Chapter 1
Circuit Variables
ECE 202: Chapter 1
1
Clemson University
Motivation
Basis for future courses
Foundational to Electrical Engineering and
Computer Engineering
Used for actual circuits and circuit models
ECE 202: Chapter 1
2
Clemson University
Chapter 6
p
Inductors and Capacitors
ECE 202: Chapter 6
1
Clemson University
Inductors
Based on Magnetic Field Phenomena
Stores energy *
Passive Element
*Moving Charge Current
If i varies with time, the magnetic field varies with time
For circuit anal
Chapter 4
Analysis Techniques
ECE 202: Chapter 4
1
Clemson University
The Heart of the Course
Nodal and Mesh analysis techniques are used to solve
circuits
Planar Circuits: Can be drawn on a plane with no
crossing branches
Planar
Nodal, Mesh
ECE 202: Chap
Chapter 7
p
1st Order RL and RC Circuits
ECE 202: Chapter 7
1
Clemson University
General Response of RL Circuits
_
VL
+
VR
+
Is
Inductor is Short
Circuit for DC
_
Steady State,
t<0
Steady State Solution/Response
Switch closes
Time signals are constant or
WELCOME TO
ECE 202: Electric Circuits I
ECE 202: Introduction
1
Clemson University
Course Syllabus on Blackboard
y
Class Time:
Tues.-Thurs.
Wait15 min.
Instructors:
Dr. William R. Harrell
Office: 205 Riggs Hall
rod.harrell@ces.clemson.edu
656 5918
656-591
Chapter 8
RLC Circuits
ECE 202: Chapter 8
1
Clemson University
Parallel RLC Circuits
Natural Response
V0, I0 Initial Voltage and Current
v is same for all elements
Write equation in terms of v, by i leaving top node
ic + iL + iR = 0
KCL
dv 1 t
v
C + vdt
Chapter 4
p
Analysis Techniques
ECE 202: Chapter 4
1
Clemson University
The Heart of the Course
Nodal and Mesh analysis techniques are used to solve
circuits
Planar Circuits: Can be drawn on a plane with no
crossing branches
Planar
Nodal, Mesh
ECE 202: Ch
Chapter 9
p
Sinusoidal Steady-State
ECE 202: Chapter 9
1
Clemson University
Brief Summary
y
AC Instead of DC.
Sinusoidal Inputs Instead of Constant Inputs.
Can Still Use Techniques Of Chapters 1 4.
q
p
Must Use Complex Numbers
Everything Else is Almo
Chapter 4
Analysis Techniques
ECE 202: Chapter 4
1
Clemson University
The Heart of the Course
Nodal and Mesh analysis techniques are used to solve
circuits
Planar Circuits: Can be drawn on a plane with no
crossing branches
Non Planar
Nodal Only
Planar
Nod
Chapter 8
RLC Circuits
ECE 202: Chapter 8
1
Clemson University
Parallel RLC Circuits
+
V (t )
+
V (t )
Natural Response
V0, I0 Initial Voltage and Current
+
V (t )
(t)
v is same for all elements
V0 = v(0+ )
Write equation in terms of v, by i leaving top n
Chapter 9
Sinusoidal Steady-State
ECE 202: Chapter 9
1
Clemson University
Brief Summary
AC Instead of DC.
Sinusoidal Inputs Instead of Constant Inputs.
Can Still Use Techniques Of Chapters 1 4.
Must Use Complex Numbers
Everything Else is Almost the S
Chapter 7
1st Order RL and RC Circuits
ECE 202: Chapter 7
1
Clemson University
General Response of RL Circuits
_
VL
+
VR
+
Is
Inductor is Short
Circuit for DC
_
Steady State,
t<0
Steady State Solution/Response
Switch closes
Time signals are constant or p
Chapter 3
p
Resistive Circuits
ECE 202: Chapter 3
1
Clemson University
Series Connection
Two Elements Connected at a Single Node
i1 = i2
KCL
C
vs = is ( R1 + R2 + R3 + R4 + R5 + R6 + R7 )
Resistors in series have the same current
vs = is Req
Req = R1 + R
Chapter 6
Inductors and Capacitors
ECE 202: Chapter 6
1
Clemson University
Inductors
Based on Magnetic Field Phenomena
Stores energy *
Passive Element
*Moving Charge Current
If i varies with time, the magnetic field varies with time
For circuit analys
Chapter 3
Resistive Circuits
ECE 202: Chapter 3
1
Clemson University
Series Connection
Two Elements Connected at a Single Node
+
R1
+
i1
+
R2
i1 = i2
i2
+
+
+
+
KCL
+
+
vs= is ( R1 + R2 + R3 + R4 + R5 + R6 + R7 )
Resistors in series have the same curren
Chapter 1
Supplemental Notes
ECE 202: Chapter 1
1
Similar to Problem 1.7
I = 1400( A)
= 1.6022 1019 (C )
q
= 0.6(cm) 0.006(m)
h =
= 9(cm) 0.09(m)
W =
h
n = 10
29
(
e
m3
)
w
I
Uniform Copper Bus Bar
Solution:
=
I
Find ve
Charge Density Velocity Area
( qn )
Chapter 3
Supplemental on Current
Division
ECE 202: Chapter 3
1
Alternate Treatment of Current and Voltage Division:
Parallel
R2 ( R1 )
R | R
= is= 1 2 is
i1
R1
( R1 + R2 ) ( R1 )
R1 ( R2 )
R | R
= is= 1 2 is
i2
R2
( R1 + R2 ) ( R2 )
R2
v2 = v s
R1 + R2
S
Chapter 3
Supplemental Notes
ECE 202: Chapter 3
1
Example:
Voltage Divider / Common Connection:
v1= 12 V
v2 = 5 V
v3= 12 V
What is the Voltage Across
Each Resistor?
R2
R2
VR2 =
24V
24V
=
RT
R1 + R2 + R3
I
24V
=
PT
RT
( 24V )
RT
2
Chapter 2: Circuit Elements
Developing models, which provide an
understanding that is imperfect, but adequate,
for solving practical problems lies at the heart of
engineering.
ECE 202: Chapter 2
1
Clemson University
Sources: Ideal Sources
Ideal Voltage So
Chapter 6:
Trapped Energy Example
Trapped Energy Example
Vb = 100e
5H
20 H
i1 (0) = 6( A)
i2 (0) = 4( A)
Results
i1 (t ) = 1.6e 2.5t + 4.4( A)
(V )
t0
Vb
16 H
i0 (t ) = 2e 2.5t ( A)
2.5t
L0 = 16 H
L1 = 5 H
L2 = 20 H
i2 (t ) = 0.4e 2.5t 4.4( A)
1 2
(t ) =
Chapter 7
Supplemental Notes
ECE 202: Chapter 3
1
You
Dr. Wont name any names.
TA
Another Use of the Time Constant:
Natural Response
Tangent to Curve at t=0
Look at
di
at t = 0+ :
dt
( )
I
di +
1
I 0 e t /
0
0 = =
dt
t = 0+
Slope, Initially.
So, Equati
Chapter 1
p
Circuit Variables
ECE 202: Chapter 1
1
Clemson University
Motivation
Basis for future courses
Foundational to Electrical Engineering and
Computer Engineering
Used for actual circuits and circuit models
ECE 202: Chapter 1
2
Clemson Universit
Chapter 6
Inductors and Capacitors
ECE 202: Chapter 6
1
Clemson University
Inductors
Based on Magnetic Field Phenomena
Stores energy *
Passive Element
*Moving Charge Current
If i varies with time, the magnetic field varies with time
For circuit analys
Chapter 2: Circuit Elements
Developing
D l i models, which provide an
d l hi h
id
understanding that is imperfect, but adequate,
for solving practical problems lies at the heart of
engineering.
engineering.
ECE 202: Chapter 2
1
Clemson University
Sources:
ECE-202-002
ANSWERS TO HOMEWORK SET #1
Due Date: Thursday, August 28, 2008
Show all of your work in neat, organized detail.
Problem
Answer
1.1
110 giga-watt hour
1.3
(a) 0.62 3-minute songs or 111.6 seconds of music
(b) 2480 bytes
1.4
108.5 minutes of vid