Transcript Slide 1

Thevenin’s theorem, Norton’s
theorem and Superposition
theorem for AC Circuits
3/2/09
1
Problem10-60(b)
1. Find TEC at terminals c-d
C
10
20V
D
-j4
4
0
4A 0
j5
0
TEC: Thevenin Equivalent Circuit
2
Problem10-60(b)
(i) To obtain VTH
+ Voc -
Vc
10
20V
Vd
-j4
4
0
4A 0
j5
0
Voc =VTH
3
Problem10-60(b)
Nodal equations:
nod c:
Vc  20
10
nod d:

Vd  Vc
4j
Vc
5j


Vd
4
Vc  Vd
4j
0
4
4
Problem10-60(b)




 Find  Vc  Vd 
Vd 

Vc
Vcd  Vc  Vd
Vcd  9.615
VTH  Vcd




Vd 

Vc
 13.333  13.333i 


 8  5.333i 
Vcd  5.333  8i
 
arg Vcd  56.31 deg
VTH  5.333  8i
5
Problem10-60(b)
ZTH
(i) To obtain ZTH
10
-j4
j5
4
ZTH =[(10//j5)+4]//-j4
6
Problem10-60(b)
In Mathcad:
First we define a function to
decribe the result of two
parallel impedances:

ZP( x y ) 

ZTH  ZP ZP( 10  5i)  4  4i
x y
x y
ZTH  2.667  4i
ZTH  4.807


arg ZTH  56.31 deg
7
Problem10-60(b)
(iii) Using TEC to find Vo
Let
+
VTH
+
-
ZTH
ZL
ZL = 10+j10
ZTH
Vo 
VTH
Z L  ZTH
Vo
-
Let’s assume,
ZL  10  10j
ZL
ZTH  2.667  4i
Vo 
 VTH
ZTH  ZL
Vo  9.701
VTH  5.333  8i
Vo  2.353  9.412i
 
arg Vo  75.964 deg
8
Problem10-60(b)
2. Find NEC at terminals c-d
C
10
20V
D
-j4
4
0
j5
4A 0
0
NEC: Norton Equivalent Circuit
9
Problem10-60(b)
(i) To obtain IN
IN
V1
10
20V
-j4
4
0
4A 0
j5
0
10
Problem10-60(b)
(i) To obtain IN
V1  20
10

V1
5j

V1
4
 
V1  Find V1
IN 
20  V1
10
IN  2
4
V1  12.923  7.385i

V1
5j
IN  0.769  1.846i
 
arg IN  112.62 deg
11
Problem10-60(b)
(ii) To obtain ZN
ZN = ZTH
ZN = ZTH= 2.667 – j4
12
Problem10-60(b)
(iii) Using NEC to find Vo
Let’s assume,
+
IN
ZN
ZL
IN  0.769  1.846i

Vo
-
Vo = (ZN//ZL)IN
ZN  2.667  4i

Vo  ZP ZN  ZL  IN
Vo  9.701
ZL = 10+j10
ZL  10  10i
Vo  2.353  9.412i
 
arg Vo  75.964 deg
13
Problem10-60(b)
3. Find Vo using superposition theorem
Vo
+
10
10
20V
j10
-
-j4
4
0
4A 0
j5
0
14
Problem10-60(b)
(i) 4A acting alone
Vo1
+
10
V1
10
j10
V2
-j4
4
4A 0
j5
0
15
Problem10-60(b)
Nodal Equations
1
1
1
1

V1  
   V1  V2  


10
5j

4j
10

10j






V2
1
1


V2  V1   4j  10  10j   4





 Find  V1  V2
V2 

V1
Vo1  V1  V2


4

V2 

V1
0
 3.548  9.267i 


 7.167  0.869i 
Vo1  3.62  10.136i
16
Problem10-60(b)
(i) 20V acting alone
Vo2
+
10
V1
10
20V
j10
V2
-j4
4
0
j5
0
17
Nodal Equations
V1  20
10
V1
1
1

 0

  V1  V2  


5j

4j
10

10j


V2  V1   4j 
1




 Find  V1  V2
V2 

V1
Vo2  V1  V2
Vo  Vo1  Vo2
Vo  9.701
V2


10  10j 
4
1



V2 

V1
0
7.747  3.91i 



 1.774  4.633i 
Vo2  5.973  0.724i
Vo  2.353  9.412i
 
arg Vo  75.964 deg
18
Find vo
1W
2H
4W
+ vo 2sin 5t
0.1F
5V
10cos 2t
This circuit operates at the different frequencies:
• w = 0 ; for the DC Voltage source
• w = 2 rad/s ; for the AC voltage source
• w = 5 rad/s ; for the AC current source
19
We must use superposition theorem
Different frequencies problem reduces to
single-frequency problem.
vo  vo1  vo 2  vo3
Due to 5V
Due to 2sin 5t A
Due to 10cos 2t V
20
(i) 5V (w = 0) acting alone
1W
4W
+ vo1 5V
w=0;
jwL = 0 ;
Short-circuited
1/jwC = infinity ; Open-circuited
1
vo1 
(5)  1V
1 4
21
(ii) 10V (w = 2 rad/s) acting alone
Convert time-domain quantities to phasor quantities:
10 cos 2t
2H
0.1F
o
100
jwL  j 4
1
  j5
j wC
22
(ii) 10V (w = 2 rad/s) acting alone
j4 W
1W
+ vo2 100
1
V02 
(100o )
1 j4  Z
 2.498  30.79o
4W
-j5 W
Z   j 5 // 4
 2.439  j1.951
In time domain:
V02(t)=2.498cos(2t-30.79)
23
(iii) 2A (w = 5 rad/s) acting alone
Convert time-domain quantities to phasor quantities
5 sin 5t
2H
0.1F
o
50
jwL  j10
1
  j2
jwC
24
(ii) 2A (w = 5 rad/s) acting alone
I1
J10 W
1W
+ vo3 20
j10
I1 
(20o )
j10  1  Z
4W
-j2 W
Z   j 2 // 4
 0.8  j1.6
 2.32810o
V03  I1 1
 2.32810o
In time domain:
V03(t)=2.328sin(2t+10)
25
Total response of the circuit:
vo  vo1  vo 2  vo3
v0(t)= -1 + 2.498cos(-30.79)+2.328sin(5t+10)
Note that we can only add the individual responses
in the time domain, not in the phasor.
26
Total response of the circuit:
vo  vo1  vo 2  vo3
v0(t)= -1 + 2.498cos(-30.79)+2.328sin(5t+10)
Note that we can only add the individual responses
in the time domain, not in the phasor.
27
Find vo
4W
1/12 F
+
vo
6 cos 4t
2H
6W
1W
2cos (4t+30)
Case #1:
This circuit operates at a single frequency.
The sources are represented by a cosine function.
28
Convert time-domain quantities to phasor quantities:
w = 4 rad/s
o
60
6 cos 4t
o
o
2
30
2 cos(4t  30 )
2H
jwL  j 8
1/12 F
1
  j3
j wC
29
A phasor circuit
-j3 W
4W
60
+
vo
-
j8 W
6W
1W
230
30
Find vo
4W
1/12 F
+
vo
-
2H
6W
1W
2sin (4t+30)
6 sin 4t
Case #2:
This circuit operates at a single frequency.
The sources are represented by a sine function.
31
Convert time-domain quantities to phasor quantities:
w = 4 rad/s
o
60
6 sin 4t
o
o
2
30
2 sin( 4t  30 )
2H
jwL  j 8
1/12 F
1
  j3
j wC
We use the sine function as the reference
for the phasor.
32
A phasor circuit
-j3 W
4W
60
+
vo
-
j8 W
6W
1W
230
33
Find vo
4W
1/12 F
+
vo
-
2H
6W
1W
2sin (4t+30)
6 cos 4t
Case #3:
This circuit operates at a single frequency.
One source is represented by a sine function;
Another source is represented by cosine function.
34
Convert time-domain quantities to phasor quantities.
Use the cosine function as a reference.
w = 4 rad/s
o
6
0
6 cos 4t
o
o
o
2


60
2 cos(4t  30  90 )
2H
jwL  j 8
1/12 F
1
  j3
j wC
35
OR
Convert time-domain quantities to phasor quantities.
Use the sine function as a reference.
w = 4 rad/s
6 sin( 4t  90 )
o
2 sin( 4t  30 )
o
2H
1/12 F
6  90o
230 o
jwL  j 8
1
  j3
j wC
36