Chapter 4 Objectives

Transcription

1 Chapter 4 Engr228 Circuit Analysis Dr Curtis Nelson Chapter 4 Objectives Understand and be able to use the node-voltage method to solve a circuit; Understand and be able to use the mesh-current method to solve a circuit; Be able to determine which technique is best for a particular circuit; Understand source transformations and be able to use them to simplify a circuit; Understand the concept of Thevenin and Norton equivalent circuits and be able to derive one; Know the condition for maximum power transfer to a resistive load and be able to calculate the value of the load resistor that satisfies this condition. Engr228 - Chapter 4, Nilsson 10E 1

2 Engr228 - Chapter 4, Nilsson 10E 2

3 Circuit Analysis As circuits get more complicated, we need an organized method of applying KVL, KCL, and Ohm s law. Nodal analysis assigns voltages to each node, and then we apply Kirchhoff's Current Law to solve for the node voltages. Mesh analysis assigns currents to each mesh, and then we apply Kirchhoff s Voltage Law to solve for the mesh currents. Review - Nodes, Paths, Loops, Branches These two networks are equivalent. There are three nodesand five branches: Node: a point at which two or more elements have a common connection point. Branch: a single path in a network composed of one simple element and the node at each end of that element. A path is a sequence of nodes. A loop is a closed path. Engr228 - Chapter 4, Nilsson 10E 3

4 Review - Kirchhoff s Current Law Kirchhoff s Current Law (KCL) states that the algebraic sum of all currents entering a node is zero. i A + i B + ( i C ) + ( i D ) = 0 Review - Kirchhoff s Voltage Law Kirchhoff s Voltage Law (KVL) states that the algebraic sum of the voltages around any closed path is zero. -v 1 + v 2 + -v 3 = 0 Engr228 - Chapter 4, Nilsson 10E 4

5 Node Example Node = every point along the same wire 6K V 10V 4K 3 nodes Nodes How many nodes are there in the circuit(s) below? Engr228 - Chapter 4, Nilsson 10E 5

6 Notes on Writing Nodal Equations All terms in the equations are in units of current. Everyone has their own style of writing nodal equations The important thing is that you remain consistent. Probably the easiest method if you are just getting started is to remember that: current entering a node = current leaving the node Current directions can be assigned arbitrarily, unless they are previously specified. The Nodal Analysis Method Assign voltages to every node relative to a reference node. Engr228 - Chapter 4, Nilsson 10E 6

7 Choosing the Reference Node By convention, the bottom node is often the reference node. If a ground connection is shown, then that becomes the reference node. Otherwise, choose a node with many connections. The reference node is most often assigned a value of 0.00 volts. Apply KCL to Find Voltages Assume reference voltage = 0.0 volts Assign current names and directions Apply KCL to node v 1 ( Σ out = Σ in) Apply Ohm s law to each resistor: v v v = 3.1 Engr228 - Chapter 4, Nilsson 10E 7

8 Apply KCL to Find Voltages Apply KCL to node v 2 ( Σ out = Σ in) Apply Ohm s law to each resistor: v1 v2 v2 0 = + ( 1.4) 5 1 We now have two equations for the two unknowns v 1 and v 2 and we can solve them simultaneously: v 1 = 5V and v 2 = 2V Example: Nodal Analysis Find the current i in the circuit below. Answer: i = 0 (since v 1 =v 2 =20 V) Engr228 - Chapter 4, Nilsson 10E 8

9 Nodal Analysis: Dependent Source Example Determine the power supplied by the dependent source. Key step: eliminate i 1 from the equations using v 1 =2i 1 v1 v2 v = v v v2 + 3i1 = 1 0 i1 = v Answer: 4.5 kw being generated Example #2 How many nodes are in this circuit? How many nodal equations must you write to solve for the unknown voltages? 4Ω -3A V2 V1 3Ω V3-8A 1Ω 5Ω -25A 0V Engr228 - Chapter 4, Nilsson 10E 9

10 Example #2 node V1-3A 4Ω 3Ω V1 V2 V3-8A 1Ω 5Ω -25A 0V At node V 1 V1 V 3 V1 V = V 1 3V V 1 4V 2 = 0 7V1 4V 2 3V 3 = 132 Example #2 node V2-3A 4Ω 3Ω V1 V2 V3-8A 1Ω 5Ω -25A 0V At node V 2 V 2 V1 V 2 V 3 V = V 2 2V V 2 3V 3 + 6V 2 = 0 2V 1+ 11V 2 3V 3 = 18 Engr228 - Chapter 4, Nilsson 10E 10

11 Example #2 node V3-3A 4Ω 3Ω V1 V2 V3-8A 1Ω 5Ω -25A 0V At node V 3 V 3 V 2 V 3 V1 V = V 3 10V 2 + 5V 3 5V V 3 = 0 5V 1 10V V 3 = 500 All 3 Equations 7V1 4V 2 3V 3 = 132 2V 1+ 11V 2 3V 3 = 18 5V 1 10V V 3 = 500 Answer: V1 = 0.956V V2 = V V3 = V Engr228 - Chapter 4, Nilsson 10E 11

12 Voltage Sources and the Supernode If there is a DC voltage source between two non-reference nodes, you can get into trouble when trying to use KCL between the two nodes because the current through the voltage source may not be known, and an equation cannot be written for it. Therefore, we create a supernode. The Supernode Analysis Technique Apply KCL at Node v 1. Apply KCL at the supernode. Add the equation for the voltage source inside the supernode. v 1 = V v 2 = 10.5V v 3 = 32.5V v1 v3 4 v1 v2 3 v v 3 2 v1 v + 3 v1 v + 4 = = 3 8 v2 v3 = Engr228 - Chapter 4, Nilsson 10E 12

13 Supernode Example -3A 4Ω 3Ω V1 V2 V3 1V -8A 1Ω 5Ω -25A 0V Supernode Example Node V1-3A 4Ω 3Ω V1 V2 V3 1V -8A 1Ω 5Ω -25A 0V At node V 1 V1 V 3 V1 V = V 1 3V V 1 4V 2 = 0 7V1 4V 2 3V 3 = 132 Engr228 - Chapter 4, Nilsson 10E 13

14 Supernode Example Nodes V2 and V3 4Ω -3A 3Ω V1 V2 V3 1V -8A 1Ω 5Ω -25A supernode At node V 2 Supernode 0V V 2 V1 V 3 V1 V 3 0 V = V 2 20V V 3 15V V V 2 = 0 35V 1+ 80V V 3 = 1680 V 2 V 3 = 1 Solving Simultaneous Equations 7V 1 4V 2 3V 3 = V 1+ 80V V 3 = 1680 V 2 V 3 = 1 V1 = V V2 = V V3 = V Engr228 - Chapter 4, Nilsson 10E 14

15 Textbook Problem 4.32 Nilsson 10E Use the node-voltage method to solve for the currents in the circuit below. Answer: i a = 0.1A i b = 0.3A i c = 0.2A Mesh Analysis: Nodal Alternative A mesh is a loop that does not contain any other loops within it. In mesh analysis, we assign mesh currents and solve using KVL. All terms in the equations are in units of voltage. Remember voltage drops in the direction of current flow except for sources that are generating power. The circuit below has four meshes: Engr228 - Chapter 4, Nilsson 10E 15

16 Mesh Example Simple resistive circuit showing three paths, which represent three mesh currents. Note that I R3 = I 1 I 2 The Mesh Analysis Method Mesh currents Branch currents Engr228 - Chapter 4, Nilsson 10E 16

17 Mesh: Apply KVL Apply KVL to mesh 1 ( Σ voltage drops = 0 ): i 1 +3(i 1 -i 2 ) = 0 Apply KVL to mesh 2 ( Σ voltage drops = 0 ): 3(i 2 -i 1 ) + 4i 2-10 = 0 i 1 = 6A i 2 = 4A Example: Mesh Analysis Determine the power supplied by the 2 V source. Applying KVL to the meshes: 5 + 4i 1 + 2(i 1 i 2 ) 2 = (i 2 i 1 ) + 5i = 0 i 1 = A i 2 = A Answer: W (the source is generating power) Engr228 - Chapter 4, Nilsson 10E 17

18 A Three Mesh Example Follow each mesh clockwise Simplify Solve the equations: i 1 = 3 A, i 2 = 2 A, and i 3 = 3 A Example Use mesh analysis to determine Vx 1Ω I2 7V I1 6V + Vx - 3Ω I3 1Ω Engr228 - Chapter 4, Nilsson 10E 18

19 Example - continued 7V I1 6V 1Ω I2 + Vx - 3Ω I3 1Ω 7 + 1( I1 I 2) ( I1 I3) = 0 3I1 I 2 2I3 = 1 Equation I 1( I 2 I1) + 2I 2 + 3( I 2 I3) = 0 I1+ 6I 2 3I3 = 0 Equation II 2( I3 I1) 6 + 3( I3 I 2) + I3 = 0 2I1 3I 2 + 6I3 = 6 Equation III I 1 = 3A, I 2 = 2A, I 3 = 3A Vx = 3(I3-I2) = 3V Current Sources and the Supermesh If a current source is present in the network and shared between two meshes, then you must use a supermesh formed from the two meshes that have the shared current source. Engr228 - Chapter 4, Nilsson 10E 19

20 Supermesh Example Use mesh analysis to evaluate Vx 1Ω I2 7V I1 7A + Vx - 3Ω I3 1Ω Supermesh Example - continued 1Ω I2 7V I1 7A + Vx - 3Ω I3 1Ω Loop 2: 1( I 2 I1) + 2I 2 + 3( I 2 I3) = 0 I1+ 6I 2 3I3 = 0 Equation I Engr228 - Chapter 4, Nilsson 10E 20

21 Supermesh Example - continued Supermesh 1Ω I2 7V I1 + Vx - 3Ω 7A I3 1Ω I 1 = 9A I 2 = 2.5A I 3 = 2A Vx = 3(I3-I2) = -1.5V 7 + 1( I1 I 2) + 3( I3 I 2) + I3 = 0 I1 4I 2 + 4I3 = 7 I1 I3 = 7 Equation II Equation III Node or Mesh: How to Choose? Use the one with fewer equations, or Use the method you like best, or Use both, as a check. Engr228 - Chapter 4, Nilsson 10E 21

22 Dependent Source Example Find i 1 Answer: i 1 = ma. Dependent Source Example Find Vx 1Ω I2 15A I1 + Vx - 3Ω 1/9 Vx I3 1Ω Engr228 - Chapter 4, Nilsson 10E 22

23 15A Dependent Source Example - continued I1 1Ω 1/9 Vx I2 + Vx - 3Ω I3 1Ω I1 = 15A Equation I 1( I 2 I1) + 2I 2 + 3( I 2 I3) = 0 I1+ 6I 2 3I3 = 0 Equation II 1 I3 I1 = Vx 9 Equation III Vx = 3( I3 I 2) Equation IV I 1 =15A, I 2 =11A, I 3 =17A Vx = 3(17-11) = 18V Textbook Problem 4.52 Hayt 7E Obtain a value for the current labeled i 10 in the circuit below I 10 = -4mA Engr228 - Chapter 4, Nilsson 10E 23

24 Textbook Problem 4.56 Nilsson 10th Find the power absorbed by the 20V source in the circuit below. Power 20V = 480 mw absorbed Linear Elements and Circuits A linear circuit element has a linear voltage-current relationship: If i(t) produces v(t), then Ki(t) produces Kv(t) If i 1 (t) produces v 1 (t) and i 2 (t) produces v 2 (t), then i 1 (t) + i 2 (t) produces v 1 (t) + v 2 (t), Resistors and sources are linear elements Dependent sources need linear control equations to be linear elements A linear circuit is one with only linear elements Engr228 - Chapter 4, Nilsson 10E 24

25 The Superposition Concept For the circuit shown below, the question is: How much of v 1 is due to source i a, and how much is due to source i b? We will use the superposition principle to answer this question. The Superposition Theorem In a linear network, the voltage across or the current through any element may be calculated by adding algebraically all the individual voltages or currents caused by the separate independent sources acting alone, i.e. with All other independent voltage sources replaced by short circuits (i.e. set to a zero value) and All other independent current sources replaced by open circuits (also set to a zero value). Engr228 - Chapter 4, Nilsson 10E 25

26 Applying Superposition Leave one source ON and turn all other sources OFF: Voltage sources: set v=0 These become short circuits. Current sources: set i=0 These become open circuits. Then, find the response due to that one source Add the responses from the other sources to find the total response Superposition Example (Part 1 of 4) Use superposition to solve for the current i x Engr228 - Chapter 4, Nilsson 10E 26

27 Superposition Example (Part 2 of 4) First, turn the current source off: i ʹ x = = 0.2 Superposition Example (Part 3 of 4) Then, turn the voltage source off: i xʹ = 6 (2) = Engr228 - Chapter 4, Nilsson 10E 27

28 Superposition Example (Part 4 of 4) Finally, combine the results: i x = i xʹ +i xʹ = =1.0 Source Transformation The circuits (a) and (b) are equivalent at the terminals. If given circuit (a), but circuit (b) is more convenient, switch them. This process is called source transformation. Engr228 - Chapter 4, Nilsson 10E 28

29 Example: Source Transformation We can find the current I in the circuit below using a source transformation, as shown. I = (45-3)/( ) = ma I = ma Textbook Problem 5.6 Hayt 8E (a) Determine the individual contributions of each of the two current sources to the nodal voltage v 1 (b) Determine the power dissipated by the resistor v 17A = 6.462V, v 14A = V, v 1tot = 4.31V, P = 3.41W Engr228 - Chapter 4, Nilsson 10E 29

30 Textbook Problem 5.17 Hayt 8E Determine the current labeled i after first transforming the circuit such that it contains only resistors and voltage sources. i = -577mA Textbook Problem 5.19 Hayt 8E Find the power generated by the 7V source. P 7v = 17.27W (generating) Engr228 - Chapter 4, Nilsson 10E 30

31 Thévenin Equivalent Circuits Thévenin s theorem: a linear network can be replaced by its Thévenin equivalent circuit, as shown below: Thévenin Equivalent using Source Transformations We can repeatedly apply source transformations on network A to find its Thévenin equivalent circuit. This method has limitations due to circuit topology, not all circuits can be source transformed. Engr228 - Chapter 4, Nilsson 10E 31

32 Finding the Thévenin Equivalent Disconnect the load; Find the open circuit voltage v oc ; Find the equivalent resistance R eq of the network with all independent sources turned off. Set voltage sources to zero volts short circuit Set current sources to zero amps open circuit Then: V TH = v oc and R TH = R eq Thévenin Example Engr228 - Chapter 4, Nilsson 10E 32

33 Example Find Thévenin s equivalent circuit and the current passing thru RL given that RL = 1Ω 10Ω 10V 3Ω RL Example - continued Find V TH 10V 6V 10Ω 6V 10V 3Ω 0V 0V 0V 3 V TH = 10 = 6V Engr228 - Chapter 4, Nilsson 10E 33

34 Example - continued 10Ω Find R TH 10V 3Ω Short voltage source 10Ω 3Ω R TH 2 3 = = 13. R TH = Example - continued Thévenin s equivalent circuit 13. 6V RL The current thru RL = 1Ω is = A Engr228 - Chapter 4, Nilsson 10E 34

35 Example: Bridge Circuit Find Thévenin s equivalent circuit as seen by RL R1=2K R3=4K 10V RL=1K + - R2=8K R4=1K Example - continued Find V TH 10V R1=2K R3=4K 10V 8V 2V R2=8K R4=1K 0V V TH = 8-2 = 6V Engr228 - Chapter 4, Nilsson 10E 35

36 Example - continued Find R TH R1=2K R3=4K R TH R2=8K R4=1K R1=2K R3=4K R1=2K R3=4K R2=8K R4=1K R2=8K R4=1K Example - continued R1=2K R3=4K R2=8K R4=1K R TH = 2K 8K + 4K 1K = 1.6K + 0.8K = 2. 4K Engr228 - Chapter 4, Nilsson 10E 36

37 Example - continued Thévenin s equivalent circuit 2.4K 6V RL Norton Equivalent Circuits Norton s theorem: a linear network can be replaced by its Norton equivalent circuit, as shown below: Engr228 - Chapter 4, Nilsson 10E 37

38 Finding the Norton Equivalent Replace the load with a short circuit; Find the short circuit current i sc ; Find the equivalent resistance R eq of the network with all independent sources turned off (same as Thévenin) Set voltage sources to zero volts short circuit; Set current sources to zero amps open circuit. Then: I N = i sc and R N = R eq Source Transformation: Norton and Thévenin The Thévenin and Norton equivalents are source transformations of each other. R TH =R N =R eq and v TH =i N R eq Engr228 - Chapter 4, Nilsson 10E 38

39 Example: Norton and Thévenin Find the Thévenin and Norton equivalents for the network faced by the 1-kΩ resistor. The load resistor This is the circuit we will simplify Example: Norton and Thévenin Thévenin Norton Source Transformation Engr228 - Chapter 4, Nilsson 10E 39

40 Thévenin Example: Handling Dependent Sources The normal technique for finding Thévenin or Norton equivalent circuits can not usually be used if a dependent source is present. In this case, we can find both V TH and I N and solve for R TH =V TH / I N Thévenin Example: Handling Dependent Sources Another situation that rarely arises, is if both V TH and I N are zero, or just I N is zero. In this situation, we can apply a test source to the output of the network and measure the resulting short-circuit (I N ) current, or open-circuit voltage (V TH ). R TH is then calculated as V TH /I N Engr228 - Chapter 4, Nilsson 10E 40

41 Thévenin Example: Handling Dependent Sources Solve: v test =0.6 V, so R TH = 0.6 Ω v test 2 + v test (1.5i) 3 i = 1 =1 Recap: Thévenin and Norton Thévenin s equivalent circuit Norton s equivalent circuit V RL RL R V TH TH = R = I N N R TH Same R value 6 = Engr228 - Chapter 4, Nilsson 10E 41

42 Textbook Problem 5.50 Hayt 7E Find the Thévenin equivalent of the circuit below. R TH = kω V TH = 83.5 V Maximum Power Transfer Thévenin s or Norton s equivalent circuit, which has an R TH connected to it, delivers a maximum power to the load R L for which R TH = R L Engr228 - Chapter 4, Nilsson 10E 42

43 Maximum Power Theorem Proof R TH V TH RL Plug it in P = I 2 R L VTH P = RTH + R L 2 VTH and I = RTH + R 2 VTH RL RL = 2 ( R + RL ) TH L dp dr L ( R = TH + R L 2 2 ) VTH VTH RL 4 ( R + R ) TH 2 L 2( R TH + R L ) = 0 Maximum Power Theorem Proof - continued dp dr L ( R = TH + R L 2 2 ) VTH VTH RL 4 ( R + R ) TH 2 L 2( R TH + R L ) = 0 ( R TH + R ) ( R L TH 2 V 2 TH R L TH = V = R 2 TH + R ) = 2R L L R L 2( R TH + R ) L For maximum power transfer Engr228 - Chapter 4, Nilsson 10E 43

44 Example Evaluate RL for maximum power transfer and find the power. 10Ω 10V 3Ω RL Example - continued Thévenin s equivalent circuit V RL RL should be set to 13. to get maximum power transfer. Max. power is 2 V R = 2 (6/ 2) 13.2 = 0.68W Engr228 - Chapter 4, Nilsson 10E 44

45 Practical Voltage Sources Ideal voltage sources: a first approximation model for a battery. Why do real batteries have a current limit and experience voltage drop as current increases? Two car battery models: Practical Source: Effect of Connecting a Load For the car battery example: V L = I L This line represents all possible R L Engr228 - Chapter 4, Nilsson 10E 45

46 Chapter 4 Summary Illustrated the node-voltage method to solve a circuit; Illustrated the mesh-current method to solve a circuit; Practiced choosing which technique is better for a particular circuit; Explained source transformations and how to use them to simplify a circuit; Illustrated the techniques of constructing Thevenin and Norton equivalent circuits; Explained the principle of maximum power transfer to a resistive load and showed how to calculate the value of the load resistor that satisfies this condition. Engr228 - Chapter 4, Nilsson 10E 46