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ECE 221 Electric Circuit Analysis I Chapter 7 Node Voltage Method

ECE 221 Electric Circuit Analysis I Chapter 7 Node Voltage Method. Herbert G. Mayer, PSU Status 10/15/2015. Syllabus. Studying ECE 221 Parallel Resistor Definitions General Circuit Problem Samples Node Voltage Method (No Vo Mo) Node Voltage Method Steps. Studying ECE 221.

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ECE 221 Electric Circuit Analysis I Chapter 7 Node Voltage Method

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  1. ECE 221Electric Circuit Analysis IChapter 7 Node Voltage Method Herbert G. Mayer, PSU Status 10/15/2015

  2. Syllabus Studying ECE 221 Parallel Resistor Definitions General Circuit Problem Samples Node Voltage Method (No Vo Mo) Node Voltage Method Steps

  3. Studying ECE 221 You learned how to construct simple circuit models Learned the 2 Kirchhoff Laws: KCL and KVL You know constant voltage and constant current sources You know dependent current and dependent voltage sources When a dependent voltage source depends on a*ixthen that factor defines the voltage, i.e. the amount of Volt generated, NOT the current! Ditto for dependent current source, depending on some voltage: A current is being defined, through a numeric voltage factor of b*vx We’ll repeat a few terms and laws

  4. Parallel Resistors Resistors R1 and R2 are in a parallel circuit: What is their resulting resistance? How to derive this? Hint: think about Siemens! Not the engineer, the conductance This seems trivial, yet will return again with inductivities in the same way And in a similar (dual) way for capacities

  5. Definitions Node (Nd): where 2 or more circuit elements come together Essential Nd: ditto, but 3 or more circuit elements come together Path: trace of >=1 basic elements w/o repeat Branch (Br): circuit path connecting 2 nodes Essential Br: path connecting only 2 essential nodes, not more Loop: path whose end-node equals start-node w/o repeat Mesh: loop not enclosing any other loops Planar Circuit: circuit that can be drawn in 2 dimensions without crossing lines Cross circuit exercise in classand pentagon to test planarity!

  6. General Circuit Problem Given n unknown currents in a circuit C1 How many equations are needed to solve the system? Rhetorical question: We know n equations! If circuit C1 also happens to have nnodes, can you solve the problem of computing the unknowns using KCL? Also rhetorical question: We know that n nodes alone will not suffice using only KCL! How then can we compute all n currents?

  7. General Circuit Problem Given n unknowns and n nodes: one can generate n-1 independent equations via KCL But NOT n equations, as electrical units at the nth node can be derived from the other n-1 equations: would be redundant Redundant equations do not help solve unknowns But one can also generate equations using KVL, to compute the remaining currents –or other unknowns

  8. Circuit for Counting Nodes etc.

  9. Sample: How Many of Each? Nodes: 5 Essential Nodes: 3 Paths: large number, since includes sub-paths Branches: 7 Essential Branches: 5 Loops: 6 Meshes: 3 Is it Planar: yes

  10. Node Voltage Method (No Vo Mo) Nodes have no voltage! So what’s up? What does this phrase mean? Nodes are connecting points of branches Due to laws of nature, expressed as KCL: nodes have a collective current of 0 Amp! So why discuss a Node Voltage Method (No Vo Mo)? No Vo Mo combines using KCL and Ohm’s law across all paths leading to any one node, from all essential nodes to a selected reference node The reference node is also an essential node Conveniently, we select the essential node with the largest number of branches as reference node Is not necessary; works with any essential node

  11. Node Voltage Steps Interestingly, the No Vo Mo –i.e. Node Voltage Method– applies also to non-planar circuits! The later to be discussed Mesh-Current Method only applies to planar circuits We’ll ignore this added power of No Vo Mo for now, and focus on planar circuits Here are the steps:

  12. Node Voltage Steps Analyze your circuit, locate and number all essential nodes; we call that number of essential nodes ne For now, view only planar circuits From these ne essential nodes, pick a reference node Best to select the one with the largest number of branches; simplifies the formulae Then for each remaining essential node, compute the voltage rises from the reference node to the selected essential node, using KCL For ne essential nodes we can generate n-1 Node Voltage equations

  13. Sample 1

  14. Node Voltage Steps for Sample 1 Using KCL: Analyze the circuit below, and generate 2 Node Voltage equations Enables us to compute 2 unknowns We see that v1 and v2 are unknown Once v1 and v2 are known then we can compute all currents

  15. Node Voltage Sample 1 In the Sample 1 circuit, use the Node Voltage Method to compute v1 and v2 There are 3 essential nodes Pick the lowest one as the reference node, since it unites the largest number of branches Once voltages v1 and v2 are known, the currents in the 5 and 10 Ohm resistors are computable Using KCL and Ohm’s Law, all other currents can be computed Note: the current through the right-most branch of Sample 1 is known to be 2 A Here is the Sample 1 circuit:

  16. Node Voltage Sample 1

  17. Node Voltage Sample 1 For node n1 compute all currents using KCL: (v1 - 10)/1 + v1/5 + (v1 - v2)/2 = 0 For node n2 compute all currents using KCL: V2/10 + (v2 - v1)/2 - 2 = 0 Students compute v1 and v2

  18. Node Voltage Sample 1: Compute v1 = 100 / 11 = 9.09 V v2 = 120 / 11 = 10.91 V

  19. Sample 2

  20. Node Voltage Sample 2 In the following sample circuit, use the Node Voltage Method to compute v1, ia, ib, and ic There are 2 essential nodes Hence we need just 1 equation to compute v1 We pick the lowest one as the reference node, since it has the largest number of branches Once voltage v1 is known, the currents are computable Using KCL, all other current can be computed

  21. Node Voltage Sample 2

  22. Node Voltage Sample 2 For node n1 compute all currents using KCL: v1/10 + (v1-50)/5 + v1/40 - 3 = 0 Students compute v1, ia, ib, and ic

  23. Node Voltage Sample 2: Compute v1/10 + (v1-50)/5 + v1/40 - 3 = 0 // *40 +3 4*v1 +8*v1 - 50*8 + v1 = 3 * 40 v1*( 4 + 8 + 1 ) - 400 = 120 13*v1 = 520 v1 = 40 V ia = ( 50 – 40 ) / 5 = 2 A ib = 40 / 10 = 4 A ic = 40 / 40 = 1 A

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