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Purdue ECE 20100 - lect8 (1)

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Slide 1Slide 2Slide 3Slide 4Slide 5Slide 6Slide 7Slide 8Slide 9Slide 10Slide 11EE201 Lecture 8 P. 1Nodal Analysis (cont.); Mesh AnalysisNodal Analysis -- Good and Bad Choices+_+_2A3V5V53452Consider the following circuitEE201 Lecture 8 P. 2Good choices: (1) labelled nodes, (2) ground at E. Therefore, only 2 variables and not four.+_+_2A3V5V53452ABCDEEE201 Lecture 8 P. 3Apply KCL to find node voltages.2 unknowns: VA and VD KCL@A: 5(VA - 5) + 2VA + 2 = 07 VA = 23KCL@D: 5VD + 4(VD - 3) - 2 = 09 VD = 14Therefore, VD = 14/9 V; VA = 23/7 VThe intelligent choices for nodal analysis led to reduction of a 5 node circuit to two unknowns and no coupled equations to solve simultaneously.EE201 Lecture 8 P. 4Bad choices: (1) ground at D (The importance of choosing the right node as ground: you do not apply KCL at ground!) +_+_2A3V5V53452ABCDESupernodeEE201 Lecture 8 P. 5Apply KCL to find node voltages.2 unknowns: VA and VB KCL@A: 5(VA - VB) + 2(VA - VE) + 2 = 0But, VE = VB - 5 7VA - 7VB + 12 = 0KCL@SN: 5(VB - VA) + 2(VE - VA) + 5VE + 4VC = 0But, VC = VB - 2 -7VA + 16VB - 43 = 0Simultaneous equations: 7VA - 7VB + 12 = 0 -7VA + 16VB - 43 = 0VA = 109/63; VB = 217/63EE201 Lecture 8 P. 6Are solutions the same?Look at voltage drop across circuit elements.Good choice: VD = 14/9; VA = 23/7Bad: VB = 217/63; VA = 109/63But what is drop across 0.2  resistor?Good choice:V5mho = VA - VB = 23/7 - 5 = -12/7 VBad:V5mho = VA - VB = 109/63 - 217/63 = -12/7 VNode voltages are different because of the different reference nodes used. It is the voltage difference between nodes that is physical and must be the same regardless of choice of ground.EE201 Lecture 8 P. 7Loop (Mesh) Analysis A circuit analysis technique based on KVL, Ohm’s law, and the existence of circulation currents.RiA-iBiAiBConsider the circuit below.TerminologyThere are 3 loops and 2 meshes in this circuit.A loop is any closed path around a circuit, but passing through a node only once.A mesh is a loop which contains no other loops.EE201 Lecture 8 P. 8i1i2Mesh currents can be defined within the circuit, as shown below.Basis of Mesh Analysis: Write KVL equations for each mesh, then solve the coupled simultaneous equations.KVL1: V1 - R1i1 - R3(i1-i2) = 0KVL2: -V2 - R2i2 - R3(i2-i1) = 0Once equations are solved for i1 and i2, the behavior of the circuit is known.R1R2__++V1V2R3EE201 Lecture 8 P. 9Mesh Analysis - guidelines1) Draw a clear circuit diagram indicating meshes.2) Assign a current to each mesh. For n meshes, there will be n equations.3) Apply KVL around each mesh (If dependent voltage source present, relate controlling variable to the mesh current).4) If circuit contains independent or dependent current sources, apply KVL to a ‘supermesh’ to bypass the current source. Relate currents through sources to the assigned mesh currents.5) Do not define any voltage variables.EE201 Lecture 8 P. 10Example: Solve by mesh analysis._+51324_+5V3A10V3 meshes for KVL, but the mesh equations for meshes 2 and 3 contain a current source. Therefore, define supermesh (or loop) that contains 2 and 3.EE201 Lecture 8 P. 11Solution: Use KVL to obtain equations.KVL-1: 10 - 5i1 - 3(i1-i2) - 2(i1-i3) = 0KVL-SM: -5 - 4i3 - 2(i3-i1) - 3(i2-i1) - 1i2 = 0I source: i3 - i2 = 3A3 equations and 3 unknowns -- solve for i1, i2, i3Once the currents are known, the circuit behavior is


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Purdue ECE 20100 - lect8 (1)

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