Chapter 5 Objectives


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1 Chapter 5 Engr228 Circuit Analysis Dr Curtis Nelson Chapter 5 Objectives State and apply the property of linearity State and apply the property of superposition Investigate source transformations Define and construct Thevenin and Norton equivalent circuits Investigate maximum power transfer to a load Engr228  Chapter 5, Hayt 8E 1
2 inear Elements and Circuits A linear circuit element has a linear voltagecurrent 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 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. Engr228  Chapter 5, Hayt 8E 2
3 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). Applying Superposition eave 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 Engr228  Chapter 5, Hayt 8E 3
4 Superposition Example (Part 1 of 4) Use superposition to solve for the current i x Superposition Example (Part 2 of 4) First, turn the current source off: i x = = 0.2 Engr228  Chapter 5, Hayt 8E 4
5 Superposition Example (Part 3 of 4) Then, turn the voltage source off: i x = 6 (2) = Superposition Example (Part 4 of 4) Finally, combine the results: i x = i x + i x = =1.0 Engr228  Chapter 5, Hayt 8E 5
6 Example: Power Ratings Determine the maximum positive current to which the source I x can be set before any resistor exceeds its power rating. Answer: I x < ma Superposition with a Dependent Source When applying superposition to circuits with dependent sources, these dependent sources are never turned off. Engr228  Chapter 5, Hayt 8E 6
7 Superposition with a Dependent Source Current source off Voltage source off i x = i x +i x =2 + ( 0.6) = 1.4 A Superposition Example Find voltage Vx. First, solve by nodal analysis. (42Vx)/6 = (Vx(10))/4 + Vx/3 Vx = 54/9 = 6 volts Engr228  Chapter 5, Hayt 8E 7
8 Example  Continued Vx (42 V ) = 9.333V (3 4) = 42 = 6 + (3 4) (12 / 7) (12 / 7) Example  Continued Vx (10 V ) = 3.333V (6 3) 2 = 10 = 10 (6 3) Engr228  Chapter 5, Hayt 8E 8
9 Example  Continued Vx = Vx (42V ) + Vx (10V ) = = 6V 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: Engr228  Chapter 5, Hayt 8E 9
10 Practical Source: Effect of Connecting a oad For the car battery example: V = I This line represents all possible R Practical Voltage Source v = v s R i s i OR vs = R s v R s v i oc sc = v = s v R s s Engr228  Chapter 5, Hayt 8E 10
11 Engr228  Chapter 5, Hayt 8E 11 Practical Current Source s sc s p oc i i i R v = = p s R v i i = Equivalent Practical Sources p s R v i i = s s s R v R v i = These two practical sources are equivalent if R S = R P S S S R v i =
12 Source Transformation and Equivalent Sources The sources are equivalent if R s =R p and v s =i s R s 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 5, Hayt 8E 12
13 Example: Source Transformation We can find the current I in the circuit below using a source transformation, as shown. I = (453)/( ) = ma I = ma Example Use source transformations to find Ix Engr228  Chapter 5, Hayt 8E 13
14 Example  continued Example  continued Engr228  Chapter 5, Hayt 8E 14
15 Example  continued I X = = = A Textbook Problem 5.24 Hayt 7E Using source transformations, determine the power dissipated by the 5.8 kω resistor mw Engr228  Chapter 5, Hayt 8E 15
16 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 2Ω resistor v 17A = 6.462V, v 14A = V, v 1tot = 4.31V, P 2Ω = 3.41W 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 = 577µA Engr228  Chapter 5, Hayt 8E 16
17 Textbook Problem 5.18 Hayt 8E Using source transformations, reduce the circuit to a voltage source in series with a resistor, both of which are in series with the 6 MΩ resistor. V s = V R s = kω Textbook Problem 5.19 Hayt 8E Find the power generated by the 7V source. P 7v = 17.27W (generating) Engr228  Chapter 5, Hayt 8E 17
18 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  not all circuits can be source transformed. Engr228  Chapter 5, Hayt 8E 18
19 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 Then: V = v oc and R = R eq Thévenin Example Engr228  Chapter 5, Hayt 8E 19
20 Example Find Thevenin s equivalent circuit and the current passing thru R given that R = 1Ω Example  continued Find V 10V 6V 6V 0V 0V 0V 3 V = 10 = 6V Engr228  Chapter 5, Hayt 8E 20
21 Example  continued Find R Short voltage source R = = = 13.2Ω + 2 R Example  continued Thevenin s equivalent circuit The current thru R = 1Ω is = A Engr228  Chapter 5, Hayt 8E 21
22 Example Find Thevenin s equivalent circuit Example  continued Find V 5V 3V 3V 0V 0V 0V V = 1 3 = 3V Engr228  Chapter 5, Hayt 8E 22
23 Example  continued Find R Open circuit current source R = 15Ω = R Example  continued Thevenin s equivalent circuit Engr228  Chapter 5, Hayt 8E 23
24 Example: Bridge Circuit Find Thevenin s equivalent circuit as seen by R Example  continued Find V 10V 8V 2V 0V V = 82 = 6V Engr228  Chapter 5, Hayt 8E 24
25 Example  continued Find R R Example  continued R = 2K 8K + 4K 1K = 1.6K + 0.8K = 2. 4K Engr228  Chapter 5, Hayt 8E 25
26 Example  continued Thevenin s equivalent circuit Thevenin s Equivalent Circuit  Recap Engr228  Chapter 5, Hayt 8E 26
27 Norton Equivalent Circuits Norton s theorem: a linear network can be replaced by its Norton equivalent circuit, as shown below: 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 Then: I N = i sc and R N = R eq Engr228  Chapter 5, Hayt 8E 27
28 Source Transformation: Norton and Thévenin The Thévenin and Norton equivalents are source transformations of each other! R =R N =R eq and v =i N R eq Example: Norton and Thévenin Find the Thévenin and Norton equivalents for the network faced by the 1kΩ resistor. The load resistor This is the circuit we will simplify Engr228  Chapter 5, Hayt 8E 28
29 Example: Norton and Thévenin Thévenin Norton Source Transformation Thévenin Example: Handling Dependent Sources One method to find the Thévenin equivalent of a circuit with a dependent source: find V and I N and solve for R =V / I N Example (Textbook p. 148): Engr228  Chapter 5, Hayt 8E 29
30 Thévenin Example: Handling Dependent Sources Finding the ratio V / I N fails when both quantities are zero! Solution: apply a test source (Textbook p. 149) Thévenin Example: Handling Dependent Sources Solve: v test =0.6 V, and so R = 0.6 Ω v test 2 + v test (1.5i) 3 i = 1 =1 Engr228  Chapter 5, Hayt 8E 30
31 Example Find Norton s equivalent circuit and the current through R if R = 1Ω Example  continued Find I N Find R total: Find I total: Current divider: (10 + 2) = 2 + = 4. 4Ω I V 10 = = = 2. A R I SC = 2.27 = 0. 45A Engr228  Chapter 5, Hayt 8E 31
32 Example  continued Find Rn R = = = 13.2Ω + 2 Norton s Equivalent Circuit The current thru R = 1Ω is = A Engr228  Chapter 5, Hayt 8E 32
33 Relationship Between Thevenin and Norton I R V = R = I N N R I SC Slope =  1/R V OC V Recap: Thevenin and Norton Thevenin s equivalent circuit Norton s equivalent circuit R V = R = I N N R Same R value 6 = Engr228  Chapter 5, Hayt 8E 33
34 Relationship Between Thevenin and Norton Norton s equivalent circuit Thevenin s equivalent circuit 0.2 x 15 = 3 Textbook Problem 5.50 Hayt 7E Find the Thevenin equivalent of the circuit below. R = kω V = 83.5 V Engr228  Chapter 5, Hayt 8E 34
35 Equivalent Circuits with Dependent Sources We cannot find R in circuits with dependent sources using the total resistance method. But we can use R = V I OC SC Example Find Thevenin and Norton s equivalent circuits Engr228  Chapter 5, Hayt 8E 35
36 Example  continued Find Voc I1 I2 KV loop1 KV loop I ( I1 I 2) = I1 4000I 2 = ( I 2 I1) I I 2 100Vx = 0 Vx = 4000( I1 I 2) I I 2 = 0 Example  continued I1 I2 Solve equations I1 = 3.697mA I2 = 3.678mA V OC = 80I 2 100Vx = 80I ( I1 I 2) = 80(3.678mA) ( ) = 7.3V Engr228  Chapter 5, Hayt 8E 36
37 Example  continued Find Isc I1 I2 I3 KV loop1 KV loop I ( I1 I 2) = I I 2 = ( I 2 I1) I ( I 2 I 3) 100Vx Vx = 4000( I1 I 2) I I 2 80I 3 = 0 = 0 KV 80( I3 I 2) + 100Vx = 0 loop I I I 3 = 0 Example  continued Find Isc I1 I2 I3 I1 = 0.632mA I2 = 0.421mA I3 = A Isc = I3 = A Engr228  Chapter 5, Hayt 8E 37
38 Example  continued R VOC 7.28 = = = 6. 94Ω I SC Thevenin s equivalent circuit Norton s equivalent circuit Maximum Power Transfer Thevenin s or Norton s equivalent circuit, which has an R connected to it, delivers a maximum power to the load R for which R = R Engr228  Chapter 5, Hayt 8E 38
39 Maximum Power Theorem Proof Plug it in P = I 2 R V P = R + R 2 V and I = R + R 2 V R R = 2 ( R + R ) dp dr ( R = + R 2 2 ) V V R 4 ( R + R ) 2 2( R + R ) = 0 Maximum Power Theorem Proof  continued dp dr ( R = + R 2 2 ) V V R 4 ( R + R ) 2 2( R + R ) = 0 ( R + R ) ( R 2 V 2 R = V = R 2 + R ) = 2R R 2( R + R ) For maximum power transfer Engr228  Chapter 5, Hayt 8E 39
40 Example Evaluate R for maximum power transfer and find the power. Example  continued Thevenin s equivalent circuit R should be set to 13.2Ω to get max. power transfer Max. power is 2 V R = (6 / 2) = 0.68W Engr228  Chapter 5, Hayt 8E 40
41 Chapter 5 Summary Stated and applied the property of linearity Stated and explored the property of superposition Investigated source transformations Defined and constructed the Thevenin and Norton equivalent circuits Investigated maximum power transfer to a load Engr228  Chapter 5, Hayt 8E 41
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