, note that if
L
no resonance.
Fig. 6.19 shows a practical series resonant circuit. The input impedance as a function of ω is
and the circuit is a pure resistive with R0 =
Z(jω) = jωL +
G2
ωC
G
−j 2
2
2
+ω C
G + ω2C 2
Condition for resonance is
ωC
ωL = 2
G + ω 2 C2
1
C
1
1
ω2 =
=
− G2
− 2 2
2
L
C
LC
C R
L
1
1−
=
LC
CR2
L
1
1−
ω=√
CR2
LC
Impedance at resonance is
Z0 =
G2
Figure 6.19
G
L
G
= C =
2
+ ωC
CR
L
L
The circuit at resonance is a purely resistive, and Z0 = R0 =
. However, note that here
CR
L
also resonance is not possible for
> 1.
CR2
In both the circuits, shown in Figs 6.18 and 6.19, resonance is achieved by varying either C or
L until the input impedance or admittance is real and this process is called tuning. For this reason
these circuits are called tuned circuits.
Series circuits
EXAMPLE
6.1
Two coils, one of R1 = 0.51 Ω, L1 = 32 mH, the other of R2 = 1.3 Ω and L2 = 15 mH and
two capacitors of 25 μF and 62 μF are all in series with a resistance of 0.24 Ω. Determine the
following for this circuit
(i) Resonance frequency
(ii) Q of each coil
Resonance
(iii) Q of the circuit
(iv) Cut off frequencies
(v) Power dissipated at resonance if E = 10 V.
SOLUTION
From the given values, we find that
Rs = 0.51 + 1.3 + 0.24 = 2.05Ω
Ls = 32 + 15 = 47 mH
25 × 62
Cs =
μF = 17.816 μF
87
(i) Resonant frequency:
ω0 = √
1
Ls Cs
1
47 ×
× 17.816 × 10−6
= 1092.8 rad/ sec
=√
(ii) Q of coils:
For Coil 1,
For Coil 2,
(iii) Q of the circuit:
10−3
ω0 L1
R1
1092.8 × 32 × 10−3
=
= 68.57
0.51
ω0 L2
Q2 =
R2
1092.8 × 15 × 10−3
=
= 12.6
1.3
Q1 =
ω0 Ls
Rs
1092.8 × 47 × 10−3
=
= 25
2.05
Q=
(iv) Cut off frequencies: Band width is,
1092.8
ω0
=
= 43.72
B=
Q
25
Considering Q > 5, the cut off frequencies,
B
= 1092.8 ± 21.856
ω2,1 = ω0 ±
2
Therefore,
ω2 = 1115 rad/ sec and ω1 = 1071 rad/ sec .
| 471
472
j
Network Analysis
(v) Power dissipated at resonance:
Given E = 10 V
We know that at resonance, only the resistance portion will come in to effect. Therefore
P = ER
2
EXAMPLE
=
102
= 48:78 W
2:05
6.2
For the circuit shown in Fig. 6.20, find the out put voltages at
(i)
(ii)
(iii)
! = !0
! = !1
! = !2
when vs (t) = 800 cos !t mV.
Figure 6.20
SOLUTION
For the circuit, using the values given, we can find that resonant frequency
!0 = 1
pLC
1
=p
312 10 1:25 10
3
Quality factor:
Band width:
As Q > 5,
Hence;
and
Q = !R0 L
1:6 106 312 10
=
62:5 103
12
= 1:6
3
=8
B = !Q0
1:6 106
=
= 0:2 106 rad= sec
8
!2;1 = !0 B2
= (1:6 0:1)106 rad= sec
!2 = 1:7 106 rad= sec
!1 = 1:5 106 rad= sec
10
6
rad= sec
Resonance
| 473
(i) Output voltage at ω0 :
Using the relationship of transfer function, we get
50Im
62.5Im
= 0.8 0◦
H(jω)|ω=ω0 =
Since the current is maximum at resonance and is same in both resistors,
vo (t) = 0.8 × 800 cos(1.6 × 106 t) mV
= 640 cos(1.6 × 106 t) mV
At ω1 and ω2 , Zin =
√
2Rs ±45◦ . Therefore,
Rout
50
=√
45◦
Zin
2 × 62.5
= 0.5657 45◦
H(jω)|ω=ω1 =
H(jω)|ω=ω2 = 0.5657 45◦
and
(ii) Out put voltage at ω = ω1
vo (t) = 0.5657 × 800 cos(1.6 × 106 t + 45◦ ) mv
= 452.55 cos(1.6 × 106 t + 45◦ ) mV
(iii) Out put voltage at ω = ω2
vo (t) = 452.55 cos(1.6 × 106 t − 45◦ ) mV
EXAMPLE
6.3
In a series circuit R = 6 Ω, ω0 = 4.1 × 106 rad/sec, band width = 105 rad/sec. Compute L, C,
half power frequencies and Q.
SOLUTION
We know that Quality factor,
Q=
Also,
Q=
Therefore,
L=
and
Hence,
Q=
C=
=
ω0
4.1 × 106
= 41
=
B
105
ω0 L
R
QR
41 × 6
=
= 60 μH
ω0
4.1 × 106
1
ω0 CR
1
ω0 QR
1
= 991.5 pF
6
4.1 × 10 × 41 × 6
| Network Analysis
474
As Q > 5,
B
105
= 4.1 × 106 ±
2
2
6
ω2 = 4.15 × 10 rad/ sec
ω2,1 = ω0 ±
That is,
ω1 = 4.05 × 106 rad/ sec
and
EXAMPLE
6.4
In a series resonant circuit, the current is maximum when C = 500 pF and frequency is 1 MHz. If
C is changed to 600 pF, the current decreases by 50%. Find the resistance, inductance and quality
factor.
SOLUTION
Case 1
Given,
C = 500 pF
I = Im
f = 1 × 106 Hz
ω0 = 2π × 106 rad/ sec
⇒
We know that
ω0 = √
1
LC
Therefore, Inductance,
1012
1
=
(2π × 106 )2 × 500
ω02 C
= 0.0507 mH
L=
Case 2
When C = 600 pF,
Im
E
=
⇒ |Z| = 2R
2
2R
√
R2 + X 2 = 2R ⇒ X = 3R
I=
X = XL − XC
= 2π × 106 × 0.0507 × 10−3 −
= 318.56 − 265.26
√
= 53.3 Ω = 3R
1012
2π × 106 × 600
Resonance
Therefore resistance,
j
475
p:3 = 30:77Ω
R = 53
3
Quality factor,
EXAMPLE
:56 = 10:35
Q = !R0 L = 318
30:77
6.5
In a series circuit with R = 50 Ω, L = 0.05 H and C = 20 F, frequency is varied till the
voltage across C is maximum. If the applied voltage is 100 V, find the maximum voltage across
the capacitor and the frequency at which it occurs. Repeat the problem for R = 10 Ω.
SOLUTION
Case 1
Given R = 50 Ω,
We know that
L = 0.05 H, C = 20 F
!0 = p 1 = p 10
= 103 rad/sec
0:05 20
LC
3 0:05
10
L
!
Q = R0 = 20 = 1
3
Using the given value of E = 100 V in the relationship
VCm = q QE 1
1
we get
4Q2
VCm = q100 1
= 115:5 V
r
!C = ! 1
r1
1
1
4
and the corresponding frequency at this voltage is
0
= 103
2
2Q2
= 707 rad= sec
Case 2
When R = 10 Ω,
0:05 = 5
10
5 100
= 502:5 V
VCm = q
1
r 1
3
Q = 10
1
4 25
!C = 103
1
50
= 990 rad= sec
476
j
Network Analysis
EXAMPLE
6.6
(i) A series resonant circuit is tuned to 1 MHz. The quality factor of the coil is 100. What is the
ratio of current at a frequency 20 kHz below resonance to the maximum current?
(ii) Find the frequency above resonance when the current is reduced to 90% of the maximum
current.
SOLUTION
(i) Let !a be the frequency 20 kHz below the resonance, Ia be the current and Za be the impedance
at this frequency.
Then
!a
!0
20 103 = 980 kHz
103
980
40:408 10 3 = 2
=
Now the ratio of current,
!a = 106
!0 = 980
!a 103
Ia = R =
1
Im Za 1 + j (2)Q
1
1 j 100(40:408
1
=
1 j 4:0408
= 0:2402 /76
=
10
(ii) Let !b be the frequency at which Ib = 0:9Im
Then
or
where
Then;
or
and
We know that
Hence
Ib I = 1 + j1(2)Q = 0:9
m
p1 + x
1
0:9
x = (2)100
1
1 + x2 =
= 1:2346
0:81
x2 = 0:2346
2
=
x = 0:4843
0:4843
200
0
0:4843
!b = 1 + 200 !0
= 1:00242 MHz
= !!b
1=
3)
Resonance
EXAMPLE
6.7
For the circuit shown in Fig. 6.21, obtain the values of ω0 and vC at ω0 .
Figure 6.21
SOLUTION
For the series circuit,
ω0 = √
1
LC
=
1
4×
1
4
×
10−6
= 103 rad/sec
At this ω0 , I = Im . Therefore,
V1 = 125Im
and the circuit equation is
1.5 = V1 + (Im − 0 · 105V1 )10 + jVL − jVC
Since VL = VC , the above equation can be modified as
1.5 = 125Im + 10Im − 1.05 × 125Im
1.5
A
3.75
1.5
4 × 106
Vc =
×
3.75
103
= 1600 V
Hence,
Im =
and
EXAMPLE
6.8
For the circuit shown in Fig. 6.22(a), obtain Zin and then find ω0 and Q.
Figure 6.22(a)
| 477
478
j
Network Analysis
SOLUTION
Taking I as the input current, we get
VR = 10I
and the controlled current source,
0:3VR = 0:3
= 3I
10I
The input impedance can be obtained using the standard formula
V
voltage
=
Zin (j!) = Applied
Input current
I
(6.6)
For futher analysis, the circuit is redrawn as shown in Fig. 6.22(b). It may be noted that the
controlled current source is transformed to its equivalent voltage source.
Figure 6.22(b)
Referring Fig. 6.22(b), the circuit equation may be obtained as
V
=
10 + j 10
3
9
! j3010!
30!
Substituting equation (6.7) in equation (6.6), we get
Zin = 10 + j
10
3
!
4
For resonance, Zin should be purely real. This gives
10
Rearranging,
3
10
9
10 Ω
9
30!
! = 4 3010
!
9
!2 = 304 10103
= 0:133 1012
9
j3
I
(6.7)
Resonance
Solving we get
j
479
p0:133 10
= 365 10 rad= sec
! = !0 =
Quality factor
12
3
Q = !R0 L
=
365
= 36:5
10 10
3
3
10
Parallel circuits
EXAMPLE
6.9
For the circuit shown in Fig. 6.23(a), find !0 , Q, BW and half power frequencies and the out put
voltage V at !0 .
Figure 6.23(a)
SOLUTION
Transforming the voltage source into current source, the circuit in Fig. 6.28(a) can be redrawn as
in Fig. 6.23(b).
Then;
!0 = p 1
LC
p400109 100
= 5 10 rad= sec
=
6
Q = !0 CR
10 100 10 100 10
!
B = = 5 10 = 10 rad= sec
=5
6
12
6
0
Q
50
5
3
= 50
Figure 6.23(b)
480
j
Network Analysis
As Q > 10,
!2;1 = B2 + !0
=5
10 102
6
5
!2 = 5:05 M rad= sec
Hence;
Output voltage,
and
V = I 80 kΩ
10 3 80 103
j 5 106 400 10
= 0:04 90 V
=
EXAMPLE
!1 = 4:95 M rad= sec
6
6.10
In a parallel RLC circuit, C = 50 F. Determine BW, Q, R and L for the following cases.
(i)
(ii)
!0 = 100; !2 = 120
!0 = 100; !1 = 80
SOLUTION
(i)
!0 = 100; !2 = 120
We know that
Rearraging we get
!0 = p !1 !2
2
!1 = !!0
2
1002
=
= 83:33 rad=sec
120
Band width
B = !2 !1
= 120
Quality factor,
Q = !B0
=
We know that
83:33 = 36:67 rad= sec
100
= 2:73
36:67
Q = !RL = !0 RC
0
(6.8)
Resonance
Rearraging equation (6.8),
j
481
R = !QC
0
2:73 106
= 546 Ω
=
100
1
L = !2C
Similarly
50
0
=
(ii)
106
=2H
1002 50
!0 = 100; !1 = 80: Solving the same way as in case (i), we get
2
= 125
!2 = 100
80
BW = B = 125 80 = 45 rad= sec
Q = 100
= 2:22
45
EXAMPLE
6.11
In the circuit shown in Fig. 6.24(a), vs (t) = 100 cos !t volts. Find resonance frequency, quality
factor and obtain i1 ; i2 ; i3 . What is the average power loss in 10 kΩ. What is the maximum stored
energy in the inductors?
10k
40k
Figure 6.24(a)
SOLUTION
The circuit in Fig. 6.24(a) is redrawn by replacing its voltage source by equivalent current source
as shown in Fig. 6.24(b).
Resonance frequency,
!0 = p 1
LC
=
p50 10 1 1:25 10
3
= 4000 rad= sec
Quality factor,
6
Figure 6.24(b)
Q = !0 C Req
= 4000
= 40
1:25 10 8 10
6
3
482
j
Network Analysis
At resonance, the current source will branch into resistors only. Hence,
v(t)
jj
= (10kΩ 40kΩ)
vs (t)
10000
= 80 cos 4000t volts
i1 (t) lags v(t) by 90 . Therefore,
i1 (t) = 50 10803 4000 sin 40000t
= 400 sin 4000t mA
i2 (t) = 40 801000 cos 4000t
= 2 cos 4000t mA
i3 (t) = i1 (t)
= 400 sin 4000t mA
Average power in 10 kΩ:
p
802
2
Pav = 10 103
= 0:32 W
Maximum stored energy in the inductance:
E = 12 LIm2
1
50 10
=
2
= 4 mJ
EXAMPLE
6.12
For the network shown in Fig. 6.25(a), obtain
frequency and quality factor.
3
(400 10
3 2
)
Yin and then use it to determine the resonance
Figure 6.25(a)
Figure 6.25(b)
SOLUTION
Considering V as the input voltage and I as the input current, it can be found that
10kΩ
IR =
V
) 10
4
IR = V
Resonance
j
483
The circuit in Fig. 6.25(a) is redrawn by replacing the controlled voltage source in to its
equivalent current source by taking s = j! and is shown in Fig. 6.25(b). Referring Fig. 6.25(b),
I
10V
1 1
= V sC + +
R1 sL
11
sL
)
I = V sC + R + sL
Input admittance, with s is being replaced by j! is
103
Yin = VI = 1014 + j!1 10 8 j 11
! 4:4
j
2500
= 10 4 + j! 10 8
!
At resonance, Yin should be purely real. This enforces that
10
Therefore;
! = 2500
!
p
!0 = 108 2500
= 500 K rad= sec
8
Quality factor:
Q = !0 RC
= 500
10 10 10
3
4
8
= 50
EXAMPLE
6.13
In a parallel RLC circuit, cut off frequencies are 103 and 118 rad/sec.
10 Ω. Find R, L and C .
jZ j at ! = 105 rad/sec is
SOLUTION
Given
!1 = 103 rad= sec
!2 = 118 rad= sec
Therefore
Resonant frequency,
B = 118
103 = 15 rad= sec
!0 = p
p !1 !2
=
118
103 = 110:245 rad= sec
484
| Network Analysis
Quality factor
ω0
B
110.245
= 7.35
=
15
Q=
Admittance,
Since
we get
Note that,
Therefore,
1
1
Y = + j ωC −
R
ωL
R
1
1 + j ωCR −
=
R
ωL
ω0 ωCR
Rω0
1
1+j
−
=
R
ω0
ωω0 L
R
Q = ω0 RC =
,
ω0 L
ω
ω0
1
1 + jQ
−
Y =
R
ω0
ω
ω
ω0
105
110.245
−
=
−
= −0.0975
ω0
ω
110.245
105
1
(1 + j7.35(−0.0975))
R
1
= (1 − j0.7168)
R
1.23
1
1 + (0.7168)2 =
⇒ |Y | =
R
R
Y =
It is given that |Z| = 10 and therefore |Y | =
get
1
1
= 1.23
10
R
1
. Putting this value of Y in equation (6.9), we
10
⇒
R = 12.3Ω
From the relationship Q = ω0 CR, we get
Therefore,
(6.12)
ω0 CR = 7.35
7.35
1
C=
×
12.3 110.245
= 5.42 μF
Resonance
Inductance,
j
485
L = !21C
0
110:
= 15:18 mH
=
EXAMPLE
2452
1
5:42
10
3
6.14
For the circuit shown in Fig. 6.26(a), find !0 , V1 at !0 , and V1 at a frequency 15 k rad/sec above
!0 .
Figure 6.26(a)
SOLUTION
Changing voltage source of Fig. 6.26(a) into its equivalent current source, the circuit is redrawn
as shown in Fig. 6.26(b).
Referring Fig. 6.26(b),
k
!0 = 1
pLC
1
=p
100 10 10 10
= 10 rad=sec
6
9
Figure 6.26(b)
6
Voltage across the inductor at !0 is,
V1 = j 106
3 10 5 10
9
3
= j 15 V
Quality factor,
Q = !0 CR
= 106
10 10 5 10
9
= 50
Given
!a = !0 + 15 k rad/sec
10 + 10
= 1:015 10 rad=sec
= 15
Now;
!a !0 = 1:015
!0 !a
3
6
6
1
= 0:03
1:015
3
486
| Network Analysis
Using this relation in the equation,
ωa ω0
1
1 + jQ
Y =
−
R
ω0 ωa
1
Y =
(1 + j50 × 0.03)
5000
= 3.6 × 10−4 56.31◦
we get
The corresponding value of V1 is
V1 = I Y −1
= jωa × 3 × 10−9 × Y −1
=
j 1.015 × 106 × 3 × 10−9
3.6 × 10−4 56.31◦
= 8.444 33.69◦ V
EXAMPLE
6.15
A parallel RLC circuit has a quality factor of 100 at unity power factor and operates at 1 kHz and
dissipates 1 Watt when driven by 1 A at 1 kHz. Find Bandwidth and the numerical values of R, L
and C.
SOLUTION
Given f = 1 kHz, P = 1 W, I = 1 A, Q = 100, cos φ = 1
B=
103 × 2π
ω0
=
= 20π rad/sec
Q
100
P = I 2R
Therefore
R=1Ω
R
L=
ω0 Q
1
20π × 100
= 159 μH
1
C= 2
ω0 L
=
=
10
(20π)2 159
= 16.9 μF
Resonance
EXAMPLE
6.16
For the circuit shown in Fig. 6.27, determine resonance frequency and the input impedance.
Figure 6.27
SOLUTION
Equation for resonance frequency is
#
!
"
$
2 − L
$ 1
RL
C
%
ωL =
2 − L
LC RC
C
=
1
0.1 × 10−3
= 98.47 rad/sec
We know that
X L = ω0 L
= 98.47 × 0.1
and
= 9.847 Ω
1
XC =
ω0 C
1
=
98.47 × 10−3
= 10.16 Ω
22 − 100
1 − 100
| 487
488
| Network Analysis
Admittance Y at resonance is purely real and is given by
Y = G1 + G2 + G3
1
2
1
=
2
2 + 5 +
3
2 + (0.1ω0 )
1 + 10
ω0
2
1
1
+ +
2
+ 9.847
5 1 + 10.162
= 0.23 S
Y =
22
and the input impedance,
Z=
EXAMPLE
1
= 4.35 Ω
Y
6.17
The impedance of a parallel RLC circuit as a function of ω is depicted in the diagram shown in
Fig. 6.28. Determine R, L and C of the circuit. What are the new values of ω0 and bandwidth if
C is increased by 4 times?
0.4
Figure 6.28
SOLUTION
It can be seen from the figure that
ω0 = 10 rad/sec
B = 0.4 rad/sec
R = 10 Ω
Then Quality factor
Q=
10
ω0
=
= 25
BW
0.4
L=
10
R
=
= 0.04 H
ω0 Q
10 × 25
We know that
Resonance
j
489
As Q = !0 CR,
C = 10 25
10 = 0:25 F
If C is increased by 4 times, the new value of C is 1 Farad. Therefore,
!0 = p 1
LC
=
p01:04 = 5
and the corresponding bandwith
1
B = RC
EXAMPLE
= 0:1
6.18
In a two branch RL RC parallel resonant circuit, L = 0:4 H and C = 40 F. Obtain resonant
frequency for the following values of RL and RC .
(i)
(ii)
(iii)
(iv)
(v)
RL = 120; RC = 80
RL = RC = 80
RL = 80; RC = 0
RL = RC = 100
RL = RC = 120
SOLUTION
As RL and RC are given separately, we can use the following formula to calculate the resonant
frequency.
!0
v
u
1 u
t
=p
LC
RL2
RC2
L
C
L
C
Let us compute the following values
LC = 0:4 40 10
= 16
10
6
6
1
pLC
= 250
(i)
RL = 120; RC = 80
Using equation (6.10),
L = 104
C
r 120
!0 = 250
2
502
104
104
!
(6.10)
490
| Network Analysis
As the result is an imaginary number resonance is not possible in this case.
(ii) RL = RC = 80
802 − 104
802 − 104
= 250 rad/sec
ω0 = 250
(iii) RL = 80; RC = 0
802 − 104
−104
= 150 rad/sec
ω0 = 250
(iv) RL = RC = 100
1002 − 104
1002 − 104
As the result is indeterminate, the circuit resonates at all frequencies.
ω0 = 250
(v) RL = RC = 120
1202 − 104
1202 − 104
= 250 rad/sec
ω0 = 250
EXAMPLE
6.19
The following information is given in connection with a two branch parallel circuit:
RL = 10 Ω, RC = 20 Ω, XC = 40 Ω, E = 120 V and frequency = 60 Hz. What are the
values of L for resonance and what currents are drawn from the supply under this condition?
SOLUTION
As the frequency is constant, the condition for resonance is
XC
XL
= 2
2
+ XL
RC + XC2
XL
40
1
= 2
=
2
2
2
20 + 40
50
10 + XL
2
RL
⇒
XL2 − 50XL + 100 = 0
⇒
Solving we get
XL = 47.913 Ω or 2.087 Ω
Then the corresponding values of inductances are
L=
XL
= 0.127 H
ω
or 5.536 mH
Resonance
| 491
The supply current is
I = EG = E (GL + GC )
10
+ 0.02 = 1.7 A for XL = 47.913 Ω
I = 120
102 + 47.9132
10
+ 0.02 = 12.7 A for XL = 2.087 Ω
I = 120
102 + 2.0872
Thus,
or
Exercise Problems
E.P
6.1
Refer the circuit shown in Fig. E.P. 6.1, where Ri is the source resistance
(a) Determine the transfer function of the circuit.
(b) Sketch the magnitude plot with Ri = 0 and Ri = 0.
Figure E.P. 6.1
Ans:
E.P
R
Vo (s)
L s
=
H(s) =
R
+R
Vi (s)
i
s2 +
s+
L
1
LC
6.2
For the circuit shown in Fig. E.P. 6.2, calculate the following:
(b) Q,
(c) fc1 ,
(d) fc2 and (e) B
(a) f0 ,
Figure E.P. 6.2
Ans:
(a) 254.65 kHz
(b) 8
(c) 239.23 kHz
(d) 271.06 kHz
(e) 31.83 kHz
492
j
E.P
Network Analysis
6.3
Refer the circuit shown in Fig. E.P. 6.3, find the output voltage, when (a) ! = !0 (b) ! = !1 , and
(c) ! = !c2 .
Figure E.P. 6.3
Ans: (a) 640 cos(1.6 106 t)mV
(b) 452.55 cos(1.5 106 t + 45 )mV
(c) 452.55 cos(1.7 106 t 45 )mV
E.P
6.4
Refer the circuit shown in Fig. E.P. 6.4. Calculate Zi (s) and then find (a) !0 and (b) Q.
Figure E.P. 6.4
Ans: (a) 364.69 krad/sec, (b) 36
E.P
6.5
Refer the circuit shown in Fig. E.P. 6.5. Show that at resonance,
Figure E.P. 6.5
jVoj
max
=
GQjV j
s
1
1
4Q2
.
Resonance
E.P
6.6
Refer the circuit given in Fig. E.P. 6.6, calculate !0 ; Q and
jVoj
j
493
max
Figure E.P. 6.6
Ans:
3513.64 rad/sec, 26.35, 316 volts.
E.P
6.7
A parallel network, which is driven by a variable frequency of 4 A current source has the following
values: R = 1 kΩ , L = 10 mH, C = 100 F. Find the band width of the network, the half
power frequenies and the voltage across the network at half-power frequencies.
Ans: 10 rad/sec, 995 rad/sec, 10005 rad/sec
E.P
6.8
For the circuit shown in Fig. E.P. 6.8, determine the expression for the magnitude response,
1
.
versus ! and Zin at !0 = pLC
Ans:
(a) jZin j =
E.P
6.9
qR
(
jZinj
Figure E.P. 6.8
!2 RLC )2 +(!L)2 ,
1+(!RC )2
(b) jZin j =
G
1
C
L
(1+R2 C
L)
A coil under test may be represented by the model of L in series with R. The coil is connected in
series with a variable capacitor. A voltage source v (t) = 10 cos 1000 t volts is connected to the
coil. The capacitor is varied and it is found that the current is maximum when C = 10F. Also,
when C = 12:5F, the current is 0.707 of the maximum value. Find Q of the coil at ! = 1000
rad/sec.
Ans:
5
494
j
E.P
Network Analysis
6.10
A fresher in the devices lab for sake of curiosity sets up a series RLC network as shown in Fig.
E.P.6.10. The capacitor can withstand very high voltages. Is it safe to touch the capacitor at
resonance? Find the voltage across the capacitor.
Figure E.P. 6.10
Ans: Not safe, jVc jmax = 1600 V
Chapter
7
7.1
Two Port Networks
Introduction
A pair of terminals through which a current may enter or leave a network is known as a port.
A port is an access to the network and consists of a pair of terminals; the current entering one
terminal leaves through the other terminal so that the net current entering the port equals zero.
There are several reasons why we should study two-ports and the parameters that describe them.
For example, most circuits have two ports. We may apply an input signal in one port and obtain
an output signal from the other port. The parameters of a two-port network completely describes
its behaviour in terms of the voltage and current at each port. Thus, knowing the parameters of a
two port network permits us to describe its operation when it is connected into a larger network.
Two-port networks are also important in modeling electronic devices and system components.
For example, in electronics, two-port networks are employed to model transistors and Op-amps.
Other examples of electrical components modeled by two-ports are transformers and transmission
lines.
Four popular types of two-ports parameters are examined here: impedance, admittance, hybrid, and transmission. We show
the usefulness of each set of parameters,
demonstrate how they are related to each
other.
Fig. 7.1 represents a two-port network.
A four terminal network is called a two-port
network when the current entering one terFigure 7.1 A two-port network
minal of a pair exits the other terminal in
the pair. For example, I1 enters terminal a and exits terminal b of the input terminal pair a-b.
We assume that there are no independent sources or nonzero initial conditions within the linear
two-port network.
496
7.2
j
Network Theory
Admittance parameters
The network shown in Fig. 7.2 is assumed to be linear
and contains no independent sources. Hence, principle of superposition can be applied to determine the
current I1 , which can be written as the sum of two
components, one due to V1 and the other due to V2 .
Using this principle, we can write
I1 = y11 V1 + y12 V2
Figure 7.2 A linear two-port network
where y11 and y12 are the constants of proportionality
with units of Siemens.
In a similar way, we can write
I2 = y21 V1 + y22 V2
Hence, the two equations that describe the two-port network are
I1 = y11 V1 + y12 V2
(7.1)
I2 = y21 V1 + y22 V2
(7.2)
Putting the above equations in matrix form, we get
I1
I2
=
y11 y12
y21 y22
Here the constants of proportionality y11 ; y12 ; y21
and y22 are called y parameters for a network. If
these parameters y11 , y12 , y21 and y22 are known,
then the input/output operation of the two-port is
completely defined.
From equations (7.1) and (7.2), we can determine
y parameters. We obtain y11 and y21 by connecting
a current source I1 to port 1 and short-circuiting port
2 as shown in Fig. 7.3, finding V1 and I2 , and then
calculating,
y11
I1 =
V1 V2 =0
V1
V2
Figure 7.3 Determination of y11 and y12
y21
I2 =
V1 V2 =0
Since y11 is the admittance at the input measured in siemens with the output short-circuited,
it is called short-circuit input admittance. Similarly, y21 is called the short-circuit transfer admittance.
Two Port Networks
j
497
Similarly, we obtain y12 and y22 by connecting a current source I2 to port 2 and shortcircuiting port 1 as in Fig. 7.4, finding I1 and V2 , and then calculating,
y12
I1 =
V2 V1 =0
y12 is called the short-circuit transfer admittance and y22 is called the shortcircuit output admittance. Collectively the
y parameters are referred to as short-circuit
admittance parameters.
Please note that y12 = y21 only when
there are no dependent sources or Op-amps
within the two-port network.
EXAMPLE
y22
I2 =
V2 V1 =0
Figure 7.4 Determination of y12 and y22
7.1
Determine the admittance parameters of the T network shown in Fig. 7.5.
Figure 7.5
SOLUTION
To find y11 and y21 , we have to short the output terminals and connect a current source I1 to the
input terminals. The circuit so obtained is shown in Fig. 7.6(a).
V1
V1
=
22
5
4+
2
+
2
I1 1
y11 =
= S
V1 V2 =0 5
I1 =
Hence;
Using the principle of current division,
I1 2
I1
I2 =
=
2 + 2 2
1 V1
)
I2 =
2 5
I2 1
S
Hence; y21 =
=
V1 V2 =0
10
Figure 7.6(a)
498
j
Network Theory
To find y12 and y22 , we have to short-circuit the input terminals and connect a current source
I2 to the output terminals. The circuit so obtained is shown in Fig. 7.6(b).
V2
I2 =
42
2+
4+2
V2
=
4
2+
3
3V2
=
10
I2 3
Hence; y22 =
= S
V2 V1 =0 10
Employing the principle of current division, we have
Figure 7.6(b)
I2 2
2+4
2I2
I1 =
6
1 3V2
I1 =
3 10
I1 1
y12 =
S
=
V2 V1 =0
10
I1 =
)
)
Hence;
It may be noted that, y12 = y21 . Thus, in matrix form we have
I = YV
3 2 1
I1
5
4
5=6
4
1
I2
10
2
)
EXAMPLE
3
1 32
V1
10 7 4
5
5
3
V2
10
7.2
Find the y parameters of the two-port network shown in Fig. 7.7. Then determine the current in a
4Ω load, that is connected to the output port when a 2A source is applied at the input port.
Figure 7.7
Two Port Networks
j
499
SOLUTION
To find y11 and y21 , short-circuit the output terminals and connect a current source I1 to the input
terminals. The resulting circuit diagram is shown in Fig. 7.8(a).
V1
V1
=
12
1Ωjj2Ω
1+2
3
I1 = V1
2 I1 3
y11 =
= S
V1 V2 =0 2
I1 =
)
Hence;
Using the principle of current division,
I2 =
)
I2 =
)
I2 =
Hence;
y21 =
I1 1
1+2
1
I1
3
1 3
V1
3 2
I2
1
=
S
V1
2
Figure 7.8(a)
To find y12 and y22 , short the input terminals and connect a current source I2 to the output
terminals. The resulting circuit diagram is shown in Fig. 7.8(b).
I2 =
=
V2
2Ωjj3Ω
V2
5V2
=
23
6
2+3
I2 5
y22 =
= S
V2 V1 =0 6
Figure 7.8(b)
Employing the current division principle,
I1 =
)
I2 3
2+3
3
I1 = I2
5
500
j
Network Theory
)
3 5V2
I1 =
5
6
1
I1 =
V2
2 I1 1
y12 =
S
=
V2 V1 =0
2
)
Hence;
Therefore, the equations that describe the two-port network are
3
1
I1 = V1
V2
2
2
1
5
I2 =
V1 + V2
2
6
(7.3)
(7.4)
Putting the above equations (7.3) and
(7.4) in matrix form, we get
2 3
6 2
4
1
2
3 2
3
1 32
I1
V1
2 74
5=4
5
5
5
V2
I2
6
Referring to Fig. 7.8(c), we find that
I1 = 2A and V2 = 4I2
Substituting I1 = 2A in equation (7.3),
we get
3
2 = V1
4
Multiplying equation (7.4) by
Figure 7.8(c)
1
V2
2
(7.5)
4, we get
20
V2
6
20
V2 = 2V1
V2
6
20
0 = 2V1
+ 1 V2
6
1
13
0=
V1 + V2
2
12
4I2 = 2V1
)
)
)
Putting equations (7.5) and (7.6) in matrix form, we get
2 3
6 2
4
1
2
3 2 3
1 32
2
V1
2 74
5=4 5
5
13
V2
0
12
(7.6)
Two Port Networks
j
501
It may be noted that the above equations are simply the nodal equations for the circuit shown
in Fig. 7.8(c). Solving these equations, we get
3
V2 = V
2
1
3
I2 =
V2 =
A
4
8
and hence,
EXAMPLE
7.3
Refer the network shown in the Fig. 7.9 containing a current-controlled current source. For this
network, find the y parameters.
Figure 7.9
SOLUTION
To find y11 and y21 short the output terminals and connect a current source I1 to the input
terminals. The resulting circuit diagram is as shown in Fig. 7.10(a) and it is further reduced
to Fig. 7.10(b).
Figure 7.10(a)
V1
22
2+2
I1 = V1
I1 y11 =
= 1S
V1 V2 =0
I1 =
)
Hence;
Figure 7.10(b)
502
j
Network Theory
Applying KCL at node A gives (Referring to Fig. 7.10(a)).
)
)
)
Hence;
I3 + I2 = 3I1
V1
+ I2 = 3I1
2
V1
+ I2 = 3V1
2
5V1
= I2
2
I2
5
y21 =
= S
V1
2
To find y22 and y12 , short the input terminals and connect a current source I2 at the output
terminals. The resulting circuit diagram is shown in Fig. 7.10(c) and further reduced to Fig.
7.10(d).
Figure 7.10(c)
I2 =
)
Hence;
I01 =
V2
2
V2
2
I1
1
=
S
y12 =
V2
2
I1 =
Applying KCL at node B gives
V2 V2
+
+ 3I1
2
2
V2
I1 =
2
V2 V2
V2
I2 =
+
3
2 2
2
I2 y22 =
= 0:5S
V2 V1 =0
I2 =
But
Hence;
)
Figure 7.10(d)
Two Port Networks
j
503
Short-cut method:
Referring to Fig. 7.9, we have KCL at node V1 :
V1 V1 V2
+
2
2
= V1 0:5V2
I1 =
Comparing with
I1 = y11 V1 + y12 V2
we get
y11 = 1S and y12 =
0:5S
KCL at node V2 :
V2 V2 V1
+
2
2
V2 V2 V1
= 3 [V1 0:5V2 ] +
+
2
2
5
I2 = V1 0:5V2
2
I2 = 3I1 +
)
Comparing with I2 = y21 V1 + y22 V2
we get
y21 = 2:5S and y22 =
EXAMPLE
0:5S
7.4
Find the y parameters for the two-port network shown in Fig. 7.11.
Figure 7.11
SOLUTION
To find y11 and y21 short-circuit the output terminals as shown in Fig. 7.12(a). Also connect a
current source I1 to the input terminals.
504
j
Network Theory
Figure 7.12(a)
KCL at node V1 :
V1 V1 Va
+
1
1
2
2Va = I1
I1 =
)
3V1
(7.7)
KCL at node Va :
V1 Va 0
+ 2V1 = 0
+
1
1
2
2Va 2V1 + Va + 2V1 = 0
Va
)
)
Va = 0
(7.8)
Making use of equation (7.8) in (7.7), we get
3V1 = I1
I1 y11 =
= 3S
V1 V2 =0
Hence;
Since Va = 0, I2 = 0,
)
y21
I2 =
= 0S
V1 V2 =0
To find y22 and y12 short-circuit the input terminals and connect a current source I2 to the
output terminals. The resulting circuit diagram is shown in Fig. 7.12(b).
Two Port Networks
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505
Figure 7.12(b)
KCL at node V2 :
V2 V2 Va
= I2
+
1
1
2
3V2 Va = I2
)
(7.9)
KCL at node Va :
Va
V2
1
)
or
+
Va 0
+0=0
1
2
3Va V2 = 0
1
Va = V2
3
(7.10)
Substituting equation (7.10) in (7.9), we get
3V2
)
Hence;
We have;
Also;
)
1
V2 = I2
3
8
V2 = I2
3
8
I2
= S
y22 =
V2
3
1
Va = V2
3
I1 + I3 = 0
I1 =
=
I3
Va
=
1
2
(7.11)
2Va
(7.12)
506
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Network Theory
Making use of equation (7.12) in (7.11), we get
I1
1
= V2
2
3 I1 2
y12 =
S
=
V2 V1 =0
3
Hence;
EXAMPLE
7.5
Find the y parameters for the resistive network shown in Fig. 7.13.
Figure 7.13
SOLUTION
Converting the voltage source into an equivalent current source, we get the circuit diagram shown
in Fig. 7.14(a).
To find y11 and y21 , the output terminals of Fig. 7.14(a) are shorted and connect a current
source I1 to the input terminals. This results in a circuit diagram as shown in Fig. 7.14(b).
Figure 7.14(a)
KCL at node V1 :
Figure 7.14(b)
V1 V1 V2
+
= I1 + 3V1
2
1
Two Port Networks
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507
Since V2 = 0, we get
)
Hence;
KCL at node V2 :
Since V2 = 0, we get
)
Hence
V1
+ V1 = I1 + 3V1
2
3
I1 =
V1
2 I1 3
y11 =
S
=
V1 V2 =0
2
V2 V1
V2
+ 3V1 +
= I2
2
1
0 + 3V1
V1 = I2
I2 = 2V1
I1
= 2S
y21 =
V2
To find y21 and y22 , the input terminals of Fig. 7.14(a) are shorted and connect a current
source I2 to the output terminals. This results in a circuit diagram as shown in Fig. 7.14(c).
Figure 7.14(c)
Figure 7.14(d)
508
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Network Theory
KCL at node V2 :
V2 V2 0
+
= I2
2
1
3
V2 = I2
2
3
I2
= S
y22 =
V2
2
)
Hence;
KCL at node V1 :
I1 =
V1 V1 V2
+
=0
2
1
Since V1 = 0, we get
I1 =
V2
I1
=
y12 =
V2
Hence;
EXAMPLE
1S
7.6
The network of Fig. 7.15 contains both a dependent current source and a dependent voltage
source. Find the y parameters.
Figure 7.15
SOLUTION
While finding y parameters, we make use of KCL equations. Hence, it is preferable to have current
sources rather than voltage sources. This prompts us to convert the dependent voltage source into
an equivalent current source and results in a circuit diagram as shown in Fig. 7.16(a).
To find y11 and y21 , refer the circuit diagram as shown in Fig. 7.16(b).
KCL at node V1 :
V1 V1 V2
+
+ 2V1 = 2V2 + I1
1
1
Two Port Networks
Figure 7.16(a)
Figure 7.16(b)
Since V2 = 0, we get
)
Hence;
KCL at node V2 :
V1 + V1 + 2V1 = I1
4V1 = I1
I1
y11 =
= 4S
V1
V2 V2 V1
+
= 2V1 + I2
1
1
Since V2 = 0, we get
)
Hence;
V1 = 2V1 + I2
3V1 = I2
I2 y21 =
=
V1 V2 =0
3S
j
509
510
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Network Theory
To find y22 and y12 , refer the circuit diagram shown in Fig. 7.16(c).
KCL at node V1 :
V1 V1 V2
+
+ 2V1 = 2V2 + I1
1
1
Since V1 = 0, we get
V2 = 2V2 + I1
)
3V2 = I1
I1 y12 =
=
V2 V1 =0
Hence;
3S
Figure 7.16(c)
KCL at node V2 :
Since V1 = 0, we get
)
V2 V2 V1
+
= 2V1 + I2
1
1
V2 + V2 = 0 + I2
2V2 = I2
I2 y22 =
= 2S
V2 V1 =0
Hence;
7.3
Impedance parameters
Let us assume the two port network shown in Fig. 7.17 is a
linear network that contains no independent sources. Then
using superposition theorem, we can write the input and output voltages as the sum of two components, one due to I1
and other due to I2 :
V1 = z11 I1 + z12 I2
V2 = z21 I1 + z22 I2
Figure 7.17
Two Port Networks
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511
Putting the above equations in matrix from, we get
V1
V2
=
z11 z12
z21 z22
I1
I2
The z parameters are defined as follows:
z11
V1 =
I1 I2 =0
z12
V1 =
I2 I1 =0
z21
V2 =
I1 I2 =0
z22
V2 =
I2 I1 =0
In the preceeding equations, letting I1 or I2 = 0 is equivalent to open-circuiting the input or
output port. Hence, the z parameters are called open-circuit impedance parameters. z11 is defined
as the open-circuit input impedance, z22 is called the open-circuit output impedance, and z12 and
z21 are called the open-circuit transfer impedances.
If z12 = z21 , the network is said to be reciprocal network. Also, if all the z-parameter are
identical, then it is called a symmetrical network.
EXAMPLE
7.7
Refer the circuit shown in Fig. 7.18. Find the z parameters of this circuit. Then compute the
current in a 4Ω load if a 24 0 V source is connected at the input port.
Figure 7.18
SOLUTION
To find z11 and z21 , the output terminals are open circuited. Also connect a voltage source V1 to
the input terminals. This gives a circuit diagram as shown in Fig. 7.19(a).
Figure 7.19(a)
512
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Network Theory
Applying KVL to the left-mesh, we get
)
12I1 + 6I1 = V1
V1 = 18I1
V1 z11 =
= 18Ω
I1 I2 =0
Hence;
Applying KVL to the right-mesh, we get
)
Hence;
V2 + 3 0 + 6I1 = 0
z21
V2 = 6I1
V2
= 6Ω
=
I1
To find z22 and z12 , the input terminals are open circuited. Also connect a voltage source V2
to the output terminals. This results in a network as shown in Fig. 7.19(b).
Figure 7.19(b)
Applying KVL to the left-mesh, we get
V1 = 12 0 + 6I2
)
V1 = 6I2
V1 z12 =
= 6Ω
I2 I1 =0
Hence;
Applying KVL to the right-mesh, we get
)
Hence;
V2 + 3I2 + 6I2 = 0
V2 = 9I2
V2 z22 =
= 9Ω
I2 I1 =0
The equations for the two-port network are, therefore
V1 = 18I1 + 6I2
(7.13)
V2 = 6I1 + 9I2
(7.14)
Two Port Networks
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513
The terminal voltages for the network shown in Fig. 7.19(c) are
V1 = 24 0
(7.15)
V2 =
(7.16)
4I2
Figure 7.19(c)
Combining equations (7.15) and (7.16) with equations (7.13) and (7.14) yields
24 0 = 18I1 + 6I2
0 = 6I1 + 13I2
Solving, we get
EXAMPLE
I2 =
0:73 0 A
7.8
Determine the z parameters for the two port network shown in Fig. 7.20.
Figure 7.20
SOLUTION
To find z11 and z21 , the output terminals are open-circuited and a voltage source is connected to
the input terminals. The resulting circuit is shown in Fig. 7.21(a).
Figure 7.21(a)
514
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Network Theory
By inspection, we find that I3 = V1
Applying KVL to mesh 1, we get
R1 (I1 I3 ) = V1
R1 I1 R1 I3 = V1
R1 I1 R1 V1 = V1
(1 + R1 ) V1 = R1 I1
V1 R1
z11 =
=
I1 I2 =0 1 + R1
)
)
)
Hence;
Applying KVL to the path V1
! R2 ! R3 ! V2, we get
V1 + R2 I3
R3 I2 + V2 = 0
Since I2 = 0 and I3 = V1 , we get
)
V1 + R2 V1
0 + V2 = 0
R2 )
R 1 I1
= (1 R2 )
1 + R1
V2 R1 (1 R2 )
z21 =
=
I1 I2 =0
1 + R1
V2 = V1 (1
Hence;
The circuit used for finding z12 and z22 is shown in Fig. 7.21(b).
Figure 7.21(b)
By inspection, we find that
I2
)
)
I3 = V1 and V1 = I3 R1
I2
I3 = (I3 R1 )
I3 (1 + R1 ) = I2
Two Port Networks
Applying KVL to the path R3
)
Hence;
515
! R2 ! R1 ! V2, we get
R3 I2 + (R2 + R1 ) I3
)
j
V2 = 0
I2
R3 I2 + (R2 + R1 )
= V2
1 + R1
R2 + R1
= V2
I2 R 3 +
1 + R1
V2 z22 =
I2 I1 =0
R2 + R1
Ω
= R3 +
1 + R1
Applying KCL at node a, we get
V1 + I3 = I2
V1
V1 +
= I2
R1
1
= I2
V1 +
R1
V1 1
z12 =
=
1
I2 I1 =0
+
R1
R1
=
1 + R1
)
)
)
EXAMPLE
7.9
Construct a circuit that realizes the following z parameters.
z=
SOLUTION
Comparing z with =
z11 z12
z21 z22
12 4
4 8
, we get
z11 = 12Ω;
z12 = z21 = 4Ω;
z22 = 8Ω
Let us consider a T network as shown in Fig. 7.22(a). Our objective is to fit in the values of
R1 ; R2 and R3 for the given z.
Applying KVL to the input loop, we get
V1 = R1 I1 + R3 (I1 + I2 )
= (R1 + R3 ) I1 + R3 I2
516
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Network Theory
Comparing the preceeding equation with
V1 = z11 I1 + z12 I2
we get
z11 = R1 + R3 = 12Ω
)
z12 = R3 = 4Ω
R1 = 12
Figure 7.22(a)
R3 = 8Ω
Applying KVL to the output loop, we get
V2 = R2 I2 + R3 (I1 + I2 )
)
V2 = R3 I1 + (R2 + R3 ) I2
Comparing the immediate preceeding
equation with
V2 = z21 I1 + z22 I2
we get
z21 = R3 = 4Ω
z22 = R2 + R3 = 8Ω
)
R2 = 8
Figure 7.22(b)
R3 = 4Ω
Hence, the network to meet the given z parameter set is shown in Fig. 7.22(b).
EXAMPLE
If z =
7.10
40 10
20 30
Ω for the two-port network, calculate the average power delivered to 50Ω
resistor.
Figure 7.23
Two Port Networks
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517
SOLUTION
We are given z11 = 40Ω z12 = 10Ω z21 = 20Ω z22 = 30Ω
Since z12 6= z21 , this is not a reciprocal network. Hence, it cannot be represented only by
passive elements. We shall draw a network to satisfy the following two KVL equations:
V1 = 40I1 + 10I2
V2 = 20I1 + 30I2
One possible way of representing a network that is non-reciprocal is as shown in Fig. 7.24(a).
Figure 7.24(a)
Now connecting the source and the load to the two-port network, we get the network as shown
in Fig. 7.24(b).
Figure 7.24(b)
KVL for mesh 1:
)
60I1 + 10I2 = 100
6I1 + I2 = 10
KVL for mesh 2:
)
)
80I2 + 20I1 = 0
4I2 + I1 = 0
I1 =
4I2
518
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Network Theory
Solving the above mesh equations, we get
)
)
24I2 + I2 = 10
23I2 = 10
10
I2 =
23
= j I2 j 2 R L
Power supplied to the load
100
50
(23)2
= 9:45 W
=
EXAMPLE
7.11
Refer the network shown in Fig. 7.25. Find the z parameters for the network. Take =
4
3
Figure 7.25
SOLUTION
To find z11 and z21 , open-circuit the output terminals as shown in Fig. 7.26(a). Also connect a
voltage source V1 to the input terminals.
Figure 7.26(a)
Applying KVL around the path V1
! 4Ω ! 2Ω ! 3Ω, we get
4I1 + 5I3 = V1
Also;
V2
V2 = 3I3 ; so I3 =
3
(7.17)
(7.18)
Two Port Networks
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519
KCL at node b leads to
I1
V2
I3 = 0
(7.19)
Substituting equation (7.18) in (7.19), we get
Hence;
V2
1
V2
= +
I 1 = V2 +
3
3
4 1
=
V2
+
3 3
V2 3
z21 =
= Ω
I1 I2 =0 5
Substituting equation (7.18) in (7.17), we get
V2
3
5
3
= 4I1 +
I1 3
5
V1
= 5Ω
z11 =
I1
V1 = 4I1 + 5
Therefore;
Since
V2
3
=
I1
5
To obtain z22 and z12 , open-circuit the input terminals as shown in Fig. 7.26(b). Also, connect
a voltage source V2 to the output terminals.
Figure 7.26(b)
KVL for the mesh on the left:
V1 + 5I4
3I2 = 0
(7.20)
V2 + 3I4
3I2 = 0
(7.21)
KVL for the mesh on the right:
Also;
I 4 = V2
(7.22)
520
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Network Theory
Substituting equation (7.22) in (7.21), we get
)
Hence;
V2 + 3V2
3I2 = 0
V2 (1 + 3) = 3I2
V2 3
z22 =
=
I2 I1 =0 1 + 3
3
3
= Ω
=
4
5
1+3
3
Substituting equation (7.22) in (7.20), we get
V1 + 5V2 = 3I2
3
Substituting V2 = I2 , we get
5
3
V1 + 5
5
I2
= 3I2
V1 Hence;
z12 =
I2 I1 =0
= 3 3
4
=3 3 =
3
Finally, in the matrix form, we can write
2
Please note that z12
EXAMPLE
z11 z12
3
2
1Ω
3
5
1
5=4 5 3 5
z=4
z21 z22
3 5
6 z21, since a dependent source is present in the circuit.
=
7.12
Find the Thevenin equivalent circuit with respect to port 2 of the circuit in Fig. 7.27 in terms of z
parameters.
Figure 7.27
Two Port Networks
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521
SOLUTION
The two port network is defined by
V1 = z11 I1 + z12 I2 ;
V2 = z21 I1 + z22 I2 ;
here, V1 = Vg Zg I1
and V2 = IL ZL = I2 ZL
To find Thevenin equivalent circuit as seen
from the output terminals, we have to remove
the load resistance RL . The resulting circuit
diagram is shown in Fig. 7.28(a).
Vt = V2 jI2 =0
= z21 I1
Figure 7.28(a)
(7.23)
With I2 = 0, we get
V1 = z11 I1
Vg I1 Zg
V1
I1 =
=
z11
z11
)
Solving for I1 , we get
I1 =
Vg
z11 + Zg
(7.24)
Substituting equation (7.24) into equation (7.23), we get
Vt =
z21 Vg
z11 + Zg
To find Zt , let us deactivate all the independent sources and then connect a voltage source
V2 across the output terminals as shown in Fig.
7.28(b).
V2 Zt =
; where V2 = z21 I1 +z22 I2
I2 Vg =0
We know that
V1 = z11 I1 + z12 I2
Substituting, V1 = I1 Zg in the preceeding equation, we get
Solving;
I1 Zg = z11 I1 + z12 I2
z12 I2
I1 =
z11 + Zg
Figure 7.28(b)
522
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Network Theory
We know that,
V2 = z21 I1 + Z22 I2
z12 I2
+ z22 I2
= z21
z11 + Zg
V2
z21 z12
Zt =
= z22
I2
z11 + Zg
Thus;
The Thevenin equivalent circuit with respect to the output terminals along with load impedance ZL is as shown in Fig. 7.28(c).
EXAMPLE
Figure 7.28(c)
7.13
(a) Find the z parameters for the two-port network shown in Fig. 7.29.
(b) Find V2 (t) for t > 0 where vg (t) = 50u(t)V.
Figure 7.29
SOLUTION
The Laplace transformed network with all initial conditions set to zero is as shown in Fig. 7.30(a).
Figure 7.30(a)
Two Port Networks
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523
(a) To find z11 and z21 , open-circuit the output terminals and then connect a voltage source
V1 across the input terminals as shown in Fig. 7.30(b).
Applying KVL to the left mesh, we get
1
V1 = s +
s
I1
V1 z11 =
I1 I2 =0
Hence;
=s+
1
1
s
=
s2 + 1
s
Also;
V2 = I1
Hence;
V2 1
z21 =
=
I1 I2 =0
s
s
To find z21 and z22 , open-circuit the input terminals and then connect a voltage source V2
across the output terminals as shown in Fig. 7.30(c).
Figure 7.30(b)
Figure 7.30(c)
Applying KVL to the right mesh, we get
V2 = s +
s
I2
V2 s2 + 1
z22 =
=
I2 I1 =0
s
1
V1 = I2
)
Also;
s V1 1
z12 =
=
I2 I1 =0
s
)
Summarizing,
1
2
z=4
z11 z12
z21 z22
3
2
s2 + 1
6
s
5=6
4
1
s
1
3
7
s
7
s2 + 1 5
s
524
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Network Theory
(b)
Figure 7.30(d)
Refer the two port network shown in Fig. 7.30(d).
V1 = Vg
)
)
I1 Zg = z11 I1 + z12 I2
Vg = (z11 + Zg ) I1 + z12 I2
V2
Vg = (z11 + Zg ) I1 + z12
ZL
V2 = z21 I1 + z22 I2
V2
V2 = z21 I1 z22
ZL
z22
1
1+
V2
I1 =
z21
ZL
and
)
)
(7.25)
(7.26)
Substituting equation (7.26) in equation (7.25) and simplifying, we get
V2
Vg
=
z21 zL
(ZL + z22 ) (z11 + Zg )
(7.27)
z12 z21
Substituting for ZL , Zg and z-parameters, we get
V2 (s)
= 2
Vg (s)
s +1
=
)
V2 (s)
=
Vg (s)
=
Hence;
V2 (s) =
1
s2
s +1
+1
+1
s
s
s
2
(s + s + 1)2 1
1
s3 + 2s2 + 3s + 2
1
(s + 1) (s2 + s + 2)
Vg (s)
(s + 1) (s2 + s + 2)
1
s2
(7.28)
Two Port Networks
The equation s2 + s + 2 = 0 gives
s1;2 =
1
2
j
V2 (s) =
This means that,
(s + 1) s +
)
Hence;
V2 (s) =
=
525
p
7
2
Vg (s)
p!
p!
1
7
7
s+ +j
j
2
2
2
1
2
vg (t) = 50u(t)
50
Vg (s) =
s
Given
j
50
1
s(s + 1) s +
2
K1
K2
+
+
s
s+1
p
7
j
2
!
p
1
7
s+ +j
2
2
!
K3
K3
p +
p
7
7
1
1
s+
s+ +j
j
2
2
2
2
By performing partial fraction expansion, we get
K1 = 25; K2 = 25; K3 = 9:45 /90
9:45 / 90
25
25
9:45 /90
p
p
+
Hence;
V2 (s) =
+
s
s+1
7
7
1
1
s+
s+ +j
j
2
2
2
2
Taking inverse Laplace transform of the above equation, we get
V2 (t) = 25
25e t + 18:9e 0:5t cos(1:32t + 90 ) u(t)V
Verification:
V2 (0) = 25
25 + 18:9 cos 90 = 0
V2 (1) = 25 + 0 + 0 = 25V
Please note that at t = 1, the circuit diagram of Fig. (7.29) looks as shown in Fig. 7.30(e).
I(1) =
50
= 25A
2
Hence; V2 (1) = VC (1) = 25V
Figure 7.30(e)
}
526
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Network Theory
EXAMPLE
7.14
The following measurements were made on a resistive two-port network:
Measurement 1: With port 2 open and 100V applied to port 1, the port 1 current is 1.125A and
port 2 voltage is 104V.
Measurement 2: With port 1 open and 50V applied to port 2, the port 2 current is 0.3A, and the
port 1 voltage is 30 V.
Find the maximum power that can be delivered by this two-port network to a resistive load at
port 2 when port 1 is driven by an ideal voltage source of 100 Vdc.
SOLUTION
V1 100
z11 =
= 88:89Ω
=
I1 I2 =0 1:125
V2 104
z21 =
= 92:44Ω
=
I1 I2 =0 1:125
z12 =
V1 30
= 100Ω
=
I2 I1 =0 0:3
V2 50
z22 =
= 166:67Ω
=
I2 I1 =0 0:3
We know from the previous example 7.12 that,
Zt = z22
z12 z21
z11 + Zg
= 166:67
92:44 100
88:89 + 0
= 166:67
103:99
= 62:68Ω
For maximum power transfer,
ZL = Zt
= 62:68Ω (For resistive load)
Vt =
=
z21 Vg
z11 + Zg
92:44 100
88:89 + 0
= 104 V
Two Port Networks
The Thevenin equivalent circuit with respect
to the output terminals with load resistance is as
shown in Fig. 7.31.
Pmax = I2t RL
2
104
62:68
=
62:68 2
= 43:14 W
EXAMPLE
Figure 7.31
7.15
Refer the network shown in Fig. 7.32(a). Find the impedance parameters of the network.
Figure 7.32(a)
SOLUTION
Figure 7.32(b)
j
527
528
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Network Theory
Referring to Fig. 7.32(b), we can write
V3 = 2 (I1 + I2 )
KVL for mesh 1:
2I1 + 2I2 + 2 (I1 + I2 ) = V1
)
4I1 + 4I2 = V1
KVL for mesh 2:
)
)
)
2 (I2
2V3 ) + 2 (I1 + I2 ) = V2
4 2 (I1 + I2 ) + 2 (I1 + I2 ) = V2
2I2
6 (I1 + I2 ) = V2
2I2
6I1
z11 =
4I2 = V2
V1 4I1 + 4I2 =
= 4Ω
I1 I2 =0
I1
I2 =0
V2 =
z21 =
I1 I2 =0
6I1 4I2 =
I1
I2 =0
6Ω
V1 4I1 + 4I2 =
= 4Ω
z12 =
I2 I1 =0
I2
I1 =0
V2 =
z22 =
I2 I1 =0
EXAMPLE
6I1 4I2 =
I2
I1 =0
4Ω
7.16
Is it possible to find z parameters for any two port network ? Explain.
SOLUTION
It should be noted that for some two-port networks, the z parameters do not exist because they
cannot be described by the equations:
V1 = I1 z11 + I2 z12
V2 = I1 z21 + I2 z22
As an example, let us consider an ideal transformer as shown in Fig. 7.33.
(7.29)
Two Port Networks
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529
Figure 7.33
The defining equations for the two-port network shown in Fig. 7.33 are:
V1 =
1
V2 I1 =
n
n I2
It is not possible to express the voltages in terms of the currents, and viceversa. Thus, the ideal
transformer has no z parameters and no y parameters.
7.4
z and y parameters by matrix partitioning
For z parameters, the mesh equations are
V1 = z11 I1 + z12 I2 + + z1n In
V2 = z21 I1 + z22 I2 + + z2n In
0 = 0 = zn1 I1 + zn2 I2 + + znn In
By matrix partitioning, the above equations can be written as
z11
z12
z1n
z21
z22
z2n
z31
z32
z3n
zn1
zn2
znn
530
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Network Theory
The above equation can be simplified as (exact analysis not required)
V1
V2
=
M
N Q
1
P
I1
I2
M NQ 1 P gives z parameters.
Similarly for y parameters,
)
M NQ
EXAMPLE
1P
I1
I2
=
M
N Q
1
P
V1
V2
gives y parameters.
7.17
Find y and z parameters for the resistive network shown in Fig. 7.34(a). Verify the result by using
Y Δ transformation.
3
I1
I3
Figure 7.34(a)
SOLUTION
For the loops indicated, the equations in matrix form,
1
2
I2
Two Port Networks
Then;
)
V1
V2
=
y=z
3 0
0 0:5
1
3:5
1:8571 0:2857
0:2857 0:4285
1
=
0:6
0:4
2
0:5
2 0:5
I1
I2
j
531
= [z]
0:4
2:5
Verification
Figure 7.34(b)
Refer Fig 7.34(b), converting T of 1, 1’, 2 into equation,
11+12+12
=5
1
Z2 = 5
5
Z3 =
2
1
5
2 = 5
Z2 =
5:5
11
Therefore,
3
2
13
y11 = ; y12 = y21 =
; y22 =
5
5
5
The values with transformed circuit is shown in Fig 7.34(c).
Z1 =
EXAMPLE
Figure 7.34(c)
7.18
Find y and z parameters for the network shown in Fig.7.35 which contains a current controlled
source.
Figure 7.35
532
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Network Theory
SOLUTION
At node 1,
0:5V2 = I1
1:5V1
At node 2,
0:5V1 + V2 = I2
3I1
In matrix form,
1:5
0:5
0:5
1
)
V1
V2
V1
V2
=
=
=
Therefore;
[z] =
[y] = [z]
7.5
1 0
3 1
1:5
0:5
0:5
1
0:4 0:4
3:2 1:2
0:4 0:4
3:2 1:2
1
I1
I2
=
1:5
4
1
I1
I2
0:5
0:5
1 0
3 1
I1
I2
Hybrid parameters
The z and y parameters of a two-port network do not always exist. Hence, we define a third set
of parameters known as hybrid parameters. In the pair of equations that define these parameters,
V1 and I2 are the dependent variables. Hence, the two-port equations in terms of the hybrid
parameters are
or in matrix form,
V1 = h11 I1 + h12 V2
(7.30)
I2 = h21 I1 + h22 V2
(7.31)
V1
I2
h11 h12
h21 h22
I1
V2
These parameters are particularly important in transistor circuit analysis. These parameters
are obtained via the following equations:
h11
V1 =
I1 V2 =0
h12
V1 =
V2 I1 =0
h21
I2 = I1 V2 =0
h22
I2 =
V2 I1 =0
The parameters h11 ; h12 ; h21 and h22 represent the short-circuit input impedance, the opencircuit reverse voltage gain, the short-circuit forward current gain, and the open-circuit output
admittance respectively. Because of this mix of parameters, they are called hybrid parameters.
Two Port Networks
EXAMPLE
j
533
7.19
Refer the network shown in Fig. 7.36(a). For this network, determine the h parameters.
Figure 7.36(a)
SOLUTION
To find h11 and h21 short-circuit the output terminals so that V2 = 0. Also connect a current
source I1 to the input port as in Fig. 7.36(b).
Figure 7.36(b)
Applying KCL at node x:
I1 +
)
)
Vx
Vx 0
+
+ I1 = 0
RB
RC
I1 [
Vx
1
1
+
RB RC
(1 )I1 RB RC
Vx =
RB + RC
1] =
534
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Network Theory
V1 h11 =
I1 V2 =0
Hence;
Vx + I1 RA =
I1
V2 =0
(1 )I1 RB RC
+ R A I1
=
(RB + RC ) I1
(1 )RB RC
+ RA
=
RB + RC
KCL at node y:
)
)
Hence;
I1 + I2 + I3 = 0
Vx 0
=0
I1 + I2 +
RC
1 (1 )I1 RB RC
=0
I1 + I2 +
RC
RB + RC
I2 h21 = I1 V2 =0
(1 )RB
= RB + RC
(RC + RB )
=
RB + RC
To find h22 and h12 open-circuit the input port so that I1 = 0. Also, connect a voltage source
V2 between the output terminals as shown in Fig. 7.36(c).
Figure 7.36(c)
KCL at node y:
V1
V1 V2
+
+ I1 = 0
RB
RC
Two Port Networks
j
535
Since I1 = 0, we get
V1
)
V1
V2
=0
RC RC
1
V2
1
=
+
V1
RB RC
RC
V1 RB
h12 =
=
V2 I1 =0 RB + RC
RB
+
Applying KVL to the output mesh, we get
V2 + RC (I1 + I2 ) + RB I2 = 0
Since I1 = 0, we get
)
EXAMPLE
R C I2 + R
B I2 = V2
I2 1
h22 =
=
V2 I1 =0 RC + RB
7.20
Find the hybrid parameters for the two-port network shown in Fig. 7.37(a).
Figure 7.37(a)
SOLUTION
To find h11 and h21 , short-circuit the output port and connect a current source I1 to the input port
as shown in Fig. 7.37(b).
Figure 7.37(b)
536
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Network Theory
Referring to Fig. 7.37(b), we find that
V1 = I1 [2Ω + (8Ωjj4Ω)]
= I1 4:67
V1 h11 =
= 4:67Ω
I1 V2 =0
Hence;
By using the principle of current division, we find that
2
I1 8
= I1
8+
3
4
I2 2
h21 = =
I1 V2 =0
3
I2 =
Hence;
To obtain h12 and h22 , open-circuit the input port and connect a voltage source V2 to the
output port as in Fig. 7.37(c).
Using the principle of voltage division,
8
2
V2 = V2
8+4
3
2
V1
=
h12 =
V2
3
V2 = (8 + 4)I2
V1 =
Hence;
Also;
)
EXAMPLE
= 12I2
I2 1
h22 =
= S
V2 I1 =0 12
Figure 7.37(c)
7.21
Determine the h parameters of the circuit shown in Fig. 7.38(a).
Figure 7.38(a)
Two Port Networks
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537
SOLUTION
Performing Δ to Y transformation, the network shown in Fig. 7.38(a) takes the form as shown in
1
Fig. 7.38(b). Please note that since all the resistors are of same value, RY = RΔ .
3
Figure 7.38(b)
To find h11 and h21 , short-circuit the output port and connect a current source I1 to the input
port as in Fig. 7.38(c).
V1 = I1 [4Ω + (4Ωjj4Ω)]
Hence;
= 6I1
V1 h11 =
= 6Ω
I1 V2 =0
Using the principle of current division,
I1
4
4+4
I1
I2 =
2 I2 1
h21 = =
I1 V2 =0
2
I2 =
)
Hence;
To find h12 and h22 , open-circuit the input port
and connect a voltage source V2 to the output port
as shown in Fig. 7.38(d).
Using the principle of voltage division, we get
V2
4
4 +4
V1 1
h12 =
=
V2 I1 =0 2
V1 =
)
Also;
)
V2 = [4 + 4] I2 = 8I2
I2 1
h22 =
= S
V2 I1 =0 8
Figure 7.38(c)
Figure 7.38(d)
538
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Network Theory
EXAMPLE
7.22
Determine the Thevenin equivalent circuit at the output of the circuit in Fig. 7.39(a).
Figure 7.39(a)
SOLUTION
To find Zt , deactivate the voltage source Vg and apply a 1 V voltage source at the output port, as
shown in Fig. 7.39(b).
Figure 7.39(b)
The two-port circuit is described using h parameters by the following equations:
V1 = h11 I1 + h12 V2
(7.32)
I2 = h21 I1 + h22 V2
(7.33)
But V2 = 1 V and V1 = I1 Zg
Substituting these in equations (7.32) and (7.33), we get
)
I1 Zg = h11 I1 + h12
h12
I1 =
Zg + h11
I2 = h21 I1 + h22
(7.34)
(7.35)
Two Port Networks
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539
Substituting equation (7.34) into equation (7.35), we get
h21 h12
h11 + Zg
h11 h22 h21 h12 + h22 Zg
=
h11 + Zg
h11 + Zg
V2
1
Zt =
=
=
I2
I2
h11 h22 h12 h21 + h22 Zg
I2 = h22
Therefore;
To get Vt , we find open circuit voltage V2 with I2 = 0. To find Vt , refer the Fig. 7.39(c).
Figure 7.39(c)
At the input port, we can write
Vg + I1 Zg + V1 = 0
V1 = Vg I1 Zg
)
(7.36)
Substituting equation (7.36) into equation (7.32), we get
Vg I1 Zg = h11 I1 + h12 V2
Vg = (h11 + Zg ) I1 + h12 V2
)
(7.37)
and substituting I2 = 0 in equation (7.33), we get
0 = h21 I1 + h22 V2
h22
I1 =
V2
h21
)
(7.38)
Finally substituting equation (7.38) in (7.37), we get
Vg = (h11 + Zg )
)
V2 = Vt =
h12 h21
h22
V2 + h12 V2
h21
Vg h21
h11 h22 Zg h22
Hence, the Thevenin equivalent circuit as seen from the
output terminals is as shown in Fig. 7.39(d).
Figure 7.39(d)
540
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Network Theory
EXAMPLE
7.23
Find the input impedance of the network shown in Fig. 7.40.
Figure 7.40
SOLUTION
For the two-port network, we can write
V1 = h11 I1 + h12 V2
(7.39)
I2 = h21 I1 + h22 V2
(7.40)
But
V2 = IL ZL =
where
ZL = 75 kΩ
I2 Z L
(7.41)
Substituting the value of V2 in equation (7.40), we get
)
I2 = h21 I1 h22 I2 ZL
h21 I1
I2 =
1 + ZL h22
(7.42)
Substituting equation (7.42) in equation (7.41), we get
V2 =
ZL h21 I1
1 + ZL h22
Substituting equation (7.43) in equation (7.39), we get
V1 = h11 I1
Hence
Zin =
V1
I1
h12 ZL h21 I1
1 + ZL h22
ZL h12 h21
1 + ZL h22
75 103 10 5 200
= 3 103
1 + 75 103 10 6
= 2:86kΩ
= h11
(7.43)
Two Port Networks
EXAMPLE
7.24
Find the voltage gain,
V2
Vg
for the network shown in Fig. 7.41.
S
Figure 7.41
SOLUTION
For the two-port network we can write,
V1 = h11 I1 + h12 V2 ;
I2 = h21 I1 + h22 V2 ;
Hence;
)
)
Also;
)
here V1 = Vg Zg I1
here V2 = ZL I2
Vg Zg I1 = h11 I1 + h12 V2
Vg = (h11 + Zg ) I1 + h12 V2
Vg h12 V2
I1 =
h11 + Zg
V2
I2 =
= h21 I1 + h22 V2
ZL
Vg h12 V2
V2
+ h22 V2
= h21
ZL
h11 + Zg
From the above equation, we find that
V2
Vg
=
h21 ZL
(h11 Zg ) (1 + h22 ZL )
h12 h21 ZL
100 50 103
=
(2 103 + 1 103 ) (1 + 10 5 50 103 ) (10
= 1250
4
100 50 103)
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541
542
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Network Theory
EXAMPLE
7.25
The following dc measurements were done on the resistive network shown in Fig. 7.42(a).
Measurement 1
V1 = 20 V
I1 = 0:8 A
V2 = 0 V
I2 = 0:4 A
Measurement 2
V1 = 35 V
I1 = 1 A
V2 = 15 V
I2 = 0 A
Find the value of Ro for maximum power transfer.
Figure 7.42(a)
SOLUTION
For the two-port network shown in Fig. 7.41, we can write:
V1 = h11 I1 + h12 V2
I2 = h21 I1 + h22 V2
From measurement 1:
V1 20
h11 =
= 25Ω
=
I1 V2 =0 0:8
I2 0:4
=
=
h21 = I1 V2 =0
0:8
0:5
From measurement 2:
)
)
Then;
)
V1 = h11 I1 + h12 V2
35 = 25 1 + h12 15
10
h12 =
= 0:67
15
I2 = h21 I1 + h22 V2
0 = h21 1 + h22 15
h21
0:5
h22 =
=
= 0:033 S
15
15
Two Port Networks
| 543
For example (7.22),
Vt =
=
=
Zt =
=
=
Vg h21
h12 h21 − h11 h22 − Zg h22
50 × (−0.5)
0.67 × (−0.5) − 25 × 0.033 − 20 × 0.033
−25
= 13.74 Volts
−1.82
h11 + Zg
h11 h22 − h12 h21 + h22 Zg
25 + 20
25 × 0.033 − 0.67 × (−0.5) + 0.033 × 20
45
= 24.72 Ω
1.82
For maximum power transfer, ZL = Zt = 24.72 Ω (Please note that, ZL is purely resistive).
The Thevenin equivalent circuit as seen from the output terminals along with ZL is shown in
Fig. 7.42(b).
Pmax = It2 × 24.72
2
13.74
=
× 24.72
24.72 + 24.72
=
(13.74)2
= 1.9 Watts
4 × 24.72
Figure 7.42(b)
EXAMPLE
7.26
Determine the hybird parameters for the network shown in Fig. 7.43.
C
Figure 7.43
544
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Network Theory
SOLUTION
To find h11 and h21 , short-circuit the output terminals so that V2 = 0. Also connect a current
source I1 to the input port as shown in Fig. 7.44(a).
Figure 7.44(a)
Applying KVL to the mesh on the right side, we get
)
)
)
Hence;
1
[I1 + I2 ] = 0
R2 [I1 + I2 ] +
j!C
1
I1 + R 2 +
I2 = 0
R2 +
j!C
j!C
[ + j!R2 C ] I1 =
I2 =
[1 + j!CR2 ] I2
[ + j!R2 C ]
I1
1 + j!R2 C
I2 h21 = I1 V2 =0
=
+ j!CR2
1 + j!R2 C
Applying KVL to the mesh on the left side, we get
V1 = R1 I1 + R2 [I1 + I2 ]
Hence;
= [R1 + R2 ] I1 + R2 I2
R2 ( + j!CR2 )
I1
= R1 + R2
1 + j!R2 C
V1 h11 =
I1 V2 =0
Two Port Networks
j
545
R2 ( + j!R2 C )
1 + j!R2 C
R1 + R2 (1 ) + j!R1 R2 C
=
1 + j!R2 C
= R1 + R2
To find h22 and h12 , open-circuit the input terminals so that I1 = 0. Also connect a voltage
source V2 to the output port as shown in Fig. 7.44(b). The dependent current source is open,
because I1 = 0.
V1 = I2 R2
V2
=
R2
1
R2 +
Hence;
j!C
V1 h12 =
V2 I1 =0
j!CR2
1 + j!CR2
V2
j!C V2
I2 =
=
1
1 + j!CR2
R2 +
j!C
I2 j!C
h22 =
=
V2 I1 =0 1 + j!CR2
=
)
7.6
Figure 7.44(b)
Transmission parameters
The transmission parameters are defined by the equations:
V1 = AV2
BI2
I1 = CV2
DI2
–
Figure 7.45 Terminal variables used to define the ABCD Parameters
546
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Network Theory
Putting the above equations in matrix form we get
V1
I1
=
A B
C D
V2
I2
Please note that in computing the transmission parameters, I2 is used rather than I2 , because
the current is considered to be leaving the network as shown in Fig. 7.45.
These parameters are very useful in the analysis of circuits in cascade like transmission lines
and cables. For this reason they are called Transmission Parameters. They are also known as
ABCD parameters. The parameters are determined via the following equations:
V1 A=
V2 I2 =0
V1 B=
I2 V2 =0
I1 C=
V2 I2 =0
I1 D=
I2 V2 =0
A, B, C and D represent the open-circuit voltage ratio, the negative short-circuit transfer
impedance, the open-circuit transfer admittance, and the negative short-circuit current ratio,
respectively. When the two-port network does not contain dependent sources, the following relation holds good.
AD
BC = 1
EXAMPLE
7.27
Determine the transmission parameters in the s domain for the network shown in Fig. 7.46.
Figure 7.46
SOLUTION
The s domain equivalent circuit with the assumption that all the initial conditions are zero is
shown in Fig. 7.47(a).
Figure 7.47(a)
Two Port Networks
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547
To find the parameters A and C, open-circuit the output port and connect a voltage source V1
at the input port. The same is shown in Fig. 7.47(b).
V1
I1 =
1+
1
1
s
=
sV1
s+1
Then
V2 = I1
Also;
1 sV1
V1
V2 =
=
ss+
1 s+1
V1 A=
=s+1
V2 I2 =0
1
V2 = I1
)
)
)
s
Figure 7.47(b)
s I1 C=
=s
V2 I2 =0
To find the parameters B and D, short-circuit
the output port and connect a voltage source V1 to
the input port as shown in Fig. 7.47(c).
Figure 7.47(c)
The total impedance as seen by the source V1 is
1
Z=1+ s
1
s
1
+1
1
s+2
=
s+1 s+1
V1
V1 (s + 1)
I1 =
=
Z
(s + 2)
=1+
(7.44)
Using the principle of current division, we have
Hence;
1
I1
I1
s
I2 =
=
1
s+1
+1
s I1 D=
=s+1
I2 V2 =0
From equation (7.44) and (7.45), we can write
V1 (s + 1)
(s + 2)
V1 B=
=s+2
I2 V2 =0
I2 (s + 1) =
Hence;
(7.45)
548
j
Network Theory
Verification
We know that for a two port network without any dependent sources,
AD
(s + 1) (s + 1)
EXAMPLE
BC = 1
s (s + 2) = 1
7.28
Determine the ABCD parameters for the two port network shown in Fig. 7.48.
Figure 7.48
SOLUTION
To find the parameters A and C, open-circuit the output port as shown in Fig. 7.49(a) and connect
a voltage source V1 to the input port.
Applying KVL to the output mesh, we get
V2 + mI1 + 0 RC + I1 RA = 0
)
V2 = I1 (m + RA )
I1 1
Hence; C =
=
V2 I2 =0 m + RA
Applying KVL to the input mesh, we get
Hence;
V1 = I1 (RA + RB )
V1 RA + RB
A=
=
V2 I2 =0
m + RA
Figure 7.49(a)
To find the parameters B and D, short-ciruit the output port and connect a voltage source V1
to the input port as shown in Fig. 7.49(b).
Two Port Networks
Figure 7.49(b)
Applying KVL to the right-mesh, we get
mI1 + RC I2 + RB (I1 + I2 ) = 0
)
)
Hence;
(m + RB ) I1 =
(RC + RB ) I2
(RC + RB )
I2
I1 =
(m + RB )
I1 (RC + RB )
D=
=
I2 V2 =0
(m + RB )
Applying KVL to the left-mesh, we get
V1 + RA I1 + RB (I1 + I2 ) = 0
)
Hence;
EXAMPLE
V1 = (RA + RB ) I1 + RB I2
(RC + RB )
= (RA + RB )
I2 + R B I2
(m + RB )
R C R A + R C R B + R B R A mR B
I2
=
m + RB
V1 B=
I2 V2 =0
R C R A + R C R B + R B R A mR B
=
m + RB
7.29
The following direct-current measurements were done on a two port network:
Port 1 open
V1 = 1 mV
V2 = 10 V
I2 = 200 A
Port 1 Short-circuited
I1 = 0:5 A
I2 = 80 A
V2 = 5 V
Calculate the transmission parameters for the two port network.
j
549
550
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Network Theory
SOLUTION
For the two port network, we can write
V1 = AV2
BI2
I1 = CV2
DI2
From I1 = 0 (port 1 open): 1 10
From V1 = 0 (Port 1 short):
3
= A 10
0=A5
B 200 10
B 80 10
6
6
Solving simultaneously yields,
From I1 = 0:
From V1 = 0:
4 10
4
; B = 25Ω
0 = C 10 D 200 10 6
0:5 10 6 = C 5 D 80 10 6
A=
Solving simultaneously yields,
C=
In summary,
5 10
S;
D=
A=
4 10
B=
25Ω
C=
D=
EXAMPLE
7
5 10
0:025
4
7
S
0:025
7.30
Find the transmission parameters for the network shown in Fig. 7.50.
Figure 7.50
SOLUTION
To find the parameters A and C, open the output port and connect a voltage source V1 to the input
port as shown in Fig. 7.51(a).
Two Port Networks
Figure 7.51(a)
Applying KVL to the input loop, we get
V1 = 1:5 103 I1 + 10
3
V2
Also KCL at node a gives
V2
=0
40 103
V2
)
I1 =
= 6:25 10 6 V2
160 103
Substituting the value of I1 in the preceeding loop equation, we get
40I1 +
)
V1 = 1:5 103
V1 =
9:375 10
Also;
3
6
V2 + 10
V2 + 10
3
8:375 10 V2
V1 = 8:375 10
A=
V2 I2 =0
=
Hence;
6:25 10
I1 C=
=
V2 I2 =0
V2
3
6:25 10
3
6
To find the parameters B and D, refer the circuit shown in Fig. 7.51(b).
Figure 7.51(b)
3
V2
j
551
552
j
Network Theory
Applying KCL at node b, we find
)
40I1 + 0 = I2
I2 = 40I1
I1 1
D=
=
I2 V2 =0
40
Hence;
Applying KVL to the input loop, we get
V1 = 1:5 103 I1
)
V1 = 1:5 103 V1 B=
I2 V2 =0
1:5
103
=
40
= 37:5Ω
Hence;
EXAMPLE
I2
40
7.31
Find the Thevenin equivalent circuit as seen from the output port using the transmission parameters for the network shown in Fig. 7.52.
Figure 7.52
SOLUTION
For the two-port network, we can write
V1 = AV2
BI2
(7.46)
I1 = CV2
DI2
(7.47)
Two Port Networks
j
553
Figure 7.53(a)
Refer the network shown in Fig. 7.53(a) to find Vt .
At the input port,
Vg
Also;
I1 Zg = V1
(7.48)
I2 = 0
(7.49)
Making use of equations (7.48) and (7.49) in equations (7.46) and (7.47) we get,
Vg
and
I1 Zg = AV2
(7.50)
I1 = CV2
(7.51)
Making use of equation (7.51) in (7.50), we get
Vg
)
CV2 Zg = AV2
V2 = Vt =
Vg
A + CZg
To find Rt , deactivate the voltage source Vg and then connect a voltage source V2 = 1 V at
the output port. The resulting circuit diagram is shown in Fig. 7.53(b).
Figure 7.53(b)
554
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Network Theory
Referring Fig. 7.53(b), we can write
V1 =
I1 Z g
Substituting the value of V1 in equation (7.46), we get
I1 Zg = AV2 BI2
B
A
V2 +
I2
I1 =
)
Zg
Zg
(7.52)
Equating equations (7.47) and (7.52) results
Hence
DI2 =
A
V2 +
B
I2
Zg
Zg
B
A
= D+
I2
V2 C +
Zg
Zg
B
D+
Zg
V2
=
Zt =
A
I2
C+
Zg
B + DZg
=
A + CZg
CV2
Figure 7.54
Hence, the Thevenin equivalent circuit as seen from the output port is as shown in Fig. 7.54.
EXAMPLE
7.32
For the network shown in Fig. 7.55(a), find RL for maximum power transfer and the maximum
power transferred.
Figure 7.55(a)
Two Port Networks
j
555
SOLUTION
From the previous example 7.31,
B + DZg
20 + 3 20
20 + 60
=
=
= 6.67Ω
A + CZg
4 + 0:4 20
4+8
Vg
100
100
Vt =
=
=
= 8.33V
A + CZg
4 + 0:4 20
12
Zt =
For maximum power transfer,
RL = Zt = 6:67Ω (purely resistive)
Hence, the Thevenin equivalent circuit as seen from
output terminals along with RL is as shown in
Fig. 7.55(b).
8:33
= 0:624A
6:67 + 6:67
(PL )max = It2 6:67
It =
= (0:624)2 6:67
Figure 7.55(b)
= 2.6Watts
EXAMPLE
7.33
Refer the bridge circuit shown in Fig. 7.56. Find the transmission parameters.
Figure 7.56
SOLUTION
Performing Δ to Y transformation, as shown in Fig. 7.57(a) the network reduces to the form as
shown in Fig. 7.57(b). Please note that, when all resistors are of equal value,
1
RY = RΔ
3
556
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Network Theory
Figure 7.57(a)
Figure 7.57(b)
To find the parameters A and D, open the output port and connect a voltage source V1 at the
input port as shown is Fig. 7.57(c).
Applying KVL to the input loop we get
I1 + 4I1 = V1
)
Also;
Also;
Hence;
)
V1 = 5I1
V2
I1 =
4
)
I1 1
C=
= S
V2 I2 =0 4
5
V1 = 5I1 = V2
4
V1 5
A=
=
V2 I2 =0 4
To find the parameters B and D, refer the circuit shown in Fig. 7.57(d).
Figure 7.57(c)
Two Port Networks
j
557
I1 4
4
= I1
4 + 1
5
I1 5
D=
=
I2 v2 =0 4
I2 =
Hence
Applying KVL to the input loop, we get
V1 + 1 I1 + 4 (I1 + I2 ) = 0
Substituting I1 =
Figure 7.57(d)
5
I2 in the preceeding equation, we get
4
V1
)
)
Hence;
5
5
I2 + 4
I2 + I2 = 0
4
4
5
V1
I2 5I2 + 4I2 = 0
4
4V1 = 9I2
V1 9
B=
= Ω
I2 v2 =0 4
Verification:
For a two port network which does not contain any dependent sources, we have
AD
5
4
7.7
54
1
4
BC = 1
25
94 = 16
9
=1
16
Relations between two-port parameters
If all the two-port parameters for a network exist, it is possible to relate one set of parameters to
another, since these parameters interrelate the variables V1 ; I1 ; V2 and I2 : To begin with let us
first derive the relation between the z parameters and y parameters.
The matrix equation for the z parameters is
)
V1
V2
=
z11 z12
z21 z22
I1
I2
V = zI
Similarly, the equation for y parameters is
)
I1
I2
y11 y12
=
y21 y22
I = yV
(7.53)
V1
V2
(7.54)
558
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Network Theory
Substituting equation (7.54) into equation (7.53), we get
V = zyV
Hence;
z=y
1
=
Δy = y11 y22
where
adj(y)
Δy
y21 y12
This means that we can obtain z matrix by inverting y matrix. It is quite possible that a
two-port network has a y matrix or a z matrix, but not both.
Next let us proceed to find z parameters in terms of ABCD parameters.
The ABCD parameters of a two-port network are defined by
V1 = AV2
)
)
BI2
I1 = CV2 DI2
1
V2 = (I1+ DI2 )
C
1
D
V2 = I1 + I2
C
C
I1 DI2
BI2
+
V1 = A
C
C
AI1
AD
=
+
B I2
C
C
(7.55)
(7.56)
Comparing equations (7.56) and (7.55) with
V1 = z11 I1 + z12 I2
V2 = z21 I1 + z22 I2
respectively, we find that
AD BC
1
D
A
z12 =
z21 =
z22 =
C
C
C
C
Next, let us derive the relation between hybrid parameters and z parameters.
z11 =
V1 = z11 I1 + z12 I2
(7.57)
V2 = z21 I1 + z22 I2
(7.58)
From equation (7.58), we can write
I2 =
z21
V2
I1 +
z22
z22
Substituting this value of I2 in equation (7.57), we get
(7.59a)
z21 I1
V2
+
V1 = z11 I1 + z12
z22
z22
z12 V2
z11 z22 z12 z21
I1 +
=
z22
z22
(7.59b)
Two Port Networks
j
559
Comparing equations (7.59b) and (7.59a) with
V1 = h11 I1 + h12 V2
I2 = h21 I1 + h22 V2
Δz
z11 z22 z12 z21
=
h11 =
z22
Z22
z12
z21
h12 =
h21 =
z22
z22
Δz = z11 z22 z12 z21
we get,
where
h22 =
1
z22
Finally, let us derive the relationship between y parameters and ABCD parameters.
I1 = y11 V1 + y12 V2
(7.60)
I2 = y21 V1 + y22 V2
(7.61)
From equation (7.61), we can write
I2
y22
V2
y21 y21
y22
1
=
V2 +
I2
y21
y21
V1 =
(7.62)
Substituting equation (7.62) in equation (7.60), we get
y11 y22
y11
V2 + y12 V2 +
I2
y21
y21
y11
Δy
V2 +
I2
=
y21
y21
I1 =
(7.63)
Comparing equations (7.62) and (7.63) with the following equations,
V1 = AV2
we get
where
y22
A=
y21
BI2
I1 = CV2 DI2
1
Δy
B=
C=
y21
y21
Δy = y11 y22 y12 y21
D=
y11
y21
Table 7.1 lists all the conversion formulae that relate one set of two-port parameters to another.
Please note that Δz, Δy, Δh, and ΔT, refer to the determinants of the matrices for z, y, hybrid,
and ABCD parameters respectively.
560
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Network Theory
Table 7.1 Parameter relationships
z
2
4
z
z11 z12
3
5
z21 z22
y
2 z
22
6 Δz
4
z21
Δz
2
h
3
z11
6 z21
6
4 1
z21
T
Δz
z21 7
7
z22 5
z21
Δz
6 z22
6
4 z21
z22
Δz = z11 z22
EXAMPLE
y22
6 Δy
6
4 y21
Δy
z12 3
Δz 7
5
z11
Δz
2
3
z12
z22 7
7
1 5
z22
y
2
2
4
3
y12
Δy 7
7
y11 5
Δy
y11 y12
y22
6 y21
6
4 Δy
y21
2
1
6 y11
6
4 y21
y11
z12 z21 ; Δy = y11 y22
1
A
6 C
6
4 1
C
2
3
5
y21 y22
2
T
2
3
ΔT
B 7
7
A 5
B
2
3
y21 7
7
y11 5
y21
3
ΔT
C 7
7
D 5
C
D
6 B
6
4 1
B
3
y12
y11 7
7
Δy 5
y11
3
4
A B
B
6 D
6
4 1
D
5
3
Determine the y parameters for a two-port network if the z parameters are
z=
SOLUTION
10 5
5 9
Δz = 10 9 5 5 = 65
z22
9
y11 =
= S
Δz
65
z12
5
y12 =
=
S
Δz
65
z21
5
y21 =
=
S
Δz
65
z11
10
y22 =
= S
Δz
65
1
6 h11
6
4 h21
h11
2
4
3
h12
h22 7
7
1 5
h22
3
h22
h11 7
7
Δh 5
h11
3
h11
h21 7
7
1 5
h21
h11 h12
3
5
h21 h22
h12 h21 ; ΔT = AD
7.34
2
Δh
6 h21
6
4 h22
h21
ΔT
D 7
7
C 5
D
y12 y21 ; Δh = h11 h22
Δh
6 h22
6
4 h21
h22
2
C D
2
h
2
BC
Two Port Networks
EXAMPLE
j
561
7.35
Following are the hybrid parameters for a network:
h11 h12
h21 h22
=
5 2
3 6
Determine the y parameters for the network.
Δh = 5 6 3 2 = 24
1
1
y11 =
= S
h11
5
h22
6
=
S
y12 =
h11
5
h21
3
= S
y21 =
h11
5
Δh
24
= S
y22 =
h11
5
SOLUTION
Reinforcement problems
R.P
7.1
The network of Fig. R.P. 7.1 contains both a dependent current source and a dependent voltage
source. Determine y and z parameters.
Figure R.P. 7.1
SOLUTION
From the figure, the node equations are
Iab =
I2
V2
2
At node a,
I1 = V1
2V2
I2
V2
2
562
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Network Theory
At node b,
V1 = V2
I2
2V1
V2
2
Simplifying, the nodal equations, we get
3
V2
2
3
3V1 + V2
2
I1 + I2 = V1
I2 =
In matrix form,
)
1 1
0 1
I1
I2
I1
I2
2
6
=4
3
=
2
y=4
Therefore;
Z=
and
R.P
1
1 1
0 1
4
3
3 3
2 7 V1
5
V2
3
2
2
3 3
1
1
6
2 7 V1
4
5
V2
3
3
2
3
3
3 5
2
2 3
14
2
3
3
3
3
5=4
4
7.2
Compute y parameters for the network shown in Fig. R.P. 7.2.
Figure R.P. 7.2
2
0:5
1
3
1
4 5
3
Two Port Networks
j
563
SOLUTION
The circuit shall be transformed into s-domain and then we shall use the matrix partitioning
method to solve the problem. From Fig 7.2, Node equations in matrix form,
I
I
R.P
s
s
s
P
V
Q
V
s
V
V
V
M N
V1
I1
= P Q N 1 M
V2
I2
1 2 s+3
s
2 1
=
s s+2
5 1
8
2 4 2 39
>
>
<
=
s+3
s
6 5 5 7
=
4
5
s s+2
>
2 1 >
:
;
5
5
s + 2:2
(s + 0:4)
y=
(s + 0:4)
s + 1:8
V
V1
V2
V1
V2
7.3
Determine for the circuit shown in Fig. R.P. 7.3(a): (a) Y1 , Y2 , Y3 and gm in terms of y parameters.
(b) Repeat the problem if the current source is connected across Y3 with the arrow pointing to the
left.
Figure R.P. 7.3(a)
SOLUTION
(a) Refering Fig. R.P. 7.3(a), the node equations are:
At node 1
I1 = Y1 V1 + (V1 V2 )Y3
= V1 (Y1 + Y3 )
Y3 Y2
(7.64)
At node 2
I2 = gm V1 + V2 Y2 + (V2
= (gm
V1 )Y3
Y3 )V1 + (Y2 + Y3 )V2
(7.65)
564
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Network Theory
Then from equations (7.64) and (7.65),
y11 = Y1 + Y3 ;
y12 =
y21 = gm
y22
Y3 ;
Y3
= Y2 + Y3
Solving,
Y3 = y12 ; Y1 = y11 + y12
Y2 = y22 + y12 ; gm = y21 y12
(b) Making the changes as given in the problem, we get the circuit shown in Fig R.P. 7.3(b).
Figure R.P. 7.3(b)
Node equations : At node 1
I1 = Y1 V1 + (V1
= (Y1 + Y3
V2 )Y3
gm )V1
gm V1
Y3 V2
(7.66)
At node 2,
I2 = V2 Y2 + (V2
= (gm
V1 )Y3 + gm V1
Y3 )V1 + V2 (Y2 + Y3 )
(7.67)
From equations (7.66) and (7.67),
y11 = Y1 + Y3
y21 = gm
gm ;
Y3 ;
Y3
y22 = Y2 + Y3
y12 =
Solving,
Y3 = y12 ; Y2 = y22 y12
gm = y21 y12
Y1 = y11 Y3 + gm
= y11 + y12 + y21 y12 = y11
y21
Two Port Networks
R.P
j
565
7.4
Complete the table given as part of Fig. R.P. 7.4. Also find the values for y parameters.
Table
Sl.no
1
2
3
4
5
Figure R.P. 7.4
V1
50
100
200
SOLUTION
From article 7.2,
I1
I2
=
y11 y12
y21 y22
V1
V2
Substituting the values from rows 1 and 2,
1 7
27 24
Post multiplying by [V]
=
y11 y12
y21 y22
For row 4:
For row 5:
I1
I2
V1
V2
V1
V2
50 100
100 50
1,
=
=
For row 3:
y11 y12
y21 y22
=
=
=
1 7
27 24
0:1
0:14
0:1
0:14
0:06
0: 2
0:1
0:14
0:06
0: 2
0:1
0:14
0:06
0: 2
0:06
0: 2
1
1
200
0
1
50 100
100 50
20
0
10
30
=
20
28
=
=
140:84
98:59
133:8
56:338
V2
100
50
0
I1
1
7
I2
27
24
20
10
0
30
566
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Network Theory
R.P
7.5
Find the condition on a and b for reciprocity for the network shown in Fig. R.P. 7.5.
Figure R.P. 7.5
SOLUTION
The loop equations are
aV2 = 3(I1 + I3 )
V2 = (I2 I3 ) bI1
I3 = V2 (V1 aV2 )
= (1 + a)V2 V1
V1
(7.68)
(7.69)
(7.70)
Substituting equation (7.70) in equations (7.68) and (7.69),
aV2 = 3I1 + 3(1 + a)V2 3V1
4V1 = (3 + 4a)V2 + 3I1
V2 = I2 [(1 + a) V2 V1 ] bI1
V1 + (2 + a) V2 = bI1 + I2
V1
)
)
Putting equations (7.71) and (7.72) in matrix form and solving
V1
V2
=
4
1
(3 + 4a)
2+a
1
1 2 + a 3 + 4a
1
4
Δ
1
M
3 + 4a
=
N
Δ 3 4b
=
For reciprocity,
3 + 4a = 3
Therefore;
a= b
4b
3
0
b 1
3
0
b 1
I1
I2
I1
I2
I1
I2
(7.71)
(7.72)
Two Port Networks
R.P
7.6
For what value of a is the circuit reciprocal? Also find h parameters.
Figure R.P. 7.6
SOLUTION
The node equations are
0:5V1
V1
I1 = V2
V2 = (I1 + I2 )2 + aI1
h=
=
1
Δ
2
6
0:5
0
0
2
1
2 0
0 0:5
2
=4 2+a
2
2
0:5
For reciprocity,
)
)
Therefore
h21
2+a
2=
2
4 = 2 + a; a = 2
2 2
h=
2 0:5
h12 =
3
7
5
1
2+a
1
1
1
2+a
1
1
(Δ =
1)
j
567
568
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Network Theory
R.P
7.7
Find y12 and y21 for the network shown in Fig. R.P. 7.7 for n = 10. What is the value of n for the
network to be reciprocal?
Figure R.P. 7.7
SOLUTION
Equations for I1 and I2 are
0:01V2
5
= 0:2V1 0:002V2
V2
V2 0:01V2
I2 =
+ n I1 +
20
50
Substituting the value of I1 from equation (7.73) in equation (7.74), we get
V2 V2 0:01V2
I2 = n(0:2V1 0:002V2 ) +
+
20
20
Simplifying the above equation with n = 10,
I1 =
V1
I2 = 2V1 + 0:0498V2
From equation (7.73),
y12 =
0:002
and from equation (7.75),
y21 = 0:2n
For reciprocity
y12 = y21
Hence;
R.P
)
0:002 = 0:2n
n = 0:01
7.8
Find T parameters (ABCD) for the two-port network shown in Fig. R.P. 7.8.
Figure R.P. 7.8
(7.73)
(7.74)
(7.75)
(7.76)
Two Port Networks
j
569
SOLUTION
Network equations are
V1
1:5V1
10I1 = V2
V2
I1
1:5V1
+ I2
25
(7.77)
V2
=0
20
(7.78)
Simplifying,
2:5V1
10I1 = V2
0:06V1 + I1 = 0:09V2
I2
In matrix form,
Therefore
2:5
0:06
T=
R.P
2:5
0:06
10
1
10
1
V1
I1
1
=
1
0
0:09 1
1
0
0:09 1
=
V2
I2
0:613 3:23
0:053 0:806
7.9
(a) Find T parameters for the active two port network shown in Fig. R.P. 7.9.
(b) Find new T parameters if a 20 Ω resistor is connected across the output.
Figure R.P. 7.9
SOLUTION
(a) With
Therefore;
Vx = 10I1 ,
0:08Vx = 0:8I1
V1
10I1 = V2
I1 + I2
= V2
V1
0:8I1 =
5(I2
0:08Vx )
5I2 + 4I1
10I1
50
(7.79)
(7.80)
570
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Network Theory
Simplying the equations (7.79) and (7.80),we get
V1
14I1 = V2
V1 + 20I1 =
and
Therefore,
1
1
T=
14
20
1
1 5
0 50
5I2
50I2
=
3:33 133:33
0:167 9:17
(b) Treating 20 Ω across the output as a second T network for
which
#
"
1 0
T=
1
1
20
Then new T-parameters,
T=
R.P
3:33 133:33
0:167 9:17
" 1 0 # 10
133:33
=
1
0:625 9:17
1
20
7.10
Obtain z parameters for the network shown in Fig. R.P. 7.10.
Figure R.P. 7.10
SOLUTION
At node 1,
V1 = (I1
= 10I1
At node 2,
V2 = I2
)
0:3V2 )10 + V2
(7.81)
2V2
V2
6
8
V2 = V1 + 10I2
3
10 + V1 = 10I2 + V1
5
V2
3
(7.82)
Two Port Networks
Putting in matrix form,
1
1
Therefore,
z=
R.P
1
1
2
−8
3
2
−8
3
V1
V2
−1 =
10
0
0 −10
10
0
0 −10
=
I1
I2
5.71 −4.286
2.143 2.143
7.11
Obtain z and y parameters for the network shown in Fig. R.P. 7.11.
Figure R.P. 7.11
SOLUTION
For the meshes indicated, the equations in matrix form is
| 571
572
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Network Theory
By matrix partitioning,
2 s+2
2
6 s
s
z=4
s+4
2
s
2s
2 s+2
2
6 s
s
=4
s+4
2
s
2s
2 s+2
2
6 s
s
=6
4 2
s+4
s
2s
2 2
s + 10s + 8
6 s(3s + 4)
6
=6
6 s2 + 6s + 8
4
s(3s + 4)
R.P
3
7
5
3
7
5
2
3
1 2s
4 1 5
3s + 4
2
2
2s
6
4
3s + 4
3
2
7
7
5
4
2s
6 3s + 4
6
s
3s + 4
3
s2 + 6s + 8
s(3s + 4) 7
7
7
2
s + 8s + 8 7
5
s(3s + 4)
1
1
2
1
1
2
1 3
2 7
5
1
4
3
s
3s + 4 7
7
8s 5
3s + 4
7.12
Find z and y parameters at ! = 108 rad/sec for the transistor high frequency equivalent circuit
shown in Fig. R.P. 7.12.
Figure R.P. 7.12
SOLUTION
In the circuit, Vx = V1 . Therefore the node equations are
5
+ j 6 10
j 10 4 V2
I2 = j 10 4 V1 + 0:01V1 + 10 4 (1 + j )V2
I1 = (10
4
)V1
Two Port Networks
Simplifying the above equations,
I1 = 10−4 [(0.1 + j6)V1 − j1V2 ]
Therefore,
I2 = 10−4 [(100 + j1)V1 + (1 + j)V2 ]
⎡
⎤
⎡
⎤
0.1
+
j6
−j1
V
1
I1
⎦ × 10−4 ⎣
⎦
=⎣
I2
100 − j1 1 + j1
V2
ωC1 = 108 × 5 × 10−12 = 5 × 10−4
ωC2 = 108 × 10−12 = 10−4
Δ = 10−8 [(0.1 + j6)(1 + j) + (100 − j1)(j1)]
= 10−8 × 106.213 /92.64◦
⎡
Therefore,
y=⎣
6 /89◦
100 /−0.6◦
⎡
z = y−1 = ⎣
Then,
⎡
=⎣
R.P
√
2 /45◦
√
2 /45◦
j1
100 /−180.6◦
6 /89◦
√
2 /45◦
j1
100 /−180.6◦
6 /89◦
⎡
=⎣
−j1
⎤
⎦ × 10−4
⎤
⎦ × 10−4 ÷ Δ
⎤
⎦×
10−4
10−8 × 106.213 /92.64◦
133.15 /−47.64◦
94.16 /−2.64◦
94.16 /86.8◦
565 /−31.6◦
⎤
⎦
7.13
Obtain TA , TB , TC for the network shown in Fig. R.P. 7.13 and obtain overall T.
A
B
Figure R.P. 7.13
C
| 573
574
j
Network Theory
SOLUTION
Using the equation for T-parameters for a T -network
Z1 + Z3
;
A=
Z3
We have for A
For B;
For C;
Overall T :
P
B=
Z1 Z3
;
Z3
C=
1
;
Z3
2 7
D=
2
3
Z2 + Z3
Z3
7
6 5
5
1
1
5
2 9 54 3
6 6 6 7
TB = 4
5
1 10
6 6
#
"
1 0
TC = 1
1
7
TA = 4
T = [TA ][TB ][TC ] =
4:709 15:93
0:962 3:46
This dervation is left as an exercise to the reader.
Verification:
Using T Δ transformation, that is changing T (3, 4, 6 ) of Fig. R.P. 7.13, in to Δ,
Zxz = 13:5
Zxy = 9
Zyz = 18
Figure R.P.7.13(a)
Putting the values in the circuit of Fig. R.P. 7.13, we get
Figure R.P.7.13(b)
Two Port Networks
Reducing, the above circuit, we get the circuit shown in Fig. R.P. 7.13c.
Figure R.P. 7.13(c)
Converting the circuit into T, we get the circuit shown in Fig. R.P. 7.13(d).
Now from Fig. R.P. 7.13(d),
3:8564 + 1:0396
= 4:709
1:0396
1:0396(3:8564 + 2:5644) + 3:8564 2:5644
B=
1:0396
= 15:93 Ω
1
1
C=
=
= 0:962
Zp
1:0396
2:5644 + 1:0396
D=
= 3:46
1:0396
A=
Figure R.P. 7.13(d)
Exercise Problems
E.P
7.1
Find the y parameters for the network shown in Fig. E.P. 7.1.
Figure E.P. 7.1
Ans:
y11 =
α + RA + RB
1
; y12 =
; y21 =
RA RB
RB
(α + RA )
1
; y22 =
:
RA RB
RB
j
575
576
j
E.P
Network Theory
7.2
Find the z parameters for the network shown in Fig. E.P. 7.2.
Figure E.P. 7.2
Ans:
E.P
z11 =
13
Ω;
7
z12 =
2
Ω;
7
z21 =
2
Ω;
7
z22 =
3
Ω:
7
7.3
Find the h parameters for the network shown in Fig. E.P. 7.3.
Figure E.P. 7.3
Ans:
h12
E.P
sC) R) R* + R) + (1
sC) R* + 1
sC) R*
;
=
sC) R* + 1
h11 =
m)R*
;
h22
7.4
Find the y parameters for the network shown in Fig. E.P. 7.4.
Figure E.P. 7.4
Ans:
y11 = y22 =
7
S;
15
y12 = y21 =
2
S.
15
sC) R* + m
.
sC) R* + 1
sC)
=
.
sC) R* + 1
h21 =
Two Port Networks
E.P
j
577
7.5
Find the y parameters for the network shown in Fig. E.P. 7.5. Give the result in s domain.
Figure E.P. 7.5
Ans:
E.P
y11 = y22 =
2s(2s + 1)
;
4s + 1
y12 = y21 =
4s2
.
4s + 1
7.6
Obtain the y parameters for the network shown in Fig. E.P. 7.6.
Figure E.P. 7.6
Ans:
E.P
y11 = 0:625 S;
y12 =
0:125 S;
y21 = 0:375 S;
y22 = 0:125 S:
7.7
Find the z parameters for the two-port network shown in Fig. E.P. 7.7. Keep the result in s domain.
Figure E.P. 7.7
Ans:
z11 =
2s + 1
;
s
z12 = z21 = 2;
z22 =
2s + 2
.
s
578
j
E.P
Network Theory
7.8
Find the h parameters for the network shown in Fig. E.P. 7.8. Keep the result in s domain.
Figure E.P. 7.8
Ans: h11 =
E.P
5s + 4
;
2(s + 2)
h12 =
s+4
;
2(s + 2)
h21 =
(s + 4)
;
2(s + 2)
h22 =
s
:
2(s + 2)
7.9
Find the transmission parameters for the network shown in Fig. E.P. 7.9. Keep the result in s
domain.
Figure E.P. 7.9
Ans: A =
E.P
3s + 4
;
s+4
B=
2s + 4
;
s+4
C=
4s
;
s+4
D=
3s + 4
:
s+4
7.10
For the same network described in Fig. E.P. 7.9, find the h parameters using the defining equations.
Then verify the result obtained using conversion formulas.
2s + 4
s+4
(s + 4)
4s
;
h12 =
;
h21 =
;
h22 =
:
Ans: h11 =
3s + 4
3s + 4
3s + 4
3s + 4
Two Port Networks
E.P
j
579
7.11
Select the values of RA , RB , and RC in the circuit shown in Fig. E.P. 7.11 so that A =1, B = 34 Ω,
C = 20 mS and D = 1.4.
I1
I2
V1
V2
Figure E.P. 7.11
Ans:
E.P
R) = 10Ω;
R* = 20Ω;
R+ = 50Ω:
7.12
Find the s domain expression for the h parameters of the circuit in E.P. 7.12.
Figure E.P. 7.12
Ans:
E.P
h11 =
1
s
C
s2 +
1
LC
;
h12 = h21
1
LC ;
=
1
2
s +
LC
h22
2
Cs
+
LC
=
1
s2 +
LC
s2
7.13
Find the y parameters for the network shown in Fig. E.P. 7.13.
Figure E.P. 7.13
Ans:
y11 = 0:04S;
y12 =
0:04S;
y21 = 0:04S;
y22 =
0:03S:
:
580
j
E.P
Network Theory
7.14
Find the two-port parameters h12 and y12 for the network shown in Fig. E.P. 7.14.
Figure E.P. 7.14
Ans: h12 = 1:2;
E.P
y12 = 0:24S:
7.15
Find the ABCD parameters for the 4Ω resistor of Fig. E.P. 7.15. Also show that the ABCD
parameters for a single 16Ω resistor can be obtained by (ABCD)4 :
Figure E.P. 7.15
Ans: Verify your answer using the relation between the parameters.
E.P
7.16
For the T -network shown in Fig. E.P. 7.16. show that AD
Figure E.P. 7.16
BC =1.
Two Port Networks
E.P
j
581
7.17
Find y21 for the network shown in Fig. E.P. 7.17.
Ans:
E.P
y21
Figure E.P. 7.17
s
=
:
4s + 1
7.18
Determine the y-parameters for the network shown in Fig. E.P. 7.18.
Figure E.P. 7.18
Ans:
E.P
y11 =
s3
s2
+ + 2s + 1
;
s(s2 + 2)
y12 = y21 =
1
;
2
s(s + 2)
y22 =
s3 + s2 + 2s + 1
:
s(s2 + 2)
7.19
Obtain the h-parameters for the network shown in Fig. 7.19.
Figure E.P. 7.19
Ans:
E.P
h11 =
30
Ω;
11
h21 =
1
;
11
h12 =
1
;
11
h22 =
4
11
7.20
The following equations are written for a two-port network. Find the transmission parameters for
the network. (Hint: use relation between y and T parameters).
I1 = 0:05V1
0:4V2
I2 =
0:4V1 + 0:1V2
582
j
Network Theory
E.P
7.21
Find the network shown in figure, determine the z and y parameters.
Figure E.P. 7.21
Ans: y11 = 4 ;
z11 = 1Ω;
E.P
y22 = 3;
4
z22 = Ω;
3
y12 = y21 =
3,
z12 = z21 = 1Ω:
7.22
Determine the z, y and Transmission parameters of the network shown in Fig. 7.22.
Figure E.P. 7.22
3
,
55
= 20Ω,
Ans: y11 =
z11
4
,
55
= 15Ω
y22 =
z12 = z21
z22
A = 55Ω
E.P
1
,
55
= 5Ω
y12 = y21 =
B = 55Ω
C = 0.2,
D = 3.
7.23
For the network shown in Fig. E.P. 7.23 determine z parameters.
Figure E.P. 7.23
Ans: z11 =
2s(5s + 1)
2s
2s3 + 5s2 + 3s + 5
=
z
=
=
,
z
,
z
12
21
22
2s2 + 5s + 1
2s2 + 5s + 1
2s2 + 5s + 1
The unit and S are same
Two Port Networks
E.P
7.24
Determine the y parameters of the two-port network shown in Fig. E.P. 7.24.
Figure E.P. 7.24
Ans:
y11 =
1
,
4
y21 =
1
,
4
y12 =
5
,
4
y22 =
4
.
3
j
583