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Chapter 4 Network Theorem
Outline：
Superposition Theorem
Substitution Theorem
Thevenin’s and Norton’s Theorems
Properties of Linear Circuits :
（1) Multiplication of all independent source voltages
and currents by a constant k increases all the current
and voltage responses by the same factor k. This is
called homogeneity.
（2）The response that is brought about by bringing
two excitations to bear on circuit at the same time is the
sum of responses that are brought about by bringing the
excitation to bear on the circuit respectively ，which is
called Superposition.
§4.1 Superposition Theorem
一、Concept：
In any linear resistive network ,the voltage across or the current
through any resistor or source may be calculated by adding
algebraically all the individual voltages and currents caused by the
separate independent sources acting alone ,with all other independent
voltage sources replaced by short circuits and all other independent
current sources replaced by open circuits.
Act alone: one source acts at a time with the other sources
inactive.( equal to zero.)
Source
deactivated
（set it equal to
zero）
voltage source（us=0) short -circuit
+
uS
–
Current source (is=0) open -circuit
is
proof of superposition theorem：find the expression of
1. Two sources act
voltage labeled u1 and current labeled i2.
synchronously
R1i1  R2 i2  uS
R1
i1 R1
u1 
uS
R1 ( i2  i S )  R2 i2  uS
R1  R2
+ u1 – i
is
2
+
1
R1
i2 
uS 
iS  R1 R2 iS uS
R2
–
R1  R2
R1  R2
R R
1
2
2.Current source
acts alone，uS=0
short-circuit
3.Voltage source
acts alone，iS=0
open-circuit
R1
i2' 
iS
R1  R2
1
i2" 
uS
R1  R2
u1'   i2' R2
u1"  R1 i2"
R1 R2

iS
R1  R2
R1

uS
R1  R2
i1'
R1
+ u1' –
i2'
R2
i1" R1
+
uS
–
+ u1"–
i2"
R2
is
Example4-1 Find current labeled i and power P of R applying
superposition theorem。
i 4
Solution
12
1
R
i
'



1
A
4
12V voltage
4  4 // 4 2
6
4
source acts alone：
+
12V
6
i"  
 1 A
–
6V voltage
4  4 // 4
i' 4
source acts alone：
i  i'i"  0 A
p  i 2 R  0W
If add the two powers applying
superposition,would be
4
+
12V
–
R
6
4
p  i' R  i" R  8W
2
i" 4
2
R
Conclusion:Superposition
theorem dose’t apply to power.
+
6V
–
4
4
6
+
6V
–
Example4-2 Find voltage Us in terms of superposition theorem
Solution:
I1 6
(1) 10V voltage source acts alone：
U S '  10 I1' U 1'
U S '  10 I 1' U 1' 10V
10
I1 
 1A
64
 10 I 1' 4 I 1'
6
(2) 4A current source acts alone：
+
–
4
I 1  
 4  1.6 A
46
I1'' 6
U
"


10
I
"

U
"
S
1
1
46


U1 
 4  9.6V
 25.6V
46
4A
Us
–
4
–
10V
–
+
I1 '
U S  U S 'U S "  19.6V
10 I1
+
 6V
U S"  10 I1"U1"
+
10 I1'
+
–
+
4 U1'
–
10 I1''
+
–
+
+
4 U1" Us''
–
–
Us'
+
4A
e.g.4-3 In above example ,determine Us
again when combining a 6V voltage
source with a 4 resistor in series.
I1
6
10V
+
(1) 10V voltage source and 4A current
–
source act synchronously ：
U
（1）
S
Solve it
group by
group
 19.6V
(2) 6V-voltage source acts alone：
6
I

 0.6 A
64
（2）
2）
U S  10I（1 2） 4 I（
6
1
（2）
1
 9.6V
US  U
（1）
S
U
（2）
S
 29.2V
+
4
+
6V
–
6
–
I1⑵ 6
–
4A
+
Us
–
+
4
+
10 I1
10 I1
–
4A
+
US⑴‫׳‬
–
10 I1 ⑵
+
–
4
+
6V
–
+
Us ⑵
-
Pay attention to the following tips when applying
superposition theorem：
1.Superposition theorem can be only applied to calculating voltage and
current in linear circuit；It cannot be applied to power(power is
source’s quadric function)。It cannot be used in nonlinear
circuit,either.
2. Circuit’s structure parameter should be identical when applying it
3. Replace inactive voltage sources with short circuit and inactive
current sources open circuit.
4. Superposition theorem can be also used in linear circuit containing
dependent sources ; the dependent sources should be reserved .
5. Pay attention to reference direction when performing algebraic addition.
6. Yon can solve the problems by grouping the sources when applying
the theorem,and the number of sources within each subcircuit can be
more than one.
二、 Homogeneity Principle
The response(voltage or current) is proportional to excitation, provided
a single excitation (independent source) involved in the
circuit.(Multiplication of all independent source voltages and currents
by a constant k increases all the current and voltage responses by the
same factor k. ).
us
R
r
kus
if k =2，it can be demonstrated by
employing superposition theorem.
+
2uS
–
+
uS
–
+
uS
–
R
kr
Application of Homogeneity Principle ：
us1
r1
us1
R
1.
us2
r2
us2
R
k1 us1
2.
k2 us2
R
R
us1
3.
us2
R
k1 r1
k2 r2
k1 us1
k2 us2
r
k us1
k us2
r 1 + r2
R
k1 r1+ k2 r2
R
kr
R
In a linear circuit,multiplication of all independent
source voltages and currents by a constant k increases all the
current and voltage responses by the same factor k.
example 4-3，according to
superposition theorem,if change 6V to
+
8V,voltage Us that is created by 8V
voltage source acting alone is：
–
(1) 10V voltage source and 4A current
source act synchronously ：
(2) 6V voltage source acts alone：
（2）
US
 9.6V

（2）
US
6
+
10 I1
4
+
6V
8V
–
U
（1）
S
–
+
Us
–
 19.6V
8V voltage source
acts alone
9.6  8

 12.8V
6

（1）
（2）
U S  U S  U S  19.6  12.8  32.4V
4A
e.g.4-2 Find UL in T-shaped circuit
A
B
2Ω
Method 1：voltage
divider&current divider
Method 2：source conversion
Method 3:homogeneity principle
20Ω
20Ω
U
– -
IL
+
+ +
12V
2Ω
2Ω
UL 20Ω
–
C
solution：
if IL =1A
U
K = Us / U
UL= K IL RL
If IL=1A U BC  22V U AC  ( 22  1 )  2  22  26.2
20
12
12
26.2 22

U'  2  (
  1 )  26.2  33.02 K 
U' 33.02
20 20
12
U L  20 I L K  20 
 7.2 68V
33.02
§4.2 Substitution Theorem
This theorem states the following :
Any branch within a circuit may be replaced by an equivalent
branch ,provided the replacement branch has the same current
through it and voltage across it as the original branch.
ik
A
+
uk
–
A
–
branch
k
A
Satisfy equivalent
conversion
+
uk
ik
conditions that should be met
when applying substitution theorem:
Notice：
1) both the original circuit and the circuit after
replacement bear one and only solution respectively。
?
2.5A
2.5A
1A
dissatisfy
2
2
10V
5V
5
10V
5V
?
1.5A
1A
A
+
1V
B
+
_1V
1A
satisfy
A
+
1V
B
1
5V
A
+
1A
?
-
dissatisfy
1A
B
2) No coupled relationship between replacement branch and the parts
of circuits 。(the branches containing controlling variables of cannot
be replaced if the controlling variables don not exist )
Discussion：substitution of the generalized branch
ik
A
+
uk
–
A
+
–
电
路
N
A
uk
ik
When controlling variable of dependent sources A is
in N,substitution cannot be used if controlling
variable do not exist after substitution.
§4-3
Thevenin’s Theorem （French telegraph engineer）
一、Introduction to the theorem
-
+
R1
R2
R3
R0
Passive
one-port
：
R4
Key
point
4
18V
+
Thevenin’s equivalent
？
2A
+
battery model
real voltage source
-
+
_
Active
oneport
Concept:
One-port：a pair of terminals at which a signal may enter
or leave a network is called a port,and a network having only
one such pair of terminals is called a one-port .
(Single-port network）。 （ One-port= Two-terminals ）
Passive linear one-port ： N0
One-port of not containing
independent source but
containing linear resistor or
dependent source.
A
Active linear one-port： NS
One-port of containing
independent source、linear
resistor or dependent source.
+ - A
+
+
-
B
B
二、 Thevenin’s Theorem （ French
telegraph engineer ， work
established in 1883）
The theorem states the following：Given any linear
circuit,rearrange it in the form of two networks A and B
connected by two wires.Define a voltage Voc as the opencircuit voltage which appears across the terminals of A when B
is disconnected.Then all currents and voltages in B will
remain unchanged if all independent voltage and current
sources in A are “zeroed out”,and an independent voltage
source Voc is connected,with proper polarity ,in series with
the inactive A network.
Active
linear
oneport
I
+
UO
C-
U= Uoc – R0 I
Uoc is open-circuit voltage of NS ;
R0 is corresponding input resistance of N0.
+
-
UOC
R0
U
External characteristic
Output characteristic
V-A characteristic
三、Proof of Thevenin’s Theorem
proof：
Step 1：apply substitution theorem
i a
substitute
+
NS u– R
b
a
Step 2 apply superposition
NS
？
suprerpose
+
u'
–
NS
b
u'= uoc
Derive, u =
u'
+
u"
N0
+
R0
a
+
u''
–
i a
+
u
i
–
b
i
b
u"= - R0 i
= uoc –i R0
+
-
i
uOC
R0
+
u
-
R
Techniques for finding Uoc and R0
Example 1:Given circuits as follows ， find I applying Thevenin’s
I
Theorem。
Ia
4
18V 2A 6
+
b
a
+
4
UOC
18V 2A
+
R0
a
+
6
UOC
b
b
Solution:（1）take off unknown branch，build one-port containing
source
（2）draw Thevenin’s equivalent circuit—model of practical voltage
source
（3）find Uoc with proper direction labeled . UOC= 4×2-18 = -10V
（4）find R0： R0= 4
（5）remove in unknown branch，derive：I =-1A
Additional example： use Thevenin’s theorem to compute UR
– 6I1 +
6
+
–
9V
3
I1
a
+
UR
3
–
b
a
R0
+
Uoc
–
solution： (1 ) remove off unknown branch
（2）draw Thevenin’s equivalent circuit
(3) compute open-circuit voltage Uoc
b
Uoc=Uab=6I1+3I1
6
+
–
9V
3
– 6I1 +
I1
a
+
Uoc
I1=9/9=1A
Uoc=9V
–
b
4-7
Additional example ： circuit as shown below ， R is a
changeable resistor，regulate R to make ammeter read
zero，and calculate the value of R。
+
6V
2Ω
+ A+
2A
12V +
R
6Ω
5Ω
-
-
+
A
UOC1
R01
+
UOC2
R02
-
solution：UOC1=6 +2×2 =10V
U OC2
R

12  10V derive, R=30Ω
R 6
4-6
Additional example ： use Thevenin’s Theorem to find UR.
– 6I1 +
6
+
–
9V
3
I1
3
a
a
+
UR
R0
+
Uoc
–
–
b
solution： (1 ) Take off unknown branch
（2）draw Thevenin’s equivalent circuit
(3) compute open-circuit voltage Uoc
6
– 6I1 + a
+
I1
9V
–
3
+
Uoc
b
Uoc=Uab=6I1+3I1
I1=9/9=1A
Uoc=9V
–
b
4-7
(4) Determine equivalent resistance: R0
R0 = U /I
method of adding external
resistance （at N0）
– 6I1 +
6
a
I1
a
R0
+
Uoc
–
I
–
(5)remove in
unknown branch
3
b
U=6I1+3I1=9I1
+
3
difficulty
U
b
notice：
+ difference
between
UR
UR and
UOC
6
2
I1=6  3 I  3 I
2
U =9 
I =6I
3
R0 = U /I=6 
3
UR 
 9  3V
63
4-8
Application Example 1
Use dc voltmeter with internal resistance RV to measure
the voltage drop between terminals b,c;and analyze the
measurement error that caused by RV.
a
R1
+
b
U
-
R0
+
Uoc
–
Real value is the open-circuit voltage
between b and c ---Uoc
S
R2
c
U is practical measured U 
V
+
b
U
RV
-
c
voltage
Relative
measurement
error：
RV U
OC
RV  R0
U  U OC
RV

1
U OC
RV  R0

 R0
 100%
RV  R0
When R1=20K，R2=30K，R0=12K
RV=500K，measurement error is-2.34%
Summary: techniques for obtainingR0 ：
（1）use method of series-parallel to find resistance R0
in a passive resistive one-port N0
（2）use R0 = U/ I ,if there is resistors as well
as dependent sources in N0，
（3）……
difficulty
四、Steps of solving problems using Thevenin’s Theorem：
1、take off unknown branches in original circuit ，build
active one-port NS，
2、draw Thevenin equivalent circuit。
3、indicate the direction of UOC at the location of
NS ;compute open-circuit voltage UOCf。
4、compute equivalent resistance R0
at N。
5、remove in unknown branches and
compute the desired variable.
U= Uoc – R0 I
Analyze the load-driven ability of power source
R0
+
UOC
-
I
R
五、Pay attention to the following tips when applying Thevenin’s
Theorem：
(1)Make sure to draw the equivalent circuit。
(2) Take off sources：set voltage sources short-circuited and
current sources open-circuited with power supply’s
internal resistance ,all other resistances ,dependent
sources and circuit construction unchanged.
(3) take care of the direction of UOC particularly .
(4) Be sure to change to N0 when finding R0; when
there is dependent sources within N0,,method of
adding external voltage is available.,,
(5) For the Complex circuits ,it is
valid to use the theorem repeatedly .
Analogue
electronic circuits
4-9
i a
NS
+
u
–
？
b
i
i
+
-
ISC
UOC
R0
R0
experiment
Thevenin’s Theorem
Norton’s Theorem
4-9
二、Norton’s Theorem
Given any linear circuit,rearrange it in the form of two
networks A and B connected by two wires.Define a current Isc as
the short-circuit current that appears when B is disconnected.Then
all currents and voltages in B will remain unchanged if all
independent voltage and current sources in are “zeroed out”,and
an independent current source Isc is connected,with proper
polarity,in paralleled with the inactive A network.
a
a
Isc
NS
Gi
b
b
It can be proved through source equivalent
conversion in according to Thevenin’s Theorem.
Example4-7 find Norton’s equivalent
circuit。
20
40V
+
40
+
40V
-
3A
20
Isc
60V
+
directio
n
Isc
Ri
solution： 
60 40 40 
i SC   3     A
20 40 20 

i SC  1 A
1
Ri 
 8
1 1 1
 
20 40 20
Notice：DO draw the
equivalent circuit when
applying Thevenin’s and
Norton’s Theorems.
Application Example
NS
i a
+
u
–
b
R0
+
UOC
-
R
given active one-port ，connected with
changeable resistor R externally，what
value of R result in maximum power
absorbed by it？And what the maximum
power is？
2
OC
R
u
pi R
2
( R0  R )
2
i
R
Regulate the value of R，maximum
dp
power occur at the condition of
0
When R=R0 ,maximum power is
obtained,the value is
dR
Derive easily,
2
oc
u
pmax 
4 R0
Deduction course：
由
Req
+
Uoc
–
uoc
dp
2
pi R(
) R
0
Req  R
dR
2
R
i
namely,
dp
 uoc
dR
( Req  R )2  2 R( Req  R )
( Req  R )
4
0
( Req  R)2  2R( Req  R)  0
Req  2 RReq  R  2 RReq  2 R  0
2
2
2
Req  R  0
2
2
Req  R
pmax
2
oc
2
oc
u
u

 Req 
2
( 2 Req )
4 Req
Additional Example： given circuit,R is changeable，what value of R result in
maximum power be absorbed by it？And what the maximum power is？
10
20
+
15V
-
20
+
5V
-
10 10
10
2A
i
5
+
85V
-
10
R
10V+
-
15  5
uoc 
 20  5  10V Req  10
20  20


uoc  2  20  10  50V Req  30
50  85
U0 
 5  85  80V
35
30  5
R0 
 4.29 Ω
35
30
5
+
85V
-
-
uo

 315W
4 R0
R
i
50V+
R  R0  4.29
2
Pmax
5
+
85V
-
2A
i
R0
+
U0
-
R
i
R
§4.4 Tellegen’s Theorem
一、 Tellegen’s Theorem 1：
For a network containing n nodes and b branches,let the vector i=(i1，
i2…..，ib ) and u=(u1，u2…..，ub) stands for branch currents and
branch voltages respectively;regulate the branch currents and
branch voltages ，then：
b
u i
k k
0
Relation between
branch voltage and
node voltage
u1 ：
 un1
KCL：
u2  un1  un 2
i1  i2  i4  0
③
u3  un 2  un 3
 i 2  i 3  i5  0
u4  un 3  un1
 i 3  i4  i6  0
u5  un 2
k 1
demonst
ration：
4
②
①
2
1
3
5
6
0
u6  un 3
6
u i
k k
 u1 i1  u2 i2  u3 i3  u4 i4  u5 i5  u6 i6
k 1
 un1i1  (un1  un 2 )i2  (un 2  un3 )i3  (un3  un1 )i4  un 2i5  un3i6
 un1 ( i1  i2  i4 )  un 2 (  i2  i3  i5 )  un 3 (  i3  i4  i6 )  0
Conversation of energy is Tellegen’s
i1  i2  i4  0
KCL：
Theorem 1 special case.
 i 2  i 3  i5  0
二、 Tellegen’s Theorem 2：
Assume two networks N and N̂ ，which are composedof
i3 different
 i4  i6  0
two-terminals components and are different in circuits;their branch
voltage and branch current vectors are expressed as :
i  ( i1 , i2 ,..........., ib )
u  ( u1 , u2 ,..........., ub )
î  ( î1 , î2 ,..........., îb )
û  ( û1 , û2 ,..........., ûb )
Regulate all the branch voltages and currents are passive sign convention. hence,
b
 u î
k k
k 1
0
b

k 1
ûk ik  0
Similarly proved as
precedent
KCL、KVL and Tellegen’s Theorem is called topological
constraints，which is used in any lumped circuit.
example4-8 find current ix 。
i1
+
10V
1A
R
N
solution：
ix
5V
+
R
i2
N̂
Assume i1 , i2 ，direction as shown
According to Tellegen’s Theorem 2，we can get：
10  (  i x )  0  i2 
b
 u î
k k
0
3
 uk iˆk  ik Rk iˆk  ik uˆ k
b
0  (  i1 )  ( 5)  1   uˆ k i k  0
3
  10i x  5
i x  0.5 A
§4.5 Reciprocal Theorem
For a linear resistive network,if there is only one excitation，the
ratio of excitation and response stay unchanged when substituting
the location of them.
一、First form:voltage source acts as excitation,the response is
current.
+
u1
–
uS
a
i1
c
Linear
resistive
networ
kN
+
–
+
u2
–
a
+
i2
1
–
b
d
b
î1û
( a)
Linear
resistive
network
N
î 2 c
+
d
+
ûûS2
– –
(b)
demonstration：
Assume the num of branch is b in total，
b
u1  uS , u2  0; û1  0, û2  ûS
u1 î1  u2 î2 
uk îk  0
 uk  Rk ik ûk  Rk îk
k 3

û1 i1  û2 i2 
b
 û i
k k
k 3
0

uk îk  Rk ik îk  ( Rk îk )ik  ûk ik

u1î1  u2 î2  û1i1  û2 i2
a i1
+
+
u
u1 S
–
–
according to u1  uS , u2  0;
uˆ1  0, uˆ2  uˆS
u S iˆ1  uˆ S i2
u S uˆ S
namely ,

i2
iˆ1
b
a
+
û1
particularly , uS  uˆS , then i2  iˆ1
î1
–
b
Linear
resistive
network
N
(a)
Linear
resistive
networ
kN
c
+
u2
–
i2
d
î 2
c
+
–
+
û S û2
d
–
(b)
Second form: current source acts as excitation,the response is voltage
a i1
iS
b
Linear
resistive
network
N
(a)
c
+
u2
–
d
a
+
û1
–
b
Linear
resistive
network
N
î 2 c
î S
d
(b)
+
a i1
i2 c
+
u2
–
d
iS
u1
–
b
(a)
demonstration：
Assume total
number of
branches is b，

u2 uˆ1
namely ,

iS iˆS
î1
Linear
resistive
network
N
î 2 c
î S
d
(b)
i1   i S , i2  0; î1  0 , î2   î S
u1î1  u2 î2  û1i1  û2 i2
u2iˆS  uˆ1iS
a
+
û1
–
b
u1 î1  u2 î2 
û1 i1  û2 i2 
b
 u î
0
 û i
0
k k
k 3
b
k k
k 3
particular ly iS  iˆS , then u2  uˆ1
+
û2
–
三、Third form: current source acts as excitation,the response is
current before Reciprocal ; voltage source acts as excitation,the
response is voltage after Reciprocal .
a i1
iS +
u1
–
b
Linear
resistive
network
N
(a)
î1
c
+
u2
–
i2
d
a
+
û1
–
b
Linear
resistive
network
N
î 2 c
+
+
û S
–
û2
–
d
(b)
demonstration ：
Assume total number i1   i S , u2  0; î1  0 , û2  ûS
of branches is b ，
u1î1  u2 î2  û1i1  û2 i2
 uˆ i  uˆ i  0
1 S
namely,
S 2
i2
uˆ1

iS
uˆ S
paticularly, iS  uˆS , then i2  uˆ1
example4-9
+
10V
–
2
I
3
2
4
solution：
2
I
I3
2
8
Find circuit I 。
Loop method，nodal method，
Thevenin’s Theorem
+ 10V –
I1
3 I2
8
10
I1 
 1A
8  ( 2 // 2  3) // 4
I2 = 0.5 I1=0.5A
4
I3 = 0.5 I2=0.25A
I= I1-I3 = 0.75A
According to Reciprocal Theorem , I= 0.75A
Given circuit, find current I1。
Additional Example ：
_ 2V
0.25A
+
R
2
I1
R
2
solution：
Reciprocal
homogeneity
0.25A
2
Notice the direction
R
+
_ 2V
10
I 1  ( 0.25)  1.25 A
2
+
10V
_
Keep in mind
a i1
uS
Linear
resistive
network
N
+
–
b
a i1
iS
b
a i1
iS
b
a
c
i2
d
î1
c
+
u2
–
d
a
+
û1
–
b
c
a
+
û1
–
b
Linear
resistive
network
N
î 2 c
î S
d
(b)
i2
d
(a)
–
û S
(b)
(a)
Linear
resistive
network
N
+
d
b
(a)
Linear
resistive
network
N
Linear
resistive
network
N
î 2 c
Linear
resistive
network
N
î 2 c
+
û S
–
d
(b)
Pay attention to following tips when employing Reciprocal Theorem ：
(1) It can only be used in single-source linear circuit to help find volt ampere relation between source branch and another branch.
(2) voltage source acts as excitation,the
response is current; current source acts as
excitation,the response is voltage
Voltage and current is
reciprocal 。
(3) If voltage source acts as excitation ， the current source is
replaced by an open circuit in the original location and let it
connected with another branch in series.
(4)If current source acts as excitation ， the current source is
replaced by an short circuit in the original location and let it
connected with another branch
(4) Pay attention to directions of voltages and currents.
(5) Generally speaking,the theorem cannot be used in network
containing dependent sources.
§4.6 Duality Principle
Duality Principle states the following ：In a circuit, if some
elements(or equations) is substituted correspondingly by their
duality elements，the resulted new relation(or equations) is
valid.
一、duality elements：
1. Component constraint：
Voltage and current
Voltage and current sources
Resistance and conductance
u  Ri
i  Gu
2. dependent sources：
rmi1
+
+ u 2
U2 = rmi1
i2 =gmu1
gmu1
i2
二、Conversion relationship duality：
Serial And Parallel Combinations
KVL与KCL
R2
R1
+
uS
G1
i
G2
+
u
-
iS
uS = R1i + R2i
iS = G1u + g2u
R1
+
uS1
-
R3
R2
① G2
i
uS2
+
② i
iS2
iS1
G1 G3
( R1  R2 )im1  R2 im 2  uS 1
( G1  G2 )un1  G2 un 2  i S 1
 R2 im1  ( R2  R3 )im 2  uS 2
 G2 un1  ( G2  G3 )un 2  i S 2
Nodal voltage and mesh current duality
• Conclusion：1、Superposition Theorem
2、Thevenin’s Theorem、
Maximum Power Transform Theorem
3、 Norton’s Theorem
homework：4—3、11、12、13、16、
Homework Problems：
1、The direction for Open-Circuit Voltage is
identical to that of equivalent source voltage.
2、 The direction for shortCircuit current is identical
to that of equivalent source
current
20
40V
+
40
+
40V
-
3A
+
4
UOC
18V 2A +
20
60V
+
a
R0
+
UOC
-
b
b
Isc
Isc
Explain：4-12 (a)
a
Ri
Homework Problems：
Practical voltage source do not have Norton equivalent circuit ;
Practical current source do not have Thevenin equivalent circuit.
explain：4-13 (a)（b)
It is indispensable to draw equivalent circuits when
applying Thevenin’s Theorem to solve problems.