1. Chapter 4 Network Theorem
Outline：
Superposition Theorem
Substitution Theorem
Thevenin’s and Norton’s Theorems

2. 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.

3. §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.) + – uS
Source deactivated（set it equal to zero）
voltage source（us=0) short -circuit is
Current source (is=0) open -circuit

4. proof of superposition theorem：find the expression of voltage labeled u1 and current labeled i2.
Two sources act
synchronously R1 R2 uS + – is i2 i1 u1
2.Current source acts alone，uS=0 short-circuit
3.Voltage source acts alone，iS=0 open-circuit R1 R2 is + – R1 R2 uS + –

5. Conclusion:Superposition theorem dose’t apply to power.
Example4-1 Find current labeled i and power P of R applying superposition theorem。
12V 4 6 6V + – i R
Solution
12V voltage
source acts alone：
6V voltage
source acts alone： 4 6 + – R
12V
If add the two powers applying superposition,would be 4 6 6V + – R
Conclusion:Superposition theorem dose’t apply to power.

6. Example4-2 Find voltage Us in terms of superposition theorem
Solution: + –
10V 6 I1 4A Us
10 I1 4
(1) 10V voltage source acts alone：
10V + – 6
I1'
10 I1' 4 -
Us'
U1'
(2) 4A current source acts alone： 6
I1'' 4A + –
Us''
10 I1'' 4
U1"

7. (1) 10V voltage source and 4A current source act synchronously ：
e.g.4-3 In above example ,determine Us again when combining a 6V voltage source with a 4 resistor in series. I1
10 I1 + – 6 4A Us 4
10V + – 6V
(1) 10V voltage source and 4A current source act synchronously ：
Solve it group by group
10 I1 + – 6 4A
US⑴׳ 4
(2) 6V-voltage source acts alone： 4 + – 6V 6
I1⑵
10 I1 ⑵
Us ⑵ -

8. 2. Circuit’s structure parameter should be identical when applying it
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.

9. 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. ). R us r R
kus kr
if k =2，it can be demonstrated by employing superposition theorem. + –
2uS + – uS

10. Application of Homogeneity Principle ：
us1 r1
r1+ r2
us1
us2 R 1. R
us2 r2 R
k1 us1
k1 r1
k2 us2
k1 r1+ k2 r2 R
k1 us1 2. R
k2 us2
k2 r2
us1
us2 r R
k us1
k us2
k r R 3.
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.

11. (1) 10V voltage source and 4A current source act synchronously ：
example 4-3，according to superposition theorem,if change 6V to 8V,voltage Us that is created by 8V voltage source acting alone is：
10 I1 + – 6 4A Us 4 + – 6V 8V
(1) 10V voltage source and 4A current source act synchronously ：
(2) 6V voltage source acts alone：
8V voltage source acts alone

12. e.g.4-2 Find UL in T-shaped circuitB 2Ω + –
12V UL
20Ω IL U - C
Method 1：voltage divider&current divider
Method 2：source conversion
Method 3:homogeneity principle
solution：
UL= K IL RL
K = Us / U
if IL =1A U
If IL=1A

13. §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. A + – uk A ik + – uk
branch k ik A
Satisfy equivalent conversion

14. ? Notice： conditions that should be met
when applying substitution theorem:
1) both the original circuit and the circuit after replacement bear one and only solution respectively。
2.5A ?
1.5A
2.5A 1A
dissatisfy 5V
10V 2
10V 5V 2 5 ? A 1A B 1V + _ A 1A 1 B B A 1A 1V + - + - ? 1V + -
satisfy
dissatisfy
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 )

15. Discussion：substitution of the generalized branch+ – uk A ik + – uk
电路N ik A
When controlling variable of dependent sources A is in N,substitution cannot be used if controlling variable do not exist after substitution.

16. §4-3 Thevenin’s Theorem （French telegraph engineer）
一、Introduction to the theorem - +
Passive one-port： R0 R2 R4 R1 R3
battery model
real voltage source
Thevenin’s equivalent
Key point 2A + - 4
18V
Active one-port + _ ？ - +

17. 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 ） N0 NS
Passive linear one-port ：
One-port of not containing independent source but containing linear resistor or dependent source.
Active linear one-port：
One-port of containing independent source、linear resistor or dependent source. - + B A + - A B - +

18. External characteristic Output characteristic
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 open-circuit 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. R0
UOC + - I U
UOC + -
Active linear one-port
U= Uoc – R0 I
External characteristic
Output characteristic
V-A characteristic
Uoc is open-circuit voltage of NS ;
R0 is corresponding input resistance of N0.

19. 三、Proof of Thevenin’s Theorem
Step 2 apply superposition
Step 1：apply substitution theorem i a b + – u NS i a b NS + – u R
substitute ？ a b N0 i + –
u'' R0 a b + – u' NS
suprerpose +
u'= uoc
u"= - R0 i i - u + R0
uOC R
Derive, u = u' + u" = uoc –i R0

20. Techniques for finding Uoc and R0
Example 1:Given circuits as follows，find I applying Thevenin’s Theorem。 I 6 4 6 2A I + -
18V a b 4 2A + -
18V a b a b
UOC + R0 -
UOC - +
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

21. Additional example： use Thevenin’s theorem to compute UR3 6 I1 + – 9V UR
6I1 a b
Uoc + – R0 a b
solution： (1 ) remove off unknown branch
（2）draw Thevenin’s equivalent circuit
(3) compute open-circuit voltage Uoc
Uoc=Uab=6I1+3I1
I1=9/9=1A
Uoc=9V 3 6 I1 + – 9V
6I1 a b + –
Uoc
4-7

22. Additional example：circuit as shown below，R is a changeable resistor，regulate R to make ammeter read zero，and calculate the value of R。 + - + - A - +
R02
UOC2 A - +
12V 6V 2Ω 6Ω 2A R 5Ω
UOC1 + -
R01
solution：
UOC1=6 +2×2 =10V
derive, R=30Ω
4-6

23. Additional example ： use Thevenin’s Theorem to find UR.3 6 I1 + – 9V UR
6I1 a b
Uoc + – R0 a b
solution： (1 ) Take off unknown branch
（2）draw Thevenin’s equivalent circuit
(3) compute open-circuit voltage Uoc 3 6 I1 + – 9V
6I1 a b
Uoc=Uab=6I1+3I1
I1=9/9=1A
Uoc=9V + –
Uoc
4-7

24. difficulty R0 = U /I a I1= b a b
(4) Determine equivalent resistance: R0
difficulty
R0 = U /I
method of adding external resistance （at N0）
U=6I1+3I1=9I1
I1= a 3 6 I1 + –
6I1 I U + – b
R0 = U /I=6 
U =9  =6I
(5)remove in unknown branch
notice：difference between UR and UOC
Uoc + – R0 a b 3 UR - +
4-8

25. b c 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. V R1 R2 a b c + - US
Real value is the open-circuit voltage between b and c ---Uoc
U is practical measured voltage
Relative measurement error： c b
Uoc + – R0 RV U -
When R1=20K，R2=30K，R0=12K RV=500K，measurement error is-2.34%

26. 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

27. 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。 R0 + -
UOC
4、compute equivalent resistance R0 at N。 R I
5、remove in unknown branches and compute the desired variable.
U= Uoc – R0 I
Analyze the load-driven ability of power source

28. Analogue electronic circuits
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

29. ？ Norton’s Theorem Thevenin’s Theorem NS i i + R0 ISC UOC - R0 a i + ub NS i + – u ？ R0 i
ISC i + - R0
UOC
experiment
Thevenin’s Theorem
Norton’s Theorem
4-9

30. 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 b NS a b Gi
Isc
It can be proved through source equivalent conversion in according to Thevenin’s Theorem.

31. Example4-7 find Norton’s equivalent circuit。
direction
40V 3A +
20 -
40
60V
Isc Ri
Isc
solution：
Notice：DO draw the equivalent circuit when applying Thevenin’s and Norton’s Theorems.

32. NS i R0 + R UOC - When R=R0 ,maximum power is obtained,the value is
Application Example
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？ a b NS i + – u R R i R0 + -
UOC
Regulate the value of R，maximum power occur at the condition of
Derive easily,
When R=R0 ,maximum power is obtained,the value is

33. Deduction course：
Uoc + –
Req R i

34. 15V 5V 2A + 20 - 10 5 85V R i 2A + 10 5 - 85V R i 10V 10 50V 30
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？
15V 5V 2A +
20 -
10 5
85V R i 2A +
10 5 -
85V R i
10V
10
50V
30 + - 5
85V R i U0 R0 + - R i

35. §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：
Relation between branch voltage and node voltage：
demonstration： 1 2 3 4 5 6 ① ③ ②
KCL：

36. Conversation of energy is Tellegen’s Theorem 1 special case.
KCL：
二、 Tellegen’s Theorem 2：
Assume two networks N and ，which are composed of different two-terminals components and are different in circuits;their branch voltage and branch current vectors are expressed as :
Regulate all the branch voltages and currents are passive sign convention. hence,
Similarly proved as precedent

37. example4-8 find current ix 。
KCL、KVL and Tellegen’s Theorem is called topological constraints，which is used in any lumped circuit.
example4-8 find current ix 。 R + -
10V 1A N i1 R + - 5V ix i2
solution：
Assume i1 , i2 ，direction as shown
According to Tellegen’s Theorem 2，we can get：

38. §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. i2
Linear resistive network N + – uS a b c d
( a) i1 c d
Linear resistive network N + – a b
(b) + – u1 u2 + –
demonstration：
Assume the num of branch is b in total，

39. Second form: current source acts as excitation,the response is voltage
Linear resistive network N + – uS a b c d
(a) i1 u1 u2 c d
Linear resistive network N + – a b
(b)
Second form: current source acts as excitation,the response is voltage
Linear resistive network N a b c d
(a) i1 + – u2 iS + – c d
Linear resistive network N a b
(b)

40. Assume total number of branches is b，+ – u1 a b c d
(a) i1 + – u2 iS + – c d
Linear resistive network N a b
(b) + –
demonstration：
Assume total number of branches is b，

41. Assume total number of branches is b ，
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 . + –
Linear resistive network N a b c d
(a) i1 iS i2 + – c d
Linear resistive network N a b
(b) + – u1
demonstration ：
Assume total number of branches is b ，

42. example4-9
Find circuit I 。 I 2 4 8 + –
10V 3
Loop method，nodal method，Thevenin’s Theorem
solution： 8 I 2 4 + –
10V 3 I1 I2 I3
I2 = 0.5 I1=0.5A
I3 = 0.5 I2=0.25A
I= I1-I3 = 0.75A
According to Reciprocal Theorem , I= 0.75A

43. Notice the direction Additional Example ：
Given circuit, find current I1。 R + _ 2V 2
0.25A R + _
10V 2 I1
solution：
Reciprocal R + _ 2V 2
0.25A
homogeneity
Notice the direction

44. Keep in mind i2 + – uS a b c d (a) i1 (b) u2 iS
Linear resistive network N + – uS a b c d
(a) i1
(b) u2 iS

45. 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.

46. §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：
2. dependent sources：
1. Component constraint：
rmi1 + - u2
Voltage and current i2
gmu1
Voltage and current sources
Resistance and conductance
U2 = rmi1
i2 =gmu1

47. 二、Conversion relationship duality：
Serial And Parallel Combinations
KVL与KCL + - u iS G1 G2 + - uS R1 R2 i
uS = R1i + R2i
iS = G1u + g2u

48. + -
uS1 R1 R2 i R3
uS2
iS1 G1 G2 i G3
iS2 ① ②
Nodal voltage and mesh current duality

49. Conclusion：1、Superposition Theorem
2、Thevenin’s Theorem、Maximum Power Transform Theorem
3、 Norton’s Theorem
homework：4—3、11、12、13、16、

50. Homework Problems： + + - - a a R0 4 - UOC UOC 2A 18V + b b
1、The direction for Open-Circuit Voltage is identical to that of equivalent source voltage. a b
UOC + R0 - 4 2A + -
18V a b
UOC - +
2、 The direction for short-Circuit current is identical to that of equivalent source current
40V 3A +
20 -
40
60V
Isc Ri
Isc
Explain：4-12 (a)

51. 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.