Chapter 4 Circuit Theorems


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1 Chapter 4 Circuit Theorems 1. Linearity and Proportionality. Source Transformation 3. Superposition Theorem 4. Thevenin s Theorem and Norton s Theorem 5. Maximum Power Transfer Theorem Mazita Sem BEL10103
2 Learning Outcomes... At the end of this topic, students should be able to: Simplify the circuit s complexity by using TheveninNorton equivalent networks and/or source transformation Analyse the circuits by using the superposition theorem, etc.
3 Linearity and Proportionality Motivation: Engineering combines the study of mathematics and natural and social sciences to direct the forces of nature for the benefit of humankind. An engineer, in the accomplishment of a task, 1. Analyses the problem. Synthesises a solution 3. Evaluates the results and possibly resynthesises a solution use models to represent the electric circuit element Source: Richard C. Dorf and James A. Svoboda
4 Linearity and Proportionality (cont.) A device or element is said to be linear if its excitation and response satisfy the properties of superposition and homogeneity. Mathematically: Superposition: then i 1 + i i i 1 v v v v Homogeneity: then i v ki kv
5 Example Consider the element represented by the relationship between current and voltage as v = Ri. Determine whether this device is linear. Solution: then v 1 + v v v 1 = = = = Ri Ri 1 Ri1 + R i ( + i ) 1 Ri satisfy the superposition property i v v then = = = = ki1 Ri Rki kv 1 1 This device is linear as it satisfies both superposition and homogeneity properties. satisfy the homogeneity property
6 Source Transformation A source transformation is a procedure for transforming one source into another while retaining the terminal characteristics of the original source. Based on the concept of equivalence, i.e. terminal characteristics still remain identical to those of the origin.
7 Source Transformation (cont.) Consider two extreme values of R L R L = 0 Ω R L = Ω Fig. i: Fig. ii: i = v R i = i s s s Fig. i: Fig. ii: v ab = v s v = i For both circuits to be equivalent, v ab must be equal. R S = R p ab s R P
8 Source Transformation (cont.) Thevenin Norton Norton Thevenin
9 Example Find the source transformation for the circuits shown below.
10 Solution For circuit i; R i P S = = R v R S S S = 7 Ω 8 = = 7 4 A
11 Solution (cont.) For circuit ii; S S = RP = 1 Ω = i R = 1 Current A is flowing down reverse the terminal polarity for voltage source R v S P = 4 V
12 Exercise Using source transformation, determine the current i for the circuit shown below. (Answer: 1.15 A)
13 Superposition Theorem Linear element holds superposition theorem, i.e. then i 1 i i + i 1 v v v The superposition principle requires that the total effect of several causes acting simultaneously is equal to the sum of the effects of the individual causes acting one at a time v
14 How to apply the principle of Superposition Theorem? The principle of superposition is only applicable for linear circuits consisting of linear elements and independent sources. Steps in applying the superposition theorem: 1. Activate only one independent source at one time and deactivate the rest of independent sources. If the dependent source is available, it should remain active.. Determine the current or voltage where necessary. 3. Repeat steps 1 until the effects of all the independent sources in the network have been analysed. 4. Add the total currents or voltages.
15 Example 1 Find the current in the 6 Ω resistor using the principle of superposition for the circuit shown below.
16 Solution Consider the effect of 6 V voltage source: i v = = = R T 6 3 A Note: 1. Set the current source to zero appears as an open circuit.. Label portion of current due to excitation by 6V source as i Do circuit analysis.
17 Solution (cont.) Consider the effect of A current source: Apply current divider: R3Ω 3 i = A = = R + R Ω 6Ω 3 A Note: 1. Set the voltage source to zero appears as a short circuit.. Label portion of current due to excitation by A source as i. 3. Do circuit analysis.
18 Solution (cont.) The total current, i = i 1 + i : i = i1 + i = = A 3 Note: 1. Check whether all the independent sources have been analysed.. If yes, total up the current value.
19 Example Find the current i using the principle of superposition for the circuit shown below.
20 Solution Consider the effect of 4 V voltage source: Apply KVL (follow the direction of current, i 1 ) 4 + 3i 1 + i 1 + 3i1 8i i 1 1 = 0 = 4 = 3 A Note: 1. Set the current source to zero appears as an open circuit. A dependent source should remain active.. Label current as i 1. Dependent source is now referred to i Do circuit analysis.
21 Solution (cont.) Consider the effect of 7 A current source: Apply KCL at node a; va 3i = i + 7 va 3i = i + 14 v = 5i + 14Lequation (1) a Apply Ohm s Law across 3 Ω resistor; v a = i 3 Lequation () Note: 1. Set the voltage source to zero appears as a short circuit. A dependent source should remain active.. Label current as i. Dependent source is now referred to i. 3. Do circuit analysis.
22 Solution (cont.) Solve equations (1) & () to get i ; i Finally, the total current, i: i 14 = 8 = = i1 + i 7 = = A A Note: 1. Check whether all the independent sources have been analysed.. If yes, total up the current value.
23 Exercise Using the principle of superposition, find the voltage v of the circuit shown below. (Answer: 4 V)
24 Thevénin s Theorem Motivation: To reduce the complexity of circuits. How? Reduce some portion of the circuit to an equivalent source and a single element Thevenin equivalent circuit
25 Thevénin s Theorem (cont.) Thevenin s theorem requires that, for any circuit of resistance elements and energy sources with an identified terminal pair, the circuit can be replaced by a series combination of an ideal source, v TH and a resistance, R TH. v TH is the opencircuit voltage at the two terminals R TH is the input/equivalent resistance at the terminals when the independent sources are turned off. Source: Richard C. Dorf & James A. Svoboda and Charles K. Alexander & Matthew N.O. Sadiku
26 Thevénin s Theorem (cont.) Summary of Thevenin Circuit Approach 1) Identify circuit A and circuit B ) Separate circuit A and circuit B 3) Replace circuit A with its Thevenin equivalent 4) Reconnect circuit B and determine the variable of interest
27 Example 1 Find the Thevenin equivalent circuit between the output terminals A and B of the circuit shown below.
28 Solution Determine the voltage, V TH Note: 1. Since there is no current flowing through R 4, therefore no voltage drop across it. ( Ignore R 4 ) V TH = = V 10. Use voltage divider rule to determine the V TH = V AB.
29 Solution (cont.) Determine the resistance, R TH Note: 1. Turn off the independent source, i.e. replace the voltage source with short circuit. R TH ( R //( R + )) = R4 + 1 R3 1 1 = = 1.41kΩ 0 1. Determine the total equivalent resistance across terminals AB.
30 Solution (cont.) The Thevenin equivalent circuit is
31 Exercise 1 For the circuit shown below, determine the Thevenin equivalent circuit as viewed from the output terminals A and B. (Answer: R TH = kω, V TH = 3.06 V)
32 Exercise Find the Thevenin equivalent of the circuit shown below. (Answer: R TH = 8 Ω, V TH = 3 V)
33 Thevenin Theorem (cont.) Please take note that: Thevenin Equivalency Depends on the Viewpoint
34 Norton s Theorem Motivation: To reduce the complexity of circuits. How? Reduce some portion of the circuit to an equivalent source and a single element Norton equivalent circuit
35 Norton s Theorem Norton and Thevenin equivalent circuits are related by a source transformation. To determine Thevenin or Norton equivalent circuit, we need to find i N, Norton current equals to the shortcircuit current at the terminals of interest R N, Norton resistance equals to the Thevenin resistance, i.e. resistance at terminals of interest when all the independent sources are off. v TH, Thevenin voltage equals to the opencircuit voltage across the terminals of interest.
36 Example Determine the Norton equivalent circuit as seen by the R L for the circuit shown below.
37 Solution Determine the current, I N Note: 1. Short circuit terminals AB.. To determine the I N, use current divider rule. R T I T = 47 + = = = 1.05 A Ω 47 Therefore, I N I = A = T
38 Solution (cont.) Determine the resistance, R N Note: 1. Turn off the independent source, i.e. replace the voltage source with short circuit. R N ( R ) = R3 + 1 // R 47 = = 13.5 Ω. Determine the total equivalent resistance across terminals AB.
39 Solution (cont.) The Norton equivalent circuit is
40 Exercise Find the Norton equivalent of the circuit shown below. (Answer: R N = 8 Ω, I N = 4 A)
41 Maximum Power Transfer Many applications of circuits require maximum power available from a source be transferred to a load resistor, R L. General problem of power transfer deals with its efficiency and effectiveness issues. Example: power utility systems to transport the power to the load with greatest efficiency by reducing the losses on the power lines Signal transmission (communication): difficulty in attaining the maximum signal strength at the load (e.g. FM radio, mobile telephone).
42 Maximum Power Transfer (cont.) It is known that a complex circuitry could be reduced to its Thevenin equivalent: Current, I is given by I = R L V + TH R TH Hence, the power to the load is P = R L V + TH R TH R L
43 Maximum Power Transfer (cont.) To obtain maximum power transfer, dp R L + RTH RL RL + R = VTH 4 drl ( RL + RTH ) Solving this: ( ) ( ) R L = R TH TH = 0 Note: Use differentiation : d dx u v = du v dx u v dv dx Power delivered to the load varies with the changes of R L values. To confirm the point is maximum: d dr L P < 0
44 Maximum Power Transfer (cont.) Hence Maximum power transfer theorem states that the maximum power delivered by a source represented by its Thevenin equivalent circuit is attained when the load R L is equal to the Thevenin resistance R TH. P V TH max = RL = R L TH V 4R L
45 Maximum Power Transfer (cont.) The efficiency of power transfer is defined as the ratio of the power delivered to the load P OUT to the power supplied by the source P IN. We know that Efficiency is η = P P IN = V OUTMAX P IN TH = I η = = V VTH 4R VTH R OUT Therefore, the efficiency obtained at maximum power condition is only ½ i.e. 50%. L TH L P P IN VTH R = 1 L V = R TH L
46 Example Find the load R L that will result in maximum power delivered to the load for the circuit shown below. Also determine the P max and its efficiency.
47 Solution First, determine the Thevenin equivalent circuit. Determine V TH Determine R TH VTH 150 = 180 = 150V R TH = = 5Ω Therefore, maximum power transfer is obtained when R L = R TH = 5 Ω.
48 Solution (cont.) Then, determine maximum power transfer and power transfer efficiency at R L = 5 Ω. P max = = VTH = 4R L W Total power delivered by the Thevenin source is P IN 150 = 150 = = 450 W The power transferred efficiency is P 5 η % = max 100 = 100 = 50% 450 P IN Note: The actual source of the circuit is 180 V and it delivers a power p = 180i 1 where i 1 is the current through 5 Ω, i.e. 3.5 A. Actual source delivers 630 W resulting in an efficiency of 35.7 %.
49 Exercise Find the load R L that will result in maximum power delivered to the load of the circuit shown below. Determine the maximum power delivered to the load. (Answer: R L = 1 Ω, P max = 3 W)
50 References Alexander Sadiku, Fundamentals of Electric Circuits, 4 th edition, McGrawHill, 009 Richard C. Dorf and James A. Svoboda, Introduction to Electric Circuits, 3 rd edition, John Wiley, 1996 Thomas L. Floyd and David M. Buchla, Electric Circuits Fundamentals, 8 th edition, Pearson, 010 James W. Nilsson & Susan A. Riedel, Electric Circuits, 9 th edition, PearsonPrentice Hall, 011
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