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1 ECE 221 Electric Circuit Analysis I Chapter 7 Node-Voltage Method Herbert G. Mayer, PSU Status 11/10/2014 For use at Changchun University of Technology.

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Presentation on theme: "1 ECE 221 Electric Circuit Analysis I Chapter 7 Node-Voltage Method Herbert G. Mayer, PSU Status 11/10/2014 For use at Changchun University of Technology."— Presentation transcript:

1 1 ECE 221 Electric Circuit Analysis I Chapter 7 Node-Voltage Method Herbert G. Mayer, PSU Status 11/10/2014 For use at Changchun University of Technology CCUT

2 2 Syllabus Current Status Current Status Parallel Resistor Parallel Resistor Definitions Definitions General Circuit Problem General Circuit Problem Samples Samples What is Node-Voltage? What is Node-Voltage? Node-Voltage Steps Node-Voltage Steps

3 3 Current Status You have 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*i x then 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 voltage of b*v x We’ll repeat a few terms and laws

4 4 Parallel Resistors Two 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 5 Definitions Node (Nd):point where 2 circuit elements come together Essential Nd:ditto, but 3 or more elements Path:trace of >=1 basic elements w/o repeat Branch (Br):path connecting 2 nodes Essential Br:path connecting only 2 essential nodes, not more Loop:path whose end node equals the start node Mesh:loop not enclosing other loops Planar Circuit:circuit that can be drawn in 2 dimensions Cross circuit exercise in class! And pentagon!

6 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 that n equations are needed If circuit C1 also happens to have n nodes, 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 can we compute all n currents?

7 7 General Circuit Problem Given n unknowns and n nodes: one can can generate only n-1 equations using KCL But NOT n equations, as the application of the n th node can be derived from the other n- 1 equations, so this would be redundancy Redundant equations do not help solve any more currents than n-1! But one can also generate equations using KVL, to compute the remaining currents –or unknowns

8 8 Circuit for Counting Nodes etc.

9 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 10 What is Node-Voltage? Nodes have no voltage! So what’s up? What does this question mean? Nodes are connections 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

11 11 Node-Voltage Steps Interestingly, the No Vo Mo –i.e. Node- Voltage Methodology– 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 12 Node-Voltage Steps Analyze your circuit, locate and number all essential nodes; we call that number of essential nodes n e For now, view only planar circuits From these n e 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 n e essential nodes we can generate n-1 Node- Voltage equations

13 13 Node-Voltage Steps 1. 1.Using KCL: 2. 2.Analyze the circuit below, and generate 2 Node-Voltage equations 3. 3.Enables us to compute 2 unknowns 4. 4.We see that v1 and v2 are unknown 5. 5.Once v1 and v2 are known then we can compute all currents

14 14 Node-Voltage Sample1 In the following sample 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 current can be computed Note: the current through the right-most branch is known to be 2 A Here is the sample circuit:

15 15 Node-Voltage Sample1

16 16 Node-Voltage Sample1 For node n 1 compute all currents using KCL: (v1 - 10)/1 + v1/5 + (v1 - v2)/2= 0 For node n 2 compute all currents using KCL: V2/10 + (v2 - v1)/2 - 2= 0 Students compute v1 and v2

17 17 Node-Voltage Sample1: Compute v1=100 / 11=9.09 V v2=120 / 11=10.91 V

18 18 Node-Voltage Sample2 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

19 19 Node-Voltage Sample2

20 20 Node-Voltage Sample2 For node n 1 compute all currents using KCL: v1/10 + (v1-50)/5 + v1/40 – 3=0 Students compute v1, ia, ib, and ic

21 21 Node-Voltage Sample2: Compute v1/10 + (v1-50)/5 + v1/40 – 3=0/*40 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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