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Transcript 
Parallel Combinatio n of Admittanc es
(COMPLEX) ADMITTANCE
Y p  Yk
1
 G  jB (Siemens)
Z
G  conductanc e
Y
k
YR  0.1S
YC 
B  Suceptanc e
1
1
R  jX
R  jX


 2
Z R  jX
R  jX R  X 2
L
C
V  jLI
1
V
I
jC
Series Combinatio n of Admittanc es
0.1S
Element Phasor Eq. Impedance
V  RI
Y p  0.1  j1( S )
1
1

Ys k Yk
R
R2  X 2
X
B 2
R  X2
G
R
1
 j1( S )
 j1
ZR
Z  jL
Z
1
jC
Admittance
1
Y  G
R
1
Y
jL
Y  jC
 j 0.1S
1
1
1


Ys 0.1  j 0.1
 10  j10
(0.1)( j 0.1) 0.1  j 0.1

0 . 1  j 0 .1 0 . 1  j 0 .1
1
10  j10
Ys 

10  j10
200
Ys  0.05  j 0.05 S
Ys 
LEARNING EXAMPLE VS  6045(V )
LEARNING EXTENSION
FIND Y p , I
Y p  YR  YL
 0.5  j 0.25
Y p  0.5  j 0.5  j1  0.25  0.75  j 0.5( S )
Y p  0.901433.69( S )
2  j4
2  j4
Yp 
 0.5  j 0.25( S )
2  j4
j8
I  Y pV  0.901433.69 1020
I  Y pV  (0.5  j 0.25)  6045( A)
I  9.01453.79( A)
Zp 
I  0.559  26.565  6045( A)
I  33.5418.435( A)
LEARNING EXAMPLE
SERIES-PARALLEL REDUCTIONS
1
2  j4
 2
2  j 4 (2)  (4) 2
1
4  j2
Y34 

4  j2
20
Y2 
Z3  4  j 2
Y4   j 0.25  j 0.5  j 0.25
Z 4  1 / Y4   j 4
1 ( j 2)
1 j2
1
Z1 
1  j 0.5
Z1 
1  j 0.5
Z1 
1  (0.5) 2
Z1  0.8  j 0.4()
Z4 
j 4  ( j 2) 8

j4  j2
j2
Y2  0.1  j 0.2( S )
Y34  0.2  j 0.1
Z2  2  j 6  j 2  2  j 4
Z34  4  j 2
Z 234 
Y234  0.3  j 0.1( S )
Z 234 
1
1
0.3  j 0.1


Y234 0.3  j 0.1
0.1
Z 2 Z34
 3  j1
Z 2  Z 34
Zeq  Z1  Z234  3.8  j 0.6  3.8478.973
LEARNING EXTENSION FIND THE IMPEDANCE ZT
Z1  4  j 6  j 4
Z1  4  j 2
Y12  Y1  Y2
( R  P ) Z1  4.47226.565
Y1  0.224  26.565
( P  R)Y1  0.200  j 0.100
Y12  Y1  Y2  0.45  j 0.35
( R  P )Y12  0.570  37.875
Z12  1.75437.875
( P  R) Z12  1.384  j1.077
Z2  2  j 2 ( R  P ) Z2  2.82845
Y2  0.354  45
( P  R)Y2  0.250  j 0.250
1
4  j2
ZT  2  (1.384  j1077)  3.383  j1.077
Y1 
 2
4  j 2 (4)  (2) 2
1
2  j2
Y2 
 2
2  j 2 (2)  (2) 2
1
1
0.45  j 0.35
Z12 


Y12 0.45  j 0.35
0.325
1
Z12 
Y12
BASIC ANALYSIS USING KIRCHHOFF’S LAWS
PROBLEM SOLVING STRATEGY
For relatively simple circuits use
Ohm' s law for AC analysis; i.e., V  IZ
The rules for combining Z and Y
KCL AND KVL
Current and voltage divider
For more complex circuits use
Node analysis
Loop analysis
Superposit ion
Source transformation
Thevenin' s and Norton' s theorems
MATLAB
PSPICE
LEARNING EXAMPLE
COMPUTE ALL THE VOLTAGES AND CURRENTS
Compute I1
Use current divider for I2 , I3
Ohm' s law for V1 , V2
V1  690 I 2
Zeq  4  ( j 6 || 8  j 4)
V1  16.2678.42(V )
24  j 48 32  j8  24  j 48
Z eq  4 

8  j2
8  j2
V2  7.2815(V )
56  j56
79.19645

 9.60430.964()
8  j2
8.24614.036
V
2460
I1  S 
 2.49829.036( A)
Zeq 9.60430.964
Z eq 
j6
690
I1 
2.49829.036( A)
8  j2
8.24614.036
8  j4
8.944  26.565
I2 
I1 
2.49829.036( A)
8  j2
8.24614.036
I3 
I1  2.529.06
I 2  2.71  11.58
V2  4  90 I3
I3  1.82105
LEARNING EXTENSION
IF VO  845, COMPUTE VS
THE PLAN...
COMPUTE I3
COMPUTE V1
COMPUTE I2 , I1
COMPUTE VS
VO
( A)  445( A)
2
V1  (2  j 2) I3  8  45  445
I3 
V1  11.3140(V )
I2 
V1 11.3140

 5.657  90( A)
j2
290
I1  I 2  I3  5.657  90  445
I1   j5.657  (2.828  j 2.828)( A)
I1  2.828  j 2.829( A)
VS  2 I1  V1  2(2.828  j 2.829)  11.3140
VS  16.97  j5.658(V )
VS  17.888  18.439
ANALYSIS TECHNIQUES
PURPOSE: TO REVIEW ALL CIRCUIT ANALYSIS TOOLS DEVELOPED FOR
RESISTIVE CIRCUITS; I.E., NODE AND LOOP ANALYSIS, SOURCE SUPERPOSITION,
SOURCE TRANSFORMATION, THEVENIN’S AND NORTON’S THEOREMS.
COMPUTE I0
V2  60
V
 20  V2  2  0
1  j1
1  j1
 1
1 
6
V2 
1

2

1  j1
1  j1
1  j1
V2
1. NODE ANALYSIS
V1
V
V
 20  2  2  0
1  j1
1 1  j1
V1  V2  60
I0 
V2
( A)
1
(1  j1)  (1  j1)(1  j1)  (1  j1) 2(1  j1)  6

(1  j1)(1  j1)
1  j1
V2
4
 8  j2
1 j
V2 
(4  j )(1  j )
2
3
5
I 0    j ( A)
2
2
I0  2.92  30.96
NEXT: LOOP ANALYSIS
2. LOOP ANALYSIS
ONE COULD ALSO USE THE SUPERMESH
TECHNIQUE
SOURCE IS NOT SHARED AND Io IS
DEFINED BY ONE LOOP CURRENT
LOOP 1 : I1  20
I2
I 0   I3
LOOP 2 : (1  j )( I1  I 2 )  60  (1  j )( I 2  I3 )  0
LOOP 3 : (1  j )( I 2  I3 )  I3  0
CONSTRAINT : I1  I 2  20
SUPERMESH : (1  j ) I1  60  ( I 2  I 3 )  0
MUST FIND I3
2 I 2  (1  j ) I3  6  (1  j )(2)
(1  j ) I 2  (2  j ) I3  0
(1  j)

/* (1  j )
/* (2)
 2(2  j ) I3  (1  j )(8  2 j )
5 3
10  6 j
I


 j ( A)
I3 
0
2 2
4
2
MESH 3 : ( I 3  I 2 )  (1  j ) I 3  0
I 0  I 2  I3
NEXT: SOURCE SUPERPOSITION
Circuit with voltage source
set to zero (SHORT CIRCUITED)
SOURCE SUPERPOSITION
I
I L2
1
L
=
V
1
L
+
Circuit with current
source set to zero(OPEN)
Due to the linearity of the models we must have
I L  I L1  I L2
VL  VL1  VL2
Principle of Source Superposition
The approach will be useful if solving the two circuits is simpler, or more convenient, than
solving a circuit with two sources
We can have any combination of sources. And we can partition any way we find convenient
VL2
3. SOURCE SUPERPOSITION
I 0'  10( A)
Z '  (1  j ) || (1  j ) 
(1  j )(1  j )
1
(1  j )  (1  j )
COULD USE SOURCE TRANSFORMATION
TO COMPUTE I"0
V1"
Z"
Z"
"
 "
60(V ) I 0  "
60( A)
Z 1 j
Z 1 j
1 j
1 j
I 0" 
6
2

j
(
1

j
)

3

j
I 0" 
6 ( A)
1 j
6 6
"
1 j
I

 j ( A)
0
2 j
4 4
5 3 
I 0  I 0'  I 0"    j ( A)
2 2 
Z"
Z "  1 || (1  j )
NEXT: SOURCE TRANSFORMATION
THEVENIN’S EQUIVALENCE THEOREM
LINEAR CIRCUIT
May contain
independent and
dependent sources
with their controlling
variables
PART A
ZTH


RTH
vTH

i
a
vO
b
_

i
a
LINEAR CIRCUIT
vO
_
LINEAR CIRCUIT
May contain
independent and
dependent sources
with their controlling
variables
PART B
b
PART B
PART A
Phasor
Thevenin Equivalent Circuit
for PART A
vTH
Thevenin Equivalent Source
RTH
Thevenin Equivalent Resistance
Impedance
5. THEVENIN ANALYSIS
Voltage Divider
VOC 
10  6 j
1 j
(8  2 j ) 
2
(1  j )  (1  j )
ZTH  (1  j ) || (1  j )  1
8 2j
I0 
53j
( A)
2
NEXT: NORTON
LEARNING EXAMPLE
Find the current i(t) in steady state
The sources have different frequencies!
For phasor analysis MUST use source superpositio
SOURCE 1: FREQUENCY 10 r/s
SOURCE 2: FREQUENCY 20r/s
Principle of superposition
Frequency domain
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