Transcript DC circuits - Department of Electrical Engineering

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