CHAPTER 4. Circuit Theorems

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1 CHAPTER 4 Circuit Theorems The growth in areas of application of electrical circuits has led to an evolution from simple to complex circuits. To handle such complexity, engineers over the years have developed theorems to simplify circuit analysis. These theorems (Thevenin s and Norton s theorems) are applicable to linear circuits which are composed of resistors, voltage and current sources Linearity Property Definition A linear circuit is a circuit whose output is linearly related (or directly proportional) to its input 1. The input and output can be any voltage or current in the circuit. When we says that the input and output are linearly related, we mean they need to satisfies two properties: (a) Homogeneous (Scaling): If the input is multiplied by a constant k, then we should observed that the output is also multiplied by k. (b) Additive: If the inputs are summed then the output are summed. Example The linear circuit below is excited by a voltage source v s, which serves as the input. The circuit is terminated by a load R. We take the current i through R as the output. Suppose v s = 10 V gives i = 2 A. According to the linearity 1 The input and output are sometimes referred to as cause and effect, respectively. 37

2 38 4. CIRCUIT THEOREMS principle, v s = 1V will give i = 0.2 A. By the same token, i = 1 ma must be due to v s = 5 mv. Example A resistor is a linear element when we consider the current i as its input and the voltage v as its output because it has the following properties: Homogeneous (Scaling): If i is multiplied by a constant k, then the output v is multiplied by k. ir = v (ki)r = kv Additive: If the inputs, i 1 and i 2, are summed then the corresponding output are summed. i 1 R = v 1, i 2 R = v 2 (i 1 + i 2 )R = v 1 + v 2 Example For the circuit below, find v o when (a) i s = 15 and (b) i s = 30. Remark: Because p = i 2 R = v 2 /R (making it a quadratic function rather than a linear one), the relationship between power and voltage (or current) is nonlinear. Therefore, the theorems covered in this chapter are not applicable to power.

3 4.2. SUPERPOSITION Superposition Superposition technique = A way to determine currents and voltages in a circuit that has multiple independent sources by considering the contribution of one source at a time and then add them up. Example Consider the following circuit. The superposition principle states that the voltage across (or current through) an element in a linear circuit is the algebraic sum of the voltages across (or currents through) that element due to each independent source acting alone. However, to apply the superposition principle, we must keep two things in mind. 1. We consider one independent source at a time while all other independent source are turned off. 2 Replace other independent voltage sources by 0 V (or short circuits) Replace other independent current sources by 0 A (or open circuits) This way we obtain a simpler and more manageable circuit. 2. Dependent sources are left intact because they are controlled by circuit variable. 2 Other terms such as killed, made inactive, deadened, or set equal to zero are often used to convey the same idea.

4 40 4. CIRCUIT THEOREMS Steps to Apply Superposition Principles: S1: Turn off all independent sources except one source. Find the output due to that active source. S2: Repeat S1 for each of the other independent sources. S3: Find the total contribution by adding algebraically all the contributions due to the independent sources. Example Using superposition theorem, find v o in the following circuit.

5 4.3. SOURCE TRANSFORMATION Keep in mind that superposition is based on linearity. Hence, we cannot find the total power from the power due to each source, because the power absorbed by a resistor depends on the square of the voltage or current and hence it is not linear (e.g. because ) Superposition helps reduce a complex circuit to simpler circuits through replacement of voltage sources by short circuits and of current sources by open circuits. However, it may very likely involve more work. For example, if the circuit has three independent sources, we may have to analyze three circuits. The advantage is that each of the three circuits is considerably easier to analyze than the original one Source Transformation We have noticed that series-parallel resistance combination helps simplify circuits. The simplification is done by replacing one part of a circuit by its equivalence. 3 Source transformation is another tool for simplifying circuits A source transformation is the process of replacing a voltage source in series with a resistor R by a current source in parallel with a resistor R or vice versa. Notice that when terminals a b are short-circuited, the short-circuit current flowing from a to b is i sc = v s /R in the circuit on the left-hand side and i sc = i s for the circuit on the righthand side. Thus, v s /R = i s in order for the two circuits to be equivalent. Hence, source transformation requires that (4.1) v s = i s R or i s = v s R. circuit. 3 Recall that an equivalent circuit is one whose v i characteristics are identical with the original

6 42 4. CIRCUIT THEOREMS Example Use source transformation to find v 0 in the following circuit: Remark: From (4.1), an ideal voltage source with R = 0 cannot be replaced by a finite current source. Similarly, an ideal current source with R = cannot be replaced by a finite voltage source.

7 4.4. THEVENIN S THEOREM Thevenin s Theorem It often occurs in practice that a particular element in a circuit is variable (usually called the load) while other elements are fixed. As a typical example, a household outlet terminal may be connected to different appliances constituting a variable load. Each time the variable element is changed, the entire circuit has to be analyzed all over again. To avoid this problem, Thevenins theorem provides a technique by which the fixed part of the circuit is replaced by an equivalent circuit. Thevenin s Theorem is an important method to simplify a complicated circuit to a very simple circuit. It states that a circuit can be replaced by an equivalent circuit consisting of an independent voltage source V T h in series with a resistor R T h, where V T h : the open circuit voltage at the terminal. R T h : the equivalent resistance at the terminals when the independent sources are turned off.

8 44 4. CIRCUIT THEOREMS Steps to Apply Thevenin s theorem. (Case I: No dependent source) S1: Find R T h : Turn off all independent sources. R T H is the input resistance of the network looking between terminals a and b. S2: Find V T h : Open the two terminals (remove the load) which you want to find the Thevenin equivalent circuit. V T h is the open-circuit voltage across the terminals. S3: Connect V T h and R T h in series to produce the Thevenin equivalent circuit for the original circuit.

9 4.4. THEVENIN S THEOREM 45 Example Find the Thevenin equivalent circuit of the circuit shown below, to the left of the terminals a-b. Then find the current through R L = 6, 16, and 36Ω. Solution:

10 We can find v THA by measuring the open-circuit voltage at the aa port in the network in Figure 3.68b. We find by inspection that v THA = 1V 46 Notice that the 1-A current flows 4. through CIRCUIT each THEOREMS of the 1- resistors in the loop containing the current source, and so v 1 is 1 V. Since there is no current in the resistor connected to the a terminal, the voltage v 2 across that resistor is 0. Thus v THA = v 1 + v 2 = 1V. Example We find R THA by measuring Determine the resistance the current looking into I inthe the aa branch port in theab network in the in circuit Figure 3.68c. The current source has been turned into an open circuit for purpose below. 1 A 1 A a b I a b FIGURE 3.67 Determining current in the branch ab. Network A Network B Solution: There are many approaches that we can take to obtain the current I. For example, we could apply the node method and determine the node voltages at nodes a and b and thereby determine the current I. However, we will find the Thvenin equivalent network for the subcircuit to the left of the aa terminal pair (Network A) and for the subcircuit to the right of the bb terminal pair (Network B), and then using these subcircuits solve 164 CHAPTER THREE network theorems for the current I. 1 A FIGURE 3.68 Finding the Thévenin equivalent for Network A. 1 A a a v THA (b) v + 1 a + - 1Ω 1Ω v a THA - + v 2 - (a) R THA (c) a a R THA 1 A FIGURE 3.69 Finding the Thévenin equivalent for Network B. b b 1 A (a) v THB R THB (b) + v b THB b - (c) R THB b b of measuring R THA. By inspection, we find that

11 4.4. THEVENIN S THEOREM Steps to Apply Thevenin s theorem.(case II: with dependent sources) S1: Find R T H : S1.1 Turn off all independent sources (but leave the dependent sources on). S1.2 Apply a voltage source v o at terminals a and b, determine the resulting current i o, then R T H = v o i o Note that: We usually set v o = 1 V. Or, equivalently, S1.2 Apply a current source i o at terminal a and b, find v o, then R T H = v o i o S2: Find V T H, as the open-circuit voltage across the terminals. S3: Connect R T H and V T H in series. Remark: It often occurs that R T H takes a negative value. In this case, the negative resistance implies that the circuit is supplying power. This is possible in a circuit with dependent sources.

12 48 4. CIRCUIT THEOREMS 4.5. Norton s Theorem Norton s Theorem gives an alternative equivalent circuit to Thevenin s Theorem. Norton s Theorem: A circuit can be replaced by an equivalent circuit consisting of a current source I N in parallel with a resistor R N, where I N is the short-circuit current through the terminals and R N is the input or equivalent resistance at the terminals when the independent sources are turned off. Note: R N = R T H and I N = V T H R T H. These relations are easily seen via source transformation. 4 4 For this reason, source transformation is often called Thevenin-Norton transformation.

13 4.5. NORTON S THEOREM 49 Steps to Apply Norton s Theorem S1: Find R N (in the same way we find R T H ). S2: Find I N : Short circuit terminals a to b. I N is the current passing through a and b. S3: Connect I N and R N in parallel. Example Find the Norton equivalent circuit of the circuit shown below, to the left of the terminals a-b.

14 50 4. CIRCUIT THEOREMS Example Find the Norton equivalent circuit of the circuit in the following figure at terminals a-b.

15 4.6. MAXIMUM POWER TRANSFER Maximum Power Transfer In many practical situations, a circuit is designed to provide power to a load. In areas such as communications, it is desirable to maximize the power delivered to a load. We now address the problem of delivering the maximum power to a load when given a system with known internal losses. The Thevenin and Norton models imply that some of the power generated by the source will necessarily be dissipated by the internal circuits within the source. Questions: (a) How much power can be transferred to the load under the most ideal conditions? (b) What is the value of the load resistance that will absorb the maximum power from the source? If the entire circuit is replaced by its Thevenin equivalent except for the load, as shown below, the power delivered to the load resistor R L is p = i 2 R L where i = V th R th + R L The derivative of p with respect to R L is given by dp = 2i di R L + i 2 dr L dr L ( ) ( V th V th Vth = 2 R th + R L (R th + R L ) 2 + R th + R L ( ) 2 ( Vth = 2R ) L + 1. R th + R L R th + R L ) 2

16 52 4. CIRCUIT THEOREMS We then set this derivative equal to zero and get R L = R T H The maximum power transfer takes place when the load resistance R L equals the Thevenin resistance R T h. As a consequence, the maximum power transferred to R L equals to ( ) 2 V th p max = R th = V th 2. R th + R th 4R th Example Find the value of R L for maximum power transfer in the circuit below. Find the maximum power.

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