CIRCUIT THEOREMS

CIRCUIT
THEOREMS
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4.6 Superposition Theorem
4.7 Thevenin’s Theorem
4.8 Norton’s Theorem
4.9 Source Transformation
4.10 Maximum Power Transfer Theorem
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4.6 Superposition Theorem
x
input
f (g)
y
output
The relationship f (x) between cause x and effect y
is linear if f (˙) is both additive and homogeneous.
definition of additive property：
If f(x1)=y1 , f(x2)=y2 then f(x1+x2)=y1+y2
definition of homogeneous property：
If f(x)=y and α is a real number then f(αx)= αy
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4.6 Superposition Theorem
Example 4.6.1
n
Assume I0 = 1 A and use linearity to find the actual
value of I0 in the circuit in figure.
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4.6 Superposition Theorem
If I 0 = 1A , then V1 = (3 + 5) I 0 = 8V
I1 =
V1
= 2A , I 2 = I1 + I 0 = 3A
4
V2 = V1 + 2 I 2 = 8 + 6 = 14V , I 3 =
V2
= 2A
7
I 4 = I 3 + I 2 = 5A ⇒ I S = 5A
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I 0 = 1A → I S = 5A , I 0 = 3A → I S = 15A
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4.6 Superposition Theorem
For a linear circuit N consisting of n inputs , namely
u1 , u2 , …… , un , then the output y can be calculated
as the sum of its components：
y = y1 + y2 + …… + yn
where
yi=f(ui) , i=1,2,……,n
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4.6 Superposition Theorem
Proof： Consider the nodal equation of the corresponding
circuit for the basic case as an example
G11 G12
G G
 21 22
M

Gn1 Gn 2
L G1n 
L G2 n 
O M 

L Gnn 
e1   I1s 
e   I 
 2  =  2 s  LLL ( A )
 M  M 
   
en   I ns 
[G ] e = I s LLLLLLLLLLLL ( B )
Let Gk = [ Gk1 Gk2 … Gkn ]T
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Then [G] = [ G1 G2 … Gn ]
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4.6 Superposition Theorem
n
Cramer’s Rule for solving Ax=b
Take n=3 as an example.
 a11
a
 21
 a31
a12
a22
a32
a13   x1   b1 
a23   x2  = b2 
a33   x1  b3 
Let
det A = △ ≠ 0
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4.6 Superposition Theorem
b1
det b2
Then
b3
x1 =
x2 =
x3 =
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a11
det a21
a31
a11
det a21
a31
a12
a22
a13
a23
a32
∆
b1
b2
b3
∆
a12
a22
a32
∆
a33
a13
a23
a33
b1
b2
b3
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4.6 Superposition Theorem
Suppose that the kth nodal voltage ek is to be found.
Then from Cramer’s rule one has
ek =
det  G 1 G 1 L I s L G n 
det [ G ]
n
=
∑
j=1
Δ jk
Δ
I js
w h ere ∆ @ det [ G ]
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∴ e k = e k1 + e k2 + L L + e kn
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4.6 Superposition Theorem
where
ek1 =
Δ 1k
I , due to I 1 s only
Δ 1s
ekn =
Δ nk
I , due to I ns only
Δ ns
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4.6 Superposition Theorem
n
Example 4.6.2
Find e2 = ?
Nodal Equation
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G1+G4+G6
-G4
-G6
-G4
G2+G4+G5
-G5
-G6
-G5
G3+G5+G6
e1
e2
e3
＝
I1S
I2S
I3S
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4.6 Superposition Theorem
By using Cramer’s rule
 G1 + G4 + G6

−G4
det 

−G6

e2 =
∆12
∆
I1S + 22 I 2 S
∆
∆
= e21 + e22 + e23
=
I1S
I2S
I3S
−G6


−G5

G3 + G5 + G6 
∆
∆
+ 32 I 3 S
∆
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4.6 Superposition Theorem
Where e21 is due to I1S only，I2S＝I3S＝0
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G1+G4+G6
-G4
-G6
-G4
G2+G4+G5
-G5
-G6
-G5
G3+G5+G6
e11
e21
e31
＝
I1S
0
0
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4.6 Superposition Theorem
G1+G4+G6
-G4
-G6
-G4
G2+G4+G5
-G5
-G6
-G5
G3+G5+G6
 G1 + G4 + G6

−G4
det 

−G6

∴ e21 =
=
I1S
0
0
∆
e11
e21
e31
＝
I1S
0
0
−G6


−G5

G3 + G5 + G6 
∆12
I1S , due to I1S only
∆
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4.6 Superposition Theorem
Similarly
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Duo to I 2 S only
Duo to I 3 S only
I1S = I 3 S = 0
I1S = I 2 S = 0
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4.7 Thevenin’s Theorem
In high school, one finds the equivalent
resistance of a two terminal resistive circuit
without sources.
Now, we will find the equivalent circuit for two
terminal resistive circuit with sources.
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4.7 Thevenin’s Theorem
Thevenin’s theorem states that a linear two-terminal
circuit can be replaced by an equivalent circuit
consisting of a voltage source VTH in series with a
resistor RTH where VTH is the open circuit voltage at
the terminals and RTH is the input or equivalent
resistance at the terminals when the independent
sources are turned off .
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4.7 Thevenin’s Theorem
I
Linear
two-terminal
circuit
a
+
V
b
Connected
circuit
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4.7 Thevenin’s Theorem
Equivalent circuit: same voltage-current relation at the
terminals.
VTH = VOC : Open circuit voltage at a-b
VTH = VOC
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4.7 Thevenin’s Theorem
RTH = RIN : input resistance of the dead circuit
Turn off all independent sources
RTH = RIN
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4.7 Thevenin’s Theorem
CASE 1
If the network has no dependent sources:
- Turn off all independent source.
- RTH : input resistance of the network looking
into a-b terminals
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4.7 Thevenin’s Theorem
CASE 2
If the network has dependent sources
-Turn off all independent sources.
-Apply a voltage source VO at a-b
R TH =
VO
IO
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4.7 Thevenin’s Theorem
-Alternatively, apply a current source IO at a-b
R TH =
VO
IO
If RTH < 0, the circuit is supplying power.
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4.7 Thevenin’s Theorem
Simplified circuit
IL =
VTH
R TH +R L
VL = R L I L =
RL
VTH
R TH +R L
Voltage divider
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4.7 Thevenin’s Theorem
Proof : Consider the following linear two terminal circuit
consisting of n+1 nodes and choose terminal b as
datum node and terminal a as node n .
L
 G 11

 M
G
 n1
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K
O
L
 V1
G 1n  
 V
M  2
 M
G n n  
Vn
  I1s 

 
 =  I2s 
  M 

 
  I ns 
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4.7 Thevenin’s Theorem
Then nodal voltage Vn when a-b terminals are open
can be found by using Cramer’s rule .
Vn =
1
∆
n
∑
k =1
∆ kn I ks
LLL
(A )
∆ is the determinant of [G] matrix
∆ ku is the corresponding cofactor of Gkn
Now connect an external resistance Ro to a-b terminals .
The new nodal voltages will be changed to e1 , e2 , … , en
respectively .
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4.7 Thevenin’s Theorem
Nodal equation
 G11 K G1n + 0 

  e1   I1s 
G
+
0
M
2
n

e   I 
 M
  2  =  2s  . . . . . . . . . ( B)
M

 M   M 
1
G L G +    
e
I
nn
 n1
Ro   n   ns 

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4.7 Thevenin’s Theorem
Note that
 G11 K G1n + 0 
 G11 K 0 




G2 n + 0 
0 
 M
 G21
 = det [G ] + det  M
det  M
M
M 




G L G + 1 
G L 1 
nn
 n1
 n1
Ro 
Ro 


= ∆ +
1
∆ nn
Ro
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4.7 Thevenin’s Theorem
Hence , en can be obtained as follows .
 G11 K I1s 


n
det  M O M 
1 n
∆ kn I ks
∑
∑ ∆ kn I ks
G L I 
Ro
∆ k =1
n1
ns 

k =1
en =
=
=
=
Vn
1
1
1 ∆ nn
Ro + RTH
∆ + ∆ nn
∆ + ∆ nn
1+
Ro
Ro
Ro ∆
∆
where R TH @ nn
∆
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4.7 Thevenin’s Theorem
TH
n
o
In other words , the linear circuit looking into terminals a-b can
be replaced by an equivalent circuit consisting of a voltage
source VTH in series with an equivalent resistance RTH , where
VTH is the open circuit voltage Vn and RTH =
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∆nn
.
∆
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4.7 Thevenin’s Theorem
Example 4.7.1
1
Ω
4
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1
Ω
2
1
Ω
6
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4.7 Thevenin’s Theorem
Example 4.7.1 (cont.)
Find open circuit voltage V2
− 2   V1   5 − 2V x 
2+ 4

 = 


−
+
V
2
2
6

  2   2V x 
2V x = 2V1
2+ 4+ 2

 −2 − 2
− 2   V1   5 
 =  

2 + 6  V2   0 
−2 
 8
∆ = det 
 = 64 − 8 = 56
8 
 −4
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4.7 Thevenin’s Theorem
Example 4.7.1 (cont.)
 8 5
det 

−4 0  20 5

∴V2 =
=
= V = VTH
56
56 14
∆
8 1
RTH = 22 =
= Ω
∆
56 7
1
Ω
7
∴ Ans.
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a
5
V
14
b
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4.7 Thevenin’s Theorem
Example 4.7.2
10Ω
20Ω
10Ω
By voltage divider principle :
open circuit voltage VTH=10V
Let independent source be zero
10
20
a
RTH=5+20=25 Ω
10
b
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4.7 Thevenin’s Theorem
Example 4.7.3
n Find the Thevenin’s equivalent circuit of the circuit
shown below, to the left of the terminals a-b. Then
find the current through RL = 6, 16, and 36 Ω.
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4.7 Thevenin’s Theorem
Example 4.7.3 (cont.)
R TH : 32V voltage source → short
2A current source
R TH = 4 P 12 +1 =
→ open
4 ×12
+ 1 = 4Ω
16
VTH
RTH
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VTH
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4.7 Thevenin’s Theorem
Example 4.7.3 (cont.)
VTH :
Mesh analysis
− 32 + 4 i 1 + 12 ( i1 − i2) = 0 , i 2 = −2A
∴i1 = 0.5A
VTH = 12 (i 1 − i 2 ) =12(0.5 + 2.0) = 30V
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4.7 Thevenin’s Theorem
Example 4.7.3 (cont.)
To get iL :
iL =
VTH
30
=
RTH + RL 4 + RL
RL = 6 → IL = 30 / 10 = 3A
RL =16 →IL = 30 / 20 =1.5A
RL = 36 →IL = 30 / 40 = 0.75A
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4.7 Thevenin’s Theorem
Example 4.7.4
Find the Thevenin’s equivalent of the following
circuit with terminals a-b.
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4.7 Thevenin’s Theorem
Example 4.7.4 (cont.)
(independent + dependent source case)
To find RTH from Fig.(a)
independent source → 0
dependent source → unchanged
Apply
vo = 1V , RTH =
vo 1
=
io io
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4.7 Thevenin’s Theorem
Example 4.7.4 (cont.)
For loop 1 , -2vx + 2(i1 − i2 ) = 0 or vx =i1 − i2
But
−4i2 = vx = i1 − i2
∴ i1 = −3i2
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4.7 Thevenin’s Theorem
Example 4.7.4 (cont.)
Loop 2 and 3 :
4i2 + 2(i2 − i1 ) + 6(i2 − i3 ) = 0
6(i3 − i2 ) + 2i3 + 1 = 0
Solving these equations gives
i3 = −
1
A
6
But io = − i3 =
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∴ RTH =
1
A
6
1V
= 6Ω
io
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4.7 Thevenin’s Theorem
Example 4.7.4 (cont.)
To find VTH from Fig.(b)
Mesh analysis
i1 = 5
− 2v x + 2(i 3 − i2 ) = 0 ⇒ vx = i 3 − i 2
4(i2 − i1 ) + 2(i2 − i3) + 6i2 = 0 ⇒12i2 − 4i1 − 2i3 = 0
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4.7 Thevenin’s Theorem
Example 4.7.4 (cont.)
But 4 (i1 − i2 ) = vx
∴ i2 =10 / 3.
VTH = voc = 6i2 = 20V
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4.7 Thevenin’s Theorem
Example 4.7.5
Determine the Thevenin’s
equivalent circuit :
Solution:
(dependent source only)
VTH = 0 , RTH =
vo
io
Nodal analysis
io + ix = 2ix +
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vo
4
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4.7 Thevenin’s Theorem
Example 4.7.5 (cont.)
But
0 − vo
v
=− o
2
2
v
v v
v
io = ix + o = − o + o = − o
4
2 4
4
or vo = −4io
ix =
Thus
RTH =
vo
= −4Ω : Supplying Power !
io
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4.8 Norton’s Theorem
n
Norton’s theorem states that 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 IN is the short-circuit current
through the terminals and RN is the input or
equivalent resistance at the terminals when the
independent sources are turned off.
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4.8 Norton’s Theorem
Linear
two-terminal
circuit
a
b
(a)
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4.8 Norton’s Theorem
Proof：
By using Mesh Analysis as an example
Assume the linear two terminal circuit is
a planar circuit and there are n meshes
when a b terminals are short circuited.
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4.8 Norton’s Theorem
Mesh equation for case 1 as an example
 R11 …… R1n   I1   V1S 


  
M   I 2   V2 S 
 M
=
O
 M
M  M   M 


  
 Rn1 LL Rnn   I n   Vns 
Hence the short circuit cuurent
1 n
In =
∑ ∆ kn Vks
∆ k =1
where ∆ = det [ Rik ]
∆ kn is the cofactor of Rkn
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4.8 Norton’s Theorem
Now connect an external resistance Ro to a , b
terminals , then all the mesh currents will be
changed to J1, J2, ‥ ‥ Jn，respectively.
 R11 …… R1n + 0   J 1   V1S 

  

R2 n + 0   J 2   V2 S 
 M
O
=
 M
 M   M 
M

  

 Rn1 LL Rnn + Ro   J n   Vns 
Note that
 R11 …… R1n + 0 
 R11 K 0 




R2 n + 0 
M
M
M 
det 
O
= ∆ + det 
O
 M

 M
M 
M




 Rn1 L Ro 
 Rn1 LL Rnn + Ro 
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= ∆ + Ro ∆ nn
52
4.8 Norton’s Theorem
Hence, one has
 R11 … V1s 


n
det  M O M 
 R L V  ∑ ∆ knVks
ns 
 n1
= k =1
Jn =
∆ + Ro ∆ nn
∆ + Ro ∆ nn
1 n
∑ ∆ knVks
∆ k =1
=
∆
1 + Ro nn
∆
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4.8 Norton’s Theorem
=
In
∆
1 + Ro nn
∆
RN
=
In
Ro + RN
where RN =
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∆
∆ nn
,
I N = In
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4.8 Norton’s Theorem
Example 4.8.1
By using the above formula
4Ω
3Ω
3Ω
3Ω
Find the short circuit current I3
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3+3
-3
-3
-3
3+3+4
-3
-3
-3
3+3
I1
I2
I3
＝
10V
0
0
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4.8 Norton’s Theorem
Example 4.8.1 (cont.)
3+3
-3
-3
-3
3+3+4
-3
-3
-3
3+3
I1
I2
I3
＝
10V
0
0
det [ Rik ] = 360 − 27 − 27 − 27 − 90 − 54 − 54 = 108
 6 − 3 10 
1
390 65

 10
I3 =
det  − 3 10 0  =
=
A = IN
( 39 ) =
108
108 18
 − 3 − 3 0  108


∆
108
36
RN =
=
=
Ω
∆ 33 60 − 9 17
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4.8 Norton’s Theorem
Example 4.8.2
Find the Norton equivalent circuit of the following circuit
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4.8 Norton’s Theorem
Example 4.8.2 (cont.)
To find RN from Fig.(a)
RN = 5 ||(8 + 4 + 8)
= 5 || 20 =
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20 × 5
= 4Ω
25
58
4.8 Norton’s Theorem
Example 4.8.2 (cont.)
To find IN from Fig.(b)
short-circuit terminal a and b
Mesh Analysis:
i1 = 2A
4Ω
2A
i
20i2 - 4i1 – 12 = 0
12V
∴ i2 = 1A = IN
8Ω
1
a
iSC
=IN
5Ω
i2
b
8Ω
(b)
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4.8 Norton’s Theorem
Example 4.8.2 (cont.)
VTH
RTH
VTH : open − circuit voltage across terminals a and b
Alternative method for IN : IN =
Mesh analysis :
i3 = 2 A , 25i4 − 4i3 −12 = 0
∴i4 = 0.8A
∴voc = VTH = 5i4 = 4V
8Ω
a
4Ω
2A
i4
i3
5Ω
VTH=vSC
12V
b
8Ω
(b)
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4.8 Norton’s Theorem
Example 4.8.2 (cont.)
Hence , IN = V TH = 4 / 4 = 1A
R TH
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4.8 Norton’s Theorem
Example 4.8.3
n
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Using Norton’s theorem, find RN and IN of the
following circuit.
62
4.8 Norton’s Theorem
Example 4.8.3 (cont.)
To find RN from Fig.(a)
Hence , io =
∴ RN =
vo 1
= = 0.2 A
5 5
vo
1
=
= 5Ω
io 0.2
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4.8 Norton’s Theorem
Example 4.8.3 (cont.)
To find IN from Fig.(b)
10
= 2.5 A
4
10V
IN =
+ 2ix
5Ω
10
= + 2(2.5) = 7 A
5
∴ IN = 7 A
ix =
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4.9 Source Transformation
R
Vs
a
v
i
N
b
The current through resistor R can be obtained
as follows :
V − v VS v
v
i= S
= − @ IS −
R
R R
R
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4.9 Source Transformation
From KCL, one can obtain the following
equivalent circuit
where I S @
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VS
R
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4.9 Source Transformation
The voltage across resistor R can be obtained as
follows :
v = ( I S − i ) R = I S R − iR @ VS − iR
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4.9 Source Transformation
From KVL, one can obtain the following
equivalent circuit
where VS @ R I S
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4.9 Source Transformation
Example 4.9.1
3Ω
a
10A
3Ω
b
a
30V
b
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4.9 Source Transformation
Example 4.9.2
n
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Find the Thevenin’s equivalent
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4.9 Source Transformation
Example 4.9.2 (cont.)
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4.10 Maximum Power Transfer Theorem
a
RL
b
n
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Problem : Given a linear resistive circuit N
shown as above, find the value of
RL that permits maximum power
delivery to RL .
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4.10 Maximum Power Transfer Theorem
Solution : First, replace N with its Thevenin
equivalent circuit.
RTH
a
VTH +-
i
RL
b
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4.10 Maximum Power Transfer Theorem
p = i2 R = (
Let
VTH
) 2 RL
RTH + RL
dp
=0 ,
dRL
Then R L =R TH and Pmax
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VTH 2
VTH2
=(
) RL =
2 RL
4 RL
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4.10 Maximum Power Transfer Theorem
Example 4.10.1
n
(a) Find RL that results in maximum power transferred to RL.
(b) Find the corresponding maximum power delivered to RL ,
namely Pmax.
(c) Find the corresponding power delivered by the 360V
source, namely Ps and Pmax/Ps in percentage.
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4.10 Maximum Power Transfer Theorem
S o lu tio n : ( a ) V
R
(b ) P
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TH
TH
150
(3 6 0 ) = 3 0 0 V
180
150 × 30
=
= 25Ω
180
=
 300 
＝ 

 50 
m ax
2
2 5 = 9 0 0W
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4.10 Maximum Power Transfer Theorem
300
× 25 = 150V
50
- (3 6 0 - 1 5 0 )
is =
= -7 A
30
P s = i s ( 3 6 0 ) = -2 5 2 0 W (d is s ip a te d )
S o lu tio n : (c ) V a b =
P m ax
900
=
= 3 5 .7 1 %
Ps
2520
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Summary
nObjective 7 : Understand and be able to use
superposition theorem.
nObjective 8 : Understand and be able to use
Thevenin’s theorem.
nObjective 9 : Understand and be able to use
Norton’s theorem.
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78
Summary
nObjective 10 : Understand and be able to use
source transform technique.
nObjective 11 : Know the condition for and be
able to find the maximum
power transfer.
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79
Summary
n Problem : 4.60
4.64
4.68
4.77
4.86
4.91
n Due within one week.
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