1. Chapter 4 Techniques of Circuit Analysis So far we have analyzed relatively simple resistive circuits by applying KVL and KCL in combination with Ohm’s law. However as circuit become more complicated and involve more elements, this direct method become cumbersome ( غير سهل, مرهقة ) Applying KVL and Ohm’s law, we have, Solving for i, we have,

2. In this chapter we introduce two powerful techniques of circuit analysis that aid in the analysis of complex circuit which they are known as Node-Voltage Method or Nodal AnalysisMesh-Current Method or Mesh Analysis In addition to these method we will develop other method such as source transformations, Thevenin and Norton equivalent circuit. However as circuit become more complicated and involve more elements, this direct method become cumbersome ( غير سهل, مرهقة )

3. 4.1 Terminology Planar Circuit : a circuit that can be drawn on a plane with no crossing branches as shown The circuit shown can be redrawn which is equivalent to the above circuit Planar Circuit

4. The circuit shown is non planar circuit because it can not be redrawn to make it planar the circuit

5. Describing a Circuit – The vocabulary Consider the following circuit Node A point where two or more circuit elements join a Essential node A node where three or more circuit elements join b path A trace of adjoining basic elements with no elements included branch A path that connects two nodes Essential branch A path that connects two essential nodes without passing through an essential node loop A path whose last node is the same as the starting node mesh A loop that does not enclose any other loops Planar Circuit A circuit that can be drawn on a plane with no crossing branches

6. Example 4.1 For the circuit identify the following

7. The nodes are a, b, c, d, e, f and g (a) All nodes ?

8. The essential nodes are b, c, e and g (b) The essential nodes ? A node where three or more circuit elements join

9. The branches are (c) All branches ? A path that connects two nodes

10. The essential branches are (d) All essential branches ? A path that connects two essential nodes without passing through an essential node

11. The meshes are (e) All meshes ?

12. (f) Two paths that are not loops or essential branches ? is a paths, but it is not a loop because it does not have the same starting and ending nodes Nor is it an essential branch because it does not connect two essential nodes Is a paths that are not loop or essential branches ?

13. (g) Two loops that are not meshes ? is a loop, but it is not a mesh because there are two loops within it

14. Simultaneous Equations- How many Let the circuit as shown There are nine branches in which the current is unknown Note that the current I is known

15. Since there are nine unknown currentTherefore we need nine independent equations Since there are 7 nodes There will be 6 independent KCL equations since the 7 th one can be written in terms of the 6 equations We still need 3 more equationsThey will come from KVL around three meshes or loops Note the meshe or loop we apply KVL around should not contain the current source I since we do not know the voltage across it

16. Since KCL at the non essential (connecting two circuit elements only) like a, d, f will have the following results Therefore we have 6 currents and 4 essential nodes (connecting three or more circuit elements) like b, c, e, g

17. Applying KCL to 3 of the 4 essential nodes since KCL on the 4 th node will have an equation that is not independent but rather can be derived from the 3 KCL equations Therefore Applying KCL to the essential nodes b, c, and e ( we could have selected any three essential nodes) we have KCL at node b we have KCL at node c we have KCL at node e we have Note applying KCL on node g Which is a linear combination of the other KCL’s

18. Therefore the remaining 3 equations will be derived from KVL around 3 meshes KVL around mesh 1 Since there are 4 meshes only, we will use the three meshes with out the current source KVL around mesh 2

19. 6 equations and 6 unknown which can be solved for the currents KCL Equations KVL Equations

20. 4.2 Node-Voltage Method Nodal analysis provide a general procedure for analyzing circuits using node voltages as the circuit variables Node-Voltage Method is applicable to both planar and nonplanar circuits Using the circuit shown, we can summarize the node-voltage methods as shown next :

21. Step 1 identify all essentials nodes 1 2 33 Do not select the non essentials nodes Step 2 select one of the essentials nodes ( 1, 2, or 3) as a reference node Although the choice is arbitrary the choice for the reference node is were most of branches, example node 3 Selecting the reference node will become apparent as you gain experience using this method (i,.e, solving problems)

22. Step 3 label all nonrefrence essentials nodes with alphabetical label as v 1, v 2 … v1v1 A node voltage is defined as the voltage rise from the reference node to a nonreference node v2v2 Step 4 write KCL equation on all labels nonrefrence nodes as shown next

23. KCL at node 1++= v1v1 v2v2 i1i1 i2i2 i3i3 By applying KVL Let us find,, similarly KVL on the middle mesh, we have

24. Therefore v1v1 v2v2 i1i1 i2i2 i3i3 ++=We have ++ = If we look at this KCL equation, we see that the current we notice that the potential at the left side of the  resistor which is tied to the + of the 10 V source is 10 V because the – is tied to the reference 10

25. v1v1 v2v2 i1i1 i2i2 i3i3 ++ = Therefore we can write KCL at node 1 without doing KVL’s as we did previously

26. v1v1 v2v2 i1i1 i3i3 i2i2 = KCL at node 2 + Similarly

27. v1v1 v2v2 10 ++ = = + Two equations and two unknowns namely v 1, v 2 we can solve and have

28. 4.3 The Node-Voltage Method and Dependent Sources If the circuit contains dependent source, the node-voltage equations must be supplemented with the constraint equations imposed by the dependent source as will be shown in the example next Example 4.3 Use the node voltage method to find the power dissipated in the  resistor

29. 12 33 The circuit has three essentials nodes 1, 2, and 3 and one of them will be the reference We need two node-voltage equations to describe the circuit We select node 3 as the reference node since it has the most branches (i.e, 4 branches)

30. v1v1 v2v2 20 The circuit has two non essentials nodes which are connected to voltage sources and will impose the constrain imposed by the value of the voltage sources on the non essential nodes voltage Applying KCL at node 1 Applying KCL at node 2

31. v1v1 v2v2 20 The second node equation contain the current i  which is related to the nodes voltages as Substituting in the second node equation, we have the following nodes equations Solving we have

32. 4.4 The Node-Voltage Method : Some Special Cases Let us consider the following circuit The circuit has three essentials nodes 1, 2, and 3 which means that two simultaneous equations are needed. From the three essentials nodes, a reference node has been chosen and the other two nodes have been labeled 12 33

33. However since the 100 V source constrains the voltage between node 1 and the reference node to 100 V. This mean that there is only one unknown node voltage namely ( v 2 ) v1v1 v2v2 100 Applying KCL at node 2 Since Knowing v 1 and v 2 we can calculate the current in every branch

34. This voltage source is connected between non reference and reference Nodal Analysis with voltage sources Some Special Cases

35. This voltage source is connected between two non reference nodes reference

36. Enclose the voltage source in a region called supper node Apply KCL to that region

37. KCL on the Supper node

38. The second equation come from the constraint on the voltage source Two equations and two unknowns, we can solve

39. Two voltage sources are connected between two non reference nodes reference

40. Two supper nodes touching or intersection

41. Dependent Voltage source supper node apply as will to dependent sources

42. Combine the two supper nodes to supper supper node

48. 4.5 Introduction to the Mesh-Current Method Consider the following planar circuit There are 4 essentials nodes

49. There are 7 essentials branches were the current is unknown Therefore to solve via the mesh-current method we must write 4 mesh -current

50. A mesh current is the current that exists only in the perimeter of a mesh as shown in the circuit below The mesh current satisfy Kirchhoff’s Current Law (KCL) At any node in the circuit, a given mesh current both enters and leaves the node Mesh current is not always equal branch current

51. The branch currentEqual the mesh current The branch currentDoesn't equal any of the mesh currents

52. Consider the following circuit were we want to solve for the currents The circuit has 3 essential branches and 2 essentials nodes Therefore to solve via the mesh-current method we must write 2 mesh -current

53. Applying KCL to the upper node (one KCL for two essentials nodes) Applying KVL around the two meshes Solving the KCL for i 3 and substituting in the KVL equations we have

54. Now consider the mesh currents i a, i b, and applying KVL around the two meshes we have The two equations are identical if we replace the currents i 1, i 2 with the mesh currents i a, i b

55. the branches currents i 1, i 2, i 3 can be written in terms of the mesh currents i a, i b Once we solve for the mesh current we can solve for any branch current Once we know the branch current we can solve for any voltage or power of interest current

56. Example 4.4

57. 4.6 The mesh-Current Method and dependent sources Example 4.5 Use the mesh-current method to determine the power dissipated in the  resistor

58. KVL on mesh 1 KVL on mesh 2 KVL on mesh 3

59. To solve, we use Cramer method as follows, we need only i 2 and i 3

60. 4.7 The Mesh-Current Method : Some Special Cases

61. Supermesh

62. The Mesh-Current Analysis of the Amplifier Circuit

63. Supermesh

64. 4.8 The Node-Voltage Method Versus the Mesh-Current Method

65. Supernode

68. Supermesh

69. 193 By inspection

70. 4.9 Source Transformations The Node-Voltage Method and the Mesh-Current Method are powerful techniques for solving circuits. We are still interested in in methods that can be used to simplify circuits like to what we did in parallel and series resistors and  to Y transformations A method called Source Transformations will allow the transformations of a voltage source in series with a resistor to a current source in parallel with resistor. The double arrow indicate that the transformation is bilateral, that we can start with either configuration and drive the other

71. Equating we have,

72. Example 4.8 (a) find the power associated with the 6 V source (b) State whether the 6 V source is absorbing or delivering power We are going to use source transformation to reduce the circuit, however note that we will not alter or transfer the 6 V source because it is the objective.

76. It should be clear if we transfer the 6V during these steps you will not be able to find the power associated with it

77. Example 4.9 (a) use source transformations to find the voltage v o ? Since the  resistor is connected across or in parallel to the 250 V source then we can remove it without altering any voltage or current on the circuit except the 250 V current which is not an objective any how Therefore we remove the  Our objective is v o Similarly the  resistor is connected in series with the 8 A source then we can remove it without altering any voltage or current on the circuit

78. Now the circuit become We now use the source transformation to replace the 250 V with the  resistor with a current source and parallel resistor We now combine the parallel resistors

79. The circuit now become

80. Let us find the Thevenin Equivalent of the following circuit 25 By inspection Therefore the Thevenin voltage for the circuit is 32 V

81. To find the R Th we deactivate the sources by shortening independent voltage sources and open independent current source

82. Let us find the Thevenin Equivalent using source transformations

83. The Thevenin Equivalent is represented as an independent voltage source V Th and a series resistor R Th Now let us put a short across the terminal and calculate the resulting current And this another method of finding R Th as

84. Let us find the i SC for the circuit shown before i SC can be found easily once we know the voltage v 2 To find v 2 select the reference as the lower node and write KCL 25 By inspection

85. Norton Equivalent

90. 4.64 Find Thevenini equivalent with respect to the terminalsand A Using short circuit current B Deactivating the independent sources

91. A Using short circuit current

92. 17.4

93. B Deactivating the independent sources

94. 4.12 Maximum Power Transfer

98. P488 Use the principle of superposition to find the voltage v in the circuit

99. 70-V Source acting alone 70 Therefore from 2 and 3 we have 1 2 3 Since 4 From 4OR Therefore from 1 and 5 we have 5 OR