Transcript ENGR-43_Lec-01a_Intro_Basic_Conceptsx

Engineering 43
Circuit Analysis
- Basic Concepts Bruce Mayer, PE
Licensed Electrical & Mechanical Engineer
[email protected]
Engineering-43: Engineering Circuit Analysis
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Circuit Analysis
 DEVELOP TOOLS FOR THE
ANALYSIS AND DESIGN OF BASIC
ELECTRIC CIRCUITS
Engineering-43: Engineering Circuit Analysis
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BASIC ANALYSIS STRATEGY
Note: Some
Devices are
NONlinear
Engineering-43: Engineering Circuit Analysis
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Potential↔Flow Circuit
 A “Circuit” describes the Relationship
between
• Indestructible-Entity
(Mass, Heat, Charge)
Movement or FLOW
• Energy-Producing
POTENTIAL (Gravity,
Pressure, Voltage,
Temperature)
Engineering-43: Engineering Circuit Analysis
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Electro-Fluidic Analogy
 Analogous Fluidic and Electrical Circuits
Fluidic
Electrical
 Examine Individual Circuit Elements
Engineering-43: Engineering Circuit Analysis
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Potential Source
 Voltage is the Electrical PRESSURE
Pressure Source → Centrifugal
Pump (with fluid slip)
Engineering-43: Engineering Circuit Analysis
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Electrical Potential Source →
Voltage Source
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Flow (Mass Movement) Source
 Current is the Electrical MASS-FLOW
Incompressible-Fluid
FlowSource
→
Positive
Displacement Pump (NO fluid
slip)
Engineering-43: Engineering Circuit Analysis
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Electrical Current Source →
Current Source
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Conductors (move mass)
 Perfect Conductors transport “Flow”
with NO LOSS of Potential
Engineering-43: Engineering Circuit Analysis
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Resistor (hinders flow)
 Resistors CAUSE a Loss of Potential
with Flow thru Them
Flow Resistor
Restriction, or
Direction
→ OrificeElectrical
Changes in Resistor → Carbon-Cylinder or
wire-coil
Engineering-43: Engineering Circuit Analysis
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Flow On/Off Control
 Valves & Switches Turn Flow On/Off
Flow On/Off → ShutOff Valve;
typically “Ball”, “Plug” or “Gate”
type
Engineering-43: Engineering Circuit Analysis
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Electrical
Current On/Off → 2-Pole,
Single-Throw (2PST) Switch
Bruce Mayer, PE
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Capacitor  Accumulates Flow
 Capacitors Produce a Flow-Change
when Pressure Changes
Q  CV
dv
i C
dt
m  C f P
dP
m  C f
dt
A Balloon would be a form of Fluidic
The Units of Electrical Capacitance are
Capacitance. The Units for the Fluidic
A∙s/V
Capacitance, Cf, are inverse m∙s2
Engineering-43: Engineering Circuit Analysis
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Inductor  Resists FLOW Chg
 Inductors Produce a Pressure-Change
when Flow Changes
Phi  Plo  P  L f
dm
dt
m
vhi  vlo  vL  L
m
i
Plo
Phi
di
dt
i
vhi
vlo
Based
on
Bernoulli-Flow
of
an
incompressible Inviscid Fluid. The Units The Units of Electrical Inductance are
for the Fluidic Inductance, Lf, are inverse V∙s/A
meters (m−1)
P 


V
2
2
W ide
2
 VNarw

V is Fluid Velocity
Engineering-43: Engineering Circuit Analysis
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Another Circuit Analogy
 Resistors in “Series
• Note that “Flow” moves from HI-potential to
LO-potential in BOTH Cases
Engineering-43: Engineering Circuit Analysis
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Electric Circuit – What is it?
L
R1
2 TERMINAL COMPONENT
a
b

NODE
characterized by the
current through it and
the voltage difference
between terminals
R2
NODE
vS
vO

+
-
C
Engineering-43: Engineering Circuit Analysis
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TYPICAL LINEAR CIRCUIT
 ELECTRIC CIRCUIT IS AN
INTERCONNECTION OF
ELECTRICAL COMPONENTS
Basic Concepts
 Learning Goals
• System Of Units: The SI Standard System;
Prefixes
• Basic Electrical Quantities: Charge,
Current, Voltage, Power, And Energy
• Circuit Elements: Active and Passive
– Passive elements typically have TWO
connections: INput & OUTput, and can only
DISSIPATE POWER
– Active elements have at LEAST 3 conntions:
Input, OUTput, and CONTROL
Engineering-43: Engineering Circuit Analysis
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International System of Units (SI)
 SI base units
Base quantity
length
Only One
“Electrical” Unit
mass
Symbol
meter
m
kilogram
kg
time
second
s
electric current
ampere
A
thermodynamic
temperature
kelvin
K
amount of substance
mole
mol
candela
cd
luminous intensity
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Name
Bruce Mayer, PE
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http://physics.nist.gov/cuu/index.html
• The SI system of Units Is Founded On
Units For Seven Base Quantities Assumed
To Be Mutually Independent, As Given
SI Base Units
Below
SI DERIVED ELECTRICAL UNITS

Other quantities, called derived quantities, are defined in terms of the seven
base quantities via a system of quantity equations. The SI units for these
derived quantities are obtained from these equations and the 7 SI base units
Derived quantity
Name
Symbol
Expression
in terms of
other SI units
Expression
in terms of
SI base units
Frequency
hertz
Hz
-
s-1
energy, work, quantity of heat
joule
J
N·m
m2·kg·s-2
power, radiant flux
watt
W
J/s
m2·kg·s-3
coulomb
C
-
s·A
volt
V
W/A
m2·kg·s-3·A-1
farad
F
C/V
m-2·kg-1·s4·A2
ohm
Ω
V/A
m2·kg·s-3·A-2
siemens
S
A/V
m-2·kg-1·s3·A2
henry
H
Wb/A
m2·kg·s-2·A-2
electric charge, quantity of electricity
electric potential difference,
electromotive force
capacitance
electric resistance
electric conductance
inductance
Engineering-43: Engineering Circuit Analysis
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SI prefixes
Factor
Name
Symbol
Factor
Name
Symbol
1024
yotta
Y
10-1
Deci
d
1021
zetta
Z
10-2
Centi
c
1018
exa
E
10-3
milli
m
1015
peta
P
10-6
micro
µ
1012
tera
T
10-9
nano
n
109
giga
G
10-12
pico
p
106
mega
M
10-15
femto
f
103
kilo
k
10-18
atto
a
102
hecto
h
10-21
zepto
z
101
deka
da
10-24
yocto
y
Engineering-43: Engineering Circuit Analysis
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Derived Units
 One Ampere Of Current Carries One
Coulomb Of Charge Every Second
AC s
• 1 Coulomb = 6.242x1018 e– e-  The Charge On An Electron
 A Volt Is A Measure Of Energy
J
V

Per Charge (Electrical Potential)
C
• Two Points Have A Potential Difference Of
One Volt If One Coulomb Of Charge Gains
One Joule Of Energy When It Is Moved
From One Point To The Other.
Engineering-43: Engineering Circuit Analysis
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Derived Units cont.
 The Ohm Is A Measure Of The
Resistance To The Flow Of Charge
• There Is One Ohm Of Resistance If It
Requires One Volt Of Electrical-Pressure
To Drive Through One Ampere Of Current
 V
A
 One Watt Of Power Is Required To
Drive One Ampere Of Current Against
An Electromotive Potential Difference
Of One Volt W  V  A
Engineering-43: Engineering Circuit Analysis
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Current And Voltage Ranges
Engineering-43: Engineering Circuit Analysis
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Bruce Mayer, PE
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Current = Charge-Flow
 Strictly Speaking Current (i) is a Basic
Quantity And Charge (q) Is Derived
• However, Physically The Electric Current is
Created By Charged-Particle Movement
 An Electric Circuit Acts a Pipeline That
Moves Charge from One Pt to Another
• The Time-Rate-of-Change for Charge
Constitutes an Electric Current;
Mathematically
dqt 
i t  
OR qt    i  d  qt   q 
dt

t
Engineering-43: Engineering Circuit Analysis
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Bruce Mayer, PE
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Water-Flow Analogy
 Water Particles in a Pipeline Move
• From HIGH-Pressure to LOW-Pressure
 In Current-Flow Electrical Charges
move
• From HIGH-Potential (Voltage) to
LOW-Potential
 Direction Convention
• Almost all Solid-State Current Flow is
Transported by ELECTRONS, Which
Move from LOW Potential to HIGH
Engineering-43: Engineering Circuit Analysis
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Bruce Mayer, PE
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Fluid↔Current Flow Analogs
 Think of the
• Voltage as the
Electrical
“Pressure”
• Current as the
Electrical Fluid
• Wire as the
Electrical “Pipe”
Engineering-43: Engineering Circuit Analysis
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Current & Electron Flow
 These Alternatives Constitute
Conventional POSITIVE Current Flow
• POSITIVE Charges Moving: Hi-v → Lo-v
• NEGATIVE Charges Moving: Lo-v → Hi-v
Vhi
+
+
+
Vhi
-
+
q(t)
-
-
q(t)
Vlo
Engineering-43: Engineering Circuit Analysis
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Vlo
Bruce Mayer, PE
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Current/Charge Calculation
 If The Charge is
Given, Determine
The Current By
Differentiation
Vhi
+
+
+
+
q(t)
 If The Current Is
Vlo
Known, Determine q(t )  4 103 sin( 120 t ) [C]
The Charge By
dq
Integration
it  
 4 10 3 120 cos(120 t ) A 
 EXAMPLE
Engineering-43: Engineering Circuit Analysis
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dt
or i (t )  480 cos(120 t )
Bruce Mayer, PE
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[mA ]
 0 t 0
i (t )   2 t
e mA t  0
Another
Example
 By MATLAB
>>
>>
in
>>
in
Find The Charge That Passes
During In The Interval 0<t<1s
t=[0:0.001:1]; % in
i_of_t = exp(-2*t); %
mA
q = trapz(t, i_of_t) %
mCoul
q =
0.4323
>> syms x t
>> q_of_t = int(exp(-2*x),
0, t)
q_of_t =
-1/2*exp(-2*t)+1/2
Engineering-43: Engineering Circuit Analysis
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1
1 2
1 2
1 0
 2
q   e d  e
  e  ( e )
0
2
2
2
Sec
0
1
1
q  (1  e  2 )
2
Units?
Find The Charge As A Function Of Time
t
t


q(t )   i ( x)dx   e  2 d
t  0  q(t )  0
t
t
1
1
t  0  q(t )   e 2 x dx   e 2 x  (1  e 2t )
2
2
0
0
And the units for the charge?...
Bruce Mayer, PE
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Charge(pC)
30
20
10
 10
DETERMINE THE
CURRENT
1 2 3 4 5 6
Current(nA )
40
30
20
10
10
 20
 10  1012  10  1012 C
m
 10  109 (C / s)
3
s
2  10  0
To determine current we
must take derivatives.
PAY ATTENTION TO UNITS
Zero currents
1 2 3 4 5 6
Engineering-43: Engineering Circuit Analysis
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Time(ms)
Here we are given the
charge flow as a function
of time.
Time(ms)
Graphical
Example
Bruce Mayer, PE
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Convention For Currents
 It Is Absolutely
Necessary To
Indicate The Direction
Of Movement Of
Charged Particles
A Positive Value For
The Current Indicates
Flow In The Direction
Of The Arrow (The
Reference Direction)
 The Universally
Accepted Convention
In Electrical
Engineering is That
Current is the Flow Of
POSITIVE Charges
A Negative Value For
The Current Indicates
Flow In The OPPOSITE
Direction Than The
Reference Direction
Engineering-43: Engineering Circuit Analysis
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• We Indicate Our
ASSUMED Direction
Of Flow For Positive
Charges
– THE REFERENCE
DIRECTION
Bruce Mayer, PE
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Double Index Notation
 If The Initial And
Terminal Node Are
Labeled One Can
Indicate Them As
SubIndices For The
Current Name
a 3A b a  3A b
I ab  3 A
I ab  3 A
a  3A b a 3A b
Iba  3 A
a
5A
b
I ab  5 A
Engineering-43: Engineering Circuit Analysis
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I ba  3 A
POSITIVE CHARGES
FLOW LEFT-RIGHT
POSITIVE CHARGES
FLOW RIGHT-LEFT
Iab   Iba
Bruce Mayer, PE
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Example
 This Example Illustrates The Various
Ways In Which The Current Notation
Can Be Used
• To Find Iab
Assuming that
Charge is
CONSERVED
– More on this
later
Engineering-43: Engineering Circuit Analysis
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I  2 A
a
2A
I
b
I cb  4 A
I ab  3A
c
3A
Bruce Mayer, PE
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Voltage Conventions
b
 One Definition For
the Volt unit
• Two Points Have A
Voltage Differential
Of One Volt If One
Coulomb Of Charge
Gains (Or Loses)
One Joule Of
Energy When It
Moves From One
Point To The Other
Engineering-43: Engineering Circuit Analysis
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 a
1C
• If The Charge GAINS
Energy Moving From a
To b, Then b Has
HIGHER Voltage Than a
• If It Loses Energy Then b
Has LOWER Voltage
Than a
Bruce Mayer, PE
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Voltage Conventions cont.
 Dimensionally, the Volt is a Derived Unit
JOULE
N m
VOLT 

COULOMB A  s
 Voltage Is Always Measured In a
RELATIVE Form As The Potential
DIFFERENCE Between Two Points
• It Is Essential That Our Notation Allows Us
To Determine Which Point Has The Higher
Electrical Potential (Voltage)
Engineering-43: Engineering Circuit Analysis
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Bruce Mayer, PE
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THE + AND - SIGNS
DEFINE THE REFERENCE
POLARITY
V
If The Number V Is Positive, Then Point A
Has V Volts More Than Point B, If The
Number V Is Negative, Then Point A Has
|V| Less Than Point B
POINT A HAS 2V MORE
THAN POINT B
Engineering-43: Engineering Circuit Analysis
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POINT A HAS 5V LESS
THAN POINT B
Bruce Mayer, PE
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2-Index Notation for Voltages
 Instead Of Showing The Reference
Polarity We Agree That The FIRST
SubIndex Denotes The Point With
POSITIVE Reference Polarity
VAB  2V
Engineering-43: Engineering Circuit Analysis
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VAB  5V
VAB  VBA
VBA  5V
Bruce Mayer, PE
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Energy = Potential/Charge
 Voltage Is A Measure Of ENERGY Per Unit CHARGE
• Charges Moving Between Points With Different Voltage
ABSORB or RELEASE Energy – They May Transfer Energy
From One Point To Another
 Converts chemical energy stored
in the battery to thermal energy in
the lamp filament which turns
incandescent and glows
EQUIVALENT CIRCUIT
Charges gain
energy here
Charges supply
Energy here
Engineering-43: Engineering Circuit Analysis
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 Battery supplies energy to
charges. Lamp absorbs energy
from charges.
 The net effect is an energy
TRANSFER
Bruce Mayer, PE
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Example
 What Energy Is Required To Move 120
[C] From Point-B To Point-A?
• Recall Potential = Work/Charge
W
V   W  VQ  240 J
Q
120 [C]
VAB  2V
• Since The Charges Move To a Point With
Higher Voltage –The Charges Gained
(Or Absorbed) Energy
Engineering-43: Engineering Circuit Analysis
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Bruce Mayer, PE
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Example
 Recall Again Potential = Work/Charge
• Work = 5 J
• Charge = 1C
W
5[J]
V  
5V
Q
1[C]
 The Voltage
Difference Is 5V
• Which Point Has
Higher Voltage?

5V

– VAB = 5V
Engineering-43: Engineering Circuit Analysis
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Bruce Mayer, PE
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Energy and Power
 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 The
Element Is 6[W]
• How Do We Recognize If
An Element SUPPLIES
Or RECEIVES Power?
Engineering-43: Engineering Circuit Analysis
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2[C/s] PASS
THROUGH
THE ELEMENT
 In General
t2
 w(t 2 , t1 )   p ( )d
t1
 P  VI
Bruce Mayer, PE
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Passive Sign Convention
 Power Received (Absorbed/Dissipated)
is Positive While Power SUPPLIED
(Generated) is Considered NEGATIVE
 Vab 
a
b
P  Vab I ab
I ab
– If Voltage And Current Are Both Positive The
Charges Move From High To Low Voltage, then
The Component Receives, or Dissipates
Energy – It Is A Passive Element
Engineering-43: Engineering Circuit Analysis
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Bruce Mayer, PE
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Passive Element Assumption
 We Assume PASSIVE Elements
• A Consequence Of This Convention Is That
The Reference Directions For Current And
Voltage Are Not Independent
– Given This Reference Potential-Polarity
 Vab 
a
b
Then the Reference
Direction For Current
Engineering-43: Engineering Circuit Analysis
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Bruce Mayer, PE
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Passive Element Convention
 The Voltage Polarity Given That The
Reference Current Flows From Pt-A to
Pt-B for the assumed Passive Element
+a
b
−
I ab
• VA POSITIVE Relative to VB  VAB > 0
– For the PASSIVE Element
Engineering-43: Engineering Circuit Analysis
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Example
 The Element Receives 20W of Power
• What is the Current?
 Select Reference
Direction Based On
Passive Sign Convention
20[ W]  Vab I ab  (10V) I ab
or
I ab  2[A]
 Vab 
a
b
I ab
Vab = -10V
Given
• 2A of Current Flows RIGHT-to-LEFT ←
Engineering-43: Engineering Circuit Analysis
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Bruce Mayer, PE
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Passive Sign Convention
 When In Doubt Label The Terminals Of
The Component
V12  12V,
or
1
1
2
2
I12  4A
P  48 W 
 Element SUPPLIES Power
Engineering-43: Engineering Circuit Analysis
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V12  4V,
or
I12  2A
P  8 W 
 Element RECEIVES Power
Bruce Mayer, PE
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Passive Sign Convention
 Select Voltage Reference Polarity Based
On CURRENT Reference Direction
I  8[ A]
VAB  4[V ]




 20[W ]  VAB  I AB
40[W ]  VAB  I AB
VAB  4V 
I AB  8A
 20W   VAB  (5 A)
40W   (5V )  I AB
 In These Examples the Unknown V & I have Polarity &
Direction OPPOSITE to the Passive Assumption
Engineering-43: Engineering Circuit Analysis
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Bruce Mayer, PE
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Assigning The Reference
 If V Given Across the Element
• Iref Flows: Vhi→Vlo
 IF I Given Thru the Element
• Vref DROPS in The Current Direction
A
A
I  5[ A]
V1  20[V ]
B
40[W ]  VAB  I AB  VAB  2A
VAB  20V  and V1  VAB
Engineering-43: Engineering Circuit Analysis
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B
 50[W ]  VAB  I AB  10V  I AB
I AB  5A and
I  I AB
Bruce Mayer, PE
[email protected] • ENGR-43_Lec-01a_Basic_Concepts.ppt
Example
 Determine Power Absorbed By Each
Circuit Element; 1, 2, and 3
• In this Passive Ckt, a single V-Source is an
Active, Power-SUPPLYING Element
– Current Runs VLo→VHi; Sets I Direction
 Assume that Voltage DROPS Across the Two Passive
Elements is In the DIRECTION of Current Flow
P3  (24V )( 2 A)
P3  (24V )( 2 A)
P3  48W 
2 A  6V 

24V
Note The SuppliedReceived Power-Balance

Engineering-43: Engineering Circuit Analysis
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P1  (6V )(2 A)  12W 

1
+
-
3
2
2A
18V

P2  (18V )(2 A)  36W 
Bruce Mayer, PE
[email protected] • ENGR-43_Lec-01a_Basic_Concepts.ppt
Engineering 43
-APPENDIXMore Examples
Applied Math
Bruce Mayer, PE
Licensed Electrical & Mechanical Engineer
[email protected]
Engineering-43: Engineering Circuit Analysis
48
Bruce Mayer, PE
[email protected] • ENGR-43_Lec-01a_Basic_Concepts.ppt
Done for Lecture/Discussion
 Please Be Ready for a LAB during the
Next Meeting.
• Mr. Phillips will have something fun &
interesting to do
 See you then… 
 See Me to Add
Engineering-43: Engineering Circuit Analysis
49
Bruce Mayer, PE
[email protected] • ENGR-43_Lec-01a_Basic_Concepts.ppt
Example
 Use Power Balance To Compute I0
P
sup
 Thus a 1A Current Flows in
the Direction Assumed in the
Figure
Engineering-43: Engineering Circuit Analysis
50
  Pabs
12  108  30  32W   176W   6V  I 0
182W   176W   6V  I 0
I 0  6VA
 1A
6V 
Bruce Mayer, PE
[email protected] • ENGR-43_Lec-01a_Basic_Concepts.ppt
Mathematical Analysis
 Develop A Set Of Mathematical
Equations That Represent The Circuit
• A Mathematical Model
 Learn How To Solve The Model To
Determine Circuit Behavior (ENGR25)
 This Course Teaches The Basic
Techniques Used To Develop
Mathematical Models For
Electric Circuits
Engineering-43: Engineering Circuit Analysis
51
Bruce Mayer, PE
[email protected] • ENGR-43_Lec-01a_Basic_Concepts.ppt
Linear Models
 The Models That Will Be Developed Have
“Nice” Mathematical Properties.
• In Particular They Will Be Linear Which Means
That They Satisfy The Principle Of
Superposition:
– Model:
y  T u 
– Principle of Superposition
 Given 1, 2, are CONSTANTS
y  T 1u1   2u2   1T u1    2T u2 
Engineering-43: Engineering Circuit Analysis
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Bruce Mayer, PE
[email protected] • ENGR-43_Lec-01a_Basic_Concepts.ppt
Sophisticated Math Tools
 This Course Applies the Tools Learned
in MTH1 & ENGR25 to Solve Complex
Physics/Math Circuit Models
• For The First Part We Will Be Expected To
Solve Systems Of Algebraic Equations
12V1  9V2  4V3  8
 4V1  16V2  V3  0
 2V1  4V2  6V3  20
– Watch for These in MTH-6
Engineering-43: Engineering Circuit Analysis
53
Bruce Mayer, PE
[email protected] • ENGR-43_Lec-01a_Basic_Concepts.ppt
Math Tools cont.
• Later The Models Will Be Differential
Equations Of The Form
dy
3  y  f t 
dt
2
d y
dy
df t 
 4  8y  3
 4 f t 
2
dt
dt
dt
– Technically These Are Linear,
Nonhomogeneous, Ordinary, Constant
Coefficient Differential Equations Of The
1st & 2nd Order
 Watch for them in MTH-4
Engineering-43: Engineering Circuit Analysis
54
Bruce Mayer, PE
[email protected] • ENGR-43_Lec-01a_Basic_Concepts.ppt
Example
 A Camcorder Battery Plate Claims That
The Unit Stores 2700 mAHr At 7.2V
• What Is The Total Charge And Energy Stored?
– Charge → The 2700 mAhr Specification Indicates
That The Unit Can Deliver 2700 mA for ONE Hour
s
C
Q  2700 10    3600 1Hr
Hr
s
3
or Q  9.72 103 [C]
– Stored Energy → The Charges Are Moved Thru a
7.2V Potential Difference
J 
W  Q[C ]  V    9.72 103  7.2[ J ] or W  6.998 10 4 [ J ]
C 
(0.02 kWhr)
Engineering-43: Engineering Circuit Analysis
55
Bruce Mayer, PE
[email protected] • ENGR-43_Lec-01a_Basic_Concepts.ppt