1. CIRCUIT ANALYSIS METHOD

2. TOPIC Node-Voltage Method Mesh-current Method Source of embodiment principle Thevenin’s Circuit Norton’s Circuit Maximum Power Transfer Superposition Principle

3. INTRODUCTION TO NODE-VOLTAGE METHOD Base on Kirchhoff’s Current Law Important step: select one node as a reference.

4. Example:Node-Voltage method

5. Previous circuit, set node 3 as a reference. By using Kirchhoff’s Current law at node 1,

6. Node-voltage equation at node2

7. Solve previous equation

8. NODE-VOLTAN METHOD THAT CONTAIN DEPENDENT SOURCE If the circuit contains dependent source, the node-voltage equation imposed by the presence of the dependent source.

9. Find power that absorb by 5Ω Resistor using node-voltage method.

10. Circuit have 3 node. Need 2 node-voltage equations. Summing the currents away from node 1 generates the equation,

11. Summing the currents away from node 2 yields

12. These two node-voltage equations contain three unknowns, namely, V 1, V 2 and iø. To eliminate iø, we must express this controlling current in terms of the node-voltage,

13. Substituting this relationship into the node 2 equation simplifies the two node- voltage equations

14. Solving for V 1 and V 2 gives, V 1 =16V V 2 = 10V

15. Then, Power absorb by 5Ω resistor

16. SPECIAL CASE When a voltage source is the only element between two essential nodes, the node- voltage method is simplified.

17. Example

18. There are three essential nodes in this circuit, which means that two simultaneous equations are needed. There is only one unknown node-voltage V 2, but V 1 =100V. Solution of this circuit thus involves only a single node- voltage equation at node 2.

19. Have V 1 =100V, and solved V 2 =125V.

20. SUPERNODE A supernode is formed by enclosing a (dependent or independent) voltage source connected between two nonreference nodes and any elements connected in parallel with it.

21. Example:supernode

22. Select node:

23. Node voltage equation at node 2 and node 3

24. Add previous equation

25. Previous equation could get when use supernode concept at node 2 and node 3.

26. Supernode

27. From 5Ω resistor

28. Pink equation was equal to green equation. Using supernode at node 2 and 3 make it simple to analyse the circuit.

29. Have V 1 =50V and V 3 can be describe with V 2,

30. Replace V 1 =50, V 3 and iø, pink equation become

31. Insert V 2

32. INTRODUCTION TO MESH-CURRENT METHOD One mesh mean a loop that no others loop inside. This mesh-current method used Kirchhoff’s voltage law to find current each mesh.

33. Example:Mesh-current

34. From Kirchhoff’s law (1) (2)

35. Use i 3 from equation (1) and insert to equation (2)

36. Mesh-current circuit with mesh current i a and i b.

37. Using KVL at those two mesh

38. After i a and i b known, then we can calculate voltage at power at each resistor.

39. MESH-CURRENT METHOD THAT HAVE DEPENDENT SOURCE When circuit have dependent source, mesh-current equation will have constant value related to dependent source.

40. Example:mesh method with dependent source

41. Find power that obserb by 4Ω resistor using mesh-current method.

42. From Kirchhoff’s voltage law

43. Have Insert equation i ø to related equation,

44. By using Cramer law, i 2 and i 3 can be calculated as below,

48. Power that absorb by 4Ω resistor

49. SPECIAL CASE (SUPERMESH) When a branch of current source can be remove and use supermesh concept (current source assume as open circuit)

51. Assume that current source as open circuit

52. Supermesh equation

53. Mesh-current equation for mesh 2

54. Known ic –ia= 5A By using Cramer law at those three equation, value for those three mesh current could be calculated.

55. SOURCE TRANSFORMATIONS Source transformation is the process of replacing a voltage source v s in series with a resistor R by a current source i s in parallel with a resistor R, or vice versa.

56. Source Transformation

57. Example:Source transformation

58. When resistor R=0, terminal a-b become close circuit. Beginning, close circuit current should be same. Therefore,

59. Close circuit current for second circuit was Is. Therefore,

60. When resistor R = ∞, these circuit become open circuit. From first circuit, we have Vab =Vs. Therefore, it was voltage for open circuit.

61. V ab for those two circuit should be same. Therefore, V s = I s R p. Replace I s

62. Summarize for Source transformation Tetapkan MethodAfterBefore

63. Tetapkan BeforeAfterMethod

64. THEVENIN EQUIVALENT CIRCUIT Introduced in 1883 by M. Leon Thevenin (1857-1926), a French telegraph engineer.

65. Thevenin’s theorem states that a linear two-terminal circuit can be replaced by an equivalent circuit consisting of a voltage source V Th in series with a resistor R Th where V Th is the open-circuit voltage at the terminals and R Th is the input or equivalent resistance at the terminals when the independent sources are turned off.

66. This theorem usually used to replace large sequence part (complex) with one simple equivalent circuit. This simple circuit makes voltage, current and circuit power could be calculated easily.

67. Thevenin equivalent circuit

68. Thevenin voltage, V Th = open circuit voltage for origin circuit. When load decrease until zero, circuit become close circuit and current become:

69. Example

70. Step 1: node-voltage equation for open circuit:

71. Step 2: replace close circuit at a-b terminal

72. Node voltage equation for close circuit:

73. Close circuit current: Thevenin resistance

74. Thevenin equivalent circuit

75. Norton equivalent circuit In 1926, about 43 years after Thevenin published his theorem, E. L. Norton, an American engineer at Bell Telephone Laboratories, proposed Norton’s theorem. This equivalent circuit have one independent source that parallel with one resistor.

76. Norton equivalent circuit could have from Thevenin equivalent circuit by source transformation.

77. Example Step 1: Source transformation

78. Step 2: Combine source and parallel resistors

79. Step 3: Source transformation, Series resistors combined, producing the Thevenin equivalent circuit

80. Step 4: Source transformation and Producing the Norton equivalent circuit

81. Norton equivalent circuit

82. TOPIC Node-Voltage method Mesh-current method Source transformation principle Thevenin equivalent circuit Norton equivalent circuit Maximum power transfer principle Superposition principle

83. MAXIMUM POWER TRANSFER Power system designed to provide power to load at high-efficiency and decrease power loss when delivered to load. Therefore, we need to decrease source resistance and delivering resistance.

84. Definition for Maximum power transfer tell that power that transfer from one source was represent by Thevenin equivalent circuit become max when load resistor R L and Thevenin resistor R Th was

85. Example

86. Power absorb by resistor R L

87. Differentiate p with R L

88. Differential was zero and p become maximum Solve

89. Therefore, for maximum power transfer, R L must equal with R TH. Maximum power transfer equation:

90. SUPERPOSITION PRINCIPLE The superposition principle states that the voltage across (or current through) an element in a linear circuit is the algebraic sum of the voltages across (or current through) that element due to each independent source acting alone.

91. Step to Apply Superposition Principle 1.Turn off all independent sources except one source. Find the output (voltage or current) due to that active.

92. 2.Repeat step 1 for each of the other independent sources. 3.Find the total contribution by adding algebraically all the contributions due to the independent sources.

93. 1.Independent voltage source become close circuit with zero volt. 2.Independent current source become open circuit. 3.If dependent source exist, it should be active while superposition process. REMEMBER !

94. Example

95. Step 1: turn off current source

96. Use voltage divider law to calculate V 0 :

97. Step 2: turn off voltage source

98. Use current divider law to calculate V 0,

99. Total V 0 : V 0 =2+5=7V.

100. Question 1

101. Answer Node 1:

102. Node 2:

105. Question 2

106. Supermesh: Mesh 3:

107. Current source known

108. Replace V 0

109. By using Cramer law

111. Current I 2 :

112. Current I 3 :

113. Question 3

114. Open circuit voltage:

115. Node-voltage equation for Voc

116. Thevenin resistance:

117. Get:

118. Question 4

119. Close circuit current

120. Norton resistance R TH = 4Ω

121. Norton equivalent circuit:

122. Question 5

123. Turn off current source

124. Turn off voltage source

125. Total V 0

126. Question 6

127. Node-voltage equation: Known: Get: V 0 =50V

128. Finally we get: V 0 =50V