3.1 Superposition Theorem

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 :

i1 ! v1

i2 ! v2 then;i1 + i2 ! v1 + v2 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 i1 will produce a response v1. Similarly, an excitation i2 will cause a response v2. Then if we use an excitation i1 + i2,we will ﬁnd a response v1 + v2. The principle of superposition has the ability to reduce a complicated problem to several easier problems each containing only a single independent source.

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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. 3.1.

Figure 3.1

SOLUTION As a ﬁrst step, set the current source to zero. That is, the current source appears as an open circuit as shown in Fig. 3.2. 6 6 i1 = = A 3+6 9

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As a next step, set the voltage to zero by replacing it with a short circuit as shown in Fig. 3.3. 2 3 6 i2 = = A 3+6 9

Figure 3.2 Figure 3.3

The total current i is then the sum of i1 and i2 12 i i i A = 1 + 2 = 9

EXAMPLE 3.2 Find io in the network shown in Fig. 3.4 using superposition.

Figure 3.4

SOLUTION As a ﬁrst step, set the current source to zero. That is, the current source appears as an open circuit as shown in Fig. 3.5.

Figure 3.5

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0 6 io = = 0:3mA (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, iR2 iA = R1 + R2

where R1 = (12 kΩjj12 kΩ) + 12 kΩ =6kΩ+12kΩ =18kΩ and R2 =12kΩ 4 103 12 103 ) iA = (12 + 18) 103 =1:6mA Figure 3.6 Again applying the current division principle, 00 iA 12 io = =0:8mA 12+12 0 00 Thus;io = io + io = 0:3+0:8=0:5mA

Figure 3.6(a)

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EXAMPLE 3.3 Use superposition to ﬁnd io in the circuit shown in Fig. 3.7.

Figure 3.7

SOLUTION 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. 3.8.

0 12 io = (2+2) 103 =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.

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Applying KVL clockwise to the upper loop, we get

3 00 3 00 2 10 io 2 10 io 6=0 00 6 ) io = = 1:5mA 4 103

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. 3.10.

Figure 3.10

Current of 2 mA splits equally.

000 Hence;io = 1mA

Applying the superposition principle, we ﬁnd that

0 00 000 io = io + io + io =3 1:5+1 = 2:5mA

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EXAMPLE 3.4 Find the current i for the circuit of Fig. 3.11.

Figure 3.11

SOLUTION 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. 3.12. Applying KVL clockwise around loop shown in Fig. 3.12, we find that

5i1 +3i1 24=0 24 ) i1 = =3A 8 As a second step, we set the voltage source to zero and determine the current i2 due to the current source. For this condition, refer to Fig. 3.13 for analysis.

Figure 3.12 Figure 3.13

Applying KCL at node 1, we get v1 3i2 i2 +7= (3.1) 2 v1 0 Noting that i2 = 3 we get, v1 = 3i2 (3.2)

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Making use of equation (3.2) in equation (3.1) leads to 3i2 3i2 i2 +7= 2 7 ) i2 = A 4 Thus, the total current i = i1 + i2 7 5 =3 A= A 4 4 EXAMPLE 3.5 For the circuit shown in Fig. 3.14, ﬁnd the terminal voltage Vab using superposition principle.

SOLUTION Figure 3.14 As a ﬁrst step in the analysis, deactivate the independent current source. This results in a circuit diagram as shown in Fig. 3.15. Applying KVL clockwise gives

4+10 0+3Vab1 + Vab1 =0

) 4Vab1 =4

) Vab1 =1V Figure 3.15 Next step in the analysis is to deactivate the independent voltage source, resulting in a circuit diagram as shown in Fig. 3.16. Applying KVL gives

10 2+3Vab2 + Vab2 =0

) 4Vab2 =20

) Vab2 =5V

Figure 3.16

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According to superposition principle,

Vab = Vab1 + Vab2 =1+5=6V

EXAMPLE 3.6 Use the principle of superposition to solve for vx in the circuit of Fig. 3.17.

Figure 3.17

SOLUTION According to the principle of superposition,

vx = vx1 + vx2

where vx1 is produced by 6A source alone in the circuit and vx2 is produced solely by 4A current source.

To ﬁnd vx1 , deactivate the 4A current source. This results in a circuit diagram as shown in Fig. 3.18. KCL at node x1 :

vx vx 4ix 1 + 1 1 =6 2 8 vx1 But ix = 1 2 v v x1 vx1 x1 4 2 Hence; + =6 2 8 vx vx 2vx ) 1 + 1 1 =6 2 8

) 4vx1 + vx1 2vx1 =48 Figure 3.18 48 ) vx = = 16V 1 3

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To ﬁnd vx2 , deactivate the 6A current source, resulting in a circuit diagram as shown in Fig. 3.19. KCL at node x2 :

vx vx (4ix ) 2 + 2 2 =4 8 2 vx vx +4ix ) 2 + 2 2 =4 (3.3) 8 2 Applying KVL along dotted path,weget

vx2 +4ix2 2ix2 =0

vx2 ) vx = 2ix or ix = (3.4) 2 2 2 2 Substituting equation (3.4) in equation (3.3), we get vx2 vx2 +4 vx2 2 + =4 8 2 vx vx 2vx ) 2 + 2 2 =4 8 2 vx vx ) 2 2 =4 8 2 ) vx2 4vx2 =32 32 ) vx = V 2 3 Hence, according to the superposition principle,

vx = vx1 + vx2 Figure 3.19 32 =16 = 5:33V 2

EXAMPLE 3.7 Which of the source in Fig. 3.20 contributes most of the power dissipated in the 2 Ω resistor ? The least ? What is the power dissipated in 2Ωresistor ?

Figure 3.20

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SOLUTION 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, 0 0 i1 = i1 + i2 0 where i1 = Contribution to i1 from 5V source alone. 0 and i2 = Contribution to i1 from 2A source alone. 0 Let us ﬁrst ﬁnd i1. This needs the deactivation of 2A source. Refer to Fig. 3.22.

0 5 i1 = =1:22A 2+2:1 0 Similarly to ﬁnd i2 we have to disable the 5V source by shorting it. Referring to Fig. 3.23, we ﬁnd that

0 2 2:1 i2 = = 1:024 A 2+2:1

Figure 3.22 Figure 3.23

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Total current,

0 0 i1 = i1 + i2 =1:22 1:024 =0:196 A 2 Thus;P2Ω =(0:196) 2 =0:0768 Watts = 76:8 mW

EXAMPLE 3.8 Find the voltage V1 using the superposition principle. Refer the circuit shown in Fig.3.24.

Figure 3.24

SOLUTION According to the superposition principle,

0 00 V1 = V1 + V1

0 00 where V1 is the contribution from 60V source alone and V1 is the contribution from 4A current source alone. 0 To ﬁnd V1, the 4A current source is opened, resulting in a circuit as shown in Fig. 3.25.

Figure 3.25

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Applying KVL to the left mesh:

30ia 60+30(ia ib)=0 (3.5) Also ib = 0:4iA = 0:4(ia)=0:4ia (3.6)

Substituting equation (3.6) in equation (3.5), we get

30ia 60+30ia 30 0:4ia =0 60 ) ia = =1:25A 48 ib =0:4ia =0:4 1:25 =0:5A 0 Hence;V1 =(ia ib) 30 =22:5V

00 To ﬁnd, V1 , the 60V source is shorted as shown in Fig. 3.26.

Figure 3.26

Applying KCL at node a:

00 Va Va V1 + =4 20 10 00 ) 30Va 20V1 = 800 (3.7)

Applying KCL at node b:

00 00 V1 V1 Va + =0:4ib 30 10 Va Also;Va =20ia ) ib = 20 00 00 V1 V1 Va 0:4Va Hence; + = 30 10 20 00 )7:2Va +8V1 =0 (3.8)

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Solving the equations (3.7) and (3.8), we ﬁnd that 00 V1 = 60V 0 00 Hence V1 = V1 + V1 =22:5+60=82:5V

EXAMPLE 3.9 (a) Refer to the circuit shown in Fig. 3.27. Before the 10 mA current source is attached to terminals x y, the current ia is found to be 1.5 mA. Use the superposition theorem to ﬁnd the value of ia after the current source is connected. (b) Verify your solution by ﬁnding ia, when all the three sources are acting simultaneously.

Figure 3.27

SOLUTION According to the principle of superposition,

ia = ia1 + ia2 + ia3 where ia1 , ia2 and ia3 are the contributions to ia from 20V source, 5 mA source and 10 mA source respectively. As per the statement of the problem,

ia1 + ia2 =1:5mA

To ﬁnd ia3 , deactivate 20V source and the 5 mA source. The resulting circuit diagram is shown in Fig 3.28. 10mA 2k ia = =1mA 3 18k+2k Hence, total current

ia = ia1 + ia2 + ia3 =1:5+1=2:5 mA

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Figure 3.28

(b) Refer to Fig. 3.29 KCL at node y:

Vy Vy 20 3 + = (10+5)10 18 103 2 103

Solving, we get Vy = 45V: Vy 45 Hence;ia = = 18 103 18 103 =2:5mA Figure 3.29 3.2 Thevenin’s theorem

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. 3.30 can be replaced by the one shown in Fig. 3.31 (The load resistor may be a single resistor or another circuit). The circuit to the left of the terminals x y in Fig. 3.31 is known as the Thevenin’s equivalent circuit.

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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 Vt in series with a resistor Rt, Where Vt is the open–circuit voltage at the terminals and Rt is the input or equivalent resistance at the terminals when the independent sources are turned off or Rt 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 Rt: Three different types of circuits may be encountered in determining the resistance, Rt : (i) If the circuit contains only independent sources and resistors, deactivate the sources and ﬁnd Rt by circuit reduction technique. Independent current sources, are deactivated by opening them while independent voltage sources are deactivated by shorting them.

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(ii) If the circuit contains resistors, dependent and independent sources, follow the instructions described below:

(a) Determine the open circuit voltage voc with the sources activated. (b) Find the short circuit current isc when a short circuit is applied to the terminals a b voc (c) Rt = isc (iii) If the circuit contains resistors and only dependent sources, then

(a) voc =0(since there is no energy source) (b) Connect 1A current source to terminals a b and determine vab. vab (c) Rt = 1 Figure 3.32

For all the cases discussed above, the Thevenin’s equivalent circuit is as shown in Fig. 3.32.

EXAMPLE 3.10 Using the Thevenin’s theorem, ﬁnd the current i through R =2Ω. Refer Fig. 3.33.

Figure 3.33

SOLUTION

Figure 3.34

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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. 3.35.

Figure 3.35

To ﬁnd Rt, we have to deactivate the independent voltage source. Accordingly, we get the circuit in Fig. 3.36.

Rt =(5Ωjj20 Ω) + 4 Ω 5 20 = +4=8Ω 5+20 Rt Referring to Fig. 3.35,

50+25I =0 ) I =2A Figure 3.36 Hence Vab = Voc = 20(I) = 40V

Thus, we get the Thevenin’s equivalent circuit which is as shown in Fig.3.37.

Figure 3.37 Figure 3.38

Reconnecting the circuit B to the Thevenin’s equivalent circuit as shown in Fig. 3.38, we get 40 i = = 4A 2+8

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EXAMPLE 3.11 (a) Find the Thevenin’s equivalent circuit with respect to terminals a b for the circuit shown in Fig. 3.39 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

SOLUTION

Figure 3.40 (a) To ﬁnd Voc : Apply KCL at node 2 : V2 V2 30 + 1:5=0 60+20 40 ) V2 = 60 Volts ;VI Hence oc = 60 V2 0 = 60 60+20 60 =60 =45V 80

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To ﬁnd isc :

Applying KCL at node 2: V2 V2 30 + 1:5=0 20 40 ) V2 = 30V V2 isc = =1:5A 20 Voc 45 Therefore;Rt = = isc 1:5 =30Ω Figure 3.40 (a)

The Thevenin equivalent circuit with respect to the terminals ab is as shown in Fig. 3.40(a). (b) Let us now ﬁnd Thevenin resistance Rt by deactivating all the independent sources,

R t Rt

Rt =60Ωjj(40 + 20) Ω 60 = = 30 Ω (veriﬁed) 2

It is seen that, if only independent sources are present, it is easy to ﬁnd Rt by deactivating all the independent sources.

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EXAMPLE 3.12 Find the Thevenin equivalent for the circuit shown in Fig. 3.41 with respect to terminals a b.

Figure 3.41

SOLUTION To ﬁnd Voc = Vab : Applying KVL around the mesh of Fig. 3.42, we get

20+6i 2i +6i =0 ) i =2A

Since there is no current ﬂowing in 10 Ω resistor, Voc =6i =12V To ﬁnd Rt: (Refer Fig. 3.43)

Since both dependent and indepen- Figure 3.42 dent sources are present, Thevenin resistance is found using the relation, voc Rt = isc Applying KVL clockwise for mesh 1:

20+6i1 2i +6(i1 i2)=0 ) 12i1 6i2 =20+2i

Since i = i1 i2,weget

12i1 6i2 =20+2(i1 i2) ) 10i1 4i2 =20

Applying KVL clockwise for mesh 2:

10i2 +6(i2 i1)=0 Figure 3.43 )6i1 +16i2 =0

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Solving the above two mesh equations, we get 120 120 i2 = A ) isc = i2 = A 136 136 voc 12 Rt = = =13:6Ω isc 120 136

EXAMPLE 3.13 Find Vo in the circuit of Fig. 3.44 using Thevenin’s theorem.

Figure 3.44

SOLUTION To ﬁnd Voc : Since we are interested in the voltage across 2 kΩ resistor, it is removed from the circuit of Fig. 3.44 and so the circuit becomes as shown in Fig. 3.45.

Figure 3.45 By inspection, i1 =4mA Applying KVL to mesh 2: 3 i i 3i 12+6 10 ( 2 1)+3 10 2 =0 3 3 3 )12+6 10 i2 4 10 +3 10 i2 =0

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Solving, we get i2 =4mA Applying KVL to the path 4kΩ ! ab ! 3kΩ,weget

3 3 4 10 i1 + Voc 3 10 i2 =0 3 3 ) Voc =4 10 i1 +3 10 i2 3 3 3 3 =4 10 4 10 +3 10 4 10 = 28V

To ﬁnd Rt : Deactivating all the independent sources, we get the circuit diagram shown in Fig. 3.46.

Figure 3.46

Rt = Rab = 4 kΩ + (6 kΩjj3kΩ)=6kΩ Hence, the Thevenin equivalent circuit is as shown in Fig. 3.47.

Figure 3.47 Figure 3.48

If we connect the 2kΩresistor to this equivalent network, we obtain the circuit of Fig. 3.48. 3 Vo = i 2 10 28 3 = 2 10 =7V (6+2) 103

EXAMPLE 3.14 The wheatstone bridge in the circuit shown in Fig. 3.49 (a) is balanced when R2 = 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 R2 to 1204 Ω ?

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Figure 3.49(a)

SOLUTION To ﬁnd Voc : We are interested in the galavanometer current. Hence, it is removed from the circuit of Fig. 3.49 (a) to ﬁnd Voc and we get the circuit shown in Fig. 3.49 (b). 120 120 i1 = = A 900 + 600 1500 120 120 i2 = = A 1204 + 800 2004 Applying KVL clockwise along the path 1204Ω ! b a ! 900 Ω,weget

1204i2 Vt 900i1 =0 ) Vt = 1204i2 900i1 120 120 = 1204 900 2004 1500 =95:8mV Figure 3.49(b) To ﬁnd Rt : 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)

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Rt = Rab = 600jj900 + 800jj1204 900 600 1204 800 = + 1500 2004 = 840:64 Ω

Hence, the Thevenin equivalent circuit consists of the 95.8 mV source in series with 840.64Ω resistor. If we connect 30Ω resistor (galvanometer resistance) to this equivalent network, we obtain the circuit in Fig. 3.50. Figure 3.50

95:8 103 iG = = 110:03 A 840:64+30Ω

EXAMPLE 3.15 For the circuit shown in Fig. 3.51, ﬁnd the Thevenin’s equivalent circuit between terminals a and b.

Figure 3.51

SOLUTION With ab shorted, let Isc = I. The circuit after transforming voltage sources into their equivalent current sources is as shown in Fig 3.52. Writing node equations for this circuit,

At a :0:2Va 0:1 Vc + I =3 At c : 0:1Va +0:3 Vc 0:1 Vb =4 At b : 0:1Vc +0:2 Vb I =1

a b Va Vb As the terminals and are shorted = Figure 3.52 and the above equations become

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0:2Va 0:1 Vc + I =3 0:2Va +0:3 Vc =4 0:2Va 0:1 Vc 1=1

Solving the above equations, we get the short circuit current, I = Isc =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

0:2Va 0:1 Vc =3 0:1Va +0:3 Vc 0:1 Vb =4 0:1Vc +0:2 Vb =1

Solving the above equations, we get

Va = 30V and Vb = 20V

Hence, Vab =30 20=10V = Voc = Vt Voc 10 Therefore Rt = = = 10Ω Isc 1 The Thevenin’s equivalent is as shown in Fig 3.53 Figure 3.53 EXAMPLE 3.16 Refer to the circuit shown in Fig. 3.54. Find the Thevenin equivalent circuit at the terminals a b.

Figure 3.54

SOLUTION To begin with let us transform 3 A current source and 10 V voltage source. This results in a network as shown in Fig. 3.55 (a) and further reduced to Fig. 3.55 (b).

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Figure 3.55(a) Again transform the 30 V source and following the reduction procedure step by step from Fig. 3.55 (b) to 3.55 (d), we get the Thevenin’s equivalent circuit as shown in Fig. 3.56.

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. 3.57.

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Figure 3.57 SOLUTION Since the circuit has no independent sources, i =0when the terminals a b are open. There- fore, Voc =0. The onus is now to find Rt. Since Voc =0and isc =0, Rt cannot be determined from Voc Rt = . Hence, we choose to connect a source of1Aattheterminals a b as shown in Fig. isc 3.58. Then, after finding Vab, the Thevenin resistance is, Vab Rt = 1 KCL at node a : Va 2i Va + 1=0 5 10 Va Also;i= 10 Va Va 2 10 Va Hence; + 1=0 5 10 50 ) Va = V 13 Va 50 Hence;Rt = = Ω 1 13 Alternatively one could find Rt by connecting a 1V source at the terminals a b and then find 1 the current from b to a. Then Rt = . The concept of finding Rt by connecting a 1A source iba 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 Vab and Vab hence, Rt = . 1 For the present problem, the Thevenin equivalent circuit as seen between the terminals a b is shown in Fig. 3.58 (a).

Figure 3.58 Figure 3.58 (a)

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EXAMPLE 3.18 Determine the Thevenin equivalent circuit between the terminals a b for the circuit of Fig. 3.59.

Figure 3.59

SOLUTION As there are no independent sources in the circuit, we get Voc = Vt =0: To ﬁnd Rt, connect a 1V source to the terminals a b and measure the current I that ﬂows from b to a. (Refer Fig. 3.60 a). R 1 t = I Ω

Figure 3.60(a)

Applying KCL at node a: Vx I =0:5Vx + 4 Since;Vx =1V 1 we get, I =0:5+ =0:75 A 4 Figure 3.60(b) 1 Hence;Rt = =1:33 Ω 0:75 The Thevenin equivalent circuit is shown in 3.60(b). Alternatively, sticking to our strategy, let us connect 1A current source between the terminals Vab a b and then measure Vab (Fig. 3.60 (c)). Consequently, Rt = = Vab Ω: 1

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Applying KCL at node a:

Vx 0:5Vx + =1) Vx =1:33V 4 Vab Vx Hence Rt = = =1:33 Ω 1 1

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 iN in parallel with resistor RN , where iN is the short-circuit current through the terminals and RN 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 RN 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 voc RN = Rt and iN = . In fact, source transformation of Thevenin equivalent circuit leads to Rt 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 RN by circuit reduction techniques. (b) Find iN with sources activated. (2) If the network contains resistors, independent and dependent sources, follow the steps given below:

(a) Determine the short-circuit current iN with all sources activated.

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(b) Find the open-circuit voltage voc. voc (c) Rt = RN = iN (3) If the network contains only resistors and dependent sources, follow the procedure described below:

(a) Note that iN =0. (b) Connect 1A current source to the terminals a b and ﬁnd vab. vab (c) Rt = 1 Note: Also, since vt = voc and iN = isc

voc Rt = = RN isc

The open–circuit and short–circuit test are sufﬁcient to ﬁnd any Thevenin or Norton equivalent.

3.3.1 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 v1 and v2 and two independent current sources i1 and i2. 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= a0i + a1v1 + a2v2 + a3i1 + a4i2 (3.9) = a0i + b0 (3.10) where b0 = a1v1 + a2v2 + a3i1 + a4i2 = contribution to the terminal voltage v by independent sources within the linear network.

Let us now evaluate the values of constants a0 and b0.

(i) When the terminals x and y are open–circuited, i =0and v = voc = vt. Making use of this fact in equation 3.10, we ﬁnd that b0 = vt.

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(ii) When all the internal sources are deactivated, b0 =0. This enforces equation 3.10 to become v = a0i = Rti ) a0 = Rt

Figure 3.62 Current-driven circuit Figure 3.63 Thevenin’s equivalent circuit of Fig. 3.62 where Rt is the equivalent resistance of the linear network as viewed from the terminals x y. Also, a0 must be Rt in order to obey the ohm’s law. Substuting the values of a0 and b0 in equation 3.10, we ﬁnd that v = Rti + v1 which expresses the voltage-current relationship at terminals x y of the circuit in Fig. 3.63. Thus, the two circuits of Fig. 3.62 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. 3.64. The current ﬂowing into the circuit can be obtained by superposition as

i = c0v + d0 (3.11) where c0v is the contribution to i due to the external voltage source v and d0 contains the contributions to i due to all independent sources within the linear circuit. The constants c0 and d0 are determined as follows : (i) When terminals x y are short-circuited, v = 0 and i = isc. Hence from equation (3.11), we ﬁnd that i = d0 = isc, where isc is the short-circuit current ﬂowing out of terminal x, which is same as Norton current iN

Thus, d0 = iN Figure 3.64 Voltage-driven circuit

(ii) Let all the independent sources within the linear network be turned off, that is d0 =0. Then, equation (3.11) becomes i = c0v

Circuit Theorems j 191

For dimensional validity, c0 must have the dimension of conductance. This enforces c0 = 1 where Rt is the equivalent resistance of the Rt linear network as seen from the terminals x y. Thus, equation (3.11) becomes 1 i = v isc Rt 1 Figure 3.65 Norton’s equivalent of = v iN voltage driven circuit Rt

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. 3.64 and 3.65 are equivalents.

EXAMPLE 3.19 Find the Norton equivalent for the circuit of Fig. 3.66.

Figure 3.66

SOLUTION As a ﬁrst step, short the terminals a b. This results in a circuit diagram as shown in Fig. 3.67. Applying KCL at node a,weget 0 24 3+isc =0 4 ) isc =9A

To ﬁnd RN , deactivate all the independent sources, resulting in a circuit diagram as shown in Fig. 3.68 (a). We ﬁnd RN in the same way as Figure 3.67 Rt in the Thevenin equivalent circuit. 4 12 RN = =3Ω 4+12

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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. 3.69. Find the value of ib using Norton equivalent circuit. Take R = 667 Ω.

Figure 3.69

SOLUTION Since we want the current ﬂowing through R, remove R from the circuit of Fig. 3.69. The resulting circuit diagram is shown in Fig. 3.70. To ﬁnd iac or iN referring Fig 3.70(a) : 0 ia = =0A 1000 12 isc = A=2mA 6000 Figure 3.70

Figure 3.70(a)

Circuit Theorems j 193

To ﬁnd RN : The procedure for ﬁnding RN is same that of Rt in the Thevenin equivalent circuit. voc Rt = RN = isc

To ﬁnd voc, make use of the circuit diagram shown in Fig. 3.71. Do not deactivate any source. Applying KVL clockwise, we get Figure 3.71 12 + 6000ia + 2000ia + 1000ia =0 4 ) ia = A 3000 4 ) voc = ia 1000 = V 3 4 v ;Roc 3 Therefore N = = 3 = 667 Ω isc 2 10 The Norton equivalent circuit along with resistor R is as shown below: isc 2mA ib = = = 1mA 2 2

Figure : Norton equivalent circuit with load R

EXAMPLE 3.21 Find Io in the network of Fig. 3.72 using Norton’s theorem.

Figure 3.72

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SOLUTION We are interested in Io, hence the 2kΩresistor is removed from the circuit diagram of Fig. 3.72. The resulting circuit diagram is shown in Fig. 3.73(a).

Figure 3.73(a) Figure 3.73(b)

To ﬁnd iN or isc: Refer Fig. 3.73(b). By inspection, V1 =12V Applying KCL at node V2 :

V2 V1 V2 V2 V1 + + =0 6kΩ 2kΩ 3kΩ Substituting V1 =12V and solving, we get

V2 =6V V1 V2 V1 isc = + =5mA 3kΩ 4kΩ

To ﬁnd RN : Deactivate all the independent sources (refer Fig. 3.73(c)).

Figure 3.73(c) Figure 3.73(d)

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Referring to Fig. 3.73 (d), we get

RN = Rab =4kΩjj [3 kΩ + (6 kΩjj2kΩ)]=2:12 kΩ

Hence, the Norton equivalent circuit along with 2kΩresistor is as shown in Fig. 3.73(e).

isc RN Io = =2:57mA R + RN Figure 3.73(e)

EXAMPLE 3.22 Find Vo in the circuit of Fig. 3. 74.

Figure 3.74

SOLUTION Since we are interested in Vo, the voltage across 4kΩresistor, remove this resistance from the circuit. This results in a circuit diagram as shown in Fig. 3.75.

Figure 3.75

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To ﬁnd isc, short the terminals a b :

Circuit Theorems j 197

Constraint equation : i1 i2 = 4mA (3.12) KVL around supermesh : 3 3 4+2 10 i1 +4 10 i2 =0 (3.13) KVL around mesh 3: 3 3 8 10 (i3 i2)+2 10 (i3 i1)=0

Since i3 = isc, the above equation becomes, 3 3 8 10 (isc i2)+2 10 (isc i1)=0 (3.14)

Solving equations (3.12), (3.13) and (3.14) simultaneously, we get isc =0:1333 mA. To ﬁnd RN : Deactivate all the sources in Fig. 3.75. This yields a circuit diagram as shown in Fig. 3.76.

Figure 3.76

RN =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 Figure 3.76(a) network shown in Fig. 3.76(b).

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Vo = isc (RN jjR)

RN R = isc RN + 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. 3.77.

Figure 3.77

SOLUTION To ﬁnd isc:

Note that vab =0when the terminals a b are short-circuited. 5 Then i = =10mA 500 Therefore, for the right–hand portion of the circuit, isc = 10i = 100 mA.

Circuit Theorems j 199

To ﬁnd RN or Rt :

Writing the KVL equations for the left-hand mesh, we get

5 + 500i + vab =0 (3.15) Also for the right-hand mesh, we get

vab = 25(10i)=250i vab Therefore i = 250 Substituting i into the mesh equation (3.15), we get vab 5 + 500 + vab =0 250 ) vab = 5V voc vab 5 RN = Rt = = =50Ω isc isc 0:1 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. 3.78.

Figure 3.78

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SOLUTION Since there are no independent sources present in the network of Fig. 3.78, iN = isc =0. To ﬁnd RN , we inject a current of 1A between the terminals a b. This is illustrated in Fig. 3.79.

Figure 3.79 Figure 3.79(a) Norton equivalent circuit

KCL at node 1: v1 v1 v2 1= + 100 50 ) 0:03v1 0:02v2 =1

KCL at node 2: v2 v2 v1 + +0:1v1 =0 200 50 ) 0:08v1 +0:025v2 =0 Solving the above two nodal equations, we get

v1 =10:64 volts ) voc =10:64 volts voc 10:64 Hence;RN = Rt = = =10:64 Ω 1 1 Norton equivalent circuit for the network shown in Fig. 3.78 is as shown in Fig. 3.79(a).

EXAMPLE 3.25 Find the Thevenin and Norton equivalent circuits for the network shown in Fig. 3.80 (a).

Figure 3.80(a)

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SOLUTION To ﬁnd Voc : Performing source transformation on 5A current source, we get the circuit shown in Fig. 3.80 (b). Applying KVL around Left mesh :

50+2ia 20+4ia =0 70 ) ia = A 6 Applying KVL around right mesh:

20+10ia + Voc 4ia =0 ) Voc = 90 V Figure 3.80(b)

To ﬁnd isc(referring Fig 3.80 (c)) : KVL around Left mesh :

50+2ia 20+4(ia isc)=0 ) 6ia 4isc =70 KVL around right mesh :

4(isc ia)+20+10ia =0 ) 6ia +4isc = 20

Figure 3.80(c)

Solving the two mesh equations simultaneously, we get isc = 11:25 A voc 90 Hence, Rt = RN = = =8Ω isc 11:25 Performing source transformation on Thevenin equivalent circuit, we get the norton equivalent circuit (both are shown below).

Thevenin equivalent circuit Norton equivalent circuit

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EXAMPLE 3.26 If an 8 kΩ load is connected to the terminals of the network in Fig. 3.81, VAB =16V.Ifa2kΩ load is connected to the terminals, VAB =8V. Find VAB if a 20 kΩ load is connected across the terminals.

SOLUTION Figure 3.81

Applying KVL around the mesh, we get (Rt + RL) I = Voc

If RL =2kΩ;I=10mA) Voc =20+0:01Rt If RL =10kΩ;I=6mA) Voc =60+0:006Rt

Solving, we get Voc = 120 V, Rt =10kΩ. V R ;I oc 120 If L =20kΩ = = 3 3 =4mA (RL + Rt) (20 10 +10 10 )

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. 3.82. Circuit A is replaced by its Thevenin equivalent circuit as seen from a and b (Fig 3.83). RL We wish to ﬁnd the value of the load such that Figure 3.82 Circuit A with load RL the maximum power is delivered to it. The power that is delivered to the load is given by

2 2 Vt p = i RL = RL (3.16) Rt + RL

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Assuming that Vt and Rt are ﬁxed for a given source, the maximum power is a function of RL. In order to determine the value of RL that maximizes p, we differentiate p with respect to RL and equate the derivative to zero. " # 2 dp 2 (Rt + RL) 2(Rt + RL) = Vt 2 =0 dRL (RL + Rt) which yields RL = Rt (3.17)

To conﬁrm that equation (3.17) is a maximum, d2p it should be shown that 2 < 0. Hence, maxi- dRL mum power is transferred to the load when RL is equal to the Thevenin equivalent resistance Rt. The maximum power transferred to the load is obtained by substituting RL = Rt in equation 3.16. Accordingly, Figure 3.83 Thevenin equivalent circuit is substituted for circuit A 2 2 Vt RL Vt Pmax = 2 = (2RL) 4RL

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 RL is equal to the Thevenin resistance Rt.

EXAMPLE 3.27 Find the load RL that will result in maximum power delivered to the load for the circuit of Fig. 3.84. Also determine the maximum power Pmax.

Figure 3.84

SOLUTION Disconnect the load resistor RL. 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.

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180 i = =1A 150+30 Voc = Vt = 150 i = 150 V

To ﬁnd Rt, deactivate the 180 V source. This results in the circuit diagram of Fig. 3.85(b).

Rt = Rab =30Ωjj150 Ω 30 150 Figure 3.85(a) = =25Ω 30 + 150 The Thevenin equivalent circuit connected to the load resistor is shown in Fig. 3.86. Maximum power transfer is obtained when RL = Rt =25Ω: Then the maximum power is V 2 2 P t (150) max = = Figure 3.85(b) 4RL 4 25 =2:25 Watts The Thevenin source Vt actually provides a total power of P i t = 150 150 = 150 25+25 = 450 Watts

Thus, we note that one-half the power is dissipated in RL. Figure 3.86 EXAMPLE 3.28 Refer to the circuit shown in Fig. 3.87. Find the value of RL for maximum power transfer. Also ﬁnd the maximum power transferred to RL.

Figure 3.87

Circuit Theorems j 205

SOLUTION Disconnecting RL, results in a circuit diagram as shown in Fig. 3.88(a).

Figure 3.88(a)

To ﬁnd Rt, deactivate all the independent voltage sources as in Fig. 3.88(b).

Figure 3.88(b) Figure 3.88(c)

Rt = Rab =6kΩjj6kΩjj6kΩ =2kΩ

To ﬁnd Vt : Refer the Fig. 3.88(d). Constraint equation :

V3 V1 =12V

By inspection, V2 =3V KCL at supernode :

V3 V2 V1 V1 V2 + + =0 6k 6k 6k V3 3 V3 12 V3 12 3 ) + + =0 6k 6k 6k Figure 3.88(d)

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) V3 3+V3 12 + V3 15=0

) 3V3 =30

) V3 =10

) Vt = Vab = V3 =10V

Figure 3.88(e)

The Thevenin equivalent circuit connected to the load resistor RL is shown in Fig. 3.88(e).

2 Pmax = i RL 2 Vt = RL 2RL =12:5mW

Alternate method : It is possible to find Pmax, without finding the Thevenin equivalent circuit. However, we have to find Rt. For maximum power transfer, RL = Rt =2kΩ. Insert the value of RL in the original circuit given in Fig. 3.87. Then use any circuit reduction technique of your choice to find power dissipated in RL. Refer Fig. 3.88(f). By inspection we find that, V2 =3V. Constraint equation :

V3 V1 =12 ) V1 = V3 12

KCL at supernode :

V3 V2 V1 V2 V3 V1 + + + =0 6k 6k 2k 6k V3 3 V3 12 3 V3 V3 12 ) + + + =0 6k 6k 2k 6k ) V3 3+V3 15+3V3 + V3 12=0

) 6V3 =30

) V3 =5 V Figure 3.88(f) 2 V3 25 Hence;Pmax = = = 12:5mW RL 2k

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EXAMPLE 3.29 Find RL for maximum power transfer and the maximum power that can be transferred in the network shown in Fig. 3.89.

Figure 3.89

SOLUTION Disconnect the load resistor RL. This results in a circuit as shown in Fig. 3.89(a).

Figure 3.89(a)

To ﬁnd Rt, let us deactivate all the independent sources, which results the circuit as shown in Fig. 3.89(b). Rt = Rab =2kΩ+3kΩ+5kΩ=10kΩ

For maximum power transfer RL = Rt =10kΩ. Let us next ﬁnd Voc or Vt. Refer Fig. 3.89 (c). By inspection, i1 = 2 mA & i2 =1mA.

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Figure 3.89(b) Applying KVL clockwise to the loop 5kΩ! 3kΩ! 2kΩ! a b,weget

5k i2 +3k(i1 i2)+2k i1 + Vt =0 3 3 3 3 3 3 3 )510 1 10 +310 2 10 1 10 +210 2 10 +Vt =0

)5 9 4+Vt =0 ) Vt =18V:

The Thevenin equivalent circuit with load resistor RL is as shown in Fig. 3.89 (d). 18 i = =0:9mA (10+10) 103 Then, 2 Pmax = PL =(0:9mA) 10 kΩ =8:1mW

Figure 3.89(c) Figure 3.89(d)

EXAMPLE 3.30 Find the maximum power dissipated in RL. Refer the circuit shown in Fig. 3.90.

Figure 3.90

Circuit Theorems j 209

SOLUTION Disconnecting the load resistor RL from the original circuit results in a circuit diagram as shown in Fig. 3.91.

Figure 3.91 As a first step in the analysis, let us find Rt. While finding Rt, we have to deactivate all the independent sources. This results in a network as shown in Fig 3.91 (a) :

Figure 3.91(a)

Rt = Rab = [140 Ωjj60 Ω] + 8 Ω 140 60 = +8=50Ω: 140+60 For maximum power transfer, RL = Rt =50Ω. Next step in the analysis is to ﬁnd Vt. Refer Fig 3.91(b), using the principle of current division,

i R2 i1 = R1 + R2 20 170 = =17A 170+30 i R1 20 30 i2 = = R1 + R2 170+30 600 = =3A 200 Figure 3.91(a)

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Applying KVL clockwise to the loop comprising of 50 Ω ! 10 Ω ! 8Ω! a b,weget

50i2 10i1 +8 0+Vt =0 ) 50(3) 10 (17) + Vt =0 ) Vt =20V

The Thevenin equivalent circuit with load resistor RL is as shown in Fig. 3.91(c). 20 Figure 3.91(c) iT = =0:2A 50+50 2 Pmax = iT 50=0:04 50 = 2W

EXAMPLE 3.31 Find the value of RL for maximum power transfer in the circuit shown in Fig. 3.92. Also ﬁnd Pmax.

Figure 3.92

SOLUTION Disconnecting RL from the original circuit, we get the network shown in Fig. 3.93.

Figure 3.93

Circuit Theorems j 211

Let us draw the Thevenin equivalent circuit as seen from the terminals a b and then insert the value of RL = Rt between the terminals a b. To ﬁnd Rt, let us deactivate all independent sources which results in the circuit as shown in Fig. 3.94.

Figure 3.94

Rt = Rab =8Ωjj2Ω 8 2 = =1:6Ω 8+2

Next step is to ﬁnd Voc or Vt. By performing source transformation on the circuit shown in Fig. 3.93, we obtain the circuit shown in Fig. 3.95.

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 60 ) i = =6A 10

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Again applying KVL clockwise to the path 2Ω! 10 V ! a b,weget 2i 10 Vt =0 ) Vt =2i 10 = 2(6) 10=2V

The Thevenin equivalent circuit with load resistor RL is as shown in Fig. 3.95 (a).

2 Pmax = iT RL 2 Vt Figure 3.95(a) Thevenin equivalent = = 625 mW 4Rt circuit

EXAMPLE 3.32 Find the value of RL for maximum power transfer. Hence ﬁnd Pmax.

Figure 3.96

SOLUTION Removing RL from the original circuit gives us the circuit diagram shown in Fig. 3.97.

Figure 3.97

To ﬁnd Voc : KCL at node A : 0 0 ia 0:9+10ia =0 ) i0 : a =01A 0 Hence;Voc =3 10ia =3 10 0:1=3V

Circuit Theorems j 213

To ﬁnd Rt, we need to compute isc with all independent sources activated. KCL at node A:

00 00 ia 0:9+10ia =0 00 ) ia =0:1A 00 Hence isc =10ia =10 0:1=1A Voc 3 Rt = = =3Ω isc 1

Hence, for maximum power transfer RL = Rt =3Ω. The Thevenin equivalent circuit with RL =3Ω inserted between the terminals ab gives the network shown in Fig. 3.97(a).

3 iT = =0:5A 3+3 2 Pmax = iT RL 2 =(0:5) 3 = 0.75 W Figure 3.97(a) EXAMPLE 3.33 Find the value of RL in the network shown that will achieve maximum power transfer, and determine the value of the maximum power.

Figure 3.98(a) SOLUTION Removing RL from the circuit of Fig. 3.98(a), we get the circuit of Fig 3.98(b). Applying KVL clockwise we get 3 0 12+2 10 i +2Vx =0 0 3 Also Vx =1 10 i

Hence; 12+2 103i +2 1 103i =0 12 i = =3mA Figure 3.98(b) 4 103

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Applying KVL to loop 1kΩ ! 2Vx ! b a,weget

3 0 1 10 i +2Vx Vt =0 ) V 3i 3i t =1 10 +2 1 10 3 3 i = 1 10 +2 10 3 3 =3 10 3 10 =9V

To find Rt, we need to find isc. While finding isc, none of the independent sources must be deactivated. Applying KVL to mesh 1:

00 12 + Vx +0=0 00 ) Vx =12 3 ) 1 10 i1 =12 ) i1 =12mA

Applying KVL to mesh 2:

3 00 1 10 i2 +2Vx =0 3 ) 1 10 i2 = 24 i2 = 24 mA Applying KCL at node a:

isc = i1 i2 =12+24=36mA Vt Voc Hence;Rt = = isc isc 9 = 36 103 = 250 Ω

For maximum power transfer, RL = Rt = 250 Ω. Thus, the Thevenin equivalent circuit with RL is as shown in Fig 3.98 (c) : 9 9 iT = = A 250 + 250 500 2 Pmax = iT 250 2 9 = 250 500 Figure 3.98 (c) Thevenin equivalent circuit = 81 mW

Circuit Theorems j 215

EXAMPLE 3.34 The variable resistor RL in the circuit of Fig. 3.99 is adjusted untill it absorbs maximum power from the circuit.

(a) Find the value of RL. (b) Find the maximum power.

Figure 3.99

SOLUTION Disconnecting the load resistor RL from the original circuit, we get the circuit shown in Fig. 3.99(a).

Figure 3.99(a) KCL at node v1 :

0 v1 100 v1 13ia v1 v2 + + =0 (3.18) 2 5 4 Constraint equations :

0 100 v1 ia = (3.19) 2 v2 v1 0 = va (applying KCL at v2) (3.20) 4 0 va = v1 v2 (potential across 4Ω) (3.21)

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From equations (3.20) and (3.21), we have

v2 v1 = v1 v2 4 ) v2 v1 =4v1 4v2 ) 5v1 5v2 =0 ) v1 = v2 (3.22)

Making use of equations (3.19) and (3.22) in (3.18), we get

(100 v1) v2 v1 100 13 v1 v1 + 2 + =0 2 5 4 (100 v1) ) 5(v1 100) + 2 v1 13 =0 2 ) 5v1 500+2v1 13 100+13v1 =0

) 20v1 = 1800 ) v1 = 90 Volts Hence;vt = v2 = v1 = 90 Volts voc vt We know that, Rt = = isc isc The short circuit current is calculated using the circuit shown below:

00 100 v1 Here ia = 2 Applying KCL at node v1 :

v1 100 v1 13ia v1 0 + + =0 2 5 4 (100 v1) v1 v1 100 13 v1 ) + 2 + =0 2 5 4

Circuit Theorems j 217

00 Solving we get v1 =80volts = va Applying KCL at node a :

0 v1 00 + isc = va 4 v1 00 ) isc = + va 4 80 = + 80 = 100 A 4 voc vt Hence;Rt = = isc isc 90 = =0:9Ω 100 Hence for maximum power transfer,

RL = Rt = 0:9Ω

The Thevenin equivalent circuit with RL =0:9Ω is as shown. 90 90 it = = 0:9+0:9 1:8 2 Pmax = it 0:9 2 90 = 0:9=2250 W 1:8

EXAMPLE 3.35 Refer to the circuit shown in Fig. 3.100 :

(a) Find the value of RL for maximum power transfer.

(b) Find the maximum power that can be delivered to RL.

Figure 3.100

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SOLUTION Removing the load resistor RL, we get the circuit diagram shown in Fig. 3.100(a). Let us proceed to ﬁnd Vt.

Figure 3.100(a) Constraint equation : 0 ia = i1 i3 KVL clockwise to mesh 1:

200+1(i1 i2) + 20 (i1 i3)+4i1 =0 ) 25i1 i2 20i3 = 200 KVL clockwise to mesh 2: 0 14ia +2(i2 i3)+1(i2 i1)=0 ) 14 (i1 i3)+2(i2 i3)+1(i2 i1)=0 ) 13i1 +3i2 16i3 =0 KVL clockwise to mesh 3:

2(i3 i2) 100+3i3 +20(i3 i1)=0 )20i1 2i2 +25i3 = 100 Solving the mesh equations, we get

i1 = 2:5A;i3 =5A Applying KVL clockwise to the path comprising of a b ! 20 Ω,weget 0 Vt 20ia =0 0 ) Vt =20ia =20(i1 i3) =20(2:5 5) = 150 V

Circuit Theorems j 219

Next step is to ﬁnd Rt. Voc Vt Rt = = isc isc

00 00 When terminals a b are shorted, ia =0. Hence, 14 ia is also zero.

KVL clockwise to mesh 1:

200+1(i1 i2)+4i1 =0 ) 5i1 i2 = 200

KVL clockwise to mesh 2:

2(i2 i3)+1(i2 i1)=0 )i1 +3i2 2i3 =0

KVL clockwise to mesh 3:

100+3i3 +2(i3 i2)=0 )2i2 +5i3 = 100

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Solving the mesh equations, we ﬁnd that

i1 = 40A;i3 = 20A; ) isc = i1 i3 = 60A Vt 150 Rt = = =2:5Ω isc 60 For maximum power transfer, RL = Rt =2:5Ω. The Thevenin equivalent circuit with RL is as shown below :

2 Pmax = i1RL 2 150 = 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 RL, with voltage vL and current iL, is connected to it. Find the values of RL, vL and iL if (a) vLiL is a maximum, (b) vL is a maximum and (c) iL is a maximum.

SOLUTION Load current calculation: r 10 10W to 250 Ω corresponds to iL = 250 = 200r mA 20 20W to 80 Ω corresponds to iL = 80 = 500 mA Using the formula for division of current between two parallel branches : i R1 i2 = R1 + R2 IN RN In the present context, 0:2= (3.23) RN + 250 IN RN and 0:5= (3.24) RN +80

Circuit Theorems j 221

Solving equations (3.23) and (3.24), we get

IN =1:7A RN =33:33 Ω

(a) If vLiL is maximum,

RL = RN =33:33 Ω 33:33 iL =1:7 33:33+33:33 = 850 mA 3 vL = iLRL = 850 10 33:33 =28:33 V (b) vL = IN (RN jjRL) is a maximum when RN jjRL is a maximum, which occurs when RL = 1. Then, iL =0and

vL =1:7 RN =1:7 33:33 =56:66 V IN RN (c) iL = is maxmimum when RL =0Ω RN + RL ) iL =1:7A and vL =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 3.101 Thevenin equivalent circuit Figure 3.102 Norton equivalent circuit

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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. 3.101 and 3.102. The Thevenin and Norton forms are equivalent if the relations

(a) Zt = ZN (b) Vt = ZN IN 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 Vt = Voc at the terminals. 3. (a) If the circuit contains only impedances and independent sources, then deactivate all the independent sources and then ﬁnd Zt by using circuit reduction techniques. (b) If the circuit contains impedances, independent sources and dependent sources, then either short–circuit the terminals and determine Isc from which

Voc Zt = Isc 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 V Z t = 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 IN = Isc. (iii) (a) If the circuit contains only impedances and independent sources, then deactivate all the independent sources and then find ZN = Zt by using circuit reduction techniques. (b) If the circuit contains impedances, independent sources and one or more dependent Voc sources, find the open–circuit voltage at the terminals, Voc, so that ZN = Zt = : Isc

Circuit Theorems j 223

EXAMPLE 3.37 Find the Thevenin and Norton equivalent circuits at the terminals a b for the circuit in Fig. 3.103.

Figure 3.103

SOLUTION As a ﬁrst step in the analysis, let us ﬁnd Vt:

Using the principle of current division, 8(4/0 ) 32 Io = = 8+j10 j5 8+j5 j320 Vt = Io(j10) = =33:92 /58 V 8+j5 To ﬁnd Zt, deactivate all the independent sources. This results in a circuit diagram as shown in Fig. 3.103 (a).

Figure 3.103(a) Figure 3.103(b) Thevenin equivalent circuit

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Zt = j10jj (8 j5) Ω (j10)(8 j5) = j10+8 j5 =10/26 Ω

The Thevenin equivalent circuit as viewed from the terminals a b is as shown in Fig 3.103(b). Performing source transformation on the Thevenin equivalent circuit, we get the Norton equivalent circuit. Figure : Norton equivalent circuit

V : I t 33 92 /58 N = = Zt 10 /26 =3:392 /32 A ZN = Zt =10/26 Ω

EXAMPLE 3.38 Find vo using Thevenin’s theorem. Refer to the circuit shown in Fig. 3.104.

Figure 3.104

SOLUTION Let us convert the circuit given in Fig. 3.104 into a frequency domain equiavalent or phasor circuit (shown in Fig. 3.105(a)). ! =1

10 cos (t 45 ) ! 10 /45 V 5 sin (t +30 ) = 5 cos (t 60 ) ! 5/60 V

L =1H! j!L= j 1 1=j1Ω 1 1 C =1F! = = j1Ω j!C j 1 1

Circuit Theorems j 225

Figure 3.105(a)

Disconnecting the capicator from the original circuit, we get the circuit shown in Fig. 3.105(b). This circuit is used for ﬁnding Vt.

Figure 3.105(b) KCL at node a :

Vt 10 /45 Vt 5/60 + =0 3 j1 Solving; Vt =4:97 /40:54 Z To ﬁnd t deactivate all the independent sources Figure 3.105(c) in Fig. 3.105(b). This results in a network as shown in Fig. 3.105(c) : Zt = Zab =3Ωjjj1Ω j3 3 = = (1 + j3) Ω 3+j 10 The Thevenin equivalent circuit along with the capicator is as shown in Fig 3.105(d). Vt Vo = (j1) Zt j1 4:97 /40:54 = (j1) 0:3(1 + j3) j1 =15:73 /247:9 V Hence;vo =15:73 cos (t + 247:9 )V Figure 3.105(d) Thevenin equivalent circuit

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EXAMPLE 3.39 Find the Thevenin equivalent circuit of the circuit shown in Fig. 3.106.

Figure 3.106

SOLUTION Since terminals a b are open,

Va = Is 10 =20/0 V

Applying KVL clockwise for the mesh on the right hand side of the circuit, we get

3Va +0(j10) + Voc Va =0

Voc =4Va =80/0 V

Let us transform the current source with 10 Ω parallel resistance to a voltage source with 10 Ω series resistance as shown in ﬁgure below :

To ﬁnd Zt, the independent voltage source is deactivated and a current source of I Ais connected at the terminals as shown below :

Circuit Theorems j 227

Applying KVL clockwise we get,

0 0 Va 3Va j10I + Vo =0 0 )4Va j10I + Vo =0 0 Since Va =10I we get 40I j10I = Vo V ; Z o j Hence t = I =40+ 10Ω

Hence the Thevenin equivalent circuit is as shown in Fig 3.106(a) : Figure 3.106(a)

EXAMPLE 3.40 Find the Thevenin and Norton equivalent circuits for the circuit shown in Fig. 3.107.

Figure 3.107

SOLUTION The phasor equivalent circuit of Fig. 3.107 is shown in Fig. 3.108. KCL at node a :

Voc 2Voc Voc 10 + =0 j10 j5 100 100 ) Voc = j = /90 V 3 3

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Figure 3.108

To ﬁnd Isc, short the terminals a b of Fig. 3.108 as in Fig. 3.108(a).

Figure 3.108 (a) Figure 3.108 (b)

Since Voc =0, the above circuit takes the form shown in Fig 3.108 (b). Isc =10/0 A 100 V /90 ; Z oc 3 10 Hence t = = = / 90 Ω Isc 10 /0 3 The Thevenin equivalent and the Norton equivalent circuits are as shown below.

Figure Thevenin equivalent Figure Norton equivalent

EXAMPLE 3.41 Find the Thevenin and Norton equivalent circuits in frequency domain for the network shown in Fig. 3.109.

Circuit Theorems j 229

Figure 3.109

SOLUTION Let us ﬁnd Vt = Vab using superpostion theorem. (i) Vab due to 100 /0

100 /0 100 I1 = = A j300 + j100 j200

Vab1 = I1 (j100) 100 = (j100) = 50 /0 Volts j200 (ii) Vab due to 100 /90

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100 /90 I2 = j100 j300

Vab2 = I2 (j300) 100 /90 = (j300) = j150 V j100 j300

Hence; Vt = Vab1 + Vab2 = 50 + j150 = 158:11 /108:43 V

To ﬁnd Zt, deactivate all the independent sources.

Zt = j100 Ωjj j300 Ω j100(j300) = = j150 Ω j100 j300 Hence the Thevenin equivalent circuit is as shown in Fig. 3.109(a). Performing source transformation on the Thevenin equivalent circuit, we get the Norton equivalent circuit. V : : I t 158 11 /108 43 : : N = = =1054 /18 43 A Zt 150 /90 ZN = Zt = j150 Ω

The Norton equivalent circuit is as shown in Fig. 3.109(b).

Figure 3.109(a) Figure 3.109(b)

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3.6 Maximum power transfer theorem

We have earlier shown that for a resistive network, maximum power is transferred from a source to the load, when the load resistance is set equal to the Thevenin resistance with Thevenin equivalent source. Now we extend this result to the ac circuits.

Figure 3.110 Linear circuit Figure 3.111 Thevenin equivalent circuit

In Fig. 3.110, the linear circuit is made up of impedances, independent and dependent sources. This linear circuit is replaced by its Thevenin equivalent circuit as shown in Fig. 3.111. The load impedance could be a model of an antenna, a TV, and so forth. In rectangular form, the Thevenin impedance Zt and the load impedance ZL are

Zt = Rt + jXt and ZL = RL + jXL

The current through the load is

Vt Vt I = = Zt + ZL (Rt + jXt)+(RL + jXL)

The phasors I and Vt are the maximump values. The corresponding RMS values are obtained by dividing the maximum values by 2. Also, the RMS value of phasor current ﬂowing in the load must be taken for computing the average power delivered to the load. The average power delivered to the load is given by

1 2 P = jIj RL 2 2 RL jVtj 2 = 2 2 (3.25) (Rt + RL) (Xt + XL)

Our idea is to adjust the load parameters RL and XL so that P is maximum. To do this, we @P @P get and equal to zero. @RL @XL

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2 @P jVtj RL (Xt + XL) = h i2 @XL 2 2 (Rt + RL) +(Xt + XL) h i 2 2 2 @P jVtj (Rt + RL) +(Xt + XL) 2RL (Rt + RL) = h i2 @RL 2 2 2 (Rt + RL) +(Xt + XL)

@P Setting = 0 gives @XL XL = Xt (3.26) @P and Setting @R = 0 gives L q 2 2 RL = Rt +(Xt + XL) (3.27) Combining equations (3.26) and (3.27), we can conclude that for maximum average power transfer, ZL must be selected such that XL = Xt and RL = Rt. That is the maximum average power of a circuit with an impedance Zt that is obtained when ZL is set equal to complex conjugate of Zt. Setting RL = Rt and XL = Xt in equation (3.25), we get the maximum average power as 2 jVtj P = 8Rt In a situation where the load is purely real, the condition for maximum power transfer is obtained by putting XL =0in equation (3.27). That is, q 2 2 RL = Rt + Xt = jZtj Hence for maximum average power transfer to a purely resistive load, the load resistance is equal to the magnitude of Thevenin impedance.

3.6.1 Maximum Power Transfer When Z is Restricted Maximum average power can be delivered to ZL only if ZL = Zt . There are few situations in which this is not possible. These situations are described below :

(i) RL and XL may be restricted to a limited range of values. With this restriction, qchoose XL as close as possible to Xt and then adjust RL as close as possible to 2 2 Rt +(XL + Xt) :

(ii) Magnitude of ZL can be varied but its phase angle cannot be. Under this restriction, greatest amount of power is transferred to the load when [ZL]=jZtj.

Zt is the complex conjugate of Zt.

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EXAMPLE 3.42 Find the load impedance that transfers the maximum power to the load and determine the maximum power quantity obtained for the circuit shown in Fig. 3.112.

Figure 3.112

SOLUTION We select, ZL = Zt for maximum power transfer.

Hence ZL =5+j6 10 /0 I = =1/0 5+5 Hence, the maximum average power transfered to the load is

1 2 P = jIj RL 2 1 2 = (1) 5=2:5W 2

EXAMPLE 3.43 Find the load impedance that transfers the maximum average power to the load and determine the maximum average power transferred to the load ZL shown in Fig. 3.113.

Figure 3.113

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SOLUTION The ﬁrst step in the analysis is to ﬁnd the Thevenin equivalent circuit by disconnecting the load ZL. This leads to a circuit diagram as shown in Fig. 3.114.

Figure 3.114

Hence Vt = Voc =4/0 3 =12/0 Volts(RMS)

To ﬁnd Zt, let us deactivate all the independent sources of Fig. 3.114. This leads to a circuit diagram as shown in Fig 3.114 (a):

Zt =3+j4Ω

Figure 3.114 (a) Figure 3.115

The Thevenin equivalent circuit with ZL is as shown in Fig. 3.115. For maximum average power transfer to the load, ZL = Zt =3 j4. 12 /0 It = =2/0 A(RMS) 3+j4+3 j4 Hence, maximum average power delivered to the load is 2 P = jItj RL =4(3)=12W 1 It may be noted that the scaling factor is not taken since the phase current is already 2 expressed by its RMS value.

Circuit Theorems j 235

EXAMPLE 3.44 Refer the circuit given in Fig. 3.116. Find the value of RL that will absorb the maximum average power.

Figure 3.116

SOLUTION Disconnecting the load resistor RL from the original circuit diagram leads to a circuit diagram as shown in Fig. 3.117.

Figure 3.117

Vt = Voc = I1 (j20) 150 /30 j20 = (40 j30 + j20) =72:76 /134 Volts:

To ﬁnd Zt, let us deactivate all the independent sources present in Fig. 3.117 as shown in Fig 3.117 (a).

Zt =(40 j30) jjj20 j20 (40 j30) = =(9:412 + j22:35) Ω j20+40 j30

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The Value of RL that will absorb the maximum average power is p 2 2 RL = jZtj = (9:412) + (22:35) =24:25 Ω

The Thevenin equivalent circuit with RL inserted is as shown in Fig 3.117 (b). Maximum average power absorbed by RL is

1 2 Figure 3.117 (a) Pmax = jItj RL 2 72:76 /134 where It = (9:412 + j22:35+24:25) =1:8 /100:2 A 1 2 ) Pmax = (1:8) 24:25 2 = 39:29 W

Figure 3.117 (b) Thevenin equivalent circuit

EXAMPLE 3.45 For the circuit of Fig. 3.118: (a) what is the value of ZL that will absorb the maximum average power? (b) what is the value of maximum power?

Figure 3.118

SOLUTION Disconnecting ZL from the original circuit we get the circuit as shown in Fig. 3.119. The ﬁrst step is to ﬁnd Vt.

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Vt = Voc = I1 (j10) 120 /0 = (j10) 10 + j15 j10

= 107:33 /116:57 V

The next step is to ﬁnd Zt. This re- quires deactivating the independent Figure 3.119 voltage source of Fig. 3.119.

Zt = (10 + j15) jj (j10)

j10 (10 + j15) = j10+10+j15

=8 j14 Ω

The value of ZL for maximum average power absorbed is

Zt =8+j14 Ω

The Thevenin equivalent circuit along with ZL =8+j14 Ω is as shown below:

107:33 / 116:57 It = 8 j14+8+j14 107:33 = /116:57 A 16 1 2 Hence;Pmax = jItj RL 2 2 1 107:33 = 8 2 16 = 180 Walts

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EXAMPLE 3.46 (a) For the circuit shown in Fig. 3.120, what is the value of ZL that results in maximum average power that will be transferred to ZL ? What is the maximum power ? (b) Assume that the load resistance can be varied between 0 and 4000 Ω and the capacitive reactance of the load can be varied between 0 and 2000 Ω. What settings of RL and XC transfer the most average power to the load ? What is the maximum average power that can be transferred under these conditions?

Figure 3.120

SOLUTION (a) If there are no constraints on RL and XL, the load indepedance ZL = Zt = (3000j4000) Ω. Since the voltage source is given in terms of its RMS value, the average maximum power delivered to the load is 2 Pmax = jItj RL 10 /0 where It = 3000 + j4000 + 3000 j4000 10 = A 2 3000 2 ) Pmax = jItj RL 100 = 3000 4 (3000)2 = 8:33 mW

(b) Since RL and XC are restricted, we ﬁrstq set XC as close to 4000 Ω as possible; hence 2 2 XC = 2000 Ω.NextwesetRL as close to Rt +(XC + XL) as possible. q 2 2 Thus, RL = 3000 +(2000 + 4000) = 3605:55 Ω Since RL can be varied between 0 to 4000 Ω, we can set RL to 3605:55 Ω. Hence ZL is adjusted to a value ZL = 3605:55 j2000 Ω:

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10 /0 It = 3000 + j4000 + 3605:55 j2000 =1:4489 /16:85 mA

The maximum average power delivered to the load is

2 Pmax = jItj RL 3 2 = 1:4489 10 3605:55 = 7:57 mW

Note that this is less than the power that can be delivered if there are no constraints on RL and XL.

EXAMPLE 3.47 A load impedance having a constant phase angle of 45 is connected across the load terminals a and b in the circuit shown in Fig. 3.121. The magnitude of ZL is varied until the average power delivered, which is the maximum possible under the given restriction.

(a) Specify ZL in rectangular form. (b) Calculate the maximum average power delivered under this condition.

Figure 3.121

SOLUTION Since the phase angle of ZL is ﬁxed at 45 , for maximum power transfer to ZL it is mandatory that jZ j jZ j L = pt = (3000)2 + (4000)2 = 5000 Ω: Hence; ZL = jZLj /45 5000 5000 = p j p 2 2

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10 /0 It = (3000 + 3535:53) + j(4000 3535:53) =1:526 /4:07 mA 2 Pmax = jItj RL 3 2 = 1:526 10 3535:53 = 8:23 mW

This power is the maximum average power that can be delivered by this circuit to a load impedance whose angle is constant at 45. Again this quantity is less than the maximum power that could have been delivered if there is no restriction on ZL. In example 3.46 part (a), we have shown that the maximum power that can be delivered without any restrictions on ZL is 8.33 mW.

3.7 Reciprocity theorem

The reciprocity theorem states that in a linear bilateral single source circuit, the ratio of excitation to response is constant when the positions of excitation and response are interchanged.

Conditions to be met for the application of reciprocity theorem : (i) The circuit must have a single source. (ii) Initial conditions are assumed to be absent in the circuit. (iii) Dependent sources are excluded even if they are linear. (iv) When the positions of source and response are interchanged, their directions should be marked same as in the original circuit.

EXAMPLE 3.48 Find the current in 2Ωresistor and hence verify reciprocity theorem.

Figure 3.122

Circuit Theorems j 241

SOLUTION The circuit is redrawn with markings as shown in Fig 3.123 (a).

Figure 3.123 (a) 1 1 1 Then;R1 =(8 +2 ) =1:6Ω R2 =1:6+4=5:6Ω 1 1 1 R3 =(5:6 +4 ) =2:3333Ω 20 Current supplied by the source = =3:16 A 4+2:3333 4 Current in branch ab = Iab =3:16 =1:32 A 4+4+1:6 8 Current in 2Ω;I1 =1:32 =1:05 A 10 Veriﬁcation using reciprocity theorem The circuit is redrawn by interchanging the position of excitation and response as shown in Fig 3.123 (b).

Figure 3.123 (b) Solving the equivalent resistances,

R4 =2Ω;R5 =6Ω;R6 =3:4286Ω Now the current supplied by the source 20 = =3:6842A 3:4286 + 2

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Therefore, 8 Icd =3:6842 =2:1053A 8+6 2:1053 I2 = =1:05A 2

As I1 = I2 =1:05 A, reciprocity theorem is veriﬁed.

EXAMPLE 3.49 In the circuit shown in Fig. 3.124, ﬁnd the current through 1:375 Ω resistor and hence verify reciprocity theorem.

Figure 3.124

SOLUTION

Figure 3.125 KVL clockwise for mesh 1: 6:375I1 2I2 3I3 =0 KVL clockwise for mesh 2: 2I1 +14I2 10I3 =0 KVL clockwise for mesh 3: 3I1 10I2 +14I3 = 10

Circuit Theorems j 243

Putting the above three mesh equations in matrix form, we get 2 3 2 3 2 3 6:375 2 3 I1 0 4 5 4 5 4 5 21410 I2 = 0 3 10 14 I3 10 Using Cramer’s rule, we get I1 = 2A Negative sign indicates that the assumed direction of current ﬂow should have been the other way. Veriﬁcation using reciprocity theorem : The circuit is redrawn by interchanging the positions of excitation and response. The new circuit is shown in Fig. 3.126.

Figure 3.126 The mesh equations in matrix form for the circuit shown in Fig. 3.126 is 2 3 2 3 2 3 0 6:375 23 I1 10 4 5 4 0 5 4 5 21410 I2 = 0 0 31014 I3 0 Using Cramer’s rule, we get 0 I3 = 2A 0 Since I1 = I3 = 2A, the reciprocity theorem is veriﬁed.

EXAMPLE 3.50 Find the current Ix in the j2Ωimpedance and hence verify reciprocity theorem.

Figure 3.127

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SOLUTION With reference to the Fig. 3.127, the current through j2Ωimpepance is found using seriesparallel reduction techniques. Total impedance of the circuit is

ZT =(2+j3)+(j5)jj(3 + j2) (j5)(3 + j2) =2+j3+ j5+3+j2 =6:537 /19:36 Ω The total current in the network is 36 /0 IT = 6:537 /19:36 =5:507 /19:36 A Using the principle of current division, we ﬁnd that IT (j5) Ix = j5+3+j2 =6:49 /64:36 A Veriﬁcation of reciprocity theorem : The circuit is redrawn by changing the positions of excitation and response. This circuit is shown in Fig. 3.128. Total impedance of the circuit shown in Fig. 3.128 is

0 ZT =(3+j2)+(2+j3) jj (j5) (2 + j3) (j5) =(3+j2) + 2+j3 j5 =9:804 /19:36 Ω

The total current in the circuit is 0 36 /0 Figure 3.128 IT = 0 =3:672 /19:36 A ZT

Using the principle of current division, 0 IT (j5) Iy = =6:49 /64:36 A j5+2+j3 It is found that Ix = Iy, thus verifying the reciprocity theorem.

EXAMPLE 3.51 Refer the circuit shown in Fig. 3.129. Find current through the ammeter, and hence verify reciprocity theorem.

Circuit Theorems j 245

Figure 3.129

SOLUTION

To ﬁnd the current through the ammeter : By inspection the loop equations for the circuit in Fig. 3.130 can be written in the matrix form as 2 3 2 3 2 3 16 1 10 I1 0 4 5 4 5 4 5 12620 I2 = 0 10 20 30 I3 50

Using Cramer’s rule, we get

I1 =4:6A I2 =5:4A

Hence current through the ammeter = I2 I1 =5:44:6=0:8A. Figure 3.130 Veriﬁcation of reciprocity theorem: The circuit is redrawn by interchanging the positions of excitation and response as shown in Fig. 3.131. By inspection the loop equations for the circuit can be written in matrix form as 2 3 2 3 2 3 0 15 0 10 I1 50 4 5 4 0 5 4 5 02520 I2 = 50 0 10 20 31 I3 0

Using Cramer’s rule we get

0 I3 =0:8A Figure 3.131

246 j Network Theory

Hence, current through the Ammeter = 0.8 A. It is found from both the cases that the response is same. Hence the reciprocity theorem is veriﬁed.

EXAMPLE 3.52 Find current through 5 ohm resistor shown in Fig. 3.132 and hence verify reciprocity theorem.

Figure 3.132

SOLUTION By inspection, we can write 2 3 2 3 2 3 12 0 2 I1 20 4 5 4 5 4 5 02+j10 2 I2 = 20 2 29 I3 0 Using Cramer’s rule, we get I3 =0:5376 /126:25 A Hence, current through 5 ohm resistor =0:5376 /126:25 A Veriﬁcation of reciprocity theorem: The original circuit is redrawn by interchanging the excitation and response as shown in Fig. 3.133.

Figure 3.133

Circuit Theorems j 247

Putting the three equations in matrix form, we get 2 3 2 3 2 3 I0 12 0 2 1 0 4 5 6 0 7 4 5 02+j10 2 4 I2 5 = 0 0 2 29 I3 20 Using Cramer’s rule, we get 0 I1 =0:3876 /2:35 A 0 I2 =0:456 /78:9 A 0 0 Hence; I2 I1 = 0:3179 j0:4335 =0:5376 /126:25 A

The response in both cases remains the same. Thus verifying reciprocity theorem.

3.8 Millman’s theorem

It is possible to combine number of voltage sources or current sources into a single equivalent voltage or current source using Millman’s theorem. Hence, this theorem is quite useful in calculating the total current supplied to the load in a generating station by a number of generators connected in parallel across a busbar. Millman’s theorem states that if n number of generators having generated emfs E1, E2; En and internal impedances Z1; Z2; Zn are connected in parallel, then the emfs and impedances can be combined to give a single equivalent emf of E with an internal impedance of equivalent value Z. E1Y1 + E2Y2 + :::+ EnYn where E = Y1 + Y2 + :::+ Yn 1 and Z = Y1 + Y2 + :::+ Yn where Y1; Y2 Yn are the admittances corresponding to the internal impedances Z1; Z2 Zn and are given by 1 Y1 = Z1 1 Y2 = Z2 . . 1 Yn = Zn

Fig. 3.134 shows a number of generators having emfs E1; E2 En connected in parallel across the terminals x and y. Also, Z1; Z2 Zn are the respective internal impedances of the generators.

248 j Network Theory

Figure 3.134

The Thevenin equivalent circuit of Fig. 3.134 using Millman’s theorem is shown in Fig. 3.135. The nodal equation at x gives

E E E E E E 1 2 n Z + Z + + Z =0 1 2 n E E E ) 1 2 n E 1 1 1 Z + Z + + Z = Z + Z + + Z 1 2 n 1 2 n ) E Y E Y E Y E 1 1 1 + 2 2 + + n n = Z Figure 3.135 where Z = Equivalent internal impedance.

or [E1Y1 + E2Y2 + + EnYn]=EY E Y E Y E Y ) E 1 1 + 2 2 + + n n = Y where Y = Y1 + Y2 + + Yn 1 1 and Z = = Y Y1 + Y2 + + Yn

EXAMPLE 3.53 Refer the circuit shown in Fig. 3.136. Find the current through 10 Ω resistor using Millman’s theorem.

Figure 3.136

Circuit Theorems j 249

SOLUTION Using Millman’s theorem, the circuit shown in Fig. 3.136 is replaced by its Thevenin equivalent circuit across the terminals PQas shown in Fig. 3.137.

E1Y1 + E2Y2 E3Y3 E = Y1 + Y2 + Y3 1 1 1 22 +48 12 5 12 4 = 1 1 1 + + 5 12 4 =10:13 Volts 1 R = Y1 + Y2 + Y3 1 = Figure 3.137 0:2+0:083+0:25 =1:88 Ω E Hence;IL = =0:853 A R +10

EXAMPLE 3.54 Find the current through (10 j3)Ω using Millman’s theorem. Refer Fig. 3.138.

Figure 3.138

SOLUTION The circuit shown in Fig. 3.138 is replaced by its Thevenin equivalent circuit as seen from the terminals, A and B using Millman’s theorem. Fig. 3.139 shows the Thevenin equivalent circuit along with ZL =10 j3Ω:

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Figure 3.139

E1Y1 + E2Y2 E3Y3 E = Y1 + Y2 + Y3 1 1 1 100 /0 +90/45 +80/30 5 10 20 = 1 1 1 + + 5 10 20 =88:49 /15:66 V

Z R 1 1 : = = Y Y Y = 1 1 1 =286 Ω 1 + 2 + 3 5 + 10 + 20

E 88:49 /15:66 I = = =6:7/28:79 A Z + ZL 2:86+10 j3 Alternately, E1Y1 + E2Y2 + E3Y3 + E4Y4 E = Y1 + Y2 + Y3 + Y4 100 51 +90 45 101 +80 30 201 = 51 +101 +201 + (10 j3)1 =70/12 V 70 /12 Therefore;I= 10 j3 =6:7/28:8 A

EXAMPLE 3.55 Refer the circuit shown in Fig. 3.140. Use Millman’s theorem to ﬁnd the current through (5+j5) Ω impedance.

Circuit Theorems j 251

Figure 3.140 SOLUTION The original circuit is redrawn after performing source transformation of5Ainparallel with 4Ω resistor into an equivalent voltage source and is shown in Fig. 3.141.

Figure 3.141

Treating the branch 5+j5Ω as a branch with Es =0V , E1Y1 + E2Y2 + E3Y3 + E4Y4 EPQ = Y1 + Y2 + Y3 + Y4 4 21 +8 31 +20 41 = 21 +31 +41 +(5 j5)1 =8:14 /4:83 V Therefore current in (5 + j5)Ω is 8:14 /4:83 I = =1:15 /40:2 A 5+j5 Alternately EPQ with (5 + j5) open E1Y1 + E2Y2 + E3Y3 EPQ = Y1 + Y2 + Y3 4 21 +8 31 +20 41 = 21 +31 +41 =8:9231V

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Equivalent resistance R =(21 +31 +41)1 =0:9231Ω Therefore current in (5 + j5)Ω is

8:9231 I = =1:15 /40:2 A 0:9231+5+j5

EXAMPLE 3.56 Find the current through 2Ωresistor using Millman’s theorem. Refer the circuit shown in Fig. 3.142.

Figure 3.142 SOLUTION The Thevenin equivalent circuit using Millman’s theorem for the given problem is as shown in Fig. 3.142(a).

E1Y1 + E2Y2 where E = Y1 + Y2 1 1 10 /10 +25/90 3+j4 5 = 1 1 + 3+j4 5 =10:06 /97:12 V 1 1 Z = = Y1 + Y2 1 1 + 3+j4 5 =2:8/26:56 Ω E 10:06 /97:12 Hence; IL = = Z +2 2:8/26:56 +2 =2:15 /81:63 A Figure 3.142(a)

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Reinforcement problems

R.P 3.1 Find the current in 2Ωresistor connected between A and B by using superposition theorem.

Figure R.P. 3.1

SOLUTION Fig. R.P. 3.1(a), shows the circuit with 2V-source acting alone (4V-source is shorted). Resistance as viewed from 2V-source is 2+R1 Ω,

3 2 where R1 = +1 12 5 (1:2+1) 12 = =1:8592 Ω 14:2 2 Hence; I a = =0:5182 A 2+1:8592 12 Then; Ib = Ia =0:438 A 12+1+1:2 3 ∴ I1 =0:438 =0:2628 A Figure R.P. 3.1(a) 5

With 4V-source acting alone, the circuit is as shown in Fig. R.P. 3.1(b).

Figure R.P.3.1(b)

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The resistance as seen by 4V-source is 3+R2 where

2 12 R2 = +1 2 14 2:7143 2 = =1:1551 Ω 4:7143 4 Hence; Ib = =0:9635 A 3+1:1551 Ib 2:7143 Thus; I2 = =0:555 A 4:7143 Finally, applying the principle of superposition,

we get, IAB = I1 + I2 =0:2628 + 0:555 =0:818 A

R.P 3.2 For the network shown in Fig. R.P. 3.2, apply superposition theorem and ﬁnd the current I.

Figure R.P. 3.2

SOLUTION Open the 5A-current source and retain the voltage source. The resulting network is as shown in Fig. R.P. 3.2(a).

Figure R.P. 3.2(a)

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The impedance as seen from the voltage source is

(8 + j10) (j2) Z =(4 j2) + =6:01 /45 Ω 8+j8 j ; I 20 : Hence a = Z =3328 /135 A Next, short the voltage source and retain the current source. The resulting network is as shown in Fig. R.P. 3.2 (b). Here, I3 =5A. Applying KVL for mesh 1 and mesh 2, we get

8I1 +(I1 5) j10 + (I1 I2)(j2) = 0 and (I2 I1)(j2) + (I2 5) (j2) + 4I2 =0

Simplifying, we get

(8 + j8)I1 + j2I2 = j50 and j2I1 +(4 j4)I2 = j10

Solving, we get

8+j8 j50

j2 j10 I I b = 2 = 8+j8 j2

j24 j4 =2:897 /23:96 A Figure R.P. 3.2(b) Since, Ia and Ib are ﬂowing in opposite directions, we have I = Ia Ib =6:1121 /144:78 A

R.P 3.3 Apply superposition theorem and ﬁnd the voltage across 1Ωresistor. Refer the circuit shown in Fig. R.P. 3.3. Take v1(t) = 5 cos (t +10 ) and i2(t) = 3 sin 2t A.

Figure R.P. 3.3

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SOLUTION To begin with let us assume v1(t) alone is acting. Accordingly, short 10V - source and open i2(t). The resulting phasor network is shown in Fig. R.P. 3.3(a). ! = 1rad=sec 5 cos (t +10 ) ! 5/10 V L1 =1H! j!L1 = j1Ω 1 C1 =1F! = j1Ω j!C1 1 1 L2 = H ! j!L2 = j Ω 2 2 1 1 Figure R.P. 3.3(a) C2 = F ! = j2Ω 2 j!C2 ∴ Va =5/10 V ) va(t) = 5 cos [t +10 ]

Let us next assume that i2(t) alone is acting. The resulting network is shown in Fig. R.P. 3.3(b). ! = 2 rad=sec 3 sin 2t ! 3/0 A 1 1 C1 =1F! = j Ω j!C1 2 L1 =1H! j!L1 = j2Ω 1 1 C2 = F ! = j1Ω 2 j!C2 1 Figure R.P. 3.3(b) L2 = H ! j!L2 = j1Ω 2

j1:5 Vb =3/0 =2:5/33:7 A 1+j1:5 ) vb(t)=2:5 sin [2t +33:7 ]A

Finally with 10V-source acting alone, the network is as shown in Fig. R.P. 3.3(c). Since ! =0, inductors are shorted and capacitors are opened. Hence, Vc =10V Applying principle of superposition, we get.

v2(t)=va(t)=vb(t)+Vc = 5 cos (t +10 )+2:5 sin (2t +33:7 ) + 10Volts Figure R.P. 3.3(c)

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R.P 3.4 Calculate the current through the galvanometer for the Kelvin double bridge shown in Fig. R.P. 3.4. Use Thevenin’s theorem. Take the resistance of the galvanometer as 30 Ω.

Figure R.P. 3.4

SOLUTION With G being open, the resulting network is as shown in Fig. R.P. 3.4(a).

Figure 3.4(a)

10 20 VA = I1 100 = 100 = V 450 9 10 I2 5 I2 = =1:66;IB = =0:1I2 45 5 45+5 1:5+ 50 Hence;VB = I2 0:5+IB 10 =2:5V 20 5 Thus;VAB = Vt = VA VB = 2:5= Volts 9 18

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To ﬁnd Rt, short circuit the voltage source. The resulting network is as shown in Fig. R.P. 3.4(b).

Figure R.P. 3.4 (b)

Transforming the Δ between B, E and F into an equivalent Y ,weget

35 10 35 5 5 10 RB = =7Ω;RE = =3:5Ω;RF = =1Ω 50 50 50

The reduced network after transformation is as shown in Fig. R.P. 3.4(c).

Figure R.P. 3.4(c)

350 100 4:5 1:5 Hence;RAB = Rt = + +7 450 6 =85:903 Ω

The Thevenin’s equivalent circuit as seen from A and B with 30 Ω connected between A and B is as shown in Fig. R.P. 3.4(d).

5 IG = 18 = 2:4mA 85:903+30 Negative sign implies that the current ﬂows from B to A. Figure R.P. 3.4(d)

R.P 3.5 Find Is and R so that the networks N1 and N2 shown in Fig. R.P. 3.5 are equivalent.

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Figure R.P. 3.5 SOLUTION Transforming the current source in N1 into an equivalent voltage source, we get N3 as shown in Fig. R.P. 3.5(a). From N3, we can write, V IR = ISR (3.28) From N2 we can write, I = 10Ia N V I Also from 2, 3= 2 a I ) V 3=2 10 I ) V 3= 5 I ) V =3 (3.29) 5

For equivalence of N1 and N2, it is requirred that equations (3.28) and (3.29) must be same. Comparing these equations, we get I IR = and ISR =3 5 3 R =0:2 Ω and IS = = 15A 0:2 Figure R.P. 3.5(a) R.P 3.6 Obtain the Norton’s equivalent of the network shown in Fig. R.P. 3.6.

Figure R.P. 3.6

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SOLUTION Terminals a and b are shorted. This results in a network as shown in Fig. R.P. 3.6(a)

Figure R.P. 3.6(a)

The mesh equations are

(i) 9I1 +0I2 6I3 =30 (3.30) (ii) 0I1 +25I2 +15I3 =30 (3.31) (iii) 6I1 +15I2 +23I3 =4VX =4(10I2) )6I1 25I2 +23I3 =0 (3.32)

Solving equations (3.30), (3.31) and (3.32), we get

IN = Isc = I3 =1:4706A

With terminals ab open, I3 =0. The corresponding equations are

9I1 = 30 and 25I2 =50 30 30 Hence;I1 = A and I2 = A 9 25 30 Then;VX =10I2 =10 =12V 25 Hence;Vt = Voc =15I2 6I1 4VX = 50 V Voc 50 Thus;Rt = = = 34 Ω Isc 1:4706 Hence, Norton’s equivalent circuit is as shown in Fig. R.P. 3.6(b).

Figure R.P. 3.6(b)

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R.P 3.7 For the network shown in Fig. R.P. 3.7, ﬁnd the Thevenin’s equivalent to show that

V1 Vt = (1 + a + b ab) 2 3 b and Zt = 2

Figure R.P. 3.7

SOLUTION V1 aV1 With xy open, I1 = 2 Hence,

V V aV I bI oc = t = 1 + 1 + 1 V1 aV1 V1 aV1 = aV1 + + b 2 2 V1 = [1 + a + b ab] 2 With xy shorted, the resulting network is as shown in Fig. R.P. 3.7(a). Figure R.P. 3.7(a)

Applying KVL equations, we get

(i) I1 +(I1 I2)=V1 aV1 ) 2I1 I2 = V1 aV1 (3.33) (ii) (I2 I1)+I2 = aV1 + bI1 )(1 + b) I1 +2I2 = aV1 (3.34)

Solving equations (3.33) and (3.34), we get

V1 (1 + a + b ab) Isc = I2 = 3 b

262 j Network Theory

Voc V1 (1 + a + b ab) Hence;Zt = = (3 b) Isc 2 V1 (1 + a + b ab) 3 b = 2

R.P 3.8 Use Norton’s theorem to determine I in the network shown in Fig. R.P. 3.8. Resistance Values are in ohms.

Figure R.P. 3.8

SOLUTION Let IAE = x and IEF = y. Then by applying KCL at various junctions, the branch currents are marked as shown in Fig. R.P. 3.8(a). Isc = 125 x = IAB on shorting A and B. Applying KVL to the loop ABCF EA,weget

0:04x +0:01y +0:02 (y 20) + 0:03 (x 105) = 0 ) 0:07x +0:03y =3:55 (3.35)

Applying KVL to the loop EDCEF,weget

(x y 30) 0:03 + (x y 55) 0:02 (y 20) 0:02 0:01y =0 ) 0:05x 0:08y =1:6 (3.36)

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Figure R.P. 3.8(a)

Solving equations (3.35) and (3.36), we get x =46:76 A Hence;Isc = IN = 120 x =78:24 A

The circuit to calculate Rt is as shown in Fig. R.P. 3.8(b). All injected currents have been opened. 0:03 0:05 Rt =0:03+0:04 + 0:08 =0:08875 Ω

Figure R.P. 3.8(b) Figure R.P. 3.8(c)

264 j Network Theory

The Norton’s equivalent network is as shown in Fig. R.P. 3.8(c).

0:08875 I =78:24 0:08875 + 0:04 =53:9A

R.P 3.9 For the circuit shown in Fig. R.P. 3.9, ﬁnd R such that the maximum power delivered to the load is3mW.

Figure R.P. 3.9

SOLUTION For a resistive network, the maximum power delivered to the load is

2 Vt Pmax = 4Rt The network with RL removed is as shown in Fig. R.P. 3.9(a). Let the opent circuit voltage between the terminals a and b be Vt. Then, applying KCL at node a,weget Figure R.P. 3.9(a) Vt 1 Vt 2 Vt 3 R + R + R =0

Simplifying we get Vt = 2 Volts With all voltage sources shorted, the resistance, Rt as viewed from the terminals, a and b is found as follows: 1 1 1 1 3 = + + = Rt R R R R R ) Rt = Ω 3

Circuit Theorems | 265

2 2 3 × −3 Hence,Pmax = R = =3 10 4 × 3 R ⇒ R =1kΩ

R.P 3.10 Refer Fig. R.P. 3.10, ﬁnd X1 and X2 interms of R1 and R2 to give maximum power dissipation in R2.

Figure R.P. 3.10

SOLUTION The circuit for ﬁnding Zt is as shown in Figure R.P. 3.10(a).

R1 (jX1) Zt = R1 + jX1 2 2 R1X1 + jR1X1 = 2 2 R1 + X1

Figure R.P. 3.10(a) For maximum power transfer, ∗ ZL = Zt 2 2 R1X1 R1X1 ⇒ R2 + jX2 = 2 2 − j 2 2 R1 + X1 R1 + X1 2 R1X1 Hence,R2 = 2 2 R1 + X1 R2 ⇒ X1 = ±R1 (3.37) R1 − R2 2 R1X1 X2 = − 2 2 (3.38) R1 + X1 Substituting equation (3.37) in equation (3.38) and simplifying, we get X2 = R2 (R1 − R2)

266 j Network Theory

Exercise Problems

E.P 3.1 Find ix for the circuit shown in Fig. E.P. 3.1 by using principle of superposition.

Figure E.P. 3.1 1 i = A Ans : x 4

E.P 3.2 Find the current through branch PQusing superposition theorem.

Figure E.P. 3.2 Ans : 1.0625 A

E.P 3.3 Find the current through 15 ohm resistor using superposition theorem.

Figure E.P. 3.3 Ans : 0.3826 A

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E.P 3.4 Find the current through 3+j4Ωusing superposition theorem.

Figure E.P. 3.4 Ans : 8.3 /85.3 A

E.P 3.5 Find the current through Ix using superposition theorem.

Figure E.P. 3.5 Ans : 3.07 /163.12 A

E.P 3.6 Determine the current through 1Ωresistor using superposition theorem.

Figure E.P. 3.6 Ans : 0.406 A

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E.P 3.7 Obtain the Thevenin equivalent circuit at terminals a b of the network shown in Fig. E.P. 3.7.

Figure E.P. 3.7 Ans : Vt =6.29 V,Rt =9.43 Ω

E.P 3.8 Find the Thevenin equivalent circuit at terminals x y of the circuit shown in Fig. E.P. 3.8.

Figure E.P. 3.8 Ans : Vt =0.192 /43.4 V, Zt =88.7/11.55 Ω

E.P 3.9 Find the Thevenin equivalent of the network shown in Fig. E.P. 3.9.

Figure E.P. 3.9 Ans : Vt =17.14 volts,Rt =4Ω

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E.P 3.10 Find the Thevenin equivalent circuit across a b. Refer Fig. E.P. 3.10.

Figure E.P. 3.10

Ans : Vt = 30 V,Rt =10kΩ

E.P 3.11 Find the Thevenin equivalent circuit across a b for the network shown in Fig. E.P. 3.11.

Figure E.P. 3.11 Ans : Verify your result with other methods.

E.P 3.12 Find the current through 20 ohm resistor using Norton equivalent.

Figure E.P. 3.12

Ans : IN =4.36 A,RN = Rt =8.8Ω,IL =1.33 A

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E.P 3.13 Find the current in 10 ohm resistor using Norton’s theorem.

Figure E.P. 3.13 100 I = 4A,R = R = Ω,I = 0.5A Ans : N t N 7 L

E.P 3.14 Find the Norton equivalent circuit between the terminals ab for the network shown in Fig. E.P. 3.14.

Figure E.P. 3.14 Ans : IN =4.98310 /5.71 A, Zt =ZN =3.6/23.1 Ω

E.P 3.15 Determine the Norton equivalent circuit across the terminals P Q for the network shown in Fig. E.P. 3.15.

Figure E.P. 3.15 Ans : IN =5A,RN = Rt =6Ω

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E.P 3.16 Find the Norton equivalent of the network shown in Fig. E.P. 3.16.

Figure E.P. 3.16

Ans : IN =8.87 A,RN = Rt =43.89 Ω

E.P 3.17 Determine the value of RL for maximum power transfer and also ﬁnd the maximum power transferred.

Figure E.P. 3.17

Ans : RL =1.92 Ω,Pmax =4.67 W

E.P 3.18 Calculate the value of ZL for maximum power transfer and also calculate the maximum power.

Figure E.P. 3.18

Ans : ZL =(7.97 + j2.16)Ω,Pmax =0.36 W

272 j Network Theory

E.P 3.19 Determine the value of RL for maximum power transfer and also calculate the value of maximum power.

Figure E.P. 3.19

Ans : RL =5.44 Ω,Pmax =2.94 W

E.P 3.20 Determine the value of ZL for maximum power transfer. What is the value of maximum power?

Figure E.P. 3.20

Ans : ZL =4.23 + j1.15 Ω,Pmax =5.68 Watts

E.P 3.21 Obtain the Norton equivalent across x y.

Figure E.P. 3.21

Ans : IN = ISC =7.35A,Rt = RN =1.52 Ω

E.P 3.22 Find the Norton equivalent circuit at terminals a b of the network shown in Fig. E.P. 3.22.

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Figure E.P. 3.22 Ans : IN =1.05 /251.6 A, Zt =ZN =10.6/45 Ω

E.P 3.23 Find the Norton equivalent across the terminals X Y of the network shown in Fig. E.P. 3.23.

Figure E.P. 3.23 Ans : IN =7A, Zt =8.19 /55 Ω

E.P 3.24 Determine the current through 10 ohm resistor using Norton’s theorem.

Figure E.P. 3.24 Ans : 0.15A

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E.P 3.25 Determine the current I using Norton’s theorem.

Figure E.P. 3.25 Ans : Verify your result with other methods.

E.P 3.26 Find Vx in the circuit shown in Fig. E.P. 3.26 and hence verify reciprocity theorem.

Figure E.P. 3.26 Ans : Vx =9.28 /21.81 V

E.P 3.27 Find Vx in the circuit shown in Fig. E.P. 3.27 and hence verify reciprocity theorem.

Figure E.P. 3.27

Ans : Vx =10.23 Volts

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E.P 3.28 Find the current ix in the bridge circuit and hence verify reciprocity theorem.

Figure E.P. 3.28 Ans : ix =0.031 A E.P 3.29 Find the current through 4 ohm resistor using Millman’s theorem.

Figure E.P. 3.29 Ans : I =2.05 A E.P 3.30 Find the current through the impedance of (10 + j10) Ω using Millman’s theorem.

Figure E.P. 3.30 Ans : 3.384 /12.6 A

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E.P 3.31 Using Millman’s theorem, ﬁnd the current ﬂowing through the impedance of (4 + j3) Ω.

Figure E.P. 3.31

Ans : 3.64 / 15.23 A