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G. Rizzoni, Principles and Applications of Electrical Engineering, 5th Edition Problem solutions, Chapter 3 Chapter 3: Resistive Network Analysis Instructor Notes Chapter 3 presents the principal topics in the analysis of resistive (DC) circuits. The presentation of node voltage and mesh current analysis is supported several solved examples and drill exercises, with emphasis placed on developing consistent solution methods, and on reinforcing the use of a systematic approach. The aim of this style of presentation, which is perhaps more detailed than usual in a textbook written for a audience, is to develop good habits early on, with the hope that the orderly approach presented in Chapter 3 may facilitate the discussion of AC and transient analysis in Chapters 4 and 5. Make The Connection sidebars (pp. introduce analogies between electrical and thermal circuit elements. These analogies are to be encountered again in Chapter 5. A brief discussion of the principle of superposition precedes the discussion of Thèvenin and Norton equivalent circuits. Again, the presentation is rich in examples and drill exercises, because the concept of equivalent circuits will be heavily exploited in the analysis of AC and transient circuits in later chapters. The Focus on Methodology boxes (p. 84 Node p. 94 Mesh pp. 115, 119 Equivalent Circuits) provide the student with a systematic approach to the solution of all basic network analysis problems. Following a brief discussion of maximum power transfer, the chapter closes with a section on nonlinear circuit elements and analysis. This section can be easily skipped in a survey course, and may be picked up later, in conjunction with Chapter 9, if the instructor wishes to devote some attention to analysis of diode circuits. Finally, those instructors who are used to introducing the as a circuit element, will find that sections 8 and 8 can be covered together with Chapter 3, and that a good complement of homework problems and exercises devoted to the analysis of the as a circuit element is provided in Chapter 8. Modularity is a recurrent feature of this book, and we shall draw attention to it throughout these Instructor Notes. The homework problems present a graded variety of circuit problems. Since the aim of this chapter is to teach solution techniques, there are relatively few problems devoted to applications. We should call the attention to the following problems: 3 on the Wheatstone 3 and 3 on 3.353 on electrical power distribution on various nonlinear resistance devices. The 5th Edition of this book includes 19 new some of the 4th Edition problems were removed, increasing the problem count from 66 to 83. Learning Objectives for Chapter 3 1. Compute the solution of circuits containing linear resistors and independent and dependent sources using node analysis. 2. Compute the solution of circuits containing linear resistors and independent and dependent sources using mesh analysis. 3. Apply the principle of superposition to linear circuits containing independent sources. 4. Compute Thévenin and Norton equivalent circuits for networks containing linear resistors and independent and dependent sources. 5. Use equivalent circuits ideas to compute the maximum power transfer between a source and a load. 6. Use the concept of equivalent circuit to determine voltage, current and power for nonlinear loads using analysis and analytical methods. 3 PROPRIETARY MATERIAL. The Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. G. Rizzoni, Principles and Applications of Electrical Engineering, 5th Edition Problem solutions, Chapter 3 Sections 3, 3, 3, 3: Nodal and Mesh Analysis Focus on Methodology: Node Voltage Analysis Method 1. 2. 3. 4. Select a reference node(usually ground). This node usually has most elements tied to it. All other nodes will be referenced to this node. Define the remaining node voltages as the independent or dependent variables. Each of the m voltage sources in the circuit will be associated with a dependent variable. If a node is not connected to a voltage source, then its voltage is treated as an independent variable. Apply KCL at each node labeled as an independent variable, expressing each current in terms of the adjacent node voltages. Solve the linear system of unknowns. Focus on Methodology: Mesh Current Analysis Method 1. 2. 3. 4. Define each mesh current consistently. Unknown mesh currents will be always defined in the clockwise known mesh currents (i., when a current source is present) will always be defined in the direction of the current source. In a circuit with n meshes and m current sources, independent equations will result. The unknown mesh currents are the independent variables. Apply KVL to each mesh containing an unknown mesh current, expressing each voltage in terms of one or more mesh currents.. Solve the linear system of unknowns. Problem 3 Note: the rightmost top resistor missing a value should be 1 Ω. Solution: Known quantities: Circuit shown in Figure P3 Find: Voltages v1 and v2 . Analysis: Applying KCL at each of the two nodes, we obtain the following equations: V1 V1 V2 0 3 1 V2 V2 V2 V1 2 2 1 Rearranging the equations, 4 V1 V2 4 3 2V2 0 Solving the equations, V1 4 V and V2 2 V 3 PROPRIETARY MATERIAL. The Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. G. Rizzoni, Principles and Applications of Electrical Engineering, 5th Edition Problem solutions, Chapter 3 Problem 3 Solution: Known quantities: Circuit shown in Figure P3 Find: Current through the voltage source. Analysis: At node 1: v1 v 2 v1 v 3 0 0 (1) At node 2: v 2 v1 v 2 0 0 0 (2) v3 v1 v 3 0 0 0 (3) At node 3: Further, we know that v 3 v 2 3 . Now we can eliminate either v2 or v3 from the equations, and be left with three equations in three unknowns: v1 v 2 v1 (v 2 3) (1) 0 0 v 2 v1 v 2 0 (2) 0 0 (v2 3) v1 (v2 3) i 0 (3) 0 0 Solving the three equations we compute i 8 Problem 3 Solution: Known quantities: Circuit shown in Figure P3 with mesh currents: I1 5 A, I2 3 A, I3 7 A. Find: The branch currents through: a) R1, b) R2, c) R3. Analysis: a) Assume a direction for the current through R1 (e., from node A to node B). Then summing currents at node A: 3 PROPRIETARY MATERIAL. The Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. G. Rizzoni, Principles and Applications of Electrical Engineering, 5th Edition Problem solutions, Chapter 3 KCL: 1 I R 1 I 3 0 I R 1 I 1 I 3 A This can also be done inspection noting that the assumed direction of the current through R1 and the direction of I1 are the same. b) Assume a direction for the current through R2 (e., from node B to node A). Then summing currents at node B: KCL: I2 IR2 I3 0 IR2 I3 I2 4 A This can also be done inspection noting that the assumed direction of the current through R2 and the direction of I3 are the same. c) Only one mesh current flows through R3. If the current through R3 is assumed to flow in the same direction, then: I R1 I 3 7 A . Problem 3 Solution: Known quantities: Circuit shown in Figure P3 with source and node voltages: VS1 VS 2 110 V , VA 103 V , VB V . Find: The voltage across each of the five resistors. Analysis: Assume a polarity for the voltages across R1 and R2 (e., from ground to node A, and from node B to ground). R1 is connected between node A and therefore, the voltage across R1 is equal to this node voltage. R2 is connected between node B and therefore, the voltage across R2 is equal to the negative of this voltage. VR 1 VA 103 V, VR 2 V The two node voltages are with respect to the ground which is given. Assume a polarity for the voltage across R3 (e., from node B to node A). Then: VA VR 3 VB 0 KVL: VR 3 VA VB 210 V Assume polarities for the voltages across R4 and R5 (e., from node A to ground , and from ground to node B): KVL: VR 4 VA 0 VR 4 VS1 VA 7 V KVL: 2 VB VR 5 0 VR 5 2 VB V 3 PROPRIETARY MATERIAL. The Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. G. Rizzoni, Principles and Applications of Electrical Engineering, 5th Edition Problem solutions, Chapter 3 Problem 3 Solution: Known quantities: Circuit shown in Figure P3 with resistance values, current and voltage source values. Find: The current, i, through the voltage source using node voltage analysis. Analysis: At node 1: v1 v1 v 2 v1 v 3 200 5 100 At node 2: v 2 v1 i 0 0 5 At node 3: v v 3 1 3 0 100 50 For the voltage source we have: v 3 v 2 50 V Solving the system, we obtain: v1 V , v 2 V , v 3 1 V and, finally, i 491 mA . Problem 3 Solution: Known quantities: The current source value, the voltage source value and the resistance values for the circuit shown in Figure P3. Find: The three node voltages indicated in Figure P3 using node voltage analysis. Analysis: At node 1: v1 v1 v 2 0 A 200 75 At node 2: v 2 v1 v 2 v 2 v 3 0 75 25 50 At node 3: v v2 v 3 3 50 100 For the voltage source we have: v 3 v 2 Solving the system, we obtain: v1 14 V , v 2 4 , v 3 V and, finally, i mA . 3 PROPRIETARY MATERIAL. The Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. G. Rizzoni, Principles and Applications of Electrical Engineering, 5th Edition Problem solutions, Chapter 3 Problem 3 Solution: Known quantities: The voltage source value, 3 V, and the five resistance values, indicated in Figure P3. Find: The current, i, drawn from the independent voltage source using node voltage analysis. Analysis: At node 1: v1 3 v1 v1 v 2 0 0 0 At node 2: v 2 v1 v 2 0 0 0 0 Solving the system, we obtain: v1 1 V , v 2 0 V v1 Therefore, i 3 A . 0 Problem 3 Solution: Known quantities: Circuit shown in Figure P3. Find: Power delivered to the load resistance. Analysis: KCL at node 1: V1 0 RI RV Or 3 V1 V2 6 ( Eq. 1) KCL at node 2: V2 ( RV ) Or 14 V2 V1 substitute Eq. 1 into Eq. 2 ( Eq. 2) V2 and voltage divider: RL VL R2 RL PL VL2 25 mW RL 3 PROPRIETARY MATERIAL. The Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. G. Rizzoni, Principles and Applications of Electrical Engineering, 5th Edition Problem solutions, Chapter 3 Problem 3 Solution: Circuit shown in Figure P3. Find: Current i1 and i2 and voltage across the resistance 10Ω . Analysis: Mesh Mesh (20 15 10)I1 I 2 0 (10 40 10)I 2 I1 50 Therefore, I1 0 A and I 2 0 A , v10Ω 10(i2 i1 ) 6 V Problem 3 Solution: Circuit shown in Figure P3. Find: Voltage across the 3Ω resistance. Analysis: Meshes 1, 2 and 3 are clockwise from the left For mesh For mesh For mesh i1 (1 2 3) i2 2) i3 3) 2 i1 2 ) i2 (2 2 1) i3 1) i1 3) i2 1) i3 (3 1 1) 0 Solving, and i1 0 A i2 0 A i 3 0 A v 3(i1 i 3) 3(0) 0 V 3 PROPRIETARY MATERIAL. The Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. G. Rizzoni, Principles and Applications of Electrical Engineering, 5th Edition Problem solutions, Chapter 3 Problem 3 Note: the mesh current should be labeled I3, not I2. Solution: Mesh (on the side) 2 2 I1 3( I1 I 2 ) 0 If we treat mesh (middle) and mesh (on the side) as a single loop containing the four resistors (but not the current source), we can write 2 3I 3 2 I 3 3( I 2 I1 ) 0 From the current source: I3 I 2 2 Solving the system of equations: I1 A I2 A I3 0 A Problem 3 Solution: Circuit shown in Figure P3. Find: Voltage across the current source. Analysis: Meshes 1, 2 and 3 go from left to right. For mesh i1 (2 3) i2 3) i3 (0) 2 For meshes and i1 3) i2 (1 3) i3 (3 2) 0 For the current source: i1 (0) i2 (1) i3 1) Solving, i3 0 and v i3 (3 2) 3 3 PROPRIETARY MATERIAL. The Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. G. Rizzoni, Principles and Applications of Electrical Engineering, 5th Edition Problem solutions, Chapter 3 Problem 3 Solution: Known quantities: Circuit of Figure P3 with voltage source, VS , current source, IS, and all resistances. Find: a. The node equations required to determine the node voltages. b. The matrix solution for each node voltage in terms of the known parameters. Analysis: a) Specify the nodes (e., A on the upper left corner of the circuit in Figure P3, and B on the right corner). Choose one node as the reference or ground node. If possible, ground one of the sources in the circuit. Note that this is possible here. When using KCL, assume all unknown current flow out of the node. The direction of the current supplied the current source is specified and must flow into node A. Vb Va Vb VS Vb 0 V VS Va Vb S a R2 R1 R1 R3 R4 KCL: KCL: 1 1 1 1 V 1 1 VS Va Va Vb Vb I S S R2 R2 R1 R1 R1 R1 R3 R4 R3 b) Matrix solution: IS Va VS R2 VS R3 1 1 R1 R2 1 R1 1 R1 1 1 V 1 1 1 VS 1 1 I S S R2 R1 R3 R4 R3 R1 R1 R3 R4 1 1 1 1 1 1 1 R1 R1 R2 R1 R3 R4 R1 R1 1 1 1 R1 R3 R4 1 V 1 IS S R2 R1 R2 1 1 VS 1 V VS I S S R3 R2 R1 R1 R2 R3 R1 Vb 1 1 1 1 1 1 1 1 1 1 R1 R1 R2 R1 R2 R1 R3 R4 R1 R1 1 1 1 1 R1 R1 R3 R4 Notes: 1. The denominators are the same for both solutions. 2. The main diagonal of a matrix is the one that goes to the right and down. 3. The denominator matrix is the matrix and has certain properties: a) The elements on the main diagonal include all the conductance connected to node b) The elements are all negative. c) The elements are all symmetric, i., the i element j element. This is true only because there are no controlled (dependent) sources in this circuit. d) The elements include all the conductance connected between node i and node j 3 PROPRIETARY MATERIAL. The Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. G. Rizzoni, Principles and Applications of Electrical Engineering, 5th Edition Problem solutions, Chapter 3 Problem 3 Solution: Known quantities: Circuit shown in Figure P3 VS1 VS2 110 V R1 500 mΩ R2 167 mΩ R3 700 mΩ R4 200 mΩ R5 mΩ Find: a. The most efficient way to solve for the voltage across R3. Prove your case. b. The voltage across R3. Analysis: a) There are 3 meshes and 3 mesh currents requiring the solution of 3 simultaneous equations. Only one of these mesh currents is required to determine, using Law, the voltage across R3. In the terminal (or node) between the two voltage sources is made the ground (or reference) node, then three node voltages are known (the ground or reference voltage and the two source voltages). This leaves only two unknown node voltages (the voltages across R1, VR1, and across R2, VR2). Both these voltages are required to determine, using KVL, the voltage across R3, VR3. A difficult choice. Choose node analysis due to the smaller number of unknowns. Specify the nodes. Choose one node as the ground node. In KCL, assume unknown currents flow out. b) VR 2 2 ) VR 2 0 VR 2 VR1 VR1 VS1 VR1 0 VR1 VR 2 KCL: KCL: R4 R3 R5 R3 R1 R2 1 1 V S1 1 1 V 1 1 1 1 VR 2 VR 2 S2 V R1 V R1 R5 R1 R3 R4 R3 R4 R3 R5 R2 R3 1 1 1 1 1 1 8 Ω R1 R3 R4 500 700 200 1 1 1 1 1 1 10 Ω R5 R2 R3 167 700 1 1 1 Ω R3 700 VS1 110 VS2 110 550 A 330 A R4 200 R5 550 VR1 10 (5731) (472) 61 V 8 (87) (2) 10 8 VR1 KVL: 550 V 85 85 VR 3 VR 2 0 VR 3 VR1 VR 2 84 V 3 PROPRIETARY MATERIAL. The Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. G. Rizzoni, Principles and Applications of Electrical Engineering, 5th Edition Problem solutions, Chapter 3 Problem 3 Solution: Known quantities: Circuit shown in Figure P3 VS 5 V AV 70 R1 2 kΩ R2 1 kΩ R3 6 k Ω R4 220 Ω Find: The voltage across R4 using KCL and node voltage analysis. Analysis: Node analysis is not a method of choice because the dependent source is a voltage source and a floating source. Both factors cause difficulties in a node analysis. A ground is specified. There are three unknown node voltages, one of which is the voltage across R4. The dependent source will introduce two additional unknowns, the current through the source and the controlling voltage (across R1) that is not a node voltage. Therefore 5 equations are required: V1 VS V1 V3 V1 V2 0 V2 V1 I CS 0 R1 R3 R2 R2 V3 V1 I CS V3 R3 R4 0 VS VR1 V1 0VR1 VS V1 V3 AV VR1 V2 0V2 V3 AV VR1 V3 AV (VS V1 ) Substitute using Equation into Equations and and eliminate V2 (because it only appears twice in these equations). Collect terms: 1 1 1 AV 1 V V A V3 I CS (0) S S V R1 R2 R1 R3 R2 R2 R3 R2 1 A 1 V A V V3 I CS S V R2 R 2 R2 R2 1 1 1 V3 I CS 0 R3 R3 R 4 1 1 1 1 10 Ω 147 10 Ω 3 R2 1 10 R3 6 10 3 1 1 1 1 702 10 Ω 3 R3 R2 6 10 1 10 3 1 1 1 1 1 AV 1 70 4 10 Ω 39 10 Ω R3 R4 6 10 3 0 10 3 R2 R2 1 10 3 1 1 1 AV 1 1 1 70 40 10 Ω R1 R3 R2 R2 2 10 3 6 10 3 1 10 3 (5)(70) (5)(70) VS AV 5 VS VS AV 194 mA 196 mA 3 3 R2 R1 R2 2 10 1 10 1 10 3 Solving, we have: VR 4 V3 5 mV Notes: 1. This solution was not difficult in terms of theory, but was terribly long and arithmetically cumbersome. This was because the wrong method was used. There are only 2 mesh currents in the the sources were voltage therefore, a mesh analysis is the method of choice. 2. In general, a node analysis will have fewer unknowns (because one node is the ground or reference node) and will, in such cases, be preferable. 3 PROPRIETARY MATERIAL. The Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. G. Rizzoni, Principles and Applications of Electrical Engineering, 5th Edition Problem solutions, Chapter 3 Problem 3 Solution: Known quantities: The values of the resistors and of the voltage sources (see Figure P3). Find: The voltage across the 10 Ω resistor in the circuit of Figure P3 using mesh current analysis. Analysis: For mesh (a): i a (50 20 20) i b (20) i c (20) 12 For mesh (b): a (20) i b (20 10) i c (10) 5 0 For mesh (c): a (20) i b (10) i c (20 10 15) 0 Solving, i a 127 mA i b mA i c 41 mA and v R 4 10 (i b i c ) 10 V . Problem 3 Solution: Known quantities: The values of the resistors, of the voltage source and of the current source in the circuit of Figure P3. Find: The voltage across the current source using mesh current analysis. Analysis: For mesh (a): i a (20 30) i b 3 For meshes (b) and (c): i a i b (10 30) i c (30 20) 0 For the current source: i c i b 0 Solving, i a mA , i b mA and i c 178 mA . Therefore, v ic (30 20 ) 8 V . 3 PROPRIETARY MATERIAL. The Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. G. Rizzoni, Principles and Applications of Electrical Engineering, 5th Edition Problem solutions, Chapter 3 Problem 3 Solution: 50 Known quantities: The values of the resistors and of the current source in the circuit of Figure P3. 10V 75 ia Find: i The current through the voltage source in the circuit of Figure P3 using mesh current analysis. 0 A Analysis: 200 ib 25 100 ic For mesh (a): i a (100) 10 0 For mesh (b): i b (200 75 25) i c 0 0 For mesh (c): i b i c (50 25) 10 Solving, i a mA , i b 148 mA and i c 183 mA . Therefore, i ic ia 283 mA Problem 3 Solution: Known quantities: The values of the resistors in the circuit of Figure P3. 1 i2 Find: The current in the circuit of Figure P3 using mesh current analysis. I i i1 Analysis: Since I is unknown, the problem will be solved in i3 terms of this current. i1 I For mesh it is obvious that: 1 For mesh i 2 i 3 0 2 1 For mesh i 2 i 0 4 3 i 2 0 Solving, i 3 0 Then, i i3 i2 and i 0 0 3 PROPRIETARY MATERIAL. The Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. G. Rizzoni, Principles and Applications of Electrical Engineering, 5th Edition Problem solutions, Chapter 3 Problem 3 Solution: Known quantities: The values of the resistors of the circuit in Figure P3. Find: The voltage gain, AV Analysis: Note that v2 , in the circuit of Figure P3 using mesh current analysis. v1 i 1 2 2 For mesh i 2 i 3(0) v1 For mesh 1 1 i 2 i 2v 4 or i 2 (2) i 3 0 For mesh i1(0) i 2 i or i1(1) i 2 i 3 (0) 0 Solving, from which and i 3 1 v 2 i 3 4 v AV 2 v1 3 PROPRIETARY MATERIAL. The Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.