EEL 3041 Sp 2012Principles of Electrical

Engineering• Instructor: Dr. Harold Klee• Office: Harris Center 206• Office Hours: tbd• Telephone: (407) 823-2270• Email: harold.klee@ucf.edu• Class Hours: M,W 12:00-13:15• Classroom: HEC 104• Textbook: A Brief Introduction to Circuit Analysis, J.

D. Irwin, Wiley, 2003

• Course Website http://www.eecs.ucf.edu/courses/eel3041/spr2012

Learning Objectives

• Fundamental laws of electrical circuits and circuit analysis, fundamentals of electronics and power systems:– 1 Course– 2 Basic Variables: voltage and current– 3 Fundamental Laws: Ohm’s, KCL, KVL

Topics/Chapters

• Basic concepts of electrical engineering

• Resistive circuit analysis

• Transient circuit analysis

• AC steady-state analysis

• Steady state power analysis

• Variable frequency network

• Electronics fundamentals

• Unannounced quizzes will be given throughout the semester and will add to your exam scores and therefore affect your final grade. NO individual makeup opportunity will be given for any missing quiz.

• Homework will be assigned on a weekly basis but will not be collected or graded. It is important that you do the homework. Some of the quiz questions will be similar or even identical to homework problems. A list of all home work problems is included in the syllabus.

Exam 1 30% Exam 2 30% Final Exam (comprehensive) 40% Quizzes

Final average will be determined fromAve = 0.25×(Exam1+Exam2+Sumof Quiz scores) + 0.4×Final

Grade System

• A 90-100

• B 80-89

• C 70-79

• D 60-69

• F 0-59

CIRCUITS 1

DEVELOP TOOLS FOR THE ANALYSIS AND DESIGN OF BASIC LINEAR ELECTRIC CIRCUITS

A FEW WORDS ABOUT ANALYSISUSING MATHEMATICAL MODELS

BASIC STRATEGY USED IN ANALYSIS

MATHEMATICAL ANALYSIS

DEVELOP A SET OF MATHEMATICALEQUATIONS THAT REPRESENT THE CIRCUIT - A MATHEMATICAL MODEL -

LEARN HOW TO SOLVE THE MODEL TO DETERMINE HOW THE CIRCUIT WILL BEHAVE IN A GIVEN SITUATION

THIS COURSE TEACHES THE BASIC TECHNIQUESTO DEVELOP MATHEMATICAL MODELS FORELECTRIC CIRCUITS

THE MATHEMATICS CLASSES - LINEAR ALGEBRA,DIFFERENTIAL EQUATIONS- PROVIDE THE TOOLSTO SOLVE THE MATHEMATICAL MODELS

THE MODELS THAT WILL BE DEVELOPED HAVENICE MATHEMATICAL PROPERTIES.IN PARTICULAR THEY WILL BE LINEAR WHICHMEANS THAT THEY SATISFY THE PRINCIPLE OFSUPERPOSITION

FOR THE FIRST PART WE WILL BE EXPECTEDTO SOLVE SYSTEMS OF ALGEBRAIC EQUATIONS

20642

0164

84912

321

321

321

VVV

VVV

VVV

LATER THE MODELS WILL BE DIFFERENTIALEQUATIONS OF THE FORM

fdt

dfy

dt

dy

dt

yd

fydt

dy

4384

3

2

2

LOW DISTORTION POWER AMPLIFIER

a b2 TERM INALS C O M PO NENT

characterized by thecurrent through it andthe vo ltage diff erencebetweeb term inals

NO DE

NO DE

ELECTRIC CIRCUIT IS AN INTERCONNECTION OF ELECTRICAL COMPONENTS

+-

L

C

1R

2R

Sv

Ov

TYPICAL LINEARCIRCUIT

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

http://physics.nist.gov/cuu/index.html

In fo rm a tio n a t th e fo u n d a tio n o fm o d e rn s c ie n c e a n d te c h n o lo g yfro m th e P h y s ic s L a b o ra to ry o f N IS T

D e ta ile d c o n te n ts

V a lu e s o f th e c o n s ta n ts a n d re la te d in fo rm a tio nS e a rc h a b le b ib lio g ra p h y o n th e c o n s ta n ts

In -d e p th in fo rm a tio n o n th e S I, th e m o d e rnm e tric s y s te m

G u id e lin e s fo r th e e x p re s s io no f u n c e rta in ty in m e a s u re m e n t

A b o u t th is referen ce. F eed b ack.

P riv a c y S ta te m e n t / S e c u rity N o tic e - N IS T D is c la im e r

SI DERIVED BASIC ELECTRICAL UNITS

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

ELECTRON ONE OF CHARGE THE IS (e)

(e) 106.28 COULOMB 1 18

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

1 ( ) 1 ( ) per ( )Volt V Joule J Coulomb C

OHM IS A MEASURE OF THE RESISTANCE TO THE FLOW OF CHARGE.THERE IS ONE OHM OF RESISTANCE BETWEEN TWO POINTS IF ONE VOLT OF ELECTROMOTIVE FORCE IS REQUIRED TO MOVE ONE AMPERE OF CURRENT BETWEEN THE TWO POINTS

1 ( ) 1 ( ) per ( )Ohm Volt V Ampere A

IT REQUIRES ONE WATT OF POWER TO MOVE ONE AMPERE OF CURRENT BETWEENTWO POINTS WITH A VOLTAGE DIFFERENCE OF ONE VOLT

1 (W) 1 ( ) 1 ( )Watt Volt V Ampere A

1 A 1 coulomb per sec

CURRENT AND VOLTAGE RANGES

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

)(tq

EXAMPLE

])[120sin(104)( 3 Cttq

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

][)120cos(480.0)( mAtti

EXAMPLE

0

00)( 2 tmAe

tti t

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

1

0

2 dxeq x)

2

1(

2

1

2

1 021

0

2 eee x

FIND THE CHARGE AS A FUNCTION OF TIME

t t

xdxedxxitq 2)()(

0)(0 tqt

t

tx edxetqt0

22 )1(2

1)(0

And the units for the charge?...

)1(2

1 2 eq Units?

1 2 3 4 5 610

102030

Charge(pC)

Time(ms)

Here we are given thecharge flow as functionof time.

)/(10100102

10101010 93

1212

sCs

Cm

1 2 3 4 5 610

102030

Time(ms)

) Current(nA

40

20

DETERMINE THE CURRENT

To determine current we must take derivatives.PAY ATTENTION TOUNITS

CONVENTION FOR CURRENTS

IT 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 NOTATION

IF 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

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

b

a

I

A3

ab

cb

I

AI

AI

4

2

A2

c

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

sA

mN

COULOMB

JOULEVOLT

VOLTAGE IS ALWAYS MEASURED IN A RELATIVE FORM AS THE VOLTAGE DIFFERENCE BETWEEN TWO POINTS

IT IS ESSENTIAL THAT OUR NOTATION ALLOWS US TO DETERMINE WHICH POINT HAS THE HIGHER VOLTAGE

THE + AND - SIGNS DEFINE THE REFERENCEPOLARITY

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

THE TWO-INDEX NOTATION FOR VOLTAGES

INSTEAD OF SHOWING THE REFERENCE POLARITY WE AGREE THAT THE FIRST SUBINDEX DENOTESTHE POINT WITH POSITIVE REFERENCE POLARITY

VVAB 2

VVAB 5 VVBA 5BAAB VV

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

2 120 240WV W V Q V C JQ

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

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

13600102700

3

3

C

HrHr

s

S

CQ

TOTAL ENERGY STOREDTHE CHARGES ARE MOVED THROUGH A 7.2VVOLTAGE DIFFERENTIAL

][10998.6

][2.71072.9][

4

3

J

JC

JVCQW

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

t

dxxpttw

PASSIVE SIGN CONVENTION

POWER RECEIVED IS POSITIVE WHILE POWERSUPPLIED IS CONSIDERED NEGATIVE

A CONSEQUENCE OF THIS CONVENTION IS THAT

THE REFERENCE DIRECTIONS FOR CURRENT AND

VOLTAGE 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

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

IVP

IVP

UNDERSTANDING PASSIVE SIGN CONVENTION

0,0 '''' BABA IV

S2 ON

V

I

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?

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

SELECT VOLTAGE REFERENCE POLARITYBASED ON CURRENT REFERENCE DIRECTION

)5(][20 AVW AB

][4VVAB

IVW )5(][40

][8 AI

SELECT HERE THE CURRENT REFERENCE DIRECTIONBASED 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

+-

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 ABS ORBED OR SUPPLIED BY EACH ELEMENT

CIRCUIT ELEMENTS

PASSIVE ELEMENTS

INDEPENDENT SOURCES

VOLTAGE DEPENDENTSOURCES

CURRENTDEPENDENTSOURCES

?,,, rg FOR UNITS

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

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