Basic Electrical Engineering

EE-111

DEPARTMENT OF ELECTRICAL ENGINEERING, UCET BZU MULTAN

INTRODUCTION

Dr. Abdul Sattar Malik

INTRODUCTIONCourse & Instructor

Department Department of Civil EngineeringDegree B.Sc. (Engineering)

Course -Title Basic Electrical Engineering

Course-Code EE-111

Contact Hours Theory 02 Practical 02

Class Timing Wed 09:40-11:20 (in the Department of Mechanical Engineering)

Lab Timing Mon 08:00-09:40 (in the Department of Electrical Engineering)

Instructor Dr. Abdul Sattar MalikEmail maliksattar777@bzu.edu.pk

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Course Description

Objectives

Students who complete the course of Electrical Engineering (EE-111) will(1)possess a working knowledge of the fundamental concepts of electrical engineering including circuit components, basic DC and AC circuit analysis techniques(2)be able to understand the power distribution systems, transformers connections, Illumination etc.(3)have a knowledge of basic elements of Electric wiring, different Electrical Measurements and instrumentation techniques

Course Structure

Lectures, Seminars, Discussions, Q&A, Presentations , Experiments in Lab

Teaching Tools Multi media and White board

INTRODUCTION

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COURSE OUTLINEPart-A: Historical Development: Charge, current, potential difference, current voltage relationship for resistance, Inductance and Capacitance, Constant Voltage and Constant Current sources. Kirchhoff’ law; Loop and node equations.Part-B: Review of power system developments: Systems of distribution, lines (over Head and underground), 2- wire DC, 3- wire DC,3-phase 3-wire, 4-wire. Calculations of distribution ring mains, interconnectors, sub mains and service mains. Cables, introduction to sub-stations, generators, transformers including their connections and particular use. Electrical safety standards. Buildings internal & external, Electrification and concept of L.T. Panels Principles of measuring physical quantities by electrical methods. Concepts of earthling and neutral grounding (importance, methods and area of use)

Recommended Books:1. Basic Engineering Circuit Analysis, by David Irwin.2. Electric Power Technology by B. L.Theraja

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BASIC CONCEPTS

LEARNING GOALS

•System of Units: The SI standard system; prefixes

•Basic Quantities: Charge, current, voltage, power and energy

•Circuit Elements: Active and Passive

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SI BASE UNITS

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Definitions of the SI base units

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SI DERIVED BASIC ELECTRICAL UNITS

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ONE AMPERE OF CURRENT CARRIES ONE COULOMB OF CHARGE EVERY SECOND.

ELECTRON ONE OF CHARGE THE IS (e)(e) 106.28 COULOMB 1 18

sCA

VOLT IS A MEASURE OF ENERGY PER CHARGE. TWO POINTS HAVE A VOLTAGE DIFFERENCE OF ONE VOLT IF ONE COULOMB OF CHARGEGAINS ONE JOULE OF CHARGE WHEN IT IS MOVED FROM ONE POINT TO THE OTHER.

CJV

OHM IS A MEASURE OF THE RESISTANCE TO THE FLOW OF CHARGE.THERE IS ONE OHM OF RESISTENCE IF IT IS REQUIRED ONE VOLT OF ELECTROMOTIVE FORCETO DRIVE THROUGH ONE AMPERE OF CURRENT

AV

IT IS REQUIRED ONE WATT OF POWER TO DRIVE ONE AMPER OF CURRENT AGAINST ANELECTROMOTIVE DIFFERENCE OF ONE VOLTS

AVW

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CURRENT AND VOLTAGE RANGES

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Strictly speaking current is a basic quantity and charge is derived. However, physically the electric current is created by a movement of charged particles.

)(tq

What is the meaning of a negative value for q(t)?

PROBLEM SOLVING TIPPROBLEM SOLVING TIP IF THE CHARGE IS GIVEN DETERMINE THE CURRENT BYDIFFERENTIATIONIF THE CURRENT IS KNOWN DETERMINE THE CHARGE BYINTEGRATION

A PHYSICAL ANALOGY THAT HELPS VISUALIZE ELECTRICCURRENTS IS THAT OF WATER FLOW. CHARGES ARE VISUALIZED AS WATER PARTICLES

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)(tq

EXAMPLE

])[120sin(104)( 3 Cttq

)(ti )120cos(120104 3 t ][A

][)120cos(480.0)( mAtti

EXAMPLE

000)( 2 tmAe

tti t

FIND THE CHARGE THAT PASSES DURING IN THE INTERVAL 0<t<1

1

0

2 dxeq x )21(2

121 02

1

0

2 eee x

FIND THE CHARGE AS A FUNCTION OF TIME

t t

xdxedxxitq 2)()(

0)(0 tqt

t

tx edxetqt0

22 )1(21)(0

And the units for the charge?...

)1(21 2 eq Units?

12

1 2 3 4 5 610

102030

Charge(pC)

Time(ms)

Here we are given the

charge flow as functionof time.

)/(1010010210101010 9

3

1212sC

sCm

1 2 3 4 5 610

102030

Time(ms)

) Current(nA40

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DETERMINE THE CURRENT

To determine current we must take

derivatives.PAY ATTENTION TO

UNITS

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CONVENTION FOR CURRENTSIT IS ABSOLUTELY NECESSARY TO INDICATETHE DIRECTION OF MOVEMENT OF CHARGED PARTICLES.

THE UNIVERSALLY ACCEPTED CONVENTION INELECTRICAL ENGINEERING IS THAT CURRENT ISFLOW OF POSITIVE CHARGES.AND WE INDICATE THE DIRECTION OF FLOWFOR POSITIVE CHARGES -THE REFERENCE DIRECTION-

A POSITIVE VALUE FORTHE CURRENT INDICATESFLOW IN THE DIRECTIONOF THE ARROW (THEREFERENCE DIRECTION)

A NEGATIVE VALUE FORTHE CURRENT INDICATESFLOW IN THE OPPOSITEDIRECTION THAN THE REFERENCE DIRECTION

a b

a

a

ab

b

b

A3

A3 A3

A3

THE DOUBLE INDEX NOTATIONIF THE INITIAL AND TERMINAL NODE ARELABELED ONE CAN INDICATE THEM AS SUBINDICES FOR THE CURRENT NAME

a bA5 AIab 5

AIab 3

AIba 3

AIab 3

AIba 3

POSITIVE CHARGESFLOW LEFT-RIGHT

POSITIVE CHARGESFLOW RIGHT-LEFT

baab II 14

This example illustrates the various waysin which the current notation can be used

b

a

I

A3

ab

cb

IAIAI42

A2

c

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CONVENTIONS FOR VOLTAGES ONE DEFINITION FOR VOLT TWO POINTS HAVE A VOLTAGE DIFFERENTIAL OFONE VOLT IF ONE COULOMB OF CHARGE GAINS (OR LOSES) ONE JOULE OF ENERGY WHEN IT MOVES FROM ONE POINT TO THE OTHER

a

b

C1

IF THE CHARGE GAINSENERGY MOVING FROMa TO b THEN b HAS HIGHERVOLTAGE THAN a.IF IT LOSES ENERGY THENb HAS LOWER VOLTAGETHAN a

DIMENSIONALLY VOLT IS A DERIVED UNIT

sAmN

COULOMBJOULEVOLT

VOLTAGE IS ALWAYS MEASURED IN A RELATIVE FORM AS THE VOLTAGE DIFFERENCE BETWEEN TWO POINTSIT IS ESSENTIAL THAT OUR NOTATION ALLOWS US TO DETERMINE WHICH POINT HAS THE HIGHER VOLTAGE

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THE + AND - SIGNS DEFINE THE REFERENCE POLARITY

V IF THE NUMBER V IS POSITIVE POINT A HAS VVOLTS MORE THAN POINT B.IF THE NUMBER V IS NEGATIVE POINT A HAS|V| LESS THAN POINT B

POINT A HAS 2V MORETHAN POINT B

POINT A HAS 5V LESSTHAN POINT B

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THE TWO-INDEX NOTATION FOR VOLTAGESINSTEAD OF SHOWING THE REFERENCE POLARITY WE AGREE THAT THE FIRST SUBINDEX DENOTESTHE POINT WITH POSITIVE REFERENCE POLARITY

VVAB 2

VVAB 5 VVBA 5

BAAB VV

WHAT ENERGY IS REQUIRED TO MOVE 120[C] FROMPOINT B TO POINT A IN THE CIRCUIT?

JVQWQWV 240

THE CHARGES MOVE TO A POINT WITH HIGHERVOLTAGE -THEY GAINED (OR ABSORBED) ENERGY

VVAB 5

V5

WHICH POINT HAS THEHIGHER VOLTAGE?

THE VOLTAGEDIFFERENCE IS 5V

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EXAMPLEA CAMCODER BATTERY PLATE CLAIMS THATTHE UNIT STORES 2700mAHr AT 7.2V.WHAT IS THE TOTAL CHARGE AND ENERGYSTORED?CHARGETHE NOTATION 2700mAHr INDICATES THATTHE UNIT CAN DELIVER 2700mA FOR ONE FULL HOUR

][1072.9

136001027003

3

C

HrHrs

SCQ

TOTAL ENERGY STOREDTHE CHARGES ARE MOVED THROUGH A 7.2VVOLTAGE DIFFERENTIAL

][10998.6

][2.71072.9][4

3

J

JCJVCQW

ENERGY AND POWER

2[C/s] PASSTHROUGH THE ELEMENT

EACH COULOMB OF CHARGE LOSES 3[J]OR SUPPLIES 3[J] OF ENERGY TO THE ELEMENT

THE ELEMENT RECEIVES ENERGY AT A RATE OF 6[J/s]

THE ELECTRIC POWER RECEIVED BY THEELEMENT IS 6[W]

HOW DO WE RECOGNIZE IF AN ELEMENTSUPPLIES OR RECEIVES POWER?

VIP IN GENERAL

2

1

)(),( 12

t

tdxxpttw

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PASSIVE SIGN CONVENTIONPOWER RECEIVED IS POSITIVE WHILE POWERSUPPLIED IS CONSIDERED NEGATIVE

A CONSEQUENCE OF THIS CONVENTION IS THATTHE REFERENCE DIRECTIONS FOR CURRENT ANDVOLTAGE ARE NOT INDEPENDENT -- IF WE ASSUME PASSIVE ELEMENTS

a b

abV

abI

ababIVP

IF VOLTAGE AND CURRENTARE BOTH POSITIVE THECHARGES MOVE FROM HIGH TO LOW VOLTAGE AND THE COMPONENTRECEIVES ENERGY --IT ISA PASSIVE ELEMENT

a b

abVGIVEN THE REFERENCE POLARITY

a babI

IF THE REFERENCE DIRECTION FOR CURRENTIS GIVEN

THIS IS THE REFERENCE FOR POLARITY

REFERENCE DIRECTION FOR CURRENT

a b

abV

VVab 10

EXAMPLE

THE ELEMENT RECEIVES 20W OF POWER.WHAT IS THE CURRENT?

abI

SELECT REFERENCE DIRECTION BASED ONPASSIVE SIGN CONVENTION

ababab IVIVW )10(][20

][2 AIab

A2

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Voltage(V) Current A - A' S1 S2positive positive supplies receivespositive negative receives suppliesnegative positive receives suppliesnegative negative supplies receives

S2S1

We must examine the voltage across the componentand the current through it

0,0 ABAB IV1S ON

0,0 '''' BABA IV2S ON

A A’

B B’

''''2

1

BABAS

ABABS

IVPIVP

UNDERSTANDING PASSIVE SIGN CONVENTION

0,0 '''' BABA IVS2 ON

V

I

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CHARGES RECEIVE ENERGY.THIS BATTERY SUPPLIES ENERGY

CHARGES LOSE ENERGY.THIS BATTERY RECEIVES THE ENERGY

WHAT WOULD HAPPEN IF THE CONNECTIONS ARE REVERSEDIN ONE OF THE BATTERIES?

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DETERMINE WHETHER THE ELEMENTS ARE SUPPLYING OR RECEIVING POWERAND HOW MUCH

a

b

a

b

WHEN IN DOUBT LABEL THE TERMINALSOF THE COMPONENT

AIab 4

VVab 2

WP 8 SUPPLIES POWER

VVab 2

A2

AIab 2

WP 4 RECEIVES POWER

1

2

1

2

AIVV 4,12 1212 AIVV 2,4 1212

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SELECT VOLTAGE REFERENCE POLARITYBASED ON CURRENT REFERENCE DIRECTION

)5(][20 AVW AB

][4VVAB

IVW )5(][40

][8 AI

SELECT HERE THE CURRENT REFERENCE DIRECTION BASED ON VOLTAGE REFERENCE POLARITY

A2

)2(][40 1 AVW

][201 VV

IVW ])[10(][50

][5 AI

WHICH TERMINAL HAS HIGHER VOLTAGE AND WHICH IS THE CURRENT FLOW DIRECTION

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+-

V24

V6

V18

A2

A2

123

P1 = 12WP2 = 36WP3 = -48W

)2)(6(1 AVP

)2)(18(2 AVP

)2)(24()2)(24(3 AVAVP

IMPORTANT: NOTICE THE POWER BALANCE IN THE CIRCUIT

COMPUTE POWER ABDORBED OR SUPPLIED BY EACH ELEMENT

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CIRCUIT ELEMENTS

PASSIVE ELEMENTS

INDEPENDENT SOURCES

VOLTAGE DEPENDENTSOURCES

CURRENTDEPENDENTSOURCES

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EXERCISES WITH DEPENDENT SOURCES

OVFIND ][40VVO OIFIND mAIO 50

DETERMINE THE POWER SUPPLIED BY THE DEPENDENT SOURCES

][40V

][80])[2])([40( WAVP

TAKE VOLTAGE POLARITY REFERENCE TAKE CURRENT REFERENCE DIRECTION

][160])[44])([10( WAVP

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POWER ABSORBED OR SUPPLIED BY EACHELEMENT

][48)4)(12(1 WAVP

][48)2)(24(2 WAVP

][56)2)(28(3 WAVP

][8)2)(4()2)(1( WAVAIP xDS

][144)4)(36(36 WAVP V

NOTICE THE POWER BALANCE

USE POWER BALANCE TO COMPUTE Io

W12

))(6( OI )9)(12(

)3)(10(

)8)(4( )11)(28(

POWER BALANCE

][1 AIO 28