Post on 14-Jan-2016
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Thevenin’s theorem, Norton’s Thevenin’s theorem, Norton’s theorem and Superposition theorem and Superposition
theorem for AC Circuitstheorem for AC Circuits
3/2/09
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Problem10-60(b)Problem10-60(b)
20V
D
0
j5
C
4A 0
4
10
0
-j4
1. Find TEC at terminals c-d
TEC: Thevenin Equivalent Circuit
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Problem10-60(b)Problem10-60(b)
(i) To obtain VTH
20V
0
j5 4A
Vc
0
4
10
0
Vd
-j4
+ Voc -
Voc =VTH
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Problem10-60(b)Problem10-60(b)
nod c: Vc 20
10
Vc
5j
Vc Vd
4j 0
nod d: Vd Vc
4j
Vd
4 4
Nodal equations:
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Problem10-60(b)Problem10-60(b)
Vc
Vd
Find Vc Vd Vc
Vd
13.333 13.333i
8 5.333i
Vcd Vc Vd Vcd 5.333 8i
Vcd 9.615 arg Vcd 56.31deg
VTH Vcd VTH 5.333 8i
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Problem10-60(b)Problem10-60(b)
j5
10
4
-j4
(i) To obtain ZTHZTH
ZTH =[(10//j5)+4]//-j4
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Problem10-60(b)Problem10-60(b)
In Mathcad:
First we define a function to decribe the result of two parallel impedances:
ZP x y( )x yx y
ZTH ZP ZP 10 5i( ) 4 4i ZTH 2.667 4i
ZTH 4.807
arg ZTH 56.31 deg
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Problem10-60(b)Problem10-60(b)
(iii) Using TEC to find Vo
+- ZL
ZTHVTH
+
Vo
-
ZL = 10+j10Let
THTHL
THo V
ZZ
ZV
ZL 10 10j ZTH 2.667 4i VTH 5.333 8i
Vo
ZL
ZTH ZLVTH Vo 2.353 9.412i
Vo 9.701 arg Vo 75.964deg
Let’s assume,
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Problem10-60(b)Problem10-60(b)
20V
D
0
j5
C
4A 0
4
10
0
-j4
2. Find NEC at terminals c-d
NEC: Norton Equivalent Circuit
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Problem10-60(b)Problem10-60(b)
(i) To obtain IN
20V
0
j5 4A 0
4
10
0
-j4
IN
V1
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Problem10-60(b)Problem10-60(b)
(i) To obtain IN
V1 20
10
V1
5j
V1
4 4
V1 Find V1 V1 12.923 7.385i
IN
20 V1
10
V1
5j IN 0.769 1.846i
IN 2arg IN 112.62deg
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Problem10-60(b)Problem10-60(b)
(ii) To obtain ZN
ZN = ZTH
ZN = ZTH= 2.667 – j4
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Problem10-60(b)Problem10-60(b)
(iii) Using NEC to find Vo
ZLZN
IN
+
Vo
-
ZL = 10+j10
Let’s assume,
Vo = (ZN//ZL)IN
IN 0.769 1.846i ZN 2.667 4i ZL 10 10i
Vo ZP ZN ZL IN Vo 2.353 9.412i
Vo 9.701 arg Vo 75.964deg
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Problem10-60(b)Problem10-60(b)
3. Find Vo using superposition theorem
20V
j10+
0
10
j5 4A 0
4
Vo
10
0
-j4
-
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Problem10-60(b)Problem10-60(b)
(i) 4A acting alone
j10+
0
10
j5 4A
V2
0
4
Vo1
10
V1
-j4
-
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Problem10-60(b)Problem10-60(b)
Nodal Equations
V11
10
1
5j
V1 V2 1
4j1
10 10j
0
V2 V1 1
4j1
10 10j
V2
4 4
V1
V2
Find V1 V2 V1
V2
3.548 9.267i
7.167 0.869i
Vo1 V1 V2 Vo1 3.62 10.136i
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Problem10-60(b)Problem10-60(b)
(i) 20V acting alone
j10+
0
10
j520V
V2
4
Vo2
10
V1
0
-j4
-
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Nodal Equations
V1 20
10
V1
5j V1 V2 1
4j1
10 10j
0
V2 V1 1
4j1
10 10j
V2
4 0
V1
V2
Find V1 V2 V1
V2
7.747 3.91i
1.774 4.633i
Vo2 V1 V2 Vo2 5.973 0.724i
Vo Vo1 Vo2 Vo 2.353 9.412i
Vo 9.701 arg Vo 75.964deg
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Find vo
2H
10cos 2t
+ vo -
1 4
0.1F 5V2sin 5t
This circuit operates at the different frequencies:
• = 0 ; for the DC Voltage source• = 2 rad/s ; for the AC voltage source• = 5 rad/s ; for the AC current source
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We must use superposition theorem
Different frequencies problem reduces to single-frequency problem.
321 oooo vvvv
Due to 5V Due to 2sin 5t A
Due to 10cos 2t V
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(i) 5V ( = 0) acting alone
+ vo1 -
1 4
5 V
= 0 ; jL = 0 ; Short-circuited
1/jC = infinity ; Open-circuited
Vvo 1)5(41
11
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(ii) 10V ( = 2 rad/s) acting alone
2H
0.1F
t2cos10 o0104jLj
51
jCj
Convert time-domain quantities to phasor quantities:
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(ii) 10V ( = 2 rad/s) acting alone
+ vo2 - 4
1 j4
-j5 100
951.1439.2
4//5
j
jZ
o
o02
79.30498.2
)010(41
1
Zj
V
In time domain:
V02(t)=2.498cos(2t-30.79)
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(iii) 2A ( = 5 rad/s) acting alone
2H
0.1F
t5sin5 o0510jLj
21
jCj
Convert time-domain quantities to phasor quantities
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(ii) 2A ( = 5 rad/s) acting alone
6.18.0
4//2
j
jZ
o
o1
10328.2
)02(110
10
Zj
jI
In time domain:
V03(t)=2.328sin(2t+10)o
103
10328.2
1
IV
J10
1 4
-j2
+ vo3 -
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Total response of the circuit:
Note that we can only add the individual responsesin the time domain, not in the phasor.
v0(t)= -1 + 2.498cos(-30.79)+2.328sin(5t+10)
321 oooo vvvv
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Total response of the circuit:
Note that we can only add the individual responsesin the time domain, not in the phasor.
v0(t)= -1 + 2.498cos(-30.79)+2.328sin(5t+10)
321 oooo vvvv
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Find vo
2H
2cos (4t+30)
+vo
-
6
1/12 F
6 cos 4t
Case #1:
This circuit operates at a single frequency.
The sources are represented by a cosine function.
1
4
29
2H
1/12 F
)304cos(2 ot
o06
8jLj
31
jCj
Convert time-domain quantities to phasor quantities:
t4cos6o302
= 4 rad/s
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A phasor circuit
230
+vo
-
6
-j3
1
j8 4
60
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Find vo
2H
2sin (4t+30)
+vo
-
6
1/12 F
6 sin 4t
Case #2:
This circuit operates at a single frequency.
The sources are represented by a sine function.
1
4
32
2H
1/12 F
)304sin(2 ot
o06
8jLj
31
jCj
Convert time-domain quantities to phasor quantities:
t4sin6o302
= 4 rad/s
We use the sine function as the reference for the phasor.
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A phasor circuit
230
+vo
-
6
-j3
1
j8 4
60
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Find vo
2H
2sin (4t+30)
+vo
-
6
1/12 F
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.
1
4
35
2H
1/12 F
)90304cos(2 oo t
o06
8jLj
31
jCj
Convert time-domain quantities to phasor quantities.Use the cosine function as a reference.
t4cos6o602
= 4 rad/s
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2H
1/12 F
)304sin(2 ot
o906
8jLj
31
jCj
Convert time-domain quantities to phasor quantities.Use the sine function as a reference.
)904sin(6 oto302
= 4 rad/s
OR