Transcript EE2003 Circuit Theory

九十七學年度第一學期 電路學（一）授課綱要
課本：
Fundamentals of Electric Circuits, 3rd edition.
by Charles K. Alexander and Matthew N. O. Sadiku
計分方式：
平時：40 % （出席：20%，小考、演習課：20%）
期中考：30 %
期末考：30 %
教師：宋皇輝
H324
分機：2178
內容：
•Basics Concepts
•Basic Laws
•Methods of Analysis
•Circuit Theorems
期中考
•Operational Amplifiers
•Capacitors and Inductors
•First-Order Circuits
•Second-Order Circuits
期末考
電路學(一)
Chapter 1
Basic Concepts
3
Basic Concepts - Chapter 1
1.1
1.2
1.3
1.4
1.5
1.6
1.7
Systems of Units.
Electric Charge.
Current.
Voltage.
Power and Energy.
Circuit Elements.
Applications.
4
1.1 System of Units (1)
Six basic units
Quantity
Basic unit
Length
meter
Mass
kilogram
Time
second
Electric current(電流) ampere
Thermodynamic
kelvin
temperature
Luminous intensity
candela
Symbol
m
Kg
s
A
K
cd
5
1.1 System of Units (2)
The derived units commonly used in electric circuit theory
Decimal multiples and
submultiples of SI units
6
1.2 Electric Charges
• Charge(電荷) is an electrical property of
the atomic particles of which matter
consists, measured in coulombs (C).
• The charge e on one electron is negative
and equal in magnitude to 1.602  10-19 C
which is called as electronic charge. The
charges that occur in nature are integral
multiples of the electronic charge.
• The law of conservation of charge.
7
1.3 Current (1)
• Electric current i = dq/dt. The unit of
ampere can be derived as 1 A = 1 C/s.
• A direct current (dc)(直流電流) is a current
that remains constant with time.
• An alternating current (ac)(交流電流) is a
current that varies sinusoidally with time.
(reverse direction)
8
1.3 Current (2)
• The direction of current flow
Positive ions
Negative ions
9
1.3 Current (3)
Example 1
A conductor has a constant current of 5 A.
How many electrons pass a fixed point on
the conductor in one minute?
Solution
Total no. of charges pass in 1 min is given by
5 A = (5 C/s)(60 s/min) = 300 C/min
Total no. of electronics pass in 1 min is given
300 C/min
21

1
.
87

10
electrons/ min
19
1.602 10 C/electron
10
1.3 Current (4)
Example 2
Determine the total charge entering a
terminal between t = 1 s and t = 2 s if the
current passing the terminal is
i = (3t2 – t) A
Q=
ò
2
t=1
idt =
ò
1
2
2
æ3 t ÷
ö
ç
(3t - t )dt = çt = 5.5 C
÷
÷
çè
2 ø1
2
2
1.4 Voltage (1)
• Voltage (or potential difference) is the energy (能量)
required to move a unit charge through an element,
measured in volts (V).
• Mathematically,
vab  dw / dq
(volt)
– w is energy in joules (J) and q is charge in coulomb (C).
• Electric voltage, vab, is always across the circuit
element or between two points in a circuit.
– vab > 0 means the potential of a is higher than potential of b.
– vab < 0 means the potential of a is lower than potential
of b.
12
1.5 Power and Energy (1)
• Power is the time rate of expending or absorbing
energy, measured in watts (W).
dw dw dq


 vi
• Mathematical expression: p 
dt
dq dt
i
i
+
+
v
v
–
–
Passive sign convention
p = +vi
p = –vi
absorbing power
supplying power
13
1.5 Power and Energy (2)
• The law of conservation of energy
p0
• Energy is the capacity to do work, measured
in joules (J).
• Mathematical expression
t
t
t0
t0
w   pdt   vidt
14
1.6 Circuit Elements (1)
Active Elements
Passive Elements
• A dependent source is an active
element in which the source quantity
is controlled by another voltage or
current.
Independent Dependant
sources
sources
• They have four different types: VCVS,
CCVS, VCCS, CCCS. Keep in minds the
signs of dependent sources.
15
1.6 Circuit Elements (2)
Example 3
Obtain the voltage v in the branch shown in the figure for i2 = 1A.
17
1.6 Circuit Elements (3)
Solution
Voltage v is the sum of the current-independent
10-V source and the current-dependent voltage
source vx.
Note that the factor 15 multiplying the control
current carries the units Ω.
Therefore, v = 10 + vx = 10 + 15(1) = 25 V
18
1.6 Circuit Elements (4)
Example 4
Calculate the power supplied or absorbed by each element in the figure.
p1 = 20(- 5) = - 100 W Supplied power
p2 = 12(5) = 60 W
Absorbed power
p3 = 8(6) = 48 W
Absorbed power
p4 = 8(- 0.2I ) = 8(- 0.2 ´ 5) = - 8 W Supplied power
p1 + p2 + p3 + p4 = - 100 + 60 + 48 - 8 = 0
1.7 Applications (1)
TV Picture Tube
Example 5
The electron beam in a TV picture carries 1015 electrons per second. As a
design engineer, determine the voltage Vo needed to accelerate the
electron beam to achieve 4 W.
dq
dn
i=
=e
= (- 1.6 ´ 10- 19 )(1015 ) = - 1.6 ´ 10- 4 A
dt
dt
p = V 0i
or
p
4
V0 = =
= 25000 V
- 4
i
1.6 ´ 10
1.8 Problem Solving (1)
•Carefully Define the problem.
•Present everything you know about the problem.
•Establish a set of Alternative solutions and
determine the one that promises the greatest
likelihood of success.
•Attempt a problem solution.
•Evaluate the solution and check for accuracy.
•Has the problem been solved Satisfactorily?