# 3.1 Superposition theorem

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1 Many electric circuits are complex, but it is an engineer s goal to reduce their complexity to analyze them easily. In the previous chapters, we have mastered the ability to solve networks containing independent and dependent sources making use of either mesh or nodal analysis. In this chapter, we will introduce new techniques to strengthen our armoury to solve complicated networks. Also, these new techniques in many cases do provide insight into the circuit s operation that cannot be obtained from mesh or nodal analysis. Most often, we are interested only in the detailed performance of an isolated portion of a complex circuit. If we can model the remainder of the circuit with a simple equivalent network, then our task of analysis gets greatly reduced and simplified. For example, the function of many circuits is to deliver maximum power to load such as an audio speaker in a stereo system. Here, we develop the required relationship betweeen a load resistor and a fixed series resistor which can represent the remaining portion of the circuit. Two of the theorems that we present in this chapter will permit us to do just that. 3.1 Superposition theorem The principle of superposition is applicable only for linear systems. The concept of superposition can be explained mathematically by the following response and excitation principle : i 1! v 1 i 2! v 2 then; i 1 + i 2! v 1 + v 2 The quantity to the left of the arrow indicates the excitation and to the right, the system response. Thus, we can state that a device, if excited by a current i 1 will produce a response v 1. Similarly, an excitation i 2 will cause a response v 2. Then if we use an excitation i 1 + i 2,we will find a response v 1 + v 2. The principle of superposition has the ability to reduce a complicated problem to several easier problems each containing only a single independent source.

2 160 j Network Theory Superposition theorem states that, In any linear circuit containing multiple independent sources, the current or voltage at any point in the network may be calculated as algebraic sum of the individual contributions of each source acting alone. When determining the contribution due to a particular independent source, we disable all the remaining independent sources. That is, all the remaining voltage sources are made zero by replacing them with short circuits, and all remaining current sources are made zero by replacing them with open circuits. Also, it is important to note that if a dependent source is present, it must remain active (unaltered) during the process of superposition. Action Plan: (i) In a circuit comprising of many independent sources, only one source is allowed to be active in the circuit, the rest are deactivated (turned off). (ii) To deactivate a voltage source, replace it with a short circuit, and to deactivate a current source, replace it with an open circuit. (iii) The response obtained by applying each source, one at a time, are then added algebraically to obtain a solution. Limitations: Superposition is a fundamental property of linear equations and, therefore, can be applied to any effect that is linearly related to the cause. That is, we want to point out that, superposition principle applies only to the current and voltage in a linear circuit but it cannot be used to determine power because power is a non-linear function. EXAMPLE 3.1 Find the current in the 6Ωresistor using the principle of superposition for the circuit of Fig Figure 3.1 As a first step, set the current source to zero. That is, the current source appears as an open circuit as shown in Fig i 1 = = 6 9 A

3 Circuit Theorems j 161 As a next step, set the voltage to zero by replacing it with a short circuit as shown in Fig i 2 = = 6 9 A Figure 3.2 Figure 3.3 The total current i is then the sum of i 1 and i 2 i = i 1 + i 2 = 12 9 A EXAMPLE 3.2 Find i o in the network shown in Fig. 3.4 using superposition. Figure 3.4 As a first step, set the current source to zero. That is, the current source appears as an open circuit as shown in Fig Figure 3.5

4 162 j Network Theory 6 i 0 o = = 0:3 ma (8 + 12) 103 As a second step, set the voltage source to zero. This means the voltage source in Fig. 3.4 is replaced by a short circuit as shown in Figs. 3.6 and 3.6(a). Using current division principle, i A = ir 2 R 1 + R 2 where R 1 = (12 kωjj12 kω) + 12 kω =6kΩ+12kΩ =18kΩ and R 2 =12kΩ ) i A = ( ) 10 3 =1:6mA Figure 3.6 Again applying the current division principle, i 00 o = i A 12 =0:8 ma Thus; i o = i 0 o + i 00 o = 0:3+0:8 =0:5 ma Figure 3.6(a)

5 Circuit Theorems j 163 EXAMPLE 3.3 Use superposition to find i o in the circuit shown in Fig Figure 3.7 As a first step, keep only the 12 V source active and rest of the sources are deactivated. That is, 2 ma current source is opened and6vvoltage source is shorted as shown in Fig i 0 o = (2+2)10 3 =3mA Figure 3.8 As a second step, keep only 6 V source active. Deactivate rest of the sources, resulting in a circuit diagram as shown in Fig. 3.9.

6 164 j Network Theory Applying KVL clockwise to the upper loop, we get i 00 o i 00 o 6=0 ) i 00 o = 6 = 1:5 ma Figure 3.9 As a final step, deactivate all the independent voltage sources and keep only 2 ma current source active as shown in Fig Current of 2 ma splits equally. Figure 3.10 Hence; i o 000 = 1mA Applying the superposition principle, we find that i o = i 0 o + i o + i o =3 1:5+1 = 2:5 ma

7 Circuit Theorems j 165 EXAMPLE 3.4 Find the current i for the circuit of Fig Figure 3.11 We need to find the current i due to the two independent sources. As a first step in the analysis, we will find the current resulting from the independent voltage source. The current source is deactivated and we have the circuit as shown as Fig Applying KVL clockwise around loop shown in Fig. 3.12, we find that 5i 1 +3i 1 24=0 ) i 1 = 24 8 =3A As a second step, we set the voltage source to zero and determine the current i 2 due to the current source. For this condition, refer to Fig for analysis. Figure 3.12 Figure 3.13 Applying KCL at node 1, we get i 2 +7= v 1 3i 2 (3.1) 2 Noting that i 2 = v we get, v 1 = 3i 2 (3.2)

8 166 j Network Theory Making use of equation (3.2) in equation (3.1) leads to i 2 +7= 3i 2 3i 2 2 ) i 2 = 7 4 A Thus, the total current i = i 1 + i 2 =3 7 4 A=5 4 A EXAMPLE 3.5 For the circuit shown in Fig. 3.14, find the terminal voltage V ab using superposition principle. As a first step in the analysis, deactivate the independent current source. This results in a circuit diagram as shown in Fig Applying KVL clockwise gives V ab1 + V ab1 =0 ) 4V ab1 =4 Figure 3.14 ) V ab1 =1V Next step in the analysis is to deactivate the independent voltage source, resulting in a circuit diagram as shown in Fig Applying KVL gives Figure V ab2 + V ab2 =0 ) 4V ab2 =20 ) V ab2 =5V Figure 3.16

9 Circuit Theorems j 167 According to superposition principle, V ab = V ab1 + V ab2 =1+5=6V EXAMPLE 3.6 Use the principle of superposition to solve for v x in the circuit of Fig Figure 3.17 According to the principle of superposition, v x = v x1 + v x2 where v x1 is produced by 6A source alone in the circuit and v x2 is produced solely by 4A current source. To find v x1, deactivate the 4A current source. This results in a circuit diagram as shown in Fig KCL at node x 1 : v x1 2 + v x 1 4i x1 =6 8 But i x1 = v x 1 2 v x1 Hence; 2 + v x 1 4 vx1 2 =6 8 ) v x1 2 + v x 1 2v x1 8 =6 ) 4v x1 + v x1 2v x1 =48 ) v x1 = 48 3 = 16V Figure 3.18

10 168 j Network Theory To find v x2, deactivate the 6A current source, resulting in a circuit diagram as shown in Fig KCL at node x 2 : v x2 8 + v x 2 ( 4i x2 ) =4 2 v x2 ) 8 + v x 2 +4i x2 =4 (3.3) 2 Applying KVL along dotted path, weget v x2 +4i x2 2i x2 =0 ) v x2 = 2i x2 or i x2 = v x 2 2 Substituting equation (3.4) in equation (3.3), we get vx2 v x2 +4 v x =4 2 v x2 ) 8 + v x 2 2v x2 =4 2 v x2 ) 8 v x 2 2 =4 ) v x2 4v x2 =32 (3.4) ) v x2 = 32 3 V Hence, according to the superposition principle, v x = v x1 + v x2 = = 5:33V Figure 3.19 EXAMPLE 3.7 Which of the source in Fig contributes most of the power dissipated in the 2 Ω resistor? The least? What is the power dissipated in 2Ωresistor? Figure 3.20

11 Circuit Theorems j 169 The Superposition theorem cannot be used to identify the individual contribution of each source to the power dissipated in the resistor. However, the superposition theorem can be used to find the total power dissipated in the 2Ωresistor. Figure 3.21 According to the superposition principle, i 1 = i i 0 2 where i 0 1 = Contribution to i 1 from 5V source alone. and i 0 2 = Contribution to i 1 from 2A source alone. Let us first find i 0 1. This needs the deactivation of 2A source. Refer to Fig i 0 1 = 2+2:1 =1:22A Similarly to find i 0 2 we have to disable the 5V source by shorting it. Referring to Fig. 3.23, we find that i 0 2 = 2 2:1 2+2:1 = 1:024 A Figure 3.22 Figure 3.23

12 170 j Network Theory Total current, i 1 = i i 0 2 =1:22 1:024 =0:196 A Thus; P 2Ω =(0:196) 2 2 = 0:0768 Watts = 76:8 mw EXAMPLE 3.8 Find the voltage V 1 using the superposition principle. Refer the circuit shown in Fig Figure 3.24 According to the superposition principle, V 1 = V1 0 + V1 00 where V is the contribution from 60V source alone and V1 is the contribution from 4A current source alone. To find V1 0, the 4A current source is opened, resulting in a circuit as shown in Fig Figure 3.25

13 Circuit Theorems j 171 Applying KVL to the left mesh: Also 30i a 60+30(i a i b )=0 (3.5) i b = 0:4i A = 0:4( i a )=0:4i a (3.6) Substituting equation (3.6) in equation (3.5), we get 30i a 60+30i a 30 0:4i a =0 ) i a = =1:25A i b =0:4i a =0:41:25 =0:5A Hence; V1 0 =(i a i b ) 30 =22:5 V To find, V1 00, the 60V source is shorted as shown in Fig Figure 3.26 Applying KCL at node a: V a 20 + V a V1 00 =4 10 ) 30V a 20V1 00 = 800 (3.7) Applying KCL at node b: V V 1 00 V a =0:4i b 10 Also; V a =20i a ) i b = V a 20 V1 00 Hence; 30 + V 1 00 V a = 0:4V a ) 7:2V a +8V1 00 =0 (3.8)

14 172 j Network Theory Solving the equations (3.7) and (3.8), we find that V1 00 = 60V Hence V 1 = V1 0 + V1 00 =22:5+60=82:5V EXAMPLE 3.9 (a) Refer to the circuit shown in Fig Before the 10 ma current source is attached to terminals x y, the current i a is found to be 1.5 ma. Use the superposition theorem to find the value of i a after the current source is connected. (b) Verify your solution by finding i a, when all the three sources are acting simultaneously. Figure 3.27 According to the principle of superposition, i a = i a1 + i a2 + i a3 where i a1, i a2 and i a3 are the contributions to i a from 20V source, 5 ma source and 10 ma source respectively. As per the statement of the problem, i a1 + i a2 =1:5 ma To find i a3, deactivate 20V source and the 5 ma source. The resulting circuit diagram is shown in Fig mA 2k i a3 = 18k+2k =1mA Hence, total current i a = i a1 + i a2 + i a3 =1:5+1=2:5 ma

15 Circuit Theorems j 173 Figure 3.28 (b) Refer to Fig KCL at node y: V y V y = (10+5)10 3 Solving, we get V y = 45V: V y Hence; i a = = =2:5mA 3.2 Thevenin s theorem Figure 3.29 In section 3.1, we saw that the analysis of a circuit may be greatly reduced by the use of superposition principle. The main objective of Thevenin s theorem is to reduce some portion of a circuit to an equivalent source and a single element. This reduced equivalent circuit connected to the remaining part of the circuit will allow us to find the desired current or voltage. Thevenin s theorem is based on circuit equivalence. A circuit equivalent to another circuit exhibits identical characteristics at identical terminals. Figure 3.30 A Linear two terminal network Figure 3.31 The Thevenin s equivalent circuit According to Thevenin s theorem, the linear circuit of Fig can be replaced by the one shown in Fig (The load resistor may be a single resistor or another circuit). The circuit to the left of the terminals x y in Fig is known as the Thevenin s equivalent circuit.

16 174 j Network Theory The Thevenin s theorem may be stated as follows: A linear two terminal circuit can be replaced by an equivalent circuit consisting of a voltage source V t in series with a resistor R t, Where V t is the open circuit voltage at the terminals and R t is the input or equivalent resistance at the terminals when the independent sources are turned off or R t is the ratio of open circuit voltage to the short circuit current at the terminal pair. Action plan for using Thevenin s theorem : 1. Divide the original circuit into circuit A and circuit B. In general, circuit B is the load which may be linear or non-linear. Circuit A is the balance of the original network exclusive of load and must be linear. In general, circuit A may contain independent sources, dependent sources and resistors or other linear elements. 2. Separate the circuit A from circuit B. 3. Replace circuit A with its Thevenin s equivalent. 4. Reconnect circuit B and determine the variable of interest (e.g. current i or voltage v ). Procedure for finding R t : Three different types of circuits may be encountered in determining the resistance, R t : (i) If the circuit contains only independent sources and resistors, deactivate the sources and find R t by circuit reduction technique. Independent current sources, are deactivated by opening them while independent voltage sources are deactivated by shorting them.

17 Circuit Theorems j 175 (ii) If the circuit contains resistors, dependent and independent sources, follow the instructions described below: (a) Determine the open circuit voltage v oc with the sources activated. (b) Find the short circuit current i sc when a short circuit is applied to the terminals a b (c) R t = v oc i sc (iii) If the circuit contains resistors and only dependent sources, then (a) v oc =0(since there is no energy source) (b) Connect 1A current source to terminals a b and determine v ab. (c) R t = v ab 1 Figure 3.32 For all the cases discussed above, the Thevenin s equivalent circuit is as shown in Fig EXAMPLE 3.10 Using the Thevenin s theorem, find the current i through R =2Ω. Refer Fig Figure 3.33 Figure 3.34

18 176 j Network Theory Since we are interested in the current i through R, the resistor R is identified as circuit B and the remainder as circuit A. After removing the circuit B, circuit A is as shown in Fig Figure 3.35 To find R t, we have to deactivate the independent voltage source. Accordingly, we get the circuit in Fig Referring to Fig. 3.35, R t =(5Ωjj20 Ω) + 4 Ω = =8Ω R t Hence 50+25I =0 ) I =2A V ab = V oc = 20(I) = 40V Figure 3.36 Thus, we get the Thevenin s equivalent circuit which is as shown in Fig Figure 3.37 Figure 3.38 Reconnecting the circuit B to the Thevenin s equivalent circuit as shown in Fig. 3.38, we get i = = 4A

19 Circuit Theorems j 177 EXAMPLE 3.11 (a) Find the Thevenin s equivalent circuit with respect to terminals a b for the circuit shown in Fig by finding the open-circuit voltage and the short circuit current. (b) Solve the Thevenin resistance by removing the independent sources. Compare your result with the Thevenin resistance found in part (a). Figure 3.39 Figure 3.40 (a) To find V oc : Apply KCL at node 2 : V V ) V 2 = 60 Volts Hence; V oc = I 60 V2 0 = :5 =0 60 = =45V

20 178 j Network Theory To find i sc : _ Applying KCL at node 2: V V :5 =0 40 ) V 2 = 30V i sc = V 2 20 =1:5A Therefore; R t = V oc = 45 i sc 1:5 =30Ω Figure 3.40 (a) The Thevenin equivalent circuit with respect to the terminals a b is as shown in Fig. 3.40(a). (b) Let us now find Thevenin resistance R t by deactivating all the independent sources, R t Rt R t =60Ωjj( ) Ω = 60 = 30 Ω (verified) 2 It is seen that, if only independent sources are present, it is easy to find R t by deactivating all the independent sources.

21 Circuit Theorems j 179 EXAMPLE 3.12 Find the Thevenin equivalent for the circuit shown in Fig with respect to terminals a b. To find V oc = V ab : Applying KVL around the mesh of Fig. 3.42, we get 20+6i 2i +6i =0 ) i =2A Figure 3.41 Since there is no current flowing in 10 Ω resistor, V oc =6i =12V To find R t : (Refer Fig. 3.43) Since both dependent and independent sources are present, Thevenin resistance is found using the relation, Applying KVL clockwise for mesh 1: R t = v oc i sc Figure i 1 2i +6(i 1 i 2 )=0 ) 12i 1 6i 2 =20+2i Since i = i 1 i 2,weget 12i 1 6i 2 =20+2(i 1 i 2 ) ) 10i 1 4i 2 =20 Applying KVL clockwise for mesh 2: 10i 2 +6(i 2 i 1 )=0 ) 6i 1 +16i 2 =0 Figure 3.43

22 180 j Network Theory Solving the above two mesh equations, we get i 2 = A ) i sc = i 2 = A R t = v oc = 12 =13:6 Ω i sc EXAMPLE 3.13 Find V o in the circuit of Fig using Thevenin s theorem. Figure 3.44 To find V oc : Since we are interested in the voltage across 2 kω resistor, it is removed from the circuit of Fig and so the circuit becomes as shown in Fig Figure 3.45 By inspection, i 1 =4mA Applying KVL to mesh 2: (i 2 i 1 ) i 2 =0 ) i i 2 =0

23 Circuit Theorems j 181 Solving, we get i 2 =4mA Applying KVL to the path 4kΩ! a b! 3kΩ, weget i 1 + V oc i 2 =0 ) V oc =410 3 i i 2 = = 28V To find R t : Deactivating all the independent sources, we get the circuit diagram shown in Fig Figure 3.46 R t = R ab = 4 kω + (6 kωjj3 kω)=6kω Hence, the Thevenin equivalent circuit is as shown in Fig Figure 3.47 Figure 3.48 If we connect the 2kΩresistor to this equivalent network, we obtain the circuit of Fig V o = i = 28 (6+2) =7V EXAMPLE 3.14 The wheatstone bridge in the circuit shown in Fig (a) is balanced when R 2 = 1200 Ω. If the galvanometer has a resistance of 30 Ω, how much current will be detected by it when the bridge is unbalanced by setting R 2 to 1204 Ω?

24 182 j Network Theory Figure 3.49(a) To find V oc : We are interested in the galavanometer current. Hence, it is removed from the circuit of Fig (a) to find V oc and we get the circuit shown in Fig (b). 120 i 1 = = A 120 i 2 = = A Applying KVL clockwise along the path 1204Ω! b a! 900 Ω,weget 1204i 2 V t 900i 1 =0 ) V t = 1204i 2 900i 1 = =95:8mV Figure 3.49(b) To find R t : Deactivate all the independent sources and look into the terminals a b to determine the Thevenin s resistance. Figure 3.49(c) Figure 3.49(d)

25 Circuit Theorems j 183 R t = R ab = 600jj jj = = 840:64 Ω Hence, the Thevenin equivalent circuit consists of the 95.8 mv source in series with Ω resistor. If we connect 30Ω resistor (galvanometer resistance) to this equivalent network, we obtain the circuit in Fig Figure 3.50 i G = 95: :64+30Ω = 110:03 A EXAMPLE 3.15 For the circuit shown in Fig. 3.51, find the Thevenin s equivalent circuit between terminals a and b. With ab shorted, let I sc = I. The circuit after transforming voltage sources into their equivalent current sources is as shown in Fig Writing node equations for this circuit, At a : 0:2V a 0:1 V c + I =3 At c : 0:1V a +0:3 V c 0:1 V b =4 At b : 0:1V c +0:2 V b I =1 Figure 3.51 As the terminals a and b are shorted V a = V b and the above equations become Figure 3.52

26 184 j Network Theory 0:2V a 0:1 V c + I =3 0:2V a +0:3 V c =4 0:2V a 0:1 V c 1=1 Solving the above equations, we get the short circuit current, I = I sc =1A. Next let us open circuit the terminals a and b and this makes I = 0. And the node equations written earlier are modified to Solving the above equations, we get 0:2V a 0:1 V c =3 0:1V a +0:3 V c 0:1 V b =4 0:1V c +0:2 V b =1 V a = 30V and V b = 20V Hence, V ab =30 20=10V = V oc = V t Therefore R t = V oc I sc = 10 1 = 10Ω The Thevenin s equivalent is as shown in Fig 3.53 Figure 3.53 EXAMPLE 3.16 Refer to the circuit shown in Fig Find the Thevenin equivalent circuit at the terminals a b. Figure 3.54 To begin with let us transform 3 A current source and 10 V voltage source. This results in a network as shown in Fig (a) and further reduced to Fig (b).

27 Circuit Theorems j 185 Figure 3.55(a) Again transform the 30 V source and following the reduction procedure step by step from Fig (b) to 3.55 (d), we get the Thevenin s equivalent circuit as shown in Fig Figure 3.55(b) Figure 3.55(c) Figure 3.55(d) Figure 3.56 Thevenin equivalent circuit EXAMPLE 3.17 Find the Thevenin equivalent circuit as seen from the terminals a b. Refer the circuit diagram shown in Fig

28 186 j Network Theory Figure 3.57 Since the circuit has no independent sources, i =0when the terminals a b are open. Therefore, V oc =0. The onus is now to find R t. Since V oc =0and i sc =0, R t cannot be determined from R t = V oc. Hence, we choose to connect a source of 1Aattheterminals a b as shown in Fig. i sc Then, after finding V ab, the Thevenin resistance is, R t = V ab 1 KCL at node a : V a 2i + V a =0 Also; i= V a 10 Hence; V a 2 V a V a 10 1=0 ) V a = V Hence; R t = V a 1 = Ω Alternatively one could find R t by connecting a 1V source at the terminals a b and then find the current from b to a. Then R t = 1. The concept of finding R t by connecting a 1A source i ba between the terminals a b may also be used for circuits containing independent sources. Then set all the independent sources to zero and use 1A source at the terminals a b to find V ab and hence, R t = V ab 1. For the present problem, the Thevenin equivalent circuit as seen between the terminals a b is shown in Fig (a). Figure 3.58 Figure 3.58 (a)

29 Circuit Theorems j 187 EXAMPLE 3.18 Determine the Thevenin equivalent circuit between the terminals a b for the circuit of Fig Figure 3.59 As there are no independent sources in the circuit, we get V oc = V t =0: To find R t, connect a 1V source to the terminals a b and measure the current I that flows from b to a. (Refer Fig a). R t = 1 I Ω Figure 3.60(a) Applying KCL at node a: I =0:5V x + V x 4 Since; V x =1V we get, I =0: =0:75 A Hence; R t = 1 0:75 =1:33 Ω Figure 3.60(b) The Thevenin equivalent circuit is shown in 3.60(b). Alternatively, sticking to our strategy, let us connect 1A current source between the terminals a b and then measure V ab (Fig (c)). Consequently, R t = V ab 1 = V ab Ω:

30 188 j Network Theory Applying KCL at node a: Hence 0:5V x + V x 4 =1) V x =1:33V R t = V ab 1 = V x 1 =1:33 Ω The corresponding Thevenin equivalent circuit is same as shown in Fig. 3.60(b) Figure 3.60(c) 3.3 Norton s theorem An American engineer, E.L. Norton at Bell Telephone Laboratories, proposed a theorem similar to Thevenin s theorem. Norton s theorem states that a linear two-terminal network can be replaced by an equivalent circuit consisting of a current source i N in parallel with 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. If one does not wish to turn off the independent sources, then R N is the ratio of open circuit voltage to short circuit current at the terminal pair. Figure 3.61(a) Original circuit Figure 3.61(b) Norton s equivalent circuit Figure 3.61(b) shows Norton s equivalent circuit as seen from the terminals a b of the original circuit shown in Fig. 3.61(a). Since this is the dual of the Thevenin circuit, it is clear that R N = R t and i N = v oc. In fact, source transformation of Thevenin equivalent circuit leads to R t Norton s equivalent circuit. Procedure for finding Norton s equivalent circuit: (1) If the network contains resistors and independent sources, follow the instructions below: (a) Deactivate the sources and find R N by circuit reduction techniques. (b) Find i N with sources activated. (2) If the network contains resistors, independent and dependent sources, follow the steps given below: (a) Determine the short-circuit current i N with all sources activated.

31 Circuit Theorems j 189 (b) Find the open-circuit voltage v oc. (c) R t = R N = v oc i N (3) If the network contains only resistors and dependent sources, follow the procedure described below: (a) Note that i N =0. (b) Connect 1A current source to the terminals a b and find v ab. (c) R t = v ab 1 Note: Also, since v t = v oc and i N = i sc R t = v oc i sc = R N The open circuit and short circuit test are sufficient to find any Thevenin or Norton equivalent PROOF OF THEVENIN S AND NORTON S THEOREMS The principle of superposition is employed to provide the proof of Thevenin s and Norton s theorems. Derivation of Thevenin s theorem: Let us consider a linear circuit having two accessible terminals x y and excited by an external current source i. The linear circuit is made up of resistors, dependent and independent sources. For the sake of simplified analysis, let us assume that the linear circuit contains only two independent voltage sources v 1 and v 2 and two independent current sources i 1 and i 2. The terminal voltage v may be obtained, by applying the principle of superposition. That is, v is made up of contributions due to the external source and independent sources within the linear network. Hence; v = a 0 i + a 1 v 1 + a 2 v 2 + a 3 i 1 + a 4 i 2 (3.9) = a 0 i + b 0 (3.10) where b 0 = a 1 v 1 + a 2 v 2 + a 3 i 1 + a 4 i 2 = contribution to the terminal voltage v by independent sources within the linear network. Let us now evaluate the values of constants a 0 and b 0. (i) When the terminals x and y are open circuited, i =0and v = v oc = v t. Making use of this fact in equation 3.10, we find that b 0 = v t.

32 190 j Network Theory (ii) When all the internal sources are deactivated, b 0 = 0. This enforces equation 3.10 to become v = a 0 i = R t i ) a 0 = R t R t V t Figure 3.62 Current-driven circuit Figure 3.63 Thevenin s equivalent circuit of Fig where R t is the equivalent resistance of the linear network as viewed from the terminals x y. Also, a 0 must be R t in order to obey the ohm s law. Substuting the values of a 0 and b 0 in equation 3.10, we find that v = R t i + v 1 which expresses the voltage-current relationship at terminals x y of the circuit in Fig Thus, the two circuits of Fig and 3.63 are equivalent. Derivation of Norton s theorem: Let us now assume that the linear circuit described earlier is driven by a voltage source v as shown in Fig The current flowing into the circuit can be obtained by superposition as i = c 0 v + d 0 (3.11) where c 0 v is the contribution to i due to the external voltage source v and d 0 contains the contributions to i due to all independent sources within the linear circuit. The constants c 0 and d 0 are determined as follows : (i) When terminals x y are short-circuited, v = 0 and i = i sc. Hence from equation (3.11), we find that i = d 0 = i sc, where i sc is the short-circuit current flowing out of terminal x, which is same as Norton current i N Thus, d 0 = i N Figure 3.64 Voltage-driven circuit (ii) Let all the independent sources within the linear network be turned off, that is d 0 =0. Then, equation (3.11) becomes i = c 0 v

33 Circuit Theorems j 191 For dimensional validity, c 0 must have the dimension of conductance. This enforces c 0 = 1 where R t is the equivalent resistance of the R t linear network as seen from the terminals x y. Thus, equation (3.11) becomes i = 1 R t v i sc = 1 R t v i N Figure 3.65 Norton s equivalent of voltage driven circuit This expresses the voltage-current relationship at the terminals x y of the circuit in Fig. (3.65), validating that the two circuits of Figs and 3.65 are equivalents. EXAMPLE 3.19 Find the Norton equivalent for the circuit of Fig Figure 3.66 As a first step, short the terminals a b. This results in a circuit diagram as shown in Fig Applying KCL at node a,weget i sc =0 ) i sc =9A To find R N, deactivate all the independent sources, resulting in a circuit diagram as shown in Fig (a). We find R N in the same way as R t in the Thevenin equivalent circuit. R N = =3Ω Figure 3.67

34 192 j Network Theory Figure 3.68(a) Figure 3.68(b) Thus, we obtain Nortion equivalent circuit as shown in Fig. 3.68(b). EXAMPLE 3.20 Refer the circuit shown in Fig Find the value of i b using Norton equivalent circuit. Take R = 667 Ω. Figure 3.69 Since we want the current flowing through R, remove R from the circuit of Fig The resulting circuit diagram is shown in Fig To find i ac or i N referring Fig 3.70(a) : i a = =0A i sc = A=2mA Figure 3.70 Figure 3.70(a)

35 Circuit Theorems j 193 To find R N : The procedure for finding R N is same that of R t in the Thevenin equivalent circuit. R t = R N = v oc i sc To find v oc, make use of the circuit diagram shown in Fig Do not deactivate any source. Applying KVL clockwise, we get Therefore; i a i a i a =0 ) i a = A ) v oc = i a 1000 = 4 3 V R N = v oc i sc = 4 3 = 667 Ω The Norton equivalent circuit along with resistor R is as shown below: i b = i sc 2 = 2mA = 1mA 2 Figure 3.71 Figure : Norton equivalent circuit with load R EXAMPLE 3.21 Find I o in the network of Fig using Norton s theorem. Figure 3.72

36 194 j Network Theory We are interested in I o, hence the 2kΩresistor is removed from the circuit diagram of Fig The resulting circuit diagram is shown in Fig. 3.73(a). Figure 3.73(a) Figure 3.73(b) To find i N or i sc : Refer Fig. 3.73(b). By inspection, V 1 =12V Applying KCL at node V 2 : V 2 V 1 6kΩ + V 2 2kΩ + V 2 V 1 3kΩ =0 Substituting V 1 =12Vand solving, we get V 2 =6V i sc = V 1 V 2 3kΩ + V 1 4kΩ =5mA To find R N : Deactivate all the independent sources (refer Fig. 3.73(c)). Figure 3.73(c) Figure 3.73(d)

37 Circuit Theorems j 195 Referring to Fig (d), we get R N = R ab =4kΩjj [3 kω + (6 kωjj2kω)]=2:12 kω Hence, the Norton equivalent circuit along with 2kΩresistor is as shown in Fig. 3.73(e). I o = i sc R N R + R N =2:57mA Figure 3.73(e) EXAMPLE 3.22 Find V o in the circuit of Fig Figure 3.74 Since we are interested in V o, the voltage across 4kΩresistor, remove this resistance from the circuit. This results in a circuit diagram as shown in Fig Figure 3.75

38 196 j Network Theory To find i sc, short the terminals a b :

39 Circuit Theorems j 197 Constraint equation : KVL around supermesh : KVL around mesh 3: i 1 i 2 = 4mA (3.12) i i 2 =0 (3.13) (i 3 i 2 ) (i 3 i 1 )=0 Since i 3 = i sc, the above equation becomes, (i sc i 2 ) (i sc i 1 )=0 (3.14) Solving equations (3.12), (3.13) and (3.14) simultaneously, we get i sc =0:1333 ma. To find R N : Deactivate all the sources in Fig This yields a circuit diagram as shown in Fig Figure 3.76 R N =6kΩjj10 kω = 6 10 =3:75 kω 6+10 Hence, the Norton equivalent circuit is as shown in Fig 3.76 (a). To the Norton equivalent circuit, now connect the 4kΩ resistor that was removed earlier to get the network shown in Fig. 3.76(b). Figure 3.76(a)

40 198 j Network Theory V o = i sc (R N jjr) R N R = i sc R N + R = 258 mv Figure 3.76(b) Norton equivalent circuit with R =4kΩ EXAMPLE 3.23 Find the Norton equivalent to the left of the terminals a b for the circuit of Fig Figure 3.77 To find i sc : Note that v ab =0when the terminals a b are short-circuited. Then i = =10mA Therefore, for the right hand portion of the circuit, i sc = 10i = 100 ma.

41 Circuit Theorems j 199 To find R N or R t : Writing the KVL equations for the left-hand mesh, we get Also for the right-hand mesh, we get i + v ab =0 (3.15) v ab = 25(10i) = 250i Therefore i = v ab 250 Substituting i into the mesh equation (3.15), we get vab v ab =0 250 ) v ab = 5V R N = R t v oc i sc = v ab i sc = 5 0:1 =50Ω The Norton equivalent circuit is shown in Fig 3.77 (a). Figure 3.77 (a) EXAMPLE 3.24 Find the Norton equivalent of the network shown in Fig Figure 3.78

42 200 j Network Theory Since there are no independent sources present in the network of Fig. 3.78, i N = i sc =0. To find R N, we inject a current of 1A between the terminals a b. This is illustrated in Fig KCL at node 1: Figure 3.79 Figure 3.79(a) Norton equivalent circuit 1= v v 1 v 2 50 ) 0:03v 1 0:02v 2 =1 KCL at node 2: v v 2 v 1 +0:1v 1 =0 50 ) 0:08v 1 +0:025v 2 =0 Solving the above two nodal equations, we get v 1 =10:64 volts ) v oc =10:64 volts Hence; R N = R t = v oc 1 = 10:64 =10:64 Ω 1 Norton equivalent circuit for the network shown in Fig is as shown in Fig. 3.79(a). EXAMPLE 3.25 Find the Thevenin and Norton equivalent circuits for the network shown in Fig (a). Figure 3.80(a)

43 Circuit Theorems j 201 To find V oc : Performing source transformation on 5A current source, we get the circuit shown in Fig (b). Applying KVL around Left mesh : 50+2i a 20+4i a =0 ) i a = 70 6 A Applying KVL around right mesh: 20+10i a + V oc 4i a =0 ) V oc = 90 V To find i sc (referring Fig 3.80 (c)) : KVL around Left mesh : 50+2i a 20+4(i a i sc )=0 ) 6i a 4i sc =70 KVL around right mesh : Figure 3.80(b) 4(i sc i a )+20+10i a =0 ) 6i a +4i sc = 20 Figure 3.80(c) Solving the two mesh equations simultaneously, we get i sc = 11:25 A Hence, R t = R N = v oc = 90 i sc 11:25 =8Ω Performing source transformation on Thevenin equivalent circuit, we get the norton equivalent circuit (both are shown below). Thevenin equivalent circuit Norton equivalent circuit

44 202 j Network Theory EXAMPLE 3.26 If an 8 kω load is connected to the terminals of the network in Fig. 3.81, V AB =16V.Ifa2kΩ load is connected to the terminals, V AB =8V. Find V AB if a 20 kω load is connected across the terminals. Figure 3.81 Applying KVL around the mesh, we get (R t + R L ) I = V oc If If R L =2kΩ;I=10mA) V oc =20+0:01R t R L =10kΩ;I=6mA) V oc =60+0:006R t Solving, we get V oc = 120 V, R t =10kΩ. If R L =20kΩ;I= V oc (R L + R t ) = 120 ( ) =4mA 3.4 Maximum Power Transfer Theorem In circuit analysis, we are some times interested in determining the maximum power that a circuit can supply to the load. Consider the linear circuit A as shown in Fig Circuit A is replaced by its Thevenin equivalent circuit as seen from a and b (Fig 3.83). We wish to find the value of the load R L such that the maximum power is delivered to it. The power that is delivered to the load is given by Figure 3.82 Circuit A with load R L p = i 2 V 2 t R L = R L (3.16) R t + R L

45 Circuit Theorems j 203 Assuming that V t and R t are fixed for a given source, the maximum power is a function of R L. In order to determine the value of R L that maximizes p, we differentiate p with respect to R L and equate the derivative to zero. dp = V 2 t dr L " # (R t + R L ) 2 2(R t + R L ) (R L + R t ) 2 =0 which yields R L = R t (3.17) To confirm that equation (3.17) is a maximum, it should be shown that d2 p dr 2 < 0. Hence, maximum power is transferred to the load when R L is L equal to the Thevenin equivalent resistance R t. The maximum power transferred to the load is obtained by substituting R L = R t in equation Accordingly, P max = V 2 t R L (2R L ) 2 = V 2 t 4R L Figure 3.83 Thevenin equivalent circuit is substituted for circuit A The 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 t. EXAMPLE 3.27 Find the load R L that will result in maximum power delivered to the load for the circuit of Fig Also determine the maximum power P max. Figure 3.84 Disconnect the load resistor R L. This results in a circuit diagram as shown in Fig. 3.85(a). Next let us determine the Thevenin equivalent circuit as seen from a b.

46 204 j Network Theory i = =1A V oc = V t = 150 i = 150 V To find R t, deactivate the 180 V source. This results in the circuit diagram of Fig. 3.85(b). R t = R ab =30Ωjj150 Ω Figure 3.85(a) = =25Ω The Thevenin equivalent circuit connected to the load resistor is shown in Fig Maximum power transfer is obtained when R L = R t =25Ω: Then the maximum power is P max = V 2 t = (150)2 4R L 4 25 =2:25 Watts The Thevenin source V t actually provides a total power of Figure 3.85(b) P t = 150 i 150 = = 450 Watts Thus, we note that one-half the power is dissipated in R L. Figure 3.86 EXAMPLE 3.28 Refer to the circuit shown in Fig Find the value of R L for maximum power transfer. Also find the maximum power transferred to R L. Figure 3.87

47 Circuit Theorems j 205 Disconnecting R L, results in a circuit diagram as shown in Fig. 3.88(a). Figure 3.88(a) To find R t, deactivate all the independent voltage sources as in Fig. 3.88(b). Figure 3.88(b) Figure 3.88(c) To find V t : Refer the Fig. 3.88(d). Constraint equation : R t = R ab =6kΩjj6 kωjj6 kω =2kΩ V 3 V 1 =12V By inspection, V 2 =3V KCL at supernode : ) V 3 3 6k V 3 V 2 6k + V k + V 1 6k + V 1 V 2 =0 6k + V k =0 Figure 3.88(d)

48 206 j Network Theory ) V 3 3+V V 3 15=0 ) 3V 3 =30 ) V 3 =10 ) V t = V ab = V 3 =10V Figure 3.88(e) The Thevenin equivalent circuit connected to the load resistor R L is shown in Fig. 3.88(e). P max = i 2 R L 2 Vt = R L 2R L =12:5 mw Alternate method : It is possible to find P max, without finding the Thevenin equivalent circuit. However, we have to find R t. For maximum power transfer, R L = R t =2kΩ. Insert the value of R L in the original circuit given in Fig Then use any circuit reduction technique of your choice to find power dissipated in R L. Refer Fig. 3.88(f). By inspection we find that, V 2 =3V. Constraint equation : V 3 V 1 =12 ) V 1 = V 3 12 KCL at supernode : ) V 3 3 6k V 3 V 2 6k + V k + V 1 V 2 6k + V 3 2k + V 1 6k =0 + V 3 2k + V 3 12 =0 6k ) V 3 3+V V 3 + V 3 12=0 ) 6V 3 =30 ) V 3 =5 V Hence; P max = V 3 2 = 25 = 12:5 mw R L 2k Figure 3.88(f)

49 Circuit Theorems j 207 EXAMPLE 3.29 Find R L for maximum power transfer and the maximum power that can be transferred in the network shown in Fig Figure 3.89 Disconnect the load resistor R L. This results in a circuit as shown in Fig. 3.89(a). Figure 3.89(a) To find R t, let us deactivate all the independent sources, which results the circuit as shown in Fig. 3.89(b). R t = R ab =2kΩ+3kΩ+5kΩ=10kΩ For maximum power transfer R L = R t =10kΩ. Let us next find V oc or V t. Refer Fig (c). By inspection, i 1 = 2 ma & i 2 =1mA.

50 208 j Network Theory Figure 3.89(b) Applying KVL clockwise to the loop 5kΩ! 3kΩ! 2kΩ! a b, weget 5k i 2 +3k(i 1 i 2 )+2ki 1 + V t =0 ) V t =0 ) V t =0 ) V t =18V: The Thevenin equivalent circuit with load resistor R L is as shown in Fig (d). 18 i = =0:9 ma (10+10)103 Then, P max = P L =(0:9 ma) 2 10 kω =8:1 mw Figure 3.89(c) Figure 3.89(d) EXAMPLE 3.30 Find the maximum power dissipated in R L. Refer the circuit shown in Fig Figure 3.90

51 Circuit Theorems j 209 Disconnecting the load resistor R L from the original circuit results in a circuit diagram as shown in Fig Figure 3.91 As a first step in the analysis, let us find R t. While finding R t, we have to deactivate all the independent sources. This results in a network as shown in Fig 3.91 (a) : Figure 3.91(a) R t = R ab = [140 Ωjj60 Ω] + 8 Ω = =50Ω: For maximum power transfer, R L = R t =50Ω. Next step in the analysis is to find V t. Refer Fig 3.91(b), using the principle of current division, i 1 = i R 2 R 1 + R 2 = =17A i 2 = i R 1 R 1 + R 2 = = =3A Figure 3.91(a)

52 210 j Network Theory Applying KVL clockwise to the loop comprising of 50 Ω! 10 Ω! 8Ω! a b, weget 50i 2 10i V t =0 ) 50(3) 10 (17) + V t =0 ) V t =20V The Thevenin equivalent circuit with load resistor R L is as shown in Fig. 3.91(c). i T = =0:2A P max = i 2 T 50=0:04 50 = 2W Figure 3.91(c) EXAMPLE 3.31 Find the value of R L for maximum power transfer in the circuit shown in Fig Also find P max. Figure 3.92 Disconnecting R L from the original circuit, we get the network shown in Fig Figure 3.93

53 Circuit Theorems j 211 Let us draw the Thevenin equivalent circuit as seen from the terminals a b and then insert the value of R L = R t between the terminals a b. To find R t, let us deactivate all independent sources which results in the circuit as shown in Fig Figure 3.94 R t = R ab =8Ωjj2 Ω = =1:6 Ω Next step is to find V oc or V t. By performing source transformation on the circuit shown in Fig. 3.93, we obtain the circuit shown in Fig Figure 3.95 Applying KVL to the loop made up of 20 V! 3Ω! 2Ω! 10 V! 5Ω! 30 V, we get 20+10i 10 30=0 ) i = =6A

54 212 j Network Theory Again applying KVL clockwise to the path 2Ω! 10 V! a b, weget 2i 10 V t =0 ) V t =2i 10 = 2(6) 10=2V The Thevenin equivalent circuit with load resistor R L is as shown in Fig (a). P max = i 2 T R L = V 2 t = 625 mw 4R t Figure 3.95(a) Thevenin equivalent circuit EXAMPLE 3.32 Find the value of R L for maximum power transfer. Hence find P max. Figure 3.96 Removing R L from the original circuit gives us the circuit diagram shown in Fig Figure 3.97 To find V oc : KCL at node A : i 0 a 0:9+10i 0 a =0 ) i 0 a =0:1 A Hence; V oc =3 10i 0 a =310 0:1 =3V

55 Circuit Theorems j 213 To find R t, we need to compute i sc with all independent sources activated. KCL at node A: i 00 a 0:9+10i 00 a =0 ) i 00 a =0:1A Hence i sc =10i 00 a =100:1=1A R t = V oc = 3 i sc 1 =3Ω Hence, for maximum power transfer R L = R t =3Ω. The Thevenin equivalent circuit with R L =3Ω inserted between the terminals a b gives the network shown in Fig. 3.97(a). i T = =0:5 A P max = i 2 T R L =(0:5) 2 3 = 0.75 W Figure 3.97(a) EXAMPLE 3.33 Find the value of R L in the network shown that will achieve maximum power transfer, and determine the value of the maximum power. Figure 3.98(a) Removing R L from the circuit of Fig. 3.98(a), we get the circuit of Fig 3.98(b). Applying KVL clockwise we get i +2V 0 x =0 Also V 0 x =110 3 i Hence; i i =0 i = =3mA Figure 3.98(b)

56 214 j Network Theory Applying KVL to loop 1kΩ! 2V 0 x! b a, weget i +2V 0 x V t =0 ) V t = i i To find R t, we need to find i sc. While finding i sc, none of the independent sources must be deactivated. Applying KVL to mesh 1: 12 + V 00 x +0=0 ) V 00 x =12 ) i 1 =12 ) i 1 =12mA Applying KVL to mesh 2: i 2 +2V 00 x =0 ) i 2 = 24 i 2 = 24 ma Applying KCL at node a: = i = =9V Hence; i sc = i 1 i 2 =12+24=36mA R t = V t = V oc i sc i sc 9 = = 250 Ω For maximum power transfer, R L = R t = 250 Ω. Thus, the Thevenin equivalent circuit with R L is as shown in Fig 3.98 (c) : 9 i T = = A P max = i 2 T = = 81 mw Figure 3.98 (c) Thevenin equivalent circuit

57 Circuit Theorems j 215 EXAMPLE 3.34 The variable resistor R L in the circuit of Fig is adjusted untill it absorbs maximum power from the circuit. (a) Find the value of R L. (b) Find the maximum power. Figure 3.99 Disconnecting the load resistor R L Fig. 3.99(a). from the original circuit, we get the circuit shown in KCL at node v 1 : Figure 3.99(a) Constraint equations : v v 1 13i 0 a 5 + v 1 v 2 4 =0 (3.18) i 0 a = 100 v 1 (3.19) 2 v 2 v 1 = v 0 a (applying KCL at v 2 ) (3.20) 4 v 0 a = v 1 v 2 (potential across 4Ω) (3.21)

58 216 j Network Theory From equations (3.20) and (3.21), we have v 2 v 1 = v 1 v 2 4 ) v 2 v 1 =4v 1 4v 2 ) 5v 1 5v 2 =0 ) v 1 = v 2 (3.22) Making use of equations (3.19) and (3.22) in (3.18), we get v ) 5(v 1 100) + 2 v 2 13 (100 v 1) 2 + v 1 v v 1 13 (100 v 1) 2 =0 =0 ) 5v v v 1 =0 ) 20v 1 = 1800 ) v 1 = 90 Volts Hence; v t = v 2 = v 1 = 90 Volts We know that, R t = v oc = v t i sc i sc The short circuit current is calculated using the circuit shown below: Here i 00 a = 100 v 1 2 Applying KCL at node v 1 : ) v v v 1 13i 00 a 5 v 1 13 (100 v 1) v =0 + v 1 4 =0

59 Circuit Theorems j 217 Solving we get v 1 =80volts = v 00 a Applying KCL at node a : Hence; 0 v i sc = v 00 a ) i sc = v v00 a = = 100 A 4 Hence for maximum power transfer, R t = v oc i sc = v t i sc = =0:9 Ω R L = R t = 0:9 Ω The Thevenin equivalent circuit with R L =0:9Ω is as shown. 90 i t = 0:9+0:9 = 90 1:8 P max = i 2 t 0: = 0:9 =2250 W 1:8 EXAMPLE 3.35 Refer to the circuit shown in Fig : (a) Find the value of R L for maximum power transfer. (b) Find the maximum power that can be delivered to R L. Figure 3.100

60 218 j Network Theory Removing the load resistor R L, we get the circuit diagram shown in Fig (a). Let us proceed to find V t. Constraint equation : Figure 3.100(a) i 0 a = i 1 i 3 KVL clockwise to mesh 1: 200+1(i 1 i 2 ) + 20 (i 1 i 3 )+4i 1 =0 ) 25i 1 i 2 20i 3 = 200 KVL clockwise to mesh 2: 14i 0 a +2(i 2 i 3 )+1(i 2 i 1 )=0 ) 14 (i 1 i 3 )+2(i 2 i 3 )+1(i 2 i 1 )=0 ) 13i 1 +3i 2 16i 3 =0 KVL clockwise to mesh 3: 2(i 3 i 2 ) 100+3i 3 +20(i 3 i 1 )=0 ) 20i 1 2i 2 +25i 3 = 100 Solving the mesh equations, we get i 1 = 2:5A;i 3 =5A Applying KVL clockwise to the path comprising of a b! 20 Ω, weget V t 20i 0 a =0 ) V t =20i 0 a =20(i 1 i 3 ) =20( 2:5 5) = 150 V

61 Circuit Theorems j 219 Next step is to find R t. R t = V oc i sc = V t i sc When terminals a b are shorted, i 00 a =0. Hence, 14 i 00 a is also zero. KVL clockwise to mesh 1: 200+1(i 1 i 2 )+4i 1 =0 ) 5i 1 i 2 = 200 KVL clockwise to mesh 2: 2(i 2 i 3 )+1(i 2 i 1 )=0 ) i 1 +3i 2 2i 3 =0 KVL clockwise to mesh 3: 100+3i 3 +2(i 3 i 2 )=0 ) 2i 2 +5i 3 = 100

62 220 j Network Theory Solving the mesh equations, we find that i 1 = 40A; i 3 = 20A; ) i sc = i 1 i 3 = 60A R t = V t = 150 i sc 60 =2:5 Ω For maximum power transfer, R L = R t =2:5Ω. The Thevenin equivalent circuit with R L is as shown below : P max = i 2 1 R L = 2:5 2:5+2:5 = 2250 W EXAMPLE 3.36 A practical current source provides 10 W to a 250 Ω load and 20 W to an 80 Ω load. A resistance R L, with voltage v L and current i L, is connected to it. Find the values of R L, v L and i L if (a) v L i L is a maximum, (b) v L is a maximum and (c) i L is a maximum. Load current calculation: 10W to 250 Ω corresponds to i L = 20W to 80 Ω corresponds to i L = r = r 200 ma = 500 ma Using the formula for division of current between two parallel branches : i 2 = i R 1 R 1 + R 2 In the present context, 0:2= I N R N R N and 0:5= I N R N R N +80 (3.23) (3.24)

63 Circuit Theorems j 221 Solving equations (3.23) and (3.24), we get (a) If v L i L is maximum, I N =1:7 A R N =33:33 Ω R L = R N =33:33 Ω 33:33 i L =1:7 33:33+33:33 = 850 ma v L = i L R L = :33 =28:33 V (b) v L = I N (R N jjr L ) is a maximum when R N jjr L is a maximum, which occurs when R L = 1. Then, i L =0and (c) i L = v L =1:7R N =1:733:33 =56:66 V I N R N is maxmimum when R L =0Ω R N + R L ) i L =1:7A and v L =0V 3.5 Sinusoidal steady state analysis using superposition, Thevenin and Norton equivalents Circuits in the frequency domain with phasor currents and voltages and impedances are analogous to resistive circuits. To begin with, let us consider the principle of superposition, which may be restated as follows : For a linear circuit containing two or more independent sources, any circuit voltage or current may be calculated as the algebraic sum of all the individual currents or voltages caused by each independent source acting alone. Figure Thevenin equivalent circuit Figure Norton equivalent circuit

64 222 j Network Theory The superposition principle is particularly useful if a circuit has two or more sources acting at different frequencies. The circuit will have one set of impedance values at one frequency and a different set of impedance values at another frequency. Phasor responses corresponding to different frequencies cannot be superposed; only their corresponding sinusoids can be superposed. That is, when frequencies differ, the principle of superposition applies to the summing of time domain components, not phasors. Within a component, problem corresponding to a single frequency, however phasors may be superposed. Thevenin and Norton equivalents in phasor circuits are found exactly in the same manner as described earlier for resistive circuits, except for the subtitution of impedance Z in place of resistance R and subsequent use of complex arithmetic. The Thevenin and Norton equivalent circuits are shown in Fig and The Thevenin and Norton forms are equivalent if the relations (a) Z t = Z N (b) V t = Z N I N hold between the circuits. A step by step procedure for finding the Thevenin equivalent circuit is as follows: 1. Identify a seperate circuit portion of a total circuit. 2. Find V t = V oc at the terminals. 3. (a) If the circuit contains only impedances and independent sources, then deactivate all the independent sources and then find Z t by using circuit reduction techniques. (b) If the circuit contains impedances, independent sources and dependent sources, then either short circuit the terminals and determine I sc from which Z t = V oc I sc or deactivate the independent sources, connect a voltage or current source at the terminals, and determine both V and I at the terminals from which Z t = V I A step by step procedure for finding Norton equivalent circuit is as follows: (i) Identify a seperate circuit portion of the original circuit. (ii) Short the terminals after seperating a portion of the original circuit and find the current through the short circuit at the terminals, so that I N = I sc. (iii) (a) If the circuit contains only impedances and independent sources, then deactivate all the independent sources and then find Z N = Z t by using circuit reduction techniques. (b) If the circuit contains impedances, independent sources and one or more dependent sources, find the open circuit voltage at the terminals, V oc, so that Z N = Z t = V oc I sc :

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