1. CHAPTER 2: DC Circuit Analysis and AC Circuit Analysis Methods of Circuit Analysis and Circuit Theorems

2. Motivation(1) & Motivation(2) Nodal Analysis (Node-Voltage Method) Mesh Analysis (Mesh-Current Method) Superposition Theorem Source Transformation Thevenin’s Theorem Norton’s Theorem Maximum Power Transfer Introduction

3. Motivation (1) If you are given the following circuit, how can we determine (1) the voltage across each resistor, (2) current through each resistor. (3) power generated by each current source, etc. What are the things which we need to know in order to determine the answers?

4. Motivation (2) Things we need to know in solving any resistive circuit with current and voltage sources only: Kirchhoff’s Current Laws (KCL) Kirchhoff’s Voltage Laws (KVL) Ohm’s Law How should we apply these laws to determine the answers?

5. Introduction (Continued…) There are four ways of solving simultaneous equations: 1. Cramer’s rule 2. Calculator (real numbers only) 3. Normal substitution and elimination (not more than two equations) 4. Computer program packages: mathcad, maple, mathematica etc.

6. Introduction Direct methods are not suitable to solve complex and large circuits or as we demand many unknowns. To aid the analysis of complex circuit structures-need to develop more powerful techniques from the basic laws by systematic approaches: Nodal Analysis and Mesh Analysis. These two techniques can be used to solve almost any kind of circuit analysis problems by means of a set of simultaneous equations.

7. Introduction (Continued…) Circuit analysis problems in this course will be limited to three linear simultaneous equations for conventional hand solutions. Circuit theorems merely developed to simplify circuit analysis applicable to linear circuits such as Thevenin’s and Norton’s theorems, superposition theorem, source transformation and maximum power transfer.

8. Introduction (Continued…) Circuit theorems are not analysis techniques, rather they add up to the list of simplifying/reduction techniques such as the series-parallel reductions and  -Y transformations. Although many computer aids facilitate us as effective mathematics tools to solve engineering problems they cannot replace the compulsory needs to study the circuit theories govern circuit behavior in performing both circuit analysis and design.

9. Circuit Analysis Method NODAL ANALYSIS (Node-Voltage Method)

10. Concept Developed based on the systematic approach of Kirchhoff’s current law (KCL) to find all circuit variables without having to sacrifice any of the elements. General procedure which is making use of node voltages in circuit analysis as key solutions.

11. Why use Node Voltage? Further reduce the number of equations to be solved simultaneously. No of independent equations = No of the marked nodes exclusive of the reference node. Element voltages and currents can be obtained in few steps using the solved node voltages.

12. Assumptions KCL is performed with current going out from a node as positive (+ve) while current entering a node as negative (- ve). in – negative (subtract) out – positive (add) All unknown currents assumed to be leaving a particular node.

13. Step to determine Node Voltages: 1. Mark all essential nodes and assign proper voltage designations except for the appointed reference node. 2. Apply KCL to each non-reference nodes. Use Ohm’s law to formulate the equation in terms of node voltages. Assume all unknown currents are directing out of the nodes. 3. Solve the resulting simultaneous equations to obtain the unknown node voltages.

14. Applying Nodal Analysis on Simple Circuit Example 1 (3 unknowns)

15. Solution: Step 1: Mark all essential nodes Assign unknown node voltages Indicate the reference node. V1V1 V2V2 V3V3

16. Solution (continued…) Step 2: Perform KCL at each marked non-reference nodes using Ohm’s law to formulate the equations in terms of the node voltages.

17. Solution (continued…) KCL V 1 : KCL V 2 : KCL V 3 : or (1) (2) (3)

18. Solution (continued…) Step 3: Solve the resulting simultaneous equations from step 2 above.

19. Solution (continued…) KCL V 1 : KCL V 2 : KCL V 3 : Simplify to (1) -2V 1 + 11V 2 - 4V 3 = 0 Simplify to (2) V 1 - V 2 = 50 Simplify to -2V 2 + 7V 3 = -30 (3)

20. Solution (continued…) Cramer’s rule: Put the equations in matrix forms. Left Matrix: Col-1:coefficients of V 1 i.e. a 1, a 2 and a 3 Col-2:coefficients of V 2 i.e. b 1, b 2 and b 3 Col-3:coefficients of V3 i.e. c 1, c 2 and c 3 Middle Matrix:Unknown variables i.e. V 1, V 2 and V 3 Right Matrix:Constants i.e. d 1, d 2 and d 3 = Col-1Col-2Col-3

21. Solution (continued…) For third-order determinants, we use shorthand methods of expansion solution. Shorthand method consists of repeating the first two columns of the determinant to the right of the determinant and then summing the products along the specific diagonals as shown below.

22. Solution (continued…) Use determinants to solve for each variable: 1 -1 -2 11 0 -2

23. Solution (continued…) Determinant  1 when coefficients for V 1 is replaced by the constants. 50 -1 0 11 -30 -2

24. Solution (continued…) Determinant  2 when coefficients for V 2 is replaced by the constants. 1 50 -2 0 0 -30

25. Solution (continued…) Determinant  3 when coefficients for V 3 is replaced by the constants. 1 -1 -2 11 0 -2

26. Solution (continued…) V 1 =  1 /  = 3330/55 = 60.55 V V 2 =  2 /  = 580/55 = 10.55 V V 3 =  3 /  = -70/55 = -1.27 V You should verify your answers using calculator for three unknowns.

27. Solution (continued…) Knowing all the node voltages, we can obtain the element voltages, element currents and even power dissipated by the passive elements such as: V R1 = V 1 – V 2 I R1 = (V 1 – V 2 )/R 1 P R1 = I R1 2 R 1 = V R1 2 /R 1 V R2 = V 2 – V 3 I R1 = (V 2 – V 3 )/R 2 P R2 = I R2 2 R 2 = V R2 2 /R 2 **V R3 = V 2 I R1 = V 2 /R 3 P R3 = I R3 2 R 3 = V R3 2 /R 3 **V R4 = V 3 I R4 = V 3 2 /R 4 P R4 = I R4 2 R 3 = V R4 2 /R 4 **In these two cases, the element voltages identical to node voltages because one of its terminals is at reference node.

28. Can you find the power dissipated by the 10 k  resistor? Need to find the element voltage of 10-k  resistor because power calculation formula uses element voltage. P 10k  = (V 1 – V 2 ) 2 /10k = (60.55 –10.55) 2 /10k = 50 2 /10k = 0.25 W

29. Applying Nodal Analysis on Simple Circuit Example 2 (2 unknowns) Question: Find the power dissipated in the 20-  resistor?

30. Solution Step 1: Mark all essential nodes Assign unknown node voltages Indicate the reference node.

31. Solution (Continued…) Step 2: Perform KCL at each marked non- reference nodes using Ohm’s law to formulate the equations in terms of node voltages. KCL V 1 : Hence KCL V 2 : (2) (1)

32. Solution (Continued…) Step 3: Solve the resulting simultaneous equations which have been simplified in step 2 above using Cramer’s rule.

33. Solution (Continued…) Hence, V 1 =  1 / = 6.8/17 = 0.4 V V 2 =  2 / = 1.7 /17 = 0.1 V P 20 = V 2 2 /20 = 0.12(20) = 0.2 W #

34. Applying Nodal Analysis on Circuit with Voltage Sources Three different effects depending on placement of voltage source in the circuit. Does the presence of a voltage source complicate or simplify the analysis?

35. Case 1: Voltage source between two non-reference essential nodes. Supernode Equation:

36. Case 2: Voltage source between a reference essential node and a non-reference essential node. Known node voltage:

37. Case 3: Voltage source between an essential node and a non- essential node. Node voltage at non-essential node:

38. Example 3 (Supernode or Known node voltage) Question: Find the power of the 10-V voltage source? Is it supplying energy to the circuit or absorbing energy from the circuit? Show your work according to the nodal analysis procedure.

39. Solution 1 (Supernode) Step 1: Mark essential nodes and assign unknown node voltages and indicate the reference node. Checklist: 3 essential nodes – 1 ref node – 1 supernode = 1 KCL Eqn. + 1 Supernode Eqn.

40. Solution 1 (Continued…) Step 2: Perform KCL at each marked non- reference nodes using Ohm’s law to formulate the equations in terms of node voltages. KCL supernode V 1 /V 2 : Hence Supernode Equation: (1) (2)

41. Solution 1 (Continued…) Step 3: Solve the resulting simultaneous equations which have been simplified in step 2 above. Solving Eqn. (1) and (2) simultaneously yields, V 1 = 30.91 V and V 2 = 20.91 V (You can check this answer by calculator or Cramer’s rule).

42. Solution 1 (Continued…) Finding current through the voltage source, KCL at V1: Hence, P 10-V = Vi= (10)(-0.636) = -6.36 W. (Delivering energy)

43. Solution 2 (Known node voltage) Step 1: Mark essential nodes and assign unknown node voltages and indicate the reference node. Checklist: 3 essential nodes – 1 ref node – 1 known node voltage = 1 KCL Eqn.

44. Solution 2 (Continued…) Step 2: Perform KCL at each marked non- reference nodes using Ohm’s law to formulate the equations in terms of node voltages. Immediately known node voltage at V 1 : KCL V 2 : (1)

45. Solution 2 (Continued…) Step 3: Solve the resulting simultaneous equations which have been simplified in step 2 above. Solving Eqn. (1) yields, Finding current through the voltage source, KCL at V 1 :

46. Solution 2 (Continued…) Hence, P 10-V = Vi= (10)(-0.636) = -6.36 W. (Delivering energy)

47. Example 4 (One of the terminals not an essential node) Question: Find the current through the 10- k  resistor. Show your work according to the nodal analysis procedure.

48. Solution Step 1: Mark essential nodes and assign unknown node voltages and indicate the reference node. For voltage sources, indicate the node voltages at both ends with respect to the assigned unknown node voltages at the essential nodes Checklist: 4 essential nodes – 1 ref node = 3 KCL Eqns.

49. Solution (Continued…) Step 2: Perform KCL at each marked non- reference nodes using Ohm’s law to formulate the equations in terms of node voltages. KCL V 1 : Hence KCL V 2 : Hence (1) (2)

50. Solution (Continued…) KCL V 3 : Hence (3)

51. Solution (Continued…) Step 3: Solve the resulting simultaneous equations which have been simplified in step 2 above. Solving Eqn. (1) till (3) simultaneously yields, V 1 = -5.43 V, V 2 = -10.17 V and V 3 = 12.13 V (You can check this answer by calculator and Cramer’s rule).

52. Solution (Continued…) Finding current through the 10-k  resistor, KCL at V 2 :

53. Applying Nodal Analysis on Circuit with Dependent Sources Circuits contain dependent sources either VCVS, CCVS, VCCS or CCCS. The presence of the dependent sources require ‘Constraint Equation’ (CE). CE describes the dependent term of the dependent sources in relation to the assigned unknown node voltages or known values at the essential nodes.

54. Example 5 (Circuit with dependent sources) Question: Use the node-voltage method to find both dependent terms i O and V x of the dependent sources of the circuit in Figure below.

55. Solution: Step 1: Mark essential nodes and assign unknown node voltages and indicate the reference node. Checklist: 4 essential nodes – 1 ref node – 1 s/node – 1 known = 1 KCL Eqn. + 1 s/node Eqn. + 2 contraint Eqns.

56. Solution (Continued…) Step 2: Perform KCL at each marked nonreference nodes using Ohm’s law to formulate the equations in terms of node voltages. Known node voltage: KCL s/node V 2 : Hence S/node equation: (1) (2)

57. Solution (Continued…) Constraint equations: (3) and (4) Hence Substituting Eqn. (3) into (2) yields (2’)

58. Solution (Continued…) Step 3: Solve the resulting simultaneous equations which have been simplified in step 2 above. Solving Eqn. (1), (2’) and (4) simultaneously yields, V 1 = -6.51 V, V 2 = 1.83 V and i o = -0.264 A (You can check this answer by calculator and Cramer’s rule).

59. Circuit Analysis Method MESH ANALYSIS (Mesh-Current Method)

60. Concept Similar to nodal analysis. Developed based on the systematic approach of Kirchhoff’s voltage law (KVL) to find all circuit variables without having to sacrifice any of the elements. General procedure which is making use of mesh current in circuit analysis as key solutions.

61. Importance terms Mesh Current: Assigned unknown current flows around the perimeter of the particular mesh/loop. Element Current: Actual current through any element or branch in the circuit.

62. Assumptions KVL is performed in clockwise direction. Voltage rise – negative (subtract) Voltage drop – positive (add)

63. Mesh Analysis Procedures: 1. Label all independent meshes and assign proper unknown mesh currents in clockwise direction. Do the checklist. 2. Formulate KVL/Supermesh/Constraint Equation. 3. Solve the resulting simultaneous equations to obtain the unknown mesh current.

64. Applying Mesh Analysis on Simple Circuit Example (2 unknowns) Question: Find power dissipated in 12  - resistor and 3  -resistor using mesh analysis.

65. Solution: 1. Label all independent meshes and assign proper unknown mesh currents in clockwise direction. Checklist: 2 meshes = 2 KVL Eqns.

66. Solution (Continued…) 2. Formulate KVL/Supermesh/Constraint Eq. KVL I 1 : 18I 1 – 12I 2 = 12 KVL I 2 : -12I 1 + 24I 2 = -8 (1) (2)

67. Solution (Continued…) 3. Solve the resulting simultaneous equations to obtain the unknown mesh current. I 1 =  1 /  I 2 =  2 / 

68. Solution (Continued…) Using calculator/Cramer’s rule we obtain: I 1 = 0.667 A and I 2 = 0 A P 12  = (I 1 -I 2 ) 2 (12) = 5.33 W P 3  = I 2 2 (3) = 0 W Notice that the branch (3  -resistor) forming the outer most boundary of the circuit will have mesh current = element current.

69. Circuit with current sources and dependent sources Two different effects depending on placement of voltage source in the circuit. Does the presence of a current source complicate or simplify the analysis? The presence of dependent source in the circuit need to impose constraint equation to describe the r/ship btw. dependent term of the dependent sources in relation to the mesh currents.

70. Case 1: Current source located at the outer most boundary Connecting mesh current immediately known. No need to apply KVL around that loop/mesh. Mesh Current = Element Current = Current Source Value Immediately known mesh current, I 3 = -I s

71. Case 2: Current source located at the boundary between 2 meshes Enclose the current source and combine the two meshes to form a SUPERMESH. KVL is performed around the supermesh – do not consider voltage across current source. Formulate supermesh equation – express the relationship between mesh currents that form the supermesh and current source that it encloses.

72. SUPERMESH KVL S/Mesh I 2 /I 3 : -12 + I 2 R 2 +I 3 R 3 + I 3 R 4= 0 S/Mesh Eq: I 3 – I 2 = 3 mA

73. Example of Supermesh Use the mesh analysis to determine i 1, i 2 and i 3.

74. Solution: Step 1: Checklist. Checklist: 3 meshes – I supermesh = 2 KVL Eqns. + 1 supernode Eqn.

75. Solution (Continued…) Step 2: Formulate KVL/supermesh equation. KVL i 3 : -2i 1 – 4i 2 + 8i 3 = 0 KVL supermesh i 1 /i 2 : 2i 1 + 12i 2 – 6i 3 = 6 Supermesh Eqn: i 1 – i 2 = 3 (1) (2) (3)

76. Solution (Continued…) Step 3: Solve the simultaneous equations using Cramer rule or by calculator. We obtain, i 1 = 3.474 A i 2 = 0.4737 A i 3 = 1.1053 A

77. Example 6 (Known current & dependent source) Find the voltage of the dependent source (CCCS).

78. Solution: Step 1: Assign mesh currents in CW direction and perform checklist. Checklist: 2 meshes – I known = 1 KVL Eqn. + 1 CE

79. Solution (Continued…) Step 2: Formulate KVL/Constraint equations. Immediately known, I 1 = 5i x KVL I 2 : 21I 2 = -8 CE:i x = I 2 (1) (2) (3)

80. Solution (Continued…) Step 3: Solve the simultaneous equations. Substitute (3) into (1) and solve (1) and (2) simultaneously, we obtain I 1 = -1.9048 AI 2 = -0.3810 A

81. Solution (Continued…) To find voltage across the CCCS, perform KVL around loop I 1. KVL I 1 : -V + 9I 1 – 8 = 0 V = 9(-1.9048) – 8 = 25.1432 V

82. TIPS: Nodal versus Mesh Analysis To select the method that results in the smaller number of equations. For example: 1.Choose nodal analysis for circuit with fewer nodes than meshes. *Choose mesh analysis for circuit with fewer meshes than nodes. *Networks that contain many series connected elements, voltage sources, or supermeshes are more suitable for mesh analysis. *Networks with parallel-connected elements, current sources, or supernodes are more suitable for nodal analysis. 2.If node voltages are required, it may be expedient to apply nodal analysis. If branch or mesh currents are required, it may be better to use mesh analysis.

83. Circuit Theorem SUPERPOSITION THEOREM

84. Concept Concept: each independent source is treated independently and the algebraic sum is found to determine a particular unknown quantity or circuit variable of the circuit under study.

85. Superposition Theorem It states that the voltage across (or current through) an element in a linear circuit is the algebraic sum of the voltage across (or currents through) that element due to EACH independent source acting alone. The principle of superposition helps us to analyze a linear circuit with more than one independent source by calculating the contribution of each independent source separately.

86. Principle of Operation Steps to apply superposition principle: Turn off all independent sources except one source. Find the output (voltage or current) due to that active source using nodal or mesh analysis. Repeat step 1 for each of the other independent sources. Find the total contribution by adding algebraically all the contributions due to the independent sources.

87. Principle of Operation (Continued…) Two things have to be keep in mind: When we say turn off all other independent sources: Independent voltage sources are replaced by 0 V (short circuit) and Independent current sources are replaced by 0 A (open circuit). Dependent sources are left intact because they are controlled by circuit variables.

88. Example 1: Find the voltage across the 12  resistor using superposition hence the power dissipated by this resistor.

89. Solution: i) Consider 12V/removed 8V.

90. Solution (Continued…) ii) Replace 8V/removed 12V

91. Solution (Continued…) Hence, V = V’ + V’’ = 6V + 2V = 8V. P 12  = V 2 /R = 8 2 /12 = 5.33W

92. Exercise: Use the superposition theorem to find v in the circuit shown below. 3A is discarded by open-circuit 6V is discarded by short-circuit Answer v = 10V

93. Circuit Theorem SOURCE TRANSFORMATION

94. What benefits from source transformation? Another tool to simplify circuit – the simpler the circuit, the easier will be the solution. How to simplify? – rearrange the resistors/sources by Source Transformation so that they end up with series/parallel connections.

95. Definition An equivalent circuit is one whose v-i characteristics are identical with the original circuit. A Source Transformation is the process of replacing a voltage source V s in series with resistor R s by a current source i s in parallel with the same resistor R s or vice versa.

96. Equivalent Circuits The connections of each case should be between the same terminals before and after transformation. (a) Independent source transform (b) Dependent source transform The arrow of the current source is directed toward the positive terminal of the voltage source. The source transformation is not possible when R = 0 for voltage source and R = ∞ for current source.

97. Example: Use series of source transformations to find i in the circuit below.

98. Solution: Transform 4A and 5  into voltage source.

99. Solution (continued…) Transform 10 A and 1  into voltage source. Transform 10 V and 40  into current source.

100. Solution (continued…) Transform 10V and 10  into current source.

101. Solution (continued…) Combine the current sources 1A and 0.25A. Combine resistors 10  and 40 . Solve for I using CDR.

102. Exercise: Use Source Transformation to find V o. (Ans:-135V)

103. Circuit Theorem THEVENIN’S THEOREM

104. Purpose Used when we are interested ONLY in the terminal behavior of the circuit particularly where a variable load is connected to. Provides a technique to replaced the fixed part of the circuit by a simple equivalent circuit. Avoid the re-do on the analysis of the entire circuit except for the changed load.

105. Thevenin’s Theorem It states that a linear two-terminal circuit (Fig. a) can be replaced by an equivalent circuit (Fig. b) consisting of a voltage source VTH in series with a resistor RTH, where VTH is the open-circuit voltage at the terminals. RTH is the input or equivalent resistance at the terminals when the independent sources are turned off.

106. Procedures to obtain V Th and R Th Step 1: Preliminary – Omitting load resistor R L (Not applicable if no load resistor) Step 2: Find R Th – setting all independent sources to zero. Find the resultant resistance between the marked terminals. Voltage source – short circuit (s.c.) Current sorce – open circuit (o.c.) Step 3: Find V Th – calculate V Th by returning all sources back to their original positions. Find the open circuit voltage between the marked terminals using the method which takes least effort.

107. Exercise 1: Find the Thevenin equivalent between terminal a-b. (Ans: V Th =32V, R Th =8  )

108. Exercise 2: Find the Thevenin equivalent circuit at the terminal a-b of the circuit below. (Ans: V Th =- 4.8V, R Th =2.4  )

109. Circuit Theorem NORTON’s THEOREM

110. Norton’s Theorem The purpose of its use is similar to the Thevenin’s theorem.

111. It states that a linear two-terminal circuit can be replaced by an equivalent circuit of a current source IN in parallel with a resistor RN, Where IN is the short circuit current through the terminals. RN is the input or equivalent resistance at the terminals when the independent sources are turned off. Norton’s Theorem The Thevenin’s and Norton equivalent circuits are related by a source transformation.

112. Procedures to obtain I N and R N Step 1: Preliminary – Omitting load resistor R L (Not applicable if no load resistor) Step 2: Find R N – setting all independent sources to zero. Find the resultant resistance between the marked terminals. Voltage source – short circuit (s.c.) Current sorce – open circuit (o.c.) Step 3: Find I N – calculate I N by returning all sources back to their original positions. Find the o.cs.c current btw the marked terminals using the method which takes least effort.

113. Exercise 1: Find the Norton Equivalent circuit with respect to the terminals a-b.

114. Exercise 2: Find the Norton equivalent circuit in Figure below. (Ans: I N =4.5A, R N =3  )

115. Circuit Theorem THEVENIN & NORTON’s THEOREMS WITH DEPENDENT SOURCE

116. Procedures to obtain V Th /I N and R Th /R N Step 1: Preliminary – Omitting load resistor R L (Not applicable if no load resistor) Step 2: Find V Th = V o.c or I N = I s.c using the method that takes the least effort. Step 3: Find R Th /R N Method 1:If circuit contains independent source. R th = R N = V o.c /I s.c = V Th /I N Method 2: If circuit contains independent source and without independent source.

117. Using Method 2 to find R Th /R N Turn off all independent sources but dependent sources left intact because they are controlled by circuit variables. Because of the presence of the dependent source, we excite the circuit with a voltage source or current source between the terminals. Set V o =1V to ease calculation since the circuit is linear. Goal? To find i o so that R Th =R N =1/i o Alternatively, we may set i o =1A. Goal? To find V o so that R Th =R N =V o /1

118. Illustration of Method 2 to find R Th /R N

119. Exercise 1: Find the Thevenin and Norton equivalent circuit for the circuit containing dependent sources below between terminals a-b. (Ans: V Th =-5V, R Th =100 , IN=-50mA)

120. Exercise 2: Find the Thevenin and Norton equivalent circuit for the circuit containing dependent source below between terminals a-b. (Ans: V Th =20V, R Th =0.625 , IN=32A)

121. Circuit Theorem MAXIMUM POWER TRANSFER

122. Introduction Power transfer from source to the load can be analyzed and discussed from two basic types of systems: 1. Efficiency – eg: power utility systems concerned with generation, transmission and distribution of large quantities of electric power. 2. Amount – eg. Comm. & instrumentation sys because in the transmission of info or data via electric signals, the power available at the transmitter or detector is limited or small. At this moment our concern is on the 2 nd type of system that is the amount of maximum power transfer in purely resistive circuit.

123. Maximum Power Transfer If the entire circuit is replaced by its Thevenin equivalent except for the load, the power delivered to the load is: For maximum power dissipated in R L, P max, for a given R TH, and V TH, The power transfer profile with different R L (1)

124. Thevenin Equivalent Circuit The Thevenin equivalent circuit is useful in finding the maximum power a linear circuit can deliver to a load. The entire circuit can be replaced by its Thevenin equivalent except for the adjustable load.

125. Proving Maximum Power Transfer Theorem Differentiate p in Eq.(1) with respect to R L and set the result equal to zero, (2)

126. Proving Maximum Power Transfer Theorem (Continued…) Implies that, 0 = (R Th + R L -2R L ) = (R Th – R L ) Yields, R L = R th Eq (3) gives the maximum power by showing that d 2 p/dR L 2 < 0. (3)

127. Maximum Power Formula Substituting Eq.(3) into (1) to obtain the maximum power transfer, Eq.(4) applies only when R L = R Th. When R L ≠ R Th, compute power from Eq.(1) (4)

128. Exercise 1: The variable resistor in the circuit below is adjusted for maximum power. Find the value of R L and the maximum power. (Ans: 5k , 45mW)

129. Exercise 2: The load resistance in both circuits below are adjusted until maximum power is delivered. Find the power delivered to the loads and the value of both R L. (Ans: 600 , 38.4mW)