DC circuits - Department of Electrical Engineering

advertisement
USC
Phasor Method
Aug 24, 2011
Outline
• Review of analysis of DC (Direct Current) circuits
• Analysis of AC (Alternating Current) circuits
– Introduction
– Challenge of analysis of AC circuits
• Phasor method
– Idea and concept
– Advantage
• Conclusions
• Next…
2
Review of Analysis of DC circuits
• DC circuits
di
u

L
Inductor:
dt
u0
Short
du
dt
i0
Open
Capacitor: i  C
u
Resistor: i 
R
L
t
•Pure Resistive
+
C
-
+
0
L
+
US
u i
R
C
US
R
3
Review of Analysis of DC circuits
• Complete solution for DC circuits
Unknown variable:
6 Voltages (b)
12 (2b)
6 Currents (b)
G
Constraint Equations:
As number of braches grows:
6 (b)
•TooElements:
many variables!
KCL: 4-1=3 (n-1)
•Too many equations!
6 (b)
Network:
+
E
–
KVL: 6-3=3 b-(n-1)
12 (2b)=12 (2b)
4
Review of Analysis of DC circuits
• Summary of DC circuits analysis methods
– Circuit simplification
• Equivalent transformation of resistors
• Equivalent
of sources
The
purposetransformation
of circuit analysis
method:
– •To
General
analytical
methods
reduce
the number
of variables and
• Node-voltage method (suitable for fewer nodes)
equations
• Mesh-current method (suitable of fewer meshs)
– Theorem
• Superposition (linear circuits)
• Thevenin and Norton equivalent
5
Introduction of AC circuits
• AC (Alternating current)
u i
• Why
AC? steady state analysis
Sinusoidal
– Generation, transmission, distribution
and consumption of electric energy
are all in steady state sinusoidal.
+
0

t
– Any signal can be thought of as
superposition of sinusoidal
signals.

f ( x)   an sin(n  n )
n 0
6
Introduction of AC circuits
• Challenge with analysis of AC circuit
u(t )  U s sin(t  s )
L u L (t )
+
u(t )
-
+
u L (t )  U L sin(t   L )
+ uC (t )
- C
di
u

L
Inductor:
dt
Capacitor: i  C
u
Resistor: i 
R
du
dt
uC (t ) operation
 U C sin(twith
 C )
+,-,*,/
trigonometric
isRnot
u R (t )  U function
) easy!
R sin(t  
RThe
KVL : u(t )  uL (t )  uC (t )
U S sin(t  S )  U L sin(t  L )  UC sin(t  C )
P  uC (t )iC (t )  U S sin(t  S ) I S sin(t  Si )
7
Review of Analysis of DC circuits
• Summary of DC circuits analysis methods
– Circuit simplification
• Equivalent transformation of resistors
• Equivalent transformation of sources
– General analytical methods
• Node-voltage method (suitable for fewer nodes)
• Mesh-current method (suitable of fewer meshs)
– Theorem
• Superposition (linear circuits)
• Thevenin and Norton equivalent
8
Introduction of AC circuits
9
Phasor Method
10sin(t  300 )  5 sin(t  600 )  20sin(t  450 )
Hint:
10
Phasor Method
Charles Proteus Steinmetz
•In 1893, he introduced the phasor method to
calculation of AC circuits
GE required him to submit a itemized invoice. They soon
received it. It included two items:
1. Marking chalk "X" on side of generator: $1.
2. Knowing where to mark chalk "X": $999.
German-American mathematician and engineer
(1865 – 1923)
11
Phasor Method
Trigonometric function
Phasor Domain
U
U sin(t   )
transform
10300
10sin(t  30 )
0
5 sin(t  600 )
Inverse
transform
5600
12
Phasor Method
Complex operation:
Sum/Subtraction:
(a1  jb1 )  (a2  jb2 )  (a1  a2 )  j (b1  b2 )
Multiplication/Division:
F11  F22  F1  F21  2 ;
F11
F2  2

F1
F2
1   2
13
Phasor Method
Time Domain
Phasor Domain
transform
Sinusoidal
expression
Phasor
(Complex)
Trigonometric
calculation
Complex
Operation
Inverse
transform
Result
(sinusoidal)
Result
(Phasor)
14
Phasor Method
Trigonometric
calculation
equivalent
Complex
Operation
10sin(t  300 )  5 sin(t  600 )
 10sin t cos300  10cost sin 300 5 sin t cos600  5 cost sin 600
0
0
0
0

(
10
sin
30

5
sin
60
) cost
 (10cos30  5 cos60 ) sin t
 a sin t  b cos t
0
0
a

10
cos
30

5
cos
60
 R sin(t   )
b  10sin 300  5 sin 600
15
Phasor Method
Trigonometric
calculation
equivalent
Complex
Operation
F  F cos  jF sin 
10sin(t  30 )  5 sin(t  60 )
10300  5600
 10cos300  j10sin 300  5 cos600  j5 sin 600
 (10cos300  5 cos600 )  j (10sin 300  5 sin 600 )
 a  jb
0
0
a

10
cos
30

5
cos
60
 R sin(t   )
 R
b  10sin 300  5 sin 600
b
2
2
R  a b ;
  arctan
a
0
0
16
Phasor Method
Example:
10sin(t  300 )  5 sin(t  600 )  20sin(t  450 )
10300  5600  20  450
 (8.66  j5)  (2.5  j 4.33)  (14.14  j14.14)
 25.3  j 4.81
 25.75  10.760
 25.75sin(t 10.760 )
17
Conclusions
• The trigonometric function involved in the
sinusoidal steady-state circuits is not
convenient to calculation.
• By projecting trigonometric function to phasor
domain, the calculation can be dramatically
simplified.
18
Quiz 1- problem1
Convert the following instantaneous currents to phasors, using cos(wt) as the reference.
Give your answer in polar form.
(1).
2).
19
20
Review of Analysis of DC circuits
• Summary of DC circuits analysis methods
– Circuit simplification
• Equivalent transformation of resistors
• Equivalent transformation of sources
– General analytical methods
• Node-voltage method (suitable for fewer nodes)
• Mesh-current method (suitable of fewer meshs)
– Theorem
• Superposition (linear circuits)
• Thevenin and Norton equivalent
21
Review of Analysis of DC circuits
• Summary of DC circuits analysis methods
– Circuit simplification
• Equivalent transformation of resistors
• Equivalent transformation of sources
– General analytical methods
• Node-voltage method (suitable for fewer nodes)
• Mesh-current method (suitable of fewer meshs)
– Theorem
• Superposition (linear circuits)
• Thevenin and Norton equivalent
22
•For the circuit shown below, compute the voltage across the load terminals.
0.1Ω
j0.5Ω
I=1250° A
+
+
-
240 0 ° V
LOAD
LOAD
-
0.1  j 0.5  0.5178.7
1250 * 0.5178.7
 63.7578.7
2400  63.7578.7
 240 12.5  j 62.5
 227.5  j 62.5
 235.93  15.36
23
USC
Power
Aug 24, 2011
Review of Phasor
Questions:
1. What is the main idea of Phasor method?
2.
10sin(t  300 )
a.1030
0
b.10  60
0
c.
10
  600
2
d.
10
300
2
25
Review of Phasor
R
L u L (t )
+
u(t )
-
+
-
-
+
u R (t )
C
+
-
uC (t )
u(t )  U s sin(t  s )
u L (t )  U L sin(t   L )
uC (t )  U C sin(t  C )
u R (t )  U R sin(t   R )
26
Power
Instantaneous Power
Average Power
Real Power
Active Power
Reactive Power
Complex Power
Apparent Power
27
Power
28
Power: Pure Resistive
29
Power: Pure Inductive
30
Power: Pure Capacitive
31
Average Power
32
Example 2.1
33
Complex Power
34
Power Triangle
35
Power Triangle
36
Download
Related flashcards

Mobile computers

37 cards

Power electronics

15 cards

Personal computers

49 cards

Radar

16 cards

Radio electronics

29 cards

Create Flashcards