Michał Tadeusiewicz

Electric Circuits

Technical University of Łódź International Faculty of Engineering

Łódź 2009

Contents

Preface…………………………………………………….. 5

1. Fundamental laws of electrical circuits…………………… 7 1.1. Introduction ............................................................. 7 1.2. Kirchhoff’s Voltage Law (KVL).............................. 8 1.3. Kirchhoff’s Current Law (KCL)................….......... 10 1.4. Independence of KCL equations………………….. 12 1.5. Independence of KVL equations………………….. 13 1.6. Tellegen’s theorem………………………………... 14

2. Circuit elements.................................................................... 16 2.1. Resistors ................................................................... 16 2.2. Independent sources………………………………. 18

3. Power and energy…………………………………………. 24

4. Simple linear resistive circuits……………………………. 26

5. Resistive circuits. DC analysis……………………………. 30 5.1. Superposition theorem and its application………… 30

6. Three terminal resistive circuits…………………………... 35

7. The Thevenin-Norton theorem……………………………. 40

8. Node method……………………………………………… 47

9. Simple nonlinear circuits………………………………….. 53

10. Controlled sources………………………………………… 59

11. Capacitor……………………………...…………………... 61 11.1. Introduction.............................................................. 61 11.2. Continuity property……………………………….. 64

11.3. Energy stored in a capacitor………………………. 65 11.4. Series connection of capacitors…………………… 66 11.5. Parallel connection of capacitors………………….. 68

12. Inductor…………………………………………………… 69 12.1. Introduction.............................................................. 69 12.2. Continuity property………………………………... 72 12.3. Hysteresis………………………………………….. 73 12.4. Energy stored in an inductor……………………… 74 12.5. Series connection of inductors…………………….. 75 12.6. Parallel connection of inductors…………………... 76

13. Operational-amplifier circuits…………………………….. 78 13.1. Description of the operational amplifier................... 78 13.2. Examples………………………………………….. 82 13.3. Finite gain model of the operational amplifier......... 85

14. First order circuits driven by DC sources…………………. 88 14.1. Resistor-inductor circuits............................……...... 88 14.2. Resistor-capacitor circuits………………………… 96

15. Sinusoidal steady-state analysis…………………………... 104 15.1. Preliminary discussion…………………………….. 104 15.2. Phasor concept…………………………………….. 108 15.3. Phasor formulation of circuit equations…………… 110 15.4. Impedance and admittance………………………... 115 15.5. Phasor diagrams…………………………………… 122 15.6. Effective value…………………………………….. 125

16. Power in sinusoidal steady state…………………………... 128 16.1. Instantaneous and average power………...……….. 128

3

16.2. Complex power……………………………….. 130 16.3. Measurement of the average power……………….. 134 16.4. Theorem on the maximum power transfer………... 135

17. Resonant circuits.................................................................. 137 17.1. Series resonant circuit……………………………... 137 17.2. Parallel resonant circuit…………………………… 144

18. Coupled inductors………………………………………… 148 18.1. Basic properties…………………………………… 148 18.2. Connections of coupled inductors………………… 151 18.3. Ideal transformer....................................................... 156

19. Three-phase systems……………………………………… 159 19.1. Introduction……………………………………….. 159 19.2. Y-connected systems……………………………… 161 19.3. Three-phase systems calculations…………………. 164 19.4. Power in three-phase circuits……………………… 168

Reference books…………………………………………... 174

4

Preface

This book presents an introductory treatment of electric circuits and

is intended to be used as a textbook for students, during the junior

years, at the International Faculty of Engineering of the Technical

University of Łódź. The book covers most of the material taught in

conventional circuit courses and gives the fundamental concepts

required to understand and tackle the electrical engineering problems.

Its prerequisites are the basic calculus, complex numbers, and some

familiarity with integral calculus and linear differential equations,

which are desirable but not essential. The objective of the book is to

feature theories and concepts of fundamental importance in electrical

engineering that are amenable to a broad range of applications.

The book includes a large number of examples. They are provided

to illustrate the concepts and to make the theory more clearer. On each

page there is a blank area where a student can note down comments,

explanations and additional examples discussed during the lectures.

The book can be thought of as consisting of three parts. Part 1

(Chapters 1-10, 13) introduces many basic concepts, laws, and

principles related to electric circuits. In addition different methods of

the DC analysis of resistive circuits are studied in detail. Part 2

(Chapters 11-12, 14) deals with simple linear dynamic circuits and

their components. The transient analysis of the first order circuits

is considered. Part 3 (Chapters 15 to 19) focuses on sinusoidal

circuits in the steady-state and discusses many different aspects of

AC analysis. At the end of this part, three-phase systems are

introduced and analysed.

I gratefully acknowledge the support and encouragement of

Dr. Tomasz Saryusz-Wolski, Head of the International Faculty of

Engineering of the Technical University of Łódź.

Łódź, 2009 Michał Tadeusiewicz

5

1. Fundamental laws of electrical circuits

1.1 Introduction

An electric circuit is an interconnection of electric devices (elements) by conducting wires. Figure 1.1 shows a circuit consisting of a voltage source, two resistors, a transistor, a capacitor, and a transformer. Any junction in the circuit where terminals of the elements are joined together is called a node. On the circuit diagrams they are marked with dots.

Fig. 1.1. An example of a circuit

1 2i

v Fig. 1.2. Reference directions of current i and voltage v

In the circuits we consider currents flowing through the elements (branches) and voltages between any two nodes. The unit for voltage is the volt (V), whereas the unit for current is the ampere (A). Figure 1.2 shows the reference direction of current i and voltage v represented by arrows.

Notes

7

If at some time current is positive, then it flows into the element by node 1. If the current is negative it flows out of the element by node 1. The reference direction of the voltage across the element is represented by an arrow v. If at some time voltage is positive, it means that the electric potential of node 1 is larger than the electric potential of node 2. If it is negative then the electric potential of node 1 is smaller than the electric potential of node 2. The reference direction of each current and each voltage can be assigned arbitrarily. When they are chosen as shown in Fig. 1.2, we say that we have chosen associated reference directions. This is the convention we will follow throughout the whole course.

1.2 Kirchhoff’s Voltage Law (KVL) The fundamental laws governing electric circuits are Kirchhoff’s voltage and current

laws and Tellegen’s theorem. In an electric circuit we consider a path traversing some branches in succession. If the starting node of a path is the same as the ending node, the path is called a loop.

Kirchhoff’s Voltage Law

For any electric circuit, for any of its loop, and at any time, the algebraic sum of the branch voltages around the loop is equal to zero. To write KVL equation we select a loop and assume arbitrarily its reference direction, clock-wise or counter clock-wise. Next we assign the plus sign to the branch voltages whose reference directions agree with that of the loop and the minus sign to the others. Example 1.1 KVL equation for the loop 1, 2, 3 in the circuit shown in Fig. 1.3:

( ) ( ) ( ) 0321 =+− tvtvtv . KVL equation for loop 1, 4, 5, 7:

( ) ( ) ( ) ( ) 07541 =−++− tvtvtvtv .

Notes

8

v6(t)

v3(t)

v2(t)

v1(t) v4(t)

v5(t)

v7(t)

Fig. 1.3. An example circuit for illustrating KVL

KVL can also be expressed in terms of voltages between nodes creating a closed node sequence. A node sequence is called a closed node sequence if it starts and ends at the some node. Kirchhoff’s Voltage Law (general version)

For any electric circuit, for any closed node sequence, and for any time, the algebraic sum of all node-to-node voltages around the chosen closed node sequence is equal to zero. Example 1.2

64 5

3 2 1

v6 v3

v2 v1

v4

v5 v2,5

Fig. 1.4. An example circuit for illustrating KVL

Notes

9

Let us consider closed node sequences: 1, 2, 5, 4, 1 and 1, 2, 3, 6, 5, 4, 1. KVL equations: 1, 2, 5, 4, 1: 013522 =++−− vvvv , . 1, 2, 3, 6, 5, 4, 1: 0135642 =++−++− vvvvvv .

1.3 Kirchhoff’s Current Law (KCL) Another fundamental law governing electric circuits is Kirchhoff’s Current Law, as follows.

For any electric circuit, for any of its nodes, and at any time the algebraic sum of all the branch currents meeting at the node is zero. In the algebraic sum we assign the plus sign to the currents leaving the node and the minus sign to the currents entering the node. Example 1.3

3

2 1

i3(t)

i 2(t)

i 1(t)

i 4(t)

Fig. 1.5. An example circuit for illustrating KCL

1: ( ) ( ) ( ) 0321 =+− tititi , or simply 0321 =+− iii .

3: 043 =−− ii .

Notes

10

To formulate KCL in a more general form we consider gaussian surface defined as a balloon-like closed surface, as illustrated in Fig. 1.6.

the gaussian surface

i2

i3

i1

Fig. 1.6. The gaussian surface

KCL (general version) For all circuits, for all gaussian surfaces, for all times t, the algebraic sum of all currents crossing the gaussian surface at time t is equal to zero. In the algebraic sum we assign the plus sign to the currents leaving the gaussian surface and the minus sign to the currents entering the surface. Example 1.4 In the circuit shown in Fig. 1.6 we write KCL equation

0321 =−+− iii .

The topological properties of a circuit can be exhibited using a graph obtained by replacing each branch by a line. Each branch of the graph has orientation indicated by an arrow on the branch. This arrow is the same as the reference direction of the current flowing through the corresponding branch of the circuit. Thus, the graph can be used to write KVL and KCL equations.

Notes

11

1.4 Independence of KCL equations

For a given circuit we can write many KCL equations. Hence, the question arises how many of them are linearly independent.

4 3

11

3

2

2

Fig. 1.7. An example graph

To answer this question we consider the graph shown in Fig. 1.7 and write KCL equations at each node

0321 =−− iii ,

0421 =++− iii ,

043 =− ii .

If we add the first two equations together, we obtain

043 =+− ii .

Multiplying both sides of this equation by ( )1− yields

043 =− ii ,

which is exactly the third equation.

Notes

12

It means that the third equation is a linear combination of the first two equations. Thus, not each equation brings new information not contained in the others and at least one equation repeats the information contained in the others. However, if we reject the third equation, then the remaining ones are linearly independent. Thus, the third equation is redundant, it is useless and can be discarded. Generally, the following independence property of KCL equations holds. For any graph with n nodes KCL equations for any ( ) of these nodes form a set of 1−n ( )1−n linearly independent equations.

1.5 Independence of KVL equations

Similarly as in the case of KCL equations the question arises how to write a set of linearly independent KVL equations. The simplest answer is as follows. We write KVL equations selecting the loops so that any equation contains at least one voltage that has not been included in any of the previous equations.

It can be shown that for a circuit having b branches and n nodes 1+− nb linearly independent equations can be formulated.

8

3

1 2

4

5

6

7

I

IV

IIIII

Fig. 1.8. A graph for illustrating independence of KVL equations

Example 1.5 Let us consider the graph shown in Fig. 1.8. In this graph we write linearly independent KVL equations using the provided rule. As a result we obtain the following set of equations

Notes

13

0271 =++ vvv ,

0432 =−−− vvv ,

0654 =++ vvv ,

0867 =++ vvv .

1.6 Tellegen’s theorem

Let us consider a graph having b branches and n nodes. Let us use the associated reference directions. Tellegen’s theorem

Let be any set of branch currents satisfying KCL at any node and let be any set of branch voltages satisfying KVL at any loop. Then it holds bi,,i,i …21

bv,,v,v …21

01

=∑=

b

kkk iv .

Note that the set of branch currents and the set of branch voltages are associated with the given graph but not necessarily with the same circuit. For example, let us consider the graph shown in Fig. 1.9 and two different circuits depicted in Figs 1.10 and 1.11 having this graph. Tellegen’s theorem enables us to write the following equations:

044332211 =+++ iviviviv ,

044332211 =+++ i~v~i~v~i~v~i~v~ ,

044332211 =+++ i~vi~vi~vi~v ,

044332211 =+++ iv~iv~iv~iv~ .

Notes

14

43

2

1

1

3

2

Fig. 1.9. A graph having three nodes and four branches

v1

v2

v4

i4

i1

i3

v3

i2

1v~

2v~

4v~4i~

1i~

3i~

3v~

2i~

Fig. 1.10. A circuit having the graph of Fig. 1.9 Fig. 1.11. A circuit having the graph of Fig. 1.9

Notes

15

2. Circuit elements

The components used to build electric circuits are called circuit elements. In this section we define simple two-terminal circuit elements: a resistor, independent voltage and current sources.

2.1 Resistors

An element is said to be a resistor if its voltage-current relation is of algebraic type.

This relation is represented graphically by a curve in iv − plane, called the characteristic of the resistor. Any resistor can be classified as linear or nonlinear.

A resistor is called linear if its characteristic is a straight line through the origin (see Fig. 2.1).

v

i0

Fig. 2.1. Characteristic of a linear resistor

It is described by the equation iRv = (2.1) or vGi = , (2.2)

Notes

16

where RG 1= . Equation (2.1) is known as Ohm’s law, R is called the resistance and G is called the conductance. The unit for resistance is the ohm ( ) and for conductance the siemens

Ω( )S . The symbol of a linear resistor is shown in Fig. 2.2.

v

i R or G

Fig. 2.2. The symbol of a linear resistor

Any resistor whose characteristic is not a straight line through the origin is

classified as a nonlinear resistor. A typical example of a nonlinear resistor is a diode, described by the equation

iv −

( )1e −= vKi λ , (2.3)

where K and λ are positive constants. Fig. 2.3 shows the symbol and the characteristic of a semiconductor diode.

0 v

i

v

i

Fig. 2.3. Symbol and characteristic of a diode

Notes

17

A general symbol of any nonlinear resistor is depicted in Fig. 2.4.

v

i

Fig. 2.4. General symbol of a nonlinear resistor

2.2 Independent sources

Voltage source

An element is called a voltage source if it maintains a prescribed voltage ( )tvS between its terminals for any current flowing through the source. Consequently, a voltage source maintains a prescribed voltage ( )tv between its terminals in an arbitrary circuit to which it is connected.

S

The symbol of the voltage source is shown in Fig. 2.5. vS(t)

Fig. 2.5. The symbol of a voltage source

Generally, the prescribed voltage is a time varying signal ( )tvS . In a special case, it is constant , called a DC voltage source. In such a case, the characteristic expressing the voltage between the terminals of the source in terms of the current flowing through the source is a horizontal line, as shown in Fig. 2.6.

SV

VS

v

i 0

Fig. 2.6. Characteristic of a DC voltage source

Notes

18

The defined voltage source is an ideal element not encountered in the physical world. A real voltage source can be represented by an equivalent circuit shown in Fig. 2.7.

RS VS

Fig. 2.7. Model of a real voltage source

Certain devices have very small and can quite effectively be approximated by the ideal voltage source.

SR

Let us consider a real voltage source terminated by a load, as shown in Fig. 2.8.

i

SRvv

RS

VS Load

Fig. 2.8. A real voltage source terminated by a load

Using KVL and Ohm’s law we write the equation

iRVv SS −= , (2.4)

that describes, on the plane, the straight line, shown in Fig. 2.9. This line is a characteristic of the real voltage source, and is called a load line.

vi −

Notes

19

slope = -RS

RVS

v

i 0

VS

Fig. 2.9. Load line of a real voltage source

Current source

A current source is an element which maintains a prescribed current ( )tiS for any voltage ( ) between its terminals. Consequently, a current source maintains a prescribed current ( ) in an arbitrary circuit to which it is connected. The symbol of a current source is shown in Fig. 2.10.

tvtiS

iS (t)

Fig. 2.10. Symbol of a current source If the prescribed current is constant, ( ) SS Iti = the current source is called a DC current source. Its vi − characteristic is shown in Fig. 2.11.

Notes

20

IS

v

i 0

Fig. 2.11. Characteristic of a DC current source

The defined current source is an ideal element. A real current source can be

represented by the circuit shown in Fig. 2.12.

vRSIS

Fig. 2.12. A real DC current source

Let us consider a real DC current source terminated by a load, as shown in Fig. 2.13.

Notes

21

i

IS v

RS

iR

Load

Fig. 2.13. A real DC current source terminated by a load

Applying KCL at the top node we write

0=−+ SR Iii . (2.5) Since

S

R Rvi = , (2.6)

then

S

S RvIi −= . (2.7)

Equation (2.7) describes characteristic of a real DC source, as shown in Fig. 2.14. This characteristic is a straight line, called a load line.

iv −

RIS v

i

0

IS

Fig. 2.14. Characteristic v - i of real DC current source

Notes

22

Let us multiply both sides of equation (2.7) by and rearrange this equation as follows

SR

iRIRv SSS −= . (2.8)

By denoting SSS IRV = we obtain equation (2.4) that describes a real voltage source. Thus, if SSS IRV = , then the real current source and the real voltage source are equivalent.

Notes

23

3. Power and energy

Let us consider a circuit and draw two wires from this circuit. As a result we obtain a two-terminal circuit called a one-port. If the one-port is supplied with a source, the current ( )ti flows into the one-port by terminal A and the same current ( )ti flows out of the one-port by terminal B. Therefore, we indicate only one current ( )ti , as shown in Fig. 3.1. The voltage ( ) between the terminals is also indicated. tv

A

B

v(t)

i(t)

Fig. 3.1. A one-port with indicated the port voltage and port current

The current ( )ti and the voltage ( )tv are called a port current and a port voltage, respectively.

The instantaneous power entering the one-port is equal to the product of the port voltage and port current

( ) ( ) ( )titvtp = , (3.1)

where ( )tv is in volts, in amperes and ( )ti ( )tp in watts (abbreviated to W). tThe energy delivered to the one-port from time to is given by the equation 0t

, ( ) ( ) ( ) ( ) τττττ divdpt,twt

t

t

t∫∫ ==00

0

Notes

24

where a variable τ means time. Hence, it holds

( )tpdtdw

= .

If the one-port is a linear resistor, specified by R or G, then

( ) ( ) ( ) ( )tRititvtp 2== , (3.2)

( ) ( ) ( ) ( )tGvtitvtp 2== . (3.3) Thus, the instantaneous power is, in this case, nonnegative for all t.

ivIf the one-port is a nonlinear resistor, represented by a − characteristic located in the first and third quadrants only, then ( ) ( ) 0≥titv and the power entering the resistor is nonnegative, its energy is a nondecreasing function of time and the resistor consumes the energy. Such a resistor is called passive. If some parts of iv − characteristic lie in the second or third quadrant, then for some t, ( ) ( ) 0<titv , the power entering the resistor is negative and the resistor delivers energy to the outside world. Such a resistor is called active.

Notes

25

4. Simple linear resistive circuits

Circuits consisting of resistors and sources are classified as resistive circuits. In particular, they can be supplied with DC sources only. The analysis of such a class of circuits is called DC analysis.

Let us consider a circuit consisting of two linear resistors connected in series, as shown in Fig. 4.1.

R2R1

v2v1

i

v Fig. 4.1. Two resistors connected in series

To analyse this circuit we apply the KVL and Ohm’s law. Since the same current traverses both resistors we write

( )iRRiRiRvvv 212121 +=+=+= . (4.1) Hence, we have

21 RRiv

+= . (4.2)

Equation (4.2) states that the series connection of two linear resistors is equivalent to resistor R,

21 RRR += . (4.3)

Voltages across the resistors are specified by the equations:

vRR

RiRv21

111 +== , v

RRRiRv

21

222 +== .

Notes

26

Hence, it follows the relation

2

1

2

1

RR

vv

= , (4.4)

which states that the series connection of resistors and can be considered as a voltage divider. The voltage v is divided in proportion to and . Formula (4.3) can be directly generalized to the series connection of n linear resistors,

1R 2R

1R 2R

nR,,R …1

nRRR ++= 1 . (4.5)

Figure 4.2 shows the circuit consisting of two linear resistors connected in parallel.

R2R1

i2i1i

v

Fig. 4.2. Two resistors connected in parallel

Voltages across resistors and are identical and equal to v. The current i according to KCL satisfies the equation:

1R 2R

21 iii += . (4.6) Using Ohm’s law

1

1 Rvi = ,

22 R

vi = (4.7)

Notes

27

we obtain

vRR

i ⎟⎟⎠

⎞⎜⎜⎝

⎛+=

21

11 , (4.8)

or

iRv = (4.9) where

21

21

21

111

RRRR

RR

R+

=+

= . (4.10)

Thus, two resistors connected in parallel are equivalent to the resistor R specified by (4.10).

Using (4.7), (4.9), and (4.10) we write

iRR

RRvi

21

2

11 +

== , iRR

RRvi

21

1

22 +

== .

Hence, it holds

1

2

2

1

RR

ii= . (4.11)

Thus, the parallel connection of resistors and can be considered as a current divider, where the currents are divided according to equation (4.11).

1R 2R

Equation (4.10) can be directly generalised to the circuit consisting of n resistors connected in parallel nR,,R …1

nRR

R11

1

1++

= . (4.12)

Figure 4.3 shows a three-terminal resistor called a potentiometer. Terminal 3 called a wiper, can be shifted along the resistor , dividing it into and . pR xR yR

Notes

28

RpRx

Ry3

R vy

i

V RpRx

Ry

Fig. 4.3. A potentiometer Fig. 4.4. Circuit containing a potentiometer

Let us consider the circuit shown in Fig. 4.4, containing a potentiometer. The circuit can be considered as a series connection of resistor and parallel connected resistors

and xR

yR R . Hence, the resistance faced by the voltage source is

RR

RRRR

y

yx ++= . (4.12)

Formula (4.12) enables us to find the current i

RRRR

R

vi

y

yx ++

= (4.13)

and then voltage yv

( ) vRRRRR

RRRR

RRiv

yyx

y

y

yy ++

=+

= . (4.14)

Notes

29

5. Resistive circuits. DC analysis

In this section we study circuits consisting of linear resistors and independent DC sources. We formulate some theorems governing these circuits, which enables us to analyse them efficiently.

5.1 Superposition theorem and its application Let us consider a linear circuit driven by n voltage sources

nSSS V,,V,V …21

, and m current sources

mSSS I,,I,I …21

.

The superposition theorem states that any branch current and any branch voltage in this circuit is given by the expression of the form

mn SmSSSnSS IkIkIkVhVhVh +++++++

2121 2121 , (5.1)

where coefficients and ( )n,,jh j …1= ( )m,,jk j …1= are constants and depend only on circuit parameters.

In other words, any branch current and any branch voltage is a linear combination of the voltage and current sources. Example 5.1 Let us consider a circuit consisting of linear resistors, a single voltage source and a single current source. We extract from this circuit the sources and an arbitrary resistor, as shown in Fig. 5.1. We wish to find the current i flowing through resistor R using the superposition theorem.

Notes

30

IS

i

RVS

Fig. 5.1. A circuit with extracted sources and a resistor

According to the superposition theorem

SS kIhVi += . (5.2)

Let and ShVi~ = SkIi~~ = , then

i~~i~i += . (5.3)

Note that if and i~i = 0=SI i~~i = if 0=SV .

If , then the branch containing the current source can be replaced by an open circuit (see Fig. 5.2). Thus, is the current flowing through resistor R in the circuit with the current source set to zero (removed). If 0

0=SIi~

=SV , then the branch containing the

voltage source can be replaced by a short circuit (see Fig. 5.3). Thus, i~~ is the current

flowing through resistor R in the circuit with the voltage source set to zero (short-circuited). The current can be regarded as a response of the circuit due to the voltage source acting alone. The current

i~

SV i~~ can be considered a response of the circuit due

to the current source acting alone. SI

Notes

31

i~

IS=0

R

VS

i~~

IS

R

VS=0

Fig. 5.2. The circuit shown in Fig. 5.1 driven Fig. 5.3. The circuit shown in Fig. 5.1. driven

by the voltage source VS by the current source IS

Generally, the response of the circuit due to several voltage and current sources is equal to the sum of the responses due to each source acting alone, that is with all other voltage sources replaced by short circuits and all other current sources replaced by open circuits. Example 5.2 Let us consider the circuit shown in Fig. 5.4, driven by a voltage source and a current source . We apply the superposition theorem to find voltage .

SV

SI 2v

R3

i2

v2R2IS

R1

VS

Fig. 5.4. A circuit driven by two sources

Notes

32

First we set the current source to zero. As a result, the circuit is driven only by the voltage source (see Fig. 5.5). SV

2i~

2v~

R3

R2

R1

VS

Fig. 5.5. Circuit of Fig. 5.4 driven only by the voltage source VS

In this circuit the same current 2i

~ traverses all the resistors. Hence, we find

321

2 RRRV

i~ S

++= ,

whereas the voltage 2v~ is, according to Ohm’s law, given by the equation

SvRRR

Ri~Rv~321

2222 ++== .

Now we set the voltage source to zero, obtaining the circuit shown in Fig. 5.6. SV

2i~~

2v~~

R3

R2IS

R1

Fig. 5.6. Circuit of Fig. 5.4 driven only by the current source IS

Notes

33

In this circuit the resistance faced by the current source equals

( )231

231

RRRRRR

+++

.

The product of this resistance and the current is voltage SI 2v~~

( )

SIRRRRRR

v~~231

2312 ++

+= .

The superposition theorem leads to the equation

( )SS I

RRRRRR

VRRR

Rv~~v~v

231

231

231

2222 ++

++

++=+= .

Notes

34

6. Three terminal resistive circuits

The three-terminal circuits shown in Figs 6.1 and 6.2 consist of three linear resistors. The circuit of Fig. 6.1 is called a Y circuit, whereas the one depicted in Fig. 6.2 is called a Δ circuit.

3Rv

1Rv2Rv

R1 R2

R3

i2

v2 i3

i1

v1

1

3

2

circuit

i2 i1 R12

R23R31

v2 v1

1

3

2

circuit

Fig. 6.1. Y circuit Fig. 6.2. Δ circuit

Notes

35

In a Y circuit it holds 213 iii += . (6.1) Hence, we find ( ) ( ) 23131213111 21

iRiRRiiRiRvvv RR ++=++=+= , (6.2) ( ) ( ) 23213213222 32

iRRiRiiRiRvvv RR ++=++=+= . (6.3)

To express and in terms of and in the Δ circuit, we apply current sources and to this circuit, as shown in Fig. 6.3.

1v 2v 1i 2i

1i 2i

i2 i1

R12

R23R31

v2 v1

1 2

3

Fig. 6.3. Δ circuit driven by two current sources

To find and we use the superposition theorem. First, we set current source to zero (see Fig. 6.4) and compute voltages

1v 2v 2i

1v~ and 2v~

Notes

36

( )

1231231

2312311 i

RRRRRR

v~+++

= , (6.4)

1231231

233123

2312

12 i

RRRRR

RRR

v~v~++

=+

= . (6.5)

i1

R12

R23R31

2v~v1

1 2

3

Fig. 6.4. The circuit shown in Fig. 6.3 with 02 =i

Next, we set current source to zero (see Fig. 6.5) and compute 1i 1v~~ and 2v~~

( )

2231231

3112232 i

RRRRRR

v~~+++

= , (6.6)

2311223

312331

3112

21 i

RRRRR

RRR

v~~v~~++

=+

= . (6.7)

Notes

37

R12

R23R31

2v~~

1 2

3

i2 1v

~~

Fig. 6.5. The circuit shown in Fig 6.3 with 01 =i

According to the superposition theorem we obtain

( )

2311223

31231

231231

231231111 i

RRRRR

iRRR

RRRv~~v~v

+++

+++

=+= , (6.8)

( )

2311223

3112231

231231

2331222 i

RRRRRR

iRRR

RRv~~v~v

+++

=++

=+= . (6.9)

The Y and Δ circuits are said to be equivalent if the sets of equations (6.4)-(6.5) and (6.8)-(6.9) are identical. In such a case, the corresponding coefficients of these equations are equal, i.e.

( )

231231

23123131 RRR

RRRRR

+++

=+ , (6.10)

311223

31233 RRR

RRR

++= , (6.11)

( )

311223

31122332 RRR

RRRRR

+++

=+ . (6.12)

Notes

38

Solving this set of equations for , , we find 1R 2R 3R

312312

31121 RRR

RRR

++= , (6.13)

312312

23122 RRR

RRR

++= , (6.14)

312312

31233 RRR

RRR

++= . (6.15)

Formulas (6.13)-(6.15) give the resistances of the Y circuit which is equivalent to the Δ circuit. Solving the set of equations (6.12)-(6.14) for , , we find 12R 23R 31R

3

212112 R

RRRRR ++= , (6.16)

1

323223 R

RRRRR ++= , (6.17)

2

131331 R

RRRRR ++= . (6.18)

Formulas (6.16)-(6.18) give the resistances of the Δ circuit which is equivalent to the Y circuit.

The Y circuit is said to be balanced if YRRRR === 321 . Using in such a case, (6.16)-(6.18), we obtain ΔRRRR === 312312 , where YΔ RR 3= . (6.19)

Notes

39

7. The Thevenin-Norton theorem

The Thevenin-Norton theorem is a very important law governing linear resistive circuits. It can be regarded as two equivalent theorems.

Let us consider an arbitrary one-port consisting of linear resistors and independent sources, as shown in Fig. 7.1.

v

i

Fig. 7.1. A linear resistive one-port The Thevenin theorem Any linear resistive one-port can be replaced by a series connection of a resistor and a voltage source , where is an input resistance across the one-port after all sources inside it are set to zero, is a voltage across the terminals of the one-port when the port is left open-circuited.

eqR

CV0 eqR

CV0

The equivalent Thevenin circuit is shown in Fig. 7.2. The elements of this circuit and can be determined as illustrated in Figs 7.3 and 7.4.

eqR

CV0

Notes

40

V0C

Req

v

i

Fig. 7.2. The Thevenin equivalent circuit

Req

i = 0

V0C

Fig. 7.3. Finding Req Fig. 7.4. Finding V0C

The Norton theorem Any linear resistive one-port can be replaced by a parallel connection of a linear resistor

and a current source . eqR SCI

eqR is defined as in the Thevenin theorem. is a current flowing through the short-circuited one-port.

SCI

Notes

41

The equivalent Norton circuit is shown in Fig. 7.5, whereas Fig. 7.6 shows the circuit enabling us to find . SCI

ISC Req v

i

ISC

i

Fig. 7.5. The equivalent Norton circuit Fig. 7.6. Finding Isc

The circuit shown in Fig. 7.6 can be replaced, on the basis of Thevenin’s theorem, by an equivalent circuit shown in Fig. 7.7.

V0C

Req ISC

Fig. 7.7. The circuit equivalent to the circuit of Fig. 7.6

Notes

42

Using KVL and Ohm’s law we write

00 =− SCeqC IRV . (7.1) Hence, it follows the equation

SC

Ceq I

VR 0= , (7.2)

showing the relation between , , and . eqR CV0 SCI Proof of the Thevenin theorem Let us consider a resistive one-port, shown in Fig. 7.8, containing n voltage sources

and m current sources . nSSS V,,V,V …

21 mSSS I,,I,I …21

v i

Fig. 7.8. A resistive one-port Fig. 7.9. The one-port of Fig. 7.8 supplied with a current source

We connect an additional current source to the one-port, as shown in Fig. 7.9.

Notes

43

On the basis of the superposition theorem we obtain

. (7.3) ikIkVhv sj

m

jjsj

n

jj 0

11++= ∑∑

==

If , then the port terminals are open-circuited and 0=i CVv 0= . Hence, we have

. (7.4) sj

m

jjsj

n

jjC IkVhV ∑∑

==

+=11

0

If we set all the sources inside the one-port to zero, that is 021

====nSSS VVV … ,

021

====mSSS III … , then equation (7.3) reduces to

ikv 0= , (7.5) hence,

eqRivk ==0 (7.6)

and equation (7.3) can be rewritten in the form

iRVv eqC += 0 , (7.7)

where and are defined as in the Thevenin theorem. Equation (7.7) describes the circuit depicted in Fig. 7.8, being Thevenin’s equivalent circuit.

CV0 eqR

V0C

Req

v

i

Fig. 7.8. The Thevenin equivalent circuit

Notes

44

Note that the Thevenin circuit shown in Fig. 7.8 is equivalent to the circuit depicted in Fig. 7.9, being the Norton circuit. Thus, the Thevenin and Norton circuits are equivalent.

eq

C

RV0

eqR SCeq

C IRV

=0

Fig. 7.9. The circuit equivalent to the circuit shown in Fig. 7.8

Example Let us consider the one-port shown in Fig. 7.10.

E2

R2

E1

R1

V0C

Req

Fig. 7.10. An example one-port Fig. 7.11. Thevenin’s circuit The equivalent Thevenin circuit is shown in Fig. 7.11. We find the elements and

of this circuit. CV0

eqRAccording to Thevenin’s theorem is the input resistance of the one-port shown in Fig. 7.12.

eqR

Notes

45

R2R1

i

V0C

E2

R2

E1

R1

Fig. 7.12. A one-port enabling us Fig. 7.13. A circuit enabling us to find Req to find V0C

Hence, we have

21

21

RRRRReq +

= . (7.8)

To find we consider a circuit with open-circuited terminals shown in Fig. 7.13. In this circuit the same current i traverses all the elements. To find this current we apply KVL and Ohm’s law

CV0

21

21

RREEi

+−

= . (7.9)

Since the current flows through resistor , we have i 2R

21

2112

21

2122220 RR

RERERREEREiREV C +

+=

+−

+=+= . (7.10)

Notes

46

8. Node method

Kirchhoff’s laws and Ohm’s law enable us to analyse simple resistive circuits. However, such an approach is inefficient in the case of more complex circuits. In this Section we develop a general, very useful and commonly applied method, called the node method.

To explain this method we consider a circuit having nodes and introduce a concept of node-to-datum voltage. For this purpose we choose arbitrarily one of these nodes as a datum node. The potential of this node is set to zero, hence, it is grounded. For the remaining 1

n

−n nodes we introduce node-to-datum voltages (or simply node voltages) , between these nodes and the datum. The reference directions of the node

voltages are shown in Fig. 8.1. 11 −ne,,e …

e1

2

en-1e2

n-1 1

n

Fig. 8.1. Reference directions of the node voltages It is easy to see that voltage between any two nodes and k j can be expressed in terms of node voltages and . ke je

jkkj eev −= . (8.1)

Notes

47

vkj

ek ej

j k

n

Fig. 8.2. Illustration of equation (8.1)

The idea of the node method will be explained using the circuit shown in Fig. 8.3, where the current sources , and the conductances - are given.

1Si 2Si 1G 5G

2Si

1Si

e3

G4

G3G2G1

i5i4

i3i2i1

v4

v3

v2

v1

v5

e2

e1

G5

1 2 3

Fig. 8.3. An example circuit

Notes

48

We choose the bottom node as a reference, introduce node voltages and write KCL equations at nodes 1, 2, 3 0

2141 =+−+ SS iiii ,

0542 =−−− iii , (8.2)

0253 =−+ Siii .

Next, we express the branch currents in terms of node voltages

( )

( )( ) ,eeGvGi

,eeGvGi,eGvGi

,eGeGvGi,eGvGi

235555

214444

33333

2222222

11111

−==−==

==−=−==

==

(8.3)

and substitute them into (8.2)

( )( ) ( )

( ) .ieeGeG,eeGeGeeG

,iieeGeG

S

SS

2

21

23533

23522214

21411

0=−+

=−−+−−

−=−+

(8.4)

Finally, we rearrange equations (8.4) as follows

( )( )( ) .ieGGeG

,eGeGGGeG

,iieGeGG

S

SS

2

21

35325

35252414

24141

0=++−

=−+++−

−=−+

(8.5)

The set of node equations (8.5) contains three unknowns , , . 1e 2e 3e

Notes

1 2

3

49

Let us replace the resistor by a voltage source , as shown in Fig. 8.4. 2G2SV

2Si

1Si

e3

G4

G3G1

i5i4

i3i2i1

v4

v3v1

v5

e2

e1

G5

1 2 3

2SV

Fig. 8.4. Circuit driven by current and voltage sources Equations written at node 1 and 3 are the same as in the previous case. Hence, we only need to write an equation at node 2. Since the current cannot be expressed in terms of the branch voltage, it is considered an additional variable. Thus, we obtain the following set of equations

2i

( )( ) ( )

( ) .ieeGeG,eeGieeG

,iieeGeG

S

SS

2

21

23533

2352214

21411

0=−+

=−−+−−

−=−+

(8.6)

Notes

1

2

3

50

This is a set of three equations with four unknown variables , , , . Therefore, we add another equation of the form

1e 2e 3e 2i

22 SVe = , (8.7)

where is the given voltage source. Substituting (8.7) into (8.6) we eliminate the variable and obtain the set of three equations in three variables , ,

2SV

2e 1e 3e 2i

( )( ) ( )

( ) .iVeGeG

,VeGiVeG

,iiVeGeG

SS

SS

SSS

22

22

212

3533

35214

1411

0

=−+

=−−+−−

−=−+

(8.8)

2Si

1Si

e3

G4

G3G2G1

i5i4

i3i2i1

v4

v3

v2

v1

v5

e2

e1

1 2 3

Fig. 8.5. Nonlinear resistor circuit

Notes 1

2

3

51

The node method can be also applied to circuits containing nonlinear resistors. It will be explained via an example circuit shown in Fig. 8.5, including a nonlinear resistor (semiconductor diode) described by the equation

( )155 −= veKi λ . (8.9)

To write the node equations we introduce temporarily the current as an additional variable

5i

( )( )

.iieG,ieGeeG

,iieeGeG

S

SS

2

21

533

522214

21411

0=+=−+−−

−=−+

(8.10)

Next we express in terms of the corresponding node voltages 5i

( ) ( )( )11 2355 −=−= −eev eKeKi λλ (8.11)

and substitute it into (8.10)

( )( ) ( )( )

( )( ) .ieKeG

,eKeGeeG

,iiveGeG

See

ee

SSS

223

23

212

1

01

33

22214

1411

=−+

=−−+−−

−=−+

−

−

λ

λ (8.12)

In this way, we obtain a set of three node equations, in three unknown variables, describing the nonlinear circuit shown in Fig. 8.5.

Notes

1

2

3

1

2

3

52

9. Simple nonlinear circuits

In this section we analyse very simple circuits consisting of nonlinear resistors by means of a graphical approach.

Series connection of resistors Figure 9.1 shows a circuit consisting of two nonlinear resistors connected in series.

i2

i1

v2

v1

i

v

Fig. 9.1. Two nonlinear resistors connected in series The resistors are specified by their characteristics depicted in Figs 9.2 and 9.3.

i1

v1

1

i2

v2

2

Fig. 9.2. The characteristic i1 – v1 Fig. 9.3. The characteristic i2 – v2

Notes

53

On the basis of KVL we write

21 vvv += . (9.1) Since the resistors are connected in series, the same current traverses each of them, hence, it holds

iii == 21 . (9.2)

On the basis of these equations we can find the characteristic vi − using a graphical approach. To trace the characteristic we add the voltages and specified by the characteristics

1v 2v

1vi − and , respectively, for each value of the current . 2vi − iThe graphical construction is illustrated in Fig. 9.4.

0 i

v2 v = v1+v2

v1

i

v1, v2, v

1

2

Fig. 9.4. Graphical construction for finding characteristic i-v

Parallel connection of resistors Figure 9.5. shows a circuit consisting of two nonlinear resistors connected in parallel.

Notes

54

i1 i2

v1 v2

i

v

Fig. 9.5. Two nonlinear resistors connected in parallel

The characteristics and are depicted in Figs 9.6 and 9.7, respectively. 11 iv − 22 iv −

v1

i1

1

i2

v2

2

Fig. 9.6. The characteristic Fig. 9.7. The characteristic 11 iv − 22 iv − Since the resistors are connected in parallel, the voltages and are identical 1v 2v 21 vvv == . (9.3)

Notes

55

Using KCL at the top node we have

21 iii += . (9.4)

Thus, the characteristic iv − can be traced in a graphical manner by adding for each value of v the corresponding currents and (see Fig. 9.8). 1i 2i

i

i1 = i2

0 v

i1, i2, i

v

2

1

Fig. 9.8. Graphical construction for finding characteristic v-i

Operating points

Nonlinear circuits driven by DC sources have constant solutions (branch voltages and current), called operating points. The basic question of the analysis of this class of circuits is finding the operating points. If a circuit is simple, the operating point can be found using a graphical approach, To illustrate this approach we consider a typical biasing circuit depicted in Fig. 9.9, including a nonlinear resistor specified by the characteristic shown in Fig. 9.10. The linear part of this circuit, consisting of the DC voltage source E and resistor R , is described by the equation

bb RiEv += . (9.5)

Notes

56

B

ia

va

ib A

Rv

vb

R

E

0

va=f(ia)

ia

Fig. 9.9. Typical biasing circuit Fig. 9.10. Characteristic of the nonlinear resistor belonging to the circuit of Fig. 9.9 We transcribe this characteristic in the bb vi − plane to the aa vi − plane. Since

ab vv = and ab ii −= , we obtain aa RiEv −= . (9.6) Equation (9.6) describes a straight line, shown in Fig. 9.11. On the same plane we plot the characteristic

( )aa ifv = .

The point of intersection ( )aa v,i is the solution of the set of equations

( )aa ifv = ,

aa RiEv −= , hence, the intersection is the operating point of the circuit shown in Fig. 9.9.

Notes

57

REai

av

E

0

va

ia

Fig. 9.11. Graphical construction for finding the operation point

Notes

58

10. Controlled sources

Controlled sources are circuit elements very useful in the modeling of electronic devices. A controlled source is a two-port consisting of two branches. A primary branch is either a short circuit or an open circuit. A secondary branch is either a voltage source or a current source. The source waveform depends on a voltage or a current of the primary branch. Thus, there are four types of controlled sources. The controlled sources can be classified as linear or nonlinear. All the controlled sources are shown in Figs 10.1 – 10.4.

1i

2i

2v1ir

01 =v

a)

1i

2i

2v( )1ifr 01 =v

b)

Fig. 10.1. Current-controlled voltage sources (CCVS) a) linear, b) nonlinear

a)

2i

2v1gv

01 =i

1v

2i

2v( )1vfg

01 =i

1v

b)

Fig. 10.2. Voltage-controlled current sources a) linear, b) nonlinear

Notes

59

1i

a)

2i

2v1iα

01 =v

1i

2i

2v ( )1ifα

01 =v

b)

Fig. 10.3. Current-controlled current sources a) linear, b) nonlinear

a)

1vμ

2i

2v1v

01 =i

b)

2i

2v( )1vfμ1v

01 =i

Fig. 10.4. Voltage-controlled voltage sources a) linear, b) nonlinear

Controlled sources are very useful in modeling electronic devices. For example the Ebers-Moll model of an npn bipolar transistor contains two linear current controlled current sources, as illustrated in Fig. 10.5.

iC

iF iR

iE

iB

BE CvBE vBC

αF iF αR iR

Fig. 10.5. The Ebers-Moll model of an npn bipolar transistor

Notes

60

11. Capacitor

11.1 Introduction

A capacitor is a two-terminal element which stores an electric charge. The simplest example of a capacitor is shown in Fig. 11.1. It is made of two flat parallel metal plates in free space.

-q(t)

q(t) i(t)

d

v(t)

Fig. 11.1. Parallel plate capacitor When a current ( )ti is applied, then a charge ( ) ( )tCvtq = is induced on the upper plate and an equal but opposite charge is induced on the lower plate. The constant of proportionality, called capacitance, is given approximately by

d

AC 0ε= ,

Notes

61

where

mF10

361 9

0−=

πε

is the dielectric constant called permittivity, A is the plate area and is the separation of the plates. The units of capacitance are farads, abbreviated to F.

d

Equation

Cvq = (11.1)

defines the qv − characteristic of the capacitor. The characteristic is a straight line through the origin with a slope C , as shown in Fig. 11.2.

v

q

Fig. 11.2. Characteristic v-q of a linear capacitor

A capacitor whose characteristic is a straight line through the origin is called a linear capacitor. Otherwise, the capacitor is said to be nonlinear. An example qv − characteristic of a nonlinear capacitor is shown in Fig 11.3.

Notes

62

v

q

Fig. 11.3. Characteristic v-q of a nonlinear capacitor

The symbols of a linear and a nonlinear capacitor and the reference directions of ( )ti and ( ) are shown in Figs 11.4 and 11.5. In these figures q is the charge on this plate which is pointed by the reference arrow of the current i .

tv

v(t)

i(t) q(t)

C

q(t)

v(t)

i(t)

Fig. 11.4. Symbol of a linear capacitor Fig. 11.5. Symbol of a nonlinear capacitor

Notes

63

The current ( )ti is given by the equation

( ) ( )ttqti

dd

= . (11.2)

Using equation (11.1) and (11.2) we obtain

( )tvC

tCv

tqi

dd

dd

dd

=== . (11.3)

Equation (11.3) expresses capacitor current in terms of capacitor voltage. To express capacitor voltage in terms of capacitor current we replace the variable t with τ obtaining

( ) ( )τττ

ddvCi = . (11.4)

Next we integrate both sides of (11.4) between and 0 t

( ) ( )( )

( )( ))0()(dd

ddd

000

vtvCvCvCitv

v

tt

−=== ∫∫∫ τττττ .

Hence, it holds

( ) ( ) ( ) ττ diC

vtvt

∫+=0

10 . (11.5)

11.2 Continuity property

Let us replace in equation (11.5) by t tt d+

( ) ( ) ( ) ττ d10dd

0∫+

+=+tt

iC

vttv (11.6)

and assume that ( )ti is bounded for all , that is there exists a constant t L , such that ( ) Lti < for all . t

Notes

64

Subtracting equation (11.5) from (11.6) we have

( ) ( ) ( ) ττ d1dd

0∫+

=−+tt

iC

tvttv . (11.7)

As , then , hence, 0d →t ( ) 0dd

0

→∫+

ττtt

i ( ) ( )tvttv →+ d .

Thus, voltage across any linear capacitor is a continuous function of time. It means that this voltage cannot jump instantaneously from one value to another.

11.3 Energy stored in a capacitor

Consider a capacitor supplied with a generator as shown in Fig. 11.6.

v(t)

i(t)

C Generator

Fig. 11.6. A capacitor supplied with a generator

The energy delivered by the generator to the capacitor from time to is given by the equation

0t t

, (11.8) ( ) ( ) ( ) ( ) τττττ dd00

0 ivpt,tAt

t

t

t∫∫ ==

where is the instantaneous power entering the capacitor. p

Notes

65

Let ( ) 00 =tv , hence, no charge is stored in the capacitor. We choose this state as the state corresponding to zero energy, i.e. ( ) 00 =tw , where at 0tt = is the initial energy of the capacitor. Let us replace t by another variable τ in equation (11.3)

( ) ( )τττ

ddvCi =

and rewrite it in the form ( ) vCi dd =ττ . (11.9)

Substituting (11.9) into (11.8) yields

( ) ( ) ( )( ) ( )

( )20

2

00 )(

21

21dd

0

tvCvCvvCivt,tAtvtvt

t

==== ∫∫ τττ . (11.10)

A capacitor is an element that stores energy, but does not dissipate it. Hence, the energy stored in the capacitor at time t is given by the equation

( ) ( ) ( ) ( )t,tAt,tAtwtw 000 =+= . (11.11)

Since ( ) 00 =tw and is specified by (11.10), we have ( )t,tA 0

( ) ( )tCvtw 2

21

= . (11.12)

11.4 Series connection of capacitors

Consider two capacitors connected in series as shown in Fig. 11.7.

i(t)

v2(t)

C2

v1(t)

C1

v(t) Fig. 11.7. Two capacitors connected in series

Notes

66

Since the same current traverses both capacitors we write, on the basis of (11.5)

( ) ( ) ( ) ττ d1001

11 ∫+=t

iC

vtv , (11.13)

( ) ( ) ( ) ττ d1002

22 ∫+=t

iC

vtv . (11.14)

Using KVL yields

( ) ( ) ( )tvtvtv 21 += . (11.15)

At equation (11.15) becomes 0=t

( ) ( ) ( )000 21 vvv += . (11.16)

We add equations (11.13) and (11.14) together and apply (11.16)

( ) ( ) ( ) ττ d110021∫⎟⎟⎠

⎞⎜⎜⎝

⎛++=

t

iCC

vtv . (11.17)

Let

CCC111

21=+ , (11.18)

then

( ) ( ) ( ) ττ d100∫+=t

iC

vtv . (11.19)

Equation (11.19) describes the equivalent capacitor of two capacitors connected in series. The initial voltage of this capacitor is specified by (11.16), whereas the capacitance is given by (11.18).

Notes

67

11.5 Parallel connection of capacitors

Figure 11.8 shows two capacitors connected in parallel. Since voltages across the capacitors are identical, both capacitors have the same initial voltage ( )0v .

C1

i2(t) i1(t) i(t)

C2 v(t)

Fig. 11.8. Two capacitors connected in parallel

Currents flowing through the capacitors are given by the equations

( ) ( )ttvCti

dd

11 = , ( ) ( )ttvCti

dd

22 = .

Applying KCL at the top node and the above equations yields

( ) ( ) ( ) ( ) ( ) ( )ttvC

ttvCCtititi

dd

dd

2121 =+=+= ,

where

21 CCC += . (11.20)

Formula (11.20) gives the capacitance of the equivalent capacitor.

Notes

68

12. Inductor

12.1 Introduction

Figure 12.1 shows an inductor made of wire wound around a core, made of a nonmetallic material.

φ i(t)

v(t)

Fig. 12.1. An example inductor When the device is supplied with a time varying current source, a magnetic flux is induced and circulates inside the core. The magnetic flux linkage ( )tφ , being the total flux linked by all turns of the coil, is proportional to the current

( ) ( )tLit =φ , (12.1) where the coefficient L is called inductance. The units of magnetic flux are webers (W), the units of inductance are henrys (H).

Notes

69

If the core shown in Fig. 12.1 is a toroid made of material having the magnetic constant (permeability) 0μ , then the inductance is given by the formula

l

ANL2

0μ= , (12.2)

where mH104 4

0−⋅= πμ , is the number of turns, N A is the cross-sectional area of the

core, is the midcircumference along the core. lThe equation

Li=φ (12.3)

defines φ−i characteristic of the inductor. The characteristic is a straight line through the origin with a slope equal to L (see Fig. 12.2). In such a case the inductor is classified as linear.

0

slope = L

i

φ

Fig. 12.2. Characteristic of a linear inductor

Otherwise, if the characteristic is not a straight line through the origin, the inductor is called nonlinear. The typical characteristic of a nonlinear inductor is shown in Fig. 12.3.

Notes

70

0 i

φ

Fig. 12.3. Characteristic of a nonlinear inductor

The symbols of linear and nonlinear inductors are shown in Fig. 12.4.

v(t)

i(t)

L v(t)

i(t)

Fig. 12.4. Symbols of linear and nonlinear inductors

According to Faraday’s induction law

( ) ( )tttv

ddφ

= . (12.4)

Notes

71

Substituting (12.3) into (12.4) yields

( )tiLLi

tv

dd

dd

== . (12.5)

To express the current flowing through a linear inductor in terms of the voltage across the inductor we replace t with

i vτ and rewrite equation (12.5) in the form

( ) ( )τττ

ddiLv = . (12.6)

Next we integrate both sides of equation (12.6) between and 0 t

( ) ( )( )

( )( ))0()(dd

ddd

000

itiLiLiLvti

i

tt

−=== ∫∫∫ τττττ .

Hence, we have

( ) ( ) ( ) ττ d100∫+=t

vL

iti . (12.7)

12.2 Continuity property

Let us replace in equation (12.7) by t tt d+

( ) ( ) ( ) ττ d10dd

0∫+

+=+tt

vL

itti (12.8)

and assume that ( )tv is bounded for all , that is, there exists a constant t M such that ( ) Mtv < for all . Subtracting equation (12.7) from (12.8) yields t

( ) ( ) ( ) ( ) ( ) ττττττ d1dd1dd

0

d

0∫∫∫++

=⎟⎟⎠

⎞⎜⎜⎝

⎛−=−+

tt

t

ttt

vL

vvL

titti .

Notes

72

Since ( ) 0d1 d→∫

+ tt

tv

Lττ as , then 0d →t ( ) ( )titti →+ d . Thus, current flowing through

any linear inductor is a continuous function of time. This means that inductor current cannot jump instantaneously from one value to another.

12.3 Hysteresis

Ferromagnetic core inductors exhibit the hysteresis phenomenon as depicted in Fig. 12.5.

1

2

3

i2

i3 i4 i1 i

φ3

φ1

φ

Fig. 12.5. Hysteresis phenomenon

The characteristic shown in Fig. 12.5 is obtained by increasing the current from to

, next decreasing this current from to and after that increasing from to . Thus, the flux decreases according to the upper branch 2 and increases according to the lower branch 3. As a result, a close curve is traced. The magnetic flux becomes zero for the negative value and the positive value of the current.

i 01i 1i 3i 3i 1i

2i 4i

Notes

73

12.4 Energy stored in an inductor

Consider an inductor supplied with a generator as shown in Fig. 12.6.

v(t)

i(t)

L Generator

Fig. 12.6. An inductor supplied with a generator

The energy delivered by the generator to the inductor from time to is given by the equation

0t t

, (12.9) ( ) ( ) ( ) ( ) τττττ dd00

0 ivpt,tAt

t

t

t∫∫ ==

where p is the instantaneous power entering the inductor. Let ( ) 00 =ti , consequently also the magnetic flux equals zero and no magnetic field exists in the inductor. Such a state can be considered a state corresponding to zero energy stored. Let us substitute (12.6) into (12.9). Then, we have

( ) ( ) ( )( )

( )

( )

( )( ))()(

21

21dd

dd

0222

0000

titiLLiiiLiiLt,tAti

ti

ti

ti

t

t

−==== ∫∫ ττττ .

Since ( ) 00 =ti the energy initially stored in the inductor is also equal to zero, ( ) 00 =tw and the energy delivered to the inductor from 0tt = to is t

( ) ( )tLit,tA 20 2

1= . (12.10)

Notes

74

An inductor is an element that stores energy, but does not dissipate it. Hence, the energy stored in the inductor at time t is given by the equation

( ) ( ) ( ) ( )tLit,tAtwtw 200 2

1=+= . (12.11)

12.5 Series connection of inductors

Consider two inductors connected in series as shown in Fig. 12.7.

i(t) v2(t)

L2

v1(t)

L1

v(t)

Fig. 12.7. Two inductors connected in series

Since identical currents flow through the inductors, both inductors have the same initial current ( )0i . KVL leads to the equation

( ) ( ) ( )tvtvtv 21 += , where

( ) ( )ttiLtv

dd

11 = , ( ) ( )ttiLtv

dd

22 = .

Hence, we have

( ) ( ) ( ) ( ) ( ) ( )ttiL

ttiLL

ttiL

ttiLtv

dd

dd

dd

dd

2121 =+=+= ,

where

21 LLL += . (12.12)

Formula (12.12) gives the inductance of the equivalent inductor.

Notes

75

12.6 Parallel connection of inductors

Figure 12.13 shows two inductors connected in parallel.

L1

i2(t) i1(t)

i(t)

L2 v(t)

Fig. 12.13. Two inductors connected in parallel

We write KCL equation

( ) ( ) ( )tititi 21 +=

and substitute into this equation

( ) ( ) ( ) ττ d1001

11 ∫+=t

vL

iti

and

( ) ( ) ( ) ττ d1002

22 ∫+=t

vL

iti .

As a result we obtain

( ) ( ) ( ) ττ d100∫⎟⎠⎞

⎜⎝⎛+=

t

vL

iti , (12.13)

where ( ) ( ) ( )000 21 iii += (12.14)

Notes

76

and

LLL111

21=+ (12.15)

or

21

21

LLLL

L+

= . (12.16)

Thus, the parallel connection of two inductors can be replaced by a single inductor having the inductance specified by (12.16), with the initial condition (12.14).

Notes

77

13. Operational-amplifier circuits

13.1 Description of the operational amplifier

An operational amplifier is a multi-terminal semiconductor device. Figure 13.1 shows the symbol of this device, including inside triangle interconnected transistors, resistors and a power supply voltage source. The terminals available for external connections are called: an inverting input, a noninverting input, an output, and external ground.

Noninvertinginput

Invertinginput

External ground

Output

Fig. 13.1. An operational amplifier

In Fig. 13.2 voltages and currents are introduced, where and denote the currents entering the operation amplifier, is the input voltage, denotes the output voltage, and the output current.

−i +i

dv 0v

0i

i+

i-

v0

vd i0

Fig. 13.2. Reference directions of voltages and currents of the operational amplifier

Notes

78

The operational amplifier is described by the following set of equations:

0=−i ,

0=+i ,

( )dvfv =0 ,

where ( )dvf is a function shown in Fig. 13.3. For belonging to a very small interval [

dv]εε ,− function ( ) is linear dvf ( ) Avf d = , where A equals at least 100 000.

0

ε -ε

vd (mV)

v0 (V)

Esat

-Esat

0.1 0.2 0.3-0.3 -0.2 -0.1

Fig. 13.3. The characteristic ( )dvfv =0 of the operational amplifier

At satEv ±=0 the function saturates. Since the slope ( )dvf A is very large we can assume that ∞=A . As a result we obtain the ideal model of the operational amplifier, having the characteristic ( ) shown in Fig. 13.4. dvfv =0

Notes

79

The characteristic is piecewise-linear and consists of three segments. They define three operating regions called: a linear region, a +saturation region, and a –saturation region.

Linear region

0

+ Saturation region

- Saturation region

vd

v0 = f (vd)

Esat

-Esat

Fig. 13.4. The characteristic of the ideal model of the operational amplifier

The linear region is described by the following equations:

0=−i ,

0=+i , (13.1)

0=dv . The output voltage in this region satisfies the inequality

satsat EvE <<− 0 (13.2) called the validating inequality.

Notes

80

In the +saturation region the operational amplifier is described by the equations

0=−i ,

0=+i , (13.3)

satEv =0 , and the output voltage satisfies the validating inequality

0>dv . (13.4) The above relationships lead to the model of the operational amplifier operating in the +saturation region, as shown in Fig. 13.5.

0v satE

0>dv

Fig. 13.5. The model of the operational amplifier operating in the +saturation region

In the –saturation region the operational amplifier is specified by the following set of equations:

0=−i ,

0=+i , (13.5)

satEv −=0 , whereas the validating inequality is

0<dv . (13.6)

Notes

81

Hence, we obtain the model of the operational amplifier operating in the –saturation region, depicted in Fig. 13.6.

0vsatE

0<dv

Fig. 13.6. The model of the operational amplifier operating in the –saturation region

13.2 Examples

To explain the method for the analysis of the circuits containing operational amplifiers we consider two examples. Example 13.1 Figure 13.7 shows a circuit called an inverter. We assume that operational amplifier operates in the linear region and express the output voltage in terms of the input voltage .

0v

iv

P

vi

i1

R2

R1

v2

i2

i+ = 0

i- = 0

v0

vd = 0 v1

Fig. 13.7. The inverter containing the operational amplifier

Notes

82

Since the operation amplifier operates in the linear region, it is described by the set of equations (13.1). Especially 0= , hence, dv

ivv =1 and 11

11 R

vRvi i== .

KCL applied at node P leads to the equation

12 ii = .

Consequently, it holds

ivRRiRiRv

1

212222 === .

Using KVL we write

20 vv −= . Thus, the output voltage equals

ivRRv

1

20 −= . (13.7)

The obtained result is valid if satisfies the validating inequality 0v

satsat EvE <<− 0 . Substituting (13.7) we obtain

satisat EvRRE <−<−

1

2 ,

or

satisat ERRvE

RR

2

1

2

1 <<− . (13.8)

Notes

83

Thus, the output voltage is specified by equation (13.7) if the input voltage is selected so that the inequality (13.8) is satisfied.

1

Example 13.2 Figure 13.8 shows a circuit called a voltage comparator.

vi (t)V = constant

i+ = 0

i- = 0

v0

vd

Fig. 13.8. The voltage comparator containing the operational amplifier

In this circuit the operational amplifier is allowed to operate in all the three regions. We wish to find the transfer characteristic ( )ivfv =0 . Using KVL we write

Vvv id −= . (13.9) Below we consider 3 cases. 1. The operational amplifier operates in the linear region.

0In this case =dv , hence using (13.9) we obtain

Vvi = .

This result is correct if the validating inequality satsat EvE <<− 0 is satisfied. As a result, we find the segment of the transfer characteristic shown in Fig. 13.9.

Notes

84

2. The operational amplifier operates in the +saturation region. 0>Thus, , and equation (13.9) gives dv satEv =0

2

3

. Vvi >

As a result, we obtain the segment of the transfer characteristic shown in Fig. 13.9.

3. The operational amplifier operates in the –saturation region. Similarly, as in case 2, we write the relations:

0<dv , satEv −=0 , Vvi < ,

which lead to segment of the characteristic (see Fig. 13.9).

1

2

3

V 0 vi

v0

Esat

-Esat

Fig. 13.9. The transfer characteristic ( )ivfv =0 of the voltage comparator shown in Fig. 13.8

13.3 Finite gain model of the operational amplifier

In this section we consider the operational amplifier model specified by the characteristic ( )dvfv =0 shown in Fig. 13.10. Unlike the ideal model the slope of the segment passing through the origin is very large, but finite.

Notes

85

A > 100000

Slope A

0 ε

-ε

vd

v0 = f (vd)

Esat

-Esat

Fig. 13.10. The characteristic ( )dvfv =0 of nonideal model of the operational amplifier

In the linear region the model is described by the equation

dAvv =0

and the validating inequality

εε <<− dv . Hence, it follows the equivalent circuit shown in Fig. 13.11.

0=+i

0=−i

0v( )dvfv =0

dv

Fig. 13.11. The equivalent circuit of the operational amplifier operating in the linear region

Notes

86

In the +saturation region the mathematical representation of the model is

satEv =0 ,

ε>dv , whereas in the –saturation region the model is described by

satEv −=0 ,

ε−<dv , The corresponding equivalent circuits are shown in Figs 13.12 and 13.13.

0=+i

0=−i

0vsatEdv ε>dv

Fig. 13.12. The equivalent circuit of the operational amplifier operating in the +saturation region

0=+i

0=−i

0vsatEdv ε−<dv

Fig. 13.12. The equivalent circuit of the operational amplifier operating in the -saturation region

Notes

87

14. First order circuits driven by DC sources

Circuits made of one inductor or one capacitor, resistors and sources are called first order dynamic circuits. In this section we study circuits driven by DC sources.

14.1 Resistor-inductor circuits

We consider a simple circuit consisting of a DC voltage source, a resistor and an

inductor, with an initial current ( ) 00 Ii = (see Fig. 14.1).

V

i

R

L vL

vR

Fig. 14.1. A first order resistor-inductor circuit Applying KVL we write

0=−− LR vvV .

Since

tiLvL d

d= ,

we obtain, after simple rearrangements

VRitiL =+

dd , ( ) 00 Ii = . (14.1)

Notes

88

Equation (14.1) can be presented in the form

VL

iti 11

dd

=+τ

, (14.2)

where

RL

=τ (14.3)

is called a time constant. We rewrite equation (14.2) in the form

ττ

τ

τ

iRVi

LV

iVLt

i −=

−=−=

11dd

and separate the variables and i t

τt

RVi

i dd−=

− . (14.4)

Next, we integrate equation (14.4)

∫∫ −=−

t

RVi

i d1dτ

,

finding

KtRVi +−=−

τln ,

where the constant K may be written as AK ln= , where A is another constant.

Notes

89

Hence, we have

τt

ARVi

−=−

ln ,

or

τt

ARVi

−+= e . (14.5)

To determine A we write equation (14.5) at 0=t and set ( ) 00 Ii =

ARVI +=0 . (14.6)

Substituting A , given by (14.6) into equation (14.5) yields

( ) τt

RVI

RVti

−⎟⎠⎞

⎜⎝⎛ −+= e0 . (14.7)

When 0=t , we have , which is the correct initial conditions. When , the

second term on the right side of equation (14.7) vanishes, hence, it holds

( ) 00 Ii = ∞→t

( )RVi =∞ .

Thus, we have

( ) ( ) ( ) ( )( ) τt

iiiti−

∞−+∞= e0 , (14.8)

where ( )0i is the initial value of ( )ti , whereas ( )∞i is the steady-state value of ( )ti . Since the steady-state current is constant, the voltage across the inductor

tiLvL d

d=

equals zero.

Notes

90

Thus, the inductor behaves like a short circuit and the circuit considered in the steady-state (as ∞→t ) consists of the voltage source V and the resistor R (see Fig. 14.2).

V

R

( )∞i

Fig. 14.2. Model of the circuit shown in Fig. 14.1 in the steady-state

The plot of ( )ti is shown in Fig. 14.3.

( ) 00 Ii =

β α

x

( )RVi =∞

0 t

i(t)

Fig. 14.3. Plot of i(t) and graphical interpretation of the time constant

The initial rate of changing of the current is given by

( ) ( )( ) ( ) ( )( )∞−−=∞−−===

−

=

iiiiti

t

t

t01e01tan

dd

00 ττα τ . (14.9)

Notes

91

On the other hand

( ) ( )( )x

iiti

t

∞−−=−=

=

0tandd

0β . (14.10)

Equalizing (14.9) and (14.10) we find

τ=x .

Thus, to find graphically the time constant, it is necessary to draw the tangent of the current at 0=t and determine its intercept with the horizontal line passing through the point ( )( )∞i,0 .

In a very special case, when 0=V , ( ) 0=∞i and equation (14.8) becomes

( ) τt

Iti−

= e0 . (14.11)

The plot of ( )ti is shown in Fig. 14.4 and the time constant x=τ .

x

0.368 I0

0

I0

t

i(t)

Fig. 14.4. Plot of i(t) in the circuit with V=0

At τ=t , ( ) 01

00 3680ee I.IIi === −−ττ

τ . Thus, in one time constant the current has declined to 0.368 of its initial value (see Fig. 14.4).

Notes

92

At τ5=t , ( ) 05

0 00670e5 I.Ii == −τ and the current is a negligible fraction of its initial value. Therefore, we usually assume that after 5 time constants the current is in the steady state. Example 14.1 The switch in the circuit shown in Fig. 14.5 is open until the steady state prevails and then it is closed. Assuming that the closing occurs at 0=t , we find the current ( ) . tiL

R1 V

iL R2

L vL

t = 0

Fig. 14.5. A first order circuit driven by a DC voltage source

When the switch is open, the circuit is in the steady state. The current is constant, hence,

the voltage across the inductor, given by tiLvL d

d= , is equal to zero. In such a case, the

inductor can be replaced by a short circuit and the circuit consists of the DC voltage source and resistors , connected in series. At the left hand side of 01R 2R =t , labeled

, the switch is open and −0

( )21

0RR

ViL +=− .

At the switch is closed. However, the current 0=t ( )ti L flowing through the inductor remains unchanged at this moment, due to the continuity property

( ) ( )21

00RR

Vii LL +== − . (14.12)

Notes

93

When the switch is closed the circuit has the same form as the circuit shown in Fig. 14.1. Hence, the current ( ) is given by equation (14.8), repeated below ti L

( ) ( ) ( ) ( )( ) τt

iiiti−

∞−+∞= e0 . (14.13)

At the circuit is in the steady state, voltage across the inductor equals zero and the current is given by the formula

∞=t

( )2R

ViL =∞ . (14.14)

Substituting (14.12) and (14.14) into equation (14.13) yields

( ) τt

L RV

RRV

RVti

−

⎟⎟⎠

⎞⎜⎜⎝

⎛−

++= e

2212 , (14.15)

where

2R

L=τ . (14.16)

The plot against is shown in Fig. 14.6. Li t

21 RRV+

2RV

0 t

iL(t)

Fig. 14.6. Plot iL(t) in the circuit shown in Fig. 14.5

Notes

94

Now we consider a general class of first order circuits consisting of DC sources, resistors, and one inductor. We extract the inductor from the circuit, obtaining the circuit shown in Fig. 14.7.

L vL

iL

Fig. 14.7. Circuit with the extracted inductor

Then, we apply the Thevenin theorem to reduce the circuit to the form shown in Fig. 14.8.

Req

V0C

L vL

iL

Fig. 14.8. The Thevenin circuit of the resistive one-port shown in Fig. 14.7

In this circuit the current is given by equation (14.8) repeated below

Notes

95

( ) ( ) ( ) ( )( ) τt

LLLL iiiti−

∞−+∞= e0 ,

where

( )eq

CL R

Vi 0=∞ and

eqRL

=τ .

14.2 Resistor-capacitor circuits

Another first order dynamic circuit is an RC circuit consisting of one capacitor,

resistors, and sources. In this section we study circuits driven by DC sources. Let us consider a simple RC circuit, driven by a DC voltage source, as shown in Fig.

14.9.

V iC

R

C vC

vR

Fig. 14.9. A first order RC circuit

To describe the circuit we write KVL equation

Vvv CR =+ ,

where CR Riv = and

t

vCi C

C dd

= .

Notes

96

Hence, we have

Vvt

vRC C

C =+d

d , (14.17)

We divide both sides of equation (14.17) by RC

VRC

vRCt

vC

C 11d

d=+

and define the time constant

RC=τ . (14.18)

As a result, we obtain the equation

Vvt

vC

C

ττ11

dd

=+ , (14.19)

with the initial condition

( ) 00 VvC = .

Thus, the circuit shown in Fig. 14.9 is described by equation (14.19). From a mathematical point of view this equation is similar to equation (14.2), therefore, its solution is as follows

( ) ( ) ( ) ( )( ) τt

CCCC vvvtv−

∞−+∞= e0 , (14.20)

where ( )∞Cv is the steady state solution, at ∞=t . Thus, the voltage at the steady state is constant, its derivative equals zero. This means that no current flows through the capacitor in the steady state. Consequently, the capacitor can be removed and the circuit, in the steady state, becomes as shown in Fig. 14.10.

Cv

Using KVL in the circuit depicted in Fig. 14.2 we find

( ) VvC =∞ .

Notes

97

( )∞Cv V R

( ) 0=∞Ci

Fig. 14.10. The circuit of Fig. 14.9 at the steady state

Figures 14.11 and 14.12 show the plot ( )tvC in two cases: when ( )0CvV > and when ( ). The graphical interpretation of the time constant is also shown in these

figures. 0CvV <

τ

( ) VvC =∞

0

vC(0)

t

vC(t)

Fig. 14.11. Plot ( )tvC , ( )0CvV >

τ

( ) VvC =∞

0

vC (0)

t

vC(t)

Fig. 14.12. Plot ( )tvC , ( )0CvV <

Notes

98

The current ( )tiC is given by

( ) ( )( ) ( ) ( ) ττ

τ

tCC

t

CCC

C Rvv

vvCt

vCi

−− −∞=⎟

⎠⎞

⎜⎝⎛∞−== e

0e1-0

dd .

Its plot is shown in Fig. 14.13.

0

iC(0)

t

iC(t)

Fig. 14.13. Plot of current iC against time

Alternatively, the current can be found as follows. Fist we compute the voltage

(see Fig. 14.9) using KVL Rv

( ) ( ) ( )( ) ( )( ) ττt

C

t

CCR vVVvVVtvVtv−−

−=⎟⎟

⎠

⎞

⎜⎜

⎝

⎛−+−=−= e0e0 .

Next, we apply Ohm’s law

( ) τt

CRC R

vVRvi

−−== e0 .

Now we consider a circuit consisting of DC sources, resistors, and one capacitor. We extract the capacitor from the circuit, obtaining the circuit shown in Fig. 14.14.

Notes

99

CvC

iC

Fig. 14.14. Circuit with the extracted capacitor

Next, we apply Thevenin’s theorem to reduce the circuit to the form shown in Fig. 14.15.

Req

V0C

C vC

iC

Fig. 14.15. The Thevenin circuit of the resistive one-port shown in Fig. 14.14

In this circuit the voltage is given by equation (14.20) repeated below ( )tvC

( ) ( ) ( ) ( )( ) τt

CCCC vvvtv−

∞−+∞= e0 (14.21)

where ( ) CC Vv 0=∞ and CReq=τ .

Notes

100

Example 14.2 In the circuit shown in Fig. 14.16 the switch has been open for a long time prior to 0 when it is closed. Find the voltage =t ( )tv for 0≥t . C

R3 = 5 Ω

R1 = 10 Ω

VS = 12 V iC (t)

R2 = 10 ΩC = 1μF

t = 0 vC (t)

Fig. 14.16. An example first order dynamic circuit

Since at , the circuit is in the steady state, no current flows through the

capacitor and the circuit becomes as shown in Fig. 14.17. −= 0t

R3

R1

VS ( ) 00 =−

Ci

( )−0Cv

Fig. 14.17. Model of the circuit shown in Fig. 14.16 at −= 0t

In the circuit of Fig. 14.17 it holds

( ) SC Vv =−0 .

Notes

101

Since the capacitor satisfies the continuity property

( ) ( ) SCC Vvv == −00 .

When the switch is closed the circuit has the form shown in Fig. 14.18.

R3

R1

VS R2 C vC (t)

Fig. 14.18 The circuit of Fig. 14.16 when the switch is closed

It consists of a resistive one-port terminated by the capacitor . Using Thevenin’s theorem we obtain the equivalent circuit shown in Fig. 14.19, where

C

Ω1021

213 =

++=

RRRRRReq ,

V6221

0 =+

= RRR

VV S

C .

Req

V0C iC

C vC (t)

Fig. 14.19. Circuit equivalent to the circuit of Fig. 14.18

Notes

102

In the circuit shown in Fig. 14.19 the voltage ( )tvC is given by equation (14.21) repeated below

( ) ( ) ( ) ( )( ) τt

CCCC vvvtv−

∞−+∞= e0 , where S . 10 5−== CReqτ

In the steady state the voltage is constant, hence, Cv 0d

d==

tv

Ci CC and

( ) V60 ==∞ CC Vv . Having ( ) V6=∞Cv , ( ) V120 == SC Vv , , we obtain S10 5−=τ

( ) ⎟⎟

⎠

⎞

⎜⎜

⎝

⎛+= −

− 510e66t

C tv .

The plot of ( )tvC is shown in Fig. 14.20.

t

vC (t)

12

6

0

Fig. 14.20. Plot of ( )tvC in the circuit shown in Fig. 14.17

Notes

103

15. Sinusoidal steady-state analysis

15.1 Preliminary discussion

The proceeding discussion was focused on the analysis of circuits driven by constant sources. However, many signals encountered in practice are time-varying. In power systems the sinusoidal signal is the most popular. In this section we concentrate on the analysis of linear circuits driven by sinusoidal sources.

Let us consider a sinusoidal signal

( ) tAtf ωcos= , (15.1)

where A is called an amplitude and ω is called an angular frequency measured in radians/second (abbreviated to sr ). The signal (15.1) repeats itself after every T seconds, where T is called a period. The product of ω and t is an angle in radians. The angle corresponding to one period is π2 radians. Therefore, we may write

Tωπ =2 or

ωπ2

=T . (15.2)

The number of periods completed each second is known as the frequency of the signal. Thus, the frequency is given by

f

T

f 1= .

The unit of frequency is hertz (abbreviated to Hz). Using equation (15.2) we have

fT

ππω 22== .

Notes

104

The waveform of the signal (15.1) is shown in Fig. 15.1. In this case, the positive peak of the signal occurs at 0=t .

one period

4T t T

43 0

2T

f (t)

T

Fig. 15.1. The waveform of signal (15.1)

A general waveform of sinusoidal signal is shown in Fig. 15.2, where the positive peak occurs at 1tt = .

t1 t 0

f (t)

Fig. 15.2. General waveform of sinusoidal signal

Notes

105

This signal can be expressed as follows

( ) ( ) ( ) ( )φωωωω +=−=−= tAttAttAtf coscoscos 11 (15.3)

where

112 tT

t πωφ −=−= (15.4)

is called an initial phase or simply a phase. Example 15.1 Let us consider a sinusoidal voltage, shown in Fig. 15.3.

5

11.51.5 t (ms) 0

v (t) (V)

Fig. 15.3. An example of a sinusoidal voltage

The quantities characterising this signal are as follows:

V5=A , ms10=T ,

Notes

106

Hz10010101

3 =⋅

=−

f ,

sr62810022 =⋅== ππω f ,

radians30511022

1 πππφ ..tT

−=−=−= .

In practice however, we use degrees for the phase, because we have a physical feel for angles measured in degrees.

°→180Since π , then °−=°⋅−→− 541803030 .. π . Using the obtained results we may write

( ) ( )°−= 54628cos5 ttv .

Note that for any integer it holds n

( ) ( )ntAtA πφωφω 2coscos ++=+ .

Hence, there are infinitely many phases specified by nπφ 2+ . To determine the phase uniquely we chose the positive peak of the sinusoidal signal closed to the origin, as in Example 15.1.

Consider two sinusoidal signals at the same frequency

( )atA)t(a φω += cos and ( )btB)t(b φω += cos .

If ba φφ ≠ , then the phase difference between them is defined as

ba φφ − . If ba φφ > then ( )ta is said to lead ( )tb by ( )ba φφ − . If ba φφ < , then ( )ta is said to lag behind ( )tb by ( )ab φφ − .

Notes

107

Example 15.2 We calculate the phase difference between the current

( ) ( )A10700cos20 °+= tti

and the voltage

( ) ( )V50700sin100 °+= ttv .

We cannot do it directly, because the current and the voltage have different forms, one is cosine whereas the other is sine. Therefore, we first convert ( )tv to a cosine expression using the trigonometric identity

( )°−= 90cossin αα .

As a result, we obtain

( ) ( ) ( )V40700cos1009050700cos100 °−=°−°+= tttv .

Hence, the phase difference between ( )tv and ( )ti is

°−=°−°−=− 501040iv φφ .

This means that ( )tv lags behind ( )ti by . °50

15.2 Phasor concept We will study linear circuits, driven by sinusoidal sources at the same frequency, in

the steady-state. In such a case, all branch currents and voltages have sinusoidal waveforms. The sinusoidal steady-state analysis is called AC analysis. An efficient approach to the AC analysis uses the concept of phasors.

Let us consider a sinusoidal signal

( )φω +tAmcos . (15.5)

To his signal we associate a complex number A called a phasor, according to the rule

Notes

108

. (15.6) φjemAA =

Thus, for the given sinusoid the phasor is specified uniquely. Conversely, knowing the frequency ω , the phasor A specifies uniquely the sinusoid by the formula

( ) ( )( ) ( )φωφωω +== + tAAA mt

mt

m coseReeRe jj . (15.7)

To illustrate graphically the relation between sinusoidal function (15.5) and associated phasor (15.6) we consider the expression

( )φωω += tm

t AA jj ee . (15.8)

In the complex plane this expression describes a vector rotating counter-clockwise at the angular velocity of ω , as shown in Fig. 15.4.

φ

( )tA ωjeIm

( )tA ωjeRe

φjemAA =

1je tA ω

2je tA ω

φcosmA0

( )φω +1cos tAm

( )φω +2cos tAm

Fig. 15.4. Rotating phasor

At , we have . This is a complex number that we may plot as a vector. For any expression (15.8) represents a vector having the same length

0=t φω jj ee mt AAA ==

0>t

mt AAA ==ωje and the phase equal to φω +t . Thus, the phase of the vector

increases, which means that it rotates counter-clockwise.

Notes

109

Next we project orthogonally the tip of the vector on the horizontal axis and label it ( ) . Hence, according to Euler’s formula, we obtain tx

( ) ( ) ( )φωφωφω +++=+ tAtAA mmt

m sinjcose j , ( ) ( )φω += tAtx mcos .

Thus, sinusoidal function (15.5) equals the orthogonal projection of the phasor rotating counter-clockwise at the angular velocity of ω .

15.3 Phasor formulation of circuit equations

Let us consider a linear circuit driven by sinusoidal sources, all at the same frequency ω . In the steady-state all branch voltages and currents in this circuit are sinusoidal at the frequency ω . Figure (15.5) shows an arbitrary node of the circuit.

i1(t)

i3(t) i4(t)

i2(t)

Fig. 15.5. A node of the circuit

The sinusoidal currents flowing through the branches meeting at this node are labeled as follows

( ) ( )11

cos1 im tIti φω += , ( ) ( )22

cos2 im tIti φω += ,

( ) ( )33

cos3 im tIti φω += , ( ) ( )44

cos4 im tIti φω += .

Notes

110

KVL applied at the node gives

( ) ( ) ( ) ( ) 04321 =+−− titititi . (15.9)

Now, we write this equation in terms of the phasors

04321 =+−− IIII . (15.10)

Below it is proved that equation (15.10) implies equation (15.9). tωjWe multiply both sides of equation (15.10) by e

(15.11) 0eeee j4

j3

j2

j1 =+−− tttt IIII ωωωω

and create the phasors of the sinusoidal currents ( ) ( )ti,,ti 41 … :

11

j1 e i

mII φ= , , , . 22

j2 e i

mII φ= 33

j3 e i

mII φ= 4

4

j4 e i

mII φ=

Hence, it holds ( ) ( ) ( ) ( ) 0eeee 4

43

32

21

1

jjjj =+−− ++++ iiii tm

tm

tm

tm IIII φωφωφωφω . (15.12)

Next, we apply Euler’s formula to each term of equation (15.12), finding after simple rearrangements

ααα jsincose j +=

( ) ( ) ( ) ( )( ) ( ) ( ) ( )[ ] .tItItItI

tItItItI

imimimim

imimimim

0sinsinsinsinj

coscoscoscos

4433211

44332211

2 =+++−+−++

++++−+−+

φωφωφωφω

φωφωφωφω (15.13)

Equation (15.13) holds if both the real and the imaginary part is equal to zero. Hence, we may write

( ) ( ) ( ) ( ) 0coscoscoscos44332211=+++−+−+ imimimim tItItItI φωφωφωφω .(15.14)

We recognize the terms of equation (15.14) as the sinusoidal currents ( ) ( )ti,,ti 41 … and rewrite this equation in the form

( ) ( ) ( ) ( ) 04321 =+−− titititi ,

Notes

111

being KCL equation (15.9). In this way, we have proved that equation (15.10) implies equation (15.9). Thus, in the sinusoidal steady-state KCL equations can be written down directly in terms of phasors, replacing any sinusoid by its phasor. Similarly, we prove that KVL equations can be written down directly in terms of phasors. Branch equations Linear resistor Let us consider a linear resistor as shown in Fig. 15.6.

R i(t)

v(t) Fig. 15.6. A linear resistor

The resistor is described by Ohm’s law

( ) ( )tRitv = , (15.15)

where we assume that and ( )ti ( )tv are sinusoids at the same frequency ω : ( ) ( )vm tVtv φω += cos , ( ) ( )im tIti φω += cos .

We form the corresponding phasors , and formulate the equation vmVV φje= i

mII φje=

RIV = . (15.16)

Below it is proved that equation (15.16) implies equation (15.15). tωjeLet us multiply both sides of equation (15.16) by

(15.17) tt RIV ωω jj ee =

and present the phasor and V I in the polar form

. tm

tm

iv RIV ωφωφ jjjj eeee =

Notes

112

After simple rearrangement we have

( ) ( )iv tm

tm RIV φωφω ++ = jj ee . (15.18)

Next, we apply Euler’s formula finding

( ) ( ) ( ) ( )imimimvm tRItRItVtV φωφωφωφω +++=+++ sinjcossinjcos . (15.19)

Since equation (15.19) holds if the real parts on the left and right side and the imaginary parts on the left and right side are identical we may write

( ) ( )imvm tRItV φωφω +=+ coscos ,

which means

( ) ( )tRitv = .

Thus, in the sinusoidal steady-state Ohm’s equation describing a linear resistor can be written down directly in terms of phasors, replacing any sinusoid by its phasor. Linear inductor Figure 15.7 shows a linear inductor. We assume that the voltage ( )tv and the current ( ) are sinusoids at the same frequency ti ω : ( ) ( )vm tVtv φω += cos , ( ) ( )im tIti φω += cos .

L i(t)

v(t)

Fig. 15.7. A linear inductor The corresponding phasors are as follows

vmVV φje= , . i

mII φje=

Notes

113

In the time domain an inductor is described by the equation

( ) ( )ttiLtv

dd

= . (15.20)

We write the corresponding equation in terms of phasors

LIV ωj= (15.21)

and prove that equation (15.21) implies equation (15.20). For this purpose we multiply both sides of equation (15.21) by tωje

. (15.22) tt LIV ωω ω jj eje =

Since 2j

ejπ

= we rewrite equation (15.22) in the form

( ) ⎟⎠⎞

⎜⎝⎛ ++

+ = 2j

j eeπφω

φω ωi

vt

mt

m LIV . (15.23)

Next, we apply Euler’s formula to the exponential functions

( )

( )

.tLItLI

tVtV

imim

vmvm

⎟⎠⎞

⎜⎝⎛ +++⎟

⎠⎞

⎜⎝⎛ ++=

=+++

2sinj

2cos

sinjcosπφωωπφωω

φωφω (15.24)

Equation (15.24) is satisfied if the real parts on the left and right side are identical. Hence, we may write

( )

( ) ( )( ) ,tIt

LtLI

tLItV

imim

imvm

φωφωω

πφωωφω

+=+−=

=⎟⎠⎞

⎜⎝⎛ ++=+

cosddsin

2coscos

or

( ) ( )ttiLtv

dd

= .

Notes

114

Thus, equation (15.21) implies equation (15.20). Linear capacitor

C i(t)

v(t) Fig.15.8. A linear capacitor

In the time domain a linear capacitor is described by the equation

( ) ( )ttvCti

dd

= . (15.25)

The corresponding equation in terms of phasors is

CVI ωj= . (15.26)

It can be rewritten in the form

IC

IC

Vωω

jj

1−== . (15.27)

The crucial point is that, in terms of phasors, branch equations (15.15), (15.21) and (15.27) become algebraic equations with complex coefficients.

15.4 Impedance and admittance

Consider the circuit shown in Fig. 15.9, where the one-port consists of linear resistors, inductors and capacitors. Let the port voltage and the port current be sinusoidal functions of time. Their phasors are indicated in this figure.

We define the impedance of the one-port as the ratio of the voltage phasor and the current phasor

VI , that is

IVZ = . (15.28)

Notes

115

V

I

Fig. 15.9. A linear one-port

Setting vVV φje= and iII φje= we have

φjeZZ = (15.29)

where

IV

Z = and iv φφφ −= .

The impedance Z given by the expression (15.28) can be transformed into rectangular representation using Euler’s formula. As a result we obtain

φφ sinjcos ZZZ += . (15.30)

Denoting

RZZ == φcos and XZZ == φsin

we have

XRZ j+= , (15.31)

Notes

116

where R is called resistance or the resistive part of the impedance and X is called reactance or the reactive part of the impedance. Both R and X are measured in ohms. On the basis of (15.31) we find the magnitude of the impedance

22 XRZ += (15.32)

and the phase angle

RX1tan−=φ . (15.33)

The relationships (15.32) and (15.33) enable us to create a triangle in the complex plane, shown in Fig. 15.10. It is called an impedance triangle.

Z

φ

0

Im (Z)

Re (Z)

X

R

Fig. 15.10. The impedance triangle

Let us consider the series connection of two circuit elements specified by their

impedances and (see Fig. 15.11). 1Z 2Z

V

Z1 I

V1

Z2

V2

Fig. 15.11. Series connection of two elements

Notes

117

We write KVL equation

21 VVV +=

and find

2121 ZZ

IVV

IVZ +=

+== .

Thus, the impedance of two elements connected in series is the sum of their impedances.

The reciprocal of the impedance Z , labeled , is called the admittance Y

Z

Y 1= . (15.34)

The units of admittance are siemens, abbreviated to S. Substituting (15.29) into (15.34) gives

φφ

jj e1

e11 −===

ZZZY . (15.35)

Thus, the magnitude of the admittance is equal to the inverse of the magnitude of the impedance

Z

Y 1= , (15.36)

whereas the phase of the admittance equals the phase of the impedance with the opposite sign ( )φ− . Plugging (15.31) into (15.34) yields

222222 jjj

1XR

XXR

RXRXR

XRY

+−

+=

+−

=+

= .

Notes

118

This expression can be presented in the compact form

BGY j+=

where 22 XRRG+

= is called conductance and 22 XRXB+

−= is called susceptance.

The units of and G B are siemens. (If the reactance is positive , the element is called inductive, if the

element is called capacitive. Since )0>X 0<X

22 XRXB+

−=

we conclude that the element is inductive when B is negative and capacitive when B is positive. On the basis of the equation

BGY j+=

we obtain 22 BGY += and

GBY 1tan −=−=∠ φ .

These relationships are summarized graphically in the complex plane as shown in Fig. 15.12. The construction is called an admittance triangle.

Y

-φ

0

Im (Y)

Re (Y)

B

G

Fig. 15.12. Admittance triangle

Notes

119

Consider two elements, specified by their admittances and , connected in parallel (see Fig. 15.13).

1Y 2Y

I2

Y2 V

I1 I

Y1

Fig. 15.13. Two elements connected in parallel

Using KCL we obtain

21 III += .

Hence, it holds

2121 YY

VII

VIY +=

+== .

The example given below explains how to use the concept of impedance and admittance in the analysis of linear circuits in the sinusoidal steady-state. The analysis of sinusoidal circuits using phasors is called the analysis in the frequency domain.

Example 15.3

C R2 vS (t)

i1 (t) R1

Fig. 15.14. An example circuit

Notes

120

Consider the circuit shown in Fig. 15.14, driven by the sinusoidal voltage source

( ) ( )V451000cos1414 °+= t.tvS .

We wish to find the steady state current ( )ti1 using the phasor concept. The phasor of ( )tvS is

. Ve1414 45j °= .VS

Figure 15.15 shows the circuit in a phasor format.

ZC Z2=R2 VS

I1 Z1=R1

Fig. 15.15. The circuit of Fig. 15.14 in phasor format

We compute the impedance faced by the voltage source. Since the elements specified by impedances and are connected in parallel we find the admittance 2Z CZ

C

CC ZZYYY 11

222 +=+= .

Hence, the impedance is CZ2

CR

CR

ZZZZ

YZ

C

C

CC

ω

ωj

j1

2

2

2

2

22

−

⎟⎠⎞

⎜⎝⎛−

=+

== .

Notes

121

Next, we compute the total impedance

Ωe246320j60

0250j40

0250j40

40 4318j-21

°=−=−

⎟⎠⎞

⎜⎝⎛−

+=+= .C .

.

.ZZZ .

Having the impedance we find

Ae2230e2463e1414 43j63.

43j18.-

45j

1°

°

°

=== ...

ZV

I S ,

hence, we have

( ) ( )A43631000cos22301 °+= .t.ti .

15.5 Phasor diagrams

Any phasor of a voltage and a current can be drawn as a vector having the length equal to the magnitude and the angle equal to the phase of the phasor. The diagram thus obtained is called a phasor diagram. For the linear inductor shown in Fig. 15.16 we have

LLL VL

VL

I 2j

e1j1 π

ωω−

== .

L i(t)

v(t)

Fig. 15.16. A linear inductor

Notes

122

Using the voltage phasor as the horizontal reference we draw the phasor vertically downward as shown in Fig. 15.17.

LV LI

VL

2πφ −=i

IL Fig. 15.17. Phasor diagram of the inductor

We say that the inductor current lags behind the inductor voltage by 2π .

For the capacitor shown in Fig 15.18 we write

CCC IC

IC

V 2j

e1j π

ωω−

=−= .

C i(t)

v(t) Fig. 15.18. A linear capacitor

Assuming that 0=iφ we draw the phasor diagram as depicted in Fig. 15.19.

IC

2πφ −=v

VC Fig. 15.19. Phasor diagram of the capacitor

Notes

123

We say that capacitor voltages lag behind the capacitor current by 2π .

Now we draw a phasor diagram showing the current and voltages in the circuit shown in Fig. 15.20.

L1

V2 V3

C3R2

VS

I

V1

Fig. 15.20. An example circuit

Using KVL we write

321 VVVVS ++= .

We use a current phasor as the horizontal reference. Since 2j

111 ejπ

ωω ILILV == , the

voltage leads the current by 1V2π . Consequently the phasor is drawn vertically

upward (see Fig. 15.21).

1V

I

V1

Fig. 15.21. A simple phasor diagram

Notes

124

The voltage IRV 22 = has the same phase as the current, hence, the phasor is horizontal, similarly as

2VI . We shift this phasor to the tip of as depicted in Fig. 15.22. 1V

V2

I

V1

Fig. 15.22. A part of the phasor diagram of the circuit shown in Fig. 15.20

Since 2j

3333 e1j

j1 π

ωωω−

=−==C

IC

IC

V , the phasor is drawn vertically downward

and shifted to the tip of . Next, the three phasors are added vectorially to obtain the voltage as illustrated in Fig. 15.23.

3V

2V

SV

VS

V3

V2

I

V1

Fig. 15.23. A phasor diagram of the circuit of Fig. 15.20

15.6 Effective value

A commonly used measure of sinusoidal voltages and currents is the effective value

also called a root mean square value (abbreviated to rms).

Notes

125

For a sinusoidal voltage

( ) ( )vm tVtv φω += cos

it is defined as follows

( )( ) ( )( )∫∫ +===T

vm

T

rmseff ttVT

ttvT

VV0

2

0

2 dcos1d1 φω . (15.37)

To integrate, we apply the trigonometric identity

( )αα 2cos121cos2 +=

finding

( ) ( )( )

( ) .ttVTV

ttVttV

T

vmm

T

vm

T

vm

∫

∫∫

++=

=++=+

0

22

0

2

0

22

d22cos21

21

d22cos121dcos

φω

φωφω

(15.38)

In the interval T−0 there are exactly two periods of the function ( )vt φω 22cos + , consequently

. ( ) 0d22cos0

=+∫T

v tt φω

Hence, we have

( ) TVtt m

T

vm2

0

22

21d22cosV =+∫ φω

and

Notes

126

22

11 2 mmrmseff

VTV

TVV === . (15.39)

Similarly we define the effective (rms) value of the sinusoidal current

( ) ( )im tIti φω += cos

finding

2m

rmseffI

II == . (15.40)

Notes

127

16. Power in sinusoidal steady state

16.1 The instantaneous and average power

Power is a very important quantity in electrical engineering, electronics, communications, and power systems. Hence, the question arises how to define power in sinusoidal circuits.

Consider a one-port consisting of resistors, inductors and capacitors (see Fig. 16.1) with sinusoidal port voltage

( ) ( )Vcos vm tVtv φω += (16.1)

and port current

( ) ( )Acos im tIti φω += . (16.2)

V

I

Fig. 16.1. A one-port consisting of resistors, inductors and capacitors

The instantaneous power received by the one-port is given by

( ) ( ) ( ) ( ) ( )ivmm ttIVtitvtp φωφω ++== coscos . (16.3)

Notes

128

Using the trigonometric identity

( ) ( )( )yxyxyx −++= coscos21coscos

we obtain

( ) ( ) ( )ivmm

ivmm IVtIVtp φφφφω −+++= cos

22cos

2 . (16.4)

Thus, the instantaneous power is the sum of two components. The first one is a sinusoid whose frequency is twice of the voltage and current. The second component is a constant. Thus, the waveform of ( )tp is a sinusoid having the angular frequency ω2 shifted upward by the value equal to the second term.

Average power, labeled , is defined as the mean value of the instantaneous power over an interval of

avPT , where T is a period of the voltage or current

( ) ttpT

PT

av d1

0∫= . (16.5)

Substituting (16.4) into (16.5) yields

( ) ( )( )∫ −+++=T

0dcos2cos

21 tt

IVT

P ivivmm

av φφφφω .

The integral of the sinusoid ( )ivt φφω ++2cos , having the period 2T , over the interval

to 0 T is zero, because the total area under the waveform of this sinusoid is equal to zero. Hence, we have

( ) φφφφ coscos2

dcos2

1 T

0effeff

mmiv

mmav IV

IVt

IVT

P ==−= ∫ , (16.6)

Notes

129

where iv φφφ −= is called a power factor. Note that ⎥⎦⎤

⎢⎣⎡−∈

22ππφ , , 0cos ≥φ and the

average power is a real, nonnegative number.

16.2 The complex power

Another power useful in the AC analysis of circuits is the complex power defined as follows

∗= IVP21 , (16.7)

where is the phasor of the port voltage whereas V ∗I is the complex conjugate of the port current phasor. Hence, it holds

( )

( ) ( )

.effIeffVeffIeffVmImVmImV

IVIV

IVIVP

VV

IVIV

φφφφ

φφφφ

φφφφ

sinjcossin21jcos

21

sin21jcos

21

e21ee

21 jj-j

+=+=

=−+−=

=== −

(16.8)

Thus, the complex power is a complex number. The real part of this number is, according to (16.6), the average power

( )PPav Re= .

Let XRZ j+= and be the impedance and admittance of the one-port, respectively (see Fig. 16.1). Then using (16.7) we obtain

BGY j+=

( ) ( )

,RIRIRI

XRIIIZPP

effm

av

222

2

21

21

j21Re

21ReRe

===

=⎥⎦⎤

⎢⎣⎡ +=⎟

⎠⎞

⎜⎝⎛== ∗

(16.9)

Notes

130

or

( )

( ) .GVGVVGVBG

VVYVYVP

effm

av

2222

21

21j

21Re

21Re

21Re

===⎥⎦⎤

⎢⎣⎡ −=

=⎥⎦⎤

⎢⎣⎡=⎥⎦

⎤⎢⎣⎡= ∗∗∗

(16.10)

The imaginary part of the complex power, labeled , is called a reactive power, that is (see (16.8))

xP

( ) φφ sinsin21Im effeffmmx IVIVPP === . (16.11)

Similarly, as in the case of the average power, we find some alternative expression for the reactive power

( ) ( )

.XIIXIXIXIR

IXRIIZPP

effm

x

22222

2

21

21j

21Im

j21Im

21ImIm

===⎥⎦⎤

⎢⎣⎡ +=

=⎥⎦⎤

⎢⎣⎡ +=⎟

⎠⎞

⎜⎝⎛== ∗

(16.12)

or

( ) ( )

( ) .BVBVVBVBG

VVYVYVPP

effm

x

2222

21

21j

21Im

21Im

21ImIm

−=−=−=⎥⎦⎤

⎢⎣⎡ −=

=⎥⎦⎤

⎢⎣⎡=⎥⎦

⎤⎢⎣⎡== ∗∗∗

(16.13)

Thus, the complex power is the complex number whose real part is the average power and the imaginary part is the reactive power

xav PPP j+= . (16.14)

Using (16.8) we obtain

Notes

131

( )φφφφ jsincos21sin

21jcos

21

+=+= IVIVIVP .

By virtue of Euler’s formula it holds

φφφ jjj ee21e

21

effeffmm IVIVIVP === . (16.15)

Expression (16.14) and (16.15) lead to the equations

222xav PPP += (16.16)

and

av

x

PP

=φtan . (16.17)

They enable us to create the geometrical construction called a power triangle. Two variants of the power triangle are shown in Fig. 16.2.

P

φ

Px > 0

Pav

a)

P

φ

Px < 0

Pav

b)

Fig. 16.2. Two variants of the power triangle

Consider the one-port consisting of resistors, inductors and capacitors, supplied with

a sinusoidal source ( ) as shown in Fig. 16.3. tv

Notes

132

v (t)

i(t)

Fig. 16.3. One-port RLC driven by a sinusoidal voltage source

It can be proved that the complex power P of the one-port is equal to the sum of the complex powers of all elements of the one-port, i.e. kP

, (16.18) ∑=

=b

kkPP

1

where is the number of the elements. bSince

xav PPP j+= and ( ) ( ) b,,k,PPP kxkavk …1j =+= ,

equation (16.18) leads to two equations

, (16.19) ( )∑=

=b

kkavav PP

1

. (16.20) ( )∑=

=b

kkxx PP

1

Notes

133

16.3 Measurement of the average power

To measure the average power we use a wattmeter. The wattmeter is an instrument having a current coil and a voltage coil. To measure the average power of a load the coils are connected as shown in Fig. 16.4.

*

*i(t)

v(t) Load

Fig. 16.4. Measurement of the average power

The current coil is connected in series to the load, whereas the voltage coil is connected in parallel to the load. There is ∗ mark on each of the coils. They are connected as shown in Fig. 16.4. The wattmeter reading is given by

( ) ( )V,IIVVI effeff ∠=∗ cosRe21 , (16.21)

where I and are phasor of the sinusoidal voltage V ( )tv and current ( )ti , respectively. For an ideal wattmeter the impedance of the current coil tends to zero and the impedance of the voltage coil tends to infinity. As a result, the ideal wattmeter does not introduce any disturbance in the circuit. Both the current and the voltage remains the same as without the wattmeter, ( ) φ=∠ V,Icos and the wattmeter reading is the average power consumed by the load.

Notes

134

16.4 Theorem on the maximum power transfer

Consider the circuit consisting of a sinusoidal voltage source, represented by phasor and impedance , and the load represented by impedance (see Fig. 16.5). We

want to find so that the average power entering the load is maximum. SV SZ LZ

LZ

ZL Zs

I Vs

Fig. 16.5. Simple circuit terminated by load ZL

Let the impedances and be given in rectangular form SZ LZ

SSS XRZ j+= , LLL XRZ j+= .

On the basis of (16.9) the average power delivered to the load is

Lav RIP 2

21

= .

Since

LS

S

ZZV

I+

=

it follows

( ) ( ).

XXRRRV

XRXR

RVRZZ

VR

ZZV

P

LSLS

LS

LLSS

LSL

LS

SL

LS

Sav

222

22

2

22

21

jj21

21

21

+++=

=+++

=+

=+

=

Notes

135

In the above expression we need to choose and to maximize . Since can either be positive or negative we choose

LR LX avP LX

SL XX −= . (16.22)

Consequently, the formula for becomes avP

( )2

2

21

SL

LSav

RRR

VP+

= . (16.23)

To determine corresponding to the optimum of we compute the derivative with respect to

LR avP

LR

( ) ( )( )4

22 2

21

dd

SL

SLLSLS

L

av

RRRRRRR

VRP

+

+−+=

and set it to zero

( )

0222

21

4

2222 =

+

−−++

SL

SLLSLSLS

RRRRRRRRR

V .

The solution of this equation is

. SL RR =

It can be proved that the optimum of means the maximum. avPThus, the maximum average power is attained when

(16.24) ∗=−= SSSL ZXRZ j

and it is

( ) S

S

S

sSav R

V

RR

VP822

1max2

22 == . (16.25)

Notes

136

17. Resonant circuits

17.1 Series resonant circuit

Let us consider the circuit shown in Fig. 17.1 consisting of an inductor, a capacitor and a resistor connected in series, driven by a sinusoidal voltage source.

L

VC VR

C

RVS

I

VL

Fig. 17.1. Series resonant circuit

Suppose that the voltage source is

( ) ( )SSS tVtvm

φω += cos ,

then the phasor associated to this voltage equals

. SmSS VV φje=

In Fig. 17.1 , , and are the phasors of the voltages across the elements whereas LV CV RVI is the phasor of the current traversing all the elements. The circuit can be considered a one-port supplied with a voltage source. The impedance of this circuit is given by

XRC

LRRZ j1 jC1j-Lj +=⎟

⎠⎞

⎜⎝⎛ −+=+=

ωω

ωω , (17.1)

Notes

137

where

( )C

LXXω

ωω 1−== . (17.2)

The reactance X of the one-port is a monotonically increasing function of ω , for ( )∞∈ ,0ω . It becomes zero at 0ωω = , i.e.

01

00 =−

CL

ωω . (17.3)

Solving equation (17.3) for 0ω yields

LC1

0 =ω . (17.4)

The frequency 0ω is called a resonant frequency and the circuit at this frequency is said to be in resonance. Since ( ) 00 =ωX , the impedance in resonance is equal to the resistance

( ) RXRZ =+= 00 j ω . (17.5)

Hence, the current in resonance is given by

R

VZV

I SS ==0

0 . (17.6)

At an arbitrary frequency ω the impedance of the one-port equals

( ) ( )ωω XRZ j+= . (17.7)

For each ω the impedance is represented by a vector in the complex plain as shown in Fig. 17.2. As ω varies in the interval ( )∞,0 the tip of this vector traces a straight line parallel to the imaginary axis. At 0ωω = the reactance of this impedance equals zero and its magnitude is the smallest.

Notes

138

Thus, all the voltage from the source is across the resistor. At low frequencies

CL

ωω 1

<< and most of the voltage is across the capacitor. At high frequencies

CL

ωω 1

>> and most of the voltage is across the inductor.

Z

φ

0

Im (Z)

Re (Z) R

Fig. 17.2. Plot of ( )ωZ in the complex plane

Since at resonance C

Lω

ω 1= , the voltage across the inductor and the capacitor are as

follows

LIVL 0jω= , IC

VC0

1jω

−= ,

hence, CL VV −= . Thus, both these voltages have the same magnitude. It is illustrated in the phasor diagram shown in Fig. 17.3.

I

VL

VR = V VC

Fig. 17.3. Phasor diagram in resonance

Notes

139

At resonance the ratio of the magnitude of the voltage across the inductor (or capacitor) to the magnitude of the source voltage is called a quality factor and labeled Q

m

m

m

m

S

C

S

L

VV

VV

Q == at 0ωω = . (17.8)

Since

mCCmLL IC

VVLIVVmm

00

1ω

ω =====

and mSS RIVV

m== ,

then

CRR

LQ

0

0 1ω

ω== . (17.9)

Now we consider the ratio of the phasor of the voltage across the resistor to the phasor of the source voltage at any frequency ω . Since

⎟⎠⎞

⎜⎝⎛ −+

==

CLR

VRRIV S

R

ωω 1j

, (17.10)

then it holds

⎟⎠⎞

⎜⎝⎛ −+

=

CLR

RVV

S

R

ωω 1j

. (17.11)

Expression (17.11) can be rearranged as follows

⎟⎟⎠

⎞⎜⎜⎝

⎛−+

=

⎟⎟⎠

⎞⎜⎜⎝

⎛−+

=

ωω

ωωω

ωω

ω 0

0

020 j1

1

j1

1

RL

RLV

V

S

R . (17.12)

Notes

140

Using (17.9) we have

⎟⎟⎠

⎞⎜⎜⎝

⎛−+

=

ωω

ωω 0

0j1

1

QVV

S

R , (17.13)

hence, it holds

( )( ) 2

0

0

21

1

⎟⎟⎠

⎞⎜⎜⎝

⎛−+

==

ωω

ωωQ

VV

VV

Sm

Rm

S

R . (17.14)

Plots of SR VV versus ω , at different quality factors, are shown in Fig. 17.4.

ω

s

R

VV

1

0 ω0

Q=20

Q=5

Fig. 17.4. Plots of SR VV against frequency ω

Notes

141

At the resonance frequency SR VV is equal to 1, whereas at 0=ω and ∞=ω the ratio equals zero. Therefore the voltage across the resistor R is the same as the source voltage at the resonant frequency. This voltage is slightly reduced for the frequencies close to the resonance frequency and considerably reduced at low and high frequencies. Such a circuit is called a bandpass filter. The passband is defined as a range of the frequencies in the neighbourhood of the resonant frequency such that at the edges of the passband

( )( ) 2

1=

Sm

Rm

VV

. (17.15)

Substituting (17.14) into (17.15) gives

2

1

1

12

0

0

2

=

⎟⎟⎠

⎞⎜⎜⎝

⎛−+ωω

ωωQ

. (17.16)

We rearrange equations (17.16) as follows

,Q 212

0

0

2 =⎟⎟⎠

⎞⎜⎜⎝

⎛−+ωω

ωω ,Q 1

20

0

2 =⎟⎟⎠

⎞⎜⎜⎝

⎛−ωω

ωω

Q10

0±=−

ωω

ωω .

Hence, we have

Q1

2

0

0

2 =−ωω

ωω

, Q1

1

0

0

1 −=−ωω

ωω

.

or

Q

2020

22

ωωωω =− ,

Q102

021

ωωωω −=− .

Notes

142

Next we subtract both sides of the above equations

( )1202

122 ωω

ωωω +=−

Q . (17.17)

Dividing equation (17.17) by ( )12 ωω + we obtain the following expression for the passband

Q

012

ωωω =− . (17.18)

The passband is illustrated in Fig. 17.5.

the passband

ω1 ω2

ω0 ω

( )( )mS

mR

S

R

VV

VV

=

1

0

21

Fig. 17.5. The passband of a resonant circuit

Notes

143

17.2 Parallel resonant circuit

Another type of a resonant circuit is the parallel resonant circuit shown in Fig. 17.6. The circuit is driven by sinusoidal source current represented by the phasor . In Fig. 17.6 , , are phasors of the currents flowing through the circuit elements and is the phasor of the voltage across the parallel connection of these elements.

SI

CI LI RI V

IS

IC IL IR

V C L R

RG 1=

Fig. 17.6. Parallel resonant circuit

The admittance of the one-port faced by the source is

⎟⎠⎞

⎜⎝⎛ −+=++=

LCG

LCGY

ωω

ωω 1j

j1j . (17.19)

The imaginary part of (susceptance) is a function of the frequency Y ω

( )L

CBω

ωω 1−= . (17.20)

The frequency 0ωω = such that ( ) 00 =ωB is called a resonant frequency. To find this frequency, we solve the equation

01

00 =−

LC

ωω , (17.21)

finding

LC1

0 =ω . (17.22)

Notes

144

At the resonance it holds

RS IGVYVI === . (17.23)

Furthermore, since

VLL

VI L00

1jj ωω

−== (17.24)

and CVIC 0jω= , (17.25)

then CL II −= or 0=+ CL II . (17.26)

Thus, all the current from the current source goes through the resistor and the currents in the inductor and capacitor add to zero. The phasor diagram is shown in Fig. 17.7.

IC

V

IL

IR=IS

IS

IC IL

IR

C L R

I=0

Fig. 17.7. The phasor diagram Fig. 17.8. Parallel resonant circuit at 0ωω = of the resonant circuit Figure 17.8 illustrates equation (17.26).

Let us consider the ratio of the currents and at an arbitrary frequency RI SI ω and rearrange it as follows

Notes

145

.RC

RCLCG

GYVGV

II

S

R

⎟⎟⎠

⎞⎜⎜⎝

⎛−+

=

=

⎟⎟⎠

⎞⎜⎜⎝

⎛−+

=⎟⎠⎞

⎜⎝⎛ −+

==

ωω

ωωω

ωω

ωωω

0

00

20

j1

1

j1

11j

(17.27)

Since RG

=1 and 2

01 ω=

LC, then we may write

( )( )

( )( )Sm

Cm

II

VGVC

GC

R

CRC ====

0

00000 1 ω

ωωωωω . (17.28)

The ratio of the current and ( )mCI ( )mSI is called a quality factor of the parallel resonant circuit and labeled . Since at the resonance Q ( ) ( )mLmC II = , we have

( )( )

( )( ) L

RCRII

II

QmS

mL

mS

mC

00 ω

ω ==== . (17.29)

Substituting (17.29) into (17.27) yields

⎟⎟⎠

⎞⎜⎜⎝

⎛−+

=

ωω

ωω 0

0j1

1

QII

S

R . (17.30)

Hence, we obtain the equation

( )( ) 2

0

0

21

1

⎟⎟⎠

⎞⎜⎜⎝

⎛−+

==

ωω

ωωQ

II

II

mS

mR

S

R , (17.31)

illustrated in Fig. 17.8 for two values of . Q

Notes

146

ω

S

R

II

1

0 ω0

Q=20

Q=5

Fig. 17.9. Plots of SR II for different values of Q

The plots shown in Fig. 17.9 have the same shapes as in Fig. 17.5 and can be similarly interpreted. Thus, the parallel resonant circuit is a bandpass filter which passes the current through the resistor at the resonant frequency, slightly reduces the current in the neighbourhood of this frequency and considerably reduces it at a low frequency and at a high frequency. Similarly, as in the case of a series resonant circuit, the passband is defined as a range of the frequencies in the neighbourhood of the resonant frequency such that at the edges of the passband it holds

2

1=

S

R

II

. (17.32)

Hence, we obtain the following expression for the passband

Q

012

ωωω =− . (17.33)

Notes

147

18. Coupled inductors

In this section we study the circuits in which two inductors are coupled magnetically. 18.1 Basic properties Let us consider two coils, placed near each other, as shown in Fig. 18.1. If we supply

the first coil with a generator so that the current varies with time and have the second coil open, then a time-varying magnetic flux

1iφ is produced. It induces a voltage in the

second coil.

i2 = 0

v2

i1

v1

φ

Fig. 18.1. Two magnetically coupled inductors

Now we consider the case where currents and flow through the inductors. If the

core is made of nonmetallic material, the following equations relate the currents and the fluxes leakages associated with the inductors

1i 2i

Notes

148

, (18.1)

iLMi,MiiL

2212

2111

+=+=

φφ

where is the self-inductance of the inductor 1, is the self-inductance of the inductor 2, and

1L 2LM is called a mutual inductance of the inductors 1 and 2. The equality

of both mutual inductances is established in physics. By Faraday’s law we have

.

ti

Lti

Mt

v

,tiM

tiL

tv

dd

dd

dd

dd

dd

dd

22

122

211

11

+==

+==

φ

φ

(18.2)

The mutual inductance M enables us to define the coefficient of coupling as follows k

21LL

Mk = . (18.3)

It can be shown that . 1≤k

φ1

φ2φ2

φ1i1i1

i2

i2

(a) (b) Fig. 18.2. Coupled inductors with dotted terminals

Notes

149

The mutual inductance M may be positive or negative. The sign of M is specified by the reference directions of the currents and the physical arrangement of the coils. To determine the sign of M , we place a dot mark in each inductor so that, when both currents of the inductors enter them through the dotted terminals, then the fluxes due to these two currents add. This is illustrated in Fig. 18.2. Next, we determine the sign of the mutual inductance M as follows. We arbitrarily assign the reference directions of the currents and . If the reference directions of these currents are such that both currents enter the inductors through the dotted terminals or leave the inductors through the dotted terminals, then

1i 2i

M is positive. If one current enters the inductor through the dotted terminal and the other leaves the inductor through the dotted terminal, then M is negative. Thus, if M is positive, then both fluxes due to the currents and add. If 1i 2i M is negative, they subtract.

According to the rule M in Fig. 18.3a is positive, whereas in Fig. 18.3b is negative.

(a) (b)

v1

v2v2

v1i1i1

i2 i2

Fig. 18.3. Coupled inductors with positive (Fig. 18.3a) and negative (Fig. 18.3b) mutual inductance M

Dotted terminals enable us to specify the sign of M without knowing the physical arrangement of the coils. Therefore, they are introduced in circuit diagrams as shown in Fig. 18.4.

Notes

150

v2

M > 0

v2

M < 0

L1

i2 i1

L2 v1 L1

i2 i1

L2 v1

(a) (b) Fig. 18.4. Coupled inductors in circuit diagrams

Two coupled inductors are described in the time domain by the equations (18.2). If we consider sinusoidal steady state we formulate the corresponding equations in the phasor form

2111 jj MIILV ωω += , (18.4)

2212 jj ILMIV ωω += . (18.5)

18.2 Connections of coupled inductors Series connection

Figure 18.5 shows two coupled inductors connected in series. We consider the sinusoidal steady state and apply the phasor concept.

L2 R2

v2 v1

M > 0 L1 R1

v

I L2 R2

v2 v1

M < 0

L1 R1

v

I

(a) (b) Fig. 18.5. Coupled inductors connected in series

Notes

151

The voltage across the connection is given by V

( ) ( )

( )[ ] .IMLLRR

MIILIRMIILIRVVV

2j

jjjj

2121

221121

++++=

=+++++=+=

ω

ωωωω

Hence, we obtain the impedance of this connection, as follows

( ) eqeq LRMLLRRIVZ ωω j2j 2121 +=++++== , (18.6)

where

21 RRReq += (18.7)

and

MLLLeq 221 ++= . (18.8)

Equation (18.6) states that the series connection shown in Fig. 18.5 can be replaced by the equivalent circuit shown in Fig. 18.6.

Leq Req

V

I

Fig. 18.6. Equivalent circuit to the circuit depicted in Fig. 18.5

Note that the equivalent inductance is given by equation (18.8), where M can be either positive or negative. Figure 18.7a shows a phasor diagram for , whereas Fig. 18.8b shows a phasor diagram for 0

0>M<M .

Notes

152

(b)

MIωj

IL2jωIL1jω

MIωj

V2V1

VR2I

R1I I

MIωj

IL2jω

IL1jω

MIωj

V2

V1

V

R2I

R1I I (a)

Fig. 18.7. Phasor diagrams of the connections shown in Fig. 18.5

Parallel connection Figure 18.8 shows two coupled inductors connected in parallel.

M

L1

I2 I1 I

L2 V

Fig. 18.8. Coupled inductors connected in parallel

Notes

153

We analyse this connection in the sinusoidal steady state using the phasor concept. As a result, we write the equations

211 jj MIILV ωω += , (18.9)

122 jj MIILV ωω += . (18.10)

Next, we solve equation (18.9) for 2I

11

2 jI

ML

MVI −=ω

(18.11)

and substitute (18.11) into (18.10)

111

22 jjj

j MIIMLL

MVLV ωωω

ω +−= .

Hence, we find the current 1I

( )

VLLM

LMV

MLL

M

ML

I21

2

2

21

2

1jj

1

−

−=

⎟⎠

⎞⎜⎝

⎛ −

−=

ωω

. (18.12)

Similarly, we obtain

( )

VLLM

LMI

212

12

j −

−=

ω . (18.13)

Taking into account (18.12) and (18.13) we have

( )

VLLM

MLLIII

212

2121

j

2

−

+−−=+=

ω . (18.14)

Notes

154

Equation (18.14) leads to

ILV eqωj= , (18.15) where

MLL

MLLLeq 221

221

−+−

= . (18.16)

Thus, the parallel connection of the coupled inductors, depicted in Fig. 18.8, is equivalent to the single inductor having the inductance specified by (18.16). To illustrate the method for the analysis of circuits containing coupled inductors, we consider the circuit shown in Fig. 18.9.

M

RLV2

R1

L1

I2 I1

V1 L2 VS

Fig. 18.9. A circuit containing coupled inductors

First we write the equations for and 1V 2V

2111 jj MIILV ωω += , (18.17)

1222 jj MIILV ωω += . (18.18)

Then, we apply KVL in the two loops

111 VIRVS += , (18.19)

022 =+VIRL . (18.20)

Notes

155

Using equations (18.17) – (18.20) we write

( ) 2111 jj MIILRVS ωω ++= , (18.21)

( ) 0jj 122 =++ MIILRL ωω . (18.22)

Next, we rearrange the set of equations (18.21) – (18.22) as follows

12

2 jj I

LRMI

L ωω+

−= , (18.23)

122

22

11 jj I

LRMLRVS ⎟⎟

⎠

⎞⎜⎜⎝

⎛

+++=

ωωω . (18.24)

Hence, the impedance 1IVZ S= is given by

22

22

11 jj

LRMLRZω

ωω+

++= . (18.25)

18.3 Ideal transformer

Let us consider two coils wound around a ferromagnetic core. In such a case, both

coils are linked by essentially the same magnetic flux. The time varying flux induces voltages across the coils which are approximately proportional to the number of turns.

v2 N1

i2 i1

v1 N2

Fig. 18.10. Symbol of the ideal transformer

Notes

156

In the ideal transformer, whose symbol is shown in Fig. 18.10, it holds

1

2

1

2

NN

vv

= , (18.26)

where and are the numbers of the coils turns. 1N 2NAnother equation describing the ideal transformer refers to the current ratio

2

1

1

2

NN

ii

−= . (18.27)

Equation (18.26) states that depending on 12 NN the voltage in the ideal transformer is stepped up or stepped down. Using equations (18.26) and (18.27) we see that a step up in voltage accompanies a step down in current. Consequently, there is no power loss in the ideal transformer.

Consider the ideal transformer, in the sinusoidal steady state, terminated by impedance , as shown in Fig. 18.11. 0Z

Z0 V2 N1

I2 I1

V1 N2

Fig. 18.11. Ideal transformer in sinusoidal steady state terminated by Z0

In such a case we use the phasor concept and rewrite equations (18.26) and (18.27) in the form

Notes

157

1

2

1

2

NN

VV

= , (18.28)

2

1

1

2

NN

II

−= . (18.29)

On the basis of (18.28) and (18.29) we obtain

2

22

2

1

1

1

IV

NN

IV

⎟⎟⎠

⎞⎜⎜⎝

⎛−= . (18.30)

Since

( )2

20 I

VZ−

= ,

then

0

2

2

1

1

11 Z

NN

IV

Z ⎟⎟⎠

⎞⎜⎜⎝

⎛== . (18.31)

Equation (18.31) expresses the impedance at the primary in terms of impedance . 1Z 0Z

Notes

158

19. Three-phase systems

19.1 Introduction Three-phase systems are commonly used in the generation, transmission and

distribution of electric power. Power in a three-phase system is constant rather than pulsating and three-phase motors start and run much better than single-phase motors. A three-phase system is a generator-load pair in which the generator produces three sinusoidal voltages of equal amplitude and frequency but differing in phase by 120° from each other.

The phase voltages , and ( )tva ( )tvb ( )tvc are as follows

( )( ) ,tVv

tVv

tVv

mc

mb

ma

240cos

120cos

cos

−=

−=

=

ω

ω

ω

(19.1)

whereas the corresponding phasors are

.VV

VV

VV

mc

mb

ma

240j

120j

e

e−

−

=

=

=

(19.2)

A three-phase system is shown in Fig 19.1. In a special case, all impedances are identical

ZZZZ cba === . (19.3)

Such a load is called a balanced load and is described by equations

ZV

IZ

VI

ZV

I cc

bb

aa === .

Notes

159

In

Ia Za Va

Ic Zc Vc

Ib Zb Vb

Fig. 19.1. Four-wire, three-phase system

Using KCL, we have

( )cbacban VVVZ

IIII ++=++=1 , (19.4)

where

( ) .VV

VVVV

mm

mcba

023j

21

23j

211240jsin240cos120jsin120cos1

ee1 240j120j

=⎟⎟⎠

⎞⎜⎜⎝

⎛+−−−=−+−+=

=⎟⎠⎞⎜

⎝⎛ ++=++ −−

Setting the above result into (19.4), we obtain

0=nI . (19.5)

Since the current flowing though the fourth wire is zero, the wire can be removed (see Fig. 19.2). The system of connecting the voltage sources and the load branches, as depicted in Fig. 19.2, is called a Y-system or a star system. Point n is called the neutral point of the generator and point n′ is called the neutral point of the load.

Notes

160

n’ n

Ia Z Va

Ic Z Vc

Ib Z Vb

Fig. 19.2. Three-wire, three-phase system

Each branch of the generator or load is called a phase. The wires connecting the

supply to the load are called lines. In the Y-system shown in Fig. 19.2 each line current is equal to the corresponding phase current, whereas the line-to-line voltages (or simply line voltages) are not equal to the phase voltages.

19.2 Y-connected systems

We consider Y-connected generator sources ( see Fig. 19.3).

Vca

c Vbc

Vab

n

a Va

Vc

bVb

Fig. 19.3. Y-connected generator sources

Notes

161

The phasors of the phase voltages can be generally written as follows

. (19.6) 240j

120j

j

e

e

e

−

−

=

=

==

VV

VV

VVV

c

b

maα

We determine the line voltages , ( see Fig. 19.3). Using KVL, we obtain abV bcV caV

.VV

VVVVV

aa

aabaab

30j33tanj

22

e3e23

23

23j

23

23j

211

1

=⎟⎟⎠

⎞⎜⎜⎝

⎛+⎟

⎠⎞

⎜⎝⎛=

=⎟⎟⎠

⎞⎜⎜⎝

⎛+=⎟⎟

⎠

⎞⎜⎜⎝

⎛++=−=

−

Thus,

30je3aab VV = . (19.7)

holds and similarly we obtain

30je3bbc VV = (19.8)

30je3cca VV = . (19.9)

The phasor diagram showing the phase and line voltages is shown in Fig. 19.4. Thus, the line voltages , , form a symmetrical set of phasors leading by 30°

the set representing the phase voltages and they are abV bcV caV

3 times greater

acabcab VVVV 3=== . (19.10)

The same conclusion is valid in the Y connected load ( see Fig. 19.5).

Notes

162

Vc

Vbc Vb

30° Va

Vab

30°

30°

Vca

Fig. 19.4. Phasor diagram

Va

c

b

a

Vbc

Vb Vc

Zc=Z

Za=Z

Zb=Z

Vab

Vca

Fig. 19.5. Y-connected load

Notes

163

19.3 Three-phase systems calculations

When the three phases of the load are not identical, an unbalanced system is produced. An unbalanced Y-connected system is shown in Fig. 19.2. The system of Fig. 19.2 contains perfectly conducting wires connecting the source to the load. Now we consider a more realistic case where the wires are represented by impedances and the neutral wire connecting and

pZn n′ is represented by impedance ( see Fig. 19.6). nZ

c

b

a

Vn

Zn

c’

a’

b’Zp

Zp

Zp

In

n’ n

Ia Za Va

Ic Zc Vc

Ib Zb Vb

Fig. 19.6. A three-phase system with not perfectly conducting wires

Using the node n as the datum, we express the currents , , and in terms of the node voltage

aI bI cI nI

nV

pa

naa ZZ

VVI

+−

= , (19.11)

pb

nbb ZZ

VVI+−

= , (19.12)

Notes

164

pc

ncc ZZ

VVI+−

= , (19.13)

n

nn Z

VI = . (19.14)

Hence, we obtain the node equation

0=+−

−+−

−+−

−pc

nc

pb

nb

pa

na

n

n

ZZVV

ZZVV

ZZVV

ZV .

Solving this equation for we have nV

pcpbpan

pc

c

pb

b

pa

a

n

ZZZZZZZ

ZZV

ZZV

ZZV

V

++

++

++

++

++

+=

1111 . (19.15)

The above relationships enable us to formulate a method for the analysis of three-phase systems. The method consists of three steps as follows: ( i ) Determine using (19.15). nV( ii ) Calculate the currents , , and applying (19.11) - (19.14). aI bI cI nI( iii ) Find the phase and line voltages using Kirchhoff’s and Ohm’s laws. When the neutral wire is removed, the system contains three connecting wires and is called a three-wire system. In such a case, we set ∞→nZ into (19.15)

pcpbpa

pc

c

pb

b

pa

a

n

ZZZZZZ

ZZV

ZZV

ZZV

V

++

++

+

++

++

+=

111 . (19.16)

Notes

165

The balanced system can be considered as a special case of the unbalanced system, where ZZZZ cba === . Using (19.16), we obtain

( )

03

1

=

+

+++

=

p

cbap

n

ZZ

VVVZZ

V . (19.17)

Consequently, the relationships (19.11)-(19.13) reduce to

p

aa ZZ

VI

+= , (19.18)

p

bb ZZ

VI

+= , (19.19)

p

cc ZZ

VI

+= . (19.20)

Since and , we have and . 120je−= ab VV 240je−= ac VV 120je−= ab II 240je−= ac IIHence, we need to calculate onlyaI using (19.18), which can be executed applying the one-phase circuit, described by equation (19.18), shown in Fig. 19.7.

n’

aV ′

n

Z

Zp Ia Va

Fig. 19.7. One-phase circuit

This means that the analysis of a balanced three-phase system can be reduced to the analysis of one-phase system depicted in Fig. 19.7.

Notes

166

Example 19.1 Let us consider three-phase system shown in Fig. 19.8. The system is supplied with a balanced three-phase generator, whereas the load is unbalanced.

c

a

bZp

Zp

Zp

Vn

caV ′bcV ′

abV ′

n’ n

Ia Za Va

Ic Zc Vc

Ib Zb Vb

Fig. 19.8. Three-phase system for Example 19.1

The effective value of the generator phase voltage is 220V, the impedance of any connecting wire is ( ) and the phase impedances of the load are

( )Ω2j2 +=pZ

Ω4j2 +=aZ , , ( )Ω2j4 −=bZ ( )Ω+= 4j2Zc . We wish to determine the line currents. Since the circuit shown in Fig. 19.8 is a three-wire system, we apply equation (19.16) to compute . The generator phase voltages are nV

( )

( ) ...VV

..VV

V

ac

ab

a

V44269j5615523j

212220e

,V44269j5615523j

212220e

,V2220

240j

120j

+−=⎟⎟⎠

⎞⎜⎜⎝

⎛+−==

−−=⎟⎟⎠

⎞⎜⎜⎝

⎛−−==

=

−

−

Notes

167

Using (19.16), we find

( ) ( )

( ) ...

....

Vn V261j597

6j41

61

6j41

6j444269j56155

644269j56155

6j42220

−=

+++

+

++−

+−−

++

=

Next, we compute the line currents using (19.11) – (19.13)

( )

( )

( ) .......ZZVV

I

......ZZVV

I

....ZZVV

I

pc

ncc

pb

nbb

pa

naa

A6354j68182j24j2

261j59744269j56155

,A7034j18422j22j4

261j59744269j56155

,A9419j49232j24j2

261j5972220

+=+++

+−+−=

+−

=

−−=++−

+−−−=

+−

=

−=++++−

=+−

=

19.4 Power in three-phase circuits

In balanced systems, the average power consumed by each load branch is the same

and given by

φcoseffeffav IVP~ = , (19.21)

where is the effective value of the phase voltage, is the effective value of the phase current and

effV effIφ is the angle of the impedance. The total average power consumed

by the load is the sum of those consumed by each branch, hence, we have

φcos33 effeffavav IVP~P == . (19.22)

In balanced Y systems, the phase current has the same effective value as the line current ( )

Leffeff II = , whereas the line voltage has the effective value ( )LeffV which is

Notes

168

3 times greater than the effective value of the phase voltage, ( ) effLeff VV 3= . Hence, using (19.22), we obtain

( ) ( ) ( ) ( ) φφ cos3cos

33

LeffLeffLeffLeff

av IVIV

P == . (19.23)

Similarly, we derive the reactive power

( ) ( ) φsin3LeffLeffx IVP = . (19.24)

In unbalanced systems, we add the powers of each phase

( ) ( ) ( ) ( ) ( ) ( ) cceffceffbbeffbeffaaeffaeffav IVIVIVP φφφ coscoscos ++= (19.25)

( ) ( ) ( ) ( ) ( ) ( ) cceffceffbbeffbeffaaeffaeffx IVIVIVP φφφ sinsinsin ++= . (19.26)

In order to measure the average power in a three-phase Y-connected load, we use three wattmeters connected as shown in Fig. 19.9. The reading of the ideal wattmeter W1 is

( ) ( ) ( ) ( ) ( ) aaaeffaeffaamamaaW PIVIVIVP ==== ∗ φφ coscos21Re

21

1 .

Similarly, W2 and W3 measure the average power of the load branch b and c, respectively. Thus, the sum of the three readings will give the total average power. This method of the average power measurement is valid for both balanced and unbalanced Y-connected loads. Note that in the case of a balanced Y-connected load all three readings are identical and therefore, we use only one wattmeter.

For measuring average power in a three-phase three-wire system, we can use a method applying two wattmeters. In this method two wattmeters are connected by choosing any one line as the common reference for the voltage coils of the wattmeters. The current coils are connected in series with the other two lines ( see Fig. 19.10) and the asterisk terminals of each wattmeter are short-circuited ( see Fig. 19.10).

Notes

169

n’

*

*

*

*

*

*

i c’

b’

a’

Vc

Vb

Zb

Ia Za

Va

Zc

Ib

W2

W1

W3

Fig. 19.9. Three-wattmeter measurement system

*

*

*

*

c Vbc

Vac b

a Ia

Ic

Ib LoadW2

W1

Fig. 19.10. Two-wattmeter measurement system

Notes

170

The indications of the wattmeters are

( )∗= aacW IVP Re21

1 , (19.27)

( )∗= bbcW IVP Re21

2 . (19.28)

The Y-connected load is shown in Fig. 19.11.

c Vbc

Vac b

a

Vc

Vb

Zb

Ia

Za

Va

Ic

Zc

Ib

Fig. 19.11. Y-connected load

Since caac VVV −= and cbbc VVV −= , we obtain

( )( ) ( )

( )( ) ( ) .IVIVIVVP

,IVIVIVVP

bcbbbcbW

acaaacaW

∗∗∗

∗∗∗

−=−=

−=−=

Re21Re

21

Re21Re

21

2

1

The sum of and gives 1WP

2WP

( )[ ]∗∗∗∗ +−+=+ bacbbaaWW IIVIVIVPP Re21

21 . (19.29)

Notes

171

The currents , , satisfy KCL aI bI cI

0=++ cba III . Hence, it holds , 0=++ ∗∗∗

cba IIIor . (19.30) 0=−=+ ∗∗∗

cba III

Substituting (19.30) into (19.29) we have

[ ] avccbbaaWW PIVIVIVPP =++=+ ∗∗∗Re21

21 . (19.31)

Equation (19.31) says that the sum of the two wattmeters readings in a Y-connected system equals the total average power consumed by the load.

Let us consider a balanced Y-connected load and calculate the instantaneous power delivered by the generator to the load

( ) ( ) ( ) ( ) ( ) ( ) ( )titvtitvtitvtp ccbbaa ++= , (19.32) where

( )( ) ( )( ) ( ) ,tVtv

,tVtv

,tVtv

mc

mb

ma

240cos

120cos

cos

−=

−=

=

ω

ω

ω

(19.33)

and

( ) ( )( ) ( )( ) ( ) ,tIti

,tIti

,tIti

mc

mb

ma

φω

φω

φω

−−=

−−=

−=

240cos

120cos

cos

(19.34)

where ( )tva , ( )tvb , are the voltages of the load branches, ( )tvc ( )tia , ( )tib , are the currents of the load branches and

( )tic

φ is the angle of the load impedance. We substitute (19.33)-( 19.34) in (19.32)

Notes

172

( ) ( ) ( ) ( )

( ) ( )]240cos240cos

120cos120coscos[cos

φωω

φωωφωω

−−−+

+−−−+−=

tt

ttttIVtp mm

and use the trigonometric identity

( ) ( )[ ]yxyxyx ++−=⋅ coscos21coscos ,

finding

( ) ( ) ( ) ( )[ ]φωφωφωφ −−+−−+−+= 4802cos2402cos2coscos321 tttIVtp mm .

Since

( ) ( ) ( ) ,ttt 04802cos2402cos2cos =−−+−−+− φωφωφω

we obtain

( ) aveffeffmm PIVIVtp === φφ cos3cos23 . (19.35)

Thus, the total instantaneous power ( )tp delivered by a three-phase generator to the balanced load is constant and equals the average power consumed by the load.

Notes

173

Reference books

[1] Chua L.O., Desoer C.A., Kuh E.S.: Linear and Nonlinear Circuits. McGraw-Hill, Inc., 1981.

[2] Cook N.P.: Electronics. A Complete Course. Pearson Education Inc., 2004.

[3] Dorf R.C.: Introduction to Electric Circuits. J. Wiley & Sons Ltd., 1989. [4] Dorf R.C.: The Electrical Engineering Handbook. CRC Press., 1997. [5] Madhu S.: Linear Circuit Analysis. Prentice-Hall, Inc., 1988. [6] Smith R.J.: Circuits, Devices, and Systems. J. Wiley & Sons, 1984.

174