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By:
DR SITI NOORAYA MOHD TAWIL
Department of Electronic Engineering
Faculty of Electrical and Electronic Engineering
Universiti Tun Hussein Onn Malaysia
ELECTRICAL AND ELECTRONIC
TECHNOLOGY
(BDU10803)
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Topic 3:
Direct Current CircuitAnalysis (III)
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Lecture Contents
Thevenins and Nortons Theorems
Superposition Theorem
Maximum Power Transfer Theorem
Circuit Theorems
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SuperpositionTheorem
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Superposition Theorem
The superposition theoremstates that
the voltage across (or current through) an
element in a circuit is the algebraic sumofthe voltages across (or currents through)
that elements due to each independent
source acting alone.
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How to apply the superposition
theorem?1. Consider one independent sourceat a
time while other independent sources are
turned-off. [short-circuit the voltagesource and open-circuit the current
source]
2. Dependent source are left intact becausethey are controlled by circuit variables.
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Example 1
Use the superposition theorem to find V in thefollowing circuit.
8
6V
4 V 2A
+
-
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SolutionStep 1:
0.5A48
6'I
circuited)-opensource(2Aoperationinsource6VOnly
4 +
TR
V
6V
8
4 V 2A
+
-6V
8
4 V
+
-I4I4
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SolutionStep 2:
V7.321.83(4)RIV 44
++
+
A83.133.15.0"' 444
4
III,currentTotal
1.33A(2)48
8"I
circuited)-shortsourceV6(operationinsource2AOnly
6V
8
4 V 2A
+
-
8
4 V
+
-b
I4I4
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Given the following circuit, calculate ixand the power
dissipated by the 10-resistor using superposition.
Example 2
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Solution
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Cont
0.6A10
6
10
VaI
625
150a
10Va15Va150
3Va12Va10Va150:x120
40
0Va
10
0Va
12
Va-1540
0VaI
12
Va-15
:a)(nodKCL
(4A)sourcecurrentcircuit-Open(a)FigureConsider
x1
x1
+
+-
-+
-
-+
V
a
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Cont
17.43Watt(10)1.32)(RIP
1.32A1.92)(0.6III
1.92A10
19.2
10
VaI
19.2V25
480-Va
048025Va:(2)()
(2)-----120I4803Va:x120
440
0-VaI
VbVa
440
0-VbI
4II
:bnod
2
x
2
xx
x2x1x
x2
2
2
2
32
-
--++
--
-
++
+
+
+
+
(1)120I22Va
120I10Va12Va:x120
I10
Va
12
Va
10
0Va
I12
Va-0
III
:a)(nodKCL
(15V)sourcevoltagecircuit-Short
(b)FigureConsider
2
2
2
2
x221
-----
-+
-+
-
+
+
a bI1 I2
I3
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Caution!
Superposition Theorem is much easier to
use in a circuit that has independent source.
If the circuit has dependent source, it is
recommended not to use this theorem.
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Equivalent circuit forms
R1 + R2
R1
R2
R2R1
V2
V1 - V2
V1
I1 I2 I1 - I2
21
21
RR
RR
+
(a)
(b)
(c)
(d)
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SOURCE TRANSFORMATION
A source transformationis the process ofreplacing a voltage Vin series with resistor
Rby a current sourceIin parallel with a
resistor R, or vice versa.
# Source transformation is another tool for simplifyingcircuits
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Cont..
V
R
I R
a
b
V
R
I R
a
b
Transformation of dependent sources
Transformation of independent sources
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Thevenins
Theorem
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Thevenins Theorem
Thevenins Theorem states that a lineartwo-terminal circuit can be replaced by an
equivalent circuit consisting of:
A voltage source VTHin SERIES withA resistor RTH
Complexcircuit VTH
RTH
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Thevenins Theorem
Where VTHis the open-circuit voltage at theterminals and,
RTHis the input or equivalent resistance at theterminals when the independent sources areturned off.
VTH
RTH
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Example 3
For the following circuit, find the Theveninsequivalent circuit seen from RB.
24
20V 3 5ARB
6
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Solution 31. Remove R
B
.
2. Find RTHfrom terminal a-b with turned-off
all independent sources
24
20V 3 5A
6
a
b
24
3
6
a
b
++
9.71(4//3)62RTH RTH
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Cont3. Find VTHfrom terminal a-b.
(2)30VyVx
056
VyVx05I
ynode
-(1)---602Vy-9Vx
2Vy2Vx4Vx3Vx60:x12
6
VyVx
3
Vx
4
Vx20
xnode
leavingIenteringInodes,forKCL
analysis,nodeusingExample
3
--------
+-
+
-+-
-+
-
+
321 III
24
20V 3 5A
6
a
bVTH
+
-
xy
I1 I2
I3
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47.14V0VyV
47.14V7
330
Vy
330270)(602
1
60
30
792
2-9
1-1
60
30
Vy
Vx
2-9
1-1
:rulesCramer'Using
TH
y
y
-
---
--
+-
-
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4. Draw the Thevenins
equivalent circuit.
BTotal
TH
R71.9
14.47
R
VI
+
VTH=47.14V
RTH=9.71
a
b
RB
A
A
58.081.71
47.14I,27Rwhenii)
17.2
21.71
47.14I,12Rwheni)
B
B
I
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Example 4
Determine RThand VThat terminals 1-2 for
the following circuits.
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Solution
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Nortons
Theorem
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Nortons Theorem
Nortons Theorem states that a linear two-terminal circuit can be replaced by
equivalent circuit consisting of:A current source I
N
in PARALLEL with
A resistor RN
Complexcircuit
IN RN
a
b
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Example 5
Find the Nortons equivalent circuit for the
following circuit seen from RB.24
20V 3 5ARB
6
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3. Find INat a-b.
(1)602Vy9Vx
2Vy2Vx4Vx3Vx60:x12 3
Vx
6
VyVx
4
Vx20
III
:xnodeleaving
Ientering
Inodes,forKCL
analysis,nodeusingExample
321
------
-+-
+-
-
+
24
20V3
5A
6
a
b
IN
x y
(2)304VyVx
3Vy3VyVx:x6
2
Vy5
6
VyVx
I5I:ynodeN3
--------+-
+-
+
I2
I1 I3
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4.85A2
9.71
2
0VyI
9.71V34
330-
Vy
-3306027030-1
609
342)(364-1
29
30-
60
Vy
Vx
4-1
29
rule;sCramer'Using
N
y
y
-
-
--
-----
-
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4. Draw the equivalent
circuit.
IN=4.85A RN=9.71 RB
a
b
B
B
N
Total
N
R71.9
09.47
)85.4(
R71.9
71.9
)I(R
RI
+
+
A
A
58.0
81.71
47.09I,27Rwhenii)
17.221.71
47.09I,12Rwheni)
B
B
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Example 6
Find the Norton equivalent with respect to
terminals a-bin the following circuits.
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Solution
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Cont
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Cont
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Example 7
Find the Norton equivalent in the following
circuit.
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Solution
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Cont
Current divider rules
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RELATION BETWEEN THEVENIN
THEOREM AND NORTON THEOREM
VTH
RTH
IN RN
a
b
TH
THN
NNTH
NTH
R
VI
RIV
RR
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IN=4.85A RN=9.71 RB
a
b
VTH=47.14V
RTH=9.71
a
b
RB
I
Thevenins equivalent circuitNortons equivalent circuit
A85.471.9
14.47
R
VI
47.14V1)(4.85)(9.7RIV9.71RR
TH
THN
NNTH
NTH
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Maximum Power
Transfer
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Maximum Power Transfer
Practicala circuit is designed to provide
power to a load.
Applicationcommunicationdesirableto maximize the power delivered to the
load.
Thevenin equivalentuseful in finding themaximum power a linear circuit can deliver
to a load.
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Cont
Consider the following circuit for maximum
power transfer.
ThR
ThV
a
b
i
LR
L
2
LTh
Th2 RRR
VRip
+
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Cont
Maximum power is transferred to the load
when the load resistance equals the
Thevenin resistance as seen from the load(RL=RTh).
C t
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Cont
For the given circuit, VTh
and RTh
are fixed.
Varying the RL, the power delivered to the
load as the graph below.
Pmax
RThRL
P
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To prove the maximum power transfer differentiate p with
respect to RLand set the results =0
(watt)4R
V
4R
VP
,RRWith
.RRwhenplacespower takemaximumthethat,Showing]R[R),R(R)2RR(R0
thatimpliesThis
0)R(R
)2RR(RV
)R(R
)R(R2R)R(RV
dR
dP
0dR
dPwhen,maximumispowerThe
L
2
Th
L
2
Thmax
ThL
ThL
LThLThLLTh
3LTh
LLTh2
Th
2LTh
LThL2
LTh2
Th
L
L
--+
+
-+
+
+-+
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FORMULA!!
TH
2
TH
L
2
THL
2
LTH
THLMAX
THL
4RV
4RVR
RRVI2RP
RR
+
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Example 8Find the value of RLfor maximum power
transfer in the following circuit and find themaximum power.
VTH=47.14V
RTH=9.71
a
b
RL
I
Thevenins equivalent circuit
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Solution
Thevenins equivalent circuit
57.17
4(9.71)
(47.14)
4R
VP
9.71RR
2
TH
2
THMAX
THL
VTH=47.14V
RTH=9.71
a
b
RL
I
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