Electrical Networks and RESONANCE SQA

7/26/2019 Electrical Networks and RESONANCE SQA.

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DG3G 34

Electrical Networks andResonance

March 2006 SQA


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Electrical Networks and Resonance DG3G 34

SQA Version 1 Developed by COLEG2

Acknowledgements

No extract from any source held under copyright by any individual or organisation hasbeen included in this publication.

Scottish Qualifications Authority Material developed by LauderCollege.This publication is licensed by SQA to COLEG for use by Scotlands colleges as commissionedmaterials under the terms and conditions of COLEGs Intellectual Property Rights document,September 2004.No part of this publication may be reproduced without the prior written consent of COLEG andSQA.


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Contents

Acknowledgements 2

Contents 3

Introduction to the unit 7

What this unit is about 7

Outcomes 7

Unit structure 7

How to use these learning materials 7

Symbols used in this unit 8

Self assessed question 8

Activity 8

Tutor assignment formative assessment 9

Other resources required 10

Assessment information 11

How you will be assessed 11

When and where you will be assessed 11

What you have to achieve 11

Opportunities for reassessment 11

Section 1: Network theorems 13

Introduction to this section 15

What this section is about 15

Outcomes, aims and objectives 15

Approximate study time 15


Assessment information for this section 16


Introduction 17

Kirchhoff's laws 17

Kirchoff's current law (KCL) 17


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Kirchhoff's voltage law (KVL) 18

Network analysis using Kirchhoff's laws 22

Superposition theorem 28

Constant voltage sources and constant current sources 33

Practical voltage source 33

Ideal voltage source 34

Practical current source 35

Ideal current source 36

Thevenins theorem 37

Nortons theorem 48

Thevenins theorem versus Nortons theorem 56

The maximum power transfer theorem 59

Maximum power transfer in a dc network 59

Calculation of maximum power transfer 59

Proof of maximum power transfer 60

Maximum power transfer in an ac network 62

Impedance matching 65

Summary of this section 68

Answers to SAQs 69

Answers to activities 86

Section 2: Resonant circuits 89

Introduction to this section 91

What this section is about 91

Outcomes, aims and objectives 91

Approximate study time 91


Assessment information for this section 92


When and where you will be assessed 92

What you have to achieve 92

Opportunities for reassessment 92


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Introduction 93

The series resonant circuit 94

Revision 94

Condition for resonance 95

Resonant frequency 96

Impedance, current and phase angle 97

Impedance and current graphs 98

Q factor 98

Damping 99

Bandwidth and selectivity 100

The parallel resonant circuit 103

Ideal and actual parallel circuits 103

Condition for resonance 105

Resonant frequency 106

Impedance, current and phase angle 107

Impedance and current graphs 108

Q factor 109

Bandwidth 110

Worked example 2.9 110

Solution 111

Summary of this section 117

Answers to SAQs 118

Answers to activities 122


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Introduction to the unit

What this unit is about

This unit is designed to enable candidates to develop knowledge and understandingand apply a range of network theorems to the solution of dc and ac electrical networkproblems. The unit also allows candidates to undertake a detailed study of series andparallel passive resonant circuits.

Outcomes

Apply network theorems to solve electrical network problems.

Solve problems involving resonating passive circuits.

Unit structure

This unit contains the following study sections:

Section number and title Approx.study time

1 Network theorems 24 hours

2 Series and parallel resonance 14.5 hours

How to use these learning materials

These learning materials consist of a number of self assessed questions (SAQs) to testyour understanding of the various topics covered along with practical activities. Theanswers and solutions to the SAQs and practical activities are given at the end of thispackage.

You should work through both sections, covering outcomes 1 and 2 before attemptingthe tutor assignment, which will prepare you for the assessment of the unit.


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Symbols used in this unit

These learning materials allow you to work on your own with tutor support. As you workthrough the course, you will encounter a series of symbols which indicate thatsomething follows which you are expected to do. You will notice that as you workthrough the study sections you will be asked to undertake a series of SAQs, activitiesand tutor assignments. An explanation of the symbols used to identify these is givenbelow.

Self assessed question

This symbol is used to indicate a self assessed question (SAQ). Most commonly, SAQsare used to check your understanding of the material that has already been covered inthe sections.

This type of assessment is self-contained; everything is provided within the section toenable you to check your understanding of the materials.

The process is simple:

you are set SAQs throughout the study section;

you respond to these by writing either in the space provided in the assessmentitself or in your notebook;

on completion of the SAQ, you turn to the back of the section to compare the modelSAQ answers to your own;

if you are not satisfied after checking your responses, turn to the appropriate part ofthe study section and go over the topic again.

Remember, the answers to SAQs are contained within the study materials. You are notexpected to guess at these answers.

Activity

This symbol indicates an activity, which is normally a task you will be asked to do thatshould improve or consolidate your understanding of the subject in general or aparticular feature of it.

The suggested responses to activities are given at the end of each section.

Remember that the SAQs and activities contained within your package are intended toallow you to check your understanding and monitor your own progress throughout thecourse. It goes without saying that the answers to these should only be checked afterthe SAQ or activity has been completed. If you refer to these answers beforecompleting the activities, you cannot expect to get maximum benefit from your course.

1

1


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Tutor assignment formative assessment

This symbol means that a tutor assignment follows. These are found at the end of eachstudy section. The aim of the tutor assignment is to cover and/or incorporate the maintopics of the section and prepare you for unit (summative) outcome assessment.

1


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Other resources required

To complete the practical activities, you will need access to a PC with a suitablesimulation program such as Electronics Workbench, Crocodile Clips, etc.

You will also need a scientific calculator and are recommended to get one that canperform coordinate transformations (polar/rectangular) such as a Casio fx-83WA orequivalent.

You may have the resources above if you have already studied unit DG54 34 SinglePhase AC Circuits.


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Assessment information

How you will be assessed

You will be formatively assessed throughout the material using the SAQs and practicalactivities. You will be summatively assessed at the end of the material by a singleclosed book written test lasting 1 hours.

When and where you will be assessed

You will sit the summative assessment once you have satisfactorily completed all theSAQs and practical activities. The location will be determined by your tutor.

What you have to achieve

To pass this unit you must achieve a minimum score of 60% in the summativeassessment. A score of 40 to 59% will require a partial reassessment and a score ofless than 40% will require a complete reassessment.

Opportunities for reassessment

Normally, you will be given one attempt to pass an assessment with one reassessmentopportunity.

Your centre will also have a policy covering 'exceptional' circumstances, for example if

you have been ill for an extended period of time. Each case will be considered on anindividual basis and is at your centre's discretion (usually via written application), andthey will decide whether or not to allow a third attempt. Please contact your tutor fordetails regarding how to apply.


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Section 1: Network theorems


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Introduction to this section

What this section is about

This section covers the use of network theorems to solve electrical network problems.

Outcomes, aims and objectives

After completing this section, you should be able to:

use Kirchhoffs laws to solve problems in dc networks;

use the superposition theorem to solve problems in dc networks;

apply the concept of constant voltage and constant current sources;

use Thevenins theorem to solve problems in both dc and ac networks;

use Nortons theorem to solve problems in both dc and ac networks;

use the maximum power transfer theorem to solve problems in dc and ac networks.

Approximate study time

24 hours.


PC with suitable simulation software. Scientific calculator.


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Assessment information for this section


There is no summative assessment for this first section.


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Introduction

The use of Ohms law in its basic form is limited to simple networks consisting of onesource and a few components. As the networks become more complex, other analysistechniques are required.

There are many methods of analysis available and in this section we shall be focusingon the following:

Kirchhoffs laws (dc circuits only)

the superposition theorem (dc circuits only)

Thevenins theorem (dc and ac circuits)

Nortons theorem (dc and ac circuits)

the maximum power transfer theorem (dc and ac circuits).

The ac circuit analysis will involve the use of complex numbers, which is also coveredin the unit DG54 34 Single Phase AC Circuits.

Kirchhoff's laws

Kirchoff's current law (KCL)

Kirchoff's current law states:

To put it another way, the total current flowing towards a junction equals the totalcurrent flowing away from the junction.

The algebraic sum of the currents at a junction in anetwork is equal to zero.


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I1= I2+ I3+ I4

I1 I2 I3 I4= 0

I= 0

Kirchhoff's voltage law (KVL)

Kirchhoff's voltage law states:

To put it another way, the sum of the energy sources is equal to the sum of thecomponent volt drops around the closed loop.

I1

I4

I3

I2

The algebraic sum of the emfs acting around aclosed loop is equal to the algebraic sum of thepotential differences around the closed loop.


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E1+ E2+ E3= IR1+ IR2+ IR3+ IR4

E= IR

Worked example 1.1

Using KCL, determine the value of Iin the diagram below.

Solution

From KCL, the total current entering the junction is equal to the total current leaving thejunction:

I+ 2 + 5 = 4 + 8

I+ 7 = 12

I= 5 A

E1 R1 E2 R2 E3 R3

R4

I

8A

5A

I

2A

4A


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Worked example 1.2

Using KVL, write down an expression for the circuit below.

Solution

From KVL, the sum of the emfs is equal to the sum of the voltage drops:

E1+ E2= 4R1+ 4R2+ 4R3

E1+ E2= 4(R1+ R2+ R3)

a) Use KCL to find the value of the current (I) in each diagram below.

(i) (ii)

4 A

R3

E1 R1 E2 R2

5A

2A

4A

I

3A

I

6A

2A

1.1


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(iii) (iv)

a) Use KVL to write down an expression for each of the following circuits.

(i)

(ii)

4A

6A

I

3A

I1

I4

I3

I

I2

I

R3

E1 R1 E2 R2

R3 R4

2A

E1 R1 E2 R2

1.2


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(iii)

(iv)

Network analysis using Kirchhoff's laws

The analysis of electrical networks makes use of both the current law and the voltagelaw. The following list is a general set of steps that should be applied to the solution ofan electrical network by Kirchhoff's laws. A worked example follows to show how theyare applied.

1. Label all network loops on the diagram.

2. Assume all batteries to discharge current and enter these onto the diagram.

3. Apply KCL.

4. Select two loops and obtain an equation from each by applying KVL.

5. Use simultaneous equations to solve for the unknown quantities.

3

I

4 6

E1 2 E2 4

2 A

E1 2 10 V R2 7 V 3


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Worked example 1.3

With reference to the circuit below, determine the magnitude and direction of the

current flowing through the 10resistor.

Solution

Steps 1, 2 and 3

4 V 4

2 V 5

10

I1 + I2

I2

D

E

F

C

B

A

4 V 4

2 V 5

10

I1


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Step 4

Consider loop ACDF

4 = 4I1+ 10(I1+ I2)

4 = 4I1+ 10I1+ 10I2

4 = 14I1+ 10I2 (eqn 1)

Consider loop BCDE

2 = 5I2+ 10(I1+ I2)

2 = 5I2+ 10I1+ 10I2

2 = 10I1+ 15I2 (eqn 2)

Consider loop ABEF

4 2 = 4I1 5I2

2 = 4I1 5I2 (eqn 3)

Although we have developed three equations here, only two will be required to solvefor the unknown currents I1and I2.

Step 5

Each equation contains two unknowns and on their own cannot be used to solve forboth I1and I2. Any two equations must be chosen and solved using the simultaneousequations method. We shall choose equations 1 and 2 for this example.

You may already be familiar with solving simultaneous equations but if not here is aquick guide in doing so. The technique involves multiplying one or both chosenequations by some number(s) which will make the coefficients of either I1or I2equal.The two equations can then be added or subtracted as appropriate to eliminate thatterm, leaving one new equation with a single term that can be solved.

Once the current (either I1or I2) has been solved from above, it is substituted back intoany of the other equations to solve for the other current. Lets put this into practice.

If we look below at the coefficients of I2, they are 10 from equation 1 and 15 fromequation 2.

4 = 14I1+ 10I2 (eqn 1)

2 = 10I1+ 15I2 (eqn 2)We shall multiply equation 1 throughout by 1.5 (remember all parts of the equationmust be multiplied). This will now make the coefficient of I2equal to 15. Since both I2coefficients are positive, to eliminate them the two equations must be subtracted fromeach other:

6 = 21I1+ 15I2 (eqn 1) 1.5

2 = 10I1+ 15I2 (eqn 2)

4 = 11I1 (subtract)

A364.0

11

41 ==I


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This value for I1can now be substituted into any of the previous equations to solvefor I2. We shall substitute I1into equation 2 which gives:

2 = 10I1+ 15I2

2 = 10(0.364) + 15I2

2 = 3.64 + 15I2

15I2= 1.64

A109.015

64.12 =

=I

Note the value of I2in this case is negative. This simply means that the assumeddirection at the start of question was incorrect. However, we carry on and use thisnegative value.

The current flow through the 10 resistor (I) is equal to I1+ I2therefore:

I= I1+ I2

I= 0.364 + (0.109)

I= 0.255 A flowing from C to D.

a) Use Kirchhoffs laws to calculate the current flowing in each branch of the circuitbelow.

12 V 2

10 V 1

2

1.3


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b) Determine the magnitude and direction of the current through the load resistorin the circuit below by Kirchoffs laws. (Hint: take care with current directions.)

c) With reference to the circuit below, use KCL and KVL to determine the voltagedrop across the load resistor.

8 V 2

12 V 1

RL= 10

9 2

4 V 6 V

1 15 RL 8


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Verification of Kirchhoffs laws

Equipment required

PC

Electronics Workbench or similar

Circuit 1: Kirchhoffs voltage law

Using the software package available to you, connect up the circuit shown below. The

symbols you use may be slightly different to the ones shown here.

Procedure

1. Connect voltmeters across the supply and across each resistor.

2. Activate the simulation and verify KVL.

Circuit 2: Kirchhoffs current law

Using the software package available to you, connect up the circuit shown below. Thesymbols you use may be slightly different to the ones shown here.

1.1


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Procedure

1. Connect ammeters into the circuit to measure the supply current and each of thebranch currents.

2. Activate the simulation and verify KCL.

Superposition theorem

The superposition theorem states:

The following list is a general set of steps that should be applied to the solution of anelectrical network using the superposition theorem. The types of circuit covered will bethe same as for Kirchhoffs laws.

In any electrical network containing two or more sources or emf (orcurrent), the current through or the potential difference across anybranch can be found by considering each source separately (with allother sources replaced by their respective internal resistances) and

adding together or superimposing their individual effects.


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1. Remove battery B, replace with its internal resistance (zero ohms unless specified)and calculate the circuit voltages and/or currents due to battery A.

2. Remove battery A, replace with its internal resistance (zero ohms unless specified)and calculate the circuit voltages/currents due to battery B.

3. Determine circuit voltages/currents by adding or superimposing the results fromthe two steps above.

Worked example 1.4

Use the superposition theorem to calculate the currents flowing through each branch ofthe network below.

A RA

B RB

RL

4 V 2

2 V 1

4


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Solution

Step 1

Remove 2 V battery.

=+=+

+= 8.28.0214

142cetanresisTotal

A429.18.2

4

currentBattery 1 ==I

)ruledivisioncurrentby(A143.114

4429.1currentBranch 2 =

+=I



+=I

Step 2

Remove 4 V battery.

I3

I2

I14 V 2

1

4

I6

I4

I5

2 V 1

2

4


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=+=+

+= 333.2333.1124

241cetanresisTotal

A857.0333.2

2currentBattery 4 ==I



+=I



+=I

Step 3

Combine or superimpose the results from steps 1 and 2.

4 V battery current = I1 I5= 1.429 0.571 = 0.858 A

The 4 V battery is discharging a current of 0.858 A into the circuit.

I3

I2

I1 4 V 2

1

4

I6

I4

I5

2 V 1

2

4


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c) With reference to the circuit below, use the superposition theorem to determinethe voltage drop across the load resistor.

Constant voltage sources and constant current sources

The next two theorems (Thevenins and Nortons) deal with circuit analysis techniquesinvolving devices known as constant voltage sources and constant current sources. Anexplanation of their function is given below.

Practical voltage source

The function of a voltage source is to deliver a known voltage to a circuit. It is shown asa voltage generator (V) connected in series with a resistor (Rint). The voltage sourcerepresents the magnitude of the voltage supplied and the resistor represents theinternal resistance of the source. For a voltage source, the value of Rintshould be aslow as possible.

Consider the diagrams below, which show two actual voltage sources, one operatingunder no load conditions and one operating under load conditions.

No load conditions Load conditions

9 2

4 V 6 V

1 15 RL 8

V

Rint

RL Vo

V

Rint

Vo


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Under no load conditions, the output voltage (Vo) will always be equal to the voltagesupplied by the generator, thus:

Vo= V

Under load conditions, a current will flow around the loop producing a potential dropacross Rint. The output voltage will drop and will be given by:

Vo= V I Rint

Under load conditions, the value of Vowill depend on the value of the load RLand willalways be less than V.

Ideal voltage source

The internal resistance of a voltage source should be as low as possible. Under ideal

conditions, the value can be considered to be zero. Let us look at how this affects thevalue of the output voltage under no load and load conditions.


Vo= V Vo= V

In this case, it can be seen that the value of the output voltage will always be the sameirrespective of RL. The source therefore supplies a constant voltage for any loadcondition. Such a source is called a constant voltage source (CVS).

V RL VoV Vo


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Example

Vo= 10 V Vo= 10 V

A110

10==I A2

5

10==I

As the load changes, the current supplied by the CVS changes but the output voltageremains constant. A CVS is therefore defined as a device that supplies a constantvoltage to a circuit irrespective of the load connected across it.

Practical current source

The function of a current source is to deliver a known current to a circuit. It is shown as

a current generator (I) connected in parallel with a resistor (Rint). The current sourcerepresents the magnitude of the current supplied and the resistor represents theinternal resistance of the source. For a current source, the value of Rintshould be ashigh as possible.

Consider the diagrams below, which show two actual current sources, one operatingunder no load conditions (load terminals on short circuit) and one operating under loadconditions.


10 V 5 Vo10 V 10 Vo

I Rint RL IoI Rint Io


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Under no load conditions, the output current (Io) will always be equal to the currentsupplied by the generator, thus:

Io= I

Under load conditions, the current from the source will split between the internalresistance and the load resistance. The output current will drop and will be given by:

Io= I IRint

Under load conditions, the value of Iowill depend on the value of the load RLand willalways be less than I.

Ideal current source

The internal resistance of a current source should be as high as possible. Under idealconditions, the value can be considered to be infinite. Let us look at how this affects thevalue of the output current under no load (load terminals on short circuit) and loadconditions.


Io= I Io= I

In this case, it can be seen that the value of the output current will always be the sameirrespective of RL. The source therefore supplies a constant current for any loadcondition. Such a source is called a constant current source(CCS).

I RL IoI Io


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Example

Io= 5 A Io= 5 A

Vo= 5 10 = 50 V Vo= 5 5 = 25 V

As the load changes, the voltage produced across the load changes but the outputcurrent remains constant. A CCS is therefore defined as a device that supplies aconstant current to a circuit irrespective of the load connected across it.

Thevenins theorem

Thevenins theorem states:

The constant voltage source is known as the Thevenin voltage (VTH)and is equal tothe open-circuit terminal voltage of the network measured between points A and B.

The resistance value is known as the Thevenin resistance (RTH)and is equal to theresistance of the network measured between points A and B with all sources replacedby their internal resistances (normally short circuit for voltage sources and open circuitfor current sources).

5 A 5 Io5 A 10 Io

Any two terminal networks can be replaced by an

equivalent circuit comprising a constant voltage source inseries with a single resistor

A

B

Networkcontaining any

number ofsources and

resistances

VTHRTH

A

B

VTH

RTH


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Once the values of VTHand RTHhave been determined, the Thevenin equivalent circuitis drawn and this can be used to determine a range of circuit quantities. The following

is a set of steps to be followed when solving network problems using Theveninstheorem:

1. Remove the resistor in question and calculate the Thevenin voltage (VTH).

2. With the resistor still removed, replace all sources with their internal resistancesand calculate the Thevenin resistance (RTH).

3. Draw the Thevenin equivalent circuit, place the original resistor across terminals ABand calculate the quantity required.

Note: When dealing with ac circuits the word resistor will be replaced with the wordimpedance. The procedures for working with Thevenins theorem remain unchanged.

Worked example 1.5

Use Thevenins theorem to calculate the voltage across the 6 resistor in the networkbelow.

Solution

Step 1

Remove the 6 resistor and calculate VTH

10 4

8 V 15 6

A

VTH

B

10 4

8 V 15


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Worked example 1.6

If the 6 resistor in the previous example is replaced by a 12 resistor, calculate thenew voltage drop across it.

Solution

This example will now show the power of Thevenins theorem. The values of VTHandRTHwill remain the same so these calculations do not need to be redone. We just needto perform step 3 again.

Step 1

VTH= 4.8 V

Step 2

RTH= 10

Step 3

Draw a Thevenin equivalent circuit and place a 12 resistor across points A and B.

V62.21210

128.46 =+=V

A

12

B

4.8V

10


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Worked example 1.7

Use Thevenins theorem to calculate the current through the 10 resistor in thenetwork below.

Solution

Step 1

Remove the 10 resistor and calculate VTH.

V15155

1520

15 =+

=V V882

810

8 =+

=V

VTH= V15 V8= 15 8 = 7 V

5 10 2

20 V 15 8 10 V

A B

5 VTH 2

20 V 15 8 10 V


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Step 2

Replace the 20 V and 10 V batteries with short circuits and calculate RTH.

If you are having difficulties in visualising how the resistors are connected together, itmay help if you redraw the network. The above may be redrawn as shown below.

35.56.175.382

82

155

155TH =+=

+

+

+

=R

Step 3


A B

5 RTH 2

15 8

BA

5 2

15 8


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A456.01035.5

7

=+=I

We shall now look at one further example based on an ac network. The rules areexactly the same as for the dc network. Complex numbers will be used in the solutionto this and all other ac networks.

Worked example 1.8

Use Thevenins theorem to calculate the current through the 100 resistor in the acnetwork below.

Solution

Step 1

Remove the 100 resistor and calculate VTH.

A

10

B

7 V

5.35

j200 100

600 j150

V012


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4.185.632

90200012

200600

200

012TH

=

= jj

V

V)6.32.1(V6.7179.3TH jVo ==

Step 2

Replace the 12 V source with a short circuit and calculate ZTH.

4.185.632

90120000150

200600

120000150

)200(600

)200(600150TH

+=

+=+

+= jj

jj

j

jjZ

=+=+= )3060()18060(1506.717.189150TH jjjjZ

A

VTH

B

j200

600 j150

V012

A

ZTH

B

j200

600 j150


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Step 3


a) Use Thevenins theorem to calculate the load current and load voltage in eachof the dc networks below.

(i)

A61023.06.108.162

6.7179.3

30j160

6.7179.3

100)30j60(

6.7179.3I o=

=

=+

=

A

100

B

(60 j30)

V6.7179.3 o

20 8

16 V 30 RL= 12

1.5


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(ii)

b) With reference to the ac network below, use Thevenins theorem to calculatethe apparent power in the load ZL.

12 V 2

10 V 1

RL= 2

ZL=j150

50 75

100 j200

100 j80

V0100


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c) Use Thevenins theorem to calculate the voltage across the load RLin the acnetwork below.

Verification of Thevenins theorem

Equipment required

PC

Electronics Workbench or similar.

Circuit


j50

10 RL= 20

j25 j35

V050

1.2


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Procedure

1. Connect an ammeter and a voltmeter to measure the current through and the

voltage across the 100 load resistor.

2. Activate the simulation and display the results.

3. Apply Thevenins theorem to the above circuit and determine values of VTHandZTH.

4. Construct the Thevenin equivalent circuit and measure the current through and the

voltage across the 100 load resistor. Compare with the answers for question 2.

Note: VTHmust be entered in polar form (magnitude and angle) and ZTHmust beentered in rectangular form (real part = resistance, imaginary part = reactance). Thereactance value must be converted back to either inductance or capacitance.

Nortons theorem

Nortons theorem states:

The constant current source is known as the Norton current (IN)and is equal to theshort-circuit terminal current of the network measured through points A and B.

The resistance value is known as the Norton resistance (RN)and is equal to theresistance of the network measured between points A and B with all sources replacedby their internal resistances (normally short circuit for voltage sources and open circuitfor current sources). This is the same as the Thevenin resistance.

Once the values of INand RNhave been determined the Norton equivalent circuit isdrawn and this can be used to determine a range of circuit quantities. The following is aset of steps to be followed when solving network problems using Nortons theorem:

5.

Any two terminal networks can be replaced by anequivalent circuit comprising a constant current source inparallel with a single resistor

A

B

Networkcontaining any

number ofsources andresistances

INRN

A

B

IN RN


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1. Remove the resistor in question, replace with a short circuit and calculate theNorton current (IN).

2. Remove the short circuit, replace all sources with their internal resistances andcalculate the Norton resistance (RN).

3. Draw the Norton equivalent circuit, place the original resistor across terminals ABand calculate the quantity required.

Note: When dealing with ac circuits the word resistor will be replaced with the wordimpedance. The procedures for working with Nortons theorem remain unchanged.

Worked example 1.9

Use Nortons theorem to calculate the voltage across the 6 resistor in the networkbelow.

Solution

Step 1

Remove the 6 resistor, replace with a short circuit and calculate IN.

16.13415

41510resistanceBattery =

+

+=R

10 4

8 V 15 6

A

B

10 4

8 V 15 IN


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A6.016.13

8currentBattery ==I

By the current division rule:

A48.0154

156.0N =

+=I

Step 2

Remove the short circuit across AB, replace the 8 V battery with a short circuit andcalculate RN.

10

1510

15104N =

+

+=R

Step 3

Draw a Norton equivalent circuit and place a 6 resistor across points A and B.

By the current division rule A3.0610

1048.0

6 =+

=I

V8.163.06 ==V

A

RN

B

10 4

15

A

B

0.48 A 10 6


51/124



Worked example 1.10

Use Nortons theorem to calculate the current through the 100 resistor in the acnetwork below.

Solution

Step 1

Remove the 100 resistor, replace with a short circuit and calculate IN.

Source impedance9050

030000600

50

30000600

)200()150(

)200()150(600

+=

+=+

+=jjj

jjZ

+=+= )600600(90600600 jZ

Source current A45014.0455.848

012

600600

012 o

jI =

=+

=

By current division rule:

A45056.09050

9020045014.0

)200()150(

20045014.0N

o

jj

jI =

=+

=

j200 100

600 j150

V012

A

B

j200

600 j150

V012 IN


52/124



Step 2

Remove the short circuit across AB, replace the 12 V source with a short circuit and

calculate ZN.

4.185.632

90120000150

200600

120000150

)200(600

)200(600150N

+=

+=+

+= jj

jj

j

jjZ

==+=+= 5.2667)3060()18060(1506.717.189150N jjjjZ

Step 3


By current division rule:

A61023.06.108.162

5.266745056.0

100)3060(

)3060(45056.0100 =

=

+

=j

jI

(60 j30)

A

ZN

B

j200

600 j150

A

B

A45056.0 o 100


53/124



a) Use Nortons theorem to calculate the load current and load voltage in each ofthe dc networks below.

(i)

(ii)

20 8

16 V 30 RL= 12

12 V 2

10 V 1

RL= 2

1.6


54/124



b) With reference to the ac network below, use Nortons theorem to calculate theapparent power in the load ZL.

c) Use Nortons theorem to calculate the voltage across the load RLin the acnetwork below.

ZL=j150

50 75

100 j200

100 j80

V0100

j50

10 RL= 20

j25 j35

V050


55/124



Verification of Nortons theorem

Equipment required

PC


Circuit


Procedure

1. Connect an ammeter and a voltmeter to measure the current through and the

voltage across the 100 load resistor.

2. Activate the simulation and display the results.

3. Apply Nortons theorem to the above circuit and determine values of INand ZN.

4. Construct the Norton equivalent circuit and measure the current through and the

voltage across the 100 load resistor. Compare with the answers for question 2.

Note: INmust be entered in polar form (magnitude and angle) and ZNmust be enteredin rectangular form (real part = resistance, imaginary part = reactance). The reactancevalue must be converted back to either inductance or capacitance.

1.3


56/124



Thevenins theorem versus Nortons theorem

You may have noticed from the examples in SAQs 1.5 and 1.6 that the same networkcan be solved by either Thevenins theorem or Nortons theorem. Some networks,depending on the types of sources used and their complexity, will favour one theoremrather than the other.

There is a simple relationship between the Thevenin equivalent circuit and the Nortonequivalent circuit and it is very easy to convert from one to the other.

A

B

VTH

RTH

TH

THN

R

VI =

A

B

RN= RTH

IN

A

B

RN

A

B

VTH= IN RN

RTH= RN


57/124



Worked example 1.11

Convert the Norton circuit below to its Thevenin equivalent and hence calculate the

voltage across a 5 resistor when it is connected across the output terminals AB.

Solution

The Thevenin voltage is VTH= IN RN= 5.5 3 = 16.5 V.

The Thevenin resistance is RTH= RN= 3 .

V3.1053

5

5.165=

+=

V

5.5 A

A

B

3

5

A

B

16.5 V

3


58/124



a) Convert each of the following Norton circuits to their Thevenin equivalents.

(i) (ii)

b) Convert each of the following Thevenin circuits to their Norton equivalents.

(i) (ii)

A

B

5 A 10 (3 +j3) A (25 j40)

A

B

100 V

20

A

B

(40 j30) V

(10 +j5)

1.7


59/124



The maximum power transfer theorem

Maximum power transfer in a dc network

The maximum power transfer theorem for a dc network states:

The simple network above shows a dc voltage source of value Vand internalresistance RSconnected across a variable load resistance RL. Maximum power will betransferred from the source to the load when:

RL= RS

This can be shown in a number of different ways: by calculation, by measurement andby mathematical proof.

Calculation of maximum power transfer

Consider a 10 V dc source of internal resistance 5 connected across a variable loadresistance.

The power transferred from a dc source to its loadwill be a maximum when the value of the loadresistance is equal to the value of the sourceinternal resistance.

RL

V

Rs


60/124



The power transferred to the load resistance can be calculated from:

L

2

LSL

2L RRR

VRIP

+==

When RL= 2 then: W08.4225

102

L

2

LSL =

+

=

+= R

RR

VP

When RL= 5 then: W5555

102

L

2

LSL =

+

=

+= R

RR

VP

Then RL= 7 then: W86.4775

102

L

2

LSL =

+

=

+= R

RR

VP

If all values of RLfrom 1 to 10 in 1 increments are calculated, the following table ofresults is obtained.

RL() 1 2 3 4 5 6 7 8 9 10

PL(W) 2.78 4.08 4.69 4.94 5 4.96 4.86 4.73 4.59 4.44

Clearly, it can be seen from the above table that the maximum power transferred to theload is when the load resistance is equal to the source resistance.

Proof of maximum power transfer

The following mathematical proof uses basic differential calculus to show thatmaximum power is transferred when the load resistance equals the source resistance.

You are not required to reproduce anything like this in any assessments but itshould not be outwith your scope to be able to follow the mathematical steps.

It was stated on the previous page that the power transferred to the load is calculated

from the expression:

RL

10V

RS= 5


61/124



L

2

LSL R

RR

VP

+=

This can be rearranged to give:

2LS

L2

L)( RR

RVP

+

=

To determine the value of RLfor maximum power transfer, PLis differentiated withrespect to RLand then the result is equated to zero (this being the standard method forfinding maximum and minimum values).

Since the term RLappears on both the top and bottom of the equation, the quotient rulemust be used which gives:

( )0

)(2)(

d

d4

LS

LSL222

LS

L

L=

+

++

=

RR

RRRVVRR

R

P

)(2)( LSL222

LS RRRVVRR +=+

)(2)( LSL2

LS RRRRR +=+

LLS 2RRR =+

SL RR =

Once again, it has been shown that maximum power is transferred when the load

resistance is equal to the source resistance.Worked example 1.12

With reference to the network below, use Thevenins theorem to reduce it to a singlesource and single resistance, and hence determine the maximum power delivered tothe load RL.

12 RL

15

3


62/124



Solution

With RLremoved, we have:

V12123

1215TH =

+=V

With 15 V supply removed and replaced with a short circuit we have:

4.2123

123TH =

+

=R

For maxiumum power transfer, the load resistance must be equal to the equivalentsource resistance, i.e. RL= RTH.

This gives the following circuit.

A5.24.24.2

12=

+=I

PL= 2.52 2.4 = 15 W

Maximum power transfer in an ac network

The condition for maximum power transfer in an ac network is slightly more involvedcompared to a dc network as there are a number of different circuit conditions that will

cause this to happen. We shall look at two cases and in each of these we will considerthe transfer of active or real power measured in watts.

Consider an ac source Vwith internal impedance ZS= RS+jXSconnected to a load ofimpedance ZL= RL+jXLas shown below. (Note: In this instance, the termXLrefers tothe reactance of the load and not inductive reactance.)

RL= 2.4

12

2.4


63/124



Condition 1: Load impedance consists of variable RLand zeroXL.

The power transferred to the load is given by:

L

2

SLSL

2

LSSL

2L

)(R

jXRR

VR

RjXR

VRIP

++=

++==

A similar mathematical proof as in the previous section can be performed on the aboveequation, which will show that maximum power is transferred from source to load whenthe load resistance is equal to the magnitude of the source impedance:

2S

2SL XRR +=

Condition 2: Load impedance consists of variable RLand variableXL.

The power transferred to the load is given by:

L

2

LSLSL

2

LLSSL

2L

)()()()(R

XXjRR

VR

jXRjXR

VRIP

+++=

+++==

A similar mathematical proof as in the previous section can be performed on the aboveequation which will show that maximum power is transferred from source to load whenthe load impedance is equal to the complex conjugate of the source impedance:

*SL ZZ =

SSLL jXRjXR =+

SLSL XXandRR ==

Worked example 1.13

With reference to the circuit below, determine the value of load resistance for maximumpower transfer. Also determine the amount of power transferred to the load.

ZL= RL+jXL

V

ZS= RS+jXS


64/124



Solution

Maximum power occurs when 252015 22L =+=R

A6.2668.26.2672.44

0120

25)2015(

0120

T

=

=++

==

jZ

VI

PL= I2RL= 2.68

2 25 = 179.6 W.

Worked example 1.14

With reference to the circuit below, determine the value of load resistance and loadinductance for maximum power transfer. Also determine the amount of powertransferred to the load resistance.

RL

V0120V o=

ZS= (15 +j20)

ZL

Hz50

V0100V o=

ZS= (20 j30)


65/124



Solution

Maximum power occurs when ZL= ZS*

ZS= (20 j30) ZL= (20 +j30)

RL= 20 andXL= 30

H0955.0502

30

2

L=

==

f

XL

A05.2040

0100

)3020()3020(

0100

T

=

=++

==

jjZ

VI

PL= I2RL= 2.5

2 20 = 125 W

Impedance matching

The purpose of this section is to show that the load resistance or load impedance of adc or an ac network can be adjusted to ensure that maximum power is transferred fromthe source to the load.

This process is called impedance matchingand is an important consideration inelectronics and communications devices that normally involve very small amounts ofpower, i.e. coupling an aerial to a receiver or a loudspeaker to an amplifier.

a) A dc source has a terminal voltage of 50 V and an internal resistance of 6 .Determine the value of load resistance that will deliver maximum power anddetermine that value of power.

b) With reference to the network below, use Thevenins theorem to reduce it to a

single source and single resistance, and hence determine the maximum powerdelivered to the load RL.

25 RL

10V

4

1.8


66/124



c) For the circuit shown below, determine the value of the source resistance RSif

maximum power is to be delivered to the15 load resistor. Also determine this

maximum power.

d) With reference to the circuit below, determine the value of load resistance andload reactance for maximum power transfer. Also determine the amount ofpower transferred to the load resistance.

1.91 mH RL= 15

kHz1

V0100=V

RS

ZL

025=V

ZS= (12 +j25)


67/124



e) Determine, for the ac network below, the values of RLandXLthat result inmaximum power being transferred across terminals AB and determine that

value of power.

RL

XL

A

B

j6

8

4050=V

2


68/124



Summary of this section

This first section has been designed to allow you to develop knowledge, understandingand skills in the application of network theorems to the solution of electrical andelectronic circuits. Such theorems underpin much of the work in electrical andelectronic engineering and so a good grasp of these theorems is essential.


69/124



Answers to SAQs

SAQ 1.1

a)

(i) I= 4 + 2 + 5 = 11 A

(ii) I+ 2 = 6 + 3

I+ 2 = 9

I= 9 2 = 7 A

(iii) I+ 4 + 3 = 6

I+ 7 = 6

I= 6 7 = 1 A

(iv) I+ I2= I1+ I3+ I4

I= I1+ I3+ I4 I2

SAQ 1.2

a)

(i) E1+ E2= IR1+ IR2+ IR3

E1+ E2= I(R1+ R2+ R3)

(ii) E2 E1= IR1+ IR2+ IR3+ IR4

E2 E1= 2(R1+ R2+ R3+ R4)

(iii) E2 E1= IR1+ IR2+ IR3+ IR4

E2 E1= I(2 + 4 + 6 + 4)

E2 E1= 16I

(iv) E1 E2+ E3= IR1+ IR2+ IR3+ IR4

E1 10 + 7 = 22 + 2R2+ 23 + 23

E1 3 = 16 + 2R2


70/124



SAQ 1.3

a)

12 = 2I1+ 2(I1+ I2)

12 = 4I1+ 2I2 (eqn 1)

10 = I2+ 2(I1+ I2)

10 = 2I1+ 3I2 (eqn 2)

12 = 4I1+ 2I2 (eqn 1)

20 = 4I1+ 6I2 (eqn 2 2)

8 = 4I2 (subtract)

I2= 2 A

12 = 4I1+ 2I2 (eqn 1)

12 = 4I1+ (22)

12 = 4I1+ 4

8 = 4I1

I1= 2 A

I1+ I2= 2 + 2 = 4 A

I1+ I2

I2

I112 V 2

10 V 1

2


71/124



b)

8 = 2I1+ 10(I1 I2)

8 = 12I1 10I2 (eqn 1)

12 = I2+ 10(I2 I1)

12 = 10I1+ 11I2 (eqn 2)

8 = 12I1 10I2 (eqn 1)

14.4 = 12I1+ 13.2I2 (eqn 2 1.2)

22.4 = 3.2I2 (add)

I2= 7 A

8 = 12I1 10I2 (eqn 1)

8 = 12I1 (107)

8 = 12I1 70

78 = 12I1

I1= 6.5 A

Load current I2 I1= 7 6.5 = 0.5 A left to right.

I1 I2I2 I1

I2

I18 V 2

12 V 1

RL= 10


72/124



c)

4 = 10I1+ 15(I1 I2)

4 = 25I1 15I2 (eqn 1)

6 = 10I2+ 15(I2 I1)

6 = 15I1+ 25I2 (eqn 2)

60 = 375I1 225I2 (eqn 1 15)

150 = 375I1+ 625I2 (eqn 2 25)

210 = 400I2 (add)

I2= 0.525 A

4 = 25I1 15I2 (eqn 1)

4 = 25I1 (150.525)

4 = 25I1 7.875

11.875 = 25I1

I1= 0.475 A

Load current I2 I1= 0.525 0.475 = 0.05 A

Load voltage = 0.05 15 = 0.75 V

I2I1

9 2

4 V 6 V

1 15 RL 8


73/124



SAQ 1.4

a)

=+

+==+

+= 222

221667.2

21

212 RR

A52

10A5.4

667.2

1241 ==== II

A5.222

25A3

21

25.4 52 =+

==+

= II

A5.222

25A5.121

15.4 63 =+==

+= II

I12V= I1 I5= 4.5 2.5 = 2 A

I10V= I4 I2= 5 3 = 2 A

I2= I3+ I6= 1.5 + 2.5 = 4 A

b)

I3

I2

I112 V 2

1

2 I6

I4

I5

10 V 1

2

2

I4

I3 12 V 1

2

RL= 10 I2

I18 V 2

1

RL= 10


74/124



=+

+==+

+= 67.2210

210191.2

110

1102 RR

A5.467.2

12A75.2

91.2

831 ==== II

A75.0102

25.4A25.0

101

175.2 42 =+

==+

= II

Iload = I4 I2= 0.75 0.25 = 0.5 A left to right.

c)

=+

+==+

+= 161015

10151016

1015

101510 RR

A375.016

6A25.0

16

431 ==== II

A15.01510

10375.0A1.0

1510

1025.0 42 =+

==+

= II

Iload= I4 I2 = 0.15 0.1 = 0.05 A

Vload= 0.05 15 = 0.75 V

9 2

I2

I1

4 V

1 15 RL 8

I3I4

9 2

6V

1 15 RL 8


75/124


76/124



(ii)

A667.021

1012

currentBattery =+

=

V667.10)1667.0(10V667.10)2667.0(12 THTH =+=== VorV

=+

= 667.0

21

21THR

A42667.0

67.10L =+=I

VL= 4 2 = 8 V

12 V

10 V 1

VTH

2

1

RTH

RL= 2

10.67 V

0.667


77/124



b)

A3.20289.03.205.346

0100

200758010050100

0100currentSource =

=++++

=

jj

V40696.198.2383.20289.0)758010050(3.20289.0TH ==+++= jV

50 75

100 j200

100 j80

V0100

VTH

50 75

100 j200

100 j80

ZTH


78/124



==

=+++++++

=

)4.613.141(5.231.154

3.205.346

4.636.2236.198.238

)200100()758010050(

)200100()758010050(

TH

TH

jZ

jj

jjZ

A8414.0328.166

4069

)150()4.613.141(

4069L =

=

+

=jj

I

V981.62901508414.0L ==V

var7.25VA907.258414.0981.62L*

LL orIVSo===

c)

A2.6886.12.689.26

050

502510

050currentSource =

=+

=

jj

ZL=j150

V4069

(141.3 j61.4)

10 VTH

-j50

j25 j35

V050


79/124



V5.13650)5.343.36(

)5.343.86(50)8.2193(50)90502.6886.1(50

TH

TH

=+=

===

VjV

jV

+=++=

+=

+=+++

+=

)2.715.34()2.365.34(35

4.4650352.689.26

8.21134535

)50()2510(

)50()2510(35

TH

TH

jjjZ

jjjj

jjjZ

V9.8315.116.527.89

0205.13650

20)2.715.34(

205.13650L =

=

++=

jV

10 ZTH

-j50

j25 j35

RL= 20

V5.13650

(34.5 +j71.2)


80/124



SAQ 1.6

a) (i)

=+

+= 3.26

308

30820cetanresisSource

A608.03.26

16currentSource ==

A48.0830

30608.0N =+

=I

RN= 20 as per SAQ 1.5.

A3.01220

2048.0L =+

=I

VL= 0.3 12 = 3.6 V

20 8

16 V 30 IN

0.48 A 20 RL= 12


81/124



(ii)

A162

12

1

10N =+=I

RN= 0.667 as per SAQ 1.5.

A42667.0

667.016L =+

=I

VL= 4 2 = 8 V

12 V 2

10 V 1

IN

16 A 0.667 RL=


82/124



b)

A4.63447.04.636.223

0100

200100

0100N =

=

=j

I

ZN= (141.3 j61.4) as per SAQ 1.5.

A8413.0328.166

5.231544.63447.0

)150()4.613.141(

4.613.1414.63447.0L =

=

+

=jj

jI

V9862901508413.0L ==V

var6.25orVA906.258413.09862L*

LL === IVS

50 75

100 j200

100 j80

V0100

IN

A4.63447.0 (141.3 j61.4) ZL=j150


83/124



c)

A6.8443.16.8435

050

9.343.3

050currentSource

)9.343.3()1.153.3()500(7.775.15)500(impedanceSource

5.808.60

90352.689.26)500(

35)2510(

)35()2510()500(impedanceSource

=

=

=

=++=+=

+=+++

+=

j

jjjj

jjj

jjj

A3.72632.05.808.60

2.689.266.8443.1

35)2510(

25106.8443.1N =

=

+++

=jj

jI

ZN= (34.5 +j71.2) as per SAQ 1.5.

A8.83556.06.527.89

1.641.793.72632.0

20)2.715.34(

2.715.343.72632.0L =

=

+++

=j

jI

V8.831.110208.83556.0L ==V

10 IN

-j50

j25 j35

V050

A3.72632.0 (34.5 +j71.2) RL= 20


84/124



SAQ 1.7

a)

(i) VTH= IN RN= 5 10 = 50 V

RTH= RN= 10

(ii) V4542.4)4025()33(TH =+= jjV

ZTH= (25 j40)

b)

(i) A520

100TH

THN === RVI

RN= RTH= 20

(ii) A5.6346.46.262.11

9.3650

510

3040N =

=

+

=j

jI

ZN= ZTH= (10 +j5)

SAQ 1.8a) RL= RS= 6

W2.104666

502

L2

L =

+

== RIP

b) V62.8254

2510TH =+

=V

=+

= 45.3254

254THR

W38.545.345.345.3

62.82

L2

L =

+

== RIP


85/124



c) For maximum power transfer RL= ZS

232

S )1091.110002(15

+= R

22S 1215 += R

== 91215 22SR

A6.2673.36.268.26

0100

15)129(

0100=

=++

=

jI

PL= I2RL= 3.73

2 15 = 208.7 W

d) For maximum power transfer ZL= ZS*ZL= (12 j25)

RL= 12 andXL= 25

A004.1024

025

)2512()2512(

025=

=++

=

jjI

PL= I2RL= 1.04

2 12 = 13 W

e) Apply Thevenins theorem to reduce network to single source and singleimpedance.

V1.349.423166.11

9.36104050

610

684050TH =

=

=j

jV

==

=

=+

= )18.071.1(9.572.13166.11

9.3620

610

1216

)68(2

)68(2TH j

j

j

j

jZ

For maximum power transfer ZL= ZS* = ZTH*

ZL= (1.71 +j0.18)

RL= 1.71 andXL= 0.18

A1.3454.12042.3

1.349.42

)18.071.1()18.071.1(

1.349.42=

=++

=

jjI

PL= I2RL= 12.54

2 1.71 = 269 W


86/124



Answers to activities

Activity 1.1

V100V100

5.625.1225100

=

++=

= IRE

A17A17

251017

leavingentering

=

++=

= II


87/124



Activity 1.2

=== 24.5016.05022L fLX

=

==

5.991032502

1

2

16C fC

X

A3.63898.05.9950

100source =

=j

I

V7.2635.89905.993.63898.0TH ==V

+=+

++= )24.3060(905.99050905.99050

)24.5020(TH jjZ

R= 60 andXL= 30.24 and L= 96 mH


88/124



Activity 1.3

=+=+++

++= 3.313.139)4.72119()5.990()24.5020(

)5.990()24.5020()050(source j

jj

jjjZ

A3.3172.03.313.139

0100source =

=I

1.5333.1)24.5020()5.990(

905.993.3172.0N =++

=

jjI

ZN= ZTH


89/124



Section 2: Resonant circuits


90/124




91/124



Introduction to this section

What this section is about

This section covers problem solving in resonating passive circuits.

Outcomes, aims and objectives

After completing this section, you should be able to:

determine resonant frequency and dynamic impedance in an RLC series circuit;

determine Q-factor and bandwidth in an RLC series circuit;

draw impedance/frequency and current/frequency graphs for the above circuits;

determine resonant frequency and dynamic impedance in an RLC parallel circuit;

determine Q-factor and bandwidth in an RLC parallel circuit;

draw impedance/frequency and current/frequency graphs for the above circuits.

Approximate study time

14.5 hours.


PC with suitable simulation software. Scientific calculator.


92/124



Assessment information for this section


You will be assessed at the end of Section 2 by a single closed book written test lasting1 hours. The assessment will cover both Sections 1 and 2.

When and where you will be assessed

You will sit the assessment once you have satisfactorily completed all the SAQs,practical activities and the tutor assignment which is at the end of Section 2. Thelocation will be determined by your tutor.

What you have to achieve

To pass this unit you must achieve a minimum score of 60% in the assessment. Ascore of 40 to 59% will require a partial reassessment and a score of less than 40% willrequire a complete reassessment.

Opportunities for reassessment

Normally, you will be given one attempt to pass an assessment with one reassessmentopportunity.

Your centre will also have a policy covering 'exceptional' circumstances, for example ifyou have been ill for an extended period of time. Each case will be considered on an

individual basis and is at your centre's discretion (usually via written application), andthey will decide whether or not to allow a third attempt. Please contact your tutor fordetails regarding how to apply.


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Introduction

This section looks at a phenomenon known as resonance, which can occur undercertain circumstances in both series and parallel RLC ac circuits. Although we shall belooking at resonance from an electrical/electronic engineering perspective, it has manyimplications in the wider context.

The Tacoma suspension bridge in Washington, USA, collapsed in 1940. Whilesubjected to a 40-mile-an-hour wind it started to oscillate and as the oscillations built upthe bridge broke up and fell into the river below.

The Millennium footbridge in London was opened in June 2000. Like all bridges, it wasdesigned to cope with a degree of movement but it soon became clear that things weregoing seriously wrong as the deck swayed back and forth. After two days of random

swaying, swinging and oscillating wildly, the bridge was closed down by embarrassedengineers. Modifications had to be made to the design and after nearly two years thebridge was reopened.

A glass has a natural resonance, a frequency at which it will vibrate easily. If the forcemaking the glass vibrate is big enough, the size of the vibration will become so largethat the glass breaks.

These are some examples of resonance in action and show that if this condition isallowed then disaster can happen. Resonance can be a problem in electrics andelectronics, but equally some types of electronic circuits rely on resonance to makethem work.

There are many different definitions of resonance depending on the context but the

following is probably the best suited to the examples above and for what we will beusing in this section.

Resonance is defined as:

The increase in amplitude of oscillation of an electric ormechanical system exposed to a periodic force whosefrequency is equal or very close to the natural undampedfrequency of the system.


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The series resonant circuit

Revision

Consider an RLC series circuit connected across an ac supply as shown below.

The supply voltage (V) will cause a current (I) to flow around the circuit, producing voltdrops across each of the individual components given by:

VR= I R(in phase with I)

VL= IXL(leading Iby 90)

VC= IXC(lagging Iby 90)

These three voltages are represented on the phasor diagram below.

I

V

VR VL VC

R L C

VL

VC

VRI


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The supply voltage will be the phasor sum of VR, VLand VC. Since VLand VCare actingin opposite directions, their phasor sum will simply be the difference between them.

This resultant phasor can then be combined with VRto give the supply voltage and thecircuit phase angle.

Since this circuit contains both inductance and capacitance, how do we know whetherthe circuit will have a lagging or a leading phase angle? This will depend on the valuesof VLand VC, and ultimately whether the difference between VLand VCacts upwards ordownwards. These two conditions are shown below.

Condition for resonance

You may have noticed that there could be a third condition where VLand VCare equal.In this case, the resultant of VLand VCwould be zero. This is a special condition in aseries ac circuit known as resonance.

I

VL

VCVR

I

VL

VC

VR

I I

V

VL VC

VR

VC VL

VR

V

I

VL

VC

VR

IV= VR


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The voltages VLand VCcan only be equal if their respective reactances are also equal.We know that reactance is dependent on frequency. Let us look at graphs ofXLandXC

against frequency.

Here we can see that at one frequency, identified as fr, the values ofXLandXCareequal and opposite and thus will cancel out. Below frXCis greater thanXLand above frXLis greater thanXC.

This frequency fris known as the resonant frequencyof the circuit and at thisfrequency certain conditions will apply. Firstly, we shall look at how to determine whatthe resonant frequency is.

Resonant frequency

It was stated earlier that the condition for resonance in a series RLC circuit is that theinductive and capacitance reactances are equal. This is the starting point fordeveloping an equation to calculate the resonant frequency.

XL=XC

CfLf rr

2

12 =

1)2( 2r =LCf

LCf

1)2( 2r =

LCf

12 r =

LCf

1

2

1r

=

Cfr21

2frL

-jXC

jXL

XL= 2fL

fCX

2

1C =

frFrequency (Hz)


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Therefore, it can be seen that the resonant frequency (expressed in Hz) depends onthe values of both the inductance (expressed in Henries) and the capacitance

(expressed in Farads).Worked example 2.1

A series ac circuit comprises a coil of resistance 10 and inductance 0.2 H connectedin series with a 47 F capacitor. Calculate the resonant frequency and the values of theinductive and capacitive reactances.

Solution

Hz521063832

1

10472.0

1

2

11

2

16r

==

== LC

f

=== 3.652.05222 rL LfX

=

==

2.651047522

1

2

16

rC

CfX

Impedance, current and phase angle

The general expression of impedance for an RLC series circuit is given by:

Z = R+j(XLXC)

Since at resonance the net reactance is zero (sinceXL=XC) then we have:

Z= R

Therefore, at resonance the circuit impedance is simply equal to the resistance and willbe a minimum value. This is sometimes referred to as the dynamic impedanceordynamic resistance.

The supply current will therefore be a maximum value and will be equal to:

R

VI=

Since the circuit impedance is purely resistive at resonance, the supply current will bein phase with the supply voltage giving a phase angle of:

= 0

Worked example 2.2

A series ac circuit comprises a coil of resistance 5 and inductance 0.1 H connectedin series with a 20 F capacitor. If the circuit is connected across a 100 V ac supplydetermine the circuit impedance and the supply current at resonance.

Solution

Z= R= 5

A205

100===

R

VI


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Impedance and current graphs

Both impedance and current will vary as a function of frequency, with impedance being

a minimum and current being a maximum at resonance. Typical impedance vs.frequency and current vs. frequency graphs are shown below.

Q factor

At resonance, if Ris much smaller thanXLandXC, the voltages VLand VCcan rise tomany times the supply voltage (see worked example). This voltage increase ormagnification is given by:

voltagesupply

oracrossvoltageresonanceationmagnificatvoltage

CL=

Resonant circuits respond to frequencies close to their own natural frequency muchstronger than they respond to other frequencies. This ratio is a measure of the qualityof a resonant circuit, i.e. how well it resonates, and is referred to as the quality factoror Q factor. The Q factor of a series circuit can also be represented in terms of thecircuit components:

Zmin = R

R

VI =max

Z

I

fr Frequency (Hz)


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R

Lf

R

X

RI

XI

V

VQ

fLLL 2==

==

SinceLC

f1

2

1r = then LC

f1

2 r =

LCR

L

R

LfQ

12 f==

C

L

RQ

1=

Worked example 2.3

A coil of resistance 5 and inductance 75 mH is connected in series with a 0.2 Fcapacitor across a 100 V variable frequency supply. Determine (a) the resonantfrequency, (b) the supply current and (c) the circuit Q factor.

Solution

a) kHz3.1102.01075

1

2

11

2

163r=

==

LCf

b) A205

100===

R

VI

c) 5.122102.0

1075

5

11

6

3

=

==

C

L

RQ

Damping

Circuits with a high Q factor will resonate with a greater amplitude (at the resonantfrequency) than circuits with a lower Q factor. The Q factor of a circuit can be altered bychanging the amount of resistance in the circuit, the higher the resistance the lower theQ factor and vice versa.

This process of increasing the circuit resistance to lower the Q factor is calleddamping. This will have an effect on the operating bandwidth of the circuit (see nextsection).

Worked example 2.4

A resistor of value 5 is connected in series with the circuit in the previous workedexample. Calculate the new Q factor and state whether the damping effect has beenincreased or decreased.

Solution

The new circuit resistance is now 5 + 5 = 10 .

25.61102.0

1075

10

116

3

=

==

C

L

RQ

Since the total circuit resistance has doubled, the Q factor has halved, which hasincreased the damping effect on the circuit.


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Bandwidth and selectivity

A typical current vs. frequency graph is shown below.

The graph shows the current at a maximum value at resonance (Imax) occurring atresonant frequency fr. Also shown are points A and B where the current is 0.707 Imaxat the frequencies f1and f2.

The difference between frequencies f2and f1is known as the bandwidth (B)and ismeasured in hertz (Hz). It is defined in this instance as the range of frequencies overwhich the current has not fallen by more than 70.7% of its maximum value. If you arestudying this unit as part of an HN Electronics course then you may already be familiarwith this term and the definition.

B= f2 f1

As was mentioned in the previous section, bandwidth and Q factor are related to oneanother, the higher the Q factor the smaller the bandwidth and vice versa. It is outwith

the scope of this unit to prove that relationship however the following equation can beused:

Q

fB r=

The term selectivityis the ability of a circuit to respond better to signals of onefrequency (or range of frequencies) than to signals of other frequencies. The responseof the circuit will become poorer the further away it operates from the resonantfrequency. High Q factor circuits will have narrow or small bandwidths and are said tobe highly selective compared to circuits with low Q factors and wide or largebandwidths.

Bandwidth

A B0.707 Imax

Imax

f1 fr f2 Frequency (Hz)


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Worked example 2.5

A series circuit has component values R= 25 , L= 50 mH and C= 100 nF. Determinethe bandwidth of the circuit.

Solution

kHz25.2101001050

1

2

11

2

193r=

==

LCf

3.2810100

1050

25

119

3

=

==

C

L

RQ

Hz5.793.28

1025.2 3r =

==Q

fB

To summarise, a series RLC circuit at resonance has the following properties:

The resonant frequency is given byLC

f1

2

1r

=

The impedance is a minimum and is given by Z= R

The supply current is a maximum and is given byR

VI=

The supply current and supply voltage are in phase, i.e. = 0

The circuit Q factor is given by C

L

RQ

1

=

The circuit bandwidth is given byQ

fB r=

The series resonant circuitEquipment required

PC


Circuit


2.1


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Procedure

1 Calculate the resonant frequency and the supply current at resonance.

2 Measure and record the supply current from 10 Hz up to 100 Hz in 10 Hz steps.

3 Plot a graph of current vs. frequency.

4 Clearly indicate on the graph the resonant frequency and the maximum current.

a) A series RLC circuit comprises a coil of resistance 10 and inductance 50 mHconnected to a 0.05 F capacitor across a 100 V variable frequency ac supply.Calculate (i) the resonant frequency, (ii) the supply current, (iii) the circuit Qfactor and (iv) the circuit bandwidth.

b) Determine the values of (i) circuit resistance and (ii) circuit capacitance thatmust be connected in series with a 50 mH pure inductor to cause a maximumcurrent of 0.12 A to flow when connected across a 24 V 40 kHz supply.

c) A series RLC circuit has the values R= 15 , L= 150 mH and C= 25 F. If thecircuit is connected across a 200 V ac supply determine the maximum voltageacross the capacitor.

d) Using the data in the table below, plot graphs of current vs. frequency andimpedance vs. frequency. Clearly show on the graphs the resonant frequency,Imax, bandwidth and Zmin.

2.1


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R= 25 R= 50

Frequency(Hz)

Impedance () Current (A) Impedance () Current (A)

10 750 0.1 750 0.1

20 330 0.3 335 0.3

30 170 0.6 175 0.55

40 75 1.4 85 1

50 25 4 50 2

60 62 1.6 75 1.4

70 110 0.9 120 0.8

80 155 0.7 160 0.6

90 200 0.5 200 0.5

100 240 0.4 240 0.4

The parallel resonant circuit

Ideal and actual parallel circuits

Consider the ideal parallel circuit shown below comprising a perfect inductor (zeroresistance) connected in parallel with a capacitor across an ac supply.

The supply current (I) will split to form two branch currents (ILand IC) given by:

IC

IL L

C

I

V


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CC

LL and

X

VI

X

VI ==

The current ILwill lag the supply voltage (V) by 90 and the current ICwill lead thesupply voltage (V) by 90. If we consider the case whereXL=XCas for the seriescircuit then the two branch currents will be equal and opposite in direction, as shown onthe phasor diagram below.

This would result in large (actually maximum) currents flowing in the parallel branches,and since I= IL+ IC, there would be no current drawn from the supply.

This was all happening when the conditionXL=XCwas met, which was the same as

the resonant condition for the series circuit. Therefore, for this parallel circuit, thiscondition must also happen when the frequency of the supply is equal to:

LCf

1

2

1r =

Clearly, this condition cannot occur in practice since all inductors will have some finiteamount of resistance, but as we shall see in the sections that follow, conditions close tothis can be achieved in actual parallel circuits, i.e. small supply current but large branchcurrents.

Consider the actual parallel circuit shown below comprising an inductor or resistance(R) and inductance (L) connected in parallel with a capacitor across an ac supply.

IC

IL

V

CIC

IRL R L

I

V


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Branch current ICwill lead the supply voltage by 90 (as in the ideal case) but due tothe resistance of the inductor which has now been included, branch current IRLwill no

longer lag the supply voltage by 90. It will lag somewhere between 0 and 90depending on circuit conditions.

Since this circuit contains both inductance and capacitance, how do we know whetherthe circuit will have a lagging or a leading phase angle? This will depend on the values

of IC, IRLand the phase angle RL. These two conditions are

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Electrical Networks and RESONANCE SQA