Shota A

Shota A. Nemsadze
Approved by the Educational
Methodical Council of the GTU
Tbilisi 1999
This textbook is meant for students of communication, information
and power engineering of technical universities.
It covers description of electric circuits and its elements, main difinitions, Ohm`s and Kirchhoff`s laws, topological descriptions of networks
It includes analysis of resistive circuits with DC drivers, steadystate performances in single-phase and multi-phase sinusoidal current
circuits, methods of their computation.
It covers also description of resonance phenomena and resonance
characteristics of electric circuits.
Examples and problems are given at the end of each section.
Reviewers: Prof. G.L.Chikovani
Ass. Prof. N.V.Tsivtsivadze
"Technical University" Publishing House, 2000
Contents
Chapter
1. Main Definitions. Elements of Electric Circuits
Electric Circuits, Voltage and Electromotive Force – Coulomb’s Law – resistance –
-Ohm’s Law - Energy Sources – Inductance – Mutual Inductance – Faraday’s Law –
-Capacitance – Units and scaling.
2.
Topological Description of Networks. Kirchhoffs’ Laws
Topological Description of Networks - Kirchhoffs’ Laws – Inductance and
Capacitance in Series and Parallel in AC Circuits.
3.
Analysis of Resistive Circuits with DC Drives
Series and Parallel Connection of Resistors – Voltage and Current Dividers –
-Voltage and Current Sources in Series and Parallel – Ladder Networks –
-Star (wye) – Delta Transformation – Bridge Networks
4.
Single-phase Sinusoidal AC Circuits
Generation of Sinusoidal emf - Root Mean Square and Average Values of
Sinusoidal Quantities - Complex Representation of Sinusoidal Quantities –
- Addition, Subtraction, Multiplication and Division of Complex Numbers- A Sinusoidal Current Through a Resistor and an Inductor and a Capacitor- Series Connected R,L and C Circuit - The Parallel G-L-C Circuit - Active,
Reactive, Apparent and Complex Powers - Two-terminal Networks
5.
Steady-State Analysis of Mufti Mesh Networks
Application of Kirchhoffs’ Laws – Mesh-Current Analysis of Networks –
- Node Voltage Analysis – Network Theorems. Superposition Theorem –
- Mutual Inductance in Series Connection of Coils – The Transformer.
Transferred Resistance and Reactance
6.
Resonance
Introduction – Voltage Resonance – Resonance Characteristics of the Series
Circuit – Bandwidth – Current Resonance – Resonance Characteristics of the
Parallel Circuit - Duality
7.
Three-phase Circuits
Introduction – Classification of the Three-phase Circuits and Systems –
- Advantages of Three-phase Systems – Star-to-star Connection – Delta -Connected Three-phase Circuits – Calculation of Unsymmetrical
Three-phase Circuits with Star-to-star Connection – Power Calculations
8.
References
Pages
Main Definitions. Elements of Electric Circuits
1.1 Electric Current, Voltage and Electromotive Force
An electric circuit is an interconnection of energy sources and energy converters(or loads) through
which an electric current can flow and where electromagnetic processes may be described by the integral
terms of current, electromotive force(emf), voltage, resistance, inductance and capacitance.
Energy sources are devices converting chemical, mechanical or other forms of energy into electrical
energy.
Energy converters or loads are devices transforming electrical energy in other forms of energy
(chemical, mechanical, heat, light)
An electric current is defined as a time rate of change of charge passing through a specified area. It’s
expressed mathematically as:
dq
1.1
,
i=
dt
where i denotes the instantaneous electric current and q the net charge, which may be positive as well as
negative kind. In metallic conductors the current is a movement of electrons in one direction. In salt
solutions the current is a movement of positive ions in one direction and negative ions in opposite
direction. In semi-conductors the current is a movement of electrons in one direction and a movement of
positively charged holes in the opposite direction.
The flow of charged particles is caused by electric field, established by the energy source of a given
circuit. Any material that allows the essentially free passage of current, when it is placed in electric field,
is called a conductor. As a rule there is a lot of free electrons in metals and that’s why gold, silver, copper,
aluminium, tin and etc. are good conductors.
Materials that possess low electrical conductivity are classified as nonconductors and are better
known as insulators or dielectrics. A vacuum is the only known perfect dielectric. All other insulating
materials, such as glass, rubber, ceramic, wood, paper, cloth, air are imperfect dielectrics.
The free electrons in conductors are forced to move in the direction of the electromotive force (emf) of
the energy source. The direction of this electron movement is from the negative terminal of the energy
source, through the conductor, to the positive terminal. Nevertheless the direction of the current in all
circuits has been standardized as going from positive to negative terminal.
The SI unit of charge is Coulomb (C), the unit of time is second (s) and the unit of current is Ampere
(A).
Since the electron has a charge of 1.6021 * 10-19 C it follows that a current of 1A corresponds to the
motion of 6.24 *1018 electrons pass any cross section area in one second.
An electromotive force is a main characteristic of an energy source and is defined as the ratio of
work to be done by the external forces to transfer the positive charge from the negative terminal to the
positive terminal of the energy source to the above charge. We note this value by “e”. The direction of
emf is indicated by the arrow. The head of the arrow corresponds to the plus terminal of the source and
the tail to the minus terminal.
The work done to move the charge on the distance “l” by action of electric field on the charge is
numerically proportional to the electric field strength “E” and the magnitude of the charge “q”
w = q * El .
1.2
The ratio of the work needed to move the charge along the path joining two points of the circuit to
the magnitude of the charge is determined as the voltage or voltage drop between these points
w
v=
1.3
q
If the work is measured in Jouls and the charge in Coulombs, then the voltage is measured in Volts.
The unit of emf is also Volt.
The voltage drop and the current have the same direction
If a differential amount of the charge dq corresponds a differential increase in energy dw, the
potential (or voltage) of the charge is increased by the amount
dw
v=
.
1.4
dq
If this voltage is multiplied by the current dq/dt, as
dw dq dw
*
=
=p
dq dt
dt
the result is seen to be a time rate of change of energy, which is the power “p”. This power is the product
of voltage and current
p = vi .
1.5
The SI unit of power is Watt. Thus Watt is the product of Volt and Ampere or Joule over the second.
Energy as a function of time is found by integrating above equation
t
w = ∫ pdt .
1.6
0
If electric current, voltage and emf are constant values, electric circuit is called as direct current
circuit (DC circuit) and these values are noted by capital letters I, V, E. In the case when they are the
time functions the small letters i, v, e are used and the electric circuit is called the alternating current
circuits (AC circuits).
1.2 Coulomb’s Law
This law is based on experiments of Coulomb and states that the force existing between two point
positive (or negative) charges q1 and q2 is
qq
1.7
F = 1 22 ,
4πεr
where
r = the distance between charges,
ε = a constant related to the surrounding medium and called the absolute electric
permittivity.
If r is in meters, q - in Coulombs, then F – is in Newtons.
We may also note that
ε = ε 0ε r ,
where εr= dimensionless value and is called the relative permittivity.
ε0 = the permittivity of free space.
Farads
1
ε0 =
≈ 8.85 * 10 −12
.
9
metre
36π * 10
1.3 Resistance
The passage of electrons through a material is accompanied by the collisions of the electrons with
other atomic particles. More over, these collisions are not elastic, and energy is lost in each collision
generating heat. This loss in energy per unit charge is interpreted as voltage drop. The property of
material to convert electrical energy to heat energy is called resistance.
Electric filament lamps, resistor furnaces and other similar loads have metal conductors carrying
electric current, in which the electric energy is converted to heat. They are characterized by the resistance
parameter “R”. The symbol of the resistor circuit element is shown in Fig. 1.1a
The SI unit of resistance is Ohm(Ω).
If the resistance parameter R doesn't depend on value of current or voltage the resistor element
V
i
V
R=const
R=var
R
i
0
a)
i
0
b)
c)
Fig. 1.1
is the linear element. The voltage-current characteristic of such resistors is represented by a straight line
shown in Fig. 1.16. An electric circuit composed of such elements is said to be linear.
If the resistance parameter depends on value of current or voltage the resistor element is said to be
nonlinear. The voltage-current characteristic of such resistors is shown in Fig.1.1c. Nonlinear resistor is
called the varistor. If among the circuit elements is at least one nonlinear element the electric circuit is
said to be nonlinear.
The reverse value of the resistance is called as the conductance and is symbolizes by letter “G”. The
SI unit of conductance is Siemen.
The resistance of a piece of metal conductor is:
l
1.8
R=ρ ,
s
where ρ = the resistivity of material of conductor,
l= the length,
S= the cross sectional area of a conductor.
The resistivity and consequently resistance of a conductor depends on the temperature of the
material. If the resistance at one temperature is known, the resistance at some other temperature may be
determined by the following formula:
RH = RL [1 + α (TH − TL )],
1.9
where RH = resistance at high temperature,
RL= resistance at lower temperature,
TH= higher temperature,
TL = lower temperature,
α = the temperature coefficient of resistance.
The temperature coefficient of metals is positive value, that’s why when the temperature is increased
the resistance is also increased.
1.4. Ohm’s Law
This law states that the value of the current in a resistor element is equal to the ratio of the applied
voltage to the resistance of the resistor element:
v
1.10
i= .
r
Using the conception of the conductance Ohm’s law may be written as
1.11
i = Gv
The instantaneous power of the resistor element is
p = vi = i 2 R = Gv 2 .
So it is always positive value. It means that this circuit element always gets energy from power
supply.
1.5. Energy Sources
Several kinds of energy sources are used in practice. Models or idealizations of such sources are
called active elements which classified by their voltage-current characteristics. Two basic models are the
voltage source and the current source.
The voltage source is assumed to deliver energy with a specified terminal voltage, v(t). In the
case of ideal voltage source the terminal voltage is independent of the current from the source. The ideal
voltage source has no internal resistance RS=0. A real voltage source may be represnted by the model
shown in Fig. 1.2a consisting of a source in series with an internal resistor.
V
R=0
e
V(t)
R=0
R
i
a)
b)
Fig.1.2
So the voltage source is characterized by emf and internal resistance. The voltage at its terminals is
v = e − Rs i.
1.12
The voltage-current characteristic of the voltage source is shown in Fig. 1.2b in the case of the ideal
source (Rs=0) and in the case of the real source (Rs≠0).
The current source is assumed to deliver energy with a specified current through terminals i(t). The
symbol for the current source differs from those for the voltage source and is shown in Fig1.3a. In the
case of an ideal current source a load current is independent of the terminal voltage and is equal to the
current is current source. In the case of a real current source the load current is:
1.13
i = i s − Gv ,
where G is the internal conductance of the current source. The current-voltage characteristic for both
cases are shown in Fig.1.3b
i
i
G =0
i
G
G =0
V
0
V
Fig.1.3
The voltage source is said to be idle if the output terminals are open such that i(t)=0. When output
terminals voltage v(t)=0 with the source turned on it is said to be in short circuit. Similarly the current
source is said to be idle when the output terminals are shorted v=0, is=i.
1.6 Inductance
It’s known that magnetic field is produced by moving elementary particles caring an electric charge.
So the magnetic field surrounds a conductor and is set up by the current flowing through the conductor.
The intensity of the magnetic field depends on the magnitude of current carried by conductor: the
grater the current the stronger the field set up by it.
The field intensity is evaluated by the vector quantity denoted by B and called as the magnetic
induction or magnetic flux density of the field.
The unit of magnetic flux density is Tesla.
The magnetic field can be visualized in terms of imaginary lines of magnetic flux density drawn so
that tangent to them at any point of their length gives the direction of the flux density vector at that point.
The magnetic lines of flux density due to a straight current-caring conductor are circles with the
centers of the circles on the axis of the conductor and is shown in Fig.1.4. The lines of the flux density
give a picture of magnetic field.
The magnetic field picture of a coil (solenoid) is shown in Fig.1.5. The direction of the magnetic lines
of force is determined by the corkscrew rule, which states that if the direction of thrust of a right-hand
corkscrew follows the current flowing through a conductor, the direction of twist of the corkscrew handle
will indicate the direction of the magnetic lines of force.
Fig. 1.6
Fig.1.5
In order to describe magnetic field a magnetic flux is used. An elementary magnetic flux through
the surface ds is:
r r
dΦ = B * ds
1.14
Fig.1.5.
The vector ds has the magnitude equal to the elementary area ds and direction of the normal
N of the surface S. The magnetic flux of the vector B through the surface S is:
r r
Φ = ∫ B ds .
1.15
The unit of flux is Weber (Wb).
In the general case the coil turns may be linked with the different fluxes as it is shown in Fig.1.5. In
this case the conception of flux is not enough to describe magnetic field and a conception of a magnetic
flux linkage must be introduced. The flux linkage of a single turn is numerically equal to the flux passing
through the surface bounded by the given turn. In more complicated case, as
shown in Fig.1.5, the flux linkage is the algebraic sum of the fluxes linked with each turn. When every
magnetic flux density lines link with all turns of a coil the flux linkage is:
ψ = wΦ,
1.16
where w-is the number of turns of the coil.
The unit of flux linkage is Weber-turns.
Experience shows that flux linkage is proportional of the current flowing through the coil
ψ = L i,
1.17
where L-is inductance.
So inductance numerically equals to the ratio of the flux linkage to the current. The SI unit of
inductance is Henry(H).
Experience shows that inductance depends on the shape and the size of conductor (geometry of
conductor) and also the medium in which the magnetic field is set up
L = f ( g , µ ),
1.18
where µ = µr µ0 - the absolute permeability of the medium,
µr = a dimensionless value and is called the relative permeability.
µ0 = the magnetic permeability of free space (the magnetic constant).
0
The symbol of inductance
is shown in Fig.1.7a
The inductance of a coil or loop with an air or ferromagnetic core is constant and independent on the
current and flux linkage involved. The main characteristic of a coil or an inductor is weber-
w
w
L
L=var
L=const
i
0
a)
i
0
b)
c)
Fig.1.7
ampere characteristic shown in Fig.1.7.b. In the above case it’s the straight line and is the characteristic of
a linear circuit element- a linear inductor.
If the coil has got the ferromagnetic core it is said to be non-linear circuit element. Its weber-ampere
characteristic is shown in Fig.1.7c. The inductance of such element is variable value and is function of
current.
1.7 Mutual Inductance
Now let’s consider two coils brought up to each other so that magnetic flux due to the current i1
in the first coil partly links the turns of the first coil (Φ11) and partly the second coil (Φ12 ):
Φ1 = Φ11 + Φ12 .
1.19
Similarly the magnetic flux due to the current i2 in the second coil partly links the turns of the
Fig.1.8
second coil (Φ22) and partly the first coil (Φ21 ).
Φ 2 = Φ 22 + Φ 21 .
1.20
The total flux linkage of the first coil is
ψ 1 = w1 (Φ1 ± Φ 21 ) = ψ 11 ± ψ 21 ,
1.21
where w1 –is the number of turns of the first coil.
The total flux linkage of the second coil is
ψ 2 = w2 (Φ 2 ± Φ12 ) = ψ 22 ± ψ 12 .
1.21
Here ψ11 and ψ22 are flux linkages of self induction of coils; ψ12 and ψ21 are flux linkages of mutual
induction of coils.
If the flux due to mutual inductance is in the same direction of the flux due to self induction in a
given coil the second term on the right hand side in the above equation should be taken positive. If not,
the term should be negative.
Experience shows that the mutual flux linkages of the first and second coils are:
ψ 21 = M 21i2 ,
ψ 12 = M 12 i1 ,
1.22
where µ21=µ12=µ are coefficients those to be depend on the number of turns, the shape, size(geometry)
and position of coils each to other. These values also depend on the magnetic permeability of the medium
in which the magnetic field is set up and is called as the mutual inductance.
µ = f ( g , µ ).
1.23
The unit of mutual inductance is Henry(H).
1.8 Faraday’s Law
In 1831 English scientist Michael Faraday discovered the phenomenon of electromagnetic induction.
Faraday’s law states that a change in a magnetic flux linking a closed electric circuit produces emf that to be directly proportional to the rate of change of flux linkage
dψ
1.24
e=−
.
dt
The negative sign in this expression is due to Lenz and is to indicate that the direction of the induced
current is always such as to oppose the action which produced it(this reaction is known as Lenz’s law).
In the case of changing a flux of self-induction the emf of self-induction is
dψ L
di
1.25
= −L .
eL = −
dt
dt
It’s supposed that L=const.
The voltage on the terminals of a coil is
di
1.26
v L = −e L = L .
dt
This equation makes relationship between voltage and current through the ideal coil (R=0). It allows
to find voltage if current is known
ψL 1 t
1.27
iL =
= ∫ v L dt + i L (0) ,
L
L0
where iL(0) is the value of current at t=0.
In the case of changing a mutual flux linkage emf of mutual inductance is produced
di
di
1.28
e1M = − M 2 ,
e2 M = − M 1 .
dt
dt
Here e1m is the emf of mutual induction induced in the first coil and e2m is the emf of mutual
induction induced in the second coil. Again it’s supposed that M=const.
In the circuit applications emf-s of mutual induction must be balanced by corresponding voltage
drops.
The inductance parameter takes into account the fact of storing energy in magnetic field that to be set
around of the circuit element.
A power of the self-induction circuit element is
di
1.29
pL = vL i = L i
dt
and the energy stored in the magnetic field will be
t
t
Li 2
1.30
.
wL ∫ p L dt = ∫ Lidi =
2
0
0
Thus in spite of a current may be positive or negative the energy of the magnetic field is always
positive because it’s proportional to the square of the current.
1.9 Capacitance
A circuit part that accumulates energy into the electric field is called a capacitor and is characterized
capacitance parameter “C”. This parameter is defined as the ratio
q
1.31
.
v
The ideal capacitor element is a capacitor with an ideal insulator. It’s the system of two isolated
bodies. Electric charge of each body is
q = C v.
1.32
If a capacitance is the constant value the capacitor is said to be linear. In the case when capacitance
depends on the voltage or electric charge the capacitor is said to be non-linear. The symbol of capacitor is
shown in Fig.1.9a.
C=
q
C = var
C = const
0
v
a)
b)
Fig.1.9
The main characteristic of a capacitor is coulomb-voltage characteristic. The coulomb-voltage
characteristic of linear and nonlinear capacitors are shown in Fig.1.9b.
In any case capacitance is a function of geometry of the conductors and absolute permittivity of the
insulator
C = f (g, ε ).
1.33
The SI unit of capacitance is Farad (F).
In order to get relationship between voltage and current in a capacitive system let’s use the
relationship of charge and current given by the equation
dq d
1.34
i=
= (Cv ) .
dt dt
If the capacitance doesn’t vary with time then
dv
1.35
i=C .
dt
Similarly starting with the equation v=q/c we find that
t
1
1.36
vC = ∫ idt + vC (0) ,
C0
where v c (0) is voltage at the time t=0.
The energy stored in the electric field of the capacitor is
Cv 2
1.37
.
2
0
0
So the capacitive parameter takes into account the fact of storing energy in electric field that to be set
around the conductor.
t
t
wC = ∫ pC dt = ∫ Cvdv =
1.10 Units and Scaling
The relationships discussed in this chapter are summarized compactly in tabular form in Table 1.1.
These equations are encountered so frequently in the study of electric circuit theory that they should be
memorized.
Table 1.1
Parameter
Symbol
Voltage–current
Energy
Basic relationship
relationship
V=Ri
R or G
ψ=Li
L or M
V=Ri; i=v/R
V=i/G; I=Gv
V=Ldi/dt;
t
1
i = ∫ vdt + i (0)
L0
w = ∫ v i dt
w=
Li 2
2
t
v=
C
q=Cv
1
idt + v(0);
C ∫0
i=C
dv
dt
Cv 2
w=
2
The units are summarized in Table 1.2 These units are part of International system of Units (SI). In
engineering applications other units which are multiples or submultiples of these units are shown in Table
1.3. For example in electric circuits capacitance is usually expressed in microfarads or in Pico farads,
inductance in millihenries, etc
Quantity
Charge
Current
Voltage
Electromotive force
Power
Energy
Resistance
Conductance
Inductance
Capacitance
Time
Frequency
Symbol
q
i,I
v,V
e,E
p,P
w,W
R
G
L
C
t
f
Unit
Coulomb
Ampere
Volt
Volt
Watt
Joule
Ohm
Siemens
Henry
Farad
second
Hertz
Table1.2
Unit abbreviation
C
A
V
V
W
J
Ω
Sm
H
F
s
Hz
Table1.3
Factor by which the unit is
multiplied
1012
109
106
103
102
10
10-1
10-2
10-3
10-6
10-9
10-12
Prefix
tere
giga
mega
kilo
hecto
deka
deci
centi
milli
micro
nano
pico
Symbol
T
G
M
K
h
da
d
c
m
µ
n
p
Problems:
1.1.Determine the resistance of a 560 m length of copper conductor (ρ=1.724.0 Ωm) whose
rectangular cross section is 40mm by 20mm.
1.2. Determine the voltage required to cause 150 µA in a 10MΩ resistor.
1.3.A coil has an inductance 10mH. Find energy of magnetic field at current 15µA.
2. Topological Description of Networks, Kirchhoffs’ Laws
2.1. Topological Description of Networks
Circuit or network analysis is made on dealing with graphic representation of the actual circuit
under consideration. Such graphic representation of the circuit is called the circuit diagram. It must be
noted that the word network is used synonymously with the term “circuit”. The circuit diagram contains
all circuit elements using standard graphical symbols. Some of them are given in the previous chapter.
The section of the electric circuit along which current remains constant is called a branch of the circuit. A
junction point where three and more branches meet is called a node. Branches with no sources of electric
energy in them are said to be passive, whereas branches containing such sources are referred to as active
branches. A close path of circuit made up of a several branches is called a mesh or loop of the circuit. As
an example a multi-mesh network’s diagram is shown in Fig.2.1.
a
•
R3
e1
e2
i3
i2
i1
L3
R1
R2
L2
C1
•
•
•
b
d
Fig.2.1
There are two nodes ”a” and “b”. Further we will note the number of nodes by “n”. So in this case
n=2. The number of branches b=3. There are branches a-c-b with circuit elements e1, R1, and C1; a-b with
elements L3 and R3; a-d-b with elements e2, R2 and L2. There are also several meshes (loops) a-c-b-a, a-bd-a, a-c-b-d-a. The number of meshes is denoted with “m”.
As it’s shown in Fig.2.1 all circuit elements included in the branch are carrying the same current and
are said to be connected in series. Branches connected a single pair of nodes are said to be in parallel and
are at one and the same voltage.
Let’s replace all circuit elements of the network with lines, constructing a skeleton of the network.
Such graphical representation of the network is called the graph and is shown in Fig.2.2a. If at the same
time indicate a reference direction of the currents by an arrow for each of the lines of the graph, then it is
known as an oriented graph. An example of an oriented graph is shown in Fig.2.2b
c
•
•
a
•
d
•
c•
•
b
a)
a
→
d
• →
b
→
b)
c•
Fig.2.2
The lines in the graph are identified as branches. The junction of three and more branches is known as
a node. Thus, graphs like electric networks are composed of nodes and branches. They are used in
describing the topological properties of the networks, which are unaffected when we stretch, twist or
otherwise distort the size and shape of the network, for example, by shortening or lengthening the
connection wires between elements. Fig 2.3 shows some topological structures occurring so frequently in
electrical engineering that they are given special names. Several of these networks are shown in Fig.2.4
Fig 2.3
and the names given are: (a) T-network, (b) π-network, (c)ladder network, (d) bridge network, (e) lattice network.
•
•
•
•
•
•
•
•
•
•
•
•
a)
b)
•
c)
•
•
•
•
e)
•
d)
Fig 2.4
Several topological terms of importance in network analysis will be outlined in this section for
application in the following chapters.
A loop (mesh) is called closed path in a circuit (graph) formed by a number of connected
branches.
A sub graph of a given graph is formed by removing branches from the original graph. A sub
graph of importance in our study is a tree. The tree of the graph contains all of the nodes of the
graph but it doesn’t contain any loops. Here are several trees of graph of Fig 2.2 in Fig 2.5
Fig.2.5
Branches removed from the graph for forming a tree are chords or links. Thus for a graph
with “b” branches and “n” nodes n - number of chords = b-(n-1).
2.2 Kirchhoff’s Laws
The usage of Kirchhoffs’ laws makes it possible to find the branch currents and voltages in circuits of
any configuration and complexity.
Kirchhoff’s first (current) law applies to circuit nodes and states that the algebraic sum of the currents meeting at
the node of an electric circuit is zero
n
∑i
k =1
k
= 0.
2.1
The currents include in the above sum differ in sign according to whether they flow to or from node.
On the basis of Kirchhoff’s current law an equation may be written for the currents at any node of a
circuit. Referring to Fig 2.1 the network equations at nodes a will be
− i1 + i 2 + i3 = 0,
2.2
where the positive sign is allocated to currents leaving the node and the negative sign to those entering the
node. From the above equation it follows that
i1 = i2 + i3 .
Thus, Kirchhoff’s current law can be stated in a different way, namely, that the total current entering
any node of a circuit equals to the total current leaving that node.
Kirchhoff’s second (voltage) law applies to loops (meshes) and states that the algebraic sum of the
e.m.f.s around a closed loop is equal to the algebraic sum of the voltage drops across the circuit elements
in that loop
n
∑e
k =1
k
= ∑ vk .
2.3
k =1
In order to write network equations in accordance with Kirchhoff’s voltage law we have to choose an
arbitrary moving direction round the loop (for example-clockwise). The terms enter the respective sum
with the positive sign if they coincide the direction of summation round the loop, and with the negative
sign, if they are in the opposite direction. Putting the e.m.f.-s on the left-hand side and the voltages on the
right-hand side of the equations for the loops acba and abda of Fig 2.1 we’ll have
t
di
1
e1 = R1i1 +
i1 dt + vC1 (0) + L1 1 + R3i3 ,
∫
C1 0
dt
2.4
di3
di2
− R3 i3 − L3
.
e 2 = R 2 i 2 + L2
dt
dt
2.3. Inductors and Capacitors in series and Parallel in AC Circuits
Suppose several inductors are connected in series as it’s shown in Fig 2.6
Fig 2.6
Applying Kirchhoff’s voltage law to the circuit in Fig 2.6
di
di
di
di
v = v1 + v 2 + L + v n = L1 + L2 + L + Ln
= Leq ,
dt
dt
dt
dt
where the equivalent inductance
2.5
n
Leq = ∑ Lk .
2.6
k =1
In the case of parallel connection of inductors as it’s shown in Fig.2.7 if the initial currents of
Fig.2.7
inductors are zero, accordance with Kirchhoff’s current law may be written as
1
1
1
1
i = i1 + i2 + L + in = ∫ vdt + ∫ vdt + L +
vdt =
vdt ,
∫
L1
L2
Ln
Leq ∫
where the inverse value of equivalent inductance is
n
1
1
=∑ .
Leq k =1 Lk
Now let’s suppose that “n” capacitors are connected in series as it is shown in Fig. 2.8
2.7
2.8
Fig.2.8
Taking into account the zero initial voltages of capacitors accordance with Kirchhoff’s voltage law
can be written as
1
1
1
1
v = v1 + v 2 + L v n =
idt +
idt + L +
idt =
idt ,
2.9
∫
∫
∫
C1
C2
Cn
C eq ∫
where
n
1
1
=∑ .
2.10
C eq k =1 C k
In the case of parallel connection of capacitors, as it’s shown in Fig.2.9, accordance with
Kirchhoff’s current law
Fig.2.9
the input current is
dv
dv
dv
dv
+ C2
+ L + Cn
= C eq
,
dt
dt
dt
dt
where the equivalent capacitance is
i = i1 + i2 + L + in = C1
2.11
n
C eq = ∑ C k .
2.12
k =1
Problems: 2.1. Assume the voltage sources in Fig.2.10 are E1=50V, E2=150V and E3=200V and
resistors are R1=5Ω, R2=15Ω, R3=20Ω, R4=10Ω. Determine current and voltages across
each resistor.
Fig.2.10
2.2. Determine the equivalent of the following series connected inductors: 0,5H, 250mH,
1.2H and 50mH.
2.3 Determine the equivalent capacitance of the following series-connected capacitors: 0,1µF,
550mF, 50⋅103pF and 0,05µF.
3. Analysis of Resistive Circuits with DC Drivers
3.1 Series and Parallel Connection of Resistors
A series circuit, such as shown in Fig. 3.1 is a closed loop of driver and resistors having the same
current through every component in the loop.
Fig. 3.1
According to Kirchhoff’s voltage law
E = IR1 + IR2 + L + IRn = IReq ,
the equivalent resistance of the series connected resistors is
where
n
Req = ∑ Rk .
3.2
k =1
If several resistors are connected in parallel as shown in Fig.3.2, they also may be replaced by the
equivalent resistor.
Fig. 3.2
According to Kirchhoff’s current law
I = I 1 + I 2 + L + I n = EG1 + EG 2 + L + EG n = EG eq ,
where the equivalent conductance of the parallel-connected resistors is
3.3
Geq = ∑ Gk .
3.4
Req = 1 Geq .
3.5
k =1
The equivalent resistance is
If two resistors are connected in parallel
Req =
RR
1
1
=
= 1 2 .
G1 + G2 1 R1 + 1 R2 R1 + R2
3.6
3.2 Voltage and Current Divider
A voltage divider, also called a potential divider, is a device used to provide an output voltage that is lower than
the input or driving voltage. A fixed voltage divider with two series-connected resistors is shown in Fig. 3.3
Fig. 3.3
The output voltage is
R2
.
R1 + R2
A simple current divider is shown in Fig. 3.4, which consists of two resistors.
Vout = IR2 = Vin
3.7
Fig. 3.4
Currents in the resistors are
I 1 = VG1 =
G1
I
G1 = I
,
Geq
G1 + G2
G2
I2 = I
.
G1 + G2
3.8
Example 3.1. In Fig. 3.3 R1=50Ω and R2=150Ω. Using the voltage divider equation determine the
output voltage if the input voltage is 100V.
Solution:
R2
150
Vout = Vin
= 100
= 75V .
R1 + R2
50 + 150
Example 3.2 In Fig. 3.4 G1=0.01Sm and G2=0.04Sm. Using current divider equation determine the
current in the first resistor if the current is 2.5A.
Solution:
I1 = I
G1
0.01
= 2.5
= 0.5 A.
G1 + G2
0.01 + 0.04
3.2 Voltage and Current Sources in Series and Parallel
If several voltage sources are connected in series they may be replaced by the equivalent voltage
source.
Fig. 3.5
The e.m.f. of the equivalent voltage source is
3.10
E eq = E1 − E 2 + E 3
and the internal resistance of the equivalent voltage source is
3.11
Req = R1 + R2 + R3 .
If several current sources are connected in parallel, as it’s shown in Fig. 3.6 they also maybe replaced
by the equivalent current source.
Fig. 3.6
According to Kirchhoff’s current law the current of the equivalent current source is
3.12
I seq = I s1 − I s 2 + I s 3
and the internal conductance of the equivalent current source is
3.13
G eq = G1 + G 2 + G3 .
In the case of parallel connection of several voltage sources in order to find the equivalent voltage
source, at first the given voltage sources must be replaced by corresponding current sources using the
following relationships:
E
1
Gk =
.
I sk = k ;
3.14
Rk
Rk
The procedure of determination of the equivalent voltage source is shown in Fig. 3.7a and Fig 3.7b
Fig3.7a
Fig3.7b
In the case of series connection of several current sources they should be replaced by corresponding
voltage sources and then voltage sources connected in series may be replaced by the equivalent voltage
source and then by the corresponding equivalent current source.
3.4 Ladder Networks
Ladder networks such as shown in Fig. 3.8a have alternative series and parallel elements. They are
called ladder networks because of their geometry. They look like ladders. In order to find the input
resistance of a ladder network, it’s better to start at the section of the network farthest from the driver,
combining elements as you progress toward the input terminals.
Fig. 3.8
According to Fig 3.8a R5 and R6 are in series and their equivalent
R56 = R5 + R6
and R4 (Fig. 3.8b) are connected in parallel, so
R4 R56
.
Rbc =
R4 + R56
According to Fig. 3.8c R3 and Rbc are in series and their equivalent
R3,bc = R3 + Rbc .
According to Fig. 3.8d R2 and R3,bc are in parallel and their equivalent
R2 R3,bc
Rac =
.
R2 + R3,bc
At last R1 and Rac (Fig. 3.8e) are in series and the input resistance of the network is
Rin = R1 + Rac
3.5 Star (Wye)-Delta Transformation
The arrangement of the three branches in network of Fig. 3.9a is known as a star (wye) connection
and the one in Fig. 3.9b as a delta connection.
Fig. 3.9
The delta connection of the resistors may be replaced by the equivalent star connection or vice-versa,
only on condition that the operation of the unconverted part of the network remains the same as before. It
means that the currents, voltages and power in this part of the network do not change.
In the case of star circuit resistors R1 and R2 are connected in series between nodes 1 and 2. In the
case of delta connection between the same nodes are in parallel resistors R12 and R23+R31. Consequently,
the resistance between nodes 1 and 2 will also be the same for both circuits
R ( R + R31 )
.
R1 + R2 = 12 23
3.15
R12 + R23 + R31
In the same way the resistance between nodes 2 and 3 is
R2 + R3 =
R23 ( R31 + R12 )
.
R12 + R23 + R31
3.16
and the resistance between nodes 3 and 1 is
R31 ( R12 + R23 )
.
3.17
R12 + R23 + R31
Solution of these three equations connection with R1 , R2 , R3 gives the formulae by which the
resistances of the star circuit can be found from the known resistors of the delta circuit
R1 + R3 =
R23 R31
R12 R23
R12 R31
:
:
.
R3 =
R2 =
3.18
R12 + R23 + R31
R12 + R23 + R31
R12 + R23 + R31
Solution of the above three equations connection with R12, R23, R31 gives the formulae by which the
resistors of the delta circuit can be found from the known resistors of the star circuit
RR
R R
RR
R31 = R3 + R1 + 3 1 .
R12 = R1 + R2 + 1 2 ;
R23 = R2 + R3 + 2 3 ;
3.19
R2
R1
R3
Example 3.3 In Fig. 3.10 80V are applied to the terminals ab. Determine the equivalent resistance
and the feeder current I.
R1 =
Fig. 3.10
The three resistors 100Ω, 60Ω and 40Ω are delta connected between nodes 1,2 and 3 as shown in Fig.
3.10. They can be converted into equivalent star connection
R1 =
60i100
= 30Ω.
100 + 40 + 60
R2 =
40i100
= 20Ω.
100 + 40 + 60
R3 =
40i 60
= 12Ω.
100 + 40 + 60
Then the network of Fig. 3.10 is reduced to a simple structure of Fig 3.11
Fig 3.11
As it is seen there are two parallel branches, one with resistance 20+80=100Ω and the other with
12+88=100Ω . Hence, their equivalent resistance is 50Ω and the equivalent resistance of the circuit
Re q = 30 + 50 = 80Ω
Then current
I=
80
= 1A.
80
3.6 Bridge Networks
Bridge networks are series-parallel circuits that have many applications in instrumentation and control. The
bridge circuit, called the Whetstone bridge, is shown in Fig 3.12. It provides a very effective means for measuring
the resistance of an unknown resistor.
Fig 3.12
The unknown resistor Rx is connected between nodes d and c, and a galvanometer is connected
between nodes b and d. The galvanometer is generally an uncalibrated micrometer. To measure the
resistance of an unknown resistor, the slide-wise rheostat is adjusted to cause the galvanometer to obtain a
null reading. It means that I1=I2, I3=IX and the potential difference between nodes b,d is equal to zero.
Then it’s possible to write that
I 1 R1 = I 3 R3
3.20
and
I 1 R2 = I 3 R x .
3.21
Dividing the eq. 3.20 by the eq. 3.21 we have the following bridge equation
R1 R3
.
=
R2 R x
3.22
The resistance of the unknown resistor
Rx =
R 2 R3
.
R1
3.23
Example 3.4. Assuming R2 and R1 have resistance values of 1kΩ an 10kΩ respectively and the
rheostat setting that balances the bridge is 1.242 kΩ calculate the resistance of the unknown resistor.
Solutions:
R X = R rheo
R2
1.0
= 1.242
= 0.1242Ω.
R1
10.0
Problems:
3.1. Determine the equivalent resistance of the following series-connected resistors 6Ω, 20Ω, 5Ω and
30Ω.
3.2. A 150V source of negligible resistance is connected in series with the following resistors: 50Ω,
200Ω, 450Ω and 1300Ω. Determine the current, voltage drop across each resistor and the power supplied
by the source.
3.3. Determine slide setting required on a 4000Ω rheostat in order to get a 9V output from a 12V
battery. Sketch the circuit.
3.4. Determine the equivalent resistance of the following parallel-connected resistors: 10Ω, 20Ω, 5Ω,
30Ω.
3.5 Using the conductance method, (Fig 3.13) find the driving current and the power drown by the
150Ω resistor.
.
Fig 3.13
Fig 3.14
3.6. Referring to Fig.3.14, determine the current in 40Ω resistors, the power in 30Ω resistor and the
voltage drop across the 10Ω resistor.
3.7. Referring to Fig.3.15, determine the driving-point (input) impedance, feeder current and the voltage drop
across the 25Ω resistor.
Fig. 3.15
Fig. 3.16
3.8. Assuming the bridge in Fig.3.16 is balanced, determine Rx.
4 Single-phase Sinusoidal Alternating Currant Circuits
4.1. Generation of Sinusoidal emf
Electric energy is produced, transmitted and distributed mainly by means of alternating current (AC) devices and
installations. This is achieved with the use of alternators, transformers, power transmission lines and AC distribution
networks.
In general, the term AC applies to any current that varies in direction and magnitude with time. If
such a current varies periodically according to the sinusoidal law, it is called a sinusoidal AC (Fig4.1a). In
certain cases the current varies periodically according to a nonsinusoidal law (Fig 4.1b)
Fig 4.1
A sinusoidal current is produced by a sinusoidal e.m.f. Sinusoidal emf is generated by an alternator (AC
generator) of special design. Fig 4.2 represents schematic drawing of a four-pole AC generator consisting of
unmoving part-stator and a rotated part-rotor. The poles are fixed on the rotor and rotated in respect to stationary
windings embedded in specially provided slots of the stator core. The pole windings are connected in series and get
supply through two metal rings fixed on the shaft of the generator. So when direct current flows through the pole
windings and the rotor rotates with constant speed the rotating multiple magnetic field is set up.
Fig 4.2
This field crossing the unmoved stator windings induces in emf The poles have special forms providing
distribution of the magnetic flux density in the air gap between the stator and the rotor poles according to the cosine
law. That’s why emf induced in the stator winding is a sinusoidal function of time shown in Fig 4.3
e = E m sin(ωt + Ψe ),
Fig 4.3
e = E m sin(ωt + Ψe ),
4.1
where e = the instantaneous value instant time t,
Em = the amplitude,
ψ = The initial phase angle.
The value T is the time occupied by one complete cycle of change, or the period. The number of
cycles within one second is the frequency.
1
4.2
f = .
T
The SI unit of frequency is Hertz (Hz.). The sinusoidal emf has frequency f=1Hz when its period is
equal to one second.
The quantity (ωt+ψe) is the phase angle of sinusoidal emf and it’s value at t=0 is called as the initial
phase angle.
The quantity ω is the angular velocity or angular frequency and may be obtained by dividing the
angle spanned in one cycle by the period of the wave. Since the angle spanned in one cycle is 2π radian
ω=
2π
rad / s.
T
4.3
The relationship between the velocity of the generator n in per revolution/min, the number of the pair
of poles and frequency f is given by the following equation
n=
60 f
.
P
4.4
If f=50 and p=1 then n=3000, if f=50 and p=2 then n=1500, etc.
In fact any sinusoidal function including currents and voltages may be represented by the similar equations:
i = I m sin(ωt + Ψi ) ,
v = V m sin(ωt + Ψvi ),
4.5
where i and v are the instantaneous values of current and voltage,
Im and Vm are the amplitudes of current and voltage,
ψi and ψv are the initial phase angles of current and voltage.
The difference between initial phase angles of voltage and current is called the phase displacement:
4.6
ϕ = ψ v −ψ i .
When ϕ=0 voltage and current are in phase, if ϕ=π they are antiphase.
The commercial (or power) frequency employed in many countries in the world is 50 Hz. In the USA
it’s 60 Hz. Sinusoidal current and voltages of relatively low frequencies (up to several kHz) are generated
by valve or semiconductor oscillators.
Direct current or voltage may be considered as the special case of a sinusoidal current or sinusoidal
voltage whose frequency is zero Hz.
4.2 Root Mean Square and Average Values of Sinusoidal Quantities
The root mean square (rms) value of an alternating current is given by that direct current which when
flowing through a given resistive circuit for a given time produces the same heat as produced by the AC
when flowing through the same circuit for the same time. It is also known as effective value of AC. Thus,
heat produced by AC in resistor R in time dt is i2Rdt, and heat developed over one cycle is
T
∫i
2
Rdt.
4.7
0
Heat due to the DC in the same circuits is rI2T. Their equating gives
T
RI T = ∫ Ri 2 dt.
2
4.8
0
Hence the root mean square value of current
T
1 2
I=
i dt .
T ∫0
4.9
Similarly, emf and voltage are
T
1 2
E=
e dt . E
T ∫0
T
1 2
v dt .
T ∫0
V=
4.10
In the case of sinusoidal quantities :
I=
Im
2
≈ 0.707 I m ;
E=
Em
2
V=
;
Vm
2
.
4.11
Alternating current ammeters and voltmeters are constructed to indicate the rms value of the
quantities being measured. The nameplate data of AC apparatus are also given as the rms values of
currents and voltages.
The average values of sinusoidal quantities over one cycle is defined as the average values during
one positive (or negative) half-cycle. Thus, the average current is
I av
2
=
T
T /2
∫
0
I m sin ωtdt =
2
π
I m ≈ 0.636 I m .
4.12
In the same way are defined average voltage and emf
T /2
2
2
V av =
V m sin ωt = V m ,
∫
π
T 0
T /2
2
2
E m sin ωt = E m .
4.13
∫
π
T 0
Due to analyses AC circuits form factor and peak or amplitude factors are used. The form factor of
sinusoidal current wave form is defined as the ratio
4.14
rms − value
I / 2
= m
= 1.11.
K if =
2
average − value
Im
E av =
π
In the same way are defined form factors of voltage or emf wave forms. The peak factor of
sinusoidal current wave form is defined as the ratio
Im
max imum _ valie
K ip =
=
2.
4.15
rms _ value
Im / 2
Peak factor of voltage or emf wave forms are defined similarly.
Example 4.2
Half-wave rectified AC has amplitude of 5A. Find the rms and average values, the form and the
peak factors.
Solution:
Half-wave rectified sinusoidal current is one whose one-half cycle has been suppressed. It’s shown in Fig
4.4, where suppressed half-cycle is shown dotted.
Fig 4.4
As said earlier, for finding rms value of current summation should be carried over the half-period
for which current actually flows, though it would be averaged for the whole cycle
I=
1
T
T /2
2
2
∫ I m Sin ωtdt =
0
Im
= 2.5 A.
2
For the same reasons the average value of current is:
I av =
1
T
T /2
∫ I m S int dt =
0
Im
π
= 1.59 A.
The form factor of half-wave rectified AC is :
K if =
I
I av
=
Im / 2
= 1.57.
Im /π
The peak factor is:
K ip =
Im
= 2.
I
4.3 Complex Representation of Sinusoidal Quantities
Complex algebra provides a relatively simple but powerful tool for obtaining the steady-state solution of electric
circuit problems. It simplifies the mathematical manipulation of sinusoidal quantities and has the advantage of
including the magnitudes and phase angles of current, voltage, emf impedance and admittance in all related
equations.
Complex algebra is associated with a two dimensional plane, called a complex plane, whose
horizontal axis is a real-number line and vertical axis is an imaginary-number line.
Fig 4.5
The convention is to mark the axis of real number with “+” and the axis of imaginary number with
“+j”. A directed line segment OA of length A (a phasor) makes an angle of α with the real axis. Its
position in the complex plane may be expressed as a complex number , which may be written in
exponential form. The projections of the phasor on the real and imaginary axis constitute the real “a” and
imaginary “b” components of the complex number. So the rectangular form of the complex number is:
.
A = a + jb.
4.16
Note that imaginary numbers cannot be added to the real numbers and vice versa. The relationships
between different forms of complex numbers are
a = A cos α ,
b = sin α ,
A = a2 + b2 ,
b
α = tn −1 .
a
4.17
The conjugate of the complex number is it is mirror image in the complex plane as shown in Fig 4.5
*
.
An asterisk on a complex number indicates it to be the conjugate. Thus , A is the conjugate of A
.
*
A = a + jb = A∠α
A = a − jb = A∠ − α .
4.18
The conjugate of complex number is used in power computation.
The angle α in Fig 4.5 is read counter-clock wise from the “+” axis. We assume that α=wt+ψ or that
.
the angle varies directly with time ”t”. Then , A rotates counterclockwise with constant angular velocity
ω. Assume that the complex function
.
A = I m cos(ωt + ψ i ) + jI m sin(ωt + ψ i ).
4.19
The term Imcos(ωt+ψi) is the real part of the complex function and Imsin(ωt+ψi) is the imaginary part
of the same function. Thus, a sinusoidal current may be represented by the imaginary part of the above
complex function
.
i = I m A = I msin(ωt + ψ i )
or, by the projection of the rotating phasor Imej(ωt+ψi) on to the “+j” axis which is the same. For uniformity,
the phasors of sinusoidal quantities are usually shown in the complex plane for a reference time t=0, then
phasor Imej(ωt+ψ) becomes
.
I m e jψ i = I m ,
4.20
where
.
I m = a complex amplitude of current, making angle ψ i with “+” axis in the complex plane.
Example 4.3
Express the sinusoidal voltage v = 150 sin(ωt + 45 o ) in phasor, exponential and rectangular forms. Write
corresponding conjugate of complex numbers.
Solution:
The phasor form is:
.
.
*
V m = 150∠45 ,
The exponential form is:
.
.
.
V m = 150∠ − 45 o ,
o
*
V m 150e − j 45
V m = 150e j 45 ,
The rectangular form is:
o
o
.
V = 150 cos 45 o + j150 sin 45 o = 106.05 + j160.05,
*
V m = 106.5 − j106.5.
4.4. Addition, Subtraction Multiplication and Division of Complex Numbers
The sum of two or more complex numbers may be obtained graphically by tip-to-tail addition or parallelogram
method in the complex plane. Fig. 4.6 shows the component complex amplitudes of currents and the complex
amplitude of resultant current obtained by the tip-to-tail addition.
Fig. 4.6
When using the complex algebra, the addition or subtraction can be done only in rectangular form.
.
.
A1 ± A2 = (a1 + jb1 ) ± (a 2 + jb2 ) = (a1 ± a 2 ) + j (b1 ± b2 ),
4.21
Multiplication and division of complex numbers becomes very simple and easy if they are represented
in polar or experimental form. Hence, product of any two complex numbers A and B is given by another
complex number equal in magnitude to A⋅B and having the phase angle equal to the sum of angles α and
β.
.
.
A⋅ B = Ae jα ⋅ Be jβ = ABe j (α +β ) .
In the case of division of complex numbers we have:
4.22
.
A
.
=
B
.
Ae jα A j (α − β )
= e
.
B
Be jβ
4.23
.
Hence A : B is another complex number having a magnitude of A:B and having a phase angle α - β .
Example 4.4
The following three complex numbers are given
Perform the following indicated operations:
.
.
.
AB
1.
B = 30∠ − 120
.
and C = 10 + j 0.
.
BC
and
.
.
A = 20 + j 20,
.
.
C
A
Solution:
Rewriting all three complex numbers in polar form, we get
.
A = 28.3∠45 o ,
.
.
C = 10∠0 o .
B = 30∠ − 120 o ,
Then
.
.
28.3∠45 o ⋅ 30∠ − 120 o
=
= 84.9∠ − 75 o .
.
o
10∠0
C
AB
.
.
BC
.
=
A
30∠ − 120 o ⋅10∠0 o
= 10.6∠ − 165 o .
o
28.3∠45
4.5 A Sinusoidal Current Through the Resistor, the Inductor and the Capacitor
Let the current through a resistor in Fig 4.7a be I m sin ωt. The resistor has resistance.
Parameter R. Voltage across the resistor is
v = R i = RI m sin ωt = Vm sin ωt ,
where Vm = RI m . Dividing this equality by
V = RI
or in complex form
.
4.24
2 we have Ohm’s law
4.25
.
V = R I,
4.26
Fig 4.7
.
.
where V and I are the complex rms value of sinusoidal voltage and the complex rms value of sinusoidal
current respectively.
The instantaneous power is the product of instantaneous voltage and instantaneous current
p = v ⋅ i = V m I m sin 2 ωt = VI (1 − cos 2ωt ).
4.27
It has an unvarying component VI and a variable component VI cos 2ωt varying at frequency 2ω. The
instantaneous values of current, voltage and power are given by plots of Fig 4.7b. Fig 4.7c shows
complex rms voltage and complex rms current phasors. Referring to Fig 4.7b,c the voltage and current of
the resistance are in phase (phase displacement ϕ = 0 ).
Now, let us consider an ideal inductor characterized by pure inductance L shown in Fig. 4.8a.
Fig 4.8
Let the current through L be I m sin ωt . Then the emf of self-induction will be
eL = −L
di
= −ωLI m cos ωt = E m sin(ωt − 90 o ).
dt
Voltage across the inductor is
v = −e l = V m cos ωt = V m sin(ωt + 90 o ),
4.28
4.29
where V m = ωLI m = X L I m .
The value X L = ωL is termed the inductive reactance. The unit of this value is Ohm.Thus, the
inductance offers to sinusoidal current an opposition equal to X L = ωL which is directly proportional to
frequency. Also, current through the inductance lags by 90o on the voltage across it.
Dividing the above equation by 2 , we have
V = IX L
4.30
or in complex form
.
.
.
V =I jXL
4.31
.
because of multiplication of complex current by j means rotation of complex current by 90o in the
counter clock wise direction.
The instantaneous value of power is:
p = vi = V m cos ωtI m sin ωt = VI sin 2ωt
4.32
So, it has only variable component varying at frequency 2ω. There is now dissipation of energy in
the circuit. It alternately draws energy from the source of supply and returns it back. Fig 4.8b shows plots
of the instantaneous current, voltage and power and Fig 4.8c shows complex rms voltage, current and
emf. Voltage across inductor leads the current by 90 o and emf lags the current by 90 o , so it is in antiphase with voltage.
Let us consider an ideal capacitor across which a sinusoidal voltage is applied. If capacitance
C=const. then current through the capacitor is:
dv
π
i=C
= CωV m cos ωt = I m sin(ωt + ).
4.33
dt
2
Vm Vm
=
.
1
Xc
ωc
The value X c = 1 / ωC is termed the capacitive reactance. Dividing the eq. 4.34 by
Im =
I=
V
Xc
4.34
2 we have
4.35
or in complex form
.
.
V
I=
− jX c
The instantaneous power
p = vi = Vm sin ωt ⋅ I m cos ωt = VI sin 2ωt.
4.36
4.37
Fig.4.9
Again it has only variable component varying at frequency 2ω.The circuit alternately draws energy
from source supply, and gives it back Fig. 4.9b shows the plots of instantaneous voltage, current and
power and Fig. 4.9c shows complex rms values. The current flowing through capacitor leads the voltage
by 90o.
4.6 Series Connected R,L and C Circuit
Series connected coil and capacitor may be represented in the equivalent circuit diagram (Fig. 4.10)
by a combination of resistive and reactive elements.
Fig.4.10
According to Kirchhoff’s voltage law a driving voltage is equal to the sum of voltages on the resistance,
inductance and capacitance.
di 1
+
idt.
4.38
dt c ∫
In the case of sinusoidal voltages and current it is possible to introduce complex rms values of
voltages and current. Then taking into account that
v = v R + v L + vC = Ri + L
.
di
= jω I ,
dt
.
I
∫ idt = jω .
instead of above integro-differential equation we have the following algebraic equation
.
.
.
1 . .
1
V = R I + jωL I +
I = I ( R + jωL +
).
jωc
jωC
The value
.
V
.
= R + j (ωL −
1
) = Z = Ze jϕ
−
ωc
4.39
4.40
4.41
I
is called the complex impedance and the expression
.
.
V
I=
Z
4.42
−
is Ohm’s law in complex form.
The magnitude or modulus of the complex impedance is the circuit impedance
Z = R 2 + (ωL −
1 2
) = R2 + X 2 ..
ωc
4.43
The symbol of circuit impedance is shown in Fig 4.11a
1
The value X = X L − X C = ωL −
is the total reactance of the circuit. The argument or phase of the
ωc
complex impedance is actually the phase displacement of the circuit and is determined as
1
ωL −
ωC .
ϕ = tn −1
4.44
R
Phasors of voltages and current are shown in the following phasor diagram of Fig. 4.11b
Fig. 4.11
The triangle abc is called the voltage triangle, from which the following values may be found
V − VC X
V
IR R
2
V = V r + (V L − VC ) ,
4.45
cos ϕ = R =
= ,
sin ϕ = L
= .
V
IZ Z
V
Z
If X L > X C as it’s shown in Fig. 4.11, the circuit reactance is positive and current lags voltage by
angle ϕ. If X L < X C , the circuit reactance is negative and current leads the applied voltage by the angle
ϕ. In the case X L = X C the circuit reactance is equal to zero and circuit impedance Z=R. It means that
the amplitudes of voltage components across inductance and capacitance are of equal value and voltages
balance each other. Therefore the reactive component of the applied voltage is zero. The applied voltage
is equal to the active voltage component and it is in phase with circuit current (ϕ=0). In this case R-L-C
circuit is said to be at voltage resonance. This phenomenon will be discussed later.
4.7 The Parallel G-L-C Circuit
Suppose a coil and a capacitor are connected in parallel as shown in Fig. 4.12
By Kirchhoff’s current law, the instantaneous value of the input current is equal to the sum of the
instantaneous value of the branch currents
i = iG + i L + iC ,
4.46
1
dv
L∫
i = Gv +
vdt + C
dt
.
In the case of sinusoidal voltage and currents the last equation in complex form will be:
Fig.4.12
.
.
I = GV +
1
.
jωL
.
.
V + jωC V = V (G +
1
jωL
+ jωC ).
The value
.
I
.
V
= Y = G − j(
−
1
− ωC ) = Ye − jϕ
ωL
4.47
is called a circuit complex admittance. So, Ohm’s law in complex form may be written as
.
.
I =VY.
−
4.48
The modulus of the complex admittance is the circuit admittance
Y = G2 + (
1
− ωC ) 2 .
ωL
4.49
The symbol of circuit admittance is shown in Fig. 4.13a. The value
BL =
1
ωL
is called an inductive susceptance and the value
4.50
B L = ωC
4.51
is called a capacitive susceptance. Then the value
1
− ωC
ωL
B = B L − BC =
4.52
is the total reactive susceptance. The unit of admittance and susceptance is Siemens. The argument of
complex admittance
ϕ = tn −1
1
− ωC
ωL
.
G
4.53
Voltage and current phasors are shown in Fig. 4.13b.
The triangle abc is called as the current triangle. If the current triangle is divided by voltage V the
conductance triangle will be obtained. From the current and conductance triangles the following
quantities can be found
I = Ia 2 + ( I L − I C ) 2 ;
y = G2 + b2 ;
4.54
Fig. 4.13
cos ϕ =
I a VG G
=
= ;
I VY Y
sin ϕ =
I L − I C B L − BC B
=
= ;
Y
Y
I
4.55
If B L > BC , B > 0 and the input current legs on the applied voltage by angle ϕ. When B L < BC ,
B < 0 and the input current leads the applied voltage by angle ϕ. In the case of B L = BC , the total
susceptance B=0. It means that the amplitudes of branch currents flowing through inductance and
capacitance are equal values and balance each other. The circuit admittance is equal to the conductance
Y=G and the input current and applied voltage are in phase (ϕ=0). In this case G-L-C circuit is said to be
at current resonance. This phenomenon will be discussed later.
Example 4.7
350µF capacitor, 10.0 Ω resistor and a coil whose inductance is 0.03H respectively are connected in
parallel (Fig 4.12) and supplied by a 120 V 50 Hz generator. Determine: a) The rms complex current
through each branch; b) The rms complex input current; c) The input admittance d) The instantaneous
value of input current.
Solution
a) BC = ωC = 0.110Sm; B L =
.
.
.
.
1
1 1
= 0.106 Sm; G = =
= 0.100 Sm; B = B L − BC = −0.0045Sm;
ωL
R 10
I a = V G = 120∠0 o ⋅ 0.1∠0 o = 12∠0 o A.
.
.
I c = V jBC = 120∠0 o ⋅ 0.11∠90 o = 13.2∠90 o A.
I L = V (− jB L ) = 120∠0 o ⋅ 0.106∠ − 90 o = 12.72∠ − 90 o A.
.
b) I = I G + I C + I L = 12∠0 o + 13.2∠90 o + 12.72∠ − 90 o = 12.01∠2.29 o A.
c) Y = G 2 + ( B L − BC ) 2 = 0.10008Sm.
d) i = 2 ⋅12.01 sin(314t + 2.29 o )
4.8. Active, Reactive, Apparent and Complex Powers
Power is the rate of energy conversion. In AC circuits the instantaneous power p=vi and the average
power over the period is
ρ=
1
T
∫
T
0
pdt =
1
T
∫
T
0
vidt.
4.56
This equation gives the average actual power and is referred to as the active power
1 T
ρ = ∫ V m sin(ωt + Ψv ) I m sin(ωt + Ψi )dt = VI cos ϕ
4.57
T 0
where the phase displacement ϕ = Ψv − Ψi . Active power is measured in Watts (W).
The value cosϕ is called the power factor. Since V cos ϕ = IR or I cos ϕ = VG the active power may
be written as P=I2R=V2G. If the circuit contains reactive (energy storing) components, the energy will
oscillate between circuit and supply. The peak of the oscillating power is given by
Q = VI sin ϕ .
4.58
It is termed the reactive power and is measured in reactive voltamperes. It is positive with a leading
voltage (sinϕ>0) and negative with lagging voltage (sinϕ<0). Since V sin ϕ = IX and I sin ϕ = VB the
reactive power is:
Q = I 2 X = V 2 B.
4.59
The product VI gives the apparent power
S = VI = I 2 Z = V 2Y .
4.60
The apparent power is measured in voltamperes (VA).The reactive, active and apparent powers may
be represented by a power triangle. In order to get complex power it's necessary to multiply the complex
voltage by the conjugate complex current
.
. *
.
S = V I = Ve jΨV ⋅ I e − jΨi = VIe j ( Ψv − Ψi ) = VIe iϕ = VI cos ϕ + jVI sin ϕ = ρ + jQ.
4.61
Thus the real part of complex power is the active power and the imaginary part of complex power in
the reactive power.
4.9. Two-terminal Networks
Consider the passive (without energy source inside) two-terminal network shown in Fig.4.14a. It
may be the network of any complexity and may be replaced by the equivalent series or parallel circuits
shown in Fig. 4.14
The input complex impedance and admittance of the network are
Z=
−
.
.
V
I
.
= R + jX = Ze jϕ ;
Y=
−
.
=
1
= G − jB = Ye − jϕ .
Z
4.62
V −
I
If X>0 (B>0), the two-terminal network is an inductive one (Fig. 4.14b,c) and if X<0 (B<0), it is a
capacitive one. (Fig. 4.14d,e). When X=0 it is a purely resistive circuit. The input impedance
(admittance) of a two-terminal network can be found either by calculation or experimentally. For an
experimental determination the circuit in Fig. 4.15 can be used, where A is an ammeter measuring the
current I, V is a voltmeter measuring the voltage applied to network and W is a wattmeter measuring
active power VIcosϕ.
Fig. 4.14
The magnitude of the input equivalent impedance of the network may be found using readings of
voltmeter and ammeter
Z eq =
V
I
4.63
and the power factor may be determined by readings of voltmeter, ammeter and wattmeter
cos ϕ =
ρ
.
VI
The equivalent resistance and conductance may be found as
Req=P/I2; Geq=P/V2.
and the equivalent reactance and susceptance are
2
2
X eq = ± Z eq − Req ;
4.64
4.65
2
2
B = ± y eq − G eq .
4.66
It is important to know the sign of the above formulae. It is determined by closing the switch «s» in
Fig. 4.15, there by bringing a small capacitor C in a circuit. If with a switch closed the ammeter gives a
decreased reading phase angle ϕ is positive and corresponding reactance is also positive. So the
equivalent complex impedance or admittance is inductive. If the ammeter gives an increased reading,
angle ϕ is negative and corresponding reactance is negative. So the equivalent complex impedance is
capacitive.
The active power may be written as
P = I 2 R = V 2G ⇒ R =
V2
G;
I2
Then the relationship between active resistance and conductance is:
4.67
Fig 4.15
G
or
y2
Similarly the reactive power is
R=
G=
R
.
Z2
V2
Q = I X = V B ⇒ X = 2 B.
I
2
2
Then the relationship between reactance and susceptance is
B
X
X= 2
or B = 2 .
Y
Z
Problems:
4.68
4.69
4.70
4.1. The equation of current wave is I=32sin314t. Determine the amplitude, the frequency, the period and
the instantaneous value of current at t=0.05s.
4.2. A certain filament lamp draws 100W when connected to a 120V 50 Hz generator. Calculate the
resistance of the lamp and write the equation of current wave.
4.3. A 440 V 50 Hz voltage wave, when applied across a resistor, dissipates energy at the rate of 10.0 kw.
Determine a) maximum and rms value of current; b) the resistance of the load and write the equations of
voltage and current waves.
4.4. A 1.2 H ideal inductor is connected across 200V 50 Hz drives. Determine the inductive reactance,
the rms current and the peak value of energy stored in the magnetic field.
4.5. A 100µF capacitor is connected across a 180V 10kHz generator. Find the capacitive reactance, the
rms current and the peak value of energy stored in the electric field.
4.6. A coil, a capacitor and a resistor are connected in series and supplied by a 1 kHz generator. The coil
has a resistance of 3Ω and an inductance of 0.3mH, the capacitive reactance is 5Ω and the resistor is 2Ω.
If the current in the circuit is 5∠30 o A, find the complex impedance, the complex applied voltage and the
complex voltage drops across each circuit elements.
4.7. A certain series circuit is composed of two generators, a 40Ω resistor, a 10Ω inductive reactance
and a 30Ω capacitive reactance. The generator emfs are 60sin(1000t+50o) and 100sin(1000t+50o).
Determine the complex impedance, the complex current and the complex power. Draw the phasor
diagram in scale showing current, applied voltage and voltage drops across each circuit elements.
4.8. A parallel circuit consisting of a 4.0Ω resistor, a 2.0Ω capacitive reactance and a 6.0Ω inductive
reactance is fed from a 50 Hz 240 V generator. Sketch the circuit diagram and find the complex
admittance, the feeder current and the time-domain equations for the emf and the feeder current. Draw
the phasor diagram (in scale) showing branch currents and driving voltage.
4.9. Determine the input impedance of a parallel circuit consisting of a 2.0 resistor branch, a (3.0+j8.0)Ω
branch and a branch containing 4.0Ω capacitive reactance.
5. Steady-state Analysis of Multi Mesh Networks
5.1. Application of Kirchhoff’s Laws
The determination of the branch currents and voltages across circuit elements in a simple series
or parallel circuit is relatively easy. However, if the network is complicated and contains more than
one source and several meshes and nodes, the determination of branch currents and voltages requires
the application of special methods of network analysis. One of them is the method based on the
application of Kirchhoff`s laws.
The network in Fig. 5.1 will be used as a guide for the development of Kirchhoff`s laws technique.
It is supposed that voltages and currents vary according to sinusoidal
Z1
I&1
d
•
Z2
I&2
I&6
Z6
E& 2
E& 6
E& 1
Z4
a •
Zc
I& 4
•
b
Z5
I&5
Z3
• c
I&3
I&s
Fig. 5.1
law. Let "b" be the total number of branches and "bc" -the number of branches containing current
sources. In the circuit of Fig. 5.1, b=7 and bc=1. The number of nodes is n=4. It is supposed that the
parameters energy sources and circuit passive elements are given. It is necessary to determine branch
currents. Since the current in the branch with current source is known, the number of unknown currents
is b-bc=7-1=6. These currents are I1 ,I2 , I3 , I4 , I5 and I6 .
Before writing down the network equations for solution, one should: 1. Assume direction of branch
currents and positive direction of summation in meshes. 2.Write ( n-1 ) network equations according to
Kirchhoff`s current law. In this case
n-1=4-1=3 and the equations are
5.1
I&1 + I&4 − I&3 + I&S = 0 for the node "a";
− I&4 − I&5 + I&6 = 0 for the node "b";
I&2 + I&5 + I&3 = 0 for the node "c";
3. Write down equations according to Kirchhoff`s voltage law. The necessary
number of equations are b – bc – ( n – 1 ) = 7 – 1 – 3 = 3.
E&1 − E& 6 = I&1 Z 1 − I&6 Z 6 − I&4 Z 4 ,
E& 6 − E&1 = I&6 Z 6 − I&2 Z 2 + I&5 Z 5 ,
0 = I&4 Z 4 − I&5 Z 5 + I&3 Z 3 .
4. Solve the network equations 5.1 and 5.2.
Example
5.1. For the network of Fig. 5.2 a E&1= 100 V; Z 1= 1Ω; E&2 = 50∠ 90° V;
Z 2= 2 Ω; Z 3= 30 + j 40 = 50∠53.13° Ω. Determine all branch currents.
Solution:
5.2
Z1
I&1
I&2
Z2
•
I&3
Z3
E&1
E& 2
•
Fig. 5.2
The set of network equations is
I&1 + I&2 − I&3 = 0,
I&1 × 1 + I&3 × 50∠53.13° = 100,
I&2 × 2 + I&3 × 50∠53.13° = 50 90°.
The solution of this set is
I&1 =
∆1
,
∆
I&2 =
∆2
,
∆
I&3 =
∆3
∆
,
where
−1
1 1
∆=1 0
0
50∠53.13
0
= −151.21∠52.52 , ∆ 1 = 100
0
0 2 50∠53.130
1
∆2 = 1
0
100
0 50∠90 0
1
0 50∠53.130 = −5770∠25.68 0 ,
50∠90 0 2 50∠53.130
−1
1 1
0
0
50∠53.13 = −5568∠26.10 , ∆ 3 = 1 0
50∠53.13 0
The branch currents are:
-5770∠25.68°
I& 1 = --------------------- = 38.16∠-26.84° ( A );
-151.21∠52.52°
5568∠26.10°
I& 2 = --------------------- = -36.82∠-26.42° ( A );
-151.21∠52.52°
-206∠14°
I& 3 = --------------------- = 1.36∠-38.52° ( A );
-151.21∠52.52°
−1
0
100
0 2 50∠90 0
= −206∠14 0 ,
5.2. Mesh –current Analysis of Networks
Mesh-current analysis also called loop-current analysis, is a method that uses Kirchhoff`s voltage law
to obtain a set of network equations that when solved, can be used to determine the branch currents.
The circuit shown in Fig. 5.3 will be used to illustrate and developed the mesh-current analysis
technique.
I&1
I&11
d
•
I&2
I&5
Z5
I&22
E& 2
E& 5
E&1
Z4
a •
I&4
•
b
I&33
Z3
I&6
Z6
• c
I&3
E& 3
Fig.5.3
In the mesh-current method a circulating current is assigned to each independent mesh of the
network being solved. Network equations are written in the mesh currents by Kirchhoff`s voltage law.
Therefore, the mesh-current method is more economical in calculation work then the one based on
Kirchhoff`s laws, since it requires a smaller number of equations to be written. The guideline for the
efficient writing of mesh-current equations is as follows:
1. Convert all current sources, if any, to voltage sources.
2. Determine the required number of mesh-current equations from the following equation
m=b–(n–1)
3. Select the meshes and the direction for each mesh current. Each mesh must contain at least one
branch that is not in any of the others.
4. Using Kirchhoff`s voltage law write the mesh-current equations .When summing the emf-s,
assign a positive sign to every emf whose arrow indicate that it is of the same direction as the
assumed mesh current. When summing voltage drops, assign a positive sign to all voltage drops that
are in the same direction as the assumed mesh current.
5. Solve the set of equations.
6. In branches which are not common to any two adjacent meshes the mesh currents will be the actual
branch currents. For common branches the actual branch currents is determined from the corresponding mesh currents.
Following this guideline the required number of network equations is
m = b-(n-1) = 6-(4-1) = 3.
Independent meshes are abda,bcdb and abca. Corresponding mesh currents are I11, I22 and I33,
their directions being shown in Fig.5.2. Branch currents are also shown in Fig.5.2.
Network equations according to Kirchhoff`s voltage law are:
I&11 ( Z 1 + Z 5 + Z 4 ) + I&22 Z 5 + I&33 Z 4 = E&1 + E& 5 ,
5.3
I&22 ( Z 2 + Z 5 + Z 6 ) + I&11 Z 5 − I&33 Z 6 = E& 2 + E& 5 ,
I&33 ( Z 3 + Z 6 + Z 4 ) + I&11 Z 4 − I&22 Z 6 = E& 3 .
Let` s introduce the following notations:
Z 11 = Z 1 + Z 5 + Z 4 ,
Z 22 = Z 2 + Z 5 + Z 6 ,
Z 33 = Z 3 + Z 6 + Z 4 ,
Z 12 = Z 21 = Z 5 , Z 23 = Z 32 = − Z 6 , Z 13 = Z 31 = Z 4 ,
5.4
E&11 = E&1 + E& 5 , E& 22 = E& 2 + E& 5 , E& 33 = E& 3 .
Z 11 , Z 22 , Z 33 are the self impedances of meshes.
Z 12 , Z 23 , Z 13 are mutual or common impedances.
E&11 , E& 22 , E& 33 are the algebraic sums of emf-s of meshes.
Note that the common impedance is positive if its mesh currents have the same direction and it is
negative if the mesh currents have opposite direction.
Then the above set of equations is transformed as
I&11 Z 11 + I&22 Z 12 + I&33 Z 13 = E&11 ,
5.5
I&11 Z 21 + I&22 Z 22 + I&33 Z 23 = E& 22 ,
I&11 Z 31 + I&22 Z 32 + I&33 Z 33 = E& 33 .
Solving this set of equations we get the mesh currents
∆
∆
∆
I&11 = 1 ,
I&22 = 2 ,
I&33 = 3 ,
∆
∆
∆
where the main determinant of the set of equations is
5.6
Z 11 Z 12 Z 13
∆ = Z 21 Z 22 Z 23 .
5.7
Z 31 Z 32 Z 33
∆1, ∆2 and ∆3 are determinants that formed by replacing the first,
the second and the third columns with the right hand side of equations of the above set.
E& 11 Z 12 Z 13
∆ 1 = E& 22 Z 22 Z 23
E& 33 Z 32 Z 33
Z 11 E& 11 Z 13
∆ = Z 21 E& 22 Z 23
Z E& Z
31
Z 11 Z 12 E& 11
∆ = Z 21 Z 22 E& 22
Z Z E&
33
33
31
32
At the same time I11, I22 and I33 are branch currents to be defined
I& = I& , I& = I& , I& = I .
11
1
22
33
2
3
5.8
33
5.9
The other branch currents are defined as the algebraic sum of corresponding mesh currents
5.10
I&6 = I&22 − I&33 .
I&4 = I&11 + I&33 , I&5 = I&11 + I&22 ,
In general case when the network contains "m" loops mesh current equations are
I&11 Z 11 + I&22 Z 12 + I&33 Z 13 + ....... + I&mm Z 1m = E&11
I& Z 21 + I& Z 22 + I& Z 23 + ....... + I& Z 2 m = E&
11
22
33
mm
22
…………………………………………..
…………………………………………..
I&11 Z m1 + I&22 Z m 2 + I&33 Z m 3 + ....... + I&mm Z mm = E& mm
The general solution for the above set of equations is
∆
∆
∆km
I&kk = E&11 k1 + E& 22 k 2 + L + E& mm
,
∆
∆
∆
5.11
5.12
where ∆ is the determinant of the set of equations and ∆KN is the cofactor, obtained from ∆ by
eliminating the k-th column and n-th row and multiplying the resultant determinant by (-1)k+n .
Branch currents may be found by corresponding mesh currents.
Example 5.2. Using mesh current method determine branch currents of the network shown in Fig.5.4 if
E=50∠30° V, Z1=( 6 - j3 )Ω, Z2 =7Ω , Z3 =( 1+ j2 )Ω ,
Z4 =8Ω and Z6 = - j5Ω .
Solution:
Z1
I&1
I&5
I&2
Z5
E&
•
•
Z4
I&4
•
I&6
Z2
Z6
Z3
I&3
•
Fig. 5.4
The self impedances are
Z 11 = Z 1 + Z 5 + Z 4 = 6 − j 3 + 8 = 14 + j
Z 22 = Z 3 + Z 4 + Z 6 = 1 − j 2 + 8 − j5 = 9 + j3
Z 33 = Z 2 + Z 5 + Z 6 = 7 − j 4 − j 5 = 7 − j
The mutual impedances are
Z12 = Z21 = -Z4 = -8
Z23 = Z32 = -Z6 = j5
Z13 = Z31 = -Z5 = -j4
The algebraic sum of emf –s of meshes are
E11 = E = 50∠30° ; E22 = -E = -50∠30° ; E33 = 0.
The mesh current equations are
I11 (14 + j ) + I22 ( -8 ) + I33 ( -j4 ) = 50∠30° ;
I11 ( -8 ) + I22 ( 9 – j3 ) + I33 (-j5 ) = -50∠30° ;
I11 ( -j4 ) + I22 ( j5 ) + I33 ( 7 – j ) = 0.
The main determinant is
14 + j
−8
− j4
∆ = − 8 9 − j3
j 5 = 676∠ − 28.16 0 ;
− j4
j5
7− j
The cofactors are
50∠30 0
− 8 − j4
14 + j 50∠30 0
∆ 1 = − 50∠30 9 − j 3 j 5 = 1188.46∠ − 37.75 ;
− j4
j5 7 − j
0
0
14 + j
∆3 =
−8
∆2 =
− j4
− 8 − 50∠30
j 5 = 1950∠ − 148.53 0 ;
− j4
0
7− j
0
50∠30 0
− 8 9 − j 3 − 50∠30 0
− j4
j5
0
= 1735.6∠108.4 0 ;
The mesh currents are
∆
1188.46∠ − 37.75 0
I&11 = 1 =
= 1.76∠ − 9.59 0 A,
∆
676∠ − 28.16 0
∆
1950∠ − 148.530
I&22 = 2 =
= 2.88∠ − 120.37 0 A,
∆
676∠ − 28.16 0
∆
1735.6∠108.4 0
I&33 = 3 =
= 2.57∠136.56 0 A,
0
∆
676∠ − 28.16
The branch currents are
I1=I11=1.76∠ - 9.59° A, I2=I33=2.57 ∠136.56° A, I3=I22=2.88 ∠ –120.37° A,
I4=I11-I22=1.76∠– 9.59° - 2.88 ∠- 120.37° =3.87 ∠34.49A,
I5=I11-I33= 1.76∠– 9.59° - 2.57 ∠136.56° = 4.15∠–29.78° A,
I6=I33-I22= 2.57 ∠136.56° - 2.88 ∠-120.37° = 4.27 ∠95.5° A;
5.3. Node Voltage Analysis
In the case of network analysis when one less than the number of nodes is less than number of
independent meshes it is better to use the node voltage analysis. This method uses Kirchhoff`s current law
and allows to determine the voltages across each branch.
The node-voltage method has excellent applications in circuits having common ground connections
such as those used in electronic circuits. The circuit shown in Fig.5.4 will be used to illustrate the node
voltage technique. The number of node equations required is equal to one less than the number of nodes.
Hence, when selecting the nodes, omit the node that connects to the greatest number
of branches. Thus in Fig.5.5. the node 3 will not be used. It has the greatest number of branches and is generally the
ground node.
The node equations will be simplified if all voltage sources are converted to equivalent current
sources and all impedances are replaced by corresponding admittances using the relationships
E
1
Is= ----- ;
Y = ------ .
Z
Z
Hnce the network shown in Fig.5.5. is identical to that of Fig.5.6.
I&1
1
•
Z5
I&5
I&2
2
•
I&4
I&3
Z2
Z1
Z4
Z3
E& 1
E& 2
•
•
3
Fig.5.5
Y 12
1
•
•
2
•
V&1
I&s1
Y1
Y3
•
V&2
Y4
•
•
•
0
I&s 2
Y2
•
Fig.5.6
It is supposed that the voltages of nodes are more than the voltage of the grounded node. The last one
is arbitrarily set at zero. It means that currents of the passive
branches are directed to the grounded node. In accordance with Kirchhoff`s current law for nodes 1 and 2
we have
V&1 Y 1 + V&1 Y 3 + (V&1 − V&2 )Y 12 = I&s1 ,
5.13
V& Y 2 + V& Y 4 + (V& − V& )Y 21 = I& .
2
2
2
1
s2
Let’s introduce the following notations
Y11=Y1+Y3+Y12 ,
Y22=Y2+Y4+Y21 .
5.14
They are called the node self admittances. The values Y12=Y21 are called as mutual or common
admittances. Then the above set of network equations will be
V&1 Y 11 − V&2 Y 12 = I&s1
5.15
− V& Y 21 + V& Y 22 = I& .
1
2
s2
The solution of this set of equations is
∆
∆
V&1 = 1 ,
V&2 = 2 ,
∆
∆
where
− Y 12
I&s1
− Y 12
Y
∆1 =
∆ = 11
;
;
− Y 21 Y 22
I&s 2
Y 22
5.16
∆2 =
Y11
− Y 21
I&s1
.
I&
5.17
s2
The branch currents may be found using Ohm’s law or Kirchhoff`s voltage law ( Fig. 5.5 )
E& − V&1
E& 1 = I&1 Z 2 + V&1 ⇒ I&1 = 1
,
Z1
E& 2 = I&2 Z 2 + V&2 ⇒
E& − V&2
I&2 = 2
,
Z2
V& − V&2
I&5 = 1
,
Z5
V&
V&
I&3 = 1 ,
I&4 = 2 .
Z3
Z4
In general case, when the network contains "n" nodes ( n-1) number of node voltages may be found using the following set of node voltage equations
V&1 − V&2 = I&5 Z 5 ⇒
5.18
V&1 Y 11 − V&2 Y 12 − V&3 Y 13 − LLL − V&n −1 Y 1,n −1 = I&s11 ,
− V&1 Y 21 + V&2 Y 22 − V&3 Y 23 − LLL − V&n −1 Y 2, n −1 = I&s 22 ,
LLLLLLLLLLLLLLLLLLLLLL
5.19
LLLLLLLLLLLLLLLLLLLLLL
− V&1 Y n −1,1 − V&2 Y n − 2, 2 − V&3 Y n −3,3 − LLL + V&n −1 Y n −1,n −1 = I&sn −1,n −1 .
The general solution for the above set of equations is
∆ k ,n −1
∆
∆
V&k = I&s11 k1 + I&s 22 k 2 + LL + I&sn −1,n −1
,
5.20
∆
∆
∆
where ∆km is the cofactor, obtained from the main determinant of the set of equations by eliminating the
k-th column and m-th row and multiplying the resultant determinant by ( -1 )k+m . Is kk is the algebraic sum
of currents of current sources being connected to the k-th node. If the current of the current source is
interring it must be written with "+" and in the case of leaving current with "-" sign.
Example 5.3. Determine all branch currents of the circuit in Fig.5.5 if E1=100∠0°V; Z1=1,0 Ω ;
E2=150∠30° V; Z2=0,5Ω; Z3=j15Ω; Z4=-j10Ω ; Z5=( 20 + j10 )Ω.
Solution:
The current sources currents and internal admittances are
E&
100∠0 0
1
1
I&s1 = 1 =
= 100∠0 0 A;
=
= 1∠0 0 Sm;
Y1 =
0
0
Z 1 1.0∠0
Z 1 1∠0
0
E&
150∠30
1
1
= 300∠0 0 A;
=
= 2∠0 0 Sm
I&s 2 = 2 =
Y2 =
0
0
Z 2
Z 2 0.5∠0
0.5∠0
The admittances all other branches are
1
1
1
1
Y3 =
=
= 0.067∠ − 90 0 Sm;
Y4 =
=
= 0.1∠90 0 Sm;
0
Z 3 15∠90
Z 4 10∠ − 90 0
1
1
Y 12 =
=
= 0.0447∠ − 26.56 0 Sm.
Z 5 20 + j10
The admittances are
Y11 =Y1+Y3+Y12=1-j0.067+0.04-j0.02=1.04-j0.087=1.044∠-4.78° Sm.
Y22 =Y2+Y4+Y21=2+j0.1+0.04-j0.02=2.04+j0.08=2.04∠2.24° Sm.
The mutual admittances are
Y12 =Y21=0.04-j0.02=0.0447∠-26.56° Sm.
The set of node voltage equations is
V11.0044∠-4.78° -V20.0447∠-26.56° =100∠0° ,
-V10.0447∠-26.56° +V22.04∠2.24° =300∠30° .
The node voltages are
∆1 100∠0° *2.04∠2.24° -300∠30° *(-0.0447∠-26.56° )
V1=--- =--------------------------------------------------------------- =102.16∠4.8° V,
∆ 1.044∠-4.78° *2.04∠2.24° -(0.0447∠-26.56° )
∆
1.044∠-4.78° *300∠30° -(0.0447∠-26.56° )*100∠0°
V2= --- = --------------------------------------------------------------- =148.49∠27.06°V
∆
1.044∠-4.78° *2.04∠2.24° -(0.0447∠-26.56° )
The branch currents are
I1=I1s=100∠0° A, I2=I2s=300∠30° A, I3=V1Y1=102.16∠4.8° .1∠0° =102.16∠4.8° A
I4=V2Y4=148.49∠27.06° *0.1∠90° =14.85∠117.06° A,
I5=(V1-V2)Y12=(102.16∠4.8° -148.49∠27.06° ).0.0447∠-25.56° =2.97∠217.16° A
5.4. Network Theorems. Superposition Theorem
The network theorems and equivalent circuits introduced in this chapter provide additional
techniques for simplifying the solution of complex circuit problems. The superposition theorem has application in circuits having two or more generators and can be applied only to circuits having linear circuit
elements.
Due to apply the superposition theorem we calculate the component currents for each branch using
one generator at a time, with all other generators replaced by their respective internal impedances. Each
branch current is the sum of the component currents contributed by each generator to that branch.
Suppose it is desired to determine currents of the network shown in Fig.5.7. Using the diagram in
Fig.5.8a, where the voltage source E2 is removed, we find the branch currents due to the E1
E&1
Z3
Z2
5.21
;
;
.
I&2′ = I&1′
I&3′ = I&1′
I&1′ =
Z1Z 3
Z2 + Z3
Z2 + Z3
Z1 +
Z1 + Z 3
E2
Using the diagram in Fig.5.8b, where the voltage source E1 is removed, we find the currents due to the
I&2′′ =
E& 2
;
Z1Z 3
Z2 +
Z1 + Z 3
I&1′′ = I&2′′
Z3
;
Z1 + Z 3
I&1
I&3′′ = I&2′′
Z1
Z1 + Z 3
5.22
I&2
•
E& 2
E&1
Z3
Z1
Z2
•
Fig.5.7
I&1
•
I&2
I&1
•
I&2
E& 2
E&1
Z3
Z2
Z1
Z3
Z1
Z2
•
b)
•
a)
Fig.5.8
Algebraic summation of the component currents in the branches gives searching branch currents
I1 = I'1 – I1 `` ; I2 = I2`` – I2`; I3 = I3` + I3``.
5.23
5.5. Thevemin’s Theorem
Thevemin’s theorem provides to replace a section of a network having one or more generators and
impedances with an equivalent circuit model that contains only one voltage source with emf Eth and
internal impedance Zth. The Thevenin equivalent section has the same current and voltage relationships at
its output terminals as does the section it replaces. The general procedure for calculating the Thevenin
equivalent section is developed with the aid of Fig.5.9a. In this figure, the section to the left of terminals a
b is to be replaced by its Thevenin equivalent as shown in Fig.5.9b.
If electric circuit contains current sources they must be replaced by the equivalent voltage sources. In
order to determine Eth we have to remove the load ZL and calculate the voltage drop across terminals
a b. Connection with the load this voltage drop is an open circuit voltage
E& 1
E& Th = V&ab =
Z 3.
5.24
Z1 + Z 2 + Z 3
Z2
Z4
•
a
•
•
a
E& Th
E& 1
Z3
ZL
ZL
Z Th
Z1
•
a)
•
b
•
b
b)
Fig.5.9
Due to determine ZTh we have to disconnect the load ZL and remove voltage source but keep its
internal impedance. In the case of Fig.5.9a we have
Z (Z + Z 2 )
Z Th = Z 4 + 3 1
5.25
Z1 + Z 2 + Z 3
The current flowing through the load is
E& Th
.
I&L =
5.26
Z Th + Z L
Very useful application of Thevenin`s theorem is the case of network analysis when only one branch
current must be found.
Example 5.4. With reference to the network of Fig.5.9, by applying Thevenin`s theorem, find the
current in the load if E1=50∠0° V, Z1=(3+j4)Ω , Z2=0, Z3=-j2Ω , Z4=1.67Ω and Zl=j2.61Ω .
Solution:
The Thevenin equivalent emf is
50∠0°
E1
Eth= -------- .Z3= ---------- (-j2)=27.7∠-123.69° V.
3+j4-j2
Z1+Z3
The Thevenin equivalent impedance is
Z3.Z1
-j2 . (3+j4)
Zth=Z4+------- = 1.67+ ------------- =(2.59-j2.61)Ω .
Z3+Z1
-j2+3+j4
The load current is
27.7∠-123.69°
Eth
Il= ------- = --------------------- =10.64∠-123.69° A.
Zth+Zl 2.59-j2.61+j2.61
5.6. Mutual Inductance in Series Connection of Coils
Fig.5.10a shows two coils connected in series aiding and Fig.5.10b two coils connected in series
opposition.
According to Kirchhoff`s voltage law for series aiding connection
I&
•
R1
I&
•
L1
∗
M
V
•
L2
∗
L1
M
V
∗
R1
•
R2
a)
∗
L2
R2
b)
Fig 5.10
v = R1i + L1
di
di
di
di
+M
+ R 2 i + L2 + M
dt
dt
dt
dt
In complex notation
V& = I&[R1 + R2 + jω ( L1 + L2 + 2M )] = I& Z aid
where
Z aid = [R1 + R2 + jω ( L1 + L2 + 2M )].
In the case of series opposition connection (Fig.5.10b) the network equation is
di
di
di
di
v = R1i + L1 − M
+ R 2 i + L2 − M
dt
dt
dt
dt
5.27
5.28
5.29
5.30
In complex notation it will be
5.31
V& = I&[R1 + R2 + jω ( L1 + L2 − 2 M )] = I& Z opp .
where
Fig.5.10 shows a phasor diagram for series aiding connection, where V is the complex voltage
across the first coil and V2 is the complex voltage across the second coil. Fig.5.11 shows a phasor
diagram for series opposition connection.
I&jωM
I&jωL2
V&
V&2
I&jωL2
I&jωM
V&1
I&R2
I&jωL1
I&jωM
V&
I&R1
I&jωM
I&jωL1
V&2
I&R2
V&1
I&R1
I&
I&
Fig.5.10
Fig.5.11
It’s possible to find mutual inductance by experiment. We shall discuss two practical methods for the
determination of mutual inductance.
Method one. We have to make two experiments. In one, we connect two coils in series aiding and
measure the current, voltage and active power at the input of in the circuit of Fig.5.12.
•
A
W
R1
L1
∗
M
V
∗
•
L2
R2
Fig.5.12
Using readings of meters and the following formulas
V
P
5.32
X aid = Z 2 − ( R1 + R2 ) 2
Z= ;
R1 + R2 = 2 ;
I
I
we’ll find the equivalent circuit reactance. At the second test the coils are connected in series
opposition and the same measurement and calculation must be done. The result of calculation is the
equivalent circuit impedance, resistance and reactance (Xopp.).
Then taking into account that
Xaid=ω ( L1+L2+2M )
5.33
and
5.34
Xopp=ω ( L1+L2-2M )
the difference is
Xaid - Xop =4ωM.
5.35
Consequently the mutual inductance is
X aid − X opp
M =
.
5.36
4ω
Method two. Connect one coil to a sinusoidal voltage source through an ammeter and the other coil to
a high-impedance voltmeter, the reading of which is V2 . The instantaneous voltage of the second coil
di
5.37
v2 = M 1 .
dt
The rms value of v2 is
V2 = ωMI1,
5.38
where V2- is the reading of voltmeter and I1- the reading of ammeter. Hence the mutual inductance is
V
M = 2 .
5.39
ωI 1
Example 5.5. Due to find the mutual inductance of two series connected coils the circuit of Fig.5.12
is used. The readings of meters in the case of aiding connection are V1 = 120V, I1 = 10A and P1=
600W. In the case of opposition connection –V2 =120 V, I2 = 12A and P2 = 864W. The frequency of AC
current is 50 Hz.
Solution:
The impedance, the resistance and the reactance of the circuit in the first case are:
V 120
P 600
Z1 = 1 =
= 12Ω ; R1 = 12 = 2 = 6Ω; X 1 = Z 12 − R12 = 12 2 − 6 2 = 10.4Ω .
I 1 10
I 1 10
The impedance, the resistance and the reactance of the circuit in the second case are
V
P
120
864
Z2 = 2 =
= 10Ω ; R2 = 22 = 2 = 6Ω; X 2 = Z 22 − R22 = 10 2 − 6 2 = 8Ω
I2
12
I 2 12
The mutual inductance is
X − X 2 10.4 − 8.0
=
M = 1
= 1.91mH .
4ω
4 ∗ 2κ ∗ 50
5.7. The Transformer. Transferred Resistance and Reactance
A transformer is a static device intended to convert an AC at one voltage to an AC at another
voltage by means of mutual induction at unchanged frequency. It is the static device because it hasn’t
moving parts. It provides electric separation between two circuits.
A transformer has at least two coils (Fig.5.13) wound on the common core. Let the permeability of
the core be constant. The parameters of one coil are R1 and L1. It is termed the primary coil and is
switched on to the terminals of source of supply. The parameters of the second coil are R2 and L2. It is
connected to the load and called as the secondary coil. The mutual inductance between coils is M and the
load impedance is ZL.
M
I&1
•
R1
R2
•
∗
V&1 ↓
L1
V&2 ↓
L2
I&2
∗
•
ZL
•
Fig.5.13
The network equations for the primary and the secondary loops are
I&1 R1 + I&1 jωL1 + I&2 jωM = V&1 ,
I& R + I& jωL + I& jωM + I& Z L = 0.
2
2
2
2
1
5.40
2
Voltage across the load is
V&2 = I&2 Z L = I&2 ( R L + jX L ) .
Hence the second equation of the above set will be
I&2 [( R2 + RL ) + j (ωL2 + X L )] + I&1 jωM = 0.
5.41
5.42
have
Solving this equation accordance with I2 and substituting it in the first equation of the above set we
⎤ &
⎡
I&1 jωM
I&1 R1 + I&1 jωL1 + jωM ⎢−
⎥ = V1
R
R
j
L
X
(
)
(
ω
)
+
+
+
1
2
2
L ⎦
⎣
5.43
or
I&1 ( R1 + ∆R ) + j ( X 1 − ∆X ) = V&1 ,
5.44
where
∆R =
( R2 + RL )ω 2 M 2
(ωL2 + X L )ω 2 M 2
,
∆
X
=
( R2 + RL ) 2 + (ωL2 + X L ) 2
( R2 + RL ) 2 + (ωL2 + X L ) 2
5.45
are transferred resistance and transferred reactance from the secondary to the primary. They are such that
to be added to the primary by connecting them in series with R1 and X1= ω L1 in order to take into
account the effect of the secondary impedance on the current in the primary circuit.
The primary current is
V&1
I&1 =
5.46
R1 + ∆R + j ( X 1 − ∆X )
Example 5.6. For the transformer of the Fig.5.13 the following values are given: R1=100Ω ; X1=10Ω
; X2=20Ω ; Xm=10Ω ; V1=100V.Find the value of the capacitive reactance of the load (RL=0) at which the
circuit draws only active power from the power supply. Draw the phazor diagram.
Solution:
If the circuit draws only active power from the source of supply then the equivalent circuit reactance
is equal to zero
(X 2 + X L )X M
= 0.
X eq = X 1 − ∆X = X 1 −
( R2 + R L ) 2 + ( X 2 + X L ) 2
In this case
XL=XC and R2=0, RL=0.
XM2
XM2
10?
X1= --------⇒
XC=X2- ---- =20 - ----- = 10Ω .
X2-XC
X1
10
Now let’s construct the phazor diagram. Because of ∆R=0 Xeq=0 the primary current
V1
100
I1 = ------ = -------- = 1 A.
R1
100
The secondary current is
jωM
j 10
I2 = -I1 ------------------------------- = -I1 --------------------- = - I1 = -1A.
( R2 + RL) + j( X2 + XL)
0 + j( 20 - 10)
The load voltage is
VL = I2 j XL = -1* ( - j10 ) = 10 ∠90° V.
The phazor diagram drawn in scale is shown in Fig.5.14
V&L
I&2
I&1
Fig.5.14
V&1
Problems:
5.1. Using Kirchhoff`s laws method find all branch currents of the circuit of
Fig.5.15, where E1= 100∠0° V; Z1= 0.5Ω; E2= 50 ∠90° V;
•
•
•
E&1
E& 2
Z1
Z3
I&s
Z3
Z1
Z2
E&1
Z2
E& 3
E& 2
•
•
•
Fig.5.15
Fig.5.16
Z2= 1.5Ω ; Z3= ( 40 + j60 )Ω ; Is= 1.2 ∠30° A.
5.2. Using mesh current method find all branch current of the circuit of Fig.5.16,
where E1=120V; E2=( 120 + j10 )V; E3= ( 70 +j25 )V; Z1=( 10 + j30)Ω ; Z2=( 15 + j40 )Ω ; Z3=
35Ω .
5.3. Using node voltage method find branch currents of the circuit of problem No.5.1. Write the power
balance equation.
5.4. Using the superposition theorem find all branch currents of the circuit of Fig.5.17 if E1= 100 V,
E2=100 ∠-30° V,
Z1= Z2 =( 50 + j30)Ω and Z3=100Ω.
•
E& 1
•
•
Z1
E& 2
Z1
•
Z2
•
Z5
E& 1 Z 3
Z3
•
Z6
E& 2
Z4
Z2
•
•
•
•
Fig.5.17
Fig.5.18
5.5 Using node voltage method determine all branch currents if E1=25V, E2=20V, Z1=Z2=100Ω ,
Z3=25Ω ,Z4=100Ω ,Z5=50Ω ,Z6=-j100Ω (Fig.5.18).
5.6. Using the Thevenin theorem determine the load current if the parameters of the circuit of Fig.5.19
are E=100V, Z1=(10-j50)Ω , Z2=-j100Ω and ZL=(5+j20)Ω .
Z1
•
E& 1
•
I&1
M
R1
R2
∗
∗
•
I&2
RL
Z2
ZL
V&1 ↓
L1
L2
V&2 ↓
CL
•
Fig.5.19
•
•
Fig.5.20
5.7. A transformer is loaded by the impedance ZL=(5-j10)Ω (Fig.5.20). Find the voltage V2 across the
terminals of the load if the driving voltage V1=120V, R1=10Ω , ω L1=42Ω , R2=15Ω , ω L2=70Ω and ω
M=20Ω .
6. Resonance
6.1. Introduction
Resonance is a physical condition that exists when the frequency of an external force applied to a
structure is equal to the natural frequency of the structure.
A structure offers little opposition to motion when it is excited to its resonance frequency by
external forces. If the resonance frequency of the driver is sustained and of sufficient amplitude,
bridges may collapse, machinery may be damaged, electric circuits may burn out. A few examples of
the many useful applications of resonance in electric circuits are radio, television, lasers and powerfactor improvement.
6.2. Voltage Resonance
In general resonance in electric circuit is defined as the steady-state sinusoidal condition that exists
in circuits containing inductance and capacitance when the input current and applied voltage are in
phase.
Assuming that R, L and C are connected in series as shown in Fig.6.1, the complex impedance
Z=R+j( ωL-1/ωC).
6.1
Fig.6.1
When the circuit is turned so that ωL= 1/ωC the impedance reduces to Z=R that is, becomes
purely resistive in its effect. The reactance is equal to zero. The current is in phase with the voltage and
the rms value of current is maximum
I = V/R
6.2
for a given applied voltage, that large voltages appear across the reactive components inside the circuit
may be many times the applied voltage. This state is termed voltage resonance or series resonance. The
condition of voltage resonance is
ω 0 = 1 ω 0 C ⇒ ω 0 = 1 LC ,
6.3
where ωo is the resonance angular frequency.
The ratio
LC
ω 0 L 1 ω 0C
6.4
=
=
=Q
R
R
R
is termed the quality factor. The quality factor of a circuit is the ability of the circuit to discriminate
between different frequencies. The circuit with the lower resistance has a higher Q. It exhibits a sharper
resonance peak and a higher value of rms current at the resonance frequency than does the circuit with
the higher resistance.
The resonance condition may be achieved by varying frequency, inductance or capacitance. The
resonance inductance and capacitance are
Lo=1/ω C ;
Co=1/ω L.
6.5
The value ρ =√ L/C has dimension of resistance and is termed the wave resistance or a
characteristic impedance of the loop.
58
Example 6.1. A coil having inductance and capacitance 1mH and 5Ω ,respectively, is connected in
series with a capacitor. The circuit is supplied from a 200V, 5kHz generator. Find the value of
capacitance that will cause the system to be in resonance and the rms value of current at the resonance
frequency. Determine quality factor of the circuit and characteristic impedance. Calculate the
maximum energy stored in the magnetic field of the coil at resonance frequency.
Solution:
Resonance capacitance is
1
1
Co= ---------- = --------------------- = 1.01µF.
4π² fo²L 4π² (50*103) *10-3
At series resonance Z=R and the rms current is
V
200
I = ----- = ------ = 40A.
R
5
The quality factor is
2πfoL 2π 5000*103
Q = ----- = ------------------- = 6.28.
R
5
The energy stored in the magnetic field of the coil
LIm² 10-3 *(40√2)²
Wm= ----- = --------------- = 1.6J.
2
2
6.3. Resonance Characteristics of the Series Circuit
The functions Z(ω ), X(ω ), ϕ(ω ) are termed the resonance characteristics of the circuit. As it's
known
R = const; XL= ω L; XC= -1/ω C, Z = R 2 + (ωL − 1 ωC ) 2 .
The corresponding curves versus frequency are shown in Fig.6.2.
Fig.6.2
59
6.6
Active resistance R doesn't depend on the frequency. Corresponding plot is the straight line being
parallel of the horizontal axis. Inductive reactance is proportional to frequency. The corresponding
curve is the straight line that goes through the origin of coordinates. Capacitive reactance is expressed
by the hyperbola. Circuit reactance is the algebraic sum of inductive and capacitive reactances.The
corresponding plot crosses the frequency axis at resonance frequency ωo. According to Fig.6.2
inductive reactance XL is seen to increase in direct proportion to frequency, and capacitive reactance
XC decreases inversely with increasing frequency.
At resonance frequency circuit impedance Z is minimum and becomes equal to resistance R. When
ω<ωo , capacitive reactance XC prevails and causes impedance to increase with decreasing frequency.
When ω >ωo, inductive reactance XL prevails and causes the impedance to increase with frequency
increase.
At frequencies below the resonance frequency the capacitive reactance prevails, so that the input
current and voltage have a negative phase displacement. The phase displacement decreases with
increasing frequency and is zero at resonance after which it becomes positive and increases with the
frequency. The plots of the resonance characteristic ϕ(ω) in the cases R=0 and R≠0 are shown in
Fig.6.3.
Fig.6.3
Now let's consider resonance characteristics G ( ω ); B ( ω ); Y ( ω ). It's known that the circuit
complex admittance
1
1
R-jX
R
X
Y = ----- = -------- = ---------- = ---------- - j ---------- = G – jB
6.7
Z
R+jX
R²+X²
R²+X²
R²+X²
So the active conductance is the function of frequency.
R
G = ------------------------ .
6.8
R² + ( ωL – 1/ωc)²
The susceptance
X
ω L – 1/ωC
B = -------- = -----------------------6.9
R²+X² R² + ( ωL – 1/ωc)²
is also a function of frequency. The corresponding curves are shown in Fig.6.4 for the cases R=0 and
R≠0. ω 1 and ω 2 are frequencies at which susceptance reaches the maximum values. At resonance the
susceptance of the ideal circuit (R=0) is equal to infinity.
The above dependence of circuit impedance and admittance on frequency determines the variation of the circuit current and voltages with frequency at constant driving voltage. Resonance charac-
60
Fig.6.4
teristics I(ω), VL(ω), VC(ω) are shown in Fig.6.5. Curve I(ω) is termed frequency response curve.
Fig.6.5
At ω =0 current is zero and then increases with increasing frequency until maximum current is
attained at ω =ωo. The further increase of frequency causes a gradual drop of current which becomes
zero at ω = ∞. Voltage across resistance changes by the similar law since it is directly proportional to
the current. When ω =0, capacitor voltage VC is equal to input voltage V by reason of the fact that the
reactance of the capacitor XC=∞ and in consequence, the circuit may be considered to be an open
circuit. Capacitor voltage increases with frequency and attains the maximum value at a frequency
slightly below the resonance frequency. After that the capacitor voltage drops with increasing
frequency and becomes zero at ω =∞. When ω=0, voltage across inductance VL=0 since the inductive
reactance XL=0. Voltage VL increases with increasing frequency and attains the maximum value when
frequency is slightly above the resonance frequency. Further increase of frequency is accompanied by
a drop of voltage VL which finally becomes equal to the driving voltage at ω =∞.
6.4. Bandwidth
The general shape of a frequency-response curve is shown in Fig.6.6
Io is rms value of current at the resonance angular frequency ωo
The range of frequencies within which the variable does not drop below 0.707≅1/√2 of its maximum
value is called the pass band or bandwidth. Thus, referring to Fig.6.6, the bandwidth is
or
BW = f2 – f1
6.10
BW = ω2 – ω1
61
Fig.6.6
The frequency extremes of the bandwidth are called the cut-off or corner frequencies. They are also
called the half-power frequencies because only half the power is drawn at these frequencies compared
with the power drawn at the resonance frequency.
The relationship between the resonance frequency and the two half-power frequencies is their geometric
mean
ω 0 = ω 1ω 2 .
611
However, for circuits whose Q ≥ 10 the resonance frequency is sufficiently centered with respect to the
two cutoff frequencies that, for all practical purposes are
ω1= ωo – BW/2 .
6.12
ω2= ωo + BW/2 ;
The bandwidth depends on the quality factor of the circuit. In the case of series connection of R L
and C
I = R 2 + (ωL − 1 ωC ) 2
At the resonance frequency I0=V/R.
According to the definition of the bandwidth at the frequencies of the pass band the ratio
I
V
R
1
1
≥
. ≥
I0
2
2
R 2 + (ωL − 1 ωC ) 2 V
6.13
6.14
or
1
⎡ω L ⎛ ω
1
−
1 + ⎢ 0 ⎜⎜
⎣ R ⎝ ω 0 ω Cω 0 L
⎞⎤
⎟⎟⎥
⎠⎦
2
=
1
⎡ ⎡
1 ⎤⎤
1 + ⎢Q ⎢η − ⎥ ⎥ 2
⎣ ⎣ η ⎦⎦
≥
1
2
,
6.15
where
ω
ωoL
η= ------;
Q = -------- .
R
ωo
Rising this inequality to the second power we have
1
1
------------------------- ≥ ----2 ≥ [ 1+ Q² ( η - 1/η )² ]
1 + Q² ( η - 1/η)²
2
Suppose
1 + Q² ( η - 1/η)² = 2
62
6.16
6.17
Then
Q² ( η - 1/η)² = 1
η - 1/η= ± 1/Q ⇒.
η=±
1
1
±
+ 1.
2Q
4Q 2
6.18
Taking into account that frequencies must be positive value then
1
1
1
1
6.19
η1 = −
+
+1 ;
η2 =
+
+1
2
2Q
2Q
4Q 2
4Q
The bandwidth
BW = ω2 – ω1 = ( η2 – η1)ωo = ωo/Q ;
BW = f2 – f1= ( η2 – η1)fo = fo/Q .
6.20
Frequency response curves of the circuit with difference values of quality factor are shown in Fig.6.7.
Fig.6.7
As indicated in Fig.6.7 the bandwidth depends on the quality factor. The circuit with Q=10 has a
smaller bandwidth BW1 compare with the bandwidth BW2 of the circuit with Q=1. Then the circuit with a
smaller bandwidth has a sharper resonance peak and greater selectivity.
Example 6.2.
A coil whose inductance and resistance are 5mH and 10Ω ,respectively, is connected in series with a
capacitor 2.25µF. Determine the resonance frequency, the quality factor, the bandwidth and cut-off
frequencies.
Solution:
The resonance frequency is
1
1
f0 =
=
= 1.5kHz
2π LC 2π 5 ⋅ 10 −3 ⋅ 2.25 ⋅ 10 −6
The quality factor is
ωoL 2π*1500*0.005
Q= ------ = -------------------- = 4.71.
R
10
The bandwidth is:
fo
1500
BW= ---- = ------ = 318.5Hz.
Q
4.71
63
The cut-off frequencies are
1
1
1
1
η1 = − +
+1 = −
+
+ 1 = 0.8995
2
2 ⋅ 4.71
2q
4 ⋅ 4.712
4Q
f1 = η1fo=0.8995*1500=1349Hz.
1
1
1
1
η2 =
+
+1 = −
+
+1
2
2 ⋅ 4.71
2q
4 ⋅ 4.712
4Q
f2 = η2 * fo = 1.1118 * 1500 = 1668 Hz.
6.5. Current resonance
Conditions produced in a circuit containing an inductor and a capacitor of equal susceptances
connected in parallel is termed current or parallel resonance.
Let consider the case where the circuit conductance is in parallel with pure inductance L and pure
capacitance C. The values of the circuit susceptances depend on frequency ω of the applied voltage.
The circuit conductance is G = 1 / R and the circuit susceptances are
Fig. 6.8
1
1
BC = ωC ;
B = BL – BC = ----- - ωC
6.20
BL = ------ ;
ωL
ωL
At current resonance we have
1
------- = ωoC ,
ωo²LC = 1
6.21
BL = BC ⇒
ωoL
and the resonance frequency
ω 0 = 1 / LC
6.23
The resonance frequency is seen to be given by the same equation as for the case of voltage
resonance and for the natural frequency of a circuit with no power losses.
Current resonance can be brought about in the same way as voltage resonance either by varying the
L and C parameters of the circuit, or by varying the source frequency. Resonance inductance and
capacitance are
Lo=1/ω²C ;
Co=1/ω²L .
6.24
The value
γc = C/L
6.25
has dimension conductance and is called the circuit characteristic admittance.
At current resonance
C/L
C
ω 0C = 1/ ω 0 L =
=
6.26
L
L
64
The ratio of inductance or capacitance current to input current is equal to the ratio of characteristic
admittance to conductance
IC
V ( 1/ωoL)
VωoC
γc
IL
------- = ------ = ----------------- = --------------- = ----- = Q
6.27
I
I
VG
VG
G
This ratio is a quality factor or Q-factor of the circuit. At resonance frequency I = IG. Q-factor indicates how much times inductive or capacitive current is more than input current.
Example 6.3. A parallel circuit having a 4.0 µF capacitor and a coil whose inductance is 10
µF, is connected to the current source having an internal conductance 1 mSm . Determine the resonance
frequency, the quality factor of the circuit, the characteristic admittance and the bandwidth.
Solution:
The resonance frequency is:
1
1
f0 =
=
= 796.2 Hz
−3
2π CL 2π 10 ⋅10 −3 ⋅ 4 ⋅ 10 −3
The quality factor of the circuit is
2π * 796.2 * 4 * 10 ⎯ 6
ωoC
Q = --------- = ----------------------------- = 20
10⎯ ³
Gs
The characteristic admittance is
C
4 ⋅ 10 −6
γ =
= 0.02Sm
=
L
10 ⋅ 10 −3
The bandwidth is BW=f0/Q=796.2/20=39.8 Hz.
6.6. Resonance Characteristics of Parallel Circuit
The functions Y(ω) G(ω) B(ω) ϕ ( ω) are called resonance characteristics of the circuit. As it’s
known:
1
1
B=
G = const ;
6.28
BC = ωC ;
Y = G2 + (
− ωC ) 2
ωL
ωL
Corresponding curves versus frequency are given in Fig. 6.9.
Fig.6.9
65
Active conductance G doesn’t depend on frequency. Corresponding plot is the straight line that to
be parallel of the frequency axis. Inductive susceptance is expressed by the hyperbola. Capacitive
susceptance is proportional of frequency and the corresponding curve is the straight line that goes
through the origin of coordinates. The circuit admittance is the algebraic sum of inductive and capacitive
susceptances. The corresponding curve crosses the frequency axis at resonance frequency ωo.
At resonance frequency the circuit admittance is minimum and becomes equal to conductance G.
When ω < ωo , inductive susceptance prevails and causes the admittance to increase with decreasing
frequency. When ω > ωo ,the capacitive susceptance prevails and causes the admittance to increase
with increasing frequency. At frequencies below resonance frequency the circuit suseptence is positive
and the phase displacement is also positive. It decreases with increasing frequency and is zero at
resonance frequency, after which it becomes negative and comes to – 90° at infinite frequency.
Fig.6.10
The curves of resonance characteristic ϕ( ω ) in the cases G = 0 and G ≠ 0 are shown in fig 6.10.
Because of the circuit admittance and susceptance depend on the frequency input current and currents
flowing through inductor and capacitor also depend on frequency. There are resonance
characteristics V(ω); IG(ω); IL(ω) and IC(ω) in Fig.6.11 when I = const.
I
V = --------------------------- ;
IG = V*G ; IL = V * (1/ωL) ; IC = V *ωC
6.29
G 2 + (1 / ωL − ωC )2
The curve V(ω) is called frequency response curve. At ω=0, voltage is zero and then increases with
increasing frequency until the maximum is attained when ω =ωo. Further increase of frequency causes a
gradual drop of voltage which becomes zero at ω =∞ the inductive susceptance is infinite (BL=∞) and
the capacitor susceptance is zero. The circuit may be considered as to be short circuited. inductive
current changes by the similar law since it is directly proportional to voltage. When ω =0, inductive
current IL is equal to input current by reason of the fact that inductive susceptance increases with the
frequency and attains the maximum value at frequency ω 1. After that the inductive current decreases
with increasing frequency. It becomes equal to the capacitor current at ω =ωo. When ω =0,the capacitor
current is zero as the capacitor susceptance is zero. This current increases with increasing frequency and
it becomes equal to the inductive current at ω =ωo. Then it reaches the maximum at ω =ω2 and
drops to I at ω =∞
66
Fig. 6.11
Now let’s consider non ideal coil and capacitor in parallel as shown in Fig.6.12
Fig.6.12
The case of current resonance takes place at equality of branch susceptances B1 and B2
XL
XC
-------------- = --------------- ,
6.30
R1² + XL²
R2² + XC²
where R1 and R2 are the resistances of the actual coil and capacitor. We get the following equation for
determination of resonance frequency:
ωo L
1/ ωoC
------------------ = -------------------- ,
6.31
R1² + ωo² L²
R2² + 1/ ωo² C²
from where
L
− R12
1
C
ω0 =
6.32
LC L − R 2
2
C
As may be seen from this equation, the resonance frequency depends on the resistance of the coil
and capacitor. If the coil and capacitor resistances are so low as to produce negligible energy losses, the
resistances may be ignored and ωo = 1/ LC .
67
Example 6.4. A tank circuit consisting of a capacitor (60nF) in parallel with a coil whose
inductance and resistance are 2mH and 100Ω, respectively, is driven by the sinusoidal current source.
The parameters of the current source are Is=2.5mA and Gs=1mSm. Determine the resonance
frequency, the quality factor, the bandwidth and the capacitor voltage.
Solution:
Resonance frequency when R2=0 is:
L / C − R12
1
=
2
2π
2π LC L / C − R2
The quality factor is:
f0 =
ω 0C
1
R2
1
1
− 12 =
LC L
2π
1
100 2
−
= 12.16 KHz
2 ⋅ 10 −3 ⋅ 60 ⋅ 10 −6 (2 ⋅ 10 −3 ) 2
2π ⋅ 12.16 ⋅ 10 3 ⋅ 60 ⋅ 10 −6
= 4.58
Gs
10 −3
Bandwidth is:
fo
12.16*10 ³
BW = ---- = ------------ = 2.65kHz.
Q
4.58
Q=
=
6.7. Duality
Several analogous situations should have been noted in the preceding discussions. The statements
of the two Kirchhoff`s laws were almost word similar with voltage substituted for current, voltage
source- for current source, loop-for node, series connection- for parallel connection, resistance- for
conductance, inductance for- capacitance, reactance for susceptance, impedance for -admittance,
voltage resonance for -current resonance.
Likewise, integro-differential equations that resulted from the application of two Kirchhoff`s laws
have been similar in appearance . This similarity with all the implications is termed the principally of
duality.
Consider the two networks completely different in physical appearance shown in Fig.6.13.
Fig.6.13
Inspection shows that the first might be analysed to advantage on the loop basis and the other on the
node basis. The network equations are
68
t
L
1
di
+ Ri + ∫ idt + vC (0) = e(t ) ,
dt
C0
t
1
dv
C
+ Gv + ∫ vdt + i L (0) = i s (0) .
dt
L0
6.33
These two equations specify identical mathematical operations, the only difference being in letter
symbols. The solution of one equation is also the solution of the other. These two network are duals.
Problems:
6.1. In series R, L, C circuit L=158µH, R=16Ω the voltage resonance takes place at frequency
fo=1MHz. Driving voltage is .8V. Determine resonance capacitance, rms current, inductor and
capacitor voltages, and active power drawn by the circuit from the source of supply.
6.2. Circuit consists of coil (R, L) connected in series with an ideal capacitor C. The driving voltage of
the circuit is 35V. Find the voltage across the coil at resonance frequency if the voltage across the
capacitor is 120V.
6.3. A series connected R, L,C circuit draws from the source of supply the active power 0.1W at the
resonance angular frequency 5000rad/s and the current 0.1A. The capacitor voltage is 200V. Determine
the circuit parameters R, L and C.
6.4. An electric circuit consists of series connected coil (R=10Ω , L=100µH) and capacitor
(C=100pF). Determine the resonance frequency, the characteristic impedance and the quality factor.
Find the value of current, voltages across the coil and the capacitor at the resonance frequency if the
driving voltage of the circuit is 1V.
6.5. Determine the quality factor of a coil operating at 10kHz, whose resistance and inductance are 10Ω
and 0.02H, respectively.
6.6. A series resonance circuit has resistance of 1kΩ and cut-off frequencies of 20kHz and 100kHz.
Determine the bandwidth, the resonance frequency, the quality factor, the inductance and the
capacitance.
6.7. A parallel circuit consisting of capacitor 5.4µF in parallel with coil (R=18Ω ,L=8mH), is driven at
resonance frequency by a 240V sinusoidal generator. Sketch the circuit and determine the resonance
frequency.
6.8. A 450V 60Hz source supplies energy a parallel circuit consisting of a 25∠30° Ω and a 12∠-40°
Ω parallel branches. Determine the inductance that, if connected in series with the source, will cause
the system to be in resonance.
6.9. Current source (Is=2.6mA, Rs=60kΩ) supplies a tank circuit whose parameters are C=105nF,
L=10.5mH, Rcoil=106Ω . Determine the resonance frequency, the quality factors of the coil and the
circuit, and the BW.
69
7. Three-phase Circuits
7.1. Introduction
The greater part of the power generated is utilized by electric motors serving all kinds of electric
drives. One of the most commonly used motors is the three-phase induction motor. It is the widespread
application of this particular type of motor in all branches of industry that has led to the rapid
development of three-phase systems of electric power supply brought about by three-phase generators,
transformers, transmission lines and distribution networks.
A polyphase circuit is a combination of single-phase circuits with sinusoidal emf`s having the same
frequency but time-displaced from one another by the angle q2π/m and supplied by a single source of
energy. According to this definition, a system of three circuits is called a three-phase system.
7.2.Clasification of Three-phase Circuits and Systems
The emf``s constituting a symmetrical three-phase system are produced by a three-phase generator
with windings spaced at 120° from one another. Fig.7.1 shows a schematic drawing of a three-phase
generator with three-phase windings on stator.
A
Z
Y
N
C
120 0
S
B
X
Fig.7.1
The generator in this case has a single pair of field poles on the rotor. The stator phase windings AX, B-Y, C-Z are identical and arranged at equal angles to one another around the stator of the
generator. Because of the fact that the phase windings have an equal number of turns and are equally
spaced the emf`s produced by the generator are of equal magnitude and time displaced by 120° .
A set of three sinusoidal emf`s of the same frequency and amplitude displaced by q*120° in time
is called a three phase symmetrical system of emf`s. The value q may be 0,1,2 . . .
If the emf`s are of different amplitude or displaced unequally in phase the set of emf`s is said to be
unsymmetrical system of emf`s. In the same way are defined symmetrical and unsymmetrical systems
of three phase voltages and currents.
An example of the three phase symmetrical system of emf`s is
eA= Emsin ωt,
7.1
eB=Emsin(ωt-120°)
70
ec= Emsin(ωt – 240°).
These relationships may be represented graphically by the sinusoidal curves and the vector diagram
shown in Fig.7.2a and Fig.7.2b
e
+j
eA
eA
E& C
eA
0
E& A
ωt
120
0
120
0
120
+
E& B
0
a.)
Fig.7.2
b.)
If the vectors EA, EB and EC of Fig.7.2b revolve counter-clockwis at an angular velocity ω they
cross the imaginary axis +j in the following order: EA—EB---EC referred to , as the ABC order of phase
sequence or phase rotation of the three-phase system. The ABC order is positive phase sequence.
Thrfore in the case, when q = 1 we have a positive phase sequence three phase symmetrical system of
emf`s. When q = 2 we have ACB order of phase sequence. It is called a negative phase-sequence
system. When q = 0 all emf`s are in phase and this set of emf`s is called a zero phase-sequence system.
Vector diagrams of above mention systems of emf`s are shon in Fig.7.3
EA
EA
E
E
E
A
EC
EB
EB
q=2
q =1
EC
B
C
q=0
Fig.7.3
In the same way are defined systems of three phase voltage and current with positive, negative and
zero phase sequences.
The order of phase rotation may be found by means of phase-sequence indicator shown in Fig.7.4
71
•
Bright
Dim
•
•
Fig.7.4
It consists of two identical lamps and a capacitor. The capacitor is chosen so that the capacitor
reactance is equal to the resistance of each lamp. When the three leads of the instrument are connected
to the phases of the three-phase system, the lamps come on, one shining bright and the other dull.
Assuming that the lead from the capacitor is connected to the phase A , the phase containing the bright
lamp will be the phase B and the phase containing the dim lamp will be the phase C.
A three-phase generator may be connected to the three-phase load in a variety ways. Thus each
phase winding of the generator could be connected to the corresponding load by two wires. This system
is called an unconnected three phase system with six wires. In practice mostly are used interconnected
three phase systems, where the number of wires is greatly reduced. The most common types of
interconnections are star connection and delta connection, applicable to both the generator and the load.
As a result, the number of wires in a system is reduced to three or four. There are a three phase star and
a three phase delta systems circuit diagrams in Fig.7.5 and 7.6 respectively.
A′
A
eA
eC
ZA
ZB
eB
0
C
0′
Fig.7.5
A′
A
eCA
C
Z A′ B′
e AB
e BC
B′
B
C′
ZC
B′
B
Z C ′ A′
Z B′ C ′
Fig. 7.6.
It`s possible to have a star-to-delta or a delta-to-star three phase systems.
Three phase systems are divided in balanced and unbalanced systems. If an instantaneous power of
the three phase system depends on time it is the unbalanced three phase system and in opposite case it
is the balanced three phase system.
72
7.3. Advantages of Three-phase Systems
The popularity of three-phase systems can be explained by the several principal advantages they
offer, namely:
It is more economical to transmit AC power three-phase than with any other number of phases.
1.
It allows to generate a revolving magnetic field which is used in the three-phase induction motor. It
2.
must be noted that the three-phase induction motor is the most commonly used type of motor.
The components of three-phase systems, such as three-phase synchronous alternators, three-phase
3.
induction motors and three-phase transformers are simple to manufacture and economical and reliable
in service.
7.4. Star-to-star Connection
Fig.7.7 shows star-to-star connected three phase circuit were the ends (X,Y,Z) of the source phase
windings are connected together to form a so-called neutral point O, leaving the beginnings (A,B,C) as
terminals for connecting the source to the load by means of three wires referred to as the line wires.
I& A
A′
A
E& A
E& C
0Z x
y
ZA
I&N
0′
ZB
E& B
ZC
I&B
C
B′
B
C′
I&C
Fig.7.7
The currents iA, iB, iC flowing through the line wires are called the line currents. Similarly the ends
of the load phases are connected together to form a neutral point O’. A wire O –O’ connected the
neutral points of the source and load is called the neutral line. Current flowing through the neutral line
is called the neutral current. The current flowing through the phase of the source or load is called the
phase current. As it`s clear from the Fig.7.7 line and phase currents are equal to each other. So in the
case of star connection Il-= Iph.
The potential difference between the line terminals and the neutral is called the phase voltage. The
phase voltages of the source are VAO, VBO, VCO and the load phase voltages are VA’O’, VB’O’VC’O’. The
potential difference between any pair of line wires is called the line voltages. The source line voltages
are VAB, VBC, VCA and the load line voltages are VA’B’, VB’C’, VC’A’. The relationship between the line
and phase voltages are established by Kirchhoff`s voltage law. For example the relationship between
source line and phase voltages are
V&AB = V&A − V&B ⎫
⎪⎪
.V&BC = V&B − V&C ⎬
7.2
⎪
V&CA = V&C − V&A ⎪⎭
The phasor diagrams of symmetrical three phase and line voltages are shown in Fig.7.8a
73
− V&B
C
V&AB
I&C
b V&A
30 0 ϕ
α
30
V&C
− V&A
V&CA &
IB
I&A
I&A
V&BC B
0
C
I&C
− V&C
60 0 I&
B
b.)
a.)
Fig.7.8
The solution of triangle ∆abc gives the relationship between phase and line voltages of the
symmetrical system
V L = 3V ph
7.3
According to the Kirchhoff`s current law for the node O we have
I&N = I& A + I&B + I&C .
7.4
There are the vector sum of line (phase) currents in Fig.7.8b. Hence in the case of the symmetrical
three-phase system the neutral current is equal to zero IN=0.The line or phase current may be found
using Ohm`s law
Iph = IL = Vph/Zph,
7.5
where Zph= the impedance of the phase of the load.
In the case of the symmetrical three phase system Z = ZA = ZB = ZC. There is no necessity in the
neutral line and it`s possible to use three phase three wire (instead of four wire) system shown in
Fig.7.5.
Above formulae are used in order to calculate three phase three or four wire symmetrical systems. It
must be noted that these system are calculated like a single phase circit.
Example 7.1.For the circuit of Fig.7.7 the three phase load impedances are ZA = 3Ω , ZB =4∠60°Ω
. Determine line and neutral currents if the circuit is supplied from a 240V, three phase, 50Hz source.
Solution:
It is supposed that the set of three phase voltages of the load is
V&A = 240 3 = 138.6∠0 0 V ,
V& = 138.6∠ − 120 0 V ,
B
V&C = 138.6∠120 0 V ,
Applying Ohm`s law the line currents will be
0
0
&
&
&I = V A = 138.6∠0 = 46.2∠0 0 A; I& = VB = 138.6∠ − 120 = −3465∠0 0 A;
A
B
ZA
ZB
3∠0 0
4∠60 0
V&
138.6∠120 0
I&C = C =
= 27.7∠30 0 A.
ZC
5∠90 0
The neutral current is
I&N = I& A + I&B + I&C = 46..2 − 34.65 + 27.∠30 0 = 38.14∠21.30 0 A.
74
7.5. Delta Connected Three-phase Circuits
Delta connection of the three phase windings of the source of supply makes a closed circuit as shown in Fig.7.9. A
similar circuit is formed at delta connection of the load phases.
A′
I& A
A
I& A′B′
E& CA
I&C ′A′
Z C ′ A′
Z A′ B′
E& AB
I&B
C
E& BC
B
B′
I&C
I&B′C ′ Z B′ C ′
Fig.7.9
The phase junctions make the line terminals by which source and load are intercconnected, thus
forming a three-phase three-wire system. There are line currents IA, IB, IC and phase currents IAB, IBC,
ICA in Fig.7.9. In this case phase and line voltages are equal VL = Vph. As to line and phase currents
using Kirchhoff`s current law for the nodes A B C may be written
I& A = I& AB − I&CA ,
7.6
I& = I& − I& ,
B
BC
AB
I&C = I&CA − I&BC .
The phasor diagram of the line and phase currents is shown in Fig.7.10.
− I&B ′C ′
C
I&C ′A′
I&C
I&B
− I& A′B ′
I& A′B ′
a
I&B ′C ′
30 0
− I&C ′A′
I& A
b
Fig.7.10
The phasors are drawn for the symmetrical three phase system. From the diagram follows that
I& A + I&B + I&C = 0 ,
7.7
I& + I& + I& = 0 .
AB
BC
CA
From the ∆abc bc = √ 3 ab. So the relationship between line and phase currents is
I L = 3I ph .
The rms value of the phase current may be found using Ohm`s law
75
7.8
I& ph =
V&ph
.
Z
These formulae are used to calculate symmetrical delta connected three-phase systems.
7.9
Example 7.2. For the circuit of Fig.7.9 the three-phase load impedances are
ZAB =5∠10° Ω , ZBC = 10∠80° Ω , ZCA = 9∠30° Ω . Determine the phase and line currents if the
circuit is supplied from a 450V three-phase 50Hz source.
Solution:
The set of three phase emf-s is
EAB = 450V, EBC = 450∠-120° V, ECA = 450∠120° V.
Applying Ohm`s law to the three phase impedances the phase currents will be
EAB 450∠0°
IAB = ---- = -------- = 90∠-10° A,
5∠10°
ZAB
Ebc 450∠-120°
IBC = ---- = -------------- = 45∠-200° A,
ZBC
10∠80°
ECA 450∠120°
ICA = ---- = ------------- = 50∠90° A.
9∠30°
ZCA
The line currents are
IA = IAB – ICA = 90∠-10° - 50∠90° = 110.3∠-36.52° A,
Ib = IBC - IAB = 45∠-200° - 90∠-10° = - 134.54∠-13.33° A,
IC = ICA – IBC = 50∠90° - 45∠-200° = 54.65∠39.30° A.
7.6. Calculation of Unsymmetrical Three-phase Circuits with
Star-to-star Connection
As may be seen from the circuit diagram shown in Fig.7.7 the star connected three phase system is
an electric circuit with tow nodes 0 and 0’.Therefore it is most convenient to calculate such a circuit
using node voltage method.
The node voltage V00′ may be found by the formula
E& Y + E& B Y B + E& C Y C
,
7.10
V&00& = A A
Y A +Y B +YC +Y N
where YA, YB, YC are phase admittances,
YN is admittance of the neutral line.
The load phase voltages will be
V&A = E& A − V&00′ ; V&B = E& B − V&00′ ; V&C = E& C − V&00′
7.11
and the phase currents are
I& A = V&A Z A ;
I&B = V&B Z B :
I&C = V&C Z C .
7.12
The current in the neutral line is
IN = V00’ /ZN,
7.13
where ZN is the impedance of the neutral line.
76
Suppose the source line and phase voltages are symmetrical three phase system of voltages. Then
the phasor diagram of voltages looks as it`s shown in Fig.7.11.
A
V&A′
V&A
V&B
V&C
V&AB
N
V&N ′N
V&CA
N′
C
B
V&C ′
V&B′
V&BC
Fig.7.11
For each phases
V&A = V&A′ + V&00′
.V&B = V&B′ + V&00′
V& = V& ′ + V& ′
C
C
00
⎫
⎪⎪
⎬
⎪
⎪⎭
7.14
In this case the load is unbalanced and the neutral wire impedance ZN other than zero. As may be
seen from the diagram, the neutral point of the load does not coincide with that of the source, points 0
and 0 ‘ are at different potentials.
Four-wire systems are used extensively for connecting power and lighting loads to 380/220 Volt
supply.
Example 7.3. For the circuit of Fig.7.8 the three-phase load impedances are ZAB=15∠30° Ω , ZBC=
12∠-30° Ω , ZCA= 25∠0° Ω . The impedance of the neutral line ZNN=∞. Determine the line currents
and the node (neutral) voltage VNN if the phase emf-s are EA= 240V, EB= 260∠-100V and EC=
180∠110° V.
Solution:
The phase admittances are
1
1
1
1
YA= ----- = --------- = 0.0667∠-30° Sm,
YB= ---- = ---------- = 0.0833∠30°Sm,
ZAB 15∠30°
ZBC 12∠-30°
1
1
YC= ---- = --------- = 0.04∠0° Sm,
YN= 0.
ZCA 25∠0°
The neutral voltage is
EAYA+ EBYB+ ECYC 240*0.0667∠-30° + 260∠-100° *0.0833∠30° +
V00’= ------------------------ = -------------------------------------------------------------0.0667∠-30° + 0.0833∠30° + 0.04∠0° + 0
YA+ YB+ YC+ YN
+ 180∠110° *0.04∠0° .
77
---------------------------- = 168.41∠-51.7° V.
The load phase voltages will be
VA= EA- VNN= 240 – 168.41∠-51.7°= 189.34∠44.3° V,
VB= EB- VNN= 260∠-100° - 168.41∠-51.7° = - 194.18∠219.6° V,
VC= EC – VNN= 180∠110 – 168.41∠-51.7° = 344.02∠118.8° V.
The line currents are
VA 189.37∠44.3°
IA= ---- = ------------------- = 12.62∠14.3°= (12.23+ j3.12)A,
15∠30°
ZAB
194.18∠219.6°
VB
IB= ---- = -------------------- = 16.18∠249.6°= (- 5.64 – j15.16) A,
12∠-30°
ZBC
VC 344.02∠118.8°
IC= ---- = ------------------- = 13.76∠118.8° =(- 6.63+ j12.04)A.
25∠0°
ZCA
The sum of line currents is
IA+ IB+ IC = 12.23+ j3.12 – 5.64 – j15.16 – 6.63+ j12.04 ≅ 0.
7.7. Power Calculations
If the load is unbalanced, the power should be found separately for each phase. As an example in phase A
PA = V A I A cos ϕ A ⎫
⎪
Q A = V A I A sin ϕ A ⎬
7.15
⎪
S A = VA I A
⎭
The powers in other phase are found similarly. The total active, reactive and apparent power are
P = PA + PB + PC ⎫
⎪
Q = Q A + QB + QC ⎬
7.16
⎪
S = S A + S B + SC ⎭
In the case of symmetrical three phase systems powers are
P = 3Pph = 3V ph I ph cos ϕ , ⎫
⎪⎪
7.17
Q = 3Q ph = 3V ph I ph sin ϕ ,⎬
⎪
S = 3S ph = 3V ph I ph.
⎪⎭
It`s possible to find powers using line voltage and line current
P = 3V L I L cos ϕ , ⎫
⎪⎪
Q = 3VL I L sin ϕ , ⎬
⎪
S = 3VL I L .
⎪⎭
78
7.18
In above formulae the reactive power of a coil is taken to be positive and that of a capacitor negative.
Problems:
7.1. A star connected three phase impedances to be equal to each other is connected to a three phase
star connected generator the phase impedance of which ZG= (0.3+j0.8)Ω .
The line voltage at the load draws active power 5,46W at cosϕ = 0.8 lagging. Find the emf of
generator.
7.2. A delta connected three phase load gets supply from the symmetrical three phase sorce having the
line voltage 220V. The phase impedance of the load Zph=(10+j10)Ω .
Find line and phase currents of the load. Determine active, reactive and apparent powers of the load.
7.3. A three phase circuit consists of a symmetrical star connected three phase generator having line
voltage 380V and a nonsimmetrical star connected three phase load phase impedances of which are
ZA= (6+j8)Ω , ZB= (25+j7)Ω and ZC=20Ω . Find current, voltage and active power of each phases.
7.4. Line voltages of the three phase system are VAB=120∠0° V, VBC= 110∠-130° V and VCA=
125∠110° V. A three-phase delta connected load having phase impedances ZAN= 25Ω , ZBC= 20 Ω
and ZCA= (16+j8)Ω is connected to the above system. Find line and phase currents.
References
1. L. Bessonov. Applied Electricity for Engineers. Mir . Moscow
2. Charles I. Huberst. Electric Circuits AC/DC. An integrated Approach. International Student
Editions.
3. F. E. Evdokimov. Fundamentals of Electricity. Mir . Moscow, 1977
4. M. E. Van Valkenburg. Network Analysis. Prentice-Hall INC. Englewoad Cliffs, New Jercey.
5. p. merabiSvili, g. xosroSvili. eleqtroteqnikis Teoriuli safuZvlebi.
“ganaTleba”, Tbilisi, 1988.
6. Г. И. Атабеков. Основы теории цепей. Энергия, Москва, 1965
7. А. Ф. Белецкий. Теория линейных электрических цепей. Радио и Связь, Москва, 1986
8. Г. В. Зевеке, П. А. Ионкин, А. В. Нетушил, С. В. Страхов. Основы теории цепей.
Энергоaтoмиздать. Москва, 1989
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Школа, 1981.
10. М. Р. Шебес, М. В. Каблукова. Задачник по теории линейных электрических цепей.
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