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Preliminaries
ECE221 Circuit Analysis I
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Specific course learning objectives
At the end of this course, students will be able to:
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analyze resistive circuits using basic laws;
perform nodal and mesh analysis;
apply circuit theorems;
analyze circuits containing ideal operational amplifiers;
understand the behavior of inductors and capacitors;
use the basic capabilities of MATLAB and LTspice
use standard laboratory equipment such as a protoboard, multimeter,
power supply, function generator, and oscilloscope
• design, analyze, simulate, and build Op Amp circuits
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General course learning objectives
At the end of this course, students will be able to:
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formulate a solution strategy;
apply first principles reasoning;
perform symbolic (as well as numerical) manipulations;
provide a physical interpretation of the mathematics;
validate problem solutions.
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Course Sequence
ECE222
circuit
analysis II
ECE221
circuit
analysis I
ECE321
electronics
I
ECE315
Signals &
Systems I
ECE223
circuit
analysis III
ECE317
Feedback
& Control
ECE331
emag I
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Career path
learn tools
study problems
already solved
(what tool to use)
solve new problems
invent new systems
boring?
exciting
Analysis
Synthesis
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Bloom’s taxonomy
of the cognitive domain
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Solution strategies
How to Solve It
George Pólya Hungarian mathematician; December 13, 1887 – September 7, 1985
Professor of mathematics (1914 – 1940) Swiss Federal Institute of Technology, Zurich
Professor of mathematics (1940 – 1953) Stanford University
Fundamental contributions to combinatorics, number theory, numerical analysis
and probability theory
Also noted for his work in heuristics and mathematics education.
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How to Solve It
• Understanding the problem
– What are the knowns? What are the unknowns?
– Draw a figure; introduce suitable notation.
• Devising a plan
– Have you seen this or a related problem before?
– If you cannot envision a plan for this problem, can you imagine a more
accessible related problem (i.e., can you simplify)?
• Carrying out the plan
– Check each step.
• Looking back
– Can you check the result? Can you derive the result differently?
– Does the answer make sense (the sanity check)?
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How to Solve It Plus One
• What did you learn from this problem?
– Can you make use of this to solve other related problems?
– How was your approach similar to/different from other approaches?
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Course fundamentals
Fundamental assumption for the linear circuits in this course: no
net power consumption/generation. For a given circuit, the sum
of all the powers (created and consumed) is zero.
“Electric circuit” can mean an actual physical circuit or a
mathematical model of it. In this course, we study “lumped‐
parameter systems.”
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Basic assumptions
• Electrical effects happen instantaneously throughout the
system
– Electrical phenomena happen at the speed of light. Any system that is
small enough that we can ignore this propagation delay is called a
“lumped‐parameter system.”
• No net charge exists on any component
• No magnetic coupling between components (can occur within
a component, however)
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How small is “small”?
• answer: less than a wavelength, 
f  60Hz,   5  106 m
• If the system is < /10 = 500km, then we’re OK.
f  109 Hz=1GHz,   30cm
• If the system is < /10 = 3cm, then we’re OK.
• BTW, what is 0.5Hz?
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Basic Concepts Overview
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SI prefixes
Definitions: current, voltage, power & energy
Passive sign convention
Circuit elements
Ideal sources
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Fundamental SI quantities
quantity
unit
symbol
length
meter
m
mass
kilogram
kg
time
second
s
electric current
Ampere
A
temperature
(degree) Kelvin
K
amount
mole
mol
luminous intensity
candela
cd
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Derived SI quantities
T
=
Tesla
=
magnetic flux density
H
=
Henry
=
inductance
V
=
Volt
=
potential difference
F
=
Farad
=
capacitance
Ω
=
Ohm
=
resistance
N
=
Newton
=
force
C
=
Coulomb =
electric charge
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Derived SI quantities
Vs
N
Wb
kg
kg
=
=
=
=
m2
Am
m2
Cs
A s2
m 2 kg
J
Wb
Vs
JCs
H =
=
=
=
=
s2 A2
A2
A
A
Cs
T =
V =
F =
W
J
=
A
As
As
J
=
V
V2
V
=
A
kg m
N =
s2
C = As
Ω =
m 2 kg
s C2
=
m 2 kg
C2
= Ωs
Nm
kg m 2
kg m 2
Nm
J
=
=
=
=
3
As
As
C s2
C
C
Ws
C
C2
C2
s 2 C2
=
=
=
=
=
V2
V
J
Nm
m 2 kg
=
=
=
J
s A2
=
kg m 2
s3 A 2
=
s4 A2
m 2 kg
=
s
Ω
Js
C2
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Derived SI quantities
Vs
N
Wb
kg
kg
=
=
=
=
m2
Am
m2
Cs
A s2
J
Wb
Vs
JCs
m 2 kg
H =
=
=
=
=
A2
A
A
Cs
s2 A2
T =
V =
F =
W
J
=
A
As
As
J
=
V
V2
V
=
A
kg m
N =
s2
C = As
Ω =
m 2 kg
C2
= Ωs
Nm
kg m 2
Nm
J
kg m 2
=
=
=
=
3
As
C s2
C
C
As
Ws
C
C2
C2
s 2 C2
=
=
=
=
=
2
V
V
J
Nm
m 2 kg
=
=
m 2 kg
s C2
=
=
J
s A2
=
kg m 2
s3 A 2
=
s4 A2
m 2 kg
=
s
Ω
Js
C2
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SI Prefixes used in Science & Engineering
Multiplier
Prefix
Symbol
Example
1012
tera
T
TB
109
giga
G
GB
106
mega
M
MHz
103
kilo
k
k
10‐3
milli
m
mH
10‐6
micro

A
10‐9
nano
n
ns
10‐12
pico
p
pF
10‐15
femto
f
fW
10‐18
atto
a
100
V
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Assessment Problem 1.2
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Assessment Problem 1.2
What do we know from the statement of the problem?
$100 billion year
…but units are not what is desired.
What do we know from past experience?
365 days
year
24 hours

day
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Assessment Problem 1.2
assessment problem 1.2
How many dollars per millisecond would the federal government
have to collect to retire a deficit of $100 billion in one year?
ans =
$1011
365 days 24 hours 3600 s 103 ms
1 year
year
day
hour
s
= $3.17 ms
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Circuit Analysis: Introduction
• Electric circuit: a collection of interconnected elements
• Charge: an electrical property of the constituents of matter,
measured in Coulombs (C)
• 1C = 6.24 x 1018 electrons
• Law of Conservation of Charge: charge cannot be created or
destroyed, merely re‐distributed
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Flashlight Circuit
switch
batteries
1.5 V
each
light bulb
Schematic Representation
Pictorial Representation
• A flashlight circuit has four circuit elements
• We use symbols to represent these components
• Symbolic representation facilitates analysis
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Electric Current
• Electric Current is the rate of change of charge, measured in
amperes (A)
• 1A = 1C/s
• Two general categories
– Direct Current (DC): Current is constant
– Alternating Current (AC): Current varies sinusoidally with time
i 
where
dq
dt
i  current in amperes (A)
q  charge in Coulombs (C)
t
 time in seconds (s)
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Notes on Current
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Current in circuits physically realized by movement of electrons
Direction of current must be specified by an arrow
Convention: current direction defined as flow of positive charge
Note positive charge not flowing physically
Electrons have negative charge
Electrons flow in opposite direction of current
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Propagation Rate
• Average drift velocity is very small (mm/s)
• Propagation is very fast (close to speed of light)
• Effects appear to be instantaneous throughout circuit
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Electron Drift Velocity
I  1600A
What is the velocity of the electrons
in this bus bar?
4mm
Known parameters:
N  1029
16cm
e m3
e  1.602  10
A=
C
s
Dimensional considerations suggest:
v 
19
(free electron density)
C e (charge on an electron)
e
m3
I
m
s
m2
C
e
 N v A e
I
 0.156mm s
NA e
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Voltage
Whenever positive and negative charges are separated
energy is expended.
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Voltage
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Voltage is the change in energy per unit charge.
Is produced by the separation of charges
Sometimes called potential difference
Analogous to pressure in a hydraulic system
Is a measure of the potential (difference) between two points
Voltage pushes charge in one direction.
We use polarity (+ and – markings) to indicate which direction
the charge is being pushed.
v
dw
dq
where
v = voltage in volts (V)
w = energy in Joules (J)
q = charge in Coulombs (C)
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Voltage Concept
• The voltage sources push current through the circuit
• The current is the rate of flow of charge (i.e. electrons)
• The light bulb (resistor) resists the flow of current
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Example 1: Hydraulics analogy
pressure
reservoir
conduit
flow
• Voltage ~ pressure
• Current ~ water flow in conduit
• Resistance ~ drag due to conduit walls, cross‐section
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Example 2: Dynamics analogy
F
• Voltage ~ force due to gravity (mg)
• Electric Current ~ raindrop velocity
• Resistance ~ aerodynamic drag (F)
mg
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Assessment Problems 1.3, 1.4
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Power
• Power: time rate of energy expenditure or generation
• Denoted by p
• By convention:
– Circuit elements absorbing power have a positive value of p
– Circuit elements generating power have a negative value of p
p
where
dw
dt
p   vi
p = power in watts (W = J/s)
w = energy in Joules (J)
t = time in seconds (s)
v = voltage in volts (V)
i = current in amperes (A)
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Energy
• Law of Conservation of Energy: net power absorbed by a
circuit is 0
– The total energy produced in a circuit is equal to the total energy
absorbed
– Every watt absorbed by an element must be produced by some other
element(s)
• Energy: capacity to do work, measured in joules (J)
t
t
t0
t0
w   p   d    v   i   d
• If current and voltage are constant (DC)
t
w   pd   t  t0  p
t0
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Passive Sign Convention
• Passive Sign Convention (PSC): Current enters the positive terminal
of an element (equivalent: current leaves negative terminal)
• Most two‐terminal circuit elements (e.g. batteries, light bulbs,
resistors, switches) are characterized by a single equation that
relates voltage to current: v = ±f(i) or i = ± g(v).
• Called the defining equation
• The PSC defines the sign of the relationship
– If PSC is satisfied, v = + f(i) or i = g(v).
– If PSC is not satisfied, v = ‐ f(i) or i = ‐ g(v).
• Convention applies to power as well
– If PSC is satisfied, p = +vi (element is consuming power)
– If PSC is not satisfied, p = ‐vi (element is producing power)
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Example 3: Passive Sign Convention
Suppose the circuit element shown below is characterized by v = ± f(i)
and i = ±g(v). Determine whether the PSC is satisfied and write the
equation for power in terms of voltage and current for each diagram
i
i
i
i




v
v
v
v




p  v  i
p  v i
p  v  i
p  v i
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Example 4: Passive Sign Convention
Find the power (absorbed) for each element
2A
2A
2A
2A
‐
+
+
‐
4V
4V
4V
4V
+
‐
‐
+
p  8W
p  8W
p  8W
p  8W
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Example 5: Passive Sign Convention
Find the total power absorbed in the circuit.
+
vd
‐
voltage
(V)
current
(A)
a
‐18
‐51
b
‐18
45
+ vc ‐
c
2
‐6
c
d
20
‐20
e
16
‐14
f
36
31
total
power
(W)
‐
va
+
element
a
ia
d
id
ic
‐
vb
+
ib
b
+
vf
‐
f
ie
+
ve
‐
if
e
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Example 5: Passive Sign Convention
 ia  51A

va  18V
va  18V
a

a

ia  51A
p  918W
ib  45A

vb  18V
b

ib  45A

vb  18V
b

p  810W
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Example 5: Passive Sign Convention
Find the total power absorbed in the circuit.
‐
va
+
+
vd
‐
element
voltage
(V)
current
(A)
power
(W)
a
‐18
‐51
‐918
b
‐18
45
‐810
+ vc ‐
c
2
‐6
‐12
c
d
20
‐20
+400
e
16
‐14
+224
f
36
31
+1,116
total
0
a
ia
d
id
ic
‐
vb
+
ib
b
+
vf
‐
f
+
ve
‐
ie
if
e
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Assessment Problem 1.6
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Assessment Problem 1.7
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Passive Sign Convention Remarks
• Key idea is that the defining equations depend upon the
voltage polarity and current direction.
• Example: p = ±vi
• You must examine how the polarity of v and the direction of i
are labeled on the circuit diagram to determine the sign.
• Failure to follow the PSC will result in a wrong equation in the
early stages of circuit analysis.
• All of the subsequent calculations will be incorrect.
• Translation: lost points on quizzes, exams.
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