elec 3202 chap 4
TRANSCRIPT
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Chapter (4)
Additional Analysis TechniquesHere we will study four additional Techniques
• Superposition
• Source transformation• Thevenin and Norton Theorems
• Maximum power principle
1. Superposition :Definition :
Whenever a linear circuit is excited by more than one independent
source, the total response is the algebraic sum of individual responses
The idea is to activate one independent source at a time to
get individual response.
Then add the individual response to get total response
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Note:
1. Dependent source are Never deactivated (always active)
2. When an independent voltage source is deactivated, it is set to zero.replaced by short circuit
3. When an independent current source is deactivated, it is set to zero.
replaced by open circuit
Example:
Use superposition to find i1,i2,i3,i4 ?
+
-
Ω4
Ω2Ω6
Ω3 A12120 V
i1
i2
i3
i4
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+
-
Ω4
Ω2Ω6
Ω3120 V
'i1
'i2
'i3
'i4
V1
open circuit for
current source
•Activate independent voltage source 120 V only
•Using KCL at V1 (nodal analysis)
0i'i'i' 321 =−−
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V30V
06
1
3
1
6
1V20
0
42
V
3
V
6
V120
1
1
111
=⇒
=
++−
=
+
−−−
A56
Vii'
A10
3
30
3
Vi'
A156
90
6
V120i'
143
12
11
===
===
==−
=
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Ω4
Ω2Ω6
Ω3
''i1
''i2
''i3
''i4
V3
short circuit for
voltage source 12 A
V 4
* Activate the independent current source only
(1)0V3V6-
0)V(V3V2V
02
VV
3
V
6
V
0i"i"i"
43
333
4333
321
KK
=+
=−−−−=
−
−−
−
=−−KCL at V3:
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0124
V2
VV
012i"i"
443
43
=−−−
=−−
(2)48V3V2
48VV2V2
43
443
KK=−
=−−
KCL at V4:
V24V
V12V
4
3
−=−=
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A1-65i"i'i
A1165i"i'i
A6410i"i'i
A17215i"i'i
444
333
222
111
=−=+=
=+=+=
=−=+=
=+=+=
A6
4
24
4
Vi"
A62
2412
2
VVi"
A43
12
3
Vi"
A26
12
6
Vi"
44
433
3
2
31
−=−
==
=+−
=−
=
−=−==
==−
=
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Use super position to find V0 ?
Activate voltage source only:
Example :
Ω01Ω02 A5
10 V+
-
Ω5
Φi
ΦV0.4
ΦV
+
-
+
-
oV
Φi2
Ω01Ω02
10 V+
-
Ω5
Φi'
ΦV'0.4
ΦV'
+
-
+
-
oV'
Φi'2
X
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0V'V'4V'
V'4)V'(0.410V'
ΦΦΦ
ΦΦΦ
=⇒=
==
V8V'
25
2010V'
0
0
=
=
Dependent current source is open
Ω5
+
-oV'
+
-
Ω20
10 V
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Activate independent current source only:
Ω01Ω
02
Ω5
Φi''
ΦV''0.4
ΦV''
+
-
+
-
oV''
Φi''2
y
5 A
Z
KCL at node (y):
(1)0V"8V"5-
0V"8V"V"4-
0V"0.420
V"
5
V-"
Φ0
Φ00
Φ
00
KK=+
=+−=+−
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V8-168V"V'V
V16
5
80-V"
5
8V"
V10V"5V"0.5
0V"0.410
V"5
000
Φ0
ΦΦ
Φ
Φ
=−=+=
−===∴
−=⇒−=
=++
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Example:
Consider the independent source only
Use superposition to find V ?
100 V+
-
4 A
Ω12
Ω2Ω5
Ω10+
-
V
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V59.23V'2214
1
10
1
5
1V'
014
V'
10
V'
5
V'22
014
V'
10
V'
5
V'100
0iii 321
=⇒=
++
=−−−
=−−−
=−−
Apply KCL at node (x) :
100 V+
- Ω12
Ω2Ω5
Ω10
+
-
'V
i3i1
i2
X
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Consider the independent source only.
4 A
Ω12
Ω2Ω5
Ω10
+
-
''V
4 A
Ω2Ω5
Ω10
+- ''V
Ω12
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Current divider
V509.2359.23V"V'V
V9.233
10iV"
A2.769
23
1012
12A4i
x
x
=−=+=
−=
−=
=
++=
4 A
Ω2Ω
310
Ω12
+- ''V
ix
iy
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2. Source Transformation:
A transformation that allow a voltage source in serieswith a resistor to be replaced by a current source in parallel
with the same resistor or vice versa
How?
We need to find Is and Vs such that VL and IL is the same in
both circuits
VL
R
R L
+
-
Vs
IL
+
-
a
bKCT 1
+
-
VL
IL
a
b
IS
KCT 2
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sL
L IR R
R
I +=
ss IR V =
In KCT 2,
For IL to be the same , we need
In KCT 1 ,
R R
VI
L
sL +=
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R
VS
+
-
a
b
R
VI s
s =
Where
ss IR V = or
a
b
IS R
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Example :
Using source transformation, find the power associated with the
6 V source.
1. Consider the 40 V source in series with (5Ω)
+
-
40 V+
-
Ω4 Ω5
Ω02Ω03
Ω6
6 V
Ω01
+ -
Ω4
Ω5Ω02Ω03
Ω6
6 V
Ω01
A85
40=
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2. Take (5// 20 Ω)
3. Consider 8A in parallel with (4Ω)
+
-
Ω4
Ω4
02//5
=Ω03
Ω6
6 V
Ω01
A8
+ -
Ω4
Ω03
Ω6
6 V
Ω01
A4
+ -
(8 A)(4)=
32 V
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4. Take (4+6+10) in series
5. Consider 32 V in series with (20Ω)
+
-
Ω
4
Ω036 V
Ω20
+
- 32 V
+
-
Ω4
Ω02Ω036 V A6.120
32
=
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6. Take (30//20 Ω)
7. Consider 1.6A in parallel with (15 Ω)
+
-
Ω4
Ω12
20//03
=
6 VA6.1
+
-
Ω4
6 V
Ω12
+
-1.6 A (12)
=19.2 V
i
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)(absorbingW4.95P
(0.825)6ivPA0.825124
619.2i
6v
6V
=
==⇒=+
−=
Example :
Use source transformation to find V0
+
-
Ω52
Ω001
Ω5
250 V
A8
+
-
Vo Ω15
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2. Consider (250 V) in series with (25 Ω)
+
-
Ω52
Ω66.16
250 V
A8+
-
Vo
1. Take (5//15)//100 = 6.66 Ω
Ω52 Ω66.16A8+
-
Vo
A10
25
250=
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V20Ω)(10A)(2R iV0 ===
3. Find equivalent
A2810i =−=
Ω10
25//16.66R
=
=+
-
Vo
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Example:
Use source transformation to find V0
+
-
60 V
Ω8
Ω5
Ω6.1
Ω02
Ω6120 V
+
-+
-
Vo
36 A
Ω8
Ω5
Ω6.1
Ω02Ω6
+
-
Vo
36 AA12
5
60=A6
20
120=
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( ) ( ) V48Ω8A6R iV
A6(30)81.62.4
2.4i
R R R
R i
320
321
12
===
=++
=++
=
Ω8
Ω6.1
+
-
VoR 1
i = 36 + 6 - 12= 30 A
Ω2.4
6//5//20
=
R 2
R 3
i1 i2
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Example : Use source transformation to find V0?
Consider (10V) in series with (1Ω)
+
Vo
--
2 A
+
-
Vs
10V 1 ohm
1 ohm1 ohm
1 ohm
1 ohm
1 ohm10 A 2 A1 ohm
1 ohm1 ohm
1 ohm
+
Vo
--
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Take (1//1)=0.5
Consider (10A) in parallel with (0.5 Ω)
1 ohm
1 ohm1 ohm
1/2 ohm 2 A10 A
+
Vo
--
+
-
5 V
1/2 ohm
2 A
1 ohm1 ohm
1 ohm
+
Vo
--
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Take 0.5Ω in series with 1 Ω
Consider 5V in series with 1.5 Ω
1 ohm
1 ohm
2 A
1.5 ohm
+
-
5 V
+
Vo
--
+
Vo
--
1.5 ohm10/3 A 2 A
1 ohm
1 ohm
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Add the current sources
7. Take (4/3A) in parallel with (3/2 Ω)
( ) V4/725.111
1V0 =
++=
4/3 A 1.5 ohm
1 ohm
1 ohm
+
Vo
--
+
-
2 V
1.5 ohm
1 ohm
1 ohm
+
Vo
--
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Thevenin and Norton Theorems
Thevenin Theorem:A portion of the circuit at a pair of nodes can be replaced
by a voltage source Voc in series with a resistor R TH, where Voc
is the open circuit voltage and R TH is the Thevenin’s equivalentresistance obtained by considering the open circuit with all
independent sources made zero
R Lcircuit
a
b
+
-R L
a
b
VOC
R TH
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Norton Theorem :
A portion of the circuit at pair of nodes can be replaced by a
current source Isc in a parallel with a resistor R TH. Isc is the shortcircuit current at the terminals, and R TH is the Thevenin’s
equivalent resistance
R Lcircuit
a
b
R L
a
b
ISC
R TH
Here we will consider ( 3 ) cases :
1. Circuit containing only independent sources.
2. Circuit containing only dependent sources.
3. Circuit containing both independent and dependent sources.
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Case (1): Circuit containing only independent sources:
• Procedure of Thevenin’s Theorm:a. Find the open circuit voltage at the terminals , Voc.
b. Find the Thevenin’s equivalent resistance, RTH at the
terminals when all independent sources are zero:
¾Replacing independent voltage sources by short circuit
¾ Replacing independent current sources by open circuit
c. Reconnect the load to the Thevenin equivalent circuit
• Procedure of Norton’s Theorm:
a. Find the short circuit current at the terminals, Isc. b. Find Thevenin’s equivalent resistance, R TH (as before).
c. Reconnect the load to Norton’s equivalent circuit.
R
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Example :
Use Thevenin’s and Norton Theorms to find V0
+
-R L
VOC
R TH
R L
ISC
R TH
Using Thevenin Theorm:
+ - + -
Ωk 2
6 V 12 V
Ωk 2Ωk 4
+
-
Vo
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V4k)(4iVAm1k 4k 2
V6i 14k Ω1 −==⇒=+=
V8V4V12Voc =−=
First find VOC:
+ - + -
Ωk 2
6 V 12 V
Ωk 4
+
-
Voc
+
-
i1
![Page 37: Elec 3202 Chap 4](https://reader030.vdocuments.site/reader030/viewer/2022021123/577ce35a1a28abf1038be7c8/html5/thumbnails/37.jpg)
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Second, find R TH
R TH = 2k//4k = 4/3 k Ω
Thevenin equivalent circuit is
( )
V4.8V
V8k
310
k 2
VR k 2
Ω
k 2V
0
oc
TH
0
=
=
+=
Ωk 2Ωk 4
R TH
+
-
VOC
R TH
+
-
Vo Ωk 2
![Page 38: Elec 3202 Chap 4](https://reader030.vdocuments.site/reader030/viewer/2022021123/577ce35a1a28abf1038be7c8/html5/thumbnails/38.jpg)
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+ - + -
Ωk 2
6 V 12 V
Ωk 4ISC
+ -
Ωk 2
6 V
Ωk 4 +
-
12 V
i1i
i2
+
-V 2 k
-
12 V
+
X
Using Norton Theorm
First find Isc
![Page 39: Elec 3202 Chap 4](https://reader030.vdocuments.site/reader030/viewer/2022021123/577ce35a1a28abf1038be7c8/html5/thumbnails/39.jpg)
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R TH is the same as before:
Am2.4m)(6k 2k
34
k 34)(I
k 2R
R I sc
TH
TH0 =
+=
+=
V4.8k)(2m)(2.4k)(2IV 00 ===
2 k
ISC
R TH
Io +
-
Vo
![Page 41: Elec 3202 Chap 4](https://reader030.vdocuments.site/reader030/viewer/2022021123/577ce35a1a28abf1038be7c8/html5/thumbnails/41.jpg)
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Example :
Use Thevenin and Norton to find V0
Using Thevenin Theorm:
+
-Ω4
Ω8Ω5
Ω20
+
-
Vo
Ω72
Ω12
![Page 42: Elec 3202 Chap 4](https://reader030.vdocuments.site/reader030/viewer/2022021123/577ce35a1a28abf1038be7c8/html5/thumbnails/42.jpg)
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KVL around the upper loop :
1. Find Voc :
+
-
Ω8Ω5
Ω20Ω72
Ω12
i1
a
b
(1)0i5i25
0)i(i5i8i12
21
2111
KK=−
=−++
![Page 43: Elec 3202 Chap 4](https://reader030.vdocuments.site/reader030/viewer/2022021123/577ce35a1a28abf1038be7c8/html5/thumbnails/43.jpg)
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A3iA,0.6i
(2)72i25i572i20)i(i5
21
21
212
==
=+− =+−KK
V64.8V
(3)20(0.6)8
i20i8V
oc
21oc
=
+=
+=
KCL around lower loop :
![Page 44: Elec 3202 Chap 4](https://reader030.vdocuments.site/reader030/viewer/2022021123/577ce35a1a28abf1038be7c8/html5/thumbnails/44.jpg)
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2. Find R TH
R TH = (8+4) // 12 = 12 // 12 = 6Ω
Ω8Ω5
Ω20
Ω12
a
b
Ω8( )
Ω4
Ω//205
=
Ω12
a b
![Page 45: Elec 3202 Chap 4](https://reader030.vdocuments.site/reader030/viewer/2022021123/577ce35a1a28abf1038be7c8/html5/thumbnails/45.jpg)
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V25.92V
(64.8)
64
4
VR 4
4V
o
oc
TH
o
=+
=
+=
3. Reconnect the load :
+
-
VOC
R TH
+
-
Vo Ωk 44Ω
![Page 46: Elec 3202 Chap 4](https://reader030.vdocuments.site/reader030/viewer/2022021123/577ce35a1a28abf1038be7c8/html5/thumbnails/46.jpg)
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Using Norton Theorm:
1. Find ISC :
KVL around upper loop :
(1)0i8i5i25
0)i(i5)i(i8i12
321
21311
KK=−−
=−+−+
+
-
Ω8Ω5
Ω20
Ω72
Ω12
i1
i3
i2
![Page 47: Elec 3202 Chap 4](https://reader030.vdocuments.site/reader030/viewer/2022021123/577ce35a1a28abf1038be7c8/html5/thumbnails/47.jpg)
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KVL around lower loop :
KVL around right loop :
(2)72i20i25i572)i(i20)i(i5
321
3212
KK=−+− =−+−
(3)0i28i20i8
0)i(i20)i(i8
321
2313
KK=+−−
=−+−
A10.8I
A10.8i,A12.72i,A6i
SC
321
=⇒
===
i d
![Page 48: Elec 3202 Chap 4](https://reader030.vdocuments.site/reader030/viewer/2022021123/577ce35a1a28abf1038be7c8/html5/thumbnails/48.jpg)
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2. Find R TH
From before , R TH = 6 Ω3. Reconnect the load
( ) V25.9210.846
64V
I4R
R Ω)(4
iΩ
)(4V
o
SC
TH
TH
2o
=
+=
+=
=
a
b
ISC
R TH
i1 i2
Ω4
+
-
Vo
![Page 49: Elec 3202 Chap 4](https://reader030.vdocuments.site/reader030/viewer/2022021123/577ce35a1a28abf1038be7c8/html5/thumbnails/49.jpg)
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Case(2) : Circuits containing only dependent sources
Here there is NO energy source in the circuit.¾ VOC is always zero and ISC is always zero
¾ So we can only find R TH
Procedure for finding R TH
1. Connect an independent voltage ( or current) source at the
terminals ,Vx (or Ix)
2. Find the corresponding current ( or voltage) at the terminal ,Io ( or Vo)
3. Find R TH = Vx /Io or R TH = Vo/Ix a
b
R TH
Example:
![Page 50: Elec 3202 Chap 4](https://reader030.vdocuments.site/reader030/viewer/2022021123/577ce35a1a28abf1038be7c8/html5/thumbnails/50.jpg)
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Example:
Find the Thevenin equivalent circuit
1. Apply voltage source at the terminals (Vx=1V)
Ωk 3
2000 Ix
Ωk 2
Ωk 4
Ix
a
b
Ωk 3
2000 Ix
Ωk 2
Ωk 4
Ix
+
-
Vx = 1Vi1i2
V1
![Page 51: Elec 3202 Chap 4](https://reader030.vdocuments.site/reader030/viewer/2022021123/577ce35a1a28abf1038be7c8/html5/thumbnails/51.jpg)
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0IIk 3
4
k 3
1I2I
0I
3000
I40001
2000
I4000I2000
Ik)(4Vwhere
XXXX
XXXX
X1
=−−+−
=−−
+−
=
KCL at node V1 :
0Ik 3
V1
k 2
VI2000
0Iii
X11X
X21
=−−
+−
=−+
Am0.1I
30001
342I
X
X
=
=
+
![Page 52: Elec 3202 Chap 4](https://reader030.vdocuments.site/reader030/viewer/2022021123/577ce35a1a28abf1038be7c8/html5/thumbnails/52.jpg)
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( ) ( ) ( )k 3
Am0.1k 41
k 3
Ik 4V
k 3
VVi
XX
1X2
−=
−=
−=
Ωk 5
Am0.2
V1
i
VR
Am0.2i
2
XTH
2
===
=
a
b
R TH
Case (3) : Circuits containing both independent and
![Page 53: Elec 3202 Chap 4](https://reader030.vdocuments.site/reader030/viewer/2022021123/577ce35a1a28abf1038be7c8/html5/thumbnails/53.jpg)
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Case (3) : Circuits containing both independent and
dependent sources
Procedure of Thevenin or Norton Theorms:
a. Find the open circuit voltage and the terminals ,VOC
b. Find the short circuit current at the terminals, ISC .
c. Compute R TH = VOC/ISC
Note :
R TH can not be found as in the case of only independent sources
d. Construct the Thevenin or Norton circuits
a
b
ISC
R TH
Norton circuitThevenin circuit
a
b
R TH
+
-Voc
Example :
![Page 54: Elec 3202 Chap 4](https://reader030.vdocuments.site/reader030/viewer/2022021123/577ce35a1a28abf1038be7c8/html5/thumbnails/54.jpg)
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UAE University Department of Electrical Engineering Dr.Hazem N.Nounou
Example :
Find the Thevenin equivalent circuit with respect to the terminals a, b
1. Find VOC :
+
-
Voc
Ω20
Ω80Ω60
4 A Ω40
160 ix
ix
a
b
+
-
Voc
Ω20
Ω80
Ω60
Ω40
160 ix
ix+
-
i240 V i1
Z
KVL d h lif l
![Page 55: Elec 3202 Chap 4](https://reader030.vdocuments.site/reader030/viewer/2022021123/577ce35a1a28abf1038be7c8/html5/thumbnails/55.jpg)
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KVL around the lift loop :
(1)240i200i80
0i40i160i80240
x
xx
KK=+
=+++−
(2)0ii2
i40i80
x1
x1
KK=−
=
KVL around right loop :
KCL at Z:
(3)0iii x1 KK=−−
A0.75i
A0.375iA1.125i
x
1
=
==
![Page 56: Elec 3202 Chap 4](https://reader030.vdocuments.site/reader030/viewer/2022021123/577ce35a1a28abf1038be7c8/html5/thumbnails/56.jpg)
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V30V
Ω)(40(0.75A)
Ω)(40iV
OC
xOC
=
=
=∴
2. Find ISC :
Since we have short circuit , 80 // 40 // 0 = 0
Ω20
Ω80Ω60
4 AΩ40
160 ix
ix
V
i 0
![Page 57: Elec 3202 Chap 4](https://reader030.vdocuments.site/reader030/viewer/2022021123/577ce35a1a28abf1038be7c8/html5/thumbnails/57.jpg)
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Current divider
( ) A34
2060
60ISC =
+
=
Ω10A3V30
IVR
SC
OCTH ===
Ω20
Ω60
4 A ISC
3. Find R TH
ix=0
160ix source is zero
a
b
R TH
+
-Voc4.
V
Example :
![Page 58: Elec 3202 Chap 4](https://reader030.vdocuments.site/reader030/viewer/2022021123/577ce35a1a28abf1038be7c8/html5/thumbnails/58.jpg)
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Example :
Use Thevenin theorem to find the Thevenin equivalent circuit with
respect to a, b
1. Find VOC
KCL at node z :
+
- Ω1
Ω5V40
ix
2 ix
i1 a
b
Z
8 A
(1)8ii3
0i8ii2
1x
1xx
KK−=−
=−++
KVL d l
![Page 59: Elec 3202 Chap 4](https://reader030.vdocuments.site/reader030/viewer/2022021123/577ce35a1a28abf1038be7c8/html5/thumbnails/59.jpg)
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(2)40ii5
0i1i540
1x
1x
KK=+
=++−
KVL around outer loop
V20i1V
A20i,A4i
1OC
1x
==⇒
==⇒
Find ISC :
+
-Ω1
Ω5V40
ix
2 ix
a
b
8 AISC
2 i
![Page 60: Elec 3202 Chap 4](https://reader030.vdocuments.site/reader030/viewer/2022021123/577ce35a1a28abf1038be7c8/html5/thumbnails/60.jpg)
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KVL around outer loop :
A8i0i540 xx =⇒=+−
( ) ( ) A32883I
I8i3
I8ii2
SC
SCx
SCxx
=+=⇒
=+=++
Ω0.62532
20
I
V
R SC
OC
TH ===
KCL at z :
3. Find R TH :
+
-
Ω5V40
ix
2 ix
8 AISC
Z
a
b
R TH
+
-Voc
Thevenin equivalent circuit is
4 Maximum Power Transfer
![Page 61: Elec 3202 Chap 4](https://reader030.vdocuments.site/reader030/viewer/2022021123/577ce35a1a28abf1038be7c8/html5/thumbnails/61.jpg)
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4. Maximum Power Transfer• A technique in which the load is selected to maximize the
power transfer.• This technique is based on the Thevenin equivalent circuit.
L
2
LTH
OC
L
2
LL
R R R
V
R iivP
+=
==
RL
RTH
+
-
Voc
i +
VL
--
![Page 62: Elec 3202 Chap 4](https://reader030.vdocuments.site/reader030/viewer/2022021123/577ce35a1a28abf1038be7c8/html5/thumbnails/62.jpg)
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( ) ( )
TH
2
OCmaxL
TH
2
OC
2
TH
TH
2
OC
TH
2
TH
OC
L
2
maxL
R 4
VP
R 4
V
R 4
R V
R
R 2
V
R iP
=
==
=
=
RL
RTH
+
-
Voc i
Example:
![Page 64: Elec 3202 Chap 4](https://reader030.vdocuments.site/reader030/viewer/2022021123/577ce35a1a28abf1038be7c8/html5/thumbnails/64.jpg)
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Example:
•Find R L for maximum Power Transfer ?
•Find the maximum Power transfer to R L ?
Ωk 4Am3+
-
10 V+ -
Ωk 01
Ωk 02
Ωk 5.2Ωk 8
R L
10 V
Let’s find Thevenin equivalent circuit
![Page 65: Elec 3202 Chap 4](https://reader030.vdocuments.site/reader030/viewer/2022021123/577ce35a1a28abf1038be7c8/html5/thumbnails/65.jpg)
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Ωk 4Am3+
-
10 V
+ -
Ωk 01
Ωk 02
Ωk 5.2Ωk 8
10 V
i1 i4
i3
i2 +
-
Voc
V1 V2
Let’s find Thevenin equivalent circuit .
KCL at node V1 :
0Ωk 8
VV
Ωk 4
VAm3
0iiAm3
211
21
=−
−−
=−−
111
![Page 66: Elec 3202 Chap 4](https://reader030.vdocuments.site/reader030/viewer/2022021123/577ce35a1a28abf1038be7c8/html5/thumbnails/66.jpg)
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(1)m3Vm0.125Vm0.375
m3Vk 8
1
k 8
1
k 4
1V
21
21
KK=−
=
−
+
KCL at node V2:
0k 12.5
V
k 20
10V
k 8
VV0iii
2221
432
=−−
−−
=−−
(2)m0.5Vm0.255Vm0.125
m0.5Vk 12.5
1
k 20
1
k 8
1
k 8
1V
21
21
KK−=−
−=
++−
V7.03V
V10.34V
2
1
=
=
ik1010V +=
![Page 67: Elec 3202 Chap 4](https://reader030.vdocuments.site/reader030/viewer/2022021123/577ce35a1a28abf1038be7c8/html5/thumbnails/67.jpg)
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( )
V4.375V
7.0312.5
1010
k 12.5
V2k 1010
ik 1010V
OC
4OC
−=
+−=
+−=
+−=
To find R TH :
Ωk 8 Ωk 5.2
Ωk 4 Ωk 02Ωk 01
R TH
( )[ ] k10//k2 5k20//k4k8R ++=
![Page 68: Elec 3202 Chap 4](https://reader030.vdocuments.site/reader030/viewer/2022021123/577ce35a1a28abf1038be7c8/html5/thumbnails/68.jpg)
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( )[ ]
( )[ ]
( )
Ωk 5R
k 10//k 10
k 10//k 2.5k 7.5
k 10//k 2.5k 20//k 12
k 10//k 2.5k 20//k 4k 8R
TH
TH
=
=
+=
+=
++=
( )
Wm0.957P
k)(54
4.375
R 4
VP
maxL
2
TH
2
OCmaxL
=
−==
R TH
+
-V
oc
R L
= R TH
Example :
![Page 69: Elec 3202 Chap 4](https://reader030.vdocuments.site/reader030/viewer/2022021123/577ce35a1a28abf1038be7c8/html5/thumbnails/69.jpg)
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UAE University Department of Electrical Engineering Dr.Hazem N.Nounou
Example :
1. Find R L for maximum Power Transfer?
2. Find max. power transfer to R L ?
First , find Thevenin equivalent:
RL
3k ohm
1k ohm
4m A
+ -
2000 Ix
2k ohm
4k ohm
Ix
4k ohm2000 Ix'
1k ohm
![Page 70: Elec 3202 Chap 4](https://reader030.vdocuments.site/reader030/viewer/2022021123/577ce35a1a28abf1038be7c8/html5/thumbnails/70.jpg)
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+
Voc
--
Using source transformation
+
Voc
--
4k ohm
2k ohm
+ -
4m A
1k ohm
3k ohm Ix’
+
-16V
4k ohm
+ -
2000 Ix'
2k ohm
4k ohm
Ix’
KVL around the loop:
![Page 71: Elec 3202 Chap 4](https://reader030.vdocuments.site/reader030/viewer/2022021123/577ce35a1a28abf1038be7c8/html5/thumbnails/71.jpg)
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UAE University Department of Electrical Engineering Dr.Hazem N.Nounou
Now, find Isc:
KVL around the loop:
-16 + 4k Ix’ - 2k Ix’ + 2k Ix’=0
Ix’= 4mA.
Voc= (2k Ω) Ix’= 8V.
4k ohm
2k ohm
+ -
2000 Ix''4k ohm
+
-
16V Ix’’ Isc
I1
V1
KCL at V1:
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UAE University Department of Electrical Engineering Dr.Hazem N.Nounou
KCL at V1:
I1 - Ix’’ – Isc = 0
04
1
2
1
4
16
=−− k
V
k
V
k
04
1
2
1
4
)''21(16 =−−−−k
V
k
V
k
kIxV
And
Where V1=2k Ix’’
Hence,
Or V1=5.333V
mA
k
V Isc 333.1
4
1==
V8V
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( )( )
Wm38P
k 24
64
k 64
8
R 4
VP
Ωk 6Am1.333
V8
I
VR
L(max)
2
TH
2
OCL(max)
SC
OCTH
=
===
===