2. Review of Last Class  Calculation of Input Resistance The Equivalent Change of electric Current source  Voltage source Key Points and Difficult Points

3. Chapter 3 General Analysis of Resistive Circuit Key mastery ： 1. Some Concepts,independent node,independent loop of graph methods 。 Branch-current methods node voltage methods loop current methods 2. Three analysis methods in circuit ： Key Points and Difficult Points

4. §3.1 graphs of circuit 1. Graph （ G ）： Complete Set of node and branch 。 ① ② ③ ④ R1R1 R2R2 R3R3 R4R4 R5R5 R6R6 iSiS i6i6 i5i5 i3i3 i2i2 i4i4 i1i1 + – uSuS ① ② ③ ④ i1i1 i2i2 i3i3 i4i4 i5i5 iSiS i6i6 ① ② ③ ④ 1 2 3 4 5 6 When voltage source and a resistance are series or current source and resistance are parallel,they can be viewed as a branch. 2. graphs of circuit ， the branch is described as abstract line 一、 Concepts 3.Directed Graph Designate the reference direction of current and the number of node in a graph 。 direction and the number of nodes must be identical to those of primary graph

5. 4. Planar Graph ： Each branch except for nodes are not intercross 。  When a node is moved out the branches which associate with it must be move out.Branch must end nodes 。 5. Connected Graph ： Every two nodes have more than one route 6. Loop ： A closed route is composed of branches 。 ① ② ③ ④ 1 2 3 4 5 6 7. Mesh ： In planar graphs a natural orifice,and it does not contain branches or nodes 。 notice ：  Move out a branch don’t mean moving out a node,so a isolated node can exist 。

6. If a planar graph contains n nodes and b branches, Branch ： n － 1 Link ： the number of branches which don’t compose tree is b-n+1 。 8.Tree （ T ） a connected subgraph,it contains all nodes and no loops. ① ② ③ ④ ① ② ③ ④ 9. Single branch loop （ basic loop) :only a branch loop 。 have (b-n+1) number of single branch of loop.

7. 13. The principle of stating independence loop ： every loop of independence loop group at least includes one branch which is not included by other loops 。 10. Independence node: the number of nodes that can state independence KCL equation is （ n － 1 ） choose a node as reference nodes ， other nodes are independence nodes 。 11. independence loop: state KVL Independent Equations 。 12.Independence loops ： a series of independence loops ① ② ③ ④ 1 2 3 4 5 6

8. 3 ① ② 1 2 Node ①② KCL equations ： Not independent When the circuit has n nodes,the number of KCL independence equations is n － 1.if the nodes don’s have equations they are reference potential node. 一、 KCL number of Independent Equations ： §3.2 KCL and KVL Number of Independent Equations conclusion ： Popularize

9. Connect two loops into one,so the loops are many but these loops are not independence 。 Basic loop ： choose a tree and add a branch to constitute a loop 。 3 ① ② ③ ④ 2 4 ⑤ 7 3 ① ② ③ ④ 1 2 4 5 6 ⑤ 8 7 1 5 6 8 Single branch loop and independence series In Plane circuit ， all meshes are a group of independence loops 。 二、 KVL number of Independent Equations ： Independence loop ？ I II Non-plane circuit

10. The number of independence KVL equation is （ b － n ＋ 1 ） KCL Independence equations （ n － 1) ＋ independence equations (b － n ＋ 1) ＝ b number of branch If circuit have n nodes, and b branches If choose branch current as Variable to state equation ， b number of Variable and b number of equation 。 Then add b number of VCR of branches, total 2b number of equation ， so call 2b method 。 一、 2b method taxing,usually not use The number of branch The number of plane circuit meshes The number of independent KCL equation is （ n － 1 ） §3.3 Branch Current Method

11. ① ③ ② 0 R1R1 R2R2 R3R3 R4R4 R5R5 R6R6 + – u S1 iS5iS5 i1i1 i6i6 i2i2 i4i4 i3i3 i5i5 + – u6u6 + – u1u1 + – u2u2 + – u4u4 + – u3u3 + – u5u5 1 2 3 KCL: basic idea ： choose branch current as variable and state ， KCL and KVL independence equations Deduce branch current equation ： KVL direction 二、 Branch Current Method

12. (1) Sign the reference direction of every branch current (voltage) ； (2) Choose (n–1) number of nodes ， state KCL equation; Steps of branch methods ： (5) Compute branch voltage and other analysis 。 (3) Choose b–(n–1) number of independence loop （ mesh or single branch loop ）， state KVL equation ； ( use the character of component (4) Solve the foregoing equation ， get b number of branch current ；

13. example1 :Given U S1 =130V ， U S2 =117V ， R 1 =1  ， R 2 =0.6  ， R 3 =24  。 Solve the power releasing by each branch current and voltage source 。 21 I1I1 I3I3 U S1 U S2 R1R1 R2R2 R3R3 b a + – + – I2I2 solution ： Node a ： –I 1 –I 2 +I 3 =0 (1) n–1=1 KCL equation ： (2) b–n+1=2 KVL equation ： R 2 I 2 +R 3 I 3 = U S2 R 1 I 1 –R 2 I 2 =U S1 –U S2 0.6I 2 +24I 3 = 117 I 1 –0.6I 2 =130–117=13 –I 1 –I 2 +I 3 =0 (3) Unit to solve I 1 =10 A I 3 = 5 A I 2 = –5 A Solve

14. P U S1ab =U S1 I 1 =130  10=1300 W P U S2re =U S2 I 2 =117  (–5)= –585 W Exam power invariable ： P R 1ab =R 1 I 1 2 =100 W P R 2ab =R 2 I 2 2 =15 W P R 3ab =R 3 I 3 2 =600 W P ab =715 W P re = P ab (4) Power Analysis P ab express absorb power P re express release power

15. Example 2 :state branch current equation of circuit shown as figure (including ideal current source) 。 b=5, n=3 KCL equation ： - i 1 - i 2 + i 3 = 0 (1) - i 3 + i 4 - i 5 = 0 (2) R 1 i 1 - R 2 i 2 = u S (3) KVL equation ： solve ： i 5 = i S (6) - R 4 i 4 + u = 0 (5) R 2 i 2 + R 3 i 3 + R 4 i 4 = 0 (4) R 1 i 1 - R 2 i 2 = u S (3) i 5 = i S (5) R 2 i 2 + R 3 i 3 + R 4 i 4 = 0 (4) i1i1 i3i3 uSuS iSiS R1R1 R2R2 R3R3 b a + – i2i2 i5i5 i4i4 c R4R4 2 1 3 + – u Also state equation group ：

16. Supplement example :State branch current equation of circuit with controlled source 。 - i 1 - i 2 + i 3 + i 4 =0 (1) - i 3 - i 4 + i 5 - i 4 =0 (2) i4i4 i1i1 i3i3 i2i2 i6i6 i5i5 uSuS  i1 i1 R1R1 R2R2 R3R3 b a + – c R4R4 +– R5R5  u 2 + – u2u2 Two steps of stating equation ： (1) View controlled source as independence source and state equation ； (2) Use variable to express controlled variable 。 solution : KCL equation ：

17. 4 1 2 3 R 1 i 1 - R 2 i 2 = u S (3) R 2 i 2 + R 3 i 3 + R 5 i 5 = 0 (4) -R 3 i 3 + R 4 i 4 =- µu 2 (5) -R 5 i 5 = -u (6) i 6 =  i 1 (7) u 2 = - R 2 i 2 (8) + – u i4i4 i1i1 i3i3 i2i2 i6i6 i5i5 uSuS  i1 i1 R1R1 R2R2 R3R3 b a + – c R4R4 +– R5R5  u 2 + – u2u2 omit equation (6) ， and get equation groups are still branch current KVL equation ： Supplement equation ：

18. Review of Last Class  1 、 Concepts ： Graph 、 Tree 、 Independence node,Independence loop  2 、 If the circuit has n nodes,b branches choose n–1 nodes,state KCL Independent Equations ； Choose b–(n–1) independence loops （ choose Mesh ）， state KVL Independent Equations ； 3 、 branch current method ：以 choose branch current as variable ， state KCL and KVL impendence equation

19. basic idea ： §3.5 Loop Current Method i1i1 i3i3 u S1 u S2 R1R1 R2R2 R3R3 b a + – + – i2i2 设独立回路电流分别为 i l 1 、 i l2 il2il2 il1il1 Choose Assumed loop current as sealed variable 。 State KVL impendence equation Given impendence loop current i l 1 、 i l2 Solve branch current from loop current i 1 = i l 1 i 3 = i l2 i 2 = i 3 - i 1 = i l2 - i l 1

20. The relation between of branch current and loop current il2il2 il1il1 i1i1 i3i3 u S1 u S2 R1R1 R2R2 R3R3 b a + – + – i2i2 u S3 + – KVL ： Deduce loop current equation ： Loop current equation Loop current method :choose loop current as sealed variable and state equation,so the number of equation is b–(n–1) For loop state KVL ， replace branch current by loop current Replace loop current by branch current

21. 看懂即可 Only need to understand Loop current equation: u l1 = u S1 - u S2 — the algebra sum of voltage of voltage source loop 1 。 R 11 =R 1 +R 2 — resistance of loop 1 。 It is equal to the sum of all resistances loop 2 。 R 12 = R 21 = –R 2 — the mutual resistance of loop 1 and loop 2 （ none controlled source ）。 u l1 = u S1 - u S2 — the algebra sum of voltage of voltage source loop 1 。 u l2 = u S2 —u S3 the algebra sum of voltage of voltage source loop2 。 R 22 =R 2 +R 3 — resistance of loop 2 。 It is equal to the sum of all resistances of loop 2 。 R 11 i l 1 + R 12 i l2 =u Sl1 R 12 i l 1 + R 22 i l2 =u Sl2 Equation standard form

22. When voltage source is written in right of equation,voltage source reference direction is same as loop current direction ， it is negative or else positive 。 Namely ： electric potential is positive il2il2 il1il1 i1i1 i3i3 u S1 u S2 R1R1 R2R2 R3R3 b a + – + – i2i2 u S3 + – When the directions of two loop current flowing through associated branch is same, the mutual resistance is positive or else negative 。 Only need to know Loop current equation R 11 i l 1 + R 12 i l2 =u Sl1 R 12 i l 1 + R 22 i l2 =u Sl2 Equation standard form Own resistance is positive

23. Generally ， for the circuit with l=b - (n - 1) loops including ： + : the direction of loop current flowing mutual resistance is same 0 :two loop is not relation or only having independence source or controlled R jk : mutual resistance Example : line network with non-controlled source R jk =R kj, matrix is balanced 。 Including controlled source ， controlled source as independence source is stated the right of equation ， it is balanced 。 R 11 i l1 +R 12 i l1 + …+R 1l i ll =u Sl1 … R 21 i l1 +R 22 i l1 + …+R 2l i ll =u Sl2 R l1 i l1 +R l2 i l1 + …+R ll i ll =u Sll R kk : own resistance (positive) k=1,2,…,l ( loop direction is same as loop current reference direction) 。 - : the direction of loop current flowing mutual resistance is reverse KVL ： the algebra sum of potential drop of loop current in loop resistance=the algebra sum of potential rise of loop voltage source Only need to know

24. General Steps of Loop Method (1) Choose l=b - (n - 1) loop ， single branch,loop current and the direction ； (2) for l independence loops ， choose branch current as variable and state ， KVL equation ； （ 3 ） replace branch current by loop current For plane circuit ， choose mesh current as independence loop ， this loop current is called mesh current ， this analysis is called mesh current method §3.4 mesh current method ：（ only use in plane ） Choose natural mesh as loop,namely mesh current method

25. State loop current equation u S1 R1R1 R2R2 R3R3 b a + – u S3 + – i l1 i l2 Solution II ： Choose loop as shown figure Solution I ： I1I1 I3I3 I2I2 U S1 – I 1 R 1 –I 2 R 2 =0 U S1 – I 1 R 1 –I 3 R 3 – U S3 =0 I 1 =I L1 + I L2 I 3 =I L2 I 2 =I L1 U S1 – （ I L1 + I L2 ） R 1 –I L1 R 2 =0 U S1 – （ I L1 + I L2 ） R 1 –I L2 R 3 – U S3 =0 The advantage Solution I ： Not need to memory formula KVL example4

26. use loop method to solve each branch current 。 solution ： (1) Singe the reference direction of branch,loop current (clockwise) (2) state KVL equation (R 1 +R 2 )I a -R 2 I b = U S1 - U S2 -R 2 I a + (R 2 +R 3 )I b - R 3 I c = U S2 -R 3 I b + (R 3 +R 4 )I c = -U S4 It is Matrix ， and mutual resistance is negative (3) Solve loop current equation I a, I b, I c (4) Solve each branch current ： IaIa IcIc IbIb + _ U S2 + _ U S1 R1R1 R2R2 R3R3 + _ U S4 R4R4 I1I1 I2I2 I3I3 I4I4 I 1 =I a, I 2 =I b -I a, I 3 =I c -I b, I 4 =-I c example5

27. （ 1 ） view VCVS as independence voltage and state equation ； （ 2 ） find out the relationship between controlled variable and loop current 。 4I a -3I b =2 （ 1 ） -3I a +6I b -I c = - 3U 2 （ 2 ） -I b +3I c =3U 2 （ 3 ） U 2 =-3I 2 =3(I b -I a ) （ 4 ） use loop method to each branch current of current including controlled voltage source 。 IaIa IbIb IcIc solution ： + _ 2V  33 U2U2 + + 3U 2 – 11 22 1  22 I1I1 I2I2 I3I3 I4I4 I5I5 The steps of solution ： Loop current equation example6

28. 4I a - 3I b =2 - 12I a +15I b - I c =0 9I a - 10I b +3I c =0 I a =1.19A I b =0.92A I c = - 0.51A Each branch current ： I 1 = I a =1.19A solve Because of including controlled source, matrix of equation is unbalanced (4) substitution system of equations I 2 = I a - I b =0.27A I 3 = I b =0.92A I 4 = I b - I c =1.43A I 5 = I c =–0.52A IaIa IbIb IcIc + _ 2V  33 U2U2 + + 3U23U2 – 11 22 1  22 I1I1 I2I2 I3I3 I4I4 I5I5

29. Solution 2 ： choose independence loop,make the branch including non-company current source belong to a single branch.this loop current is I S 。 Loop current is variable 。 I1=ISI1=IS - R 2 I 1 +(R 2 +R 4 +R 5 )I 2 +R 5 I 3 = - U S2 R 1 I 1 +R 5 I 2 +(R 1 +R 3 +R 5 )I 3 =U S1 I1I1 I2I2 I3I3 _ + _ U S1 U S2 R1R1 R2R2 R5R5 R3R3 R4R4 ISIS + assignments ： 3--3 、 7 、 8 、 10 、 12

30. ① 0 ③ ② + _ u S3 i S1 R1R1 R2R2 R3R3 R6R6 R5R5 i S6 R4R4 §3.6 Node Voltage Method---KCL Choose node voltage as unknown variable and state KCL equation 。 i6i6 i5i5 i4i4 i2i2 i1i1 i3i3 Node voltage ： choose reference nodes ， the voltage of other nodes compared to reference nodes ： u nj u n1 u n2 u n3 KCL ： basic idea ：

31. Deduce node voltage equation ： node ① node ② node ③ Use node voltage to express branch current and get node voltage equation ① 0 ③ ② + _ u S3 i S1 R1R1 R2R2 R3R3 R6R6 R5R5 i S6 R4R4 i6i6 i5i5 i4i4 i2i2 i1i1 i3i3

32. Use node method to solve each branch current 。 (1) State node voltage equation (0.05+0.025+0.1)U A - 0.1U B = 6 - 0.1U A +(0.1+0.05+0.025)U B = - 6 solution 二： 20k  10k  40k  20k  40k  +120V - 240V A B I4I4 I2I2 I1I1 I3I3 I5I5 solution 一： example8 Not need to memory formula

33. (1) Choose reference node, sign n - 1 independence nodes ； (2) For n - 1 independence nodes,choose voltage nodes as unknown variable,state KCL equation ； (3) Solve preceding equations,get n - 1 node voltage ； (5) Other analysis 。 (4) Solve each branch current (use node voltage to express) The steps of node methods:

34. i1i1 i6i6 i5i5 i4i4 i3i3 i2i2 ① 0 ③ ② + _ u S3 i S1 R1R1 R2R2 R3R3 R6R6 R5R5 i S6 R4R4 Deduce node method formula,only need to know

35. G 11 =G 1 +G 4 +G 6 — own conductance of node 1 G 22 =G 2 +G 4 +G 5 — own conductance of node 2 G 12 =G 21 = - G 4 — mutual conductance between node 1 and node 2 (no controlled source) G 33 =G 3 +G 5 +G 6 — own conductance of node 3 G 13 =G 31 = - G 6 — mutual conductance between node 1 and node 3 (no controlled source) G 23 =G 32 = - G 5 —mutual conductance between node2 and node 3 (no controlled source) i Sn1 =i S1 - i S6 — the algebra sum of all current of current source flow in node 1 。 i Sn2 = 0 — the algebra sum of all current of current source flow in node 2 。 i Sn3 =G 3 u S3 + i S6 —the algebra sum of all current of current source flow in node3 。 Only need to know

36. i1i1 i6i6 i5i5 i4i4 i3i3 i2i2 ① 0 ③ ② + _ u S3 i S1 R1R1 R2R2 R3R3 R6R6 R5R5 i S6 R4R4 i3i3 R3R3 ③ 0 Only need understand

37. i1i1 i6i6 i5i5 i4i4 i3i3 i2i2 ① 0 ③ ② + _ u S3 i S1 R1R1 R2R2 R3R3 R6R6 R5R5 i S6 R4R4 Directly State equation ： （ 1 ） own conductance ： it is equal to the algebra sum of all conductance of branches which connect with nodes.It is positive 。 Because the electric potential of reference node is zero, while the electric potential of independence node is large zero 。 （ 2)mutual conductance: it is equal to the algebra sum of all conductance of branches which connect between two nodes.it is negative. Because between neighbor nodes, the branch current must flow in one node,and flow out the other node. （ 3 ）： current sources are written in the right of equation, current of current source flowing in node is positive while flowing out node is negative 。 看懂即可 Only need understand

38. G 11 u n1 +G 12 u n2 +…+G 1,n - 1 u n,n - 1 =i Sn1 G 21 u n1 +G 22 u n2 +…+G 2,n-1 u n,n-1 =i Sn2        G n - 1,1 u n1 +G n - 1,2 u n2 +…+G n-1,n u n,n - 1 =i Sn,n - 1 G ii —own conductance ， it is equal to the algebra sum of all conductance of branches which connect with node I (including the branch that voltage source series connect with resistance 。 It is positive 。 i Sni —it is equal to the algebra sum of all current of current source which flow in node I (including the equivalent current source that voltage source series connect with resistance 。 G ij = G ji —mutual conductance ， it is equal to the algebra sum of all conductance of branches which connect with node I and node j ， it is negative 。（ no controlled source ） KCL ： the algebra sum of current flowing out nodes=the algebra sum of current flowing in nodes 看懂即可 Generally ： （ if n nodes in circuit ） Only need understand

39. ① ② i S1 R1R1 R3R3 R2R2 gmuR2gmuR2 +uR2uR2 _ R5R5 R4R4 R6R6 Supplement example ： state the equation of node voltage of VCCS circuit 。 (1) First choose controlled source as independence,then state equations ； (2) Use node voltage to express controlled variable ； The steps of solution ： Choose appropriate node to compute Notice ： when state node voltage,do not need to consider the resistances series connection with current source

40. (G 1 +G 2 )U 1 - G 1 U 2 +I =0 - G 1 U 1 +(G 1 +G 3 + G 4 )U 2 - G 4 U 3 =0 - G 4 U 2 +(G 4 +G 5 )U 3 - I =0 U 1 - U 3 = U S U 1 = U S - G 1 U 1 +(G 1 +G 3 +G 4 )U 2 - G 3 U 3 =0 - G 2 U 1 - G 3 U 2 +(G 2 +G 3 +G 5 )U 3 =0 G3G3 G1G1 G4G4 G5G5 G2G2 + _ Us I ① ③ ② G3G3 G1G1 G4G4 G5G5 G2G2 + _ ① ③ ② 补例 Supplement example: State voltage equation of nodes of circuit with None-company voltage Source 。 Solve 1:add voltage source current as variable to KCL equation ， at the same time add a relation equation of node voltage and voltage source 。 Solve 2 ： choose appropriable reference nodes

41. branch loop node n-1n-1 b-n+1b-n+1 0 0 n-1n-1 b-n+1b-n+1 n-1n-1 b-n+1b-n+1 b the conclusion of Compare among branch method, loop method and node method ： (1) The compare of the number of equations KCL equationKVL equationTotal number (2) For non-plane circuit,it is difficult to choose independence loop but choose independence node is easily 。 (3) It is easy to program with loop method, and node methods 。 At present use computer to analyze network with node methods 。 Assignment ： 3--18 、 20 、 23

42. Must draw designs must sign reference direction Notice the difference from physical methods Problems in assignments ： Teach 3-18 （ a) ， Especially notice the resistances series connection with current source

43. (R 1 +R 2 )I 1 - R 2 I 2 =U S1 +U S2 +U i - R 2 I 1 +(R 2 +R 4 +R 5 )I 2 - R 4 I 3 = - U S2 - R 4 I 2 +(R 3 +R 4 )I 3 = - U i I S =I 1 - I 3 I1I1 I2I2 I3I3 _ + UiUi _ + _ U S1 U S2 R1R1 R2R2 R5R5 R3R3 R4R4 ISIS + 讲： solve ： state loop current equation of branch circuit with None- company Current Source 。 Solution 1 ： choose voltage of current source as variable ， increase relation equation of mesh current and current of current source mesh current is variable