# CIRCUIT THEOREMS

CIRCUIT
THEOREMS
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4.6 Superposition Theorem
4.6 Superposition Theorem
4.7 Thevenin’s Theorem
4.7 Thevenin’s Theorem
4.8 Norton’s Theorem
4.8 Norton’s Theorem
4.9 Sou 4.9 Sour rce ce T Tr ran ansf sfo orm rmation ation
4.10 4.10 M Maximum Powe aximum Power r T Tr ran ansf sfe er r T Th heo eor rem em
C C..T T.. Pan Pan 2 2 2 24.6 Superposition Theorem
4.6 Superposition Theorem
x y
x y
f g
( )
input output
input output
The rela The relattionship ionship f f (x) be (x) bettween cause x a ween cause x an nd effect y d effect y
is is llinear if f inear if f ( (˙˙) ) is both is both a additi dditiv ve e a and homogeneous. nd homogeneous.
definition of additive property：
definition of additive property：
I If f f(x f(x )=y )=y , , f f(x (x )=y )=y then f(x then f(x +x +x )=y )=y +y +y
1 1 1 1 2 2 2 2 1 1 2 2 1 1 2 2
definition of homogeneous property：
definition of homogeneous property：
I If f f( f(x)=y and x)=y and αα is a real nu is a real num mber then ber then f( f(ααx)= x)= ααy y
C C..T T.. Pan Pan 3 3 3 3
4.6 Superposition Theorem
4.6 Superposition Theorem
n Example 4.6.1
n Example 4.6.1
Assume Assume I I = 1 A = 1 A and use li and use lin ne ea arity to find the rity to find the a actual ctual
0
0
value of value of I I in the circuit in the circuit iin n fig figure ure..
0 0
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4.6 Superposition Theorem
IfI= 1A , thenVI = (3+= 5) 8V
0 10
V
1
I== 2A , I =II+= 3A
1 2 10
4
V
2
V=V+2II =8+6=14V , == 2A
2 123
7
I=I+II = 5A = 5A
4 32 S
I=1AﬁI= 5A , II = 3Aﬁ=15A
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00SS
4.6 Superposition Theorem
4.6 Superposition Theorem
For a linear circuit N consisting of n inputs , namely
For a linear circuit N consisting of n inputs , namely
u u , u , u , , …… …… , u , u , then the , then the output y output y can be can be c calcu alcullate ated d
1 2 n
1 2 n
as the sum of its components：
as the sum of its components：
y = y + y + …… + y
y = y + y + …… + y
1 1 2 2 n n
where
where
y y =f(u =f(u ) , i=1,2, ) , i=1,2,…… ……,n ,n
i i
i i
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4.6 Superposition Theorem
Proof： Consider the nodal equation of the corresponding
Proof： Consider the nodal equation of the corresponding
circuit for the basic ca circuit for the basic cas se e a as an ex s an exa ample mple
G G L GIe
Ø ø Øø Øø
11 12 11 ns
1
Œ œ Œœ Œœ
G G L G eI
21 22 2ns 22
Œ œ Œœ Œœ
= LLL A
( )
Œ œ Œœ Œœ
M O MM M
Œ œ Œœ Œœ
G G L GI e
ºn12n nnß ºßn ºß ns
G e =IB LLLLLLLLLLLL
[ ] ( )
s
T T
Let G = [ G G … G ]
Let G = [ G G … G ]
k k k1 k1 k2 k2 kn kn
Then [G] = [ G G … G ]
Then [G] = [ G G … G ]
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4.6 Superposition Theorem
4.6 Superposition Theorem
nn Cram Cramer er’’s s R Rule for sol ule for solv ving Ax=b ing Ax=b
T Take ake n n=3 as a =3 as an n e ex xam ampl ple. e.
a a a xb
Ø øØ ø Øø
11 12 13 11
Œ œŒ œ Œœ
a a a xb =
21 22 23 22
Œ œŒ œ Œœ
Œa a a œŒxb œ Œœ
º 31 32 33ßº13 ß ºß
Let
Let
det A = △≠ 0
det A = △≠ 0
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4.6 Superposition Theorem
b aa
1 12 13
Then
Then
detb aa
2 22 23
b aa
3 32 33
x =
1
D
a ba
11 1 13
deta ba
21 2 23
a ba
31 3 33
x =
2
D
a ab
11 121
deta ab
21 222
a ab
31 323
x =
3
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D
4.6 Superposition Theorem
4.6 Superposition Theorem
Suppose that the kth nodal voltage e is to be found.
Suppose that the kth nodal voltage e is to be found.
k k
Then Then f fr ro om Cra m Cram mer er’’s s rule one has rule one has
detØø G GLLIG
1 1 sn
ºß
e =
k
detG
[ ]
n
Δ
jk
=I

js
Δ
j=1
where D @ detG
[ ]
\e =e +e+LL +e
k k1 k2 kn
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4.6 Superposition Theorem
where
Δ
1k
e = I , due to I only
k1 11 ss
Δ
Δ
nk
e = I , due to I only
kn ns ns
Δ
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4.6 Superposition Theorem
4.6 Superposition Theorem
nn E Example 4.6.2 xample 4.6.2
Finde ? =
2
Nodal Equation Nodal Equation
G G G +G +G +G +G +G +G - - -G G G - - -G G G
G G G +G +G +G +G +G +G - - -G G G - - -G G G e e e e e e I I I I I I
1 1 1 1 1 1 4 4 4 4 4 4 6 6 6 6 6 6 4 4 4 4 4 4 6 6 6 6 6 6 1 1 1 1S 1S 1S
1 1 1 1S 1S 1S
- - -G G G G G G +G +G +G +G +G +G - - -G G G ＝＝
- - -G G G G G G +G +G +G +G +G +G - - -G G G e e e I I I
e e e I I I
4 4 4 4 4 4 2 2 2 2 2 2 4 4 4 4 4 4 5 5 5 5 5 5 5 5 5 5 5 5
2 2 2 2 2 2 2S 2S 2S 2S 2S 2S
- - -G G G - - -G G G G G G +G +G +G +G +G +G
- - -G G G - - -G G G G G G +G +G +G +G +G +G
e e e e e e I I I I I I
6 6 6 6 6 6 5 5 5 5 5 5 3 3 3 3 3 3 5 5 5 5 5 5 6 6 6 6 6 6
3 3 3 3 3 3 3S 3S 3S 3S 3S 3S
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4.6 Superposition Theorem
By using Cramer’s rule
By using Cramer’s rule
G+G+- GIG

1 4 616 S

det-- G IG
4 25 S

-G I G++ GG
Łł 6 3S 3 56
e =
2
D
DD D
32
12 22
= I++II
1S 23 SS
D DD
=e++ ee
21 22 23
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4.6 Superposition Theorem
4.6 Superposition Theorem
Where e is due to I only，I ＝I ＝0
Where e is due to I only，I ＝I ＝0
21 1S 2S 3S
21 1S 2S 3S
G G G G G G +G +G +G +G +G +G +G +G +G +G +G +G - - - - - -G G G G G G - - - - - -G G G G G G e e e I I I
e e e I I I
1 1 1 1 1 1 4 4 4 4 4 4 6 6 6 6 6 6 4 4 4 4 4 4 6 6 6 6 6 6 11 11 11 11 11 11 1S 1S 1S 1S 1S 1S
- - -G G G G G G +G +G +G +G +G +G - - -G G G ＝ ＝
- - -G G G G G G +G +G +G +G +G +G - - -G G G e e e 0 0 0
e e e 0 0 0
4 4 4 4 4 4 2 2 2 2 2 2 4 4 4 4 4 4 5 5 5 5 5 5 5 5 5 5 5 5
21 21 21
21 21 21
- - - - - -G G G G G G - - - - - -G G G G G G G G G G G G +G +G +G +G +G +G +G +G +G +G +G +G
e e e e e e 0 0 0 0 0 0
6 6 6 6 6 6 5 5 5 5 5 5 3 3 3 3 3 3 5 5 5 5 5 5 6 6 6 6 6 6
31 31 31 31 31 31
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4.6 Superposition Theorem
G G G +G +G +G +G +G +G - - -G G G - - -G G G
G G G +G +G +G +G +G +G - - -G G G - - -G G G e e e e e e I I I I I I
1 1 1 4 4 4 6 6 6 4 4 4 6 6 6
1 1 1 4 4 4 6 6 6 4 4 4 6 6 6 11 11 11 11 11 11 1S 1S 1S 1S 1S 1S
- - - - - -G G G G G G G G G G G G +G +G +G +G +G +G +G +G +G +G +G +G - - - - - -G G G G G G ＝ ＝
e e e e e e 0 0 0 0 0 0
4 4 4 4 4 4 2 2 2 2 2 2 4 4 4 4 4 4 5 5 5 5 5 5 5 5 5 5 5 5
21 21 21 21 21 21
- - -G G G - - -G G G G G G +G +G +G +G +G +G
- - -G G G - - -G G G G G G +G +G +G +G +G +G
e e e 0 0 0
6 6 6 6 6 6 5 5 5 5 5 5 3 3 3 3 3 3 5 5 5 5 5 5 6 6 6 6 6 6 e e e 0 0 0
31 31 31 31 31 31
G+G+- GIG

1 4 616 S

det0-- GG
45

-G0 G++ GG
Łł 6 3 56
\= e
21
D
D
12
= I , due to I only
11 SS
D
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4.6 Superposition Theorem
4.6 Superposition Theorem
Similarly
Similarly
Duo to I only Duo to I only
2S 3S
II==0 II==0
13 SS 12 SS
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4.7 Thevenin’s Theorem
In high In high school, school, one one f fin ind ds the s the e eq qui uiv va alle ent nt
resis resistta anc nce of a e of a t two te wo ter rm miin nal al r res esiisti stiv ve e c ciir rc cu uit it
witho withou utt sources. sources.
Now, Now, w we w e wiill ll f fiind t nd th he e e equi quiv va ale lent nt ci cir rc cui uit for t for two two
tter erm miinal nal resis resisttiiv ve e ci cir rc cuit uit with sour with sourc ces. es.
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4.7 Thevenin’s Theorem
4.7 Thevenin’s Theorem
Thevenin’s theorem states that a linear two-terminal
Thevenin’s theorem states that a linear two-terminal
c circ ircui uitt c can an b be r e re ep plla ac ced ed b by an y an eq equ uiiv va alle en ntt c circ ircui uit t
consisting of a voltage source V in series with a
consisting of a voltage source V in series with a
TH TH
resistor R where V is the open circuit voltage at
resistor R where V is the open circuit voltage at
TH TH
TH TH
tth he e tte er rm miin nals a als an nd R d R is is tth he inp e inpu ut or t or equ equiiv va ale len nt t
T TH H
resistance at the terminals when the independent
resistance at the terminals when the independent
sources are turned off .
sources are turned off .
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4.7 Thevenin’s Theorem
I
a
Linear
+
Connected
V
two-terminal
circuit
-
circuit
b
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4.7 Thevenin’s Theorem
4.7 Thevenin’s Theorem
Equival Equivale ent circu nt circuit it: sa : sam me e v voltage oltage- -c cu ur rrent rela rent relattion at ion at the the
terminals.
terminals.
V V = V = V : Open circuit : Open circuit voltage voltage a at a t a- -b b
TH OC
TH OC
V = V
V = V
TH TH OC OC
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4.7 Thevenin’s Theorem
R R = R = R : inp : input ut resista resistan nce of the dead circuit ce of the dead circuit
TH IN
TH IN
Turn off all independent sources
Turn off all independent sources
R R = R = R
TH IN
TH IN
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4.7 Thevenin’s Theorem
4.7 Thevenin’s Theorem
CASE 1
CASE 1
I If the network f the network h has no d as no de ependent pendent sources: sources:
- - T Tur urn o n of ff all f all indepe indepen ndent so dent sou urce. rce.
- R : input resistance of the network looking
- R : input resistance of the network looking
TH TH
into a-b terminals
into a-b terminals
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4.7 Thevenin’s Theorem
CASE 2
CASE 2
If the network has dependent sources
If the network has dependent sources
-Turn off all independent sources.
-Turn off all independent sources.
-Apply a voltage source V at a-b
-Apply a voltage source V at a-b
O
O
V
O
R=
TH
I
O
C C..T T.. Pan Pan 23 23
4.7 Thevenin’s Theorem
4.7 Thevenin’s Theorem
- -Al Alttern ernat ativ ive ely, ly, a apply pply a a c curre urren nt sour t sourc ce I e I a at a t a- -b b
O O
V
O
R=
TH
I
O
If R < 0, the circuit is supplying power.
If R < 0, the circuit is supplying power.
TH
TH
C C..T T.. Pan Pan 24 244.7 Thevenin’s Theorem
4.7 Thevenin’s Theorem
Simplified circuit
Simplified circuit
V
TH
I=
L
R +R
THL
R
L
V =R I=V
L LL TH
R +R
THL
Voltage divider
Voltage divider
C C..T T.. Pan Pan 25 25
4.7 Thevenin’s Theorem
4.7 Thevenin’s Theorem
Pro Proo of : Consi f : Consid der the foll er the follo owing li wing lin near two terminal ear two terminal circuit circuit
consi consis sti tin ng of n+1 nodes a g of n+1 nodes an nd choos d choose e termin termina all b as b as
datum node and terminal a as node n .
datum node and terminal a as node n .
L
V I

1 1 s
GG K

111 n
VI
22 s

M OM =

M M

GG L

Łł n1 nn
V I
Łł n ns
Łł
C C..T T.. Pan Pan 26 264.7 Thevenin’s Theorem
4.7 Thevenin’s Theorem
Then nodal vo Then nodal volltta age V ge V w when a hen a- -b terminal b terminals s are open are open
n
n
can be found by using C can be found by using Cr ramer amer’’s s r rule . ule .
n
1
VI =D LLL A ( )

n kn ks
D
k=1
is the determinant of [G] matrix
D is the determinant of [G] matrix
D is the is the c co or rresponding cofactor of responding cofactor of G G
kn
ku kn
Now connect an external resistance R to a-b terminals .
Now connect an external resistance R to a-b terminals .
o o
The new nodal voltages will be changed to e , e , … , e
The new nodal voltages will be changed to e , e , … , e
1 1 2 2 n n
respectively .
respectively .
C C..T T.. Pan Pan 27 27
4.7 Thevenin’s Theorem
4.7 Thevenin’s Theorem
Nodal equation
Nodal equation
G+0
G K
11 1n
e I

1 1s
M G+0
2n

eI
22s

MM = . . . . . . . . . B
( )

M M
1

GGL+
e I
n1 nn
Łł n Łł ns

R
o
Łł
C C..T T.. Pan Pan 28 284.7 Thevenin’s Theorem
4.7 Thevenin’s Theorem
Note th Note that at
G K G +0 G K 0

11 1n 11

M G + 0 G 0
2n 21

detMM =+ det G detMM
[ ]

11

GLL GG +
n11 nnn

RR
oo
Łł Łł

1
= D+D

nn
R
o
C C..T T.. Pan Pan 29 29
4.7 Thevenin’s Theorem
4.7 Thevenin’s Theorem
Hence , e Hence , e can b can be e obtain obtaine ed as fol d as folllows . ows .
n
n
GI K

111s

nn
det M OM
1

DDII
kn ks kn ks
GI L
R
D
Łł n1 ns
k== 11ko
eV = = ==
nn
11 1D
RR+
nn
o TH
D+D D+D1+
nn nn
RR R D
oo o
D
nn
R @
where
where TH
D
C C..T T.. Pan Pan 30 304.7 Thevenin’s Theorem
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 V in series with an equivalent resistance R , where
TH TH
D
nn
R =
V is the open circuit voltage V and .
TH
TH n
D
C C..T T.. Pan Pan 31 31
4.7 Thevenin’s Theorem
4.7 Thevenin’s Theorem
Example 4.7.1
Example 4.7.1
1
W
1
1
2
W
W
4
6
C C..T T.. Pan Pan 32 324.7 Thevenin’s Theorem
4.7 Thevenin’s Theorem
Example 4.7.1 (cont.)
Example 4.7.1 (cont.)
Find open circuit vo Find open circuit volltta age V ge V
2
2
2+- 42 V 52- V

x
1
=

-+ 2 262 VV
ŁłŁł 2 Łł x
22 VV =
x 1
V
2+4+- 225

1
=

-2-+ 2 260 V
Ł łŁł 2 Łł
82-

D= det = 64-=8 56

-48
Łł
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4.7 Thevenin’s Theorem
4.7 Thevenin’s Theorem
Example 4.7.1 (cont.)
Example 4.7.1 (cont.)
85

det

-40 205
Łł
\V= = == VV
2 TH
56 56 14
D 81
22
R= = =W
TH
D 567
1
W
7
a
∴ Ans.
∴ Ans.
5
V
14
b
C C..T T.. Pan Pan 34 344.7 Thevenin’s Theorem
4.7 Thevenin’s Theorem
Example 4.7.2
Example 4.7.2
20W
10W
10W
By vo By volltta age divi ge divid der principle : er principle :
open circuit open circuit v voltage V oltage V =10V =10V
TH
TH
Let ind Let inde ependent pendent source be zero source be zero
20
10
a
R =5+20=25 W
TH
10
b
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4.7 Thevenin’s Theorem
4.7 Thevenin’s Theorem
Example 4.7.3
Example 4.7.3
nn Find the Find the Theven Thevenin in’’s s equivalent equivalent c ciircuit of the circu rcuit of the circuiit t
shown below, to the left of the terminals a-b. Then
shown below, to the left of the terminals a-b. Then
find the current through R = 6, 16, and 36 W.
find the current through R = 6, 16, and 36 W.
L L
C C..T T.. Pan Pan 36 364.7 Thevenin’s Theorem
4.7 Thevenin’s Theorem
Example 4.7.3 (cont.)
Example 4.7.3 (cont.)
R : 32V voltage source ﬁ short
TH
2A current source ﬁ open
4·12
R =4P12 +1= +14 =W
TH
16
V
TH
R
TH
V
TH
C C..T T.. Pan Pan 37 37
4.7 Thevenin’s Theorem
4.7 Thevenin’s Theorem
Example 4.7.3 (cont.)
Example 4.7.3 (cont.)
:
:
V
VTH
TH
Mesh analysis
Mesh analysis
- + + - = = -
- + + - = = -
32 32 4 4ii 12 12 ( (ii ii ) ) 0 0 ,, ii 2 2A A
1 1 2 2
1 1 2 2
\ \ = =
i 0.5A
i 0.5A
1 1
= - = + =
= - = + =
V V 12 12 ( (ii ii ) ) 12 12( (0 0..5 5 2 2..0 0) ) 30 30V V
TH 1 2
TH 1 2
C C..T T.. Pan Pan 38 384.7 Thevenin’s Theorem
4.7 Thevenin’s Theorem
Example 4.7.3 (cont.)
Example 4.7.3 (cont.)
To get i :
To get i :
L L
V 30
V 30
TH
TH
= =
= =
ii
L L
+ +
+ +
R R R R 4 4 R R
TH L L
TH L L
= = =
= ﬁ ﬁ = =
R R 6 6 I I 30 30//10 10 3 3A A
L L L L
= ﬁ ﬁ
=
R R 16 16 = =
= =
I I 30 30//20 20 1 1..5 5A A
L L
L L
= =
ﬁ ﬁ = =
= = I I 30 30//40 40 0 0..75 75A A
R 36
R 36
L
L
L L
C C..T T.. Pan Pan 39 39
4.7 Thevenin’s Theorem
4.7 Thevenin’s Theorem
Example 4.7.4
Example 4.7.4
Find the Thevenin’s equivalent of the following
Find the Thevenin’s equivalent of the following
ci cir rcuit cuit with te with ter rminal minals s a a- -b. b.
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4.7 Thevenin’s Theorem
E Example 4.7.4 xample 4.7.4 ( (cont.) cont.)
(independent + dependent source case)
(independent + dependent source case)
To find To find R R fr fro om Fig.(a) m Fig.(a)
TH
TH
indepe indepen ndent so dent sou urce rce →→ 0 0
dependent source → unchanged
dependent source → unchanged
v
1
o
Apply v=1VR , ==
Apply
o TH
ii
oo
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4.7 Thevenin’s Theorem
4.7 Thevenin’s Theorem
E Example 4.7.4 xample 4.7.4 ( (cont.) cont.)
For loop 1 ,
For loop 1 , -2v+ 2(i-i)=- 0 or = v ii
xx 12 12
But
But
-4iv= =-ii
2 x 12
\ii =-3
12
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4.7 Thevenin’s Theorem
E Example 4.7.4 xample 4.7.4 ( (cont.) cont.)
Loop 2 and 3:
4i+ 2(i-i)+ 6(ii-=)0
2 21 23
6(i-ii )+2 += 10
3 23
Solving these equations gives
Solving these equations gives
1
iA =-
3
6
1
But i =-=iA
o 3
6
1V
\ R = =W 6
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TH
i
o
4.7 Thevenin’s Theorem
4.7 Thevenin’s Theorem
E Example 4.7.4 xample 4.7.4 ( (cont.) cont.)
To find To find V V f fr ro om Fig.(b) m Fig.(b)
TH TH
Mesh a Mesh an nalys alysis is
=
i 5
1
= -
v i i
- + - =
2v 2(i i ) 0 x 3 2
x 3 2
- + - + =
- - =
4(i i ) 2(i i ) 6i 0
12i 4i 2i 0
3
2 1 2 2
2 1 3
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4.7 Thevenin’s Theorem
E Example 4.7.4 xample 4.7.4 ( (cont.) cont.)
4
B But ut 4( (ii ii ) ) v v
- =
- =
1 1 2 2 x x
= =
\i 10/3.
\i 10/3.
2 2
= = =
= = =
V V v v 6 6ii 20 20V V
TH TH
oc 2
oc 2
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4.7 Thevenin’s Theorem
4.7 Thevenin’s Theorem
Example 4.7.5
Example 4.7.5
Determine the Thevenin’s
Determine the Thevenin’s
equivalent circuit :
equivalent circuit :
Solution:
Solution:
(dependent source only) (dependent source only)
v
o
VR== 0 ,
TH TH
i
o
Nodal analysis
Nodal analysis
v
o
i+ii=+ 2
o xx
4
C C..T T.. Pan Pan 46 46 46 464.7 Thevenin’s Theorem
4.7 Thevenin’s Theorem
Example 4.7.5 (cont.)
Example 4.7.5 (cont.)
But
But
0 -vv
oo
i= =-
x
22
v v vv
o o oo
ii= + =- + =-
ox
4 2 44
or 4 vi =-
oo
Thus
Thus
v
o
R = =-W 4 : Suppl : Supply ying Power ! ing Power !
TH
i
o
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4.8 Norton’s Theorem
4.8 Norton’s Theorem
n Norton’s theorem states that a linear two-terminal
n Norton’s theorem
circuit can be replaced by an equivalent circuit
consisting of a current source I in parallel with a
N
resistor R where I is the short-circuit current
N N
through the terminals and R is the input or
N
equivalent resistance at the terminals when the
independent sources are turned off.
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4.8 Norton’s Theorem
a
Linear
two-terminal
circuit
b
(a)
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4.8 Norton’s Theorem
4.8 Norton’s Theorem
Proof：
Proof：
By using Mesh Analysis as an example
By using Mesh Analysis as an example
Assume the linear two terminal circuit is
Assume the linear two terminal circuit is
a planar circuit a planar circuit a and there are n meshes nd there are n meshes
when a b termi when a b termin nals are als are s short circuited. hort circuited.
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4.8 Norton’s Theorem
Mesh equation for case 1 as an example
I V
RR ……
1 1S
111n

MM IV
22S

O =

MM M M

RR LL
I V
Łł n1 nn Łł n Łł ns
Hence the short circuit cuurent
n
1
IV =D
n kn ks
D
k=1
whereR D= det
[ ]
ik
D is the cofactor ofR
kn kn
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4.8 Norton’s Theorem
4.8 Norton’s Theorem
Now connect an external resistance R to a , b
Now connect an external resistance R to a , b
o o
terminals , then all the mesh currents will be
terminals , then all the mesh currents will be
changed changed to J to J , J , J , , ‥‥‥‥ J J，，respec respecttively. ively.
1 2 n
1 2 n
RR ……+ 0 J V

1 1S
111n

R + 0 JV
M
2nS 22

O =

M MM M

R LL R +RV J
Łn1 nnołŁłn Łł ns
Note that
RR ……+0 R K 0

111n 11

+ 0
M R MM
2n

detOO = D+ det

M M MM

R LL RR + RR L
no 1
Łł n1 nno Łł
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= D+D R
o nn4.8 Norton’s Theorem
4.8 Norton’s Theorem
Hence, one has
Hence, one has
RV …

111s

n
det M OM

DV
kn ks
RV L
Łł n1 ns
k =1
J==
n
D+RR D D+D
o nn o nn
n
1
D V
kn ks
D
k =1
=
D
nn
1+ R
o
D
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4.8 Norton’s Theorem
4.8 Norton’s Theorem
I
n
=
D
nn
1+R
o
D
R
N
= I
n
RR +
oN
D
where R== , II
N Nn
D
nn
C C..T T.. Pan Pan 54 54 54 544.8 Norton’s Theorem
4.8 Norton’s Theorem
Example 4.8.1 By using the above formula
Example 4.8.1 By using the above formula
4W
3W
3W
3W
Find the short circuit current I
Find the short circuit current I
3 3
I 10V
3+3 3+3 - -3 3 - -3 3 I 10V
1 1
I ＝ 0
- -3 3 3+3+4 3+3+4 - -3 3 I 0
2 2
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I 0
- -3 3 - -3 3 3+3 3+3 I 0
3
3
4.8 Norton’s Theorem
4.8 Norton’s Theorem
Example 4.8.1 (cont.)
Example 4.8.1 (cont.)
3+3 -3 -3 I 10V
3+3 -3 -3 I 10V
1 1
-3 3+3+4 -3 I I ＝ 0 0
-3 3+3+4 -3
2
2
-3 -3 3+3 I I 0 0
-3 -3 3+3
3
3
det R = 360- 27- 27- 27-- 90 54-= 54 108
[ ]
ik
6 -3 10

1 10 390 65

I= det-3 100= 39= == AI
( )
3 N

108 108 108 18

-- 3 30
Łł
D 108 36
R= = =W
N
D- 60 9 17
33
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4.8 Norton’s Theorem
E Example 4.8.2 xample 4.8.2
Find the Norton equivalent circuit of the following circuit
Find the Norton equivalent circuit of the following circuit
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4.8 Norton’s Theorem
4.8 Norton’s Theorem
Example 4.8.2 (cont.)
Example 4.8.2 (cont.)
To find R from Fig.(a)
To find R from Fig.(a)
N N
R= 5||(8++ 4 8)
N
205 ·
= 5 || 204 = =W
25
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4.8 Norton’s Theorem
Example 4.8.2 (cont.)
Example 4.8.2 (cont.)
To find I from Fig.(b)
To find I from Fig.(b)
N N
short short- -ci cir rcuit cuit ter term minal a inal a a an nd b d b
Mesh Analys Mesh Analysiis: s:

a
i = 2A
i = 2A
1 1

2A
i
SC
i
20i - 4i – 12 = 0 1 5Ω
20i - 4i – 12 = 0
=I
N
2 2 1 1 i
2
12V
∴∴ ii = = 1A = I 1A = I
2 N
2 N
8Ω b
(b)
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4.8 Norton’s Theorem
4.8 Norton’s Theorem
Example 4.8.2 (cont.)
Example 4.8.2 (cont.)
VTH
Alternative method for INN : I=
RTH
- -
V V :: op open en c ciir rc cuit uit volt volta age ge ac acr ross oss tte er rmin mina als ls a a a and nd b b
TH TH

Mesh analysis: a
Mesh analysis:

i=2A , 25ii-4-= 120
2A
3 43
i
i 4
3 5Ω V =v
TH SC
\=iA 0.8 12V
4
\v=V == 54 iV b

oc TH 4
(b)
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4.8 Norton’s Theorem
Example 4.8.2 (cont.)
Example 4.8.2 (cont.)
V VTH TH
,,
Hence Hence
= = = = = =
I 4 / 4 1A
IN N 4 / 4 1A
RTH
RTH
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4.8 Norton’s Theorem
4.8 Norton’s Theorem
Example 4.8.3
Example 4.8.3
nn Using Norton’s theorem, find R and I of the
Using Norton’s theorem, find R and I of the
N N
N N
following circuit.
following circuit.
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4.8 Norton’s Theorem
Example 4.8.3 (cont.)
Example 4.8.3 (cont.)
To find R from Fig.(a)
To find R from Fig.(a)
N N
v 1
o
iA = == 0.2
Hence Hence ,,
o
55
v 1
o
\ 5 R= = =W
N
i 0.2
o
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4.8 Norton’s Theorem
4.8 Norton’s Theorem
Example 4.8.3 (cont.)
Example 4.8.3 (cont.)
To find I from Fig.(b)
To find I from Fig.(b)
N N
10
iA == 2.5
x
4
10V
Ii =+ 2
Nx
5W
10
=+= 2(2.5)7A
5
\=IA 7
N
C C C C. .. .T T T T. .. . Pan Pan Pan Pan 64 64 64 644.9 Source Transformation
4.9 Source Transformation
R
a i
N
V v
s
b
The current through resistor R can be obtained
The current through resistor R can be obtained
as follows :
as follows :
V -vVvv
SS
iI = =-- @
S
R RRR
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4.9 Source Transformation
4.9 Source Transformation
From From K KCL, one CL, one can obt can obta aiin the fol n the folllowing owing
equivalent circuit
equivalent circuit
V
S
whereI @
S
R
C C..T T.. Pan Pan 66 66 66 664.9 Source Transformation
4.9 Source Transformation
The voltage across resistor R can be obtained as
The voltage across resistor R can be obtained as
follows :
follows :
v=() I-iR=IR-- iR @V iR
S SS
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4.9 Source Transformation
4.9 Source Transformation
From From K KVL, one VL, one can obt can obta aiin the fol n the folllowing owing
equivalent circuit
equivalent circuit
where V @RI
SS
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4.9 Source Transformation
Example 4.9.1
Example 4.9.1

a
a
10A 3Ω
30V
b
b
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4.9 Source Transformation
4.9 Source Transformation
Example 4.9.2
Example 4.9.2
nn Find the Find the Theven Thevenin in’’s equivalent s equivalent
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4.9 Source Transformation
Example 4.9.2 (cont.)
Example 4.9.2 (cont.)
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4.10 Maximum Power Transfer Theorem
4.10 Maximum Power Transfer Theorem
a
R
L
b
nn Problem : Given a linear resistive circuit N
Problem : Given a linear resistive circuit N
shown as above, find the value of
shown as above, find the value of
R that permits maximum power
R that permits maximum power
L L
delivery delivery to R to R
L L . .
C C..T T.. Pan Pan 72 724.10 Maximum Power Transfer Theorem
4.10 Maximum Power Transfer Theorem
Solution : First, replace N with its Thevenin
Solution : First, replace N with its Thevenin
equivalent circuit.
equivalent circuit.
R
TH
a
i
+
V R
TH L
-
b
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4.10 Maximum Power Transfer Theorem
4.10 Maximum Power Transfer Theorem
V
22
TH
p== iRR ()
L
RR+
THL
dp
0 ,
Let =
dR
L
2
VV
2
TH TH
Then R =R and PR== ()
L TH max L
24 RR
LL
C C..T T.. Pan Pan 74 744.10 Maximum Power Transfer Theorem
4.10 Maximum Power Transfer Theorem
nExample 4.10.1
(a) Find R that results in maximum power transferred to R .
L L
(b) Find the corresponding maximum power delivered to R ,
L
namely P .
max
(c) Find the corresponding power delivered by the 360V
source, namely P and P /P in percentage.
s max s
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4.10 Maximum Power Transfer Theorem
4.10 Maximum Power Transfer Theorem
150
So lu tion : (a ) VV TH== (360 ) 300
180
150· 30
R TH== 25Ω
180
2
3 00

(b ) PW m ax＝ 25= 900

50
Łł
C C..T T.. Pan Pan 76 764.10 Maximum Power Transfer Theorem
4.10 Maximum Power Transfer Theorem
300
Solution : (c) V ab= · 25= 150V
50
- 360 - 150
( )
is= = -7A
30
Pss =i 360 = -2520W (dissipated)
( )
P m ax 900
= = 35.71%
P s 2520
C C..T T.. Pan Pan 77 77
Summary
Summary
nnObj Obje ec cti tive 7 ve 7 : Unders : Understta an nd and d and be be a ab ble le to to use use
s superpos uperposiition tion tthe heor ore em. m.
nObjective 8 : Understand and be able to use
nObjective 8 : Understand and be able to use
Thevenin’s theorem.
Thevenin’s theorem.
nnObj Obje ec cti tive 9 ve 9 : Unders : Understta an nd and d and be be a ab ble le to to use use
Norton Norton’’s theo s theore rem. m.
C C..T T.. Pan Pan 78 78Summary
Summary
nnObj Obje ec cti tive 10 ve 10 :: Understand Understand and and be be ab ablle e t to use o use
source source t transform ransform tte ech chn niq iqu ue. e.
nObjective 11 : Know the condition for and be
nObjective 11 : Know the condition for and be
able to find the maximum
able to find the maximum
power transfer.
power transfer.
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Summary
Summary
nn Probl Proble em m : 4.60 : 4.60
4.64 4.64
4.68 4.68
4.77 4.77
4.86 4.86
4.91 4.91
nn Due wit Due with hin on in one e week. week.
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