# Node-Voltage Method Activity

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ENGR 313—CIRCUITS AND INSTRUMENTATIONNode-Voltage Method ActivityINTRODUCTIONCombining resistive elements into equivalent resistances and using Kirchhoff’s voltage and current lawsare powerful tools for circuit analysis. For more complicated circuits containing multiple power supplies, dependent power supplies, and multiply connected resistors, a more systematic analysis approachis required. In this activity, you will learn the node-voltage method of circuit analysis. This methoduses a system of independent equations based on Kirchhoff’s current law to analyze the circuit.Work in groups of up to three to complete the following exercises. Each group must turn in one completed worksheet.LEARNING OBJECTIVES Identify nodes, essential nodes, branches, and essential branches in a circuitDetermine the number of independent simultaneous equations required to analyze a circuit using thenode-voltage method.Use the node-voltage method to analyze and solve circuits.EXERCISE 1—TERMINOLOGYExamine the following three examples.Example 1 The nodes of this circuit are: a, b, c Note that four circuit elements are connected tonode b — R1, R2, R3, and I1 (the dashed ovals areintended to illustrate a single node — they will beremoved in future examples.) Note that four elements are connected to node c— V1, R2, R3, and I1The essential nodes of this circuit are: b, cThe branches of this circuit are: R1, R2, R3, V1, I1Example 2 ENGR 313The nodes of this circuit are: a, b, c, d Node b connects three circuit elements Node c connects three circuit elements Node d connects four circuit elementsThe essential nodes of this circuit are: b, c, dThe branches of this circuit are: R1, R2, R3, R4, V1, I1The essential branches of this circuit are: R1 V1, R2,R3, R4, I1Two (2) node-voltage equations are required for thiscircuit.Node-Voltage Activity1

Example 3 The nodes of this circuit are: a, b, c, d, e, f, g Node b connects three circuit elements Node c connects three circuit elements Node e connects three circuit elements Node g connects three circuit elementsThe essential nodes of this circuit are: b, c, e, gThe branches of this circuit are: R1, R2, R3, R4, R5, R6, V1, V2, I1The essential branches of this circuit are: R1 V1, R5, I1, R2 R3, V2 R4, R6Three (3) node-voltage equations are required to solve this circuit.EXERCISE 1 QUESTIONS1. Based on the examples above, define the following circuit terminology : Node— Essential node— Branch— Essential Branch—2. What is the minimum number of circuit elements that are connected to an essential node?3. What is the relationship between the number of required node-voltage equations and the number ofessential nodes, ne, in the circuit?4. For the following circuit, identify the nodes, essential nodes, branches, essential branches, and thenumber of node-voltage equations required for full analysis of the circuit.ENGR 313 Nodes: Essential Nodes: Branches: Essential Branches: Number of node-voltage equations:Node-Voltage Activity2

EXERCISE 2—USING THE NODE-VOLTAGE METHODExamine the following two examples:Example 4Determine the current through R4, the 10 Ω resistor, using the node-voltage method. This circuit has three essential nodes; therefore, twonode-voltage equations are required.Label the essential nodes as a, b, and c.The essential branches are R1 V1, R2, R3, R4, and I1Arbitrarily choose an essential node as a referencenode. Here c will be the reference node and designate with a signal ground symbol. All node voltages will be referenced to this point.Define the voltage, with respect to the referencenode, at the essential nodes a and b as va and vb,respectively.Use Kirchoff’s current law to write conservationequations for each node, in terms of node voltages.Isolate node a and apply Kirchhoff’s current law to this node:i1 i2 i3 0 Use Ohm’s law to compute the current through each branch in terms ofnode voltages:va V1 va v b va vc va 10 V va v b va 0 V 0AR1R2R31 2 5 17va 5vb 100 V Using algebra, the first node-voltage equation is: Next, isolate node b and apply Kirchhoff’s current law to this node:i2 i4 i5 0 Using Ohm’s law and recognizing that i5 -2A because of the currentprovided by supply I1:v b va v b vcv v v 0V I1 b a b 2A 0 AR2R42 10 Multiplying by 10 Ω and rearranging gives the second node-voltage equation: 5va 6vb 20 V Solving the two simultaneous node-voltage equations (using substitution, matrices, and/or a calculator) gives: va 9.09 V and vb 10.9 V.v10 .9 VKnowing vb and using Ohm’s law, the current through R4 can bei4 b 1.09 Acomputed to be 1.09 A.R10 4ENGR 313Node-Voltage Activity3

Example 5Determine the voltage at node b with respect to the signal ground of the following circuit. This circuit has two essential nodes (a and c);therefore, one node-voltage equation is required.The essential branches are R1 R2 V1, R3 R4 I1, and R5 R6.Choose node c to be the reference node.Let i1 be the current away from node a through the essential branch R1 R2 V1. Similarly, let i2and i3 represent the current away from node a through the essential branches R3 R4 I1 and R5 R6, respectively. Applying Kirchhoff’s current law to essential node a gives:i1 i 2 i3 0 v a V1v vcva 5 Vva 0 V I1 a 0.25 A 0AR1 R2R5 R6 100 22 100 120 Rearranging and solving for va gives va 22.8 VWith the voltage at node a known, and knowing that 250mA of current flows from node b to node a,we can use Ohm’s law to compute the potential difference in the branch between node a and node b.vb va vb 22.8 V i2 R3 R4 0.25 A 15 110 Solving, we find that vb 54.05 V 54.1 V with respect to the signal ground (essential node c).EXERCISE 2 QUESTIONS6. When applying Kirchhoff’s law to the essential nodes, what assumption is made about the directionof the current flow?7. What is the assumed potential of the arbitrarily selected reference node?8. How is the current in each essential branch represented and/or calculated?9. Why were the currents from the current supplies in the examples written as negative values whendeveloping the node voltage equations?ENGR 313Node-Voltage Activity4

EXERCISE 2 QUESTIONS, CONTINUED10. Determine the node voltage equations for the following circuit and solve for the voltages at nodes aand b of the circuit, with respect to the signal ground shown. Clearly show all of your steps and useadditional paper if necessary.abAnswers:va 15.4 Vvb 7.39 VENGR 313Node-Voltage Activity5

EXERCISE 3—SPECIAL CASES FOR THE NODE-VOLTAGE METHODExamine the following four examples:Example 6This circuit, from Question 10, has three essential nodes and requires the two independent node-voltageequations shown to completely specify the circuit.abvbvav a 10 V va v b 0A22 150 vb vavb 0 V 0.02 A 0 A150 100 120 0.3 A Example 7This circuit has three essential nodes. By recognizing that va 10 V, only one node-voltage equation isneeded.abvbvav b 10 Vvb 0 V 0.02 A 0 A150 100 120 Example 8The dependent voltage supply, V1, in the circuit below produces a voltage output that is 300 times thecurrent through resistor R6 (i.e., a gain of 300 V/A). This circuit has three essential nodes and requirestwo node-voltage equations and an additional constraint equation.a300iR6ENGR 313bvbvaiR6v a 300 i R6 v a v b 0A22 150 vb vavb 0 V 0.02 A 0 A150 100 120 vb 0 Vi R6 100 120 0.3 A Node-Voltage Activity6

Example 9The dependent voltage supply, V1, in the circuit below produces a voltage output that is 300 times thecurrent through resistor R6. This circuit has three essential nodes. Recognizing that the voltage at nodea with respect to the reference node is the voltage produced by the dependent voltage supply,va 300iR6, one node-voltage equation and one constraint equation is required.abvbva300iR6v b 300iR6vb 0 V 0.02 A 0 A150 100 120 vb 0 ViR6 100 120 iR6EXERCISE 3 QUESTIONS11. How does the number of required equations change when an essential node is connected to the reference node with a branch containing only a voltage supply (dependent or independent)? Why?12. How does the number of required equations change when a dependent source is included in the circuit? Why?Note that this is true when a dependent voltage or current source (not shown in the examples) is addedto the circuit.13. What is a constraint equation? How is it determined?ENGR 313Node-Voltage Activity7

EXERCISE 4—SUPERNODESExamine the following two examples:Example 10—Supernode with Independent Voltage SupplyWhen an ideal voltage supply directly connects two nodes, neither of which are the zero-voltage reference node, a supernode exists. Nodes a and b in the following circuit form a supernode. Based on theschematic, the supply V2 creates a 3 V potential between nodes a and b; however, we do not know thepotential of node a or b with respect to the reference node, c.To solve for the unknown node voltages in this circuit,begin by applying Kirchhoff's current law at node a.i R1 i R 2 iV 2 0Using Ohm’s Law, the current through R1 and R2 can beexpressed in terms of the unknown node voltage at nodea. The voltage supply does not have a resistance, butcurrent will leave (or enter) the node through the voltagesupply. If we arbitrarily assume that current leaves node a through the voltage supply, and we call thisunknown current ix, we can rewrite the current balance at node a asva 20 V va 0 V ix 06,000 2,000 Next, applying Kirchhoff’s Current Law to node b givesiV 2 iR3 iI 1 0The unknown current through the voltage supply from node a, ix, must enter (or leave) node b. Basedon the assumption that the unknown current exited node a, we must assume that it enters node b. Thecurrent from the ideal current supply, I1, is also directed towards the node. Therefore, the current balance at node b can be rewritten as ix vb 0 V 0.006 A 04,000 The current balance equations for nodes a and b both contain the unknown current, ix, and can be combined to eliminate this as a variable, providing the supernode equationva 20 V va 0 V vb 0 V 0.006 A6,000 2,000 4,000 Unfortunately, there is now one equation for the two unknown variables, va and vb. To solve this circuitnetwork, we need an additional independent equation. Luckily, we know from the schematic that thevoltage supply V2 creates a 3 V potential between nodes a and b. This can be written as a constraintequation, which provides the second equation necessary to solve for the unknown voltages at the essential nodes. Observing the polarity of the voltage supply (node a will be at a higher potential, since the symbol is connected to this node), the constraint equation is written asva vb 3 VSolving the system of 2 equations for two unknowns givesva 11 Vvb 8 VENGR 313Node-Voltage Activity8

Example 11—Supernode with Dependent Voltage SupplyIn this example, V2 is a dependent voltage supply that creates a potential that is proportional to the current, iC, through resistor R3. This voltage supply provides a voltage of 1,500 V for every ampere of current through R3 (i.e., 1,500 V/A). Because V2 joins node a and b, neither of which are the reference node,this connection forms a supernode. As in Example 10, a current balance can be performed on nodes aand b, resulting in the supernode equation ofva 20 V va 0 V vb 0 V6,000 2,000 4,000 0.006 AAgain, the creation of this supernode equation reducesthe number of independent equations available to completely solve the system. A constraint equation thatdefines the potential developed between nodes a and bmust be developed. Based on the orientation of thevoltage supply, node a must be at a higher potentialthan node b, and the potential difference must 1,500iC.Therefore, a constraint equation is va vb 1,500 iCUnfortunately, the value of iC is not known, so yet another constraint equation is necessary to solve thissystem. This constraint equation must relate the controlling current of the dependent supply, iC, to theunknown node voltages. Luckily, we can apply Ohm’s law to determine this constraint equation. Noting that the designated direction (given in the diagram) of iC is away from node b, the potential at node bmust be higher than the potential at node c, which has been selected as the 0 V reference node. Therefore, the second constraint equation becomesv 0ViC b4,000 Now, there are three equations to solve for the three unknown variables—va, vb, and iC. Solving the system of equations givesva 11 Vvb 8 ViC 0.002 A 2 mAEXERCISE 4 QUESTIONS14. What is a supernode? When does it occur? How does it change the number of equations necessaryto completely solve the system?15. Outline the process for solving a circuit containing a supernode.ENGR 313Node-Voltage Activity9

16. In the circuit below, the output of the dependent voltage supply is proportional to the currentthrough resistor R1 with a gain of 1 V/A. How many independent equations are required to analyzethis circuit? Write the node-voltage and constraint equations and solve for the unknown node voltages. Clearly show your work and all steps involved. Use additional paper if necessary.abiR1ciR1dAnswers:va 15 Vvb 8 Vvc 10 ViR1 2 AENGR 313Node-Voltage Activity10

potential of node a or b with respect to the reference node, c. To solve for the unknown node voltages in this circuit, begin by applying Kirchhoff's current law at node a. Using Ohm’s Law, the current through R 1 and R 2 can be expressed in terms of the unknown node voltage at node

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