Fundamentals of Electric Circuit - Chapter 4: Circuit theorems

) while other
elements are fixed.
Linear 
two-terminal 
circuit
Load
b
a
V
-
+I
VTh
RTh
b
a
V
-
+
Load
I
 Each time the variable element is changed, need
to be analyzed all over again  use Thevenin’s
theorem to avoid this problem.
 Thevenin’s theorem: A linear two terminal circuit can
be replaced by an equivalent circuit consisting of a voltage source VTh in series
with a resister RTh where:
 VTh is the open-circuit voltage at the terminals
 RTh is the input or equivalent resistance at the terminals when the
independent sources are turned off.
Fundamentals of Electric Circuits – Viet Son Nguyen - 2011
20
V. Thevenin’s theorem
Chapter 4: Circuit theorems
 Finding VTh: VTh is the open-circuit voltage across
the terminals.
Linear 
two-terminal 
circuit
b
a
-
+
Linear circuit with 
all independent 
sources set equal to 
zero
b
a
Th OC
V V
Th in
R R
 Finding RTh:
 Network has no dependent sources.
 Network has dependent sources.
Linear circuit with 
all independent 
sources set equal to 
zero
b
a
i0
v0
Linear circuit with 
all independent 
sources set equal to 
zero
b
a
i0
-
+
v0
Th
v
R
i
0
0
Th
v
R
i
0
0

Fundamentals of Electric Circuits – Viet Son Nguyen - 2011
21
V. Thevenin’s theorem
Chapter 4: Circuit theorems
 Thevenin’s theorem is very important in circuit analysis:
 Help simplify a circuit: Replace a large circuit by a single independent
voltage source and a single resistor.
 Easily to determine the current and voltage on the load
Linear 
circuit
b
a
Th
L
Th L
V
I
R R


RL
IL
b
a
RL
IL
VTh
L
L L L Th
Th L
R
V R I V
R R
 

Fundamentals of Electric Circuits – Viet Son Nguyen - 2011
R 2
R 1 R 3
22
V. Thevenin’s theorem
Chapter 4: Circuit theorems
Ex 4.10: Find the Thevenin equivalent circuit of the
circuit. Find the current through RL = 6, 16, 36Ω
b
a
32V 2A
12Ω
1Ω
4Ω
RL
Th
R R
R R
R R
1 2
3
1 2
4   

i1
i2
Th Th
Th
V V
V V
R
1
32
2 30
12

   
Th
L
Th L L
V
I
R R R
30
4
 
 
 Calculating VTh: Applying nodal analysis gives
 Calculating RTh:
 When RL = 6Ω:
 The current through RL :
L
I A
30
3
10
 
 When RL = 16Ω: LI A
30
1,5
20
 
 When RL = 36Ω: LI A
30
0,75
40
 
Fundamentals of Electric Circuits – Viet Son Nguyen - 2011
R 2R 1
R 3 R 4
23
V. Thevenin’s theorem
Chapter 4: Circuit theorems
Ex 4.11: Find i by using the Thevenin’s theorem
12V 2A
6Ω
1Ω4Ω
Th
R R R R
1 2 3
12.4
( ) / / 3
12 4
    

 Calculating VTh: Applying nodal analysis gives:
 Calculating RTh:
 The current through R4 :
b
a6Ω
i
V V
V V
12
2 15
6 10

    Th
V
V R V
R R
3
2 3
15
.4 6
6 4
   
 
Th
Th
V
i A
R R
4
6
1,5
1 3
  
 
Fundamentals of Electric Circuits – Viet Son Nguyen - 2011
R 4R 2
R 3
+
-
R 4R 2
R 3
+
-
24
V. Thevenin’s theorem
Chapter 4: Circuit theorems
Ex 4.12: Find the equivalent of the circuit
2vx
5A +
2Ω
4Ω Connect to the terminal a voltage source v0 =
1V, and we find i0 through the terminal.
 To find RTh, set the independent source equal
to zero, but leave the dependent source alone
 The current through R4 :
b
a
6Ω
Th
v
R
i i
0
0 0
1
 
x
v i i
i i i i i
i i i
1 2
2 2 1 2 3
3 2 3
2 2( ) 0
4 2( ) 6( ) 0
6( ) 2 1 0
   

    
    
2Ω
-
vx
v0 = 1V
2Ω
6Ω
2Ω
+
4Ω
-
vx
2vx
i3
i2
i1
 Applying mesh analysis to loop 1, 2, 3:
i A
0
1
6
 
Th
R
i
0
1
6   
Fundamentals of Electric Circuits – Viet Son Nguyen - 2011
R 1 R 2
25
V. Thevenin’s theorem
Chapter 4: Circuit theorems
Ex 4.12: Find the equivalent of the circuit
4Ω Since the circuit has no independent sourcesVTh = 0
b
a
2Ω
ix
 From these two equations, we have:
  
x x
v
i i i 0
0
2
4
2ix
 In order to find RTh , apply a current source i0 at the terminal
R 1 R 2
4Ω
b
a
2Ω
ix
2ix
i0
 Applying nodal analysis gives: v0
 Applying Ohm’s law:

  
x
v v
i 0 0
0
2 2
         
x
v v v v
i i v i0 0 0 0
0 0 0
4
4 2 4 4
    
Th
v
R
i
0
0
4
 Note that:
 The negative value of RTh means that the circuit is supplying power by the
dependent source.
 This example shows how a dependent source and resistors could be used
to simulated negative resistance.
Fundamentals of Electric Circuits – Viet Son Nguyen - 2011
26
VI. Norton’s theorem
Chapter 4: Circuit theorems
 Norton’s theorem: A linear two terminal circuit can be
replaced by an equivalent circuit consisting of a current
source IN in parallel with a resistor RN, where:
 Finding RN:
RN
b
a
IN
 IN is the short circuit current through the terminals
 RN is the input or equivalent resistance at the terminals
when the independent sources are turned off.

N Th
R R

N sc
I i
Linear 
two-terminal 
circuit
b
a
 Finding IN:
Linear 
two-terminal 
circuit
b
a
isc = IN
 Source transformation: Relationship between Norton’s and Thevenin’s theorems:
 Th
N
Th
V
I
R
Fundamentals of Electric Circuits – Viet Son Nguyen - 2011
27
VI. Norton’s theorem
Chapter 4: Circuit theorems
 In order to determine the Thevenin or Norton equivalent circuit, we need to find:
 The open-circuit voltage voc across terminals a and b
   oc
Th oc N sc Th N
sc
v
V v I i R R
i
; ;
 The short-circuit current isc at terminals a and b
 The equivalent or input resistance Rin at terminals a and b when all
independent sources are turned off.
Fundamentals of Electric Circuits – Viet Son Nguyen - 2011
R 4
R 2
R 1
1 2 V
R 3
2 A
R 2
R 1
1 2 V
R 3
2 A
R 4
R 2
R 1
1 2 V
R 3
2 A
28
VI. Norton’s theorem
Chapter 4: Circuit theorems
Ex 4.13: Find the Norton equivalent circuit for the circuit.
 
     
  
N
R R R R R
1 2 3 4
5(4 8 8)
( ) / / 4
5 4 8 8
 Finding RN in the same way RTh
4Ω
b
a8Ω
5Ω
8Ω
 Finding IN by shortening circuit terminals a and b
i1
i2
   
  
sc N
i A
i i I A
i i
1
2
2 1
2
1
20 4 12 0
i3
i3
VTh = voc
b
a
-
+
 By another way, we can find IN by the source transform
equation:

 
  
i A
i A
i i
3
4
4 3
2
0,8
25 4 12 0
   
oc Th
v V i V
4
5 4
  Th
N
N
V
I A
R
1
Fundamentals of Electric Circuits – Viet Son Nguyen - 2011
R 2R 1
1 5 V
R 3
4 A
R 2R 1
1 5 V
R 3
4 A
29
VI. Norton’s theorem
Chapter 4: Circuit theorems
Ex 4.14: Find the Norton equivalent circuit for the circuit.
    

N
R R R R
1 2 3
6.6
( ) / / 3
6 6
 Finding RN b
a3Ω
6Ω
 Finding IN by shortening circuit terminals a and b 
applying the mesh analysis gives:
i1
i2
 
      
   
sc N
i A
i i i I A
i i i
1
2 2
1 2 2
4 27
6 27 4,5
3( ) 3 15 0 6
3Ω
3Ω
6Ω
3Ω
Fundamentals of Electric Circuits – Viet Son Nguyen - 2011
30
VI. Norton’s theorem
Chapter 4: Circuit theorems
Ex 4.15: Using Norton’s theorem, find RN and IN at terminals a-b

 
x
v
i A0 0,2
5
 Finding RN: set the independent voltage source equal to
zero and connect a voltage source of v0 = 1V to a-b
 Finding IN: Shorting – circuit terminals a and b
  
x
i A
R
2
10 10
2
5
R 2
R 1
1 0 V
b
a5Ω
2Ix
4Ω
ix
R 2
R 1
1 V
5Ω
2Ix
4Ω
ix
i0
     
x x x
i i i i A
0
2 3 0,6
     
 
N
v
R
i
0
0
1
1,67
0,6
R 2
R 1
1 0 V
5Ω
2Ix
4Ω
ix
iSC = IN    SC x xi i i A2 2 4 6
  
N SC
I i A6
Fundamentals of Electric Circuits – Viet Son Nguyen - 2011
31
VI. Norton’s theorem
Chapter 4: Circuit theorems
Ex 4.15: Using Norton’s theorem, find RN and IN at terminals a-b

 
x
v
i A0 0,2
5
 Finding RN: set the independent voltage source equal to
zero and connect a voltage source of v0 = 1V to a-b
 Finding IN: Shorting – circuit terminals a and b
  
x
i A
R
2
10 10
2
5
R 2
R 1
1 0 V
b
a5Ω
2Ix
4Ω
ix
R 2
R 1
1 V
5Ω
2Ix
4Ω
ix
i0
     
x x x
i i i i A
0
2 3 0,6
     
 
N
v
R
i
0
0
1
1,67
0,6
R 2
R 1
1 0 V
5Ω
2Ix
4Ω
ix
iSC = IN    SC x xi i i A2 2 4 6
  
N SC
I i A6
Fundamentals of Electric Circuits – Viet Son Nguyen - 2011
32
VII. Maximum power transfer
Chapter 4: Circuit theorems
 In many practical situations, a circuit is designed to provide power to a load:
     
 
L ThR RTh
L L
Th L
V
p i R R p p
R R
2
2
max
 Electric utilities: Minimizing power losses in the process distribution
 Communications: Maximize the power delivered to a load.
aRTh
VTh
i
RL
b
 Problem: Delivering pmax to a load when given a system with
known internal losses.
 Assuming that the load resistance RL can be adjusted
 Replacing entire circuit by Thevenin equivalent circuit
Linear 
circuit
b
a
RL
IL
Maximum power is transferred to the load when the load 
resistance equals the Thevenin resistance as seen from the 
load (RL = RTh).
 Th
Th
V
p
R
2
max
4
Fundamentals of Electric Circuits – Viet Son Nguyen - 2011
R 1
R 2
1 2 V
R 3 R 4
33
VI. Norton’s theorem
Chapter 4: Circuit theorems
Ex 4.16: Finding the value of RL for maximum power transfer. Find the maximum
power.
 Finding RTh:

   
 
    
i i i A
i A
i A
1 1 1
2
2
2
6 12 12
3
2
2
2Ω
RL12Ω
2A
b
     ThR R R R R1 2 3 4/ / 9
    
Th Th
i i V V V
1 2
6 3 12 22
3Ω6Ω
a
i1 i2
 Finding VTh:
 Applying KVL around the outer loop to get VTh:
 For maximum power transfer:   
L Th
R R 9
 The maximum power is:   Th
L
V
p W
R
2 2
max
22
13,44
4. 4.9
Fundamentals of Electric Circuits – Viet Son Nguyen - 2011
R 1
R 2
9 V
R 3
+
-
34
VI. Norton’s theorem
Chapter 4: Circuit theorems
Ex 4.17: Finding the value of RL for maximum power transfer. Find the maximum
power.
 Finding RTh:
2Ω
RL
3vx
4Ω
1Ω
vx
 Finding VTh:
-+