chap01 8 e basic circuit analysis
DESCRIPTION
UCE&T BZU MULTANTRANSCRIPT
Electrical EngineeringEE-111
DEPARTMENT OF ELECTRICAL ENGINEERING, UCET BZU MULTAN
Lecture No. 1
Dr. Abdul Sattar Malik
INTRODUCTIONCourse & Instructor
Department Department of Mechanical Engineering
Degree B.Sc. (Engineering)
Course -Title 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 Malik
Email [email protected]
Course Description
Objectives
Students who complete the course of Electrical Engineering (EE-111) will(2)possess a working knowledge of the fundamental concepts of electrical engineering including circuit components, basic DC and AC circuit analysis techniques(3)be able to understand the Working principle of Electric Motors, Generators, Transformers and their applications(4)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
INTRODUCTION
COURSE OUTLINEIntroduction to D.C. circuits:
Series and parallel circuits, DC circuit theorems
Theory of Alternating Current: Series and parallel circuits. Resistance, Inductance and capacitance of AC circuits, Power factor, resonance in RLC circuits, single phase and poly-phase circuits, power and power factor measurement. Current and voltage relations in phase and line circuits
DC Machines:Generators and motor types construction and characteristics, motor starters, Testing and efficiency of machines
AC Machines: Types, characteristics and testing of A.C motors, motors starters and switch gear. Electric traction and breaking
Transformers: voltage and current relationship of primary and secondary types of transformers, Losses and efficiency
House wiring: Elements of house and power wiring, Electricity rules and act testing of house wiring
Strain gauges: types and working, Electrical gauging methods for displacement, vibration and pressure
Recommended Books:1. Basic Engineering Circuit Analysis, by David Irwin.2. Electric Machinery Fundamentals by S. Chapman.3. Electric Power Technology by Theodre Wild.
COURSE PLANContents Tentative duration
Introduction to DC circuits: Series and parallel circuits, DC circuit theorems 07-weeks
Theory of Alternating Current: Series and parallel circuits. Resistance, Inductance and capacitance of AC circuits, Power factor, resonance in RLC circuits, single phase and poly-phase circuits, power and power factor measurement. Current and voltage relations in phase and line circuits
06-weeks
DC Machines: Generators and motor types construction and characteristics, motor starters, Testing and efficiency of machines
06-weeksAC Machines: Types, characteristics and testing of A.C motors, motors starters and switch gear. Electric traction and breaking
Transformers: voltage and current relationship of primary and secondary types of transformers, Losses and efficiency
04-weeks
House wiring: Elements of house and power wiring, Electricity rules and act testing of house wiring
03-weeksStrain gauges: types and working, Electrical gauging methods for displacement, vibration and pressure
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×=
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.
C
JV =
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
A
V=Ω
IT IS REQUIRED ONE WATT OF POWER TO DRIVE ONE AMPER OF CURRENT AGAINST ANELECTROMOTIVE DIFFERENCE OF ONE VOLTS
AVW ×=
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 π−×=
=)(ti )120cos(120104 3 tππ×× − ][A
][)120cos(480.0)( mAtti ππ=
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
••==
COULOMBJOULE
VOLT
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 5=BAAB VV −=
ENERGYVOLTAGE IS A MEASURE OF ENERGY PER UNIT CHARGE…
CHARGES MOVING BETWEEN POINTS WITH DIFFERENT VOLTAGE ABSORB ORRELEASE ENERGY – THEY MAY TRANSFER ENERGY FROM ONE POINT TO ANOTHER
BASIC FLASHLIGHT Converts energy stored in batteryto thermal energy in lamp filamentwhich turns incandescent and glows
The battery supplies energy to charges.Lamp absorbs energy from charges.The net effect is an energy transfer
EQUIVALENT CIRCUIT
Charges gainenergy here
Charges supplyEnergy here
WHAT ENERGY IS REQUIRED TO MOVE 120[C] FROMPOINT B TO POINT A IN THE CIRCUIT?
VVAB 2=
JVQWQ
WV 240==⇒=
THE CHARGES MOVE TO A POINT WITH HIGHERVOLTAGE -THEY GAINED (OR ABSORBED) ENERGYTHE CIRCUIT SUPPLIED ENERGY TO THE CHARGES
ENERGYVOLTAGE IS A MEASURE OF ENERGY PER UNIT CHARGE…CHARGES MOVING BETWEEN POINTS WITH DIFFERENT VOLTAGE ABSORB ORRELEASE ENERGY
VVAB 5=
−
+V5
WHICH POINTHAS THE HIGHER 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 ABDORBED OR SUPPLIED BY EACH ELEMENT
CIRCUIT ELEMENTS
PASSIVE ELEMENTS
INDEPENDENT SOURCES
VOLTAGE DEPENDENTSOURCES
CURRENTDEPENDENTSOURCES
?,,, βµ rg FOR UNITS