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Chapter 3

Chapter 3. Resistive Network Analysis. COMBINING INDEPENDENT SOURCES:. Voltage sources in series add algebraically:. R 2. R 1. +. +. -6V. 10V. -. -. I. 2V. R 3. -. +. Equivalent. R 2. R 1. +. 18V. -. I. R 3. COMBINING INDEPENDENT SOURCES:.

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Chapter 3

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  1. Chapter 3 Resistive Network Analysis Grossman/Melkonian

  2. COMBINING INDEPENDENT SOURCES: • Voltage sources in series add algebraically: R2 R1 + + -6V 10V - - I 2V R3 - + Equivalent R2 R1 + 18V - I R3 Grossman/Melkonian

  3. COMBINING INDEPENDENT SOURCES: • Current sources in parallel add algebraically: V1 3mA 1mA 5mA -4mA R2 R1 Equivalent V1 5mA R2 R1 Grossman/Melkonian

  4. NODE VOLTAGE ANALYSIS: Section 3.2 • The Node Voltage Method is based on defining the voltage at each node as an independent variable. • A reference node is selected and all other node voltages are referenced to this node. • The Node Voltage Method defines each branch current in terms of one or more node voltages. This is done by using Ohm’s Law and KCL. • Since branch currents are defined in terms of node voltages, currents do not explicitly enter into the equations. Grossman/Melkonian

  5. NODE VOLTAGE ANALYSIS: • As mention previously, node voltage equations are written from KCL equations. KCL at Node B: i1 - i2 - i3 = 0 VA -VB VB -VC VB -VRef Applying Ohm’s Law: - - R1 R3 R2 i1 i2 i3 + R1 - VB VA + R3 - VC + i3 i1 R2 i2 - VRef Grossman/Melkonian

  6. NODE VOLTAGE ANALYSIS: Node Voltage Procedure: 1. Select and label a reference node. All other nodes are referenced to this node. 2. Label all N-1 remaining node(s). 3 Apply KCL to each node (N-1 node(s)). Writing equations in terms of node voltages. 4. Solve N-1 equations for V’s. 5. Calculate I’s using V, I, and R relationships. Grossman/Melkonian

  7. NODE VOLTAGE ANALYSIS – CURRENT SOURCES: Example 1: Use Node Voltage Analysis to determine V4, V6, i1, and i2. + i2 i1 + 3A 2A 4 6 - - Grossman/Melkonian

  8. NODE VOLTAGE ANALYSIS – CURRENT SOURCES: Example 1 cont.: 1. Identify and label a reference node. All other nodes will be referenced to this node. 2. Label all N-1 remaining node(s). VA + + i2 i1 3A 2A 4 6 - - VRef Grossman/Melkonian

  9. NODE VOLTAGE ANALYSIS – CURRENT SOURCES: Example 1 cont.: 3. Apply KCL to each N-1 node. Writing equation in terms of node voltage. VA - 0 VA - 0 -2A + + 3A = 0 + 4 6 VA + + i2 i1 3A 2A 4 6 - - VRef Grossman/Melkonian

  10. NODE VOLTAGE ANALYSIS – CURRENT SOURCES: Example 1 cont.: 4. Solve equation for VA. VA = -2.4V VA + + i2 i1 3A 2A 4 6 - - VRef Grossman/Melkonian

  11. NODE VOLTAGE ANALYSIS – CURRENT SOURCES: Example 1 cont.: 5. Solve for i1 and i2. Using Ohm’s Law and the calculated value for VA: i1 = VA/4 = -2.4V/4 i2 = VA/6 = -2.4V/6 i1 = -0.6A i2 = -0.4A VA + + i2 i1 3A 2A 4 6 - - VRef Grossman/Melkonian

  12. NODE VOLTAGE ANALYSIS – CURRENT SOURCES: Combining Independent Sources: • Referring to the circuit in example 1, calculate VA by transforming the circuit to an equivalent circuit with one current source and one resistor: VA + + i2 i1 Original Circuit 3A 2A 4 6 - - VRef Grossman/Melkonian

  13. NODE VOLTAGE ANALYSIS – CURRENT SOURCES: • Since the current sources are in parallel, they can be combined into a single equivalent current source: VA VA = iT • 2.4 = -1A • 2.4 + iT 1A 2.4 VA = -2.4V - Note: iT is equal to the sum of i1 and i2 of original circuit. i1 = -0.6A iT = -1A and i2 = -0.4A Grossman/Melkonian

  14. NODE VOLTAGE – CURRENT SOURCES: Example 2: • Using node voltage analysis calculate i1 and i2: 120 i1 4mA 6mA 200 400 i2 VRef Grossman/Melkonian

  15. NODE VOLTAGE – CURRENT SOURCES: Example 2 cont.: 1. Identify and label a reference node. All other nodes will be referenced to this node. 2. Label all N-1 remaining node(s). V2 V1 120 i1 4mA 6mA 200 400 i2 VRef Grossman/Melkonian

  16. NODE VOLTAGE – CURRENT SOURCES: Example 2 cont.: 3. Apply KCL to each N-1 node. Writing equation in terms of node voltage. V1 V1 - V2 = 0 Node 1: + -4mA + 120 200 V2 V1 120 i1 4mA 6mA 200 400 i2 VRef Grossman/Melkonian

  17. NODE VOLTAGE – CURRENT SOURCES: Example 2 cont.: V2 V2 – V1 Node 2: + 6mA = 0 + 400 120 V2 V1 120 i1 4mA 6mA 200 400 i2 VRef Grossman/Melkonian

  18. NODE VOLTAGE – CURRENT SOURCES: Example 2 cont.: • Solving equations for V1 and V2: V1 = -88.89mV V2 = -622.22mV V2 V1 120 i1 4mA 6mA 200 400 i2 VRef Grossman/Melkonian

  19. NODE VOLTAGE – CURRENT SOURCES: Example 2 cont.: i1 = V1/200 = -88.89mV/200 = -0.445mA i2 = 0 – V2/400 = 622.22mV/400 = 1.556mA V2 V1 120 i1 4mA 6mA 200 400 i2 VRef Grossman/Melkonian

  20. NODE VOLTAGE – VOLTAGE SOURCES: Example 3: • Use node voltage analysis to calculate VA and VB: VB VA 1k + 3mA 6V - 2k 1.5k VRef Grossman/Melkonian

  21. NODE VOLTAGE – VOLTAGE SOURCES: Example 3 cont.: VA VA - VB = 0 + Node A: -3mA + 1k 2k VB VA 1k + 3mA 6V - 2k 1.5k VRef Grossman/Melkonian

  22. NODE VOLTAGE – VOLTAGE SOURCES: Example 3 cont.: Node B: VB = 6V VB VA 1k + 3mA 6V - 2k 1.5k VRef Grossman/Melkonian

  23. NODE VOLTAGE – VOLTAGE SOURCES: Example 3 cont.: • Solving equations for VA and VB: VA = 6V VB = 6V VB VA 1k + 3mA 6V - 2k 1.5k VRef Grossman/Melkonian

  24. NODE VOLTAGE ANALYSIS - SUPERNODE: • A Supernode is needed when neither the positive nor the negative terminal of a voltage source is connected to the reference node. • We use the fact that KCL applies to the currents penetrating this boundary • to write a node equation at the supernode. • We then write node equations at the remaining non-reference nodes in the • usual way. • We now have regular node equations plus one supernode equation, leaving • us one equation short of the N - 1 required. Using the fundamental property • of node voltages, we can write; VA - VB = Vs • The voltage source inside the supernode constrains the difference between the • node voltages at nodes A and B. The voltage source constraint provides the • additional relationship needed to write N - 1 independent equations. Vs B A - + Grossman/Melkonian

  25. NODE VOLTAGE ANALYSIS - SUPERNODE: Example 4: • Using node voltage analysis, calculate i1, i2, andV4A: 2V - + i2 i1 + 4A 2A V4A 2 1 - Grossman/Melkonian

  26. NODE VOLTAGE ANALYSIS - SUPERNODE: Example 4 cont.: 1. Identify and label a reference node. All other nodes will be referenced to this node. 2. Label all N-1 remaining node(s). 3. Apply KCL to each N-1 node. Writing equation in terms of node voltage. VB VA - + 2V i2 + i1 4A 2A V4A 2 1 - VRef Grossman/Melkonian

  27. NODE VOLTAGE ANALYSIS - SUPERNODE: Example 4 cont.: Supernode: 2A + VA/1 +VB /2 + 4A = 0 VA + 0.5VB = - 6A (one equation, two unknowns) Supernode VB VA - + 2V i2 + i1 4A 2A V4A 2 1 - VRef Grossman/Melkonian

  28. NODE VOLTAGE ANALYSIS - SUPERNODE: Example 4 cont.: • But, we know VA - VB = 2V (gives us second equation) Supernode VB VA - + 2V i2 + i1 4A 2A V4A 2 1 - VRef Grossman/Melkonian

  29. NODE VOLTAGE ANALYSIS - SUPERNODE: Example 4 cont.: VA + 0.5VB = - 6A VA = -3.334V VA - VB = 2V VB = -5.334V Supernode VB VA - + 2V i2 + i1 4A 2A V4A 2 1 - VRef Grossman/Melkonian

  30. NODE VOLTAGE ANALYSIS - SUPERNODE: Example 4 cont.: Calculate i1, i2, andV4A: i1 = -3.334A i1 = VA/1 = -3.334V/1 i2 = -2.667A i2 = VB/2 = -5.334V/2 V4A = VB VB = -5.334V Supernode VB VA - + 2V i2 + i1 4A 2A V4A 2 1 - VRef Grossman/Melkonian

  31. NODE VOLTAGE ANALYSIS - SUPERNODE: Example 5: • Calculate i1 and V1mA using node voltage analysis: 200 + 2V - + + 3mA 4V - V1mA 1mA 150 - i1 Grossman/Melkonian

  32. NODE VOLTAGE ANALYSIS - SUPERNODE: Example 5 cont.: • Identify and label all nodes. VA VC 200 + 2V - + VB + 3mA 4V - 1mA V1mA i1 - 150 VRef Grossman/Melkonian

  33. NODE VOLTAGE ANALYSIS - SUPERNODE: Example 5 cont.: Supernode: -1mA + VB/150 + (VA - VC)/ 200 = 0 Node C: VC = 4V VA - VB = 2V VC VA 200 Supernode + 2V - + + VB 3mA 4V - 1mA V1mA i1 - 150 VRef Grossman/Melkonian

  34. NODE VOLTAGE ANALYSIS - SUPERNODE: Example 5 cont.: VA = 2.943V VB = 0.943V VC = 4V i1 = VB/150 = 0.943V/150 i1 = 6.286mA VC VA 200 Supernode + 2V - + + 3mA VB 4V - 1mA V1mA i1 - 150 VRef Grossman/Melkonian

  35. NODE VOLTAGE ANALYSIS - SUPERNODE: Example 5 cont.: KVL: -V1mA + 2V + V150 = 0 -V1mA + 2V + 0.943V = 0 V1mA = 2.943V VC 200 VA + 2V - KVL + + 3mA VB + 4V - 1mA V1mA V150 150 - - VRef Grossman/Melkonian

  36. MESH CURRENT ANALYSIS: Section 3.3 • The Mesh Current Method is based on writing independent mesh current equations with mesh currents as the independent variable. • Each mesh is identified and a direction for the mesh current is selected (e.g. clockwise). • KVL is applied to each mesh containing an unknown mesh current. Using Ohm’s law, the voltage across each resistor is written in terms of one or more mesh currents and a resistance. • Since element voltages are defined in terms of mesh currents, voltages do not explicitly enter into the equations as variables. • The equations are solved to find the mesh currents. Individual branch currents are then calculated using the mesh currents. Grossman/Melkonian

  37. MESH CURRENT ANALYSIS: Mesh Current Procedure: 1. Identify and label mesh currents for each mesh. Define each mesh current consistently. Typically, mesh currents are defined in the clockwise direction. 2. Apply KVL to each mesh containing an unknown current. Using Ohm’s Law, express the voltages in terms of one or more mesh currents. 3. Solve the linear equations for the mesh currents. Grossman/Melkonian

  38. MESH CURRENT ANALYSIS – VOLTAGE SOURCES: Example 6: • Use Mesh Analysis to solve for ix, is, V1, and V2: 75 100 - + V1 iS ix - + - V2 10V 50 2V - + + - + 4V Grossman/Melkonian

  39. MESH CURRENT ANALYSIS – VOLTAGE SOURCES: Example 6 cont.: 1. Identify and label mesh currents for each mesh. 100 75 - + V1 iS ix - + - i1 V2 50 i2 10V 2V - + + - + 4V Grossman/Melkonian

  40. MESH CURRENT ANALYSIS – VOLTAGE SOURCES: Example 6 cont.: 2. Apply KVL to each mesh containing an unknown current. Use Ohm’s Law to express the voltages in terms of one or more mesh currents. Mesh 1: -10V + 75(i1) + 50(i1 - i2) + 4V = 0 mesh 1  i1 – i2 75 100 - + V1 iS ix - + - V2 i1 50 i2 10V 2V - + + - + 4V Grossman/Melkonian

  41. MESH CURRENT ANALYSIS – VOLTAGE SOURCES: Example 6 cont.: Mesh 2: 50(i2 - i1) + 100(i2) - 2V = 0 mesh 2  i2 – i1 75 100 - + V1 iS ix - + - i1 V2 50 i2 10V 2V - + + - + 4V Grossman/Melkonian

  42. MESH CURRENT ANALYSIS – VOLTAGE SOURCES: Example 6 cont.: • Solving for mesh currents i1 and i2: Note: These are mesh currents. i1 = 61.54mA i2 = 33.85mA 75 100 - + V1 iS ix - + - i1 V2 50 i2 10V 2V - + + - + 4V Grossman/Melkonian

  43. MESH CURRENT ANALYSIS – VOLTAGE SOURCES: Example 6 cont.: • Calculate ix and is: ix = (i1 - i2) = 61.54mA - 33.85mA ix = 27.69mA is = i2 = 33.85mA 75 100 - + V1 iS ix - + - i1 V2 50 i2 10V - + + - + 4V Grossman/Melkonian

  44. MESH CURRENT ANALYSIS – VOLTAGE SOURCES: Example 6 cont.: • Solving for V1, and V2: V1 = 75•i1 = 75•61.54mA V1 = 4.62V V2 = 50(i2 – i1) = 50(33.85mA - 61.54mA) V2 = -1.38V 75 100 - + V1 iS ix - + - V2 i1 50 i2 10V 2V - + + - + 4V Grossman/Melkonian

  45. MESH CURRENT ANALYSIS – VOLTAGE SOURCES: Example 6 cont.: • Check results using KVL: Mesh 1: -10V + V1 - V2 + 4V = 0 -10V + 4.62V - (-1.38V) + 4V = 0 75 100 - + V1 iS ix - + - V2 50 KVL 10V KVL 2V - + + - + 4V Grossman/Melkonian

  46. MESH CURRENT ANALYSIS – VOLTAGE SOURCES: Example 6 cont.: Mesh 2: V2 + V3 - 2V = 0 -1.38V + 100•33.85mA - 2V = 0 75 100 - - + + V1 V3 iS ix - + - V2 50 KVL 10V KVL 2V - + + - + 4V Grossman/Melkonian

  47. MESH CURRENT ANALYSIS – CURRENT SOURCES: Example 7: • Use Mesh Analysis to solve for ix, V1, and V2: 1. Identify and label mesh currents for each mesh. 1.5k 1k - + V1 ix + + V2 2k i1 20V i2 - 2mA - Grossman/Melkonian

  48. MESH CURRENT ANALYSIS – CURRENT SOURCES: Example 7 cont.: 2. Apply KVL to each mesh containing an unknown currents. Use Ohm’s Law to express the voltages in terms of one or more mesh currents. Mesh 1: -20V + 1.5k(i1) + 2k(i1 – i2) = 0 1.5k 1k - - + + V1 V3 ix + + V2 2k KVL 20V KVL 2mA - - Grossman/Melkonian

  49. MESH CURRENT ANALYSIS – CURRENT SOURCES: Example 7 cont.: Mesh 2: i2 = -2mA • Solving for mesh currents i1 and i2: i2 = -2mA i1 = 4.57mA 1.5k 1k - - + + V1 V3 ix + + V2 2k KVL 20V KVL 2mA - - Grossman/Melkonian

  50. MESH CURRENT ANALYSIS – CURRENT SOURCES: Example 7 cont.: Calculate: ix = i2 – i1 = (-2mA) – (4.57mA) ix = -6.57mA V1 = 1.5k(i1) = 1.5k(4.57mA) V1 = 6.86V V2 = 2k(-ix) = 2k(6.57mA) V2 = 13.14V 1.5k 1k - - + + V1 V3 ix + + V2 2k KVL 20V KVL 2mA - - Grossman/Melkonian

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