Transcript Chapter 1 Introduction Scott 2008.9

Electrical Engineering and Electronics II
Chapter 1
Introduction
Scott
2008.9
•Main Contents
1. Recognize interrelationships of electrical
engineering with other fields of science
and engineering.
2. List the major subfields of electrical
engineering.
3. List several important reasons for
studying electrical engineering.
•Main Contents
4. Define current, voltage, and power,
including their units.
5. Calculate power and energy, as well as
determine whether energy is supplied or
absorbed by a circuit element.
6. State and apply basic circuit laws.
7. Solve for currents, voltages, and powers
in simple circuits.
1.1 Overview of Electrical Engineering
Electrical systems have two main objectives:

To gather, store, process, transport, and
present information

To distribute and convert energy between
various forms
•Electrical Engineering Subdivisions




Communication
systems
Computer systems
Control systems
Electromagnetics




Electronics
Photonics
Power systems
Signal processing
•Why Study Electrical Engineering?




To pass the Fundamentals of Engineering (FE)
Examination
So you can lead projects in your own field
To be able to operate and maintain electrical
systems
To communicate with electrical engineering
consultants
1.2 Circuits, Currents and Voltages
•Electrical Circuit
•国标符号
i
uR
u
uL
uC
a
+
E
R0
R
L
c
I
+
U
R
d
b
C
RB
+
VCC
_
IB
IC
RC
+
UBE
 I B UCE
_
+
VCC
_
•Electrical Current

Electrical current is the time rate of
flow of electrical charge through a
conductor or circuit element.

The units are amperes (A), which are
equivalent to coulombs per second
(C/s).
•Reference Directions
•Double-Subscript Notation for Currents
•Direct Current
Alternating Current
When a current is constant with time,
we say that we have direct current,
abbreviated as DC. On the other hand, a
current that varies with time, reversing
direction periodically, is called
alternating current, abbreviated as AC.
.
Sinusoidal current
三角波
方波
•Electrical Current
Actual direction：
电流的实际方向为正电荷的运动方向。
dq (t )
i (t ) 
dt
t
q (t )   i (t )dt  q (t0 )
t0
Example 1.1 Determining Current Given Charge
q(t )(C )
q( t )  0
2
t0
q( t )  2  2 e
0
100t
t0
t
400
dq(t )
Solution : i (t ) 
dt
0 t0
 200e 100t
i(t )  200e 100t
t 0
200
t0 0
t (ms)
•Voltages
The voltage associated with a
circuit element is the energy
transferred per unit of charge that
flows through the element. The
units of voltage are volts (V), which
are equivalent to joules per
coulomb (J/C).
•Voltages
•Voltages’ Directions
1.3 POWER AND ENERGY
p(t )  v (t )i (t )
Unit:
Watt
Unit:
Joule
t2
w   p(t )dt
t1
•Power
•Passive Reference Configuration
I
a
＋
U
b
－
P  UI
U  IR
The current reference enters the
positive polarity of the voltage.
•Function of Element电源与负载的判别方法
a
Passive reference direction:
p  ui
i
+
u
-
p
if
P  UI  0
then
positive
power
absorb
if
P  UI  0
then
negative
power
sup ply b
Non passive reference direction:
a
p  ui
i
if
P  UI  0
then
positive
power
absorb
if
P  UI  0
then
negative
power
supply b
p
u
+
Absorbing
energy,
being
charged
Supplying
energy,
being
discharged
Supplying
energy,
being
discharged
Calculate the power for each element. If it is a battery,
is it being charged or discharged?
•Example
如图电路，各元件上电压、电流参考方向采用关
联参考方向。确定各元件的功率，指出哪些是电源、
哪些是负载？
U1  30V, U 2  20V, U 3  60V, U 4  30V, U 5  80V
I1  3A, I 2  1A, I 3  2A, I 4  3A, I 5  1A
Solution：box 1
P1  U1  I1  30V  3A  90W>0
Box 2
U1
U2
1
2
Load: Absorb
4
P2  U 2  I 2  20V 1A  20W>0
Load: Absorb
U4
3
U3
5
U5
U1  30V, U 2  20V, U 3  60V, U 4  30V, U 5  80V
I1  3A, I 2  1A, I 3  2A, I 4  3A, I 5  1A
Box3
U1
U2
P3  U 3  I 3  60V  (2)A  120W<0
1
2
source
Box4
P4  U 4  I 4  30V  3A  90W>0
Box5
load
4
U4
3
U3
P5  U 5  I 5  80V  (1)A  80W<0
source
注 意：
P1  P2  P3  P4  P5  90  20  120  90  80  0
功率平衡！电路中所有元件的功率之和为 0 ！
常用作对分析结果的检验准则。
5
U5
•Concepts
支路——branch
节点——node
回路——loop
I1
I2
a
c
+ R1
E1
-
B=3
d
R2 +
R3
I3
b
E2
-
N=2
L=3
2Ω
6Ω
a
I 3Ω
4Ω
Otherwise，
2Ω
a
6Ω
b
10V
4Ω
3Ω
10V
Congsidering
current I，
N=4,B=6,L=6
N=3,B=5,L=6
1.4 KIRCHHOFF’S CURRENT LAW (KCL)

The net(网络，总的) current entering
a node is zero.

Alternatively, the sum of the currents
entering a node equals the sum of
the currents leaving a node.
I1
I2
a
c
d
+ R1
I1+ I2-I3=0
或 -I1- I2+I3=0
U1
-
R2 +
R3
I3
b
也可写成：
I1+ I2 = I3
（入） （出）
U2
-
•推广到广义节点上使用，任一闭合面
IA+IB+IC=0
IA
A
【Prove】
IAB
IA= IAB- ICA
IB= IBC- IAB
IC= ICA- IBC
IB
IC
Add all three equations：
IA+IB+IC=0
or
∑I=0
B
ICA
IBC
C
N、B、L=？ I、Uab=？
a
2Ω
+
6V
5Ω
+
12V
I
-
1Ω
N=2,B=3,L=2
I=0,Uab=0
b
-
1Ω
5Ω
•Example
图示电路中，已知I1=11mA，I4=12mA，I5=6mA。
求I2，I3和I6。
solution：
I6
I3=I1－I5=11－6=5（mA)
I2=I3－I4=5－12=－7（mA)
I2
R1
R2
I1
I6=I1－I2=11－(－7)=18（mA)
I6=I4+I5=12+6=18（mA)
I4
R3
I3
I5
•Series Circuits
Series: Elements are connected end to end, no other
element can be connected to their common node.
ia=ib=ic
The current is identical.
Use KCL to determine the values of the
unknown currents shown in Fig. 1.21
Answer:
ia=4A
ib=-2A
ic=-8A
Identify the groups of the circuit elements that
are connected in series shown in Fig. 1.21
Answer:
A、B
；
E、F、G
1.5 KIRCHHOFF’S VOLTAGE LAW(KVL)
The algebraic sum of the voltages equals zero
for any closed path (loop) in an electrical
circuit.
•Example
电路如图所示，已知UAB=5V，UBC=－4V，
UDA=－3V。试求：（1）UCD；（2）UCA
解： （1）
UAB+UBC+UCD+UDA=0
A
-
5+（－4）+UCD+（－3）=0 UDA
+
UCD=2V
D
（2）ABCA is not a closed loop
UAB+UBC+UCA=0
UAB - B
+
-
+
UBC
-
UCA
+
-
UCD
+
C
UCA=－（UAB+UBC）＝－1V
•Parallel Circuits
Two components are in parallel if both ends of one
element are directly connected to the corresponding ends
of the other element.
•Parallel Circuits
va=vb=-vc
The voltage is identical.
•Use repeated application of KVL to find the
values of vC、ve for the circuit of Fig. 1.29.
Answer:
vC=8V; ve=-2V
1.6 Introduction to Circuit Elements
导线
 Voltage source 电压源
 Current source 电流源
 Resistors 电阻器
 Conductors
•Conductors
Open circuit
An unbroken line
Short circuit
•Independent Voltage Source
Maintains a specified voltage across its terminal,
regardless of other currents or voltages of any other
elements.
Circle
enclosing
reference
polarity
•Voltage Sources in Series
Substitute several series voltage sources for
one equivalent voltage source.

+
+
US1
+ US
US2
+
+
US1
US
US2
+
+
US
-
+
US
-
U S  U S1  U S 2
U S  U S1  U S 2
Named as equivalent conversion
•Ideal Circuit Elements versus Reality
•Dependent/Controlled Voltage Source
2, unitless
3, unit of ohm
•Independent Current Source
Maintains a specified current flowing through itself,
regardless of other currents or voltages of any other
elements.
Circle
enclosing an
arrow
•Current Sources in Parallel
Substitute
several parallel current
sources for a equivalent current source.
IS
IS
1
2
IS
IS
1
2
US
IS U S
I S  I S1  I S 2
US
IS U S
I S  I S1  I S 2
•Dependent/Controlled Current Source
3, unit of Siemens
2, unitless
The Summary of Dependent Sources

Voltage-controlled voltage sources—VCVS

Current-controlled voltage sources—CCVS

Voltage-controlled current sources—VCCS

Current-controlled current sources—CCCS
•Resistors


The voltage v across an ideal/linear resistor is
proportional to the current i through the resistor.
The constant of proportionality is the resistance R.
L
R
A
R——resistance value
ρ——resistivity
L——length of conductor
A——cross sectional area
•Ohm’s Law

It shows the proportional relationship between
current and voltage.
Passive reference
configuration
v  iR
vab  iab R
i
1
R
0
v
•Conductance
1
G
R
i  Gv
•Power of Resistor
Absorbing
power：
2
v
p  vi  Ri 
R
2
As a DC circuit, we use capital letters.
DC Power：
V
I 
R
2
V
P  VI  RI 
R
2
1.7 Introduction to Circuits
•Circuit Analysis Using Arbitrary Reference
P1.48 Find the current iR, the power for each element in the
circuit. Which one is supplying power? Page 43
v2
iR=2A
v2=5iR=10V
v1
v1=v2+10=20V
电流源功率:
电压源功率:
电阻功率:
Pi=-2v1=-40W
Pv=10iR=20W
PR=5i2R=20W
Source
Load
Load
P1.50 Use KVL, KCL, and Ohm’s Law to find Vx. Page 43
Answer: Vx=17.5V
•Using KVL, KCL, and Ohm’s Law to Solve a Circuit
Example 1.7 Solve for the source voltage Vs in the circuit.
Solution:
15 V
iy 
3A
5
ix  0.5ix  i y
ix  2 A
vx  10ix  20 V
Vs  vx  15
Vs  35 V
•Homework 1





1.29
1.32
1.49
1.52
1.58
Teamwork 1


Page 32 Practical application 1.1
Using resistance to measure strain
Finding other practical application of
resistance in industry and daily life