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Chapter 1 1.1 Circuit Concepts and Network Simplification Techniques Introduction Today we live in a predominantly electrical world. Electrical technology is a driving force in the changes that are occurring in every engineering discipline. For example, surveying is now done using lasers and electronic range finders. Circuit analysis is the foundation for electrical technology. An indepth knowledge of circuit analysis provides an understanding of such things as cause and effect, feedback and control and, stability and oscillations. Moreover, the critical importance is the fact that the concepts of electrical circuit can also be applied to economic and social systems. Thus, the applications and ramifications of circuit analysis are immense. In this chapter, we shall introduce some of the basic quantities that will be used throughout the text. An electric circuit or electric network is an interconnection of electrical elements linked together in a closed path so that an electric current may continuously flow. Alternatively, an electric circuit is essentially a pipe-line that facilitates the transfer of charge from one point to another. 1.2 Current, voltage, power and energy The most elementary quantity in the analysis of electric circuits is the electric charge. Our interest in electric charge is centered around its motion results in an energy transfer. Charge is the intrinsic property of matter responsible for electrical phenomena. The quantity of charge q can be expressed in terms of the charge on one electron. which is 1:602 10 19 coulombs. Thus, 1 coulomb is the charge on 6:24 1018 electrons. The current flows through a specified area A and is defined by the electric charge passing through that area per unit time. Thus we define q as the charge expressed in coulombs. Charge is the quantity of electricity responsible for electric phenomena. 2 j Network Theory The time rate of change constitutes an electric current. Mathemetically, this relation is expressed as i(t) = dq (t) q (t) = or (1.1) Z dt t 1 i(x)dx (1.2) The unit of current is ampere(A); an ampere is 1 coulomb per second. Current is the time rate of flow of electric charge past a given point . The basic variables in electric circuits are current and voltage. If a current flows into terminal a of the element shown in Fig. 1.1, then a voltage or potential difference exists between the two terminals a and b. Normally, we say that a voltage exists across Figure 1.1 Voltage across an element the element. The voltage across an element is the work done in moving a positive charge of 1 coulomb from first terminal through the element to second terminal. The unit of voltage is volt, V or Joules per coulomb. We have defined voltage in Joules per coulomb as the energy required to move a positive charge of 1 coulomb through an element. If we assume that we are dealing with a differential amount of charge and energy, v= then dw (1.3) dq Multiplying both the sides of equation (1.3) by the current in the element gives vi = dw dq dq dt ) dw dt =p (1.4) which is the time rate of change of energy or power measured in Joules per second or watts (W ). p could be either positive or negative. Hence it is imperative to give sign convention for power. If we use the signs as shown in Fig. 1.2., the current flows out of the terminal indicated by x, Figure 1.2 An element with the current which shows the positive sign for the voltage. In leaving from the terminal this case, the element is said to provide energy with a positive voltage sign to the charge as it moves through. Power is then provided by the element. Conversely, power absorbed by an element is p = vi, when i is entering through the positive voltage terminal. Circuit Concepts and Network Simplification Techniques j 3 Energy is the capacity to perform work. Energy and power are related to each other by the following equation: Z Energy = w = EXAMPLE t 1 p dt 1.1 Consider the circuit shown in Fig. 1.3 with v = 8e t V and i = 20e t A for t 0. Find the power absorbed and the energy supplied by the element over the first second of operation. we assume that v and i are zero for t < 0: Figure 1.3 SOLUTION The power supplied is p = vi = (8e = 160e 2t t )(20e t ) W The element is providing energy to the charge flowing through it. The energy supplied during the first seond is Z 1 w= Z 0 = 80(1 1.3 1 p dt = 160e 2t dt 0 e 2 ) = 69.17 Joules Linear, active and passive elements A linear element is one that satisfies the principle of superposition and homogeneity. In order to understand the concept of superposition and homogeneity, let us consider the element shown in Fig. 1.4. Figure 1.4 An element with excitation i and response v The excitation is the current, i and the response is the voltage, v . When the element is subjected to a current i1 , it provides a response v1 . Furthermore, when the element is subjected to a current i2 , it provides a response v2 . If the principle of superposition is true, then the excitation i1 + i2 must produce a response v1 + v2 . Also, it is necessary that the magnitude scale factor be preserved for a linear element. If the element is subjected to an excitation i where is a constant multiplier, then if principle of homogencity is true, the response of the element must be v . We may classify the elements of a circuir into categories, passive and active, depending upon whether they absorb energy or supply energy. 4 | 1.3.1 Network Theory Passive Circuit Elements An element is said to be passive if the total energy delivered to it from the rest of the circuit is either zero or positive. Then for a passive element, with the current flowing into the positive (+) terminal as shown in Fig. 1.4 this means that t vi dt ≥ 0 w= −∞ Examples of passive elements are resistors, capacitors and inductors. 1.3.1.A Resistors Resistance is the physical property of an element or device that impedes the flow of current; it is represented by the symbol R. Figure 1.5 Symbol for a resistor R Resistance of a wire element is calculated using the relation: ρl (1.5) R= A where A is the cross-sectional area, ρ the resistivity, and l the length of the wire. The practical unit of resistance is ohm and represented by the symbol Ω. An element is said to have a resistance of 1 ohm, if it permits 1A of current to flow through it when 1V is impressed across its terminals. Ohm’s law, which is related to voltage and current, was published in 1827 as v = Ri (1.6) v or R= i where v is the potential across the resistive element, i the current through it, and R the resistance of the element. The power absorbed by a resistor is given by v v2 = (1.7) p = vi = v R R Alternatively, (1.8) p = vi = (iR)i = i2 R Hence, the power is a nonlinear function of current i through the resistor or of the voltage v across it. The equation for energy absorbed by or delivered to a resistor is t w= −∞ t pdτ = −∞ i2 R dτ (1.9) Since i2 is always positive, the energy is always positive and the resistor is a passive element. Circuit Concepts and Network Simplification Techniques j 5 1.3.1.B Inductors Whenever a time-changing current is passed through a coil or wire, the voltage across it is proportional to the rate of change of current through the coil. This proportional relationship may be expressed by the equation v=L di (1.10) dt Where L is the constant of proportionality known as inductance and is measured in Henrys (H). Remember v and i are both funtions of time. Let us assume that the coil shown in Fig. 1.6 has N turns and the core material has a high permeability so that the magnetic fluk is connected within the area A. The changing flux creates an induced voltage in each turn equal to the derivative of the flux , so the total voltage v across N turns is v=N Figure 1.6 Model of the d inductor (1.11) dt Since the total flux N is proportional to current in the coil, we have N = Li (1.12) Where L is the constant of proportionality. Substituting equation (1.12) into equation(1.11), we get v=L di dt The power in an inductor is p = vi = L di dt i The energy stored in the inductor is Z w= t 1 Z =L p d i(t) i( 1) i di = 1 2 Li Joules 2 (1.13) Note that when t = 1; i( 1) = 0. Also note that w(t) 0 for all i(t); so the inductor is a passive element. The inductor does not generate energy, but only stores energy. 6 j Network Theory 1.3.1.C Capacitors A capacitor is a two-terminal element that is a model of a device consisting of two conducting plates seperated by a dielectric material. Capacitance is a measure of the ability of a deivce to store energy in the form of an electric field. Capacitance is defined as the ratio of the charge stored to the voltage difference between the two conducting plates or wires, C= C 1.7 Circuit symbol for a capacitor q v The current through the capacitor is given by i= dq dt =C dv (1.14) dt The energy stored in a capacitor is Zt vi d w= 1 Remember that v and i are both functions of time and could be written as v (t) and i(t). i=C Since dv dt Zt w= we have v C 1 dv d d Zv(t) v dv = =C v( Since the capacitor was uncharged at t = 1) 1 2? ?v(t) Cv ? 2 v ( 1) 1, v( 1) = 0. w = w (t) Hence 1 = Cv 2 (t) Joules 2 (1.15) 1 2 q (t) Joules 2C (1.16) Since q = Cv; we may write w (t) = Note that since w(t) element. 0 for all values of v(t), the element is said to be a passive Circuit Concepts and Network Simplification Techniques 1.3.2 | 7 Active Circuit Elements (Energy Sources) An active two-terminal element that supplies energy to a circuit is a source of energy. An ideal voltage source is a circuit element that maintains a prescribed voltage across the terminals regardless of the current flowing in those terminals. Similarly, an ideal current source is a circuit element that maintains a prescribed current through its terminals regardless of the voltage across those terminals. These circuit elements do not exist as practical devices, they are only idealized models of actual voltage and current sources. Ideal voltage and current sources can be further described as either independent sources or dependent sources. An independent source establishes a voltage or current in a circuit without relying on voltages or currents elsewhere in the circuit. The value of the voltage or current supplied is specified by the value of the independent source alone. In contrast, a dependent source establishes a voltage or current whose value depends on the value of the voltage or current elsewhere in the circuit. We cannot specify the value of a dependent source, unless you know the value of the voltage or current on which it depends. The circuit symbols for ideal independent sources are shown in Fig. 1.8.(a) and (b). Note that a circle is used to represent an independent source. The circuit symbols for dependent sources are shown in Fig. 1.8.(c), (d), (e) and (f). A diamond symbol is used to represent a dependent source. Figure 1.8 (a) An ideal independent voltage source (b) An ideal independent current source (c) voltage controlled voltage source (d) current controlled voltage source (e) voltage controlled current source (f) current controlled current source 8 | 1.4 Network Theory Unilateral and bilateral networks A Unilateral network is one whose properties or characteristics change with the direction. An example of unilateral network is the semiconductor diode, which conducts only in one direction. A bilateral network is one whose properties or characteristics are same in either direction. For example, a transmission line is a bilateral network, because it can be made to perform the function equally well in either direction. 1.5 Network simplification techniques In this section, we shall give the formula for reducing the networks consisting of resistors connected in series or parallel. 1.5.1 Resistors in Series When a number of resistors are connected in series, the equivalent resistance of the combination is given by R = R1 + R2 + · · · + Rn (1.17) Thus the total resistance is the algebraic sum of individual resistances. Figure 1.9 Resistors in series 1.5.2 Resistors in Parallel When a number of resistors are connected in parallel as shown in Fig. 1.10, then the equivalent resistance of the combination is computed as follows: 1 1 1 1 (1.18) + + ....... + = R R1 R2 Rn Thus, the reciprocal of a equivalent resistance of a parallel combination is the sum of the reciprocal of the individual resistances. Reciprocal of resistance is conductance and denoted by G. Consequently the equivalent conductance, G = G1 + G2 + · · · + Gn Figure 1.10 Resistors in parallel Circuit Concepts and Network Simplification Techniques 1.5.3 | 9 Division of Current in a Parallel Circuit Consider a two branch parallel circuit as shown in Fig. 1.11. The branch currents I1 and I2 can be evaluated in terms of total current I as follows: IR2 IG1 = R1 + R2 G1 + G2 IR1 IG2 I2 = = R1 + R2 G1 + G2 I1 = (1.19) (1.20) Figure 1.11 Current division in a parallel circuit That is, current in one branch equals the total current multiplied by the resistance of the other branch and then divided by the sum of the resistances. EXAMPLE 1.2 The current in the 6Ω resistor of the network shown in Fig. 1.12 is 2A. Determine the current in all branches and the applied voltage. Figure 1.12 SOLUTION Voltage across 6Ω = 6 × 2 = 12 volts 10 j Network Theory Since 6Ω and 8Ω are connected in parallel, voltage across 8Ω = 12 volts. 12 Therefore, the current through = 1:5 A = 8Ω (between A and B) 8 Total current in the circuit = 2 + 1:5 = 3:5 A Current in the 4Ω branch = 3.5 A Current through 8Ω (betwen C and D) = 3:5 20 20 + 8 = 2.5 A Therefore, current through 20Ω = 3:5 2:5 = 1A Total resistance of the circuit Therefore applied voltage, 6 8 8 20 + 6 + 8 8 + 20 = 13:143Ω =4+ V = 3:5 13:143 (∵ V = IR) = 46 Volts EXAMPLE 1.3 Find the value of R in the circuit shown in Fig. 1.13. Figure 1.13 SOLUTION Voltage across 5Ω = 2:5 5 = 12:5 volts Hence the voltage across the parallel circuit = 25 Current through 12.5 = 12.5 volts 20Ω = I1 or I2 12:5 = 0:625A = 20 Circuit Concepts and Network Simplification Techniques Therefore, current through R = I3 = I = 2:5 Hence; 1.6 I1 0:625 j 11 I2 0:625 = 1:25 Amps 12:5 R= = 10Ω 1:25 Kirchhoff’s laws In the preceeding section, we have seen how simple resistive networks can be solved for current, resistance, potential etc using the concept of Ohm’s law. But as the network becomes complex, application of Ohm’s law for solving the networks becomes tedious and hence time consuming. For solving such complex networks, we make use of Kirchhoff’s laws. Gustav Kirchhoff (1824-1887), an eminent German physicist, did a considerable amount of work on the principles governing the behaviour of eletric circuits. He gave his findings in a set of two laws: (i) Figure 1.14 A simple resistive network current law and (ii) voltage law, which together for difining various circuit are known as Kirchhoff’s laws. Before proceeding terminologies to the statement of these two laws let us familarize ourselves with the following definitions encountered very often in the world of electrical circuits: (i) Node: A node of a network is an equi-potential surface at which two or more circuit elements are joined. Referring to Fig. 1.14, we find that A,B,C and D qualify as nodes in respect of the above definition. (ii) Junction: A junction is that point in a network, where three or more circuit elements are joined. In Fig. 1.14, we find that B and D are the junctions. (iii) Branch: A branch is that part of a network which lies between two junction points. In Fig. 1.14, BAD,BCD and BD qualify as branches. (iv) Loop: A loop is any closed path of a network. Thus, in Fig. 1.14, ABDA,BCDB and ABCDA are the loops. (v) Mesh: A mesh is the most elementary form of a loop and cannot be further divided into other loops. In Fig. 1.14, ABDA and BCDB are the examples of mesh. Once ABDA and BCDB are taken as meshes, the loop ABCDA does not qualify as a mesh, because it contains loops ABDA and BCDB. 12 j Network Theory 1.6.1 Kirchhoff’s Current Law The first law is Kirchhoff’s current law(KCL), which states that the algebraic sum of currents entering any node is zero. Let us consider the node shown in Fig. 1.15. The sum of the currents entering the node is ia + ib ic + id = 0 Note that we have ia since the current ia is leaving the node. If we multiply the foregoing equation by 1, we obtain the expression ia ib + ic id = 0 which simply states that the algebraic sum of currents leaving a node is zero. Alternately, we can write the equation as ib + id = ia + ic which states that the sum of currents entering a node is equal to the sum of currents leaving the node. If the sum of the currents entering a node were not equal to zero, then the charge would be accumulating at a node. However, a node is a perfect conductor and cannot accumulate or store charge. Thus, the sum of currents entering a node is equal to zero. Figure 1.15 Currents at a node 1.6.2 Kirchhoff’s Voltage Law Kirchhoff’s voltage law(KVL) states that the algebraic sum of voltages around any closed path in a circuit is zero. In general, the mathematical representation of Kirchhoff’s voltage law is N X vj (t) = 0 j =1 where vj (t) is the voltage across the j th branch (with proper reference direction) in a loop containing N voltages. In Kirchhoff’s voltage law, the algebraic sign is used to keep track of the voltage polarity. In other words, as we traverse the circuit, it is necessary to sum the increases and decreases in voltages to zero. Therefore, it is important to keep track of whether the voltage is increasing or decreasing as we go through each element. We will adopt a policy of considering the increase in voltage as negative and a Figure 1.16 Circuit with three closed paths decrease in voltage as positive. Circuit Concepts and Network Simplification Techniques j 13 Consider the circuit shown in Fig. 1.16, where the voltage for each element is identified with its sign. The ideal wire used for connecting the components has zero resistance, and thus the voltage across it is equal to zero. The sum of voltages around the loop incorporating v6 ; v3 ; v4 and v5 is v6 v3 + v4 + v5 = 0 The sum of voltages around a loop is equal to zero. A circuit loop is a conservative system, meaning that the work required to move a unit charge around any loop is zero. However, it is important to note that not all electrical systems are conservative. Example of a nonconservative system is a radio wave broadcasting system. EXAMPLE 1.4 Consider the circuit shown in Fig. 1.17. Find each branch current and voltage across each branch when R1 = 8Ω; v2 = 10 volts i3 = 2A and R3 = 1Ω. Also find R2 . Figure 1.17 SOLUTION Applying KCL (Kirchhoff’s Current Law) at node A, we get i1 = i2 + i3 and using Ohm’s law for R3 , we get v3 = R3 i3 = 1(2) = 2V Applying KVL (Kirchhoff’s Voltage Law) for the loop EACDE, we get ) 10 + v1 + v3 = 0 v1 = 10 v3 = 8V 14 | Network Theory Ohm’s law for R1 is v1 = i1 R1 v1 i1 = = 1A R1 i2 = i1 − i3 ⇒ Hence, = 1 − 2 = −1A From the circuit, v2 = R2 i2 −10 v2 = = 10Ω R2 = i2 −1 ⇒ EXAMPLE 1.5 Referring to Fig. 1.18, find the following: (a) ix if iy = 2A and iz = 0A (b) iy if ix = 2A and iz = 2iy (c) iz if ix = iy = iz Figure 1.18 SOLUTION Applying KCL at node A, we get 5 + iy + iz = ix + 3 (a) ix = 2 + iy + iz = 2 + 2 + 0 = 4A (b) iy = 3 + ix − 5 − iz = −2 + 2 − 2iy ⇒ iy = 0A (c) This situation is not possible, since ix and iz are in opposite directions. The only possibility is iz = 0, and this cannot be allowed, as KCL will not be satisfied (5 = 3). EXAMPLE 1.6 Refer the Fig. 1.19. (a) Calculate vy if iz = −3A (b) What voltage would you need to replace 5 V source to obtain vy = −6 V if iz = 0.5A? Circuit Concepts and Network Simplification Techniques Figure 1.19 SOLUTION (a) vy = 1 (3 vx + iz ) vx = 5V and iz = Since vy = 3(5) we get 3 = 12V vy = 1 (3 vx + iz ) = (b) 6 = 3 vx + 0.5 ) 3 vx = vx = Hence, EXAMPLE 3A; 6:5 2.167 volts 1.7 For the circuit shown in Fig. 1.20, find i1 and v1 , given R3 = 6Ω. Figure 1.20 SOLUTION Applying KCL at node A, we get i1 From Ohm’s law, ) Hence; i2 + 5 = 0 12 = i2 R3 12 12 i2 = = = 2A R3 6 i1 = 5 i2 = 3A j 15 16 j Network Theory Applying KVL clockwise to the loop CBAC, we get ) v1 6i1 + 12 = 0 v1 = 12 = 12 EXAMPLE 6i1 6(3) = 6volts 1.8 Use Ohm’s law and Kirchhoff’s law to evaluate (a) vx , (b) iin , (c) Is and (d) the power provided by the dependent source in Fig 1.21. Figure 1.21 SOLUTION (a) Applying KVL, (Referring Fig. 1.21 (a)) we get ) 2 + vx + 8 = 0 vx = 6V Figure 1.21(a) Circuit Concepts and Network Simplification Techniques j 17 (b) Applying KCL at node a, we get vx 8 = 4 2 6 Is + 4( 6) =4 4 Is 24 1:5 = 4 Is + 4vx + ) ) ) Is = 29:5A (c) Applying KCL at node b, we get 2 vx + Is + 2 4 6 iin = 1 + 29:5 4 iin = ) 6 6 = 23A (d) The power supplied by the dependent current source = 8 (4vx ) = 8 4 6 = EXAMPLE 1.9 Find the current i2 and voltage v for the circuit shown in Fig. 1.22. Figure 1.22 SOLUTION From the network shown in Fig. 1.22, i2 = v 6 The two parallel resistors may be reduced to Rp = 36 = 2Ω 3+6 Hence, the total series resistance around the loop is Rs = 2 + Rp + 4 = 8Ω 192W 18 j Network Theory Applying KVL around the loop, we have 21 + 8i 3i2 = 0 (1.21) Using the principle of current division, i2 = = iR2 R1 + R2 3i i 9 i = 3i2 ) = = i 3 3+6 3 (1.22) Substituting equation (1.22) in equation (1.21), we get 21 + 8(3i2 ) Hence; 3i2 = 0 i2 = 1A v = 6i2 = 6V and EXAMPLE 1.10 Find the current i2 and voltage v for resistor R in Fig. 1.23 when R = 16Ω. Figure 1.23 SOLUTION Applying KCL at node x, we get 4 Also; i1 + 3i2 i1 = v 4+2 i2 = Hence; and 4 ) i2 = 0 v v 6 v R = = +3 6 v 16 v v =0 16 16 v = 96volts v 96 i2 = = = 6A 6 16 Circuit Concepts and Network Simplification Techniques EXAMPLE j 19 1.11 A wheatstone bridge ABCD is arranged as follows: AB = 10Ω, BC = 30Ω, CD = 15Ω and DA = 20Ω. A 2V battery of internal resistance 2Ω is connected between points A and C with A being positive. A galvanometer of resistance 40Ω is connected between B and D. Find the magnitude and direction of the galvanometer current. SOLUTION Applying KVL clockwise to the loop ABDA, we get 10ix + 40iz ) 10ix 20iy = 0 20iy + 40iz = 0 (1.23) Applying KVL clockwise to the loop BCDB, we get ) 30(ix iz ) 15(iy + iz ) 40iz = 0 30ix 85iz = 0 15iy (1.24) Finally, applying KVL clockwise to the loop ADCA, we get 20iy + 15(iy + iz ) + 2(ix + iy ) 2 = 0 ) 2ix + 37iy + 15iz = 2 Putting equations (1.23),(1.24) and 2 10 20 4 30 15 2 37 (1.25) in matrix form, we get 3 2 3 32 0 40 ix 85 5 4 iy 5 = 4 0 5 iz 2 15 Using Cramer’s rule, we find that iz = 0.01 A (Flows from B to D) (1.25) 20 1.7 j Network Theory Multiple current source networks Let us now learn how to reduce a network having multiple current sources and a number of resistors in parallel. Consider the circuit shown in Fig. 1.24. We have assumed that the upper node is v (t) volts positive with respect to the lower node. Applying KCL to upper node yields ) ) i1 (t) i2 (t) i3 (t) + i4 (t) i5 (t) i6 (t) = 0 i1 (t) i3 (t) + i4 (t) i6 (t) = i2 (t) + i5 (t) (1.26) io (t) = i2 (t) + i5 (t) (1.27) Figure 1.24 Multiple current source network where io (t) = i1 (t) i3 (t) + i4 (t) i6 (t) is the algebraic sum of all current sources present in the multiple source network shown in Fig. 1.24. As a consequence of equation (1.27), the network of Fig. 1.24 is effectively reduced to that shown in Fig. 1.25. Using Ohm’s law, the currents on the right side of equation (1.27) can be expressed in terms of the voltage and Figure 1.25 Equivalent circuit individual resistance so that KCL equation reduces to 1 1 io (t) = v (t) + R1 R2 Thus, we can reduce a multiple current source network into a network having only one current source. 1.8 Source transformations Source transformation is a procedure which transforms one source into another while retaining the terminal characteristics of the original source. Source transformation is based on the concept of equivalence. An equivalent circuit is one whose terminal characteristics remain identical to those of the original circuit. The term equivalence as applied to circuits means an identical effect at the terminals, but not within the equivalent circuits themselves. Circuit Concepts and Network Simplification Techniques j 21 We are interested in transforming the circuit shown in Fig. 1.26 to a one shown in Fig. 1.27. Figure 1.26 Voltage source connected Figure 1.27 Current source connected to an external resistance R to an external resistance R We require both the circuits to have the equivalence or same characteristics between the terminals x and y for all values of external resistance R. We will try for equivanlence of the two circuits between terminals x and y for two limiting values of R namely R = 0 and R = 1. When R = 0, we have a short circuit across the terminals x and y . It is obligatory for the short circuit to be same for each circuit. The short circuit current of Fig. 1.26 is is = vs (1.28) Rs The short circuit current of Fig. 1.27 is is . This enforces, is = vs (1.29) Rs When R = 1, from Fig. 1.26 we have vxy = vs and from Fig. 1.27 we have vxy = is Rp . Thus, for equivalence, we require that vs = is Rp Also from equation (1.29), we require is = vs Rs vs = ) (1.30) . Therefore, we must have vs Rs Rs = Rp Rp (1.31) Equations(1.29) and (1.31) must be true simulaneously for both the circuits for the two sources to be equivalent. We have derived the conditions for equivalence of two circuits shown in Figs. 1.26 and 1.27 only for two extreme values of R, namely R = 0 and R = 1. However, the equality relationship holds good for all R as explained below. 22 j Network Theory Applying KVL to Fig. 1.26, we get vs = iRs + v Dividing by Rs gives vs Rs =i+ v Rs (1.32) If we use KCL for Fig. 1.27, we get is = i + v Rp (1.33) Thus two circuits are equal when is = vs Rs and Rs = Rp Transformation procedure: If we have embedded within a network, a current source i in parallel with a resistor R can be replaced with a voltage source of value v = iR in series with the resistor R. The reverse is also true; that is, a voltage source v in series with a resistor R can be v in parallel with the resistor R. Parameters replaced with a current source of value i = R within the circuit are unchanged under these transformation. EXAMPLE 1.12 A circuit is shown in Fig. 1.28. Find the current i by reducing the circuit to the right of the terminals x y to its simplest form using source transformations. Figure 1.28 Circuit Concepts and Network Simplification Techniques j 23 SOLUTION The first step in the analysis is to transform 30 ohm resistor in series with a 3 V source into a current source with a parallel resistance and we get: Reducing the two parallel resistances, we get: The parallel resistance of 12Ω and the current source of 0.1A can be transformed into a voltage source in series with a 12 ohm resistor. Applying KVL, we get ) ) 5i + 12i + 1:2 5=0 17i = 3:8 i = 0:224A 24 j Network Theory EXAMPLE 1.13 Find current i1 using source transformation for the circuit shown Fig. 1.29. Figure 1.29 SOLUTION Converting 1 mA current source in parallel with 47kΩ resistor and 20 mA current source in parallel with 10kΩ resistor into equivalent voltage sources, the circuit of Fig. 1.29 becomes the circuit shown in Fig. 1.29(a). Figure 1.29(a) Please note that for each voltage source, “+” corresponds to its corresponding current source’s arrow head. Using KVL to the above circuit, 47 + 47 103 i1 4i1 + 13:3 103 i1 + 200 = 0 Solving, we find that i1 = EXAMPLE 4.096 mA 1.14 Use source transformation to convert the circuit in Fig. 1.30 to a single current source in parallel with a single resistor. Circuit Concepts and Network Simplification Techniques j 25 Figure 1.30 SOLUTION The 9V source across the terminals a0 and b0 will force the voltage across these two terminals to be 9V regardless the value of the other 9V source and 8Ω resistor to its left. Hence, these two components may be removed from the terminals, a0 and b0 without affecting the circuit condition. Accordingly, the above circuit reduces to, Converting the voltage source in series with 4Ω resistor into an equivalent current source, we get, Adding the current sources in parallel and reducing the two 4 ohm resistors in parallel, we get the circuit shown in Fig. 1.30 (a): Figure 1.30 (a) 26 j Network Theory 1.8.1 Source Shift The source transformation is possible only in the case of practical sources. ie Rs 6= 1 and Rp 6= 0, where Rs and Rp are internal resistances of voltage and current sources respectively. Transformation is not possible for ideal sources and source shifting methods are used for such cases. Voltage source shift (E shift): Consider a part of the network shown in Fig. 1.31(a) that contains an ideal voltage source. Figure 1.31(a) Basic network Since node b is at a potential E with respect to node a, the network can be redrawn equivalently as in Fig. 1.31(b) or (c) depend on the requirements. Figure 1.31(b) Networks after E-shift Figure 1.31(c) Network after the E-shift Current source shift (I shift) In a similar manner, current sources also can be shifted. This can be explained with an example. Consider the network shown in Fig. 1.32(a), which contains an ideal current source between nodes a and c. The circuit shown in Figs. 1.32(b) and (c) illustrates the equivalent circuit after the I - shift. Circuit Concepts and Network Simplification Techniques j 27 Figure 1.32(a) basic network Figure 1.32(b) and (c) Networks after I--shift EXAMPLE 1.15 Use source shifting and transformation techiniques to find voltage across 2Ω resistor shown in Fig. 1.33(a). All resistor values are in ohms. Figure 1.33(a) 28 j Network Theory SOLUTION The circuit is redrawn by shifting 2A current source and 3V voltage source and further simplified as shown below. Thus the voltage across 2Ω resistor is V =3 2 1 1 +4 1 +4 1 =3V Circuit Concepts and Network Simplification Techniques EXAMPLE | 29 1.16 Use source mobility to calculate vab in the circuits shown in Fig. 1.34 (a) and (b). All resistor values are in ohms. Figure 1.34(a) Figure 1.34(b) SOLUTION (a) The circuit shown in Fig. 1.34(a) is simplified using source mobility technique, as shown below and the voltage across the nodes a and b is calculated. b Voltage across a and b is Vab = 1 3−1 + 10−1 + 15−1 =2V a 30 j Network Theory (b) The circuit shown in Fig. 1.34 (b) is reduced as follows. Figure 1.34(c) Figure 1.34(d) Figure 1.34(e) From Fig. 1.34(e), Vbc = 12 1 6 12 1 + 10 1 + 15 1 12 = 24 V Applying this result in Fig. 1.34(b), we get vab = vac = 60 EXAMPLE vbc 24 = 36 V 1.17 Use mobility and reduction techniques to solve the node voltages of the network shown in Fig. 1.35(a). All resistors are in ohms. Circuit Concepts and Network Simplification Techniques j 31 Figure 1.35(a) SOLUTION The circuit shown in Fig. 1.35(a) can be reduced by using desired techniques as shown in Fig. 1.35(b) to 1.35(e). Figure 1.35(b) From Fig. 1.35(e) i= 34 =2A 17 Using this value of i in Fig. 1.35(e), Va = and 92= Ve = Va 22 18 V 20 = 42V 32 j Network Theory Figure 1.35(c) Figure 1.35(d) Figure 1.35(e) From Fig 1.35(a) Vd = Ve + 30 = 42 + 30 = 12V Circuit Concepts and Network Simplification Techniques j 33 Then at node b in Fig. 1.35(b), Vb 45 + 2 Vd Vb 8 =0 Using the value of Vd in the above equation and rearranging, we get, Vb ) 1 1 + 2 8 12 8 Vb = 69:6 V = 45 At node c of Fig. 1.35(b) Vc 5 ) EXAMPLE + 45 + Vc Ve =0 10 42 1 1 = 45 + Vc 5 10 10 Vc = 164 V 1.18 Use source mobility to reduce the network shown in Fig. 1.36(a) and find the value of Vx . All resistors are in ohms. Figure 1.36(a) SOLUTION The circuit shown in Fig. 1.36(a) can be reduced as follows and Vx is calculated. Thus 5 18 = 3:6V Vx = 25 34 1.9 j Network Theory Mesh analysis with independent voltage sources Before starting the concept of mesh analysis, we want to reiterate that a closed path or a loop is drawn starting at a node and tracing a path such that we return to the original node without passing an intermediate node more than once. A mesh is a special case of a loop. A mesh is a loop that does not contain any other loops within it. The network shown in Fig. 1.37(a) has four meshes and they are identified as Mi , where i = 1; 2; 3; 4. Circuit Concepts and Network Simplification Techniques j 35 Figure 1.37(a) A circuit with four meshes. Each mesh is identified by a circuit The current flowing in a mesh is defined as mesh current. As a matter of convention, the mesh currents are assumed to flow in a mesh in the clockwise direction. Let us consider the two mesh circuit of Fig. 1.37(b). We cannot choose the outer loop, v ! R1 ! R2 ! v as one mesh, since it would contain the loop v ! R1 ! R3 ! v within it. Let us choose two mesh currents i1 and i2 as shown in the figure. Figure 1.37(b) A circuit with two meshes We may employ KVL around each mesh. We will travel around each mesh in the clockwise direction and sum the voltage rises and drops encountered in that particular mesh. We will adpot a convention of taking voltage drops to be positive and voltage rises to be negative . Thus, for the network shown in Fig. 1.37(b) we have Mesh 1 : v + i1 R1 + (i1 i2 )R3 = 0 (1.34) Mesh 2 : R3 (i2 i1 ) + R2 i2 = 0 (1.35) Note that when writing voltage across R3 in mesh 1, the current in R3 is taken as i2 . Note that the mesh current i1 is taken as ‘+ve’ since we traverse in clockwise i1 direction in mesh 1, On the other hand, the voltage across R3 in mesh 2 is written as R3 (i2 i1 ). The current i2 is taken as +ve since we are traversing in clockwise direction in this case too. Solving equations (1.34) and (1.35), we can find the mesh currents i1 and i2 . Once the mesh currents are known, the branch currents are evaluated in terms of mesh currents and then all the branch voltages are found using Ohms’s law. If we have N meshes with N mesh currents, we can obtain N independent mesh equations. This set of N equations are independent, and thus guarantees a solution for the N mesh currents. 36 j Network Theory EXAMPLE 1.19 For the electrical network shown in Fig. 1.38, determine the loop currents and all branch currents. Figure 1.38 SOLUTION Applying KVL for the meshes shown in Fig. 1.38, we have Mesh 1 : Mesh 2 : Mesh 3 : ) ) 0:2I1 + 2(I1 I3 ) + 3(I1 I2 ) 5:2I1 3(I2 I1 ) + 4(I2 3I2 ) 2I3 = 10 I1 ) + 4(I3 2I1 4I3 = 15 5:2 3 4 3 7:2 2 4 32 4I2 + 11I3 = 0 3 2 I1 = 0:11A and 3 2 10 I1 4 5 4 I2 5 = 4 15 5 I3 11 0 Using Cramer’s rule, we get I2 = 2:53A I3 = 0:9A (1.37) I2 ) = 0 Putting the equations (1.36) through (1.38) in matrix form, we have 2 (1.36) I3 ) + 0:2I2 + 15 = 0 3I1 + 7:2I2 5I3 + 2(I3 10 = 0 (1.38) Circuit Concepts and Network Simplification Techniques j 37 The various branch currents are now calculated as follows: Current through 10V battery = I1 = 0:11A Current through 2Ω resistor = I1 I3 = 1:01A Current through 3Ω resistor = I1 I2 = 2:64A Current through 4Ω resistor = I2 I3 = 1:63A Current through 5Ω resistor = I3 = 0:9A Current through 15V battery = I2 = 2:53A The negative sign for I2 and I3 indicates that the actual directions of these currents are opposite to the assumed directions. 1.10 Mesh analysis with independent current sources Let us consider an electrical circuit source having an independent current source as shown Fig. 1.39(a). We find that the second mesh current i2 = is and thus we need only to determine the first mesh current i1 , Applying KVL to the first mesh, we obtain (R1 + R2 )i1 i2 = Since we get R2 i2 = v is ; (R1 + R2 )i1 + is R2 = v ) i1 = v is R2 Figure 1.39(a) Circuit containing both independent voltage and current sources R1 + R2 As a second example, let us take an electrical circuit in which the current source is is common to both the meshes. This situation is shown in Fig. 1.39(b). By applying KCL at node x, we recognize that, i2 i1 = is The two mesh equations (using KVL) are M esh 1 : M esh 2 : R1 i1 + vxy (R2 + R3 )i2 v=0 vxy = 0 Figure 1.39(b) Circuit containing an independent current source common to both meshes 38 j Network Theory Adding the above two equations, we get R1 i1 + (R2 + R3 )i2 = v Substituting i2 = i1 + is in the above equation, we find that R1 i1 + (R2 + R3 )(i1 + is ) = v (R2 + R3 )is R1 + R2 + R3 In this manner, we can handle independent current sources by recording the relationship between the mesh currents and the current source. The equation relating the mesh current and the current source is recorded as the constraint equation. ) EXAMPLE i1 = v 1.20 Find the voltage Vo in the circuit shown in Fig. 1.40. Figure 1.40 SOLUTION Constraint equations: I1 = 4 I2 = 10 3 A 2 10 3 A Applying KVL for the mesh 3, we get 4 103 [I3 I2 ] + 2 103 [I3 I1 ] + 6 103I3 3=0 Substituting the values of I1 and I2 , we obtain I3 = 0:25 mA Hence; 103I3 3 = 6 103 (0:25 10 Vo = 6 = 1:5 V 3 ) 3 Circuit Concepts and Network Simplification Techniques 1.11 j 39 Supermesh A more general technique for mesh analysis method, when a current source is common to two meshes, involves the concept of a supermesh. A supermesh is created from two meshes that have a current source as a common element; the current source is in the interior of a supermesh. We thus reduce the number of meshes by one for each current source present. Figure 1.41 shows a supermesh created from the two meshes that have a current source in common. Figure 1.41 Circuit with a supermesh shown by the dashed line EXAMPLE 1.21 Find the current io in the circuit shown in Fig. 1.42(a). Figure 1.42(a) SOLUTION This problem is first solved by the techique explained in Section 1.10. Three mesh currents are specified as shown in Fig. 1.42(b). The mesh currents constrained by the current sources are 10 i3 = 4 10 i=2 i2 3 A 3 A The KVL equations for meshes 2 and 3 respetively are 2 103 i2 + 2 103 (i2 i1 ) 6 + 1 103 i3 + vxy + 1 103 (i3 vxy = 0 i1 ) = 0 40 j Network Theory Figure 1.42(b) Figure 1.42(c) Adding last two equations, we get 6 + 1 103 i3 + 2 103 i2 + 2 103 (i2 Substituting i1 = 2 10 we get 6 + 1 103 i2 3A 4 10 and i3 = i2 4 10 3 i1 ) + 1 3A 103(i3 4 10 (1.39) in the above equation, + 2 103 i2 + 2 103 i2 +1 103 i2 i1 ) = 0 3 2 10 2 10 3 3 =0 Solving we get i2 = Thus; 10 mA 3 io = i1 i2 =2 10 3 = 4 mA 3 The purpose of supermesh approach is to avoid introducing the unknown voltage vxy . The supermesh is created by mentally removing the 4 mA current source as shown in Fig. 1.42(c). Then applying KVL equation around the dotted path, which defines the supermesh, using the orginal mesh currents as shown in Fig. 1.42(b), we get 6 + 1 103 i3 + 2 103 i2 + 2 103 (i2 i1 ) + 1 103(i3 i1 ) = 0 Note that the supermesh equation is same as equation 1.39 obtained earlier by introducing vxy , the remaining procedure of finding io is same as before. Circuit Concepts and Network Simplification Techniques EXAMPLE 1.22 For the network shown in Fig. 1.43(a), find the mesh currents i1 ; i2 and i3 . Figure 1.43(b) Figure 1.43(a) SOLUTION The 5A current source is in the common boundary of two meshes. The supermesh is shown as dotted lines in Figs.1.43(b) and 1.43(c), the branch having the 5A current source is removed from the circuit diagram. Then applying KVL around the dotted path, which defines the supermesh, using the original mesh currents as shown in Fig. 1.43(c), we find that 10 + 1(i1 i3 ) + 3(i2 i3 ) + 2i2 = 0 Figure 1.43(c) For mesh 3, we have 1(i3 i1 ) + 2i3 + 3(i3 i2 ) = 0 Finally, the constraint equation is i1 i2 = 5 Then the above three eqations may be reduced to Supemesh: 1i1 + 5i2 4i3 = 10 Mesh 3 : 1i1 3i2 + 6i3 = 0 current source : i1 i2 = 5 Solving the above simultaneous equations, we find that, i1 = 7.5A, i2 = 2.5A, and i3 = 2.5A j 41 42 j Network Theory EXAMPLE 1.23 Find the mesh currents i1 ; i2 and i3 for the network shown in Fig. 1.44. Figure 1.44 SOLUTION Here we note that 1A independent current source is in the common boundary of two meshes. Mesh currents i1 ; i2 and i3 , are marked in the clockwise direction. The supermesh is shown as dotted lines in Figs. 1.45(a) and 1.45(b). In Fig. 1.45(b), the 1A current source is removed from the circuit diagram, then applying the KVL around the dotted path, which defines the supermesh, using original mesh currents as shown in Fig. 1.45(b), we find that 2 + 2(i1 i3 ) + 1(i2 i3 ) + 2i2 = 0 Figure 1.45(a) Figure 1.45(b) . For mesh 3, the KVL equation is 2(i3 i1 ) + 1i3 + 1(i3 i2 ) = 0 Circuit Concepts and Network Simplification Techniques j 43 Finally, the constraint equation is i1 i2 = 1 Then the above three equations may be reduced to Supermesh : 2i1 + 3i2 3i3 = 2 Mesh 3 : 2i1 + i2 4i3 = 0 Current source: i1 i2 = 1 Solving the above simultaneous equations, we find that i1 = 1.55A, i2 = 0.55A, i3 = 0.91A 1.12 Mesh analysis for the circuits involving dependent sources The persence of one or more dependent sources merely requires each of these source quantites and the variable on which it depends to be expressed in terms of assigned mesh currents. That is, to begin with, we treat the dependent source as though it were an independent source while writing the KVL equations. Then we write the controlling equation for the dependent source. The following examples illustrate the point. EXAMPLE 1.24 (a) Use the mesh current method to solve for ia in the circuit shown in Fig. 1.46. (b) Find the power delivered by the independent current source. (c) Find the power delivered by the dependent voltage source. Figure 1.46 SOLUTION (a) We mark two mesh currents i1 and i2 as shown in Fig. 1.47. We find that i = 2:5mA. Applying KVL to mesh 2, we find that 2400(i2 0:0025) + 1500i2 150(i2 0:0025) = 0 (∵ ia = i2 2:5 mA) ) 3750i2 = 6 0:375 = 5:625 ) i2 = 1:5 mA ia = i2 2:5 = 1:0mA 44 Network Theory j (b) Applying KVL to mesh 1, we get vo + 2:5(0:4) 2:4ia = 0 ) vo = 2:5(0:4) 2:4( 1:0) = 3:4V Pind.source = 3:4 2:5 10 3 = 8.5 mW(delivered) (c) Pdep.source = 150ia (i2 ) = 150( 1:0 10 = EXAMPLE 3 )(1:5 10 3) 0.225 mW(absorbed) Figure 1.47 1.25 Find the total power delivered in the circuit using mesh-current method. Figure 1.48 SOLUTION Let us mark three mesh currents i1 , i2 and i3 as shown in Fig. 1.49. KVL equations : Mesh 1: 17:5i1 + 2:5(i1 i3 ) +5(i1 i2 ) = 0 ) 25i1 5i2 2:5i3 = 0 Mesh 2: 125 + 5(i2 i1 ) +7:5(i2 i3 ) + 50 = 0 ) 5i1 + 12:5i2 7:5i3 = 75 Constraint equations : i3 = 0:2Va Va = 5(i2 Thus; i3 = 0:2 i1 ) 5(i2 i1 ) = i2 i1 : Figure 1.49 Circuit Concepts and Network Simplification Techniques j 45 Making use of i3 in the mesh equations, we get Mesh 1 : 25i1 ) Mesh 2 : 5i2 2:5(i2 i1 ) = 0 27:5i1 5i1 + 12:5i2 ) 7:5i2 = 0 7:5(i2 i1 ) = 75 2:5i1 + 5i2 = 75 Solving the above two equations, we get i1 = 3:6 A; i2 = 13:2 A i3 = i2 and i1 = 9:6 A Applying KVL through the path having 5Ω ! 2:5Ω ! vcs ! 125V source, we get, 5(i2 i1 ) + 2:5(i3 i1 ) + vcs 125 = 0 ) vcs = 125 = 125 5(i2 48 i1 ) 2:5(i3 2:5(9:6 i1 ) 3:6) = 62 V Pvcs = 62(9:6) = 595:2W (absorbed) P50V = 50(i2 i3 ) = 50(13:2 9:6) = 180W (absorbed) P125V = 125i2 = 1650W (delivered) EXAMPLE 1.26 Use the mesh-current method to find the power delivered by the dependent voltage source in the circuit shown in Fig. 1.50. Figure 1.50 SOLUTION Applying KVL to the meshes 1, 2 and 3 shown in Fig 1.51, we have Mesh 1 : ) 5i1 + 15(i1 i3 ) + 10(i1 30i1 i2 ) 10i2 660 = 0 15i3 = 660 46 j Network Theory Mesh 2 : Mesh 3 : ) ) ) 20ia + 10(i2 i1 ) + 50(i2 i3 ) = 0 10(i2 i1 ) + 50(i2 i3 ) = 20ia 10i1 + 60i2 15(i3 i1 ) + 25i3 + 50(i3 15i1 50i3 = 20ia i2 ) = 0 50i2 + 90i3 = 0 Figure 1.51 Also ia = i2 i3 Solving, i1 = 42A, i2 = 27A, i3 = 22A, ia = 5A. Power delivered by the dependent voltage source = P20ia = (20ia )i2 = 2700W (delivered) 1.13 Node voltage anlysis In the nodal analysis, Kirchhoff’s current law is used to write the equilibrium equations. A node is defined as a junction of two or more branches. If we define one node of the network as a reference node (a point of zero potential or ground), the remaining nodes of the network will have a fixed potential relative to this reference. Equations relating to all nodes except for the reference node can be written by applying KCL. Refering to the circuit shown in Fig.1.52, we can arbitrarily choose any node as the reference node. However, it is convenient to choose the node with most connected branches. Hence, node 3 is chosen as the reference node here. It is seen from the network of Fig. Figure 1.52 Circuit with three nodes where the 1.52 that there are three nodes. lower node 3 is the reference node Circuit Concepts and Network Simplification Techniques j 47 Hence, number of equations based on KCL will be total number of nodes minus one. That is, in the present context, we will have only two KCL equations referred to as node equations. For applying KCL at node 1 and node 2, we assume that all the currents leave these nodes as shown in Figs. 1.53 and 1.54. Figure 1.53 Simplified circuit for Figure 1.54 Simplified circuit for applying KCL at node 1 applying KCL at node 2 Applying KCL at node 1 and 2, we find that i1 + i2 + i4 = 0 (i) At node 1: va v1 ) R1 ) v1 1 R1 + + v2 v1 R2 1 R2 + v1 + R4 1 0 v2 R4 1 R2 =0 = va (1.40) R1 i2 + i3 + i5 = 0 (ii) At node 2: v1 v2 ) ) v1 1 R2 R2 + + v2 1 R2 vb v2 R3 + 1 R3 + + v2 R5 1 R5 =0 = vb (1.41) R3 Putting equations (1.40) and (1.41) in matrix form, we get 2 1 6 R1 6 6 4 + 1 R2 + 32 1 1 R4 R2 1 1 R2 R2 + 1 R3 + 76 76 76 1 54 R5 v1 v2 3 2 v a 7 6 R1 7=6 7 6 5 4 vb 3 7 7 7 5 R3 The above matrix equation can be solved for node voltages v1 and v2 using Cramer’s rule of determinants. Once v1 and v2 are obtainted, then by using Ohm’s law, we can find all the branch currents and hence the solution of the network is obtained. 48 j Network Theory EXAMPLE 1.27 Refer the circuit shown in Fig. 1.55. Find the three node voltages va , vb and vc , when all the conductances are equal to 1S. Figure 1.55 SOLUTION At node a: (G1 + G2 + G6 )va G2 v b G6 v c = 9 At node b: G2 va + (G4 + G2 + G3 )vb G4 v c = 3 At node c: G6 v a 3 G4 vb + (G4 + G5 + G6 )vc = 7 Substituting the values of various conductances, we find that 3va vb vc = 6 va + 3vb va Putting the above equations 2 3 4 1 1 vc = 3 vb + 3vc = 7 in matrix form, we see that 3 2 3 32 1 1 6 va 5 4 5 4 vb 3 1 = 3 5 vc 1 3 7 Solving the matrix equation using cramer’s rule, we get va = 5:5V; vb = 4:75V; vc = 5:75V The determinant Δ used for computing va , vb and vc in general form is given by P a G G = Gab Gac where P i Gab P b G Gbc Gbc P G Gac c G is the sum of the conductances at node i, and Gij is the sum of conductances conecting nodes i and j . Circuit Concepts and Network Simplification Techniques j 49 The node voltage matrix equation for a circuit with k unknown node voltages is Gv = is ; 2 6 6 va vb v=6 . 4 .. where; 3 7 7 7 5 vk is the vector consisting of k unknown node voltages. 2 6 6 is1 is2 ia = 6 . 4 .. The matrix 3 7 7 7 5 isk is the vector consisting of k current sources and isk is the sum of all the source currents entering the node k . If the k th current source is not present, then isk = 0. EXAMPLE 1.28 Use the node voltage method to find how much power the 2A source extracts from the circuit shown in Fig. 1.56. Figure 1.56 SOLUTION Applying KCL at node a, we get 2+ va 4 + va ) 55 5 =0 va = 20V P2Asource = 20(2) = 40W (absorbing) Figure 1.57 50 Network Theory j EXAMPLE 1.29 Refer the circuit shown in Fig. 1.58(a). (a) Use the node voltage method to find the branch currents i1 to i6 . (b) Test your solution for the branch currents by showing the total power dissipated equals the power developed. Figure 1.58(a) SOLUTION (a) At node v1 : v1 110 2 ) + v2 v1 8 11v1 + v3 v1 2v2 16 =0 v3 = 880 At node v2 : v2 ) v1 8 + v2 + v2 v3 =0 3 24 3v1 + 12v2 v3 = 0 At node v3 : v3 + 110 v3 v2 v3 v1 + + =0 2 24 16 ) 3v1 2v2 + 29v3 = 2640 Solving the above nodal equations,we get v1 = 74:64V; v2 = 11:79V; v3 = Hence; i1 = 110 v1 = 17:68A 2 v2 = 3:93A i2 = 3 Figure 1.58(b) 82:5V Circuit Concepts and Network Simplification Techniques v3 + 110 i3 = v1 i4 = v2 i5 = v1 i6 = 2 8 v2 v3 24 16 v3 j 51 = 13:75A = 7:86A = 3:93A = 9:82A (b) Total power delivered = 110i1 + 110i3 = 3457:3W Total power dissipated = i21 2 + i22 3 + i23 2 + i24 8 + i25 24 + i26 16 = 3457.3 W EXAMPLE 1.30 (a)Use the node voltage method to show that the output volatage vo in the circuit of Fig 1.59(a) is equal to the average value of the source voltages. (b) Find vo if v1 = 150V, v2 = 200V and v3 = 50V. Figure 1.59(a) SOLUTION Applying KCL at node a, we get vo v1 R + v2 vo R ) Hence; vo + v3 vo R + + vo = R =0 + v = [v1 + v2 + + v ] n nvo = v1 + v2 + n 1 n = (b) vn vo 1 (150 + 200 3 n 1X n vk k =1 50) = 100V Figure 1.59(b) 52 j Network Theory EXAMPLE 1.31 Use nodal analysis to find vo in the circuit of Fig. 1.60. Figure 1.60 Figure 1.61 SOLUTION Referring Fig 1.61, at node v1 : v1 + 6 ) ) 6 + 3 + v1 + 3 2 v1 v1 v1 + + = 6 3 2 v1 = 2:5 V v1 =0 2:5 1 2+1 2:5 = 1 3 = 0:83volts vo = EXAMPLE v1 1.32 Refer to the network shown in Fig. 1.62. Find the power delivered by 1A current source. Figure 1.62 Circuit Concepts and Network Simplification Techniques j 53 SOLUTION Referring to Fig. 1.63, applying KVL to the path va ! 4Ω ! 3Ω, we get va = v1 v 3 v2 = 12V At node v1 : ) At node v2 : ) ) v1 4 v1 4 v3 3 + v2 v1 + v1 2 12 2 1=0 1=0 v1 = 9:33 V + v2 v3 v3 3 2 + +1=0 Figure 1.63 v3 12 2 Hence; +1=0 v3 = 6V va = 9:33 6 = 3:33 volts 1 = 3:33 1 = 3:33W (delivering) P1A source = va 1.14 Supernode Inorder to understand the concept of a supernode, let us consider an electrical circuit as shown in Fig. 1.64. Applying KVL clockwise to the loop containing R1 , voltage source and R2 , we get va = vs + vb ) va vb = vs (Constraint equation) (1.42) To account for the fact that the source voltage is known, we consider both va and vb as part of one larger node represented by the dotted ellipse as shown in Fig. 1.64. We need a larger node because va and vb are dependent (see equation 1.42). This larger node is called the supernode. Applying KCL at nodes a and b, we get va and R1 vb R2 ia = 0 + ia = is Figure 1.64 Circuit with a supernode incorporating va and vb . 54 j Network Theory Adding the above two equations, we find that va ) vb = is R1 R2 va G1 + vb G2 = is + (1.43) Solving equations (1.42) and (1.43), we can find the values of va and vb . When we apply KCL at the supernode, mentally imagine that the voltage source vs is removed from the the circuit of Fig. 1.63, but the voltage at nodes a and b are held at va and vb respectively. In other words, by applying KCL at supernode, we obtain va G1 + va G2 = is The equation is the same equation (1.43). As in supermesh, the KCL for supernode eliminates the problem of dealing with a current through a voltage source. Procedure for using supernode: 1. Use it when a branch between non-reference nodes is connected by an independent or a dependent voltage source. 2. Enclose the voltage source and the two connecting nodes inside a dotted ellipse to form the supernode. 3. Write the constraint equation that defines the voltage relationship between the two non-reference node as a result of the presence of the voltage source. 4. Write the KCL equation at the supernode. 5. If the voltage source is dependent, then the constraint equation for the dependent source is also needed. EXAMPLE 1.33 Refer the electrical circuit shown in Fig. 1.65 and find va . Figure 1.65 Circuit Concepts and Network Simplification Techniques SOLUTION The constraint equation is, vb ) va = 8 vb = va + 8 The KCL equation at the supernode is then, va + 8 500 + (va + 8) 125 Therefore; EXAMPLE 12 va 12 250 va + =0 500 va = 4V + Figure 1.66 1.34 Use the nodal analysis to find vo in the network of Fig. 1.67. Figure 1.67 SOLUTION Figure 1.68 j 55 56 j Network Theory The constraint equation is, v2 ) v1 = 12 v1 = v2 12 KCL at supernode: v2 ) ) 12 (v2 12) v3 v2 v2 v3 + + + =0 3 3 3 1 10 1 10 1 10 1 103 4 10 3 v2 2 10 3 v3 = 24 10 3 4v2 2v3 = 24 At node v3 : ) v3 (v2 12) v3 v2 + = 2 10 3 1 103 1 103 2 10 3 v2 + 2 10 3 v3 = 10 10 2v2 + 2v3 = Solving we get 10 v2 = 7V v3 = 2V Hence; EXAMPLE vo = v3 = 2V 1.35 Refer the network shown in Fig. 1.69. Find the current Io . Figure 1.69 SOLUTION Constriant equation: v3 = v1 12 3 Circuit Concepts and Network Simplification Techniques Figure 1.70 KCL at supernode: v1 ) ) 12 v1 v1 v2 + + =0 3 3 3 10 2 10 3 103 7 1 10 3 v1 10 3v2 = 4 10 6 3 7 1 v1 v2 = 4 6 3 3 KCL at node 2: v2 3 ) v1 103 1 3 ) + v2 3 103 10 3 v1 + + 4 10 2 10 3 1 v1 + 3 3 3 =0 v2 = 2 v2 = 3 4 10 3 4 Putting the above two nodal equations in matrix form, we get 2 7 6 6 6 4 1 3 32 3 2 3 1 v1 4 7 7 6 6 7 3 76 7=6 7 5 4 5 2 54 v2 4 3 Solving the above two matrix equations using Cramer’s rule, we get ) Io = v1 2 v1 = 2V 103 = 2 = 1mA 2 103 j 57 58 j Network Theory EXAMPLE 1.36 Refer the network shown in Fig. 1.71. Find the power delivered by the dependent voltage source in the network. Figure 1.71 SOLUTION Refer Fig. 1.72, KCL at node 1: v1 80 5 where ia = ) v1 + v1 v1 50 + v1 + 75ia 25 50 80 v1 + + 5 50 Solving we get v1 + 75 ) Also; =0 v 25 1 50 =0 v1 = 50V v1 Figure 1.72 50 = 1A 50 50 v1 ( 75ia ) i1 = (10 + 15) v1 + 75ia = (10 + 15) 50 + 75 1 = = 5A (10 + 15) P75ia = (75ia )i1 ia = = = 75 1 5 = 375W (delivered) EXAMPLE 1.37 Use the node-voltage method to find the power developed by the 20 V source in the circuit shown in Fig. 1.73. Circuit Concepts and Network Simplification Techniques Figure 1.73 SOLUTION Figure 1.74 Constraint equations : v2 va = 20 v1 31ib = v3 v2 ib = Node equations : (i) Supernode: v1 20 v1 ) ) 20 v1 20 + + v1 v1 2 20 2 20 + + + v1 v1 20 2 + 40 v2 v3 4 35ib ) 4 (v1 35 v2 40 4 v2 v2 + + + v3 80 (v1 v1 + 3:125va = 0 35ib ) + 3:125(20 80 35 80 v2 ) = 0 v2 40 + 3:125(20 v2 ) = 0 j 59 60 j Network Theory (ii) At node v2 : v2 + 40 v2 ) 40 v2 ) 40 + v2 v1 + 4 (v1 4 v2 + v3 v2 35ib ) 35 + v2 40 4 + v2 20 1 v2 20 1 v2 20 1 =0 =0 =0 Solving the above two nodal equations, we get v1 = 20:25V; v3 = v1 Then 35ib = v1 = Also; ig = v2 = 10V 35 v2 40 29V v1 20 2 + v2 20 1 20 + 20:25 (20 10) + 2 1 = 30.125 A = P20V = 20ig = 20(30:125) = 602.5 W (delivered) EXAMPLE 1.38 Refer the circuit shown in Fig. 1.75(a). Determine the current i1 . Figure 1.75(a) Circuit Concepts and Network Simplification Techniques j 61 SOLUTION Constraint equation: Applying KVL clockwise to the loop containing 3V source, dependent voltage source, 2A current source and 4Ω resitor, we get v1 ) Substituting i1 = 4v1 v2 0:5i1 + v2 = 0 3 v1 v2 = 0:5i1 3 4 , the above equation becomes 2 3v2 = 8 Figure 1.75(b) KCL equation at supernode: v1 4 + v2 4 2 = 2 ) v1 + 2v2 = 0 Solving the constraint equation and the KCL equation at supernode simultaneously, we find that, v2 = 727:3 mV v1 = = Then; i1 = = 2v2 1454:6 mV 4 2 1:636A v2 62 j Network Theory EXAMPLE 1.39 Refer the network shown in Fig. 1.76(a). Find the node voltages vd and vc . Figure 1.76(a) SOLUTION From the network, shown in Fig. 1.76 (b), by inspection,vb = 8 V, i1 = Constraint equation: va = 6i1 + vd ) vb va KCL at supernode: va 2 1 1 + 2 2 + va 2 + 1 1 vb + [vd 2 2 Figure 1.76(b) vc vd 2 vc vb 2 = 3vc vc ] = 3vc (1.44) Circuit Concepts and Network Simplification Techniques j 63 Substituting vb = 8 V in the constrained equation, we get (vb vc ) va = 6 + vd 2 = 3(vb vc ) + vd vc ) + vd = 3(8 (1.45) Substituting equation (1.45) into equation (1.44), we get 1 1 [3(8 vc ) + vd ] (8) + [vd vc ] = 3vc 2 2 ) 24 3vc + vd 4 + 21 vd 21 vc = 3vc ) 6:5vc + 1:5vd = 20 vb vc KCL at node c: 2 vc Substituting vb = 8V, we have 8 2 ) ) ) vc + + vd vc 2 vc 2 8 + vc 2vc vc vd =4 =4 vd = 8 vd = 16 0:5vd = 8 Solving equations (1.46) and (1.47), we get EXAMPLE vc = 1.14V vd = 18.3V 1.40 For the circuit shown in Fig. 1.77(a), determine all the node voltages. Figure 1.77(a) (1.46) (1.47) 64 j Network Theory SOLUTION Refer Fig 1.77(b), by inspection, v2 = 5V Nodes 1 and 3 form a supernode. Constraint equation: v3 = 6 v1 KCL at super node: v1 v2 10 + v3 1 +2=0 Substituting v2 = 5V, we get v1 ) ) v1 5 + v3 = 10 1 5 + 10v3 = 2 20 v1 + 10v3 = 15 Figure 1.77(b) Solving the constraint and the KCL equations at supernode simultaneously, we get v1 = 4.091V v3 = 1.909V KCL at node 4 : v4 2 + v2 v4 4 2=0 Substituting v2 = 5V, we get v4 2 Solving we get, 1.15 + v4 5 4 2=0 v4 = 4:333V: Brief review of impedance and admittance Let us consider a general circuit with two accessible terminals, as shown in Fig. 1.78. If the time domain voltage and current at the terminals are given by v = vm sin(!t + v ) i = im sin(!t + i ) then the phasor quantities at the terminals are Figure 1.78 General phasor circuit Circuit Concepts and Network Simplification Techniques j 65 V = Vm v I = Im i We define the ratio of V to I as the impedence of the circuit, which is denoted as Z. That is, V Z= I It is very important to note that impedance Z is a complex quantity, being the ratio of two complex quantities, but it is not a phasor. That is, it has no corresponding sinusoidal time-domain function, as current and voltage phasors do. Impedence is a complex constant that scales one phasor to produce another. The impedence Z is written in rectangular form as Z = R + jX where R = Real[Z] is the resistance and X = Im[Z] is the reactance. Both R and X , like Z, are measured in ohms. p 2 The magnitude of Z is written as |Z| = R2+ X and the angle of Z is denoted as Z = tan 1 X R . The relationships are shown graphically in Fig. 1.79. The table below gives the various forms of Z for different combinations of R; L and C . Figure 1.79 Graphical representation of impedance Type of the circuit Impedance Z 1. Purely resistive Z=R 2. Purely inductive Z = j!L = jXL 3. Purely capactive Z= 4. RL Z = R + j!L = R + jXL 5. RC Z=R 6. RLC Z = R + j!L j !C The reciprocal of impendance is denoted by Y= 1 Z = jXC j !C =R j !C ‘ jXC = R + j (XL XC ) 66 j Network Theory is called admittance and is analogous to conductance in resistive circuits. Evidently, since Z is a complex number, so is Y. The standard representation of admittance is Y = G + jB The quantities G = Re[Y] and B = Im[Y] are respectively called conductance and suspectence. The units of Y, G and B are all siemens. 1.16 Kirchhoff’s Laws: Applied to alternating circuits If a complex excitation, say vm ej (!t+) , is applied to a circuit, then complex voltages, such as v1 ej (!t+1 ) ; v2 ej (!t+2 ) and so on, appear across the elements in the circuit. Kirchhoff’s voltage law applied around a typical loop results in an equation such as v1 e j(ωt+θ1 ) + v2 e j(ωt+θ2 ) + : : : + vN e j (ωt+θN ) =0 Dividing by ej!t , we get ) where v1 ej1 + v2 ej2 + : : : + vN ejN = 0 V1 + V2 + : : : + VN = 0 Vi = Vi i ; i = 1; 2; N are the phasor voltage around the loop. Thus KVL holds good for phasors also. A similar approach will establish KCL also. At any node having N connected branches, I 1 + I2 + + IN = 0 Ii = Ii i ; i = 1; 2 N where Thus, KCL holds good for phasors also. EXAMPLE 1.41 Determine V1 and V2 , the node voltage phasors using nodal technique for the circuit shown in Fig. 1.80. Figure 1.80 Circuit Concepts and Network Simplification Techniques j 67 SOLUTION First step in the analysis is to convert the circuit of Fig. 1.80 into its phasor version (frequency domain representation). 1 F 2 ) j !C 5 / 0 ; ! = 2 rad=s 1 j!L = j Ω 2 ) ) 5 cos 2t 1 H 4 = j 1Ω; ) 1F j !C = 1 2 j Ω Figure 1.80(a) Figure 1.80(b) Fig. 1.80(a) and (b) are the two versions of the phasor circuit of Fig. 1.80. Z1 = j 1Ωjj j1 = j1 1 j Ω 2 1 j 2 1 j 2 = j 1Ω 68 j Network Theory 1 Z2 = j Ωjj1Ω 2 1 (1) j 1 + j2 2 = = Ω 1 5 j +1 2 KCL at node V1 : 5 /0 ) + V1 V1 V2 + =0 j1 j1 (2 + j 2)V1 j 1V2 = 10 2 (V1 ) KCL at node V2 : V1 V2 = 5∠0 + 1 + j2 j1 5 j V1 + V2 2j V2 = 5 V2 ) ) j V2 j 1V1+ (1 j 1)V2 = 5 Putting the above equations in a matrix form, we get 2 4 2 + j2 j1 j1 1 32 54 j1 V1 3 5=4 V2 Solving V1 and V2 by Cramer’s rule, we get V1 = 2 j1 V V2 = 2 + j 4 V In polar form, V1 = p p5 / 26:6 V V2 = 2 5 /63:4 V In time domain, v1 = F F5 cos(2t 2 26.6 ) V v2 = 2 5 cos(2t + 63.4 ) V 10 5 3 5 Circuit Concepts and Network Simplification Techniques EXAMPLE j 69 1.42 Find the source voltage Vs shown in Fig. 1.81 using nodal technique. Take I = 3 /45 A: Figure 1.81 SOLUTION Refer to Fig. 1.81(a). KCL at node 1: V1 ) Vs V1 V1 V2 + + =0 10 j5 5 + j2 (11 + j 12)V1 (5 + j 2)Vs = 10V2 (1.48) Figure 1.81(a) KCL at node 2: V2 V1 V2 +I+ =0 5 + j2 8 + j3 ) Also; (8 + j 3)V1 = (13 + j 5)V2 + (34 + j 31)I V2 = 4I = 4(3 /45 ) = 12 /45 p p = 6 2 + j6 2 (1.49) (1.50) 70 j Network Theory Substituting equations (1.48) and (1.50) in equation (1.49), we get ) (8 + j 3)V1 = 74:24 + j 290:62 300 /75:7 V1 = 8:54 /20:6 = 35:1 /55:1 = 20:1 + j 28:8 V Substituting V1 and V2 in equation (1.48) yields (5 + j 2)Vs = Therefore; EXAMPLE 209:4 + j 473:1 517:4 /113:9 Vs = = 96.1 /92.1 V 5:38 /21:8 1.43 Find the voltage v (t) in the network shown in Fig. 1.82 using nodal technique. Figure 1.82 SOLUTION Converting the circuit diagram shown in Fig. 1.82 into a phasor circuit diagram, we get Figure 1.83 Circuit Concepts and Network Simplification Techniques j 71 At node V1 : V1 ) ( 1 + j ) V1 V1 V2 + + =0 j2 2 j2 V1 j V2 = 1 + j V1 V2 At nodeV2 : j2 Also + Ic = 2Ix = V2 Hence; ) V2 j2 (1.51) Ic = 0 2( 1 + j ) = j2 V1 V2 + = j2 j2 j V1 + j 2V2 = 1 j 1 j 2 j2 Solving equations (1.51) and (1.52) using Cramer’s rule we get V2 = 2 /135 V v(t) = v2 (t) = Therefore; EXAMPLE p F 2 cos(4t + 135 ) V 1.44 Refer to the circuit of Fig. 1.84. Using nodal technique, find the current i. Figure 1.84 SOLUTION 1 1 F capacitor = = 5 j!C 1 1 j 5000 10 5 The parallel combinations of 2kΩ and j 1kΩ is Reactance of Zp = 2 2 103 ( j 103 ) = (1 3 3 2 10 j 10 5 = 6 j 1kΩ j 2)kΩ (1.52) 72 j Network Theory Figure 1.85 The phasor circuit of Fig. 1.84 is as shown in Fig. 1.85. Constraint equation : V2 = V1 + 3000I KCL at supernode : V1 4 /0 + 2 500 (1 5 V1 j 2) 103 + V2 =0 j 1) 103 (2 Substituting V2 = V1 + 3000I in the above equation, we get V1 4 /0 + 2 500 (1 5 V1 j 2) 103 + V1 + 3000I =0 (2 j 1) 103 Also, I= V1 4 /0 500 (1.53) Hence, V1 4∠0 + 2 500 (1 5 V1 j 2) 103 4 V1 V1 + 3000 500 + (2 j 1) 103 =0 Solving for V1 and substituting the same in equation (1.53), we get I = 24 /53:1 mA Hence, in time-domain, we have i = 24 cos(5000t + 53.1 )mA Circuit Concepts and Network Simplification Techniques EXAMPLE j 73 1.45 Use nodal analysis to find Vo in the circuit shown in Fig. 1.86. Figure 1.86 SOLUTION The voltage source and its two connecting nodes form the supernode as shown in Fig. 1.87. Figure 1.87 Constraint equation: Applying KVL clockwise to the loop formed by 12 /0 source, j 2Ω and 12 /0 + Vo ) V1 = Vo V1 = 0 12 /0 KCL at supernode: V1 V1 V2 Vo V2 Vo + + + =0 j2 1 1 j4 j 4Ω we get 74 j Network Theory Substituting V1 = Vo we get, j 2 (Vo 12 in the above equation 12) + (Vo ) Vo 12 j 2 V2 ) + Vo +1+1+ j + V2 ( 1 4 ) j V2 + Vo = 0 4 1 j 4 Vo 2 1) = 12 j6 2V2 = 12 j6 V1 V2 V2 Vo + + =0 1 2 1 Substituting V1 = Vo 12 /0 in the above equation 1 we get, V2 (Vo 12 /0 ) + V2 + V2 Vo = 0 2 5 ) 2Vo + V2 = 12 /0 2 Solving the two nodal equations,we get V2 KCL at V2 : Vo = 11.056 EXAMPLE j8.09 = 13.7 / 36.2 V 1.46 Find i1 in the circuit of Fig. 1.88 using nodal analysis. Figure 1.88 SOLUTION The phasor equivalent circuit is as shown in Fig. 1.88(a). KCL at node V1 : V1 ) KCL at node V2 : 20 /0 V1 V1 V2 + + =0 10 j 2:5 j4 (1 + j 1:5)V1 + j 2:5V2 = 20 V1 V2 j4 But I1 = + V2 = 2I1 j2 V1 j 2:5 Circuit Concepts and Network Simplification Techniques j 75 Figure 1.88(a) V1 V2 V2 2V1 = j4 j2 j 2:5 j 0:55V1 j 0:75V2 = 0 Hence; ) + Multiplying throughout by j 20, we get 11V1 + 15V2 = 0 Putting the two nodal equations in matrix form, we get 2 4 1 + j 1:5 j 2:5 11 15 32 54 V1 3 2 5=4 V2 20 3 5 0 Solving the matrix equation, we get V1 = 18:97 /18:43 V V2 = 13:91 / 161:56 V The current I1 = V1 = 7:59 /108:4 A j 2:5 Transforming this to the time-domain, we get i1 = 7.59 cos(4t + 108.4 )A EXAMPLE 1.47 Use the node-voltage method to find the steady-state expression for vo (t) in the circuit shown in Fig. 1.89 if vg 1 = 10 cos(5000t + 53:13 )V vg 2 = 8 sin 5000t V 76 j Network Theory Figure 1.89 SOLUTION The first step is to convert the circuit of Fig. 1.89 into a phasor circuit. 10 cos(5000t + 53:13 )V; ! = 5000rad=sec 8 sin 5000t = 8 cos(5000t 90 )V L = 0:4 mH C = 50F ) ) ) ) 10 /53:13 = 6 + j 8V 8 / 90 = j 8V j!L = j 2Ω 1 j!C = j 4Ω The phasor circuit is shown in Fig. 1.89(a). KCL at node 1: Vo (6 + j 8) Vo + j2 6 + Vo ( j 8) j4 =0 Solving we get Vo = 12 /0 V Figure 1.89(a) Hence, the steady-state expression is vo (t) = 12 cos 5000t EXAMPLE 1.48 Solve the example (1.47) using mesh-current method. SOLUTION Refer Fig. 1.90. KVL to mesh 1 : [6 + j 2]I1 KVL to mesh 2 : 6I1 + (6 6I2 = 10 /53:13 j 4)I2 = 8 / 90 Circuit Concepts and Network Simplification Techniques j 77 Figure 1.90 Putting the above equations in matrix form, we get 2 4 6 + j2 6 32 6 6 j4 54 I1 I2 3 " 5= 10 /53:13 # 8 / 90 Solving for I1 and I2 , we get I1 = 4 + j 3 I2 = 2 + j 3 Now; Vo = (I1 Hence in time domain, vo = 12 cos 5000t Volts EXAMPLE I2 )6 = 12 = 12 /0 V 1.49 Determine the current Io in the circuit of Fig. 1.91 using mesh analysis. Figure 1.91 SOLUTION Refer Fig 1.92 KVL for mesh 1 : ) (8 + j 10 j 2)I1 ( j 2)I2 j 10I3 = 0 (8 + j 8)I1 + j 2I2 = j 10I3 (1.54) 78 j Network Theory KVL for mesh 2 : j2 (4 ) j 2)I2 ( j 2)I3 + 20 /90 = 0 ( j 2)I1 j 2I1 + (4 j 4)I2 + j 2I3 = For mesh 3; j 20 I3 = 5 (1.56) Sustituting the value of I3 in the equations (1.54) and (1.55), we get (8 + j 8)I1 + j 2I2 = j 50 j 2I1 + (4 j 4)I2 = j 20 = j 30 j 10 Putting the above equations in matrix form,we get 2 4 8 + j8 j2 j2 4 j4 32 54 I1 3 2 j 50 5=4 3 5 j 30 I2 Figure 1.92 Using Cramer’s rule,we get I2 = 6:12 / 35:22 A The required current: Io = I2 = 6.12 /144.78 A EXAMPLE 1.50 Find Voc using mesh technique. Figure 1.93 SOLUTION Applying KVL clockwise for mesh 1 : ) 600I1 (600 j 300(I1 I2 ) (1.55) 9=0 j 300)I1 + j 300I2 = 9 Circuit Concepts and Network Simplification Techniques j 79 Figure 1.94 Applying KVL clockwise for mesh 2 : j 300(I2 2Va + 300I2 Also; Va = Hence; ) 2( j 300(I1 I1 ) = 0 j 300(I1 I2) j 300 (I2 I2 )) + 300I2 j 3I1 + (1 I1 ) = 0 j 3)I2 = 0 Putting the above two mesh equations in matrix form, we get 2 4 600 j 300 j 300 j3 1 j3 32 54 I1 I2 3 " 5= 9 # 0 Using Cramer’s rule, we find that I2 = 0:0124 / 16 A Voc = 300I2 = 3.72 / 16 V Hence; EXAMPLE p 1.51 Find the steady current i1 when the source voltage is vs = 10 2 cos(!t + 45 ) V and the current source is is = 3 cos !t A for the circuit of Fig. 1.95. The circuit provides the impedence in ohms for each element at the specified ! . Figure 1.95 80 j Network Theory SOLUTION Figure 1.96 The first step is to convert the circuit of Fig. 1.95 into a phasor circuit. The phasor circuit is shown in Fig. 1.96. p vs = 10 2 cos(!t + 45 ) is = 3 cos !t ) ) p Vs = 10 2 /45 = 10(1 + j ) Is = 3 /0 Figure 1.96(a) Constraint equation: I2 I1 = Is = 3 /0 Applying KVL clockwise around the supermesh we get I1 Z1 + I2 (Z2 + Z3 ) I2 = I1 + Is Substituting we get, ) ) Vs = 0 (from the constraint equation) I1 Z1 + (I1 + Is )(Z2 + Z3 ) = Vs (Z1 + Z2 + Z3 )I1 = Vs (Z2 + Z3 )Is (10 + j 10) (2 Vs (Z2 + Z3 )Is = I1 = Z1 + Z2 + Z3 2 = 2 + j 8 = 8:25 /76 A Hence in time domain, i1 = 8.25 cos(ωt + 76 ) A j 2)3 Circuit Concepts and Network Simplification Techniques EXAMPLE j 81 p 1.52 Find the steady-state sinusoidal current i1 for the circuit of Fig. 1.97, when vs = 10 2 cos (100t + 45 ) V: Figure 1.97 SOLUTION The first step is to convert the circuit of Fig. 1.97 int to a phasor circuit. The phasor circuit is shown in Fig. 1.98. p vs = 10 2 cos(100t + 45 ) L = 30 mH C = 5 mF p Vs = 10 2 /45 ; ) ) ! = 100 rad= sec XL = j!L = j 100 30 10 1 XC = ) 3 = j 3Ω j!C = 1 j 100 5 10 3 = j 2Ω KVL for mesh 1 : (3 + j 3)I1 j 3I2 = 10 + j 10 KVL for mesh 2 : (3 j 3)I1 + (j 3 j 2)I2 = 0 Putting the above two mesh equations in matrix form, we get 2 4 3 + j3 3 j3 j3 j1 32 54 I1 I2 3 2 5=4 10 + j 10 0 3 5 82 j Network Theory Using Cramer’s rule,we get I1 = 1:05 /71:6 A Thus the steady state time response is, i1 = 1.05 cos(100t + 71.6 )A Figure 1.98 EXAMPLE 1.53 Determine Vo using mesh analysis. Figure 1.99 SOLUTION Figure 1.100 Circuit Concepts and Network Simplification Techniques | 83 From Fig. 1.100, we find by inspection that, I1 = 2Ia = 2(I2 − I3 ) I2 = 4 mA Applying KVL clockwise to mesh 3, we get 1 × 103 (I3 − I2 ) + 1 × 103 (I3 − I1 ) + 2 × 103 I3 = 0 Substituting I1 = 2(I2 − I3 ) and I2 = 4 mA in the above equation and solving for I3 , I3 = 2 mA we get, Vo = 2 × 103 I3 Then, = 4V EXAMPLE 1.54 Find Vo in the network shown in Fig. 1.101 using mesh analysis. Figure 1.101 SOLUTION Figure 1.102 84 j Network Theory By inspection, we find that I2 = 2 /0 A. Applying KVL clockwise to mesh 1, we get 12 + I1 (2 j 1) + (I1 I2 )(4 + j 2) = 0 Substituting I2 = 2 /0 in the above equation yields, 12 + I1 (2 ) Hence j 1 + 4 + j 2) 20 + j 4 I1 = 2(4 + j 2) = 0 = 3:35 /1:85 A 6 + j1 Vo = 4(I1 I2 ) = 5:42 /4:57 V Wye Delta transformation For reducing a complex network to a single impedance between any two terminals, the reduction formulas for impedances in series and parallel are used. However, for certain configurations of network, we cannot reduce the interconnected impedances to a single equivalent impedance between any two terminals by using series and parallel impedance reduction techniques. That is the reason for this topic. Consider the networks shown in Fig. 1.103 and 1.104. Figure 1.103 Delta resistance network Figure 1.104 Wye resistance network It may be noted that resistors in Fig. 1.103 form a Δ (delta), and resistors in Fig. 1.104. form a Υ (Wye). If both these configurations are connected at only the three terminals a, b and c, it would be very advantageous if an equivalence is established between them. It is possible to relate the resistances of one network to those of the other such that their terminal characteristics are the same. The relationship between the two configurations is called Υ Δ transformation. We are interested in the relationship between the resistances R1 , R2 and R3 and the resitances Ra , Rb and Rc . For deriving the relationship, we assume that for the two networks to be equivalent at each corresponding pair of terminals, it is necessary that the resistance at the corresponding terminals be equal. That is, for example, resistance at terminals b and c with a open-circuited must be same for both networks. Hence, by equating the resistances for each corresponding set of terminals, we get the following set of equations : Circuit Concepts and Network Simplification Techniques j 85 Rab (Υ) = Rab (Δ) (i) ) Ra + Rb = R2 (R1 + R3 ) (1.57) R2 + R1 + R3 Rbc (Υ) = Rbc (Δ) (ii) ) Rb + Rc = R3 (R1 + R2 ) (1.58) R3 + R1 + R2 Rca (Υ) = Rca (Δ) (iii) ) Rc + Ra = R1 (R2 + R3 ) (1.59) R1 + R2 + R3 Solving equations (1.57), (1.58) and (1.59) gives Ra = Rb = Rc = R1 R2 (1.60) R1 + R2 + R3 R2 R3 (1.61) R1 + R2 + R3 R1 R3 (1.62) R1 + R2 + R3 Hence, each resistor in the Υ network is the product of the resistors in the two adjacent Δ branches, divided by the sum of the three Δ resistors. To obtain the conversion formulas for transforming a wye network to an equivalent delta network, we note from equations (1.60) to (1.62) that Ra Rb + Rb Rc + Rc Ra = R1 R2 R3 (R1 + R2 + R3 ) (R1 + R2 + R3 )2 = R1 R2 R3 R1 + R2 + R3 (1.63) Dividing equation (1.63) by each of the equations (1.60) to (1.62) leads to the following relationships : R1 = R2 = R3 = Ra Rb + Rb Rc + Ra Rc Rb Ra Rb + Rb Rc + Ra Rc Rc Ra Rb + Rb Rc + Ra Rc Ra (1.64) (1.65) (1.66) Hence each resistor in the Δ network is the sum of all possible products of Υ resistors taken two at a time, divided by the opposite Υ resistor. Then Υ and Δ are said to be balanced when R1= R2 = R3 = RΔ and Ra = Rb = Rc = RΥ 86 j Network Theory Under these conditions the conversions formula become 1 RΔ 3 RΔ = 3RΥ RΥ = and EXAMPLE 1.55 Find the value of resistance between the terminals a Fig. 1.105. b of the network shown in Figure 1.105 SOLUTION Let us convert the upper Δ to Υ (6k)(18k) = 3 kΩ 6k + 12k + 18k (6k)(12k) = = 2 kΩ 6k + 12k + 18k (12k)(18k) = 6 kΩ = 6k + 12k + 18k Ra 1 = Rb1 R c1 Figure 1.106 The network shown in Fig. 1.106 is now reduced to that shown in Fig. 1.106(a) Circuit Concepts and Network Simplification Techniques Hence; Rab = 4 + 3 + 7:875 + 2 = 16.875kΩ Figure 1.106(a) EXAMPLE 1.56 Find the resistance Rab using Υ Δ transformation. Figure 1.107 SOLUTION Figure 1.108 j 87 88 j Network Theory Let us convert the upper Δ between the points a1 , b1 and c1 into an equivalent Υ. 6 18 = 3:6Ω 6 + 18 + 6 66 = = 1:2Ω 6 + 18 + 6 6 18 = 3:6Ω = 6 + 18 + 6 Ra 1 = Rb1 R c1 Figure 1.108 now becomes jj Rab = 5 + 3:6 + 7:2 27:6 7:2 27:6 = 8:6 + 7:2 + 27:6 = 14:31Ω EXAMPLE 1.57 Obtain the equvivalent resistance Rab for the circuit of Fig. 1.109 and hence find i. Figure 1.109 Circuit Concepts and Network Simplification Techniques j 89 SOLUTION Let us convert Υ between the terminals a, b and c into an equivalent Δ. Rab = Ra Rb + Rb Rc + Rc Ra Rc 10 20 + 20 5 + 5 10 = 70Ω = 5 Ra Rb + Rb Rc + Rc Ra Rbc = Ra 10 20 + 20 5 + 5 10 = 35Ω = 10 Ra Rb + Rb Rc + Rc Ra Rca = Rb 10 20 + 20 5 + 5 10 = = 17:5Ω 20 The circuit diagram of Fig. 1.109 now becomes the circuit diagram shown in Fig. 1.109(a). Combining three pairs of resistors in parallel, we obtain the circuit diagram of Fig. 1.109(b). Figure 1.109(a) 70 30 = 21Ω 70 + 30 12:5 17:5 = 7:292Ω 12:5jj17:5 = 12:5 + 17:5 15 35 = 10:5Ω 15jj35 = 15 + 35 Rab = (7:292 + 10:5)jj21 = 9:632Ω 70jj30 = Thus; i= vs Rab = 12:458 A Figure 1.109(b) 90 j Network Theory Nodal versus mesh analysis The analysis of a complex circuit can usually be accomplished by either the node voltage or mesh current method. One may ask : Given a network to be analyzed, how do we know which method is better or more efficient? The choice is dictated by two factors. When a circuit contains only voltage sources, it is probably easier to use the mesh current method. Conversely, when the circuit contains only current sources, it will be easier to use the node voltage method. Also, a circuit with fewer nodes than meshes is better analyzed using nodal analysis, while a circuit with fewer meshes than nodes is better analyzed using mesh analysis. In other words, the best technique is one which gives smaller number of equations. Another point to consider while choosing between the two methods is, what information is required. If node voltages are required, it may be advantageous to apply nodal analysis. On the other hand, if you need to know several currents, it may be wise to proceed directly with mesh current analysis. It is often advantageous if we know both the techniques. The first advantage lies in the fact that the second method can verify the results of the first method. Also, both the methods have limitations. For example, while analysing a transistor circuit, only mesh method is suited and while analysing an Op-amp circuit, nodal method is only applicable. Mesh technique is applicable for planar1 networks. However, nodal method suits to both planar and nonplanar 2 networks. Reinforcement Problems R.P 1.1 Find the power dissipated in the 80Ω resistor using mesh analysis. Figure R.P.1.1 1 A planar network can be drawn on a plane without branches crossing each other. A nonplanar network is one in which crossover is identified and cannot be eliminated by redrawing the branches. 2 Circuit Concepts and Network Simplification Techniques j 91 SOLUTION KVL clockwise to mesh 1 : 14I1 4I2 8I3 = 230 KVL clockwise to mesh 2 : 4I1 + 22I2 16I3 = 260 KVL clockwise to mesh 3 : 8I1 16I2 + 104I3 = 0 Putting the above mesh equations in matrix form, we get 2 6 6 6 4 14 4 4 22 8 16 8 32 76 76 54 I1 3 2 7 7 5 6 6 4 230 3 7 7 5 16 7 6 I2 7 = 6 260 7 104 I3 0 The current I3 is found from the above matrix equation by using Cramer’s rule. Thus; R.P P80 = I3 2 I3 R80 = 5A = 52 80 = 2000W(dissipated) 1.2 Refer the circuit shown in Fig. R.P. 1.2. The current io = 4A. Find the power dissipated in the 70 Ω resistor. Figure R.P.1.2 SOLUTION By inspection, we find that the mesh current i3 = io = 4A KVL clockwise to mesh 1 : 75i1 70i2 5i3 = 180 92 j Network Theory Substituting i3 = 4A; we get 75i1 70i2 = 200 KVL clockwise to mesh 2 : 70i1 + 88i2 10i3 = 0 70i1 + 88i2 = 40 Substituting the value i3 = 4A; we get Puting the two mesh equations in matrix from, we get 75 70 70 88 i1 i2 = 200 40 Using Cramer’s rule, we get i1 = 12A; i2 = 10A i2 )2 70 = 4 P70 = (i1 70 = 280 W (dissipated) R.P 1.3 Solve for current I in the circuit of Fig. R.P. 1.3 using nodal analysis. Figure R.P.1.3 SOLUTION KCL at node V1 : V1 ) 20 / 90 V1 V1 V2 + + + 5 /0 = 0 2 j2 j1 (0:5 j 0:5)V1 + j V2 = 5 j 10 KCL at node V2 : V2 V1 j1 Also, Hence, + V2 4 I= V1 V2 j1 + 2I V1 j2 V2 2 + V1 4 j2 5 /0 = 0 5 /0 = 0 Circuit Concepts and Network Simplification Techniques ) ) (0:25 j )V2 = 5 5 0:25 j Making use of V2 in the nodal equation at node V1 ; we get j5 = 0:5(1 j )V1 5 j 10 0:25 j j 40 ) (1 j )V1 = 10 j 20 1 j4 ) V1 = 15:81 / 46:5 V 15:81 / 46:5 V1 = Hence; I= j2 2 / 90 = 7:906 /43:5 A R.P V2 = 1.4 Find Vo shown in the Fig. R.P. 1.4 using Nodal technique. Figure R.P.1.4 Figure R.P.1.4(a) . j 93 94 j Network Theory SOLUTION We find from Fig RP 1.4(a) that, V1 = Vo Constraint equation: Applying KVL clockwise along the path consisting of voltage source, capacitor, and 2Ω resistor, we find that 12 /0 + V2 V1 = 0 V1 = V2 + 12 /0 ) or V2 = V1 12 KCL at Supernode : V1 ) (2 V3 V1 V2 V2 V3 + + =0 j2 2 j4 4 j 2)V1 + (1 + j )V2 + ( 1 + j 2)V3 = 0 + KCL at node 3 : V1 V3 V2 V3 0:2Vo = 0 (1.67) j 2)V1 + V2 + ( 1 + j 2)V3 = 0 (1.68) j2 + 4 Substituting Vo = V1 , we get (0:8 Subtracting equation (1.68) from (1.67), we get 1:2V1 + j V2 = 0 Substituting V2 = V1 12 (from the constraint equation), we get ) Hence R.P 1:2V1 + j (V1 12) = 0 j 12 V1 = = Vo 1:2 + j Vo = 7.68 /50.2 V 1.5 Solve for i using mesh analysis. Figure R.P. 1.5 (1.69) Circuit Concepts and Network Simplification Techniques j SOLUTION The first step in the analysis is to draw the phasor circuit equivalent of Fig. R.P.1.5. Figure R.P. 1.5(a) ! = 2 6 sin 2t = 6 cos(2t 10 /0 V ) ) ) ) 10 cos 2t 90) L = 2H C = 0:25F 6 / 90 = j 6V XL = j!L = j 4Ω XC = 1 j!C 1 = j 2Ω = j2 1 4 Applying KVL clockwise to mesh 1 : ) j 2)I1 + j 2I2 = 0 10 + (4 (2 j 1)I1 + j I2 = 5 Applying KVL clockwise to mesh 2 : j 2I1 + (j 4 j 2)I2 + ( j 6) = 0 I1 + I2 = 3 Putting the above mesh equations in a matrix form, we get " j 2 1 j #" I1 I2 1 # " = 5 # 3 Using Cramer’s rule, we get I1 = 2 + j 0:5; I2 = 1 Io = I1 Hence j 0:5; I2 = 1 + j = 1:414 /45 io (t) = 1:414 cos (2t + 45 ) A 95 96 j Network Theory R.P 1.6 Refer the circuit shown in Fig. R.P. 1.6. Find I using mesh analysis. Figure R.P.1.6 SOLUTION Figure R.P.1.6(a) Constraint equation: ) ) I3 I2 = 2I I3 I2 = 2(I1 I3 = 2I1 Also, for mesh 4, I2 ) I2 I4 = 5 A Applying KVL clockwise for mesh 1 : ) j 2)I1 + j 2I2 = 0 ( j 20) + (2 (1 j )I1 + j I2 = j 10 (1.70) Circuit Concepts and Network Simplification Techniques j 97 Applying KVL clockwise for the supermesh : j 2)I2 + j 2I1 + 4I3 (j Substituting I3 = 2I1 I2 j I4 = 0 and I4 = 5A (8 + j 2)I1 we get (4 + j )I2 = j 5 (1.71) Putting equations (1.70) and (1.71) in matrix form, we get 1 j 8 + j2 j (4 + j ) I1 I2 = j 10 j5 Solving for I1 and I2 , we get I1 = (5:44 + j 4:26) A I2 = (11:18 + j 9:7) A I = I1 I2 = 5:735 + j 5:44 = 7.9 /43.49 A R.P 1.7 Calculate Vo in the circuit of Fig. R.P. 1.7 using the method of source transformation. Figure R.P. 1.7 SOLUTION Transform the voltage source to a current source and obtain the circuit shown in Fig. R.P.1.7(a). Is = 20 / 90 = 4 / 90 A 5 Figure R.P.1.7(a) Zp = 5Ωjj3 + j 4 = 5 (3 + j 4) = 2:5 + j 1:25Ω 5 + (3 + j 4) 98 j Network Theory Converting the current source in Fig. R.P. 1.7(b) to a voltage source gives the circuit as shown in Fig. R.P. 1.7(c). Figure R.P.1.7(b) Figure R.P.1.7(c) 4j (2:5 + j 1:25) Vs = Is Zp = j 10V =5 Vo = 10I Vs 10 = Zp + Z2 + 10 5 j 10 = [2:5 + j 1:25 + 4 j 13 + 10] = 5.519 / 28 V R.P 10 1.8 Find vx and ix in the circuit shown in Fig. R.P. 1.8. Figure R.P. 1.8 SOLUTION Constraint equation: ) i2 i1 = 3 + vx 4 i2 = i1 + 3 + v3 4 The above equation becomes very clear if one writes KCL equation at node B of Fig. R.P. 1.8(a). Circuit Concepts and Network Simplification Techniques j 99 Figure R.P. 1.8(b) Figure R.P. 1.8(a) Applying KVL clockwise to the supermesh in Fig. R.P. 1.8(b), we get 50 + 10i1 + 5i2 + 4ix = 0 But ix = i1 . Hence, ) 50 + 10i1 + 5i2 + 4i1 = 0 Making use of vx = (i1 14i1 + 5i2 = 50 i2 ) 2 in the constraint equation, we get (i1 i2 ) 2 i =i +3+ 2 ) ) ) ) (1.72) 1 i2 = i1 + 3 + 2i2 = 2i1 + 6 + i1 3i1 4 i2 i1 2 i2 3i2 + 6 = 0 i1 i2 = 2 Solving equations (1.72) and (1.73) gives i1 = 2:105 A; i2 = 4:105 A Thus; vx = 2(i1 i2 ) = 4 V ix = i1 = 2:105 A and R.P 1.9 Obtain the node voltages v1 , v2 and v3 for the following circuit. (1.73) 100 j Network Theory SOLUTION We have a supernode as shown in Fig. R.P. 1.9(a). By inspection, we find that V2 = 12V. Refer Fig. R.P. 1.9(b) for further analysis. Figure R.P.1.9(a) Figure R.P.1.9(b) . KVL clockwise to mesh 1 : v1 10 + 12 = 0 ) v1 = 2 KVL clockwise to mesh 2 : ) v3 = 8V v1 = 2 V, v2 = 12 V, v3 = Hence; R.P 12 + 20 + v3 = 0 8V 1.10 Find the equivalent resistance Rab for the circuit shown in Fig. R.P.1.10. Figure R.P. 1.10 Circuit Concepts and Network Simplification Techniques j 101 SOLUTION The circuit is redrawn marking the nodes c to j in Fig. R.P. 1.10(a). It can be seen that the network consists of four identical stars : (i) ae; ef; cb (ii) ac; cf; cd (iii) dg; gf; gj (iv) bh; f h; hj Converting each stars in to its equivalent delta, the network is redrawn as shown in Fig. R.P. 1.10(b), noting that each resistance in delta is 100 3 = 300Ω, eliminating nodes c, e, g , h. Figure R.P.1.10(a) Figure R.P.1.10(b) . Reducing the parallel resistors, we get the circuit as in Fig. R.P. 1.10(c). Figure R.P.1.10(c) Hence, there are two identical deltas af d and bf j . Converting them to their equivalent stars, we get the circuit as shown in Fig. R.P.1.10(d). 102 j Network Theory 300 150 = 75 Ω 600 1502 = Rf l = = 37:5 Ω 600 Rak = Rbl = Rkd = Rlj = Rkf Figure R.P.1.10(d) Figure R.P.1.10(e) The circuit is further reduced to Fig. R.P. 1.10(e) and then to Fig. R.P. 1.10(f) and (g). Then the equivalent resistance is Rab = Figure R.P.1.10(f) R.P 214:286 300 = 125 Ω 514:286 Figure R.P.1.10(g) 1.11 Obtain the equivalent resistance Rad for the circuit shown in Fig. R.P.1.11. a d Figure R.P.1.11 Figure R.P.1.11(a) Circuit Concepts and Network Simplification Techniques j 103 SOLUTION The circuit is redrawn as shown Fig. 1.11(a), marking the nodes a to f to identify the deltas in it. It contains 3 deltas abc, bde and def with 3 equal resistors of 30 Ω each. For 30 = 10Ω. Then the each delta, their equivalent star contains 3 resistors each of value 3 circuit becomes as shown in Fig. R.P. 1.11(b) where f is isolated. On simplification, we get the circuit as shown in Fig. R.P.1.11(c) and further reduced to Fig. R.P.1.11(d). Figure R.P.1.11(b) Figure R.P.1.11(c) Figure R.P.1.11(d) Then the equivalent ressitance, Rad = 10 + 13:33 + 10 = 33:33 Ω R.P 1.12 Draw a network for the following mesh equations in matrix form : 2 5 + j5 j5 6 j5 8 + j8 4 0 6 32 3 2 3 0 I1 30 / 0 7 7 6 7 6 0 6 5 4 I2 5 = 4 5 I3 20 / 0 10 104 j Network Theory SOLUTION The general form of the mesh equations in matrix form for a network having three mashes is given by 3 2 32 3 2 Z12 Z13 Z11 I1 V1 1 7 6 Z 6 7 6 Z22 Z23 7 21 4 5 4 I2 5 = 4 V2 2 5 Z31 Z32 Z33 V3 3 I3 Z11 = Z10 + Z12 + Z13 and, where Z10 = Sum of the impedances confined to mesh 1 alone Z12 = Sum of the impedances common to meshes 1 and 2 Z13 = Sum of the impedances common to meshes 1 and 3 Similiar difenitions hold good for Z22 and Z33 : Also, Zij = Zji For the present problem, Z11 = 5 + j 5Ω Z12 = Z21 = j 5Ω Z13 = Z31 = 0Ω Z23 = Z32 = 6Ω We know that, ) ) Similarly; Finally; ) ) ) ) Z11 = Z10 + Z12 + Z13 5 + j 5 = Z10 + j 5 + 0 Z10 = 5Ω Z22 = Z20 + Z21 + Z23 8 + j 8 = Z20 + j 5 + 6 Z20 = 2 + j 3Ω Z33 = Z30 + Z31 + Z32 10 = Z30 + 0 + 6 Z30 = 4Ω Making use of the above impedances, we can configure a network as shown below : Circuit Concepts and Network Simplification Techniques R.P j 105 1.13 Draw a network for the following nodal equations in matrix form. 2 6 6 4 1 1 + j 10 10 1 10 3 3 2 3 2 10 /0 7 Va 74 5 5 4 5 Vb = 1 0 1 10 1 (1 5 j) + 10 SOLUTION The general form of the nodal equations in matrix form for a network having two nodes is given by Y11 Y21 Y12 Y22 V1 V2 = I1 1 I2 2 Y11 = Y10 + Y12 and Y22 = Y20 + Y21 : where Y10 = sum of admittances connected at node 1 alone. Y12 = Y21 = sum of admittances common to nodes 1 and 2. Y20 = sum of admittances connected at node 2 alone: For the present problem, 1 1 + S j 10 10 1 S Y12 = Y21 = 10 1 Y22 = (1 j ) + 10 S 5 Y11 = We know that, Y11 = Y10 + Y12 ) ) Similarly; ) ) Y22 1 1 1 + = Y10 + j 10 10 10 1 Y10 = S j 10 = Y20 + Y21 1 1 1 (1 j ) + = Y20 + 5 10 10 1 Y20 = (1 j ) S 5 106 j Network Theory Making use of the above admittances, we can configure a network as shown below : Exercise problems E.P 1.1 Refer the circuit shown in Fig. E.P.1.1. Using mesh analysis, find the current delivered by the source. Verify the result using nodal technique. Figure E.P. 1.1 Ans : 5A E.P 1.2 For the resistive circuit shown in Fig. E.P. 1.2. by using source transformation and mesh analysis, find the current supplied by the 20 V source. Figure E.P. 1.2 Ans : 2.125A Circuit Concepts and Network Simplification Techniques E.P j 107 1.3 Find the voltage v using nodal technique for the circuit shown in Fig. E.P. 1.3. Figure E.P. 1.3 Ans : v = 5V E.P 1.4 Refer the network shown in Fig. E.P. 1.4. Find the currents i1 and i2 using nodal analysis. Figure E.P. 1.4 Ans : E.P i1 = 1 A, i2 = 1A 1.5 For the network shown in Fig. E.P. 1.5, find the currents through the resistors R1 and R2 using nodal technique. Figure E.P. 1.5 Ans : 3.33A, 6.67A 108 j E.P Network Theory 1.6 Use the mesh-current method to find the branch currents i1 ; i2 and i3 in the circuit of Fig. E.P. 1.6. Figure E.P. 1.6 Ans : i1 = E.P 1.7 1.72A, i2 = 1.08A , i3 = 2.8A Refer the network shown in Fig. E.P. 1.7. Find the power delivered by the dependent voltage source in the network. Figure E.P. 1.7 Ans : E.P 375 Watts 1.8 Find the current Ix using (i) nodal analysis and (ii) mesh analysis. Figure E.P. 1.8 Ans : Ix = 150(3 + j4) A 95 + j30 Circuit Concepts and Network Simplification Techniques E.P j 109 1.9 Determine the current ix in the circuit shown in Fig. E.P. 1.9 Figure E.P. 1.9 Ans : ix = 3A E.P 1.10 Determine the resistance between the terminals a 1.10. b of the network shown in Fig. E.P. Figure E.P. 1.10 Ans : 23.6 Ω E.P 1.11 Determine the resistance between the points A and B in the network shown in Fig. E.P. 1.11. Figure E.P. 1.11 Ans : 4.23 Ω 110 j E.P Network Theory 1.12 Determine the current in the galvanometer branch of the bridge network shown in Fig. E.P. 1.12. Figure E.P. 1.12 Ans : 10.62μA 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 i2 then; i1 + i2 ! ! ! v1 v2 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. 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 6 Ω resistor 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 = A i1 = 3+6 9 Circuit Theorems j 161 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 i EXAMPLE = i1 + i2 = 12 A 9 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 3.5 162 j Network Theory 6 = 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, 0 = iA = io where R1 iR2 R1 + R2 jj = (12 kΩ 12 kΩ) + 12 kΩ = 6 kΩ + 12 kΩ = 18 kΩ and = 12 kΩ 4 10 3 12 103 iA = (12 + 18) 103 = 1:6 mA Again applying the current division principle, 12 iA 00 io = = 0:8 mA 12 + 12 ) R2 Thus; io = io 0 + io 00 = 0:3 + 0:8 = 0:5 mA Figure 3.6(a) Figure 3.6 Circuit Theorems EXAMPLE j 163 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 and 6 V voltage source is shorted as shown in Fig. 3.8. 0 io 12 (2 + 2) 103 = 3 mA = 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. 164 j Network Theory Applying KVL clockwise to the upper loop, we get ) 2 10 3 io 00 2 10 io 3 00 = io 4 00 6=0 6 = 1:5 mA 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. Hence; io 000 = 1mA Applying the superposition principle, we find that io = io 0 + io 00 + io 000 =3 1:5 + 1 = 2:5 mA Circuit Theorems EXAMPLE j 165 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 current source. For this condition, refer to Fig. 3.13 for analysis. Figure 3.12 i2 due to the Figure 3.13 Applying KCL at node 1, we get i2 +7= Noting that i2 = we get, v1 = 3i2 v1 2 v1 3 3i2 (3.1) 0 (3.2) 166 j Network Theory 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 EXAMPLE = i1 + i2 7 5 =3 A= A 4 4 3.5 For the circuit shown in Fig. 3.14, find the terminal voltage Vab using superposition principle. Figure 3.14 SOLUTION 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+3 ) ) Vab1 + Vab1 = 0 4Vab1 = 4 Vab1 = 1V 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+3 ) ) Vab2 Figure 3.15 + Vab2 = 0 4Vab2 = 20 Vab2 = 5V Figure 3.16 Circuit Theorems j 167 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 = vx 1 + vx 2 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 1 2 + 4ix1 vx 1 8 But Hence; ) ) ) ix1 vx 1 2 vx 1 2 + + v 4 x21 vx 1 8 2vx1 vx 1 8 4vx1 + vx1 vx 1 =6 = vx 1 2 =6 =6 2vx1 = 48 = 48 = 16V 3 Figure 3.18 168 j Network Theory To find vx2 , deactivate the 6A current source, resulting in a circuit diagram as shown in Fig. 3.19. KCL at node x2 : ( 4ix2 ) =4 8 2 vx 2 vx + 4ix2 + 2 =4 8 2 vx 2 ) + v x2 (3.3) Applying KVL along dotted path, we get vx 2 ) vx 2 = + 4ix2 2ix2 or 2ix2 = 0 ix2 = vx 2 (3.4) 2 Substituting equation (3.4) in equation (3.3), we get vx 2 ) ) ) ) 8 vx 2 +4 + vx 2 2 =4 2 vx 2 8 + 2vx2 vx 2 vx 2 8 vx 2 2 vx 2 2 =4 =4 4vx2 = 32 32 vx 2 = V 3 Hence, according to the superposition principle, vx EXAMPLE = vx 1 + vx 2 32 = 16 = 5:33V 2 Figure 3.19 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 2 Ω resistor ? Figure 3.20 Circuit Theorems j 169 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 2 Ω resistor. Figure 3.21 According to the superposition principle, i1 = i01 + i02 where i01 = Contribution to i1 from 5V source alone. and i02 = Contribution to i1 from 2A source alone. Let us first find i01 . This needs the deactivation of 2A source. Refer to Fig. 3.22. 5 = 1:22A 2 + 2:1 Similarly to find i02 we have to disable the 5V source by shorting it. 0 i1 Referring to Fig. 3.23, we find that 0 i2 Figure 3.22 = = 2 2:1 = 2 + 2:1 1:024 A Figure 3.23 170 j Network Theory Total current, i1 = i01 + i02 = 1:22 1:024 = 0:196 A Thus; P2Ω = (0:196)2 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, V1 = V10 + V100 where V10 is the contribution from 60V source alone and V100 is the contribution from 4A current source alone. To find V10 , the 4A current source is opened, resulting in a circuit as shown in Fig. 3.25. Figure 3.25 Circuit Theorems j 171 Applying KVL to the left mesh: 30ia 60 + 30 (ia ib ) Also ib =0 (3.5) = 0:4iA = 0:4 ( ia ) = 0:4ia (3.6) Substituting equation (3.6) in equation (3.5), we get ) 60 + 30ia 30 0:4ia = 0 60 ia = = 1:25A 48 1:25 ib = 0:4ia = 0:4 30ia = 0:5A 0 Hence; V1 = (ia ib ) 30 = 22:5 V To find, V100 , the 60V source is shorted as shown in Fig. 3.26. Figure 3.26 Applying KCL at node a: Va + Va 20 30Va ) 00 V1 =4 10 20V100 = 800 (3.7) Applying KCL at node b: 00 V1 30 Also; Va = 20ia V1 ) Va V1 10 ) ib = 0:4ib = Va 20 0:4Va + = 30 10 20 7:2Va + 8V100 = 0 00 Hence; 00 + 00 V1 Va (3.8) 172 j Network Theory Solving the equations (3.7) and (3.8), we find that 00 V1 Hence V1 = 60V = V10 + V100 = 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 = 1:5 mA To find ia3 , deactivate 20V source and the 5 mA source. The resulting circuit diagram is shown in Fig 3.28. 10mA 2k = 1 mA ia3 = 18k + 2k Hence, total current ia = ia1 + ia2 + ia3 = 1:5 + 1 = 2:5 mA Circuit Theorems j 173 Figure 3.28 (b) Refer to Fig. 3.29 KCL at node y: Vy 18 103 + Vy 2 20 = (10+5) 10 103 Solving, we get Hence; ia = 3 = 45V: 45 = 3 10 18 103 = 2:5 mA Vy Vy 18 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. 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 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. Circuit Theorems 175 j (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 (c) Rt = b voc 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 . (c) Rt = vab 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 = 2 Ω. Refer Fig. 3.33. Figure 3.33 SOLUTION Figure 3.34 176 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. 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 jj = (5 Ω 20 Ω) + 4 Ω = 5 20 +4=8Ω 5 + 20 Rt Referring to Fig. 3.35, 50 + 25I = 0 Hence Vab ) I = 2A Figure 3.36 = 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 i = 40 = 4A 2+8 Circuit Theorems EXAMPLE j 177 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 ) Hence; 30 40 V2 = 60 Volts 60 + 20 Voc + V2 V 60 0 =I 2 1:5 = 0 60 60 + 20 60 = 60 = 45 V 80 = 178 j Network Theory To find isc : _ Applying KCL at node 2: ) Therefore; V2 20 + 30 V2 1:5 = 0 40 V2 = 30V isc = Rt = V2 20 Voc isc = 1:5A = 45 1:5 = 30 Ω 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 Rt by deactivating all the independent sources, Rt Rt Rt jj = 60 Ω (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. Circuit Theorems EXAMPLE 3.12 Find the Thevenin equivalent for the circuit shown in Fig. 3.41 with respect to terminals a 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 = 12 V To find Rt : (Refer Fig. 3.43) Since both dependent and independent sources are present, Thevenin resistance is found using the relation, Figure 3.42 Rt = voc isc Applying KVL clockwise for mesh 1 : 20 + 6i1 ) 2i + 6 (i1 12i1 Since i = i1 ) i2 , i2 ) =0 6i2 = 20 + 2i we get 12i1 6i2 = 20 + 2 (i1 10i1 4i2 = 20 i2 ) Applying KVL clockwise for mesh 2 : ) j 10i2 + 6 (i2 i1 ) =0 6i1 + 16i2 = 0 Figure 3.43 b. 179 180 j Network Theory Solving the above two mesh equations, we get EXAMPLE i2 = Rt = ) 120 A 136 voc isc = isc = i2 = 120 A 136 12 = 13:6 Ω 120 136 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, Applying KVL to mesh 2 : ) i1 = 4 mA 12 + 6 12 + 6 10 3 10 3 i2 4 (i2 10 + 3 103 i2 = 0 3 3 i1 ) +3 10 i2 =0 Circuit Theorems Solving, we get i2 j 181 = 4 mA ! a b ! 3 kΩ, we get + 3 10 = 0 = 4 10 + 3 10 = 4 10 4 10 + 3 10 4 10 Applying KVL to the path 4 kΩ 4 ) 10 3 i1 Voc Voc 3 3 i2 3 i1 3 i2 3 3 3 = 28V To find Rt : Deactivating all the independent sources, we get the circuit diagram shown in Fig. 3.46. Figure 3.46 jj = Rab = 4 kΩ + (6 kΩ 3 kΩ) = 6 kΩ Rt Hence, the Thevenin equivalent circuit is as shown in Fig. 3.47. Figure 3.47 Figure 3.48 If we connect the 2 kΩ resistor to this equivalent network, we obtain the circuit of Fig. 3.48. Vo 10 28 2 10 = (6 + 2) 10 =i 2 3 3 EXAMPLE 3 = 7V 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 ? 182 j Network Theory 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 = A i1 = 900 + 600 1500 120 120 i2 = = A 1204 + 800 2004 Applying KVL clockwise along the path 1204Ω b a 900 Ω, we get ! 1204i2 ) ! Vt 900i1 = 0 900i1 120 900 = 1204 2004 = 95:8 mV Vt = 1204i2 120 1500 Figure 3.49(b) To find Rt : Deactivate all the independent sources and look into the terminals Thevenin’s resistance. Figure 3.49(c) a b Figure 3.49(d) to determine the Circuit Theorems Rt jj iG jj = 95:8 10 3 840:64 + 30 Ω Figure 3.50 = 110:03 A 3.15 For the circuit shown in Fig. 3.51, find the Thevenin’s equivalent circuit between terminals 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, 0:1 Vc + I = 3 At a : 0:2Va At c : 0:1Va + 0:3 Vc At b : 183 = Rab = 600 900 + 800 1204 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. EXAMPLE j 0:1 Vb = 4 0:1Vc + 0:2 Vb I =1 As the terminals a and b are shorted Va = and the above equations become Vb Figure 3.52 a and b. 184 j Network Theory 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 = 1 A. 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 = 10 V = Voc = Vt 10 Therefore Rt = = = 10Ω Isc 1 The Thevenin’s equivalent is as shown in Fig 3.53 Voc 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). Circuit Theorems j 185 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(d) Figure 3.55(c) Figure 3.56 Thevenin equivalent circuit EXAMPLE 3.17 Find the Thevenin equivalent circuit as seen from the terminals a shown in Fig. 3.57. b. Refer the circuit diagram 186 j Network Theory Figure 3.57 SOLUTION Since the circuit has no independent sources, i = 0 when the terminals a b are open. Therefore, Voc = 0. The onus is now to find Rt . Since Voc = 0 and isc = 0, Rt cannot be determined from Rt = Voc isc . Hence, we choose to connect a source of 1 A at the terminals a b as shown in Fig. 3.58. Then, after finding Vab , the Thevenin resistance is, Rt KCL at node a : 2i Va 5 Also; i Va Hence; ) 2 5 Va 10 = + = + Vab 1 Va 1=0 10 Va 10 Va 10 50 Va = V 13 1=0 50 Ω 1 13 Alternatively one could find Rt by connecting a 1V source at the terminals a b and then find 1 . The concept of finding Rt by connecting a 1A source the current from b to a. Then Rt = Hence; Rt = Va = 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 is shown in Fig. 3.58 (a). Figure 3.58 Figure 3.58 (a) b Circuit Theorems EXAMPLE j 187 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). 1 Ω Rt = I Figure 3.60(a) Applying KCL at node a: I Since; Vx we get, I Hence; = 0:5Vx + Vx 4 = 1V = 0:5 + 1 = 0:75 A 4 Figure 3.60(b) 1 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 a b and then measure Vab (Fig. 3.60 (c)). Consequently, Rt = Vab 1 = Vab Ω: 188 j Network Theory Applying KCL at node a: 0:5Vx + Hence Rt = Vx 4 Vab 1 =1 = ) Vx 1 Vx = 1:33V = 1:33 Ω The corresponding Thevenin equivalent circuit is same as shown in Fig. 3.60(b) 3.3 Figure 3.60(c) 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 . In fact, source transformation of Thevenin equivalent circuit leads to RN = Rt and iN = 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. Circuit Theorems j 189 (b) Find the open-circuit voltage voc . (c) Rt = RN = voc 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 (c) Rt = b vab and find vab . 1 Note: Also, since vt = voc and iN = isc Rt = voc isc = RN 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 = a0 i + a1 v1 + a2 v2 + a3 i1 + a4 i2 (3.9) = a0 i + b0 where b0 (3.10) = a1 v1 + a2 v2 + a3 i1 + a4 i2 = 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, this fact in equation 3.10, we find that b0 = vt . i = 0 and v = voc = vt . Making use of 190 j Network Theory (ii) When all the internal sources are deactivated, become v = a0 i = Rt i b0 ) = 0. This enforces equation 3.10 to a0 = Rt Rt Vt Figure 3.62 Figure 3.63 Thevenin’s equivalent circuit of Fig. 3.62 Current-driven circuit 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 = Rt i + v1 which expresses the voltage-current relationship at terminals Thus, the two circuits of Fig. 3.62 and 3.63 are equivalent. x y of the circuit in Fig. 3.63. 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 = c0 v + d 0 (3.11) where c0 v 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 = c0 v 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 Thus, equation (3.11) becomes i = = 1 Rt 1 Rt v isc v iN y. 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. 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, we get 0 24 4 3 + isc = 0 ) 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 Rt in the Thevenin equivalent circuit. 4 12 RN = =3Ω 4 + 12 Figure 3.67 192 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. 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 = 0A 1000 12 isc = A = 2 mA 6000 ia = Figure 3.70 Figure 3.70(a) Circuit Theorems To find RN : The procedure for finding RN is same that of Rt in the Thevenin equivalent circuit. Rt = RN = voc 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 ) ) Therefore; 12 + 6000ia + 2000ia + 1000ia = 0 4 ia = A 3000 4 voc = ia 1000 = V 3 4 voc 3 RN = = = 667 Ω isc 2 10 3 Figure 3.71 The Norton equivalent circuit along with resistor R is as shown below: ib = isc 2 = 2mA = 1mA 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 j 193 194 j Network Theory SOLUTION We are interested in Io , hence the 2 kΩ resistor 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 = 12 V Applying KCL at node V2 : V2 V1 6 kΩ + V2 2 kΩ + V2 V1 3 kΩ =0 Substituting V1 = 12 V and solving, we get V2 = 6V isc = V1 V2 3 kΩ + V1 4 kΩ = 5 mA To find RN : Deactivate all the independent sources (refer Fig. 3.73(c)). Figure 3.73(c) Figure 3.73(d) Circuit Theorems j 195 Referring to Fig. 3.73 (d), we get RN jj jj = Rab = 4 kΩ [3 kΩ + (6 kΩ 2 kΩ)] = 2:12 kΩ Hence, the Norton equivalent circuit along with 2 kΩ resistor is as shown in Fig. 3.73(e). Io = EXAMPLE isc R RN + RN = 2:57mA Figure 3.73(e) 3.22 Find Vo in the circuit of Fig. 3. 74. Figure 3.74 SOLUTION Since we are interested in Vo , the voltage across 4 kΩ resistor, remove this resistance from the circuit. This results in a circuit diagram as shown in Fig. 3.75. Figure 3.75 196 j Network Theory To find isc , short the terminals a b : Circuit Theorems Constraint equation : i2 i1 KVL around supermesh : 4+2 KVL around mesh 3 : 8 10 ( 3 i3 = 4mA 10 i2 ) 3 +2 i1 j 197 (3.12) 10 3 +4 i2 =0 i3 i1 ) =0 isc i1 ) =0 10 ( 3 (3.13) Since i3 = isc , the above equation becomes, 8 10 ( 3 isc i2 ) +2 10 ( 3 (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 jj = 6 kΩ 10 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 4 kΩ resistor that was removed earlier to get the network shown in Fig. 3.76(b). Figure 3.76(a) 198 j Network Theory Vo = isc (RN = isc jj R) RN R RN +R = 258 mV Figure 3.76(b) Norton equivalent circuit with R = 4 kΩ 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 = 0 when the terminals a b are short-circuited. 5 = 10 mA 500 Therefore, for the right–hand portion of the circuit, isc = Then i = 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 Therefore i = = 25(10i) = 250i vab 250 Substituting i into the mesh equation (3.15), we get ) RN 5 + 500 vab + vab = 0 250 = 5V 5 = 50 Ω 0:1 vab = Rt voc isc = vab isc = 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 3.78 200 j Network Theory 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: 1= ) v1 100 0:03v1 KCL at node 2: ) v2 200 v2 + v1 v2 50 0:02v2 = 1 v1 + 0:1v1 = 0 50 0:08v1 + 0:025v2 = 0 + Solving the above two nodal equations, we get = 10:64 volts ) voc = 10:64 volts 10:64 Hence; R N = 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). v1 voc EXAMPLE 3.25 Find the Thevenin and Norton equivalent circuits for the network shown in Fig. 3.80 (a). Figure 3.80(a) Circuit Theorems j 201 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 = =8Ω Hence, Rt = RN = 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 202 j Network Theory EXAMPLE 3.26 If an 8 kΩ load is connected to the terminals of the network in Fig. 3.81, VAB = 16 V. If a 2 kΩ 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 = 2 kΩ; If RL = 10 kΩ; I ) = 6 mA ) = 10 mA I Voc = 20 + 0:01Rt Voc = 60 + 0:006Rt Solving, we get Voc = 120 V, Rt = 10 kΩ. If 3.4 RL = 20 kΩ; I = Voc (RL + Rt ) = (20 120 + 10 103 10 ) = 4 mA 3 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). We wish to find the value of the load RL such that the maximum power is delivered to it. The power that is delivered to the load is given by p 2 =i RL = Vt Rt + RL Figure 3.82 Circuit A with load RL 2 RL (3.16) Circuit Theorems j 203 Assuming that Vt and Rt are fixed for a given source, the maximum power is a function of In order to determine the value of RL that maximizes p, we differentiate p with respect to RL and equate the derivative to zero. RL . " dp = Vt2 dRL which yields RL # (Rt + RL )2 2 (Rt + RL ) =0 (RL + Rt )2 = Rt (3.17) To confirm that equation (3.17) is a maximum, 2 it should be shown that d p 2 dRL < 0. Hence, maxi- 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, 2 2 Pmax = Vt R L (2RL )2 = Figure 3.83 Thevenin equivalent circuit is substituted for circuit A Vt 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. 204 j Network Theory 180 = 1A 150 + 30 i = 150 V Voc = Vt = 150 i = To find Rt , deactivate the 180 V source. This results in the circuit diagram of Fig. 3.85(b). Rt jj = Rab = 30 Ω 150 Ω 30 150 = = 25 Ω 30 + 150 Figure 3.85(a) The Thevenin equivalent circuit connected to the load resistor is shown in Fig. 3.86. Maximum power transfer is obtained when RL = Rt = 25 Ω: Then the maximum power is (150)2 4RL 4 25 = 2:25 Watts The Thevenin source Vt actually provides a total power of 2 Pmax Pt = Vt = Figure 3.85(b) 150 = 150 25 + 25 = 150 i = 450 Watts Thus, we note that one-half the power is dissipated in RL . EXAMPLE Figure 3.86 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 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) Rt Figure 3.88(c) jj jj = Rab = 6 kΩ 6 kΩ 6 kΩ = 2 kΩ To find Vt : Refer the Fig. 3.88(d). Constraint equation : V3 V1 By inspection, KCL at supernode : = 12 V V2 =3V V2 V3 6k ) V3 6k 3 + + 12 V3 6k V1 6k + + V3 V1 V2 =0 6k 12 3 =0 6k Figure 3.88(d) j 205 206 Network Theory j ) ) ) ) V3 3 + V3 12 + V3 15 = 0 3V3 = 30 = 10 V3 = Vab = V3 = 10 V Vt Figure 3.88(e) The Thevenin equivalent circuit connected to the load resistor RL is shown in Fig. 3.88(e). Pmax = i2 RL = Vt 2 RL 2RL = 12:5 mW 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 = 2 kΩ. 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 = 3 V. Constraint equation : V3 ) V1 = 12 V1 = V3 V1 V2 12 KCL at supernode : V2 V3 6k ) ) ) ) V3 3 6k V3 + V3 + 12 6k 3 + V3 6k 3 + + V3 V3 2k + V3 6k 12 =0 + =0 2k 6k 15 + 3V3 + V3 12 = 0 6V3 = 30 V3 2 Hence; V1 Pmax = V3 RL = =5 V 25 = 12:5 mW 2k Figure 3.88(f) Circuit Theorems EXAMPLE j 207 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 = 2 kΩ + 3 kΩ + 5 kΩ = 10 kΩ For maximum power transfer RL = Rt = 10 kΩ. Let us next find Voc or Vt . Refer Fig. 3.89 (c). By inspection, i1 = 2 mA & i2 = 1 mA. 208 j Network Theory Figure 3.89(b) ! 3 kΩ ! 2 kΩ ! , we get ) + 2k + = 0 5k + 3k ( 1 10 2 10 +3 10 2 10 +2 10 Applying KVL clockwise to the loop 5 kΩ ) 5 10 1 10 ) ) 3 i1 i2 3 a i2 i1 3 3 5 9 b Vt 3 3 3 4 + Vt = 0 Vt = 18 V: The Thevenin equivalent circuit with load resistor RL is as shown in Fig. 3.89 (d). 18 = 0:9 mA i = (10 + 10) 103 Then, Pmax = PL = (0:9 mA)2 10 kΩ = 8:1 mW Figure 3.89(c) EXAMPLE Figure 3.89(d) 3.30 Find the maximum power dissipated in RL . Refer the circuit shown in Fig. 3.90. Figure 3.90 + Vt = 0 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 jj = Rab = [140 Ω 60 Ω] + 8 Ω 140 60 = + 8 = 50 Ω: 140 + 60 For maximum power transfer, RL = Rt = 50 Ω. Next step in the analysis is to find Vt . Refer Fig 3.91(b), using the principle of current division, i1 i2 + 20 170 = = 17 A 170 + 30 = 20 30 = = = i R1 R2 R2 i R1 R1 + R2 600 = 3A 200 170 + 30 Figure 3.91(a) 210 j Network Theory Applying KVL clockwise to the loop comprising of 50 Ω 50i2 10i1 + 8 ) 50(3) ) 0+ Vt EXAMPLE b, we get 10 (17) + Vt = 0 Vt = 20 V 20 = 0:2A 50 + 50 2 Pmax = iT 50 = 0:04 50 = 2 W = a =0 The Thevenin equivalent circuit with load resistor as shown in Fig. 3.91(c). iT ! 10 Ω ! 8 Ω ! RL is Figure 3.91(c) 3.31 Find the value of find Pmax . RL for maximum power transfer in the circuit shown in Fig. 3.92. Also 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 jj =8Ω 2Ω 8 2 = = 1:6 Ω 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 ) 20 + 10i ! 3 Ω ! 2 Ω ! 10 V ! 5 Ω ! 30 V, we get 10 30 = 0 60 i = = 6A 10 212 j Network Theory Again applying KVL clockwise to the path 2 Ω 2i 10 Vt = 0 ) Vt = 2(6) = 2i ! 10 V ! a b, we get 10 10 = 2 V The Thevenin equivalent circuit with load resistor RL is as shown in Fig. 3.95 (a). Pmax = i2T RL 2 = EXAMPLE Vt 4Rt = 625 mW Figure 3.95(a) Thevenin equivalent circuit 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 : ) Hence; 0 ia 0:9 + 10i0a = 0 0 ia Voc = 0:1 A = 3 10i0a =3 10 0 1 = 3 V : Circuit Theorems j 213 To find Rt , we need to compute isc with all independent sources activated. KCL at node A: ) Hence isc ia 00 0:9 + 10ia 00 = 0 ia = 10ia = 10 = = 0:1 A 01=1A 00 Rt 00 : Voc = isc 3 =3Ω 1 Hence, for maximum power transfer RL = Rt = 3 Ω. The Thevenin equivalent circuit with RL = 3 Ω inserted between the terminals a b gives the network shown in Fig. 3.97(a). iT Pmax = 3 = 0:5 A 3+3 = i2T RL = (0:5)2 3 = 0.75 W EXAMPLE Figure 3.97(a) 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 12 + 2 103 i + 2Vx0 = 0 0 Also Hence; Vx 10 10 12 = 4 10 12 + 2 i =1 3i 3 3 i +2 1 10 = 3 mA 3i =0 Figure 3.98(b) 214 j Network Theory !2 ! 1 10 Applying KVL to loop 1 kΩ Vx 3 ) Vt a, b i we get + 2Vx0 Vt =0 10 + 2 1 10 = 1 10 + 2 10 = 3 10 3 10 3 =1 3 i 3 3 3 i i 3 =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: 12 + Vx 00 + 0 = 0 ) ) 00 = 12 i1 = 12 Vx 1 10 3 ) i1 = 12 mA Applying KVL to mesh 2: 1 ) 10 3 i2 1 + 2Vx 00 = 0 10 3 i2 = 24 i2 = 24 mA Applying KCL at node a: isc = i1 i2 = 12 + 24 = 36 mA Hence; Rt = Vt isc = Voc isc 9 = 36 10 = 250 Ω 3 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 = A 250 + 250 500 2 Pmax = iT 250 iT = 9 2 500 = 81 mW = 250 Figure 3.98 (c) Thevenin equivalent circuit Circuit Theorems EXAMPLE j 215 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 Fig. 3.99(a). RL from the original circuit, we get the circuit shown in Figure 3.99(a) KCL at node v1 : 100 v1 2 + v1 13i0a + 5 v2 v1 4 =0 (3.18) Constraint equations : 0 ia v1 v2 4 0 va = 100 v1 = va0 = v1 (3.19) 2 v2 (applying K C L at v2 ) (3.20) (potential across 4 Ω) (3.21) 216 j Network Theory From equations (3.20) and (3.21), we have v2 ) ) ) v1 4 v2 v1 5v1 = v1 v2 = 4v1 4v2 5v2 = 0 v1 (3.22) = v2 Making use of equations (3.19) and (3.22) in (3.18), we get 100 v1 2 ) ) ) ) 13 v2 + v1 ) (100 2 + 5 100) + 2 5 (v1 13 v1 500 + 2v1 5v1 13 (100 v1 v1 4 v1 ) =0 =0 2 100 + 13v1 = 0 20v1 = 1800 Hence; We know that, v1 = 90 Volts vt = v2 = v1 = 90 Volts Rt = voc isc = vt isc The short circuit current is calculated using the circuit shown below: 00 Here Applying KCL at node v1 : ia 100 v1 2 ) 100 v1 2 + v1 = + 100 v1 2 13ia 0 v1 + =0 5 4 (100 v1 ) 13 v1 2 + =0 5 4 v1 Circuit Theorems Solving we get v1 = 80 volts = va00 Applying KCL at node a : 0 v1 4 ) + isc = va00 v1 + va00 4 80 = + 80 = 100 A 4 voc vt = Rt = isc Hence; = isc isc 90 = 0:9 Ω = 100 Hence for maximum power transfer, RL = Rt = 0:9 Ω The Thevenin equivalent circuit with RL = 0:9 Ω is as shown. 90 90 = 0:9 + 0:9 1:8 2 Pmax = it 0:9 it = = EXAMPLE 90 2 1:8 0 9 = 2250 W : 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 j 217 218 j Network Theory 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 25i1 i3 ) + 4i1 = 0 i2 20i3 = 200 KVL clockwise to mesh 2 : ) ) 14i0a + 2 (i2 14 (i1 i3 ) + 2 (i2 i3 ) + 1 (i2 i1 ) =0 i3 ) + 1 (i2 i1 ) =0 13i1 + 3i2 16i3 = 0 KVL clockwise to mesh 3 : ) 2 (i3 i2 ) 100 + 3i3 + 20 (i3 20i1 i1 ) =0 2i2 + 25i3 = 100 Solving the mesh equations, we get i1 = 2:5A; i3 = 5A Applying KVL clockwise to the path comprising of a ) Vt 20i0a = 0 Vt = 20i0a = 20 (i1 i3 ) = 20 ( 2:5 = 150 V 5) b ! 20 Ω, we get Circuit Theorems Next step is to find Rt . Rt When terminals a b Voc = isc = Vt isc are shorted, i00a = 0. Hence, 14 i00a 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 + 3i2 i1 ) =0 2i3 = 0 KVL clockwise to mesh 3 : ) 100 + 3i3 + 2 (i3 i2 ) =0 2i2 + 5i3 = 100 j 219 220 j Network Theory Solving the mesh equations, we find that i1 ) = 40A; i3 = 20A; = i1 i3 = 60A 150 Rt = = = 2:5 Ω isc 60 For maximum power transfer, RL = Rt = 2:5 Ω. The Thevenin equivalent circuit with RL is as shown below : isc Vt Pmax = i21 RL 150 = 2:5 + 2:5 = 2250 W EXAMPLE 2 25 : 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) vL iL is a maximum, (b) vL is a maximum and (c) iL is a maximum. SOLUTION Load current calculation: r 10 250 =r 200 mA 20 20W to 80 Ω corresponds to iL = 80 = 500 mA Using the formula for division of current between two parallel branches : 10W to 250 Ω corresponds to iL = i2 = In the present context, 0:2 = and 0:5 = i R1 R1 + R2 IN R N RN + 250 IN R N RN + 80 (3.23) (3.24) Circuit Theorems j 221 Solving equations (3.23) and (3.24), we get IN RN = 1:7 A = 33:33 Ω (a) If vL iL is maximum, = RN = 33:33 Ω 33:33 iL = 1:7 33:33 + 33:33 = 850 mA RL vL (b) vL = IN (RN . RL = Then, iL = 0 and 1 jj = iL RL = 850 10 33 33 3 : = 28:33 V RL ) is a maximum when vL jj RN RL is a maximum, which occurs when = 1 7 33 33 = 1:7 RN : : = 56:66 V (c) 3.5 iL = IN R N RN ) + RL is maxmimum when RL = 0 Ω iL = 1:7A and vL = 0 V 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 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. 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 Zt = Voc 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 Zt = 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 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 EXAMPLE j 223 3.37 Find the Thevenin and Norton equivalent circuits at the terminals Fig. 3.103. a b for the circuit in 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 = 8 + j 10 j 5 8 + j5 j 320 Vt = Io (j 10) = = 33:92 /58 V 8 + j5 Io = 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 224 j Network Theory jj Zt = j 10 (8 j 5) Ω (j 10)(8 j 5) = j 10 + 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 Vt 33:92 /58 = Zt 10 /26 = 3:392 /32 A IN = 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 ) 5 sin (t + 30 ) = 5 cos (t 60 ) L C ! = 1F ! = 1H j !L 1 j !C ! ! =j = 10 / 45 V 5 / 60 V 11= 1Ω 1 11 = 1Ω j j j 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 Vt = 4:97 / 40:54 Solving; To find Zt deactivate all the independent sources in Fig. 3.105(b). This results in a network as shown in Fig. 3.105(c) : Zt = Zab = 3Ω j 1 Ω j3 3 = = (1 + j 3) Ω 3+j 10 Figure 3.105(c) jj The Thevenin equivalent circuit along with the capicator is as shown in Fig 3.105(d). Vt ( j 1) Zt j 1 4:97 / 40:54 ( = 0:3(1 + j 3) j 1 = 15:73 /247:9 V Vo = Hence; vo j 1) = 15:73 cos (t + 247:9 ) V Figure 3.105(d) Thevenin equivalent circuit 226 j Network Theory 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 (j 10) + 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 A is connected at the terminals as shown below : Circuit Theorems Applying KVL clockwise we get, V0a ) 3V0a 0 4Va j 10I + Vo = 0 j 10I + Vo = 0 V0a = 10I Since = Vo Vo = 40 + j 10Ω Zt = I we get 40I Hence; j 10I Hence the Thevenin equivalent circuit is as shown in Fig 3.106(a) : EXAMPLE Figure 3.106(a) 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 j 10 Voc = Voc =0 j5 100 100 j = / 90 V 3 3 10 + j 227 228 j Network Theory 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 / 90 Voc 10 3 Hence; Zt = = = / 90 Ω Isc 10 /0 3 The Thevenin equivalent and the Norton equivalent circuits are as shown below. Figure Thevenin equivalent EXAMPLE Figure Norton equivalent 3.41 Find the Thevenin and Norton equivalent circuits in frequency domain for the network shown in Fig. 3.109. Circuit Theorems Figure 3.109 SOLUTION Let us find Vt = Vab using superpostion theorem. (i) Vab due to 100 /0 100 /0 100 = A j 300 + j 100 j 200 = I1 (j 100) 100 (j 100) = 50 /0 Volts = j 200 I1 = Vab1 (ii) Vab due to 100 /90 j 229 230 j Network Theory 100 /90 j 100 j 300 Vab2 = I2 ( j 300) 100 /90 = ( j 100 j 300 Vt = Vab1 + Vab2 I2 = Hence; = j 300) = j 150 V 50 + j 150 = 158:11 /108:43 V To find Zt , deactivate all the independent sources. jj Zt = j 100 Ω j 300 Ω j 100( j 300) = = j 150 Ω j 100 j 300 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. 158:11 /108:43 Vt = = 1:054 /18:43 A Zt 150 /90 ZN = Zt = j 150 Ω IN = The Norton equivalent circuit is as shown in Fig. 3.109(b). Figure 3.109(a) Figure 3.109(b) Circuit Theorems 3.6 j 231 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 ZL = RL + jXL and The current through the load is I= Vt Vt = Zt + ZL (Rt + jXt ) + (RL + jXL ) p The phasors I and Vt are the maximum values. The corresponding RM S values are obtained by dividing the maximum values by 2. Also, the RM S 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 P = = jj 1 2 I RL 2 jV j t 2 2 RL 2 (Rt + RL ) (Xt + XL )2 (3.25) Our idea is to adjust the load parameters RL and XL so that P is maximum. To do this, we get @P @RL and @P @XL equal to zero. 232 j Network Theory @P @XL @P @RL = = h j j Vt 2 RL (Xt + XL ) (Rt + RL )2 + (Xt + XL )2 j j Vt 2 h i2 (Rt + RL )2 + (Xt + XL )2 h 2 (Rt + RL )2 + (Xt + XL )2 @P Setting @XL XL @P and Setting @RL RL i 2RL (Rt + RL ) i2 = 0 gives = (3.26) Xt = 0 gives q = Rt2 + (Xt + XL )2 (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 P j j = Vt 2 8Rt In a situation where the load is purely real, the condition for maximum power transfer is obtained by putting XL = 0 in equation (3.27). That is, RL = q Rt2 j j + Xt2 = Zt 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, choose q XL as close as possible to Xt and then adjust RL as close as possible to Rt2 + (XL + Xt )2 : (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 ] = Zt . j j Zt is the complex conjugate of Zt . Circuit Theorems EXAMPLE j 233 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. ZL = 5 + j 6 10 /0 = 1 /0 I= 5+5 Hence Hence, the maximum average power transfered to the load is P EXAMPLE jj 1 2 I RL 2 1 = (1)2 5 = 2:5 W 2 = 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 3.113 234 j Network Theory 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 Vt = Voc = 4 /0 Hence 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 + j 4 Ω 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 j 4. It = 12 /0 3 + j4 + 3 j4 = 2 /0 A(RMS) Hence, maximum average power delivered to the load is P = jj It 2 RL = 4(3) = 12 W 1 It may be noted that the scaling factor is not taken since the phase current is already 2 expressed by its RM S value. Circuit Theorems EXAMPLE j 235 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 (j 20) 150 /30 j 20 = (40 j 30 + j 20) = 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). jj Zt = (40 j 30) j 20 j 20 (40 j 30) = = (9:412 + j 22:35) Ω j 20 + 40 j 30 236 Network Theory j The Value of RL that will absorb the maximum average power is RL j j = Zt = p (9:412)2 + (22:35)2 = 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 Pmax where ) = 1 2 jj It 2 Figure 3.117 (a) RL 72:76 /134 (9:412 + j 22:35 + 24:25) = 1:8 /100:2 A 1 2 Pmax = (1:8) 24:25 2 = 39:29 W It = 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 . Circuit Theorems Vt = Voc = I1 ( = j 10) 120 /0 ( 10 + j 15 j 10 j 10) = 107:33 / 116:57 V The next step is to find Zt . This requires deactivating the independent voltage source of Fig. 3.119. jj Zt = (10 + j 15) ( = =8 Figure 3.119 j 10) j 10 (10 j 10 + j 15) + 10 + j 15 j 14 Ω The value of ZL for maximum average power absorbed is Zt = 8 + j 14 Ω The Thevenin equivalent circuit along with ZL = 8 + j 14 Ω is as shown below: 107:33 / 116:57 8 j 14 + 8 + j 14 107:33 = / 116:57 A 16 1 2 Pmax = It R L 2 1 107:33 2 8 = 2 16 = 180 Walts It = Hence; jj j 237 238 j Network Theory 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 = (3000 j 4000) Ω. Since the voltage source is given in terms of its RM S value, the average maximum power delivered to the load is Pmax RL 10 /0 3000 + j 4000 + 3000 10 = A 2 3000 2 Pmax = It RL 100 = 3000 4 (3000)2 = 8:33 mW jj ) (b) Since = 2 It = where XC jj = It RL and XC Thus, are restricted, we firstqset 2000 Ω. Next we set RL as close to q RL = j 4000 Rt2 XC as close to 4000 Ω as possible; hence 2 + (XC + XL ) as possible. 30002 + ( 2000 + 4000)2 = 3605:55 Ω Since RL can be varied between 0 to 4000 Ω, we can set adjusted to a value ZL = 3605:55 j 2000 Ω: RL to 3605:55 Ω. Hence ZL is Circuit Theorems 10 /0 3000 + j 4000 + 3605:55 = 1:4489 / 16:85 mA It = j 239 j 2000 The maximum average power delivered to the load is Pmax jj = It 2 RL = 1:4489 10 3 2 = 7:57 mW 3605 55 : Note that this is less than the power that can be delivered if there are no constraints on and XL . EXAMPLE RL 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 that jZ j = pjZ j t L = (3000)2 + (4000)2 = 5000 Ω: Hence; j j p ZL = ZL / 45 5000 5000 j = 2 2 p 45 , for maximum power transfer to ZL it is mandatory 240 j Network Theory It = 10 /0 (3000 + 3535:53) + j (4000 3535:53) = 1:526 / 4:07 mA Pmax jj = It 2 RL = 1:526 10 3 2 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 2 Ω resistor 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) Then; 1 +2 1 ) 1 R1 = (8 R2 = 1:6 + 4 = 5:6Ω = 1:6Ω = (5:6 1 + 4 1 ) 1 = 2:3333Ω 20 = 3:16 A Current supplied by the source = 4 + 2:3333 4 = 1:32 A Current in branch ab = Iab = 3:16 4 + 4 + 1: 6 8 = 1:05 A Current in 2Ω; I1 = 1:32 10 R3 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 + 2 242 j Network Theory Therefore, Icd = 3:6842 8 +8 6 = 2 1053A : I2 = 2:1053 = 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 Putting the above three mesh equations in matrix form, we get 2 4 6:375 2 3 2 14 10 32 54 3 10 14 I1 I2 3 2 5=4 I3 0 0 10 j 243 3 5 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 4 6:375 2 3 2 3 14 10 10 14 32 54 0 I1 0 I2 3 2 5=4 0 I3 10 0 0 3 5 Using Cramer’s rule, we get 0 I3 Since I1 = I30 = EXAMPLE = 2A 2 A, the reciprocity theorem is verified. 3.50 Find the current Ix in the j 2 Ω impedance and hence verify reciprocity theorem. Figure 3.127 244 j Network Theory SOLUTION With reference to the Fig. 3.127, the current through j 2 Ω impepance is found using series parallel reduction techniques. Total impedance of the circuit is jj ZT = (2 + j 3) + ( j 5) (3 + j 2) ( j 5)(3 + j 2) = 2 + j3 + j5 + 3 + j2 = 6:537 /19:36 Ω The total current in the network is 36 /0 6:537 /19:36 = 5:507 / 19:36 A IT = Using the principle of current division, we find that IT ( j 5) j5 + 3 + j2 = 6:49 / 64:36 A Ix = 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 jj Z0T = (3 + j 2) + (2 + j 3) ( j 5) (2 + j 3) ( j 5) = (3 + j 2) + 2 + j3 j5 = 9:804 /19:36 Ω The total current in the circuit is I0T = 36 /0 0 ZT = 3:672 / 19:36 A Figure 3.128 Using the principle of current division, I0T ( j 5) = 6:49 / 64:36 A j5 + 2 + j3 It is found that Ix = Iy , thus verifying the reciprocity theorem. Iy = EXAMPLE 3.51 Refer the circuit shown in Fig. 3.129. Find current through the ammeter, and hence verify reciprocity theorem. Circuit Theorems j 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 4 16 1 10 1 26 20 10 20 30 32 54 I1 I2 3 2 5=4 I3 0 0 50 3 5 Using Cramer’s rule, we get I1 = 4:6 A I2 = 5:4 A Hence current through the ammeter = I2 I1 = 5:4 4:6 = 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 4 15 0 10 0 25 20 10 20 31 32 54 0 I1 0 I2 0 I3 3 2 5=4 50 50 0 3 5 Using Cramer’s rule we get 0 I3 = 0:8 A Figure 3.131 245 246 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 4 12 0 0 2 + j 10 2 2 2 2 9 32 54 I1 I2 I3 3 2 5=4 20 20 0 3 5 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 Putting the three equations in matrix form, we get 2 4 12 0 0 2 + j 10 2 2 2 2 9 32 5 64 I01 I02 I03 3 2 75 = 4 0 0 20 j 247 3 5 Using Cramer’s rule, we get I01 = 0:3876 / 2:35 A I02 = 0:456 / 78:9 A I02 Hence; I01 = 0:3179 j 0:4335 = 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. En Millman’s theorem states that if n number of generators having generated emfs E1 , E2 ; 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. E1 Y1 + E2 Y2 + : : : + En Yn Y1 + Y2 + : : : + Yn 1 Z= Y1 + Y2 + : : : + Yn E= where and where Y1 ; Y2 Yn are the admittances corresponding to the internal impedances Z1 ; Z2 and are given by 1 Z1 1 Y2 = Z2 .. . 1 Yn = Zn Z n Y1 = Fig. 3.134 shows a number of generators having emfs E1 ; E2 En connected in parallel Zn are the respective internal impedances of the across the terminals x and y . Also, Z1 ; Z2 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 ) ) E1 E E2 E En E + + + =0 Z1 Z2 Zn 1 1 E E En 1 1 2 =E + + + + + + Z1 Z2 Zn Zn Z1 1 Z2 + En Yn = E E1 Y1 + E2 Y2 + Z Figure 3.135 where Z = Equivalent internal impedance. + E Y ] = EY E Y + E Y + + E Y E= Y Y = Y + Y + + Y 1 1 Z= = Y Y + Y + + Y [E1 Y1 + E2 Y2 + or ) 1 where and EXAMPLE 1 1 n 2 n 2 n n 2 1 n 2 n 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 P Q as shown in Fig. 3.137. E= E1 Y1 + E2 Y2 E3 Y3 Y1 + Y2 + Y3 22 1 1 12 1 1 1 + + 5 12 4 = 10:13 Volts 1 R= Y1 + Y2 + Y3 1 = 0:2 + 0:083 + 0:25 = 1:88 Ω = 5 + 48 12 Hence; EXAMPLE 1 4 Figure 3.137 IL = E R + 10 = 0:853 A 3.54 Find the current through (10 j 3)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 j 3 Ω: 250 j Network Theory Figure 3.139 E= E1 Y1 + E2 Y2 E3 Y3 Y1 + Y2 + Y3 1 100 /0 5 = + 90 /45 1 10 + 80 /30 1 20 1 1 1 + + 5 10 20 = 88:49 /15:66 V Z=R= 1 = Y1 + Y2 + Y3 1 1 5 + 1 10 + 1 20 = 2:86 Ω I= 88:49 /15:66 E = = 6:7 /28:79 A Z + ZL 2:86 + 10 j 3 E= E1 Y1 + E2 Y2 + E3 Y3 + E4 Y4 Y1 + Y2 + Y3 + Y4 Alternately, = 100 5 1 + 90 45 10 1 + 80 30 5 1 + 10 1 + 20 1 + (10 j 3) 20 1 1 = 70 /12 V Therefore; I = 70 /12 10 j 3 = 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+j 5) Ω impedance. Circuit Theorems j 251 Figure 3.140 SOLUTION The original circuit is redrawn after performing source transformation of 5 A in parallel with 4 Ω resistor into an equivalent voltage source and is shown in Fig. 3.141. Figure 3.141 Treating the branch 5 + j 5Ω as a branch with Es = 0V , E1 Y1 + E2 Y2 + E3 Y3 + E4 Y4 EP Q = Y1 + Y2 + Y3 + Y4 4 2 1 + 8 3 1 + 20 4 1 = 1 2 + 3 1 + 4 1 + (5 j 5) 1 = 8:14 /4:83 V Therefore current in (5 + j 5)Ω is I= 8:14 /4:83 = 1:15 / 40:2 A 5 + j5 Alternately EP Q with (5 + j 5) open E1 Y1 + E2 Y2 + E3 Y3 Y1 + Y2 + Y3 4 2 1 + 8 3 1 + 20 = 2 1+3 1+4 1 = 8:9231V EP Q = 4 1 252 j Network Theory Equivalent resistance R = (2 1 + 3 Therefore current in (5 + j 5)Ω is I EXAMPLE = 1 +4 1) 1 = 0:9231Ω 8:9231 = 1:15 / 40:2 A 0:9231 + 5 + j 5 3.56 Find the current through 2 Ω resistor using Millman’s theorem. in Fig. 3.142. Refer the circuit shown Figure 3.142 SOLUTION The Thevenin equivalent circuit using Millman’s theorem for the given problem is as shown in Fig. 3.142(a). E= where E1 Y1 + E2 Y2 Y1 + Y2 10 /10 = 1 + 25 /90 3 + j4 1 1 + 3 + j4 5 1 5 = 10:06 /97:12 V 1 1 Z= = 1 1 Y1 + Y2 + 3 + j4 5 = 2:8 /26:56 Ω Hence; IL = E 10:06 /97:12 = Z + 2 2:8 /26:56 + 2 = 2:15 /81:63 A Figure 3.142(a) Circuit Theorems Reinforcement problems R.P 3.1 Find the current in 2 Ω resistor 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, where Hence; Then; ∴ R1 = 3 2 +1 12 5 (1:2 + 1) 12 = 1:8592 Ω = 14:2 2 Ia = = 0:5182 A 2 + 1:8592 12 Ib = Ia = 0:438 A 12 + 1 + 1:2 3 I1 = 0:438 = 0:2628 A 5 Figure R.P. 3.1(a) With 4V-source acting alone, the circuit is as shown in Fig. R.P. 3.1(b). Figure R.P.3.1(b) j 253 254 j Network Theory The resistance as seen by 4V-source is 3 + R2 where R2 = 2 12 +1 2 14 2:7143 2 = 1:1551 Ω = 4:7143 4 Hence; Ib = = 0:9635 A 3 + 1:1551 Ib 2:7143 = 0:555 A Thus; I2 = 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) Circuit Theorems j 255 The impedance as seen from the voltage source is Z = (4 j 2) + (8 + j 10) ( 8 + j8 j 2) Ia = = 6:01 / 45 Ω j 20 = 3:328 /135 A Z 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 Hence; 8I1 + (I1 and (I2 I1 ) ( I2 ) ( 5) j 10 + (I1 j 2) + (I2 5) ( j 2) j 2) =0 + 4I2 = 0 Simplifying, we get (8 + j 8)I1 + j 2I2 = j 50 and j 2I1 + (4 j 4)I2 = j 10 Solving, we get 8 + j 8 j2 Ib = I2 = 8 + j 8 j 50 j 10 j2 4 j4 = 2:897 / 23:96 A j2 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 1 Ω resistor. 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. 3.3 256 j Network Theory 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 /10 = 1H ! 1 = 1F ! 1 = H! 2 1 1 = F! 2 5 cos (t + 10 ) L1 C1 L2 C2 ∴ V = j1 Ω j !L1 j !C1 j !L2 j !C2 = j1 =j 1 Ω 2 = j2 va (t) Let us next assume that in Fig. R.P. 3.3(b). ! i2 (t) C2 L2 ) = 5 cos [t + 10 ] alone is acting. The resulting network is shown = 2 rad=sec ! 3 /0 1 = 1F ! = 1H ! 1 1 = F! 2 1 = H! 2 3 sin 2t L1 Figure R.P. 3.3(a) Ω Va = 5 /10 V ) C1 Ω A 1 Ω j !C1 2 j !L1 = j 2 Ω j !C2 j !L2 = j = j1 Figure R.P. 3.3(b) = j1 Ω Vb = 3 /0 vb (t) Ω 1 +1 51 5 = 2 5 /33 7 j : j : : : A = 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 = 10 V 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) Circuit Theorems R.P j 257 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) VA I2 10 100 = 450 100 = 209 V 5 =01 10 = = = 1 66 45 5 45 + 5 15+ 50 = 05+ 10 = I1 : ; IB I2 : Hence; VB I2 : IB = 2:5 V Thus; VAB = Vt = VA VB = 20 9 2:5 = 5 Volts 18 : I2 258 j Network Theory 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 , we get RB = 35 10 = 7 Ω 50 ; RE = 35 5 = 3:5 Ω; 50 RF = 5 10 = 1 Ω 50 The reduced network after transformation is as shown in Fig. R.P. 3.4(c). Figure R.P. 3.4(c) Hence; RAB 350 100 4:5 1:5 + +7 450 6 = 85:903 Ω = Rt = 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 18 IG = = 2:4mA 85:903 + 30 Negative sign implies that the current flows from B to A. R.P Figure R.P. 3.4(d) 3.5 Find Is and R so that the networks N1 and N2 shown in Fig. R.P. 3.5 are equivalent. Circuit Theorems j 259 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). V I R = IS R (3.28) From N3 , we can write, From N2 we can write, Also from N2 , I V ) ) ) V V V = 10Ia 3= 2Ia 3= 3= I 5 2 I 10 I 5 =3 (3.29) For equivalence of N1 and N2 , it is requirred that equations (3.28) and (3.29) must be same. Comparing these equations, we get IR R R.P = I 5 = 0:2 Ω and and IS R IS =3 = 3 = 15A 0:2 3.6 Obtain the Norton’s equivalent of the network shown in Fig. R.P. 3.6. Figure R.P. 3.6 Figure R.P. 3.5(a) 260 j Network Theory 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 9I1 + 0I2 6I3 = 30 (3.30) (ii) 0I1 + 25I2 + 15I3 = 30 (3.31) (iii) 6I1 + 15I2 + 23I3 = 4VX = 4 (10I2 ) (i) ) 6I1 25I2 + 23I3 = 0 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 Hence; Then; Hence; 9I1 = 30 30 A I1 = 9 25I2 = 50 30 and I2 = A 25 30 VX = 10I2 = 10 = 12 V 25 6I1 4VX Vt = Voc = 15I2 = Thus; Rt = and 50 V Voc Isc = 50 = 1:4706 34 Ω Hence, Norton’s equivalent circuit is as shown in Fig. R.P. 3.6(b). Figure R.P. 3.6(b) (3.32) Circuit Theorems R.P j 261 3.7 For the network shown in Fig. R.P. 3.7, find the Thevenin’s equivalent to show that and Vt = Zt = V1 2 3 (1 + a + b ab) b 2 Figure R.P. 3.7 SOLUTION With xy open, I1 = Hence, Voc aV1 V1 2 = Vt = aV1 + I1 + bI1 = aV1 + = V1 2 aV1 V1 2 [1 + a + b +b V aV1 1 2 ab] 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) (ii) ) ) I1 + (I1 2I1 (I2 I1 ) I2 ) = V1 aV1 I2 = V1 aV1 (3.33) + I2 = aV1 + bI1 (3.34) (1 + b) I1 + 2I2 = aV1 Solving equations (3.33) and (3.34), we get Isc = I2 = V1 (1 + a + b 3 b ab) 262 j Network Theory Hence; Zt = = R.P Voc Isc 3 = (1 + a + b 2 V1 (1 + a + b V1 ab) ab) (3 b) b 2 3.8 Use Norton’s theorem to determine are in ohms. I in the network shown in Fig. R.P. 3.8. Resistance Values 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 ABC F EA, we get 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 EDC EF , we get (x y 30) 0:03 + (x ) y 55) 0:02 (y 20) 0:02 0:05x 0:01y = 0 0:08y = 1:6 (3.36) Circuit Theorems j 263 Figure R.P. 3.8(a) Solving equations (3.35) and (3.36), we get x Hence; Isc = 46:76 A = IN = 120 x = 78:24 A The circuit to calculate opened. Rt is as shown in Fig. R.P. 3.8(b). All injected currents have been Rt = 0:03 + 0:04 + 0:03 0:05 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). I = 78:24 0 08875 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 is 3 mW. Figure R.P. 3.9 SOLUTION For a resistive network, the maximum power delivered to the load is 2 Pmax = Vt 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, we get 1 Vt R + 2 Vt R + Simplifying we get Vt = 2 Volts With all voltage sources shorted, the resistance, found as follows: 1 ) Rt Rt = = 1 R R 3 + Ω 1 R 3 Vt R Rt + Figure R.P. 3.9(a) =0 as viewed from the terminals, 1 R = 3 R a and b is Circuit Theorems 22 3 = = 3 × 10−3 R R 4× 3 R = 1 kΩ Hence, Pmax = ⇒ R.P | 265 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). Zt = = R1 (jX1 ) R1 + jX1 R1 X12 + jR12 X1 R12 + X12 Figure R.P. 3.10(a) For maximum power transfer, ZL = Z∗t ⇒ Hence, R2 + jX2 = R2 = ⇒ R1 X12 R12 X1 − j R12 + X12 R12 + X12 R1 X12 R12 + X12 X1 = ±R1 X2 = − R2 R1 − R2 R12 X1 R12 + X12 Substituting equation (3.37) in equation (3.38) and simplifying, we get X2 = R2 (R1 − R2 ) (3.37) (3.38) 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. Ans : ix = E.P 1 A 4 Figure E.P. 3.1 3.2 Find the current through branch P Q using superposition theorem. Figure E.P. 3.2 Ans : E.P 1.0625 A 3.3 Find the current through 15 ohm resistor using superposition theorem. Figure E.P. 3.3 Ans : 0.3826 A Circuit Theorems E.P 3.4 Find the current through 3 + j 4 Ω using superposition theorem. Figure E.P. 3.4 Ans : E.P 8.3 /85.3 A 3.5 Find the current through Ix using superposition theorem. Figure E.P. 3.5 Ans : E.P 3.07 / 163.12 A 3.6 Determine the current through 1 Ω resistor using superposition theorem. Figure E.P. 3.6 Ans : 0.406 A j 267 268 j E.P Network Theory 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 : E.P Vt = 6.29 V, Rt = 9.43 Ω 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 : E.P Vt = 0.192 / 43.4 V, Zt = 88.7 /11.55 Ω 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 = 4 Ω Circuit Theorems E.P 3.10 Find the Thevenin equivalent circuit across a b. Refer Fig. E.P. 3.10. Figure E.P. 3.10 Ans : E.P Vt = 30 V, Rt = 10 kΩ 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 : E.P Verify your result with other methods. 3.12 Find the current through 20 ohm resistor using Norton equivalent. Figure E.P. 3.12 Ans : IN = 4.36 A, RN = Rt = 8.8 Ω, IL = 1.33 A j 269 270 j E.P Network Theory 3.13 Find the current in 10 ohm resistor using Norton’s theorem. Figure E.P. 3.13 Ans : E.P IN = 4 A, Rt = RN 100 = Ω, IL = 7 0.5 A 3.14 Find the Norton equivalent circuit between the terminals a b for the network shown in Fig. E.P. 3.14. Figure E.P. 3.14 Ans : E.P IN = 4.98310 / 5.71 A, Zt = ZN = 3.6 /23.1 Ω 3.15 Determine the Norton equivalent circuit across the terminals Fig. E.P. 3.15. Figure E.P. 3.15 Ans : IN = 5 A, RN = Rt = 6 Ω P Q for the network shown in Circuit Theorems E.P j 271 3.16 Find the Norton equivalent of the network shown in Fig. E.P. 3.16. Figure E.P. 3.16 Ans : E.P IN = 8.87 A, RN = Rt = 43.89 Ω 3.17 Determine the value of RL for maximum power transfer and also find the maximum power transferred. Figure E.P. 3.17 Ans : E.P RL = 1.92 Ω, Pmax = 4.67 W 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 E.P Network Theory 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 : E.P RL = 5.44 Ω, Pmax = 2.94 W 3.20 Determine the value of ZL for maximum power transfer. What is the value of maximum power? Figure E.P. 3.20 Ans : E.P ZL = 4.23 + j1.15 Ω, Pmax = 5.68 Watts 3.21 Obtain the Norton equivalent across x y. Figure E.P. 3.21 Ans : E.P IN = ISC = 7.35A, Rt = RN = 1.52 Ω 3.22 Find the Norton equivalent circuit at terminals a b of the network shown in Fig. E.P. 3.22. Circuit Theorems j 273 Figure E.P. 3.22 Ans : E.P IN = 1.05 /251.6 A, Zt = ZN = 10.6 /45 Ω 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 : E.P IN = 7A, Zt = 8.19 / 55 Ω 3.24 Determine the current through 10 ohm resistor using Norton’s theorem. Figure E.P. 3.24 Ans : 0.15A 274 j E.P Network Theory 3.25 Determine the current I using Norton’s theorem. Figure E.P. 3.25 Ans : E.P Verify your result with other methods. 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 : E.P Vx = 9.28 /21.81 V 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 Circuit Theorems E.P 3.28 Find the current ix in the bridge circuit and hence verify reciprocity theorem. Figure E.P. 3.28 Ans : E.P ix = 0.031 A 3.29 Find the current through 4 ohm resistor using Millman’s theorem. Figure E.P. 3.29 Ans : E.P I = 2.05 A 3.30 Find the current through the impedance of (10 + j 10) Ω using Millman’s theorem. Figure E.P. 3.30 Ans : 3.384 /12.6 A j 275 276 j E.P Network Theory 3.31 Using Millman’s theorem, find the current flowing through the impedance of (4 + j 3) Ω. Figure E.P. 3.31 Ans : 3.64 / 15.23 A 4.1 Introduction There are many reasons for studying initial and final conditions. The most important reason is that the initial and final conditions evaluate the arbitrary constants that appear in the general solution of a differential equation. In this chapter, we concentrate on finding the change in selected variables in a circuit when a switch is thrown open from closed position or vice versa. The time of throwing the switch is considered to be t = 0, and we want to determine the value of the variable at t = 0 and at + t = 0 , immediately before and after throwing the switch. Thus a switched circuit is an electrical circuit with one or more switches that open or close at time t = 0. We are very much interested in the change in currents and voltages of energy storing elements after the switch is thrown since these variables along with the sources will dictate the circuit behaviour for t > 0. Initial conditions in a network depend on the past history of the circuit (before t = 0 ) and structure of the network at t = 0+ , (after switching). Past history will show up in the form of capacitor voltages and inductor currents. The computation of all voltages and currents and their derivatives at t = 0+ is the main aim of this chapter. 4.2 Initial and final conditions in elements 4.2.1 The inductor The switch is closed at t = 0. Hence t = 0 corresponds to the instant when the switch is just open and t = 0+ corresponds to the instant when the switch is just closed. The expression for current through the inductor is given by i = 1 L Zt Figure 4.1 Circuit for explaining vd switching action of an inductor 278 j Network Theory i i(t) = 1 L Z0 vd = i(0 ) + 1 L + 1 L Zt Zt vd 0 vd 0 Putting t = 0+ on both sides, we get i(0 + i(0 + ) = i(0 ) + 1 L Z0+ vd 0 ) = i(0 ) The above equation means that the current in an inductor cannot change instantaneously. Consequently, if i(0 ) = 0, we get i(0+ ) = 0. This means that at t = 0+ , inductor will act as an open circuit, irrespective of the voltage across the terminals. If i(0 ) = Io , then i(0+ ) = Io . In this case at t = 0+ , the inductor can be thought of as a current source of Io A. The equivalent circuits of an inductor at t = 0+ is shown in Fig. 4.2. t = 0+ Figure 4.2 The initial-condition equivalent circuits of an inductor The final-condition equivalent circuit of an inductor is derived from the basic relationship v =L di dt di Under steady condition, = 0. This means, v = 0 and hence L acts as short at t = (final dt or steady state). The final-condition equivalent circuits of an inductor is shown in Fig.4.3. t=¥ Io Io Figure 4.3 The final-condition equivalent circuit of an inductor Initial Conditions in Network j 279 4.2.2 The capacitor The switch is closed at t = 0. Hence, t = 0 corresponds to the instant when the switch is just open and t = 0+ corresponds to the instant when the switch is just closed. The expression for voltage across the capacitor is given by = v v (t) v (t) = 1 C 1 C Zt Z0 id = v (0 ) + + switching action of a Capacitor Evaluating the expression at t = v (0 Figure 4.4 Circuit for explaining id + 1 C 0+ , ) = v (0 ) + 1 C Z Zt id 0 t id 0 we get 1 C Z0+ id v (0 + ) = v (0 ) 0 Thus the voltage across a capacitor cannot change instantaneously. If v (0 ) = 0, then v (0+ ) = 0. This means that at t = 0+ , capacitor C acts as short circuit. q0 q0 Conversely, if v (0 ) = then v (0+ ) = . These conclusions are summarized in Fig. 4.5. C C Figure 4.5 Initial-condition equivalent circuits of a capacitor The final–condition equivalent network is derived from the basic relationship i dv =C dv dt Under steady state condition, = 0. This is, at t = , i = 0. This means that t = dt or in steady state, capacitor C acts as an open circuit. The final condition equivalent circuits of a capacitor is shown in Fig. 4.6. 280 j Network Theory Figure 4.6 Final-condition equivalent circuits of a capacitor 4.2.3 The resistor The cause–effect relation for an ideal resistor is given by v = Ri. From this equation, we find that the current through a resistor will change instantaneously if the voltage changes instantaneously. Similarly, voltage will change instantaneously if current changes instantaneously. 4.3 Procedure for evaluating initial conditions There is no unique procedure that must be followed in solving for initial conditions. We usually solve for initial values of currents and voltages and then solve for the derivatives. For finding initial values of currents and voltages, an equivalent network of the original network at t = 0+ is constructed according to the following rules: (1) Replace all inductors with open circuit or with current sources having the value of current flowing at t = 0+ . (2) Replace all capacitors with short circuits or with a voltage source of value vo = is an initial charge. q0 C if there (3) Resistors are left in the network without any changes. EXAMPLE 4.1 Refer the circuit shown in Fig. 4.7(a). Find for t < 0. i1 (0 +) and iL (0 + ). The circuit is in steady state = = t=0 Figure 4.7(a) 1W 1W Initial Conditions in Network j 281 SOLUTION The symbol for the switch implies that it is open at t = 0 and then closed at t = 0+ . The circuit is in steady state with the switch open. This means that at t = 0 , inductor L is short. Fig.4.7(b) shows the original circuit at t = 0 . Using the current division principle, 1W 1W Figure 4.7(b) 2 1 = 1A iL (0 ) = 1+1 Since the current in an inductor cannot change instantaneously, we have iL (0 + ) = iL (0 ) = 1A At t = 0 , i1 (0 ) = 2 1 = 1A. Please note that the current in a resistor can change instantaneously. Since at t = 0+ , the switch is just closed, the voltage across R1 will be equal to zero because of the switch being short circuited and hence, i1 (0 + ) = 0A Thus, the current in the resistor changes abruptly form 1A to 0A. EXAMPLE 4.2 Refer the circuit shown in Fig. 4.8. Find vC (0+ ). Assume that the switch was in closed state for a long time. Figure 4.8 SOLUTION The symbol for the switch implies that it is closed at t = 0 and then opens at t = 0+ . Since the circuit is in steady state with the switch closed, the capacitor is represented as an open circuit at t = 0 . The equivalent circuit at t = 0 is as shown in Fig. 4.9. 282 j Network Theory vC (0 ) = i(0 )R2 Using the principle of voltage divider, vC (0 )= VS R1 + R2 R2 = 5 1 = 2:5 V 1+1 Since the voltage across a capacitor cannot change instaneously, we have vC (0 Figure 4.9 + ) = vC (0 ) = 2:5V That is, when the switch is opened at t = 0, and if the source is removed from the circuit, still + vC (0 ) remains at 2.5 V. EXAMPLE 4.3 Refer the circuit shown in Fig 4.10. Find iL (0+ ) and vC (0+ ). The circuit is in steady state with the switch in closed condition. Figure 4.10 SOLUTION The symbol for the switch implies, it is closed at t = 0 and then opens at t = 0+ . In order to find vC (0 ) and iL (0 ) we replace the capacitor by an open circuit and the inductor by a short circuit, as shown in Fig.4.11, because in the steady state L acts as a short circuit and C as an open circuit. Figure 4.11 5 =1A 2+3 Using the voltage divider principle, we note that iL (0 )= vC (0 Then we note that: vC (0 )= + iL (0 + 5 3 =3V 3+2 ) = vC (0 ) = 3 V ) = iL (0 ) = 2 A Initial Conditions in Network EXAMPLE j 283 4.4 In the given network, K is closed at t = 0 with zero current in the inductor. Find the values of i; di dt , 2 d i dt2 at t = 0+ if R = 8Ω and L = 0:2H. Refer the Fig. 4.12(a). SOLUTION Figure 4.12(a) The symbol for the switch implies that it is open at t = 0 and then closes at t = 0+ . Since the current i through the inductor at t = 0 is zero, it implies that i(0+ ) = i(0 ) = 0. To find + di(0 ) dt and 2 + d i(0 ) dt2 R = 8W + : Applying KVL clockwise to the circuit shown in Fig. 4.12(b), we get +L Ri 8i + 0:2 di dt di dt Figure 4.12(b) = 12 = 12 At t = 0+ , the equation (4.1) becomes 8i(0+ ) + 0:2 8 0+02 : di(0 +) dt + di(0 dt + di(0 ) ) dt = 12 = 12 12 0:2 = 60 A=sec = Differentiating equation (4.1) with respect to t, we get 8 di dt 2 d i dt2 =0 2 + d i(0 ) dt2 =0 + 0:2 At t = 0+ , the above equation becomes 8 Hence di(0 +) dt 8 + 0:2 60 + 0 2 : 2 + d i(0 ) dt2 2 + d i(0 ) dt2 L = 0.2H =0 = – 2400 A=sec2 (4.1) 284 j Network Theory EXAMPLE 4.5 In the network shown in Fig. 4.13, the switch is closed at t = 0. Determine i; di dt , 2 d i dt2 at t = 0+ . Figure 4.13 SOLUTION The symbol for the switch implies that it is open at t = 0 and then closes at t = 0+ . Since there is no current through the inductor at t = 0 , it implies that i(0+ ) = i(0 ) = 0. R = 10W L = 1H C = 1mF Figure 4.14 Writing KVL clockwise for the circuit shown in Fig. 4.14, we get Ri +L di dt Ri + 1 Zt = 10 (4.2) + vC (t) = 10 (4.2a) i( )d C +L 0 di dt Putting t = 0+ in equation (4.2a), we get Ri + 0 +L R 0+ di + vC 0+ = 10 dt 0+ L 0+ di + 0 = 10 dt + 0 di dt = 10 L = 10 A/sec Differentiating equation (4.2) with respect to t, we get R di dt +L 2 d i dt2 + i(t) C =0 Initial Conditions in Network At t = 0+ , the above equation becomes R EXAMPLE t dt Hence at 0+ di R +L 2 d i 10 + 100 + 0+ + i 0+ 2 d i 0+ dt2 =0 dt2 =0 + 0 285 dt2 C 2 + d i 0 0 + L 2 dt C 2 d i = 0+ ; j =0 = 100 A/sec2 4.6 Refer the circuit shown in Fig. 4.15. The switch K is changed from position 1 to position 2 at t = 0. Steady-state condition having been reached at position 1. Find the values of i, and 2 d i dt2 di dt , at t = 0+ . SOLUTION Figure 4.15 The symbol for switch K implies that it is in position 1 at t = 0 and in position 2 at t = 0+ . Under steady-state condition, inductor acts as a short circuit. Hence at t = 0 , the circuit diagram is as shown in Fig. 4.16. 20 = 2A 10 Since the current through an inductor cannot change instantaneously, i 0+ = i (0 ) = 2A. Since there is no initial charge on the capacitor, vC (0 ) = 0. Since the voltage across a capacitor cannot change instanta+ neously, vC 0 = vC (0 ) = 0. Hence at t = 0+ the circuit diagram is as shown in Fig. 4.17(a). For t 0+ , the circuit diagram is as shown in Fig. 4.17(b). i 0 = Figure 4.17(a) Figure 4.16 Figure 4.17(b) 286 j Network Theory Applying KVL clockwise to the circuit shown in Fig. 4.17(b), we get Ri(t) +L di(t) + dt Ri(t) +L Zt 1 C i( )d =0 (4.3) + vC (t) = 0 (4.3a) 0+ di(t) dt At t = 0+ equation (4.3a) becomes + 0 Ri +L R 0+ di dt 2+ L + vC 0+ = 0 0+ di +0=0 dt 20 + 0+ di dt + 0 di =0 = dt 20 A/sec Differentiating equation (4.3) with respect to t, we get R At t = 0+ , we get R Hence; EXAMPLE 0+ di dt R ( +L di dt 2 d i 20) + +L 0+ 2 d i dt2 + + i i C 0+ dt2 C 2 + d i 0 2 + L 2 dt C 2 + d i 0 dt2 =0 =0 =0 2 106 A=sec2 4.7 In the network shown in Fig. 4.18, the switch is moved from position 1 to position 2 at t = 0. The steady-state has been reached before switching. Calculate i, Figure 4.18 di dt , and 2 d i dt2 at t = 0+ . Initial Conditions in Network j 287 SOLUTION The symbol for switch K implies that it is in position 1 at t = 0 and in position 2 at t = 0+ . Under steady-state condition, a capacitor acts as an open circuit. Hence at t = 0 , the circuit diagram is as shown in Fig. 4.18(a). We know that the voltage across a capacitor cannot change instantaneously. This means that Figure 4.18(a) + = v (0 ) = 40 V. vC 0 C At t = 0 , inductor is not energized. This means that i (0 ) = 0. Since current in an inductor cannot change instantaneously, i 0+ = i (0 ) = 0. Hence, the circuit diagram at t = 0+ is as shown in Fig. 4.18(b). The circuit diagram for t 0+ is as shown in Fig.4.18(c). Figure 4.18(b) Figure 4.18(c) Applying KVL clockwise, we get Ri +L di dt Ri + 1 C +L Zt i( )d =0 (4.4) 0+ di + vC (t) = 0 dt At t = 0+ , we get Ri(0 + )+L 20 di(0 +) dt 0+1 + vC (0+ ) = 0 di(0 +) dt + 40 = 0 + di(0 ) dt = 40A/ sec Diferentiating equation (4.4) with respect to t, we get R di dt +L 2 d i dt2 + i C =0 288 j Network Theory Putting t = 0+ in the above equation, we get R + di(0 ) dt R ( +L 2 + d i(0 ) dt2 40) + L + i(0 +) C 2 + d i(0 ) dt2 + 0 C + 2 d i(0 dt2 Hence EXAMPLE ) =0 =0 = 800A/ sec2 4.8 t In the network shown in Fig. 4.19, v1 (t) = e is initially uncharged, determine the value of 0 and is zero for all for t 2 d v 2 dt2 and 3 d v 2 dt3 t< 0. If the capacitor at t = 0+ . Figure 4.19 SOLUTION Since the capacitor is initially uncharged, + v2 (0 ) = 0 Referring to Fig. 4.19(a) and applying KCL at node v2 (t): v2 (t) v1 (t) R1 1 R1 + +C dv2 (t) dt 1 R2 v2 (t) + +C v2 (t) R2 dv2 (t) dt =0 = v1 (t) R1 0:15v2 + 0:05 Figure 4.19(a) dv2 dt = 0:1e t Putting t = 0+ , we get 0:15v2 (0+ ) + 0:05 0:15 0 + 0 05 : dv2 (0 dt dv2 (0 +) +) dt dv2 (0 dt +) = 0:1 = 0:1 = 0:1 = 2 Volts= sec 0:05 (4.5) Initial Conditions in Network j 289 Differentiating equation (4.5) with respect to t, we get 0:15 dv2 dt + 0:05 2 d v2 dt2 = t 0:1e (4.6) Putting t = 0+ in equation (4.6), we find that 2 + d v2 (0 ) dt2 0:1 0:3 = 8 Volts/ sec2 0:05 Again differentiating equation (4.6) with respect to t, we get 0:15 = 2 d v2 dt2 + 0:05 3 d v2 dt3 = 0:1e Putting t = 0+ in equation (4.7) and solving for 3 + d v2 (0 ) dt3 EXAMPLE = t 3 d v2 + (0 ), dt3 (4.7) we find that 0:1 + 1:2 = 26 Volts/ sec3 0:05 4.9 Refer the circuit shown in Fig. 4.20. The circuit is in steady state with switch K closed. At t = 0, the switch is opened. Determine the voltage across the switch, vK and dvK dt at t = 0+ . Figure 4.20 SOLUTION The switch remains closed at t = 0 and open at t = 0+ . Under steady condition, inductor acts as a short circuit and hence the circuit diagram at t = 0 is as shown in Fig. 4.21(a). Therefore; For t 0 + vK (0 + ) = vK (0 ) =0V the circuit diagram is as shown in Fig. 4.21(b). Figure 4.21(a) Figure 4.21(b) 290 j Network Theory i(t) At (t) = 0+ , =C dvK dt we get i(0 + )=C dvK (0 +) dt Since the current through an inductor cannot change instantaneously, we get i(0 Hence; dvK (0 + ) = i(0 ) = 2A + dvK (0 ) 2=C +) dt EXAMPLE dt = 2 C = 2 1 2 = 4V/ sec 4.10 In the given network, the switch K is opened at t = 0. At t = 0+ , solve for the values of v; and 2 d v dt2 dv dt if I = 2 A; R = 200 Ω and L = 1 H Figure 4.22 SOLUTION The switch is opened at t = 0. This means that at t = 0 , it is closed and at t = 0+ , it is open. Since iL (0 ) = 0, we get iL (0+ ) = 0. The circuit at t = 0+ is as shown in Fig. 4.23(a). Figure 4.23(a) Figure 4.23(b) v (0 + ) = IR =2 200 = 400 Volts Initial Conditions in Network Refer to the circuit shown in Fig. 4.23(b). For t 0+ , the KCL at node v (t) gives I = v (t) + R (4.8) v ( )d 0+ Differentiating both sides of equation (4.8) with respect to t, we get 1 dv (t) 1 0= + v (t) R At t = 0+ , we get 1 R dv (0 +) dt + dt + 1 L 1 dv (0 ) 1 + 200 dt 1 L v (0 291 Zt 1 L j + (4.8a) )=0 400 = 0 dv (0 +) dt = 8 104 V/ sec Again differentiating equation (4.8a), we get 2 d v (t) R dt2 1 At t = 0+ , we get EXAMPLE + 1 dv (t) L dt =0 1 d2 v (0+ ) 1 dv (0+ ) + =0 200 dt2 1 dt 2 + d v (0 ) = 200 2 dt 8 10 4 = 16 106 V/ sec2 4.11 In the circuit shown in Fig. 4.24, a steady state is reached with switch K open. At t = 0, the switch is closed. For element values given, determine the values of va (0 ) and va (0+ ). Figure 4.24 292 j Network Theory SOLUTION At t = 0 , the switch is open and at t = 0+ , the switch is closed. Under steady conditions, inductor L acts as a short circuit. Also the steady state is reached with switch K open. Hence, the circuit diagram at t = 0 is as shown in Fig.4.25(a). iL (0 )= 5 5 2 + = A 30 10 3 Using the voltage divider principle: va (0 )= 5 20 10 = V 10 + 20 3 Since the current in an inductor cannot change instantaneously, iL (0 + ) = iL (0 ) = 2 A: 3 At t = 0+ , the circuit diagram is as shown in Fig. 4.25(b). Figure 4.25(a) Figure 4.25(b) Refer the circuit in Fig. 4.25(b). KCL at node a: va (0 +) 5 + va (0 +) + va (0 10 10 1 1 1 + + + va (0 ) 10 10 20 +) vb (0 +) =0 20 1 5 + vb (0 ) = 20 10 KCL at node b: vb (0 +) va (0 +) vb (0 +) 5 2 + =0 10 3 1 5 1 1 + + vb (0+ ) = + va (0 ) 20 20 10 10 20 + 2 3 Initial Conditions in Network j 293 Solving the above two nodal equations, we get, va (0 EXAMPLE + )= 40 V 21 4.12 Find iL (0+ ); vC (0+ ); dvC (0 dt +) and diL (0 dt +) for the circuit shown in Fig. 4.26. Assume that switch 1 has been opened and switch 2 has been closed for a long time and steady–state conditions prevail at t = 0 . SOLUTION Figure 4.26 At t = 0 , switch 1 is open and switch 2 is closed, whereas at t = 0+ , switch 1 is closed and switch 2 is open. First, let us redraw the circuit at = 0 by replacing the inductor with a short circuit and the capacitor with an open circuit as shown in Fig. 4.27(a). From Fig. 4.27(b), we find that iL (0 ) = 0 t Figure 4.27(a) Figure 4.27(b) 294 j Network Theory Applying KVL clockwise to the loop on the right, we get vC (0 ) 2+1 0=0 Hence, at t + =0 : vC (0 iL (0 vC (0 The circuit diagram for Fig. 4.27(c). + + t )= 2V ) = iL (0 ) = 0A 2V ) = vC (0 ) = 0 + is shown in Figure 4.27(c) Applying KVL for right–hand mesh, we get + iL = 0 vC vL At t = 0+ , we get vL (0 + ) = vC (0+ ) = We know that 2 =L vL iL (0 0= + ) 2V diL dt At t = 0+ , we get diL (0 +) dt = Applying KCL at node X , 2 Consequently, at t = 0+ iC (0 Since + )= iC dvC (0 EXAMPLE +) dt 4.13 For the circuit shown in Fig. 4.28, find: (a) i(0+ ) and v (0+ ) (b) di(0 +) dt and dv (0 dt (c) i( ) and v ( ) +) = 2 = 1 2A/ sec + iC + iL = 0 vC (0 10 +) 2 =C = +) L 10 vC We get, vL (0 iL (0 + )=6 dvC dt + iC (0 C ) = 6 1 2 = 12V/ sec 0=6A Initial Conditions in Network j 295 Figure 4.28 SOLUTION (a) From the symbol of switch, we find that at t = 0 , the switch is closed and t = 0+ , it is open. At t = 0 , the circuit has reached steady state so that the equivalent circuit is as shown in Fig.4.29(a). 12 i(0 ) = = 2A 6 v (0 ) = 12 V Therefore, we have i(0 + ) = i(0 ) = 2A (b) For t 0 +, v (0 + ) = v (0 ) = 12V we have the equivalent circuit as shown in Fig.4.29(b). Figure 4.29(a) Figure 4.29(b) Applying KVL anticlockwise to the mesh on the right, we get vL (t) v (t) + 10i(t) = 0 Putting t = 0+ , we get vL (0 + ) vL (0 v (0 + ) + ) + 10i(0+ ) = 0 12 + 10 2=0 vL (0 + )= 8V 296 Network Theory j The voltage across the inductor is given by vL vL (0 di(0 + =L )=L +) dt = 1 L di dt di(0 +) dt vL (0 + ) 1 ( 8) = 10 Similarly, the current through the capacitor is = iC dv (0 or +) dt = iC (0 = 10 +) C 2 10 6 =C = = 0.8A/ sec dv dt i(0 +) C 0.2 106 V/ sec (c) As t approaches infinity, the switch is open and the circuit has attained steady state. The equivalent circuit at t = is shown in Fig.4.29(c). ) = 0 () = 0 Figure 4.29(c) i( v EXAMPLE 4.14 Refer the circuit shown in Fig.4.30. Find the following: (a) (b) (c) v (0 +) dv (0 and i(0+ ) +) dt and di(0 +) dt ) and () v( i Figure 4.30 SOLUTION From the definition of step function, u(t) = 1; 0; 0 t<0 t> Initial Conditions in Network j 297 From Fig.4.31(a), u(t) = 0 at t = 0 . Similarly; u( t) = or From Fig.4.31(b), we find that u( u( t) t) = 1; 0; 1; 0; 0 t<0 t> 0 t>0 t< = 1, at t = 0 . t = 0+ t = 0 Figure 4.31(a) Figure 4.31(b) Due to the presence of u( t) and u(t) in the circuit of Fig.4.30, the circuit is an implicit switching circuit. We use the word implicit since there are no conventional switches in the circuit of Fig.4.30. The equivalent circuit at t = 0 is shown in Fig.4.31(c). Please note that at t = 0 , the independent current source is open because u(t) = 0 at t = 0 and the circuit is in steady state. Figure 4.31(c) 40 = 5A 3+5 v (0 ) = 5i(0 ) = 25V i(0 Therefore i(0 v (0 )= + ) = i(0 ) = 5A + ) = v (0 ) = 25V (b) For t 0+ ; u( t) = 0. This implies that the independent voltage source is zero and hence is represented by a short circuit in the circuit shown in Fig.4.31(d). 298 j Network Theory Figure 4.31(d) Applying KVL at node a, we get dv 4+i=C At t = 0+ , We get 4 + i(0+ ) = C +) +) dt 4 + 5 = 0:1 dv (0 dt dv (0 v + dv (0 5 + +) dt v (0 +) 5 25 + 5 = 40V/ sec dt Applying KVL to the left–mesh, we get 3i + 0:25 di dt +v =0 Evaluating at t = 0+ , we get 3i(0+ ) + 0:25 3 di(0 +) dt 5 + 0 25 : + v (0+ ) = 0 di(0 di(0 +) dt + dt ) = + 25 = 0 40 1 4 = 160A/ sec (c) As t approaches infinity, again the circuit is in steady state. The equivalent circuit at t = is shown in Fig.4.31(e). Figure 4.31(e) Initial Conditions in Network Using the principle of current divider, we get j 299 4 5 = i( ) = 3+5 v ( ) = (i( ) + 4) 5 2.5A = ( 2:5 + 4)5 = 7.5V EXAMPLE 4.15 Refer the circuit shown in Fig.4.32. Find the following: (a) i(0+ ) and v (0+ ) (b) di(0 +) dt and dv (0 +) dt (c) i( ) and v ( ) Figure 4.32 SOLUTION Here the function u(t) behaves like a switch. Mathematically, u(t) = 1; 0; t> t< 0 0 The above expression means that the switch represented by u(t) is open for t < 0 and remains closed for t > 0. Hence, the circuit diagram of Fig.4.32 may be redrawn as shown in Fig.4.33(a). Figure 4.33(a) For t < 0, the circuit is not active because switch is in open state, This implies that all the initial conditions are zero. That is, iL (0 ) = 0 and vC (0 ) = 0 for t 0 +, the equivalent circuit is as shown in Fig.4.33(b). 300 j Network Theory Figure 4.33(b) From the circuit diagram of Fig.4.33(b), we find that vC i= 5 + At t = 0 , we get + vC (0 ) vC (0 ) 0 + i(0 ) = = = = 0A 5 5 5 Also v = 15iL Evaluating at t = 0+ , we get + + v (0 ) = 15iL (0 ) = 15iL (0 ) = 15 (b) The equivalent circuit at t = We find from Fig.4.33(c) that 0+ iC (0 + 0 = 0V is shown in Fig.4.33(c). ) = 5A Figure 4.33(c) From Fig.4.33(b), we can write vC dvC dt = 5i =5 di dt Multiplying both sides by C , we get C dvC dt = 5C di dt Initial Conditions in Network di(0 = 5C +) 1 + iC (0 ) 5C 1 5 = 5 14 = 4A/ sec dt Also dv dt dv dt dv dt At t = 0+ , we find that dv (0 +) dt dt = v 301 di iC Putting t = 0+ , we get j = 15iL = 15 diL dt = 15 1 diL dt = 15vL = 15vL (0+ ) From Fig.4.33(b), we find that vL (0+ ) = 0 + dv (0 ) = 15 0 Hence; dt EXAMPLE = 0V/ sec 4.16 In the circuit shown in Fig. 4.34, steady state is reached with switch K open. The switch is closed at t = 0. di1 di2 Determine: i1 ; i2 ; and at t = 0+ dt dt Figure 4.34 302 j Network Theory SOLUTION At t = 0 , switch K is open and at t = 0+ , it is closed. At t = 0 , the circuit is in steady state and appears as shown in Fig.4.35(a). 20 = 1:33A 10 + 5 1:33 = 13:3V vC (0 ) = 10i2 (0 ) = 10 i2 (0 Hence; )= Since current through an inductor cannot change instantaneously, i2 (0+ ) = i2 (0 ) = 1.33 A. Also, vC (0+ ) = vC (0 ) = 13:3V. The equivalent circuit at t = 0+ is as shown in Fig.4.35(b). i1 (0 + )= 20 6:7 13:3 = = 0.67A 10 10 Figure 4.35(a) For t 0 +, Figure 4.35(b) the circuit is as shown in Fig.4.35(c). Writing KVL clockwise for the left–mesh, we get 10i1 + 1 C Zt i1 ( )d = 20 0+ Differentiating with respect to t, we get 10 di1 dt + 1 C i1 =0 Putting t = 0+ , we get 10 di1 (0 dt +) = di1 (0 +) dt 1 10 + 1 10 1 C Figure 4.35(c) i1 (0 + i (0 6 1 + )=0 )= 0.67 105 A/ sec Initial Conditions in Network Writing KVL equation to the path made of 20V K 10Ω 2di2 10i2 + = 20 j 303 2H, we get dt 0+ , At t = the above equation becomes 10i2 (0+ ) + EXAMPLE 10 2di2 (0+ ) 1 33 + 2 : dt di2 (0 +) dt di2 (0 dt +) = 20 = 20 = 3.35A/ sec 4.17 Refer the citcuit shown in Fig.4.36. The switch K is closed at t = 0. Find: (a) v1 and v2 at t = 0+ (b) v1 and v2 at t = (c) (d) dv1 and dt 2 d v1 dt2 dv2 dt at t = 0+ at t = 0+ Figure 4.36 SOLUTION (a) The circuit symbol for switch conveys that at t = 0 , the switch is open and t = 0+ , it is closed. At t = 0 , since the switch is open, the circuit is not activated. This implies that all initial conditions are zero. Hence, at t = 0+ , inductor is open and capactor is short. Fig 4.37(a) shows the equivalent circuit at t = 0+ . Figure 4.37(a) 304 j Network Theory 10 = 1A 10 + + v1 (0 ) = 0; i2 (0 ) = 0 i1 (0 + Applying KVL to the path, 10V source (b) At = , switch conditions, capacitor circuit at = . t )= 10Ω 10Ω 2mH, we get K 10 + 10i1 (0+ ) + v1 (0+ ) + v2 (0+ ) = 0 10 + 10 + 0 + v2 (0+ ) = 0 v2 (0 K C + )=0 remains closed and circuit is in steady state. Under steady state is open and inductor L is short. Fig. 4.37(b) shows the equivalent t ) = 10 10+ 10 = 0.5A () = 0 () = 0 5 10 = 5V () = 0 i2 ( i1 v1 v2 (c) For t 0 +, : Figure 4.37(b) the circuit is as shown in Fig. 4.37(c). Figure 4.37(c) Initial Conditions in Network i2 Zt 1 = L v2 ( )d = 0+ j v1 (t) R2 Differentiating with respect to t, we get v2 1 = L dv1 R2 dt Evaluating at t = 0+ we get dv1 (0 +) dv1 (0 R2 = dt +) L2 v2 (0 + ) = 0V/ sec dt Applying KVL clockwise to the path 10 V source C F, K we get Zt 1 10 + 10i + 10Ω 4 [i( ) i2 ( )]d =0 0+ Differentiating with respect to t, we get 10 di dt 1 + C [i i2 ] =0 Evaluating at t = 0+ , we get di(0 +) dt = = = i2 (0 +) C i(0 10 0 1 10 4 10 +) 2 3 ∵ i(0+ ) = i1 (0+ ) + i2 (0+ ) 4 5 =1+0 = 1A 6 25000A/ sec Applying KVL clockwise to the path 10 V source we get 10Ω 10Ω 2 mH, K 10 + 10i + 10i2 + v2 = 0 10i + v1 + v2 = 10 Differentiating with respect to t, we get 10 di dt + dv1 dt + dv2 dt =0 305 306 j Network Theory At t = 0+ , we get 10 di(0 +) dt + dv1 (0 +) dt + 10( 25000) + 0 + dv2 (0 dt dv2 (0 dt dv2 (0 +) =0 +) =0 +) = 25 104 V/ sec dt (d) From part (c), we have Zt 1 L v2 ( )d 0+ = v1 10 Differentiating with respect to t twice, we get 1 dv2 L dt = 1 d2 v1 10 dt2 At t = 0+ , we get 1 dv2 (0 L dt 2 d v 1 (0 dt2 Hence; EXAMPLE +) +) = 1 d2 v1 (0+ ) 10 dt2 = 125 107 V/ sec2 4.18 Refer the network shown in Fig. 4.38. Switch K is changed from a to b at t = 0 (a steady state having been established at position a). Figure 4.38 Show that at t = 0+ . i1 = i2 = V R1 + R2 + R3 ; i3 =0 Initial Conditions in Network j 307 SOLUTION The symbol for switch indicates that at t = 0 , it is in position a and at t = 0+ , it is in position b. The circuit is in steady state at t = 0 . Fig 4.39(a) refers to the equivalent circuit at t = 0 . Please remember that at steady state C is open and L is short. iL1 (0 ) = 0; iL2 (0 ) = 0; vC2 (0 ) = 0; vC1 (0 )=0 Applying KVL clockwise to the left-mesh, we get V + vC3 (0 ) + 0 R2 +0=0 vC3 (0 ) = V volts: Figure 4.39(a) Since current in an inductor and voltage across a capacitor cannot change instantaneously, the equivalent circuit at t = 0+ is as shown in Fig. 4.39(b). Figure 4.39(b) i1 (0 i3 (0 + + ) = i2 (0+ ) since ) = 0 since Applying KVL to the path v C3 (0+ ) V iL1 (0 iL2 (0 + + )=0 )=0 R2 R3 R1 K we get, + R2 i1 (0+ ) + R3 i2 (0+ ) + R1 i1 (0+ ) = 0 308 j Network Theory Since i1 (0+ ) = i2 (0+ ), the above equation becomes = [R1 + R2 + R3 ] i1 (0+ ) V + + i1 (0 ) = i2 (0 ) = A R1 + R2 + R3 V Hence; EXAMPLE 4.19 Refer the circuit shown in Fig. 4.40. The switch K is closed at t = 0. Find (a) di1 (0 dt +) and (b) di2 (0 +) dt Figure 4.40 SOLUTION The circuit symbol for the switch shows that at + t = 0 , it is open and at t = 0 , it is closed. Hence, at t = 0 , the circuit is not activated. This implies that all initial conditions are zero. That is, vC (0 ) = 0 and iL (0 ) = i2 (0 ) = 0. The equivalent circuit at t = 0+ keeping in mind that vC (0+ ) = vC (0 ) and iL (0+ ) = iL (0 ) is as shown in Fig. 4.41 (a). i1 (0 + Figure 4.41(a) ) = 0 and i2 (0+ ) = 0: 0 +. Figure. 4.41(b) shows the circuit diagram for t Vo sin !t = i1 R + 1 C Zt i1 ( )d 0+ Differentiating with respect to t, we get Vo ! cos !t =R di1 dt + i1 C Initial Conditions in Network j 309 At t = 0+ , we get =R Vo ! + di1 (0 ) Also; Vo sin !t = dt di1 (0 +) dt + i1 (0 +) C V0 ω A/ sec R = i2 R + L di2 dt Evaluating at t = 0+ , we get 0 = i2 (0+ )R + L di2 (0 +) dt EXAMPLE di2 (0 +) dt Figure 4.41(b) = 0A/ sec 4.20 In the network of the Fig. 4.42, the switch K is opened at steady state with the switch closed. (a) Find the expression for vK at t = 0+ . t (b) If the parameters are adjusted such that i(0+ ) = 1, and the derivative of the voltage across the switch at t = = 0 after the network has attained di(0 0+ , +) dt dvK dt = 1, what is the value of + (0 ) ? Figure 4.42 SOLUTION At t = 0 , switch is in the closed state and at + t = 0 , it is open. Also at t = 0 , the circuit is in steady state. The equivalent circuit at t = 0 is as shown in Fig. 4.43(a). i(0 )= V R2 and vC (0 ) = 0 Figure 4.43(a) 310 j Network Theory For t 0+ , the equivalent circuit is as shown in Fig. 4.43(b). From Fig. 4.43 (b), vK At t = 0+ , vK vK (0 +) = R1 i + Zt 1 C i( )d 0+ = R1 i + vC (t) = R1 i(0+ ) + vC (0+ ) vK (0 + ) = R1 V + vC (0 ) R2 = R1 V volts R2 Figure 4.43(b) (b) vK dvK dt = R1 i + = R1 di dt Zt 1 C + i( )d 0+ i C Evaluating at t = 0+ , we get dvK (0 dt +) = R1 = R1 1 = C di(0 +) dt ( + 1) + i(0 +) C 1 C R1 volts/ sec Initial Conditions in Network j 311 Reinforcement Problems R.P 4.1 Refer the circuit shown in Fig RP.4.1(a). If the switch is closed at 2 + d iL (0 ) dt2 t = 0, find the value of at t = 0+ . Figure R.P.4.1(a) SOLUTION The circuit at t = 0 is as shown in Fig RP 4.1(b). Since current through an inductor and voltage across a capacitor cannot change instantaneously, it implies that iL (0+ ) = 18A and vC (0+ ) = 180 V. The circuit for t 0+ is as shown in Fig. RP 4.1 (c). Figure R.P.4.1(b) Figure R.P.4.1(c) Referring Fig RP 4.1 (c), we can write 2 10 3 diL dt + 60iL + 288 10 3 Zt iL (t)dt 0+ =0 (4.9) 312 j Network Theory At t = 0+ , we get diL (0 +) = dt = 60 18 + 180 2 10 3 450 103 A= sec Differentiating equation (4.9) with respect to t, we get 2 10 2 3 d iL dt2 diL + 60 dt + 288 10 3 At t = 0+ , we get 2 + d iL (0 ) 2 dt R.P iL =0 60(450)103 288 103 (18) 2 10 3 = 1:0908 1010 A= sec2 = 4.2 For the circuit shown in Fig. RP 4.2, determine 2 + d vC (0 ) dt2 and 3 + d vC (0 ) : dt3 Figure R.P.4.2 SOLUTION Given i(t) = 2u(t) = 2; 0; t t 0 0 + Hence, at t = 0 , vC (0 ) = 0 and iL (0 ) = 0. For t 0+ , the circuit equations are 1 dvC (t) 1 + 64 dt 2 Zt = 2 (4.10) 1 dvC (t) + iL (t) = 64 dt 2 (4.11) vL (t)dt 0+ Initial Conditions in Network [Note : At t = iC 0+ , + iL = j 313 2 because of the capacitor polarity] equation (4.10) gives 1 dvC (0+ ) + iL (0+ ) = 64 dt 2 Since, iL (0+ ) = iL (0 ) = 0, we get 1 dvC (0+ ) +0= 64 dt + dvC (0 ) = 2 128 volts= sec dt Differentiating equation (4.10) with respect to t we get 1 d2 vC (t) 1 + vL (t) = 0 64 dt 2 vL vC Also; 24 At t = 0+ , we get vC (0 1 = 2 +) (4.12) Zt vL dt = iL (4.13) 0+ vL (0 +) = iL (0+ ) 24 Since vC (0+ ) = 0 and iL (0+ ) = 0, we get vL (0+ ) = 0. At t = 0+ , equation (4.12) becomes 1 d2 vC (0+ ) 1 + vL (0+ ) = 0 64 dt2 2 1 d2 vC (0+ ) 1 + 0=0 64 dt2 2 2 + d vC (0 ) =0 2 dt Differentiating equation (4.12) with respect to t we get 1 d3 vC 1 dvL + =0 3 64 dt 2 dt Differentiating equation (4.13) with respect to t, we get dvC dvL dt dt 24 = 1 vL 2 (4.14) 314 j Network Theory At t = 0+ , we get dvC (0 +) dvL (0 dt dt 24 +) dvL (0 128 = +) dt 24 dvL (0 dt 1 + vL (0 ) 2 =0 +) = 128 volts= sec At t = 0+ , equation (4.14) becomes 1 d3 vC (0+ ) 1 dvL (0+ ) + =0 64 dt3 2 dt 3 + d vC (0 ) = 4096 volts= sec3 3 R.P dt 4.3 In the network of Fig RP 4.3 (a), switch K is closed at t = 0. At t = 0 all the capacitor voltages and all the inductor currents are zero. Three node-to-datum voltages are identified as v1 , v2 and + v3 . Find at t = 0 : (i) (ii) v1 , v2 dv1 dt , and v3 dv2 dt and dv3 dt Figure R.P.4.3(a) SOLUTION The network at t = 0+ is as shown in Fig RP-4.3 (b). Since vC and iL cannot change instantaneously, we have from the network shown in Fig. RP-4.3 (b), + v1 (0 ) = 0 v2 (0 v3 (0 + + )=0 )=0 Initial Conditions in Network j 315 Figure R.P.4.3(b) For t 0 +, the circuit equations are = v C1 = v C2 = v C3 Zt 1 C1 i1 dt 0+ Zt 1 C2 i2 dt 0+ Zt 1 C3 i3 dt 0+ 9 > > > > > > > > > > > > > = (4.15) > > > > > > > > > > > > > ; From Fig. RP-4.3 (b), we can write i1 (0 i2 (0 and i3 (0 + + + )= )= v (0 +) R1 v1 (0 ; +) v2 (0 +) R2 )=0 Differentiating equation (4.15) with respect to t, we get dvC1 dt = i1 C1 ; dvC2 dt = i2 and C2 dvC3 dt = i3 C3 At t = 0+ , the above equations give dv1 (0 +) dt dv2 (0 +) dt and dv3 (0 dt +) = = = i1 (0 +) C1 + i2 (0 C2 + i3 (0 C3 ) ) = = v (0 +) R 1 C1 + v1 (0 =0 ) v2 (0 R 2 C2 +) =0 316 j Network Theory R.P 4.4 For the network shown in Fig RP 4.4 (a) with switch K open, a steady-state is reached. The circuit paprameters are R1 = 10Ω, R2 = 20Ω, R3 = 20Ω, L = 1 H and C = 1F. Take V = 100 volts. The switch is closed at t = 0. (a) Write the integro-differential equation after the switch is closed. (b) Find the voltage Vo across C before the switch is closed and give its polarity. (c) Find i1 and i2 at t = 0+ . (d) Find di1 dt and di2 dt at t = 0+ . (e) What is the value of t= ? di1 dt at Figure R.P.4.4(a) SOLUTION The switch is in open state at t = 0 . The network at t = 0 is as shown in Fig RP 4.4 (b). Figure R.P.4.4(b) 100 10 = A R1 + R2 30 3 10 200 VC (0 ) = i1 (0 )R2 = 20 = volts 3 3 Note that L is short and C is open under steady-state condition. For t 0+ (switch in closed state), i1 (0 )= V = we have and 20i1 + 6 Zt 20i2 + 10 di1 dt i2 dt 0+ = 100 (4.16) = 100 (4.17) Initial Conditions in Network j 317 10 A 3 200 + and VC (0 ) = VC (0 ) = Volts 3 From equation (4.16) at t = 0+ , Also i1 (0 + ) = i1 (0 ) = di1 (0 we have +) dt = 100 = 1 100 i2 (0 ) = 20 103 100 A=sec 3 From equation (4.17), at t = 0+ , we have + 20 200 5 = A 3 3 Differentiating equation (4.17), we get 20 di2 dt + 106 i2 = 0 (4.18) From equation (4.18) at t = 0+ , we get 20di2 (0+ ) At t = dt + 106 i2 (0+ ) = 0 di2 (0 dt , +) ) = 100 =5A 20 () = 0 i1 ( di1 dt R.P 106 53 20 106 = A= sec 12 = 4.5 For the network shown in Fig RP 4.5 (a), find 2 + d i1 (0 ) . dt2 The switch K is closed at t = 0. Figure R.P.4.5(a) 318 j Network Theory SOLUTION At t = 0 , we have vC (0 ) = 0 and i2 (0 ) = iL (0 ) = 0. Because of the switching property of L and C , we have vC (0+ ) = 0 and i2 (0+ ) = 0. The network at t = 0+ is as shown in Fig RP 4.5 (b). Figure R.P.4.5(b) Referring Fig RP 4.5 (b), we find that i1 (0 The circuit equations for t 0 + R2 i2 + v (0 )= +) R1 are R1 i1 and + 1 C | + Zt 1 C Zt (i1 = v (t) (4.19) =0 (4.20) 0+ (i2 0+ i2 )dt i1 )dt +L {z di2 dt } vC (t) At t = 0+ , equation (4.20) becomes R2 i2 (0 + ) + vC (0+ ) + L di2 (0 dt di2 (0 +) +) dt =0 (4.21) =0 Differentiating equation (4.19), we get R1 di1 dt + 1 C i2 ) (i1 = dv (t) (4.22) dt Letting t = 0+ in equation (4.22), we get R1 di1 (0 dt +) + 1 C i1 (0 + ) i2 (0 + di1 (0 dt ) = +) = dv (0 +) dt 1 R1 dv (0 dt +) v (0 +) R1 C (4.23) Initial Conditions in Network Differentiating equation (4.22) gives R1 2 d i1 dt2 + 1 R.P + 1 di1 (0 C +) di2 (0 dt di1 di2 dt dt C Letting t = 0+ , we get 2 + d i1 (0 ) R1 dt2 +) dt 2 + d i1 (0 ) R1 dt2 2 + d i1 (0 ) dt2 = = j 319 2 d v (t) dt2 2 + d v (0 ) dt2 = = 1 di1 (0 C +) dt 1 R1 C 1 R1 + 2 + d v (0 ) dt2 dv (0 dt +) 1 2C R1 v (0 + ) + 2 + d v (0 ) dt2 4.6 Determine va (0 ) and va (0+ ) for the network shown in Fig RP 4.6 (a). Assume that the switch is closed at t = 0. Figure R.P.4.6(a) SOLUTION Since L is short for DC at steady state, the network at t = 0 is as shown in Fig. RP 4.6 (b). Applying KCL at junction a, we get va (0 ) 10 5 + va (0 ) vb (0 20 ) =0 Since vb (0 ) = 0, we get va (0 ) 10 5 ) 0 =0 20 0:5 10 va (0 ) = = volts 0:1 + 0:05 3 + va (0 Figure R.P.4.6(b) 320 j Network Theory Also; For t ) = iL (0+ ) = iL (0 0 +, va (0 20 2 = A 3 ) + 5 10 we can write 5 va 10 va vb and + 20 va 10 + + vb va 5 vb 10 20 =0 + iL = 0 Simplifying at t = 0+ , we get 1 + va (0 ) 4 1 + va (0 ) + 20 and Solving we get, va (0 +) = 1 1 + vb (0 ) = 20 2 3 1 + vb (0 ) = 20 6 40 = 1:905 volts 21 Exercise problems E.P 4.1 Refer the circuit shown in Fig. E.P. 4.1 Switch K is closed at t = 0. Find i(0+ ), di(0 +) dt and 2 + d i(0 ) . dt2 Figure E.P.4.1 Ans: i(0+ ) = 0.2A, di(0+ ) = dt 2 103 A/ sec, d2 i(0+ ) = 20 106 A/ sec2 dt2 Initial Conditions in Network E.P dt2 321 4.2 Refer the circuit shown in Fig. E.P. 4.2. Switch K is closed at t = 0. Find the values of i; 2 d i j di dt and at t = 0+ . Figure E.P.4.2 Ans: i(0+ ) = 0, E.P di(0+ ) d2 i(0+ ) = = 10 A/ sec, dt dt2 1000 A/ sec2 4.3 Refering to the circuit shown in Fig. E.P. 4.3, switch is changed from position 1 to position 2 at t = 0. The circuit has attained steady state before switching. Determine i, di dt and Figure E.P.4.3 Ans: i(0+ ) = 0, di(0+ ) = dt 40 A/ sec, d2 i(0+ ) = 800 A/ sec2 dt2 2 d i dt2 at t = 0+ . 322 Network Theory j E.P 4.4 In the network shown in Fig. E.P.4.4, the initial voltage on v1 (0 ) = Va and v2 (0 ) = Vb . Find the values of dv1 dt and C1 dv2 dt is Va and on C2 is Vb such that at t = 0+ . Figure E.P.4.4 Ans: E.P dv1 (0+ ) Vb Va = V/ sec, dt C1 R dv2 (0+ ) Va Vb = V/ sec dt C2 R 4.5 In the network shown in Fig E.P. 4.5, switch K is closed at t = 0 with zero capacitor voltage and zero inductor current. Find 2 d v2 dt2 at t = 0+ . Figure E.P.4.5 Ans: d2 v2 (0+ ) R2 Va = V/ sec2 dt2 R1 L1 C1 Initial Conditions in Network E.P j 323 4.6 In the network shown in Fig. E.P. 4.6, switch K is closed at t = 0. Find 2 d v1 dt2 at t = 0+ . Figure E.P.4.6 Ans: E.P d2 v1 (0+ ) = 0 V/ sec2 dt2 4.7 The switch in Fig. E.P. 4.7 has been closed for a long time. It is open at + dv (0 ) dt t = 0. Find + di(0 ) , i( ) and v ( ). i Figure E.P.4.7 Ans: di(0+ ) = 0A/ sec, dt dv(0+ ) = 20A/ sec, i() = 0A, v() = 12V dt dt , 324 j E.P Network Theory 4.8 In the circuit of Fig E.P. 4.8, calculate iL (0+ ), diL (0 + ) dv (0+ ) C dt , dt , vR ( ), vC ( ) and iL ( ). Figure E.P.4.8 Ans: iL (0+ ) = 0 A, diL (0+ ) = 0 A/ sec dt dvC (0+ ) = 2 V/ sec, dt E.P vR () = 4V, vC () = 20V, iL () = 1A 4.9 Refer the circuit shown in Fig. E.P. 4.9. Assume that the switch was closed for a long time for + diL (0 ) and iL (0+ ). Take v (0+ ) = 8 V. t < 0. Find dt Figure E.P.4.9 Ans: iL (0+ ) = 4 A, E.P diL (0+ ) = 0 A/ sec dt 4.10 Refer the network shown in Fig. E.P. 4.10. A steady state is reached with the switch K closed and + dv2 (0 ) . with i = 10A. At t = 0, switch K is opened. Find v2 (0+ ) and dt Initial Conditions in Network j 325 Figure E.P.4.10 Ans: v2 (0+ ) = 0, E.P 10Ra Rc dv2 (0+ ) = V= sec. dt Ca (Ra + Rb )(Ra + Rc ) 4.11 Refer the network shown in Fig. E.P. 4.11. The network is in steady state with switch K closed. + dvk (0 ) . The switch is opened at t = 0. Find vk (0+ ) and dt Figure E.P.4.11 Ans: vk (0+ ) = Va Rc Volts, Ra + Rb + Rc dvk (0+ ) Va (Ca + Cb ) = V/ sec dt (Ra + Rb + Rc )(Ca Cd + Cb Ca + Cb Cd ) E.P 4.12 Refer the network shown in Fig. E.P. 4.12. Find 2 + d i1 (0 ) . dt2 326 j Network Theory d2 i1 (0+ ) 1 Ans: = 2 dt Ra E.P Figure E.P.4.12 10 10 + 2 2 A/ sec2 Ra Ca 4.13 Refer the circuit shown in Fig. E.P. 4.13. Find steady state at t = 0 . di1 (0 dt +) . Assume that the circuit has attained Figure E.P.4.13 di1 (0+ ) 10 Ans: A/ sec = dt RA E.P 4.14 Refer the network shown in Fig. E.P.4.14. The circuit reaches steady state with switch K closed. + 2 + dv1 (0 ) d v2 (0 ) At a new reference time, t = 0, the switch K is opened. Find . and 2 dt Figure E.P.4.14 dt Initial Conditions in Network Ans: E.P dv1 (0+ ) 10 = V/ sec, dt Ca (Ra + Rb ) j 327 10Rb d2 v2 (0+ ) = V/ sec2 dt2 La Ca (Ra + Rb ) 4.15 The switch shown in Fig. E.P. 4.15 has been open for a long time before closing at t = 0. Find: + + i0 (0 ), iL (0 ) i0 (0 ), iL (0 ), i0 ( ), iL ( ) and vL ( ). Figure E.P.4.15 Ans: i(0 ) = 0, iL (0 ) = 160mA, i0 (0+ ) = 65mA, iL (0+ ) = 160mA, i0 () = 225mA, iL () = 0, vL () = 0 E.P 4.16 The switch shown in Fig. E.P. 4.16 has been closed for a long time before opeing at t = 0. Find: i1 (0 ), i2 (0 ), i1 (0+ ), i2 (0+ ). Explain why i2 (0 ) = i2 (0+ ). $ Figure E.P.4.16 Ans: i1 (0 ) = i2 (0 ) = 0.2mA, i2 (0+ ) = i1 (0+ ) = 0.2mA 328 j E.P Network Theory 4.17 The switch in the circuit of Fig E.P.4.17 is closed at t = 0 after being open for a long time. Find: (a) i1 (0 ) and i2 (0 ) (b) i1 (0+ ) and i2 (0+ ) (c) Explain why i1 (0 ) = i1 (0+ ) $ (d) Explain why i2 (0 ) = i2 (0+ ) Figure E.P.4.17 Ans: i1 (0 ) = i2 (0 ) = 0.2 mA, i1 (0+ ) = 0.2 mA, i2 (0+ ) = 0.2mA Chapter 5 5.1 Laplace Transform Introduction In this chapter, we will introduce Laplace transform. This is an extremely important technique. For a given set of initial conditions, it will give the total response of the circuit comprising of both natural and forced responses in one operation. The idea of Laplace transform is analogous to any familiar transform. For example, Logarithms are used to change a multiplication or division problem into a simpler addition or subtraction problem and Antilogs are used to carry out the inverse process. This example points out the essential feature of a transform: They are designed to create a new domain to make mathematical manipulations easier. After evaluating the unknown in the new domain, we use inverse transform to get the evaluated unknown in the original domain. The Laplace transform enables the circuit analyst to convert the set of integrodifferential equations describing a circuit to the complex frequency domain, where thay become a set of linear algebraic equations. Then using algebraic manipulations, one may solve for the variables of interest. Finally, one uses the inverse transform to get the variable of interest in time domain. Also, in this chapter, we express the impedance in s domain or complex frequency domain. Hence, we may analyze a circuit using one of the reduction techniques such as Thevenin theorem or source transformation discussed in earlier chapters. 5.2 Definition of Laplace transorm A transform is a change in the mathematical description of a physical variable to facilitate computation. Keeping this definition in mind, Laplace transform of a function f (t) is defined as Lff (t)g = F (s) = Z1 f (t)e st dt (5.1) 0 Here the complex frequency is s = + j! . Since the argument of the exponent e in equation (5.1) must be dimensionless, it follows that s has the dimensions of frequency and units of inverse seconds (sec 1 ). 330 j Network Theory The notation implies that once the integral has been evaluated, f (t), a time domain function is transformed to F (s), a frequency domain function. If the lower limit of integration in equation (5.1) is 1, then it is called the bilateral Laplace transform. However for circuit applications, the lower limit is taken as zero and accordingly the transform is unilateral in nature. The lower limit of integration is sometimes chosen to be 0 to permit f (t) to include (t) or its derivatives. Thus we should note immediately that the integration from 0 to 0+ is zero except when an impulse function or its derivatives are present at the origin. Region of convergence The Laplace transform of a signal f (t) as seen from equation (5.1) is an integral operation. It exists if f (t)e t Z1 f (t)e is absolutely integrable. That is t dt < 1. Cleary, only typical 0 choices of will make the integral converge. The range of that ensures the existence of X (s) defines the region of convergence (ROC) of the Laplace transform. As an example, let us take x(t) = e3t , t 0. Then Z1 X (s) = x(t)e ( +j! )t dt 0 Z1 = e( +3)t e j!t dt 0 The above integral converges if and only if + 3 < 0 or > 3. Thus, > 3 defines the ROC of X (s). Since, we shall deal only with causal signals(t 0) we avoid explicit mention of ROC. Due to the convergence factor, e t , a number of important functions have Laplace transforms, even though Fourier transforms for these functions do not exist. But this does not mean that every mathematical function has Laplace transform. The reader should be aware that, for 2 example, a function of the form et does not have Laplace transform. The inverse Laplace transform is defined by the relationship: L 1 fF (s)g = f (t) = 1 2j Z +j j 1 1 F (s)est ds (5.2) where is real. The evaluation of integral in equation (5.2) is based on complex variable theory, and hence we will avoid its use by developing a set of Laplace transform pairs. Laplace Transform 5.3 j 331 Three important singularity functions The three important singularity functions employed in circuit analysis are: (i) unit step function, u(t) (ii) delta function, (t) (iii) ramp function, r (t). They are called singularity functions because they are either not finite or they do not possess finite derivatives everywhere. The mathematical definition of unit step function is u(t) = 0; 1; t<0 t>0 (5.3) The step function is not defined at t = 0. Thus, the unit step function u(t) is 0 for negative values of t, and 1 for positive values of t. Often it is advantageous to define the unit step function as follows: u(t) = 1; 0; t 0+ t0 Figure 5.1 The unit step function A discontinuity may occur at time other than t = 0; for example, in sequential switching, the unit step function that occurs at t = a is expressed as u(t a). Figure 5.2 The step function occuring at t = a Figure 5.3 Thus; u(t a) = 0; 1; t t Similarly, the unit step function that occurs at t = Thus; u(t + a) = 0; 1; The step function occuring at t a < 0 or t < a a > 0 or t > a a is expressed as u(t + a). t + a < 0 or t < t + a > 0 or t > a a = a 332 j Network Theory We use step function to represent an abrupt change in voltage or current, like the changes that occur in the circuits of control engineering and digital systems. For example, the voltage v (t) = 0; K; t<a t>a may be expressed in terms of the unit step function as v (t) = Ku(t a) (5.4) The derivative of the unit step function u(t) is the unit impulse function (t). That is; (t) = 8 < 0; t<0 d undefined; t = 0 u(t) = : dt 0; t>0 (5.5) The unit impulse function also known as dirac delta fucntion is shown in Fig. 5.4. The unit impulse may be visualized as very short duration pulse of unit area. This may be expressed mathematically as: Z0+ (t)dt = 1 (5.6) 0 where t = 0 denotes the time just before t = 0 and t = 0+ denotes the time just after t = 0. Since the area under the unit impulse is unity, it is a practice to write ‘1’ beside the arrow that is used to symbolize the unit impulse function as shown in Fig. 5.4. When the impulse has a strength other than unity, the area of the impulse function is equal to its strength. For example, an impulse function 5 (t) has an area of 5 units. Figure 5.5 shows impulse functions, 2 (t + 2), 5 (t) and 2 (t 3). Figure 5.5 Three impulse functions Figure 5.4 The circuit impulse function Laplace Transform j 333 An important property of the unit impulse function is what is often called the sifting property; which is exhibited by the following integral: Zt2 f (t) (t t0 )dt = t1 f (t0 ); t1 < t0 < t2 0; t 1 > t0 > t2 for a fintie t0 and any f (t) continuous at t0 . Integrating the unit step function results in the unit ramp function r (t). Zt r (t) = 1 or r (t) = u( )d = tu(t) 0; t; (5.7) t0 t0 Figure 5.6 shows the ramp function. Figure 5.6 The unit ramp function In general, a ramp is a function that changes at a constant rate. Figure 5.7 The unit ramp function delayed by t0 Figure 5.8 The unit ramp function advanced by t0 A delayed ramp function is shown in Fig. 5.7. Mathematically, it is described as follows: r (t t0 ) = t t0 t0 ; t t0 0; t 334 j Network Theory An advanced ramp function is shown in Fig. 5.8. Mathematically, it is described as follows: r (t + t0 ) = 0; t t + t0 ; t t0 t0 It is very important to note that the three sigularity functions are related by differentiation as (t) = or by integration as u(t) = Zt u(t) = 5.4 du(t) ; dt 1 dr (t) dt Zt (t)dt; r (t) = 1 u( )d Functional transforms A functional transform is simply the Laplace transform of a specified function of t. Here we make an assumption that f (t) is zero for t < 0. 5.4.1 Decaying exponential function f (t) = e at u(t), where a > 0 and u(t) is the unit step function. L e at Z1 u(t) = F (s) = f (t)dt 0 Z1 e = at 0 e st dt 1 e (s+a)t (s + a) = t=0 1 = s+a 5.4.2 Unit step function f (t) = u(t) Lfu(t)g = F (s) = Z1 e 0 st dt = 1 s Laplace Transform j 335 5.4.3 Impulse function f (t) = (t) Lf (t)g = F (s) = Z1 (t)e st dt = e st t=0 =1 0 Please note that we have used the sifting property of an impulse function. 5.4.4 Sinusoidal function t0 f (t) = sin !t; sin !t = Since ej!t 2j Lfe at g = s +1 a and we have 1 Lfsin !tg = F (s) = 21j Z1 e ej!t 0 1 1 = 2j s j! = ! s2 j!t e j!t e 1 s + j! st dt + !2 Table 5.1 gives a list of important Laplace transform pairs. It includes the functions of most interest in an introductory course on circuit applications. Table 5.1 Important transform pairs f (t)(t 0) F (s) (t) 1 1 u(t) t e at sin !t cos !t s 1 s2 1 s+a ! s2 + ! 2 s 2 s + !2 336 Network Theory j f (t)(t 0) F (s) n! tn te sn+1 1 (s + a)2 at e at sin !t e at cos !t ! (s + a)2 + ! 2 s+a (s + a)2 + ! 2 All functions in the above table are represented without multiplied by u(t), since we have explicity declared that t 0. 5.5 Operational transforms (properties of Laplace transform) Operational transforms indicate how mathematical operations performed on either f (t) or F (s) are converted into the opposite domain. Following operations are of primary interest. Note: The symbol means “by the definition”. 5.5.1 Linearity Lff1 (t)g = F1 (s) and Lff2 (t)g = F2 (s) If Lfa1 f1 (t) + a2 f2 (t)g = a1 F1 (s) + a2 F2 (s) then Proof : Lfa1 f1 (t) + a2 f2 (t)g Z1 0 Z1 = a1 f1 (t)e st dt + a2 0 5.1 bt u(t)). SOLUTION We have the transform pair Lfu(t)g = 1s and Lfe bt f2 (t)e 0 Find the Laplace transform of f (t) = (A + Be u(t)g = dt Z1 = a1 F1 (s) + a2 F2 (s) EXAMPLE st [a1 f1 (t) + a2 f2 (t)]e 1 s+b st dt Laplace Transform Thus, using linearity property, Lff (t)g = F (s) = LfAu(t)g + LfBe = = bt u(t)g B A + s s+b (A + B)s + Ab s(s + b) 5.5.2 Time shifting If Lfx(t)g = X (s), then for any real number t0 , Lfx(t t0 )u(t Proof : Lfx(t t0 )u(t u(t Lfx(t we get; X (s) Z1 t0 )g x(t 0 Since; t0 s t0 )g = e t0 )u(t t0 ) = Z1 t0 )g = 1; 0; t0 )u(t t t t0 )e x(t t0 )e st dt Using the transformation of variable, t = + t0 Lfx(t Z1 t0 )u(t t0 )g = x( )e 0 EXAMPLE =e t0 s =e t0 s 5.2 Find the Laplace transform of x(t), shown in Fig. 5.9. Figure 5.9 dt t0 > 0 or t > t0 t0 < 0 or t < t0 t0 we get; st s( +t0 ) Z1 x( )e 0 X (s) d s d j 337 338 j Network Theory SOLUTION Figure 5.10(a) Figure 5.10(b) Using Figs. 5.10(a) and 5.10(b), we can write x(t) = x1 (t) + x2 (t) = u(t We know that, Lfu(t)g = 1 s u(t 2) and using time shifting property, we have Lfx(t)g = X (s) = 1s e X(s) = 1 (e s 1 2s 2s s e 4s e 4s ) 5.5.3 Shifting in s domain (Frequency-domain shifting) If Lfx(t)g = X (s), then 4) Lfes t x(t)g = X (s s0 ) 0 Proof : Lfe s0 t Z1 x(t)g es0 t x(t)e st dt 0 Z1 x(t)e = (s s0 )t 0 = X (s s0 ) dt Laplace Transform EXAMPLE j 5.3 Find the Laplace transform of x(t) = Ae at cos(!0 t + )u(t). SOLUTION Given x(t) = Ae at cos(!0 t + )u(t) = Ae at [cos !0 t cos = A cos e at sin !0 t sin ]u(t) cos !0 tu(t) A sin e at sin !0 tu(t) We know the transform pairs, Lfcos !0 tu(t)g = Lfsin !0 tu(t)g = and s s2 s2 + !02 !0 + !02 Applying frequency shifting property, we get L L and e e at cos !0 tu(t) = 2 s + !2 s 0 s at sin !0 tu(t) = 2 s + !2 !s +a s+a = (s + a)2 + !02 !0 0 s !s +a !0 = (s + a)2 + !02 Finally, applying linearity property, we get LfAe at cos(!0 t + )u(t)g = A cos Lfe at cos !0 tu(t)g A sin Lfe A cos (s + a) !0 = A sin 2 2 (s + a) + !0 (s + a)2 + !02 A[(s + a) cos θ ω0 sin θ] = (s + a)2 + ω02 5.5.4 Time scaling If Lfx(t)g = X (s), then Lfx(at)g = a1 X s a at sin !0 tu(t)g 339 340 j Network Theory Proof : Lfx(at)g Z1 st x(at)e dt 0 at = put ) adt = d Lfx(at)g = Hence Z1 0 = EXAMPLE s τa x( )e 1 Z1 x( )e a 1 a d s a d = 0 1 a 5.4 Find the Laplace transform of x(t) = sin(2!0 t)u(t). SOLUTION We know the transform pair, Lfsin !0 tu(t)g = s2 Applying scaling property, !0 + !02 3 2 Lfsin 2!0 tu(t)g = 12 64 s !2 0 + !02 2 2ω0 = 2 s + 4ω02 7 5 5.5.5 Time differentiation If Lfx(t)g = X (s), then L dx(t) dt = sX (s) x(0) Proof : y (t) = Let Then dx(t) dt Lfy(t)g = Y (s) Z1 = 0 Z1 y (t)e 0 dx(t) e dt st dt st dt X s a Laplace Transform j 341 Integrating by parts yields Z1 1 Y (s) = e st x(t)0 0 = lim [e t L Hence; dx(t) dt !1 =0 st st x(t)( se x(t)] )dt Z1 x(0) + s x(t)e st dt 0 x(0) + sX (s) = Y (s) = sX (s) x(0) Therefore, differentiation in time domain is equivalent to multiplication by s in the s domain. Whenever x(t) is discontinuous at t = 0 (like a step function), then x(0) should be read as x(0 ). The differentiation property can be extended to yield L dn x(t) dtn = sn X (s) sn 1 x(0) xn 1 (0) When x(t) is discontinuous at the origin, the argument 0 on the right side of the above equation should be read as 0 . Accordingly for a discontinuous function x(t) at the origin, we get L EXAMPLE dn x(t) dtn = sn X (s) sn 1 x(0 ) xn 1 (0 ) 5.5 Find the Laplace transform of x(t) = sin2 !0 tu(t). SOLUTION We find that, x(0) = 0 We know that, dx(t) = 2!0 sin !0 t cos !0 tu(t) dt = !0 sin 2!0 tu(t) Lfsin !0 tu(t)g = 2 !0 2 s + !0 (5.8) 342 j Network Theory Applying time scaling property, 3 2 Lfsin 2!0 tu(t)g = 12 64 s !2 0 + !02 2 = s2 7 5 2!0 + 2(!0 )2 Taking Laplace transform on both the sides of equation (5.8), we get L ) sX (s) EXAMPLE dx(t) dt = !0 Lfsin !0 tu(t)g 2!02 s2 + (2!0 )2 2ω02 X(s) = s[s2 + (2ω0 )2 ] x(0) = 5.6 Solve the second order linear differential equation 0 y (t) + 5y (t) + 6y (t) = x(t) with the initial conditions, y (0) = 2, y 0 (0) = 1 and x(t) = e t u(t). SOLUTION Taking Laplace transform on both the sides of the given differential equation, we get 2 s Y (s) sy (0) 0 y (0) + 5 jsY (s) y (0)j + 6Y (s) = X (s) X (s) = Lfe t u(t)g = where Substituting the initial conditions, we get 1 + 2s + 11 s+1 2s2 + 13s + 12 Y (s) = (s + 1)(s + 2)(s + 3) (s2 + 5s + 6)Y (s) = ) Using partial fraction expansion, we get 1 1 1 +6 Y (s) = 2 s+1 s+2 9 1 2 s+3 Taking inverse Laplace transform, we get y(t) = 1 t e + 6e 2 2t 9 e 2 3t , t0 1 s+1 Laplace Transform j 343 5.5.6 Integration in time domain For a causal signal x(t), Zt x( )d; y (t) = If 0 Lfy(t)g = Y (s) = Xs(s) then Proof : LBx(t)C = X(s) Z1 x(t)e st dt 0 Dividing both sides by s yields X (s) = s Z1 x(t) st e s 0 dt Integrating the right-hand side by parts, we get Z1 1 X (s) e ts = y (t) s s t=0 y (t) 0 e 1 Z1 X (s) e st = y (t) + y (t)e s s t=0 ts s st ( s)dt dt 0 The first term on the right-hand side evaluates to zero at both limits, because e 1=0 Z0 and y (0) = x( )d = 0 0 Hence; Y (s) = X (s) s Thus, integration in time domain is equivalent to division by s in the s domain. EXAMPLE 5.7 Consider the RC circuit shown in Fig. 5.11. The input is the rectangular pulse shown in Fig. 5.12. Find i(t) by assuming circuit is initially relaxed. 344 j Network Theory Figure 5.11 Figure 5.12 SOLUTION Applying KVL to the circuit represented by Fig. 5.11, we get Ri(t) + ) Ri(t) + Zt 1 i( )d = v (t) C 0 Zt 1 i( )d = Vo [u(t C a) u(t b)] 0 Taking Laplace transforms on both the sides, we get RI(s) + ) Vo (e as e bs ) Cs s Vo R I(s) = (e as e 1 s+ RC 1 I(s) = bs ) We know the transform pair, Lfe at u(t)g = 1 s+a and then using the time-shift property, we can find inverse of I(s). Vo i(t) = e R That is; ) i(t) = Vo h R t RC e u(t) (t a) RC t !t u(t vo e R a a) 5.5.7 Differentiation in the s domain For a signal x(t), t 0, we have Lf tx(t)g = dX (s) ds e t RC (t u(t) b) RC t !t u(t b b) i Laplace Transform Proof : For a causal signal, x(t), the Laplace transform is given by Lfx(t)g = X (s) = Z1 x(t)e st dt 0 Differentiating both the sides with respect to s, we get Z1 dX (s) = ds In general; EXAMPLE )dt [ tx(t)]e st dt 0 0 dX (s) or Lftx(t)g = ds n X (s) Lftn x(t)g = ( 1)n d ds n Lf Hence; st Z1 dX (s) = ds ) x(t)( te tx(t)g = 5.8 Find the Laplace transform of x1 (t) = te 3t u(t). SOLUTION We know that, Hence Lfe at u(t)g = Lfe 3t u(t)g = Using the differentiation in s domain property, 1 s+a 1 s+3 Lfx1 (t)g = X1 (s) = dsd s +1 3 = 1 (s + 3)2 5.5.8 Convolution If and then Lfx(t)g = X (s) Lfh(t)g = H (s) Lfx(t) h(t)g = X (s)H (s) where indicates the convolution operator. dX (s) ds j 345 346 j Network Theory Proof : Z1 x(t) h(t) = x( )h(t 1 )d Since x(t) and h(t) are causal signals, the convolution in this case reduces to Z1 x(t) h(t) = x( )h(t )d 0 2 Z1 Z1 Lfx(t) h(t)g = 4 x( )h(t Hence; 3 )d 5 e st dt 0 0 Interchanging the order of integrals, we get Lfx(t) h(t)g = Z1 2 x( ) 4 0 Using the change of variable = t 3 Z1 h(t )e st dt5 d 0 in the inner integral, we get Lfx(t) h(t)g = 2 Z1 x( )e s 0 4 Z1 h()e s 3 d5 d 0 = X (s)H (s) Please note that this theorem reduces the complexity of evaluating the convolution integral to a simple multiplication. EXAMPLE 5.9 Find the convolution of h(t) = e t and f (t) = e 2t . SOLUTION h(t) f (t) = L 1 fH (s)F (s)g 1 1 1 =L s+1 s+2 1 1 1 =L + s+1 s+2 t 2 t e , t0 =e 1 Laplace Transform EXAMPLE j 347 5.10 Find the convolution of two indentical rectangular pulses. Each rectangular pulse has unit amplitude and duration equal to 2T seconds. Also, the pulse is centered at t = T . SOLUTION Let the pulse be as shown in Fig. 5.13. From the Fig. 5.13, we can write x(t) = u(t) u(t 2T ) Taking Laplace transform, we get X (s) = 1 1 s s 1 = (1 Figure 5.13 e 2T s ) s y (t) = x(t) x(t) Let Then; Y (s) = X 2 (s) = ) 2T s e T T Y (s) = 1 2T s 2 e s 1 2 s2 s2 e 2T s + 1 s2 e 4T s Taking inverse Laplace transform, we get y (t) = tu(t) = r (t) 2(t 2T )u(t 2r (t y(t) = x(t) x(t) 5.5.9 2T ) + (t 2T ) + r (t 4T )u(t 4T ) 4T ) Figure 5.14 Initial-value theorem The initial-value theorem allows us to find the initial value x(0) directly from its Laplace transform X (s). If x(t) is a causal signal, x(0) = lim sX (s) then; s !1 (5.9) Proof : To prove this theorem, we use the time differentiation property. L dx(t) dt Z1 = sX (s) x(0) = 0 dx e dt st dt The problems with are better understood after the inverse Laplace transforms are studied. (5.10) 348 j Network Theory If we let s ! 1, then the integral on the right side of equation (5.10) vanishes due to damping factor, e st . Thus; lim [sX (s) s ) EXAMPLE !1 x(0)] = 0 x(0) = lim sX (s) s !1 5.11 Find the initial value of F (s) = s+1 (s + 1)2 + 32 SOLUTION f (0) = lim sX (s) = lim s s !1 s s2 + s s2 1 + !1 s2 = lim s (s + 1)2 + 32 !1 (s + 1) 2 + 32 = lim s !1 s+1 1+ 2 s 1 s + 10 =1 s2 We know the transform pair: Lfe bt cos atg = s+b (s + b)2 + a2 Hence, inverse Laplace transform of F (s) yields f (t) = e t cos 3t At t = 0, we get f (0) = 1. This verifies the theorem. 5.5.10 Final-value theorem The final-value theorem allows us to find the final value x(1) directly from its Laplace transform X (s). If x(t) is a causal signal, then Proof : The Laplace transform of lim x(t) = lim sX (s) t !1 s dx(t) is given by dt sX (s) !0 Z1 x(0) = 0 dx(t) e dt st dt Laplace Transform Taking the limit s ! 0 on both the sides, we get lim [sX (s) s !0 Z1 x(0)] = lim !0 0 Z1 s = 0 Z1 = dx(t) e dt st dx(t) h lim e s 0 dt ! j 349 dt st i dt dx(t) dt dt = x(t)j1 0 0 = x(1) Since; lim [sX (s) s we get; !0 x(1) Hence; x(0) x(0)] = lim [sX (s)] !0 x(0) = lim [sX (s) s !0 x(1) = lim [sX (s)] s !0 s x(0) x(0)] This proves the final value theorem. The final value theorem may be applied if, and only if, all the poles of X (s) have a real part that is negative. The final value theorem is very useful since we can find x(1) from X (s). However, one must be careful in using final value theorem since the function x(t) may not have a final value a as t ! 1. For example, consider x(t) = sin at having X (s) = 2 . Now we know s + a2 lim sin at does not exit. However, if we uncarefully use the final value theorem in this case, we t!1 would obtain: a =0 lim sX (s) = lim s 2 s !0 s !0 s + a 2 Note that the actual function x(t) does not have a limiting value as t ! 1. The final value theorem has failed because the poles of X (s) lie on the j! axis. Therefore, we conclude that for final value theorem to give a valid result, poles of X (s) should not lie to right side of the s-plane or on the j! axis. EXAMPLE 5.12 Find the final value of X (s) = Consider a function, X(s) 10 (s + 1)2 + 102 P (s) . The roots of the denomoniator polymial, Q(s) are called poles () and the Q(s) roots of the numerator polynomial, P (s) are called zeros (O). = 350 j Network Theory SOLUTION lim x(t) = x(1) t !1 s !0 s s10 !0 (s + 1)2 + 102 = 0 = lim [sX (s)] = lim We know the Laplace transform pair L Hence; e at sin bt = b (s + a)2 + b2 x(t) = L 1 =L 1 fX (s)g 10 (s + 1)2 + 102 =e t sin 10t x(1) = 0 Thus; This verifies the result obtained from final-value theorem. 5.5.11 Time periodicity Let us consider a function x(t) that is periodic as shown in Fig. 5.15. The function x(t) can be represented as the sum of time-shifted functions as shown in Fig. 5.16. Figure 5.15 A periodic function Hence; Figure 5.16 Decomposition of periodic function x(t) = x1 (t) + x2 (t) + x3 (t) + = x1 (t) + x1 (t T )u(t T ) + x1 (t 2T )u(t 2T ) + (5.11) where x1 (t) is the waveform described over the first period of x(t). That is, x1 (t) is the same as the function x(t) gated over the interval 0 < t < T . gating means the function x(t) is multiplied by 1 over the interval 0 t T and elsewhere by 0. Laplace Transform j 351 Taking the Laplace transform on both sides of equation (5.11) with the time-shift property applied, we get Ts X (s) = X1 (s) + X1 (s)e ) X (s) = X1 (s)(1 + e 1 + a + a2 + = But Hence, we get 1 1 a +e 2T s 2T s + + ) jaj < 1 ; X (s) = X1 (s) Ts + X1 (s)e 1 1 e (5.12) Ts In equation (5.12), X1 (s) is the Laplace transform of x(t) defined over first period only. Hence, we have shown that the Laplace transform of a periodic function is the Laplace transform evaluated over its first period divided by 1 e T s . EXAMPLE 5.13 Find the Laplace transform of the periodic signal x(t) shown in Fig. 5.17. Figure 5.17 SOLUTION From Fig. 5.17, we find that T = 2 Seconds. The signal x(t) considered over one period is donoted as x1 (t) and shown in Fig. 5.18(a). Figure 5.18(a) Figure 5.18(b) Figure 5.18(c) 352 j Network Theory The signal x1 (t) may be viewed as the multiplication of xA (t) and g (t). That is; x1 (t) = xA (t)g (t) = [ t + 1][u(t) ) x1 (t) = u(t tu(t) + tu(t 1)] 1) + u(t) = tu(t) + (t 1 + 1)u(t = tu(t) + (t 1)u(t u(t 1) + u(t) 1) + u(t = u(t) tu(t) + (t 1)u(t = u(t) r (t) + r (t 1) 1) u(t 1) + u(t) 1) u(t 1) 1) Taking Laplace Transform, we get X1 (s) = = Hence; 5.6 X(s) = 1 1 s s s2 + 1+e 1 s2 e s s s2 X1 (s) (s 1 + e s ) = 2 sT 1 e s (1 e 2s ) Inverse Laplace transform The inverse Laplace transform of X (s) is defined by an integral operation with respect to variable s as follows: 1 x(t) = 2 Z +j j 1 1 X (s)est ds (5.13) Since s is complex, the solution requries a knowledge of complex variables. In otherwords, the evaluation of integral in equation (5.13) requires the use of contour integration in the complex plane, which is very difficult. Hence, we will avoid using equation (5.13) to compute inverse Laplace transform. In many situations, the Laplace transform can be expressed in the form X (s) = where P (s) Q(s) P (s) = bm sm + bm n Q(s) = an s + an (5.14) m 1 1s n 1 1s + + b1 s + b0 + + a1 s + a0 ; an 6= 0 The function X (s) as defined by equation (5.14) is said to be rational function of s, since it is a ratio of two polynomials. The denominator Q(s) can be factored into linear factors. Laplace Transform A partial fraction expansion allows a strictly proper rational function j 353 P (s) to be expressed as a Q(s) factor of terms whose numerators are constants and whose denominator corresponds to linear or a combination of linear and repeated factors. This in turn allows us to relate such terms to their corresponding inverse transform. For performing partial fraction technique on X (s), the function X (s) has to meet the following conditions: (i) X (s) must be a proper fraction. That is, m < n. When X (s) is improper, we can use long division to reduce it to proper fraction. (ii) Q(s) should be in the factored form. EXAMPLE 5.14 Find the inverse Laplace transform of X (s) = 2s + 4 s2 + 4s + 3 SOLUTION 2s + 4 + 4s + 3 2(s + 2) K1 K2 = = + (s + 1)(s + 3) s+1 s+3 K1 = (s + 1)X (s)js= 1 X (s) = where; s2 2(s + 2) = (s + 3) s= 1 =1 K2 = (s + 3)X (s)js= Hence; We know that: Therefore; EXAMPLE 3 2(s + 2) = =1 (s + 1) s= 3 1 1 + X (s) = s+1 s+3 1 t Lfe u(t)g = s + x(t) = [e t + e 3t ]u(t) 5.15 Find the inverse Laplace transform of X (s) = s2 + 2s + 5 (s + 3)(s + 5)2 354 j Network Theory SOLUTION K1 K2 K3 + + s+3 s+5 (s + 5)2 K1 = (s + 3)X (s)js= 3 X (s) = Let where s2 + 2s + 5 = (s + 5)2 s= 3 =2 1 d [(s + 5)2 X (s)] K2 = 1! ds s= 5 + 2s + 5 d = ds s+3 s= 5 2 s + 6s + 1 = = 1 (s + 3)2 s= 5 K3 = (s + 5 )2 X (s)s= 5 s2 + 2s + 5 = = 10 (s + 3) s2 s= 5 1 s+5 2 X (s) = s+3 Then 10 (s + 5)2 Taking inverse Laplace transform, we get x(t) = 2e 3t x(t) = (2e or e 3t 5t e 10te 5t 5t t0 ; 10te 5t )u(t) Reinforcement problems R.P 5.1 Find the Laplace transform of: (a) cosh(at) (b) sinh(at) SOLUTION 1 (a) cosh(at) = [eat + e at ] 2 We know the Laplace transform pair: Lfe and at g= 1 s+a 1 Lfeat g = s a Laplace Transform Applying linearity property, we get, Lfcosh(at)g = 12 Lfeat g + 12 Lfe = = 1 1 1 + 2 s a s+a g s s2 1 (b) sinh at = [eat e at ] 2 Applying linearity property, a2 Lfsinh(at)g = 12 s 1 a a = 2 s a2 R.P at 1 s+a 5.2 Find the Laplace transform of f (t) = cos(!t + ). SOLUTION f (t) = cos(!t + ) Given = cos cos !t sin sin !t Applying linearity property, we get, Lff (t)g = F (s) = cos Lfcos !tg = cos = R.P s s2 + !2 sin Lfsin !tg sin s cos θ ω sin θ s2 + ω 2 ! s2 + !2 5.3 Find the Laplace transform of each of the following functions: (a) x(t) = t2 cos(2t + 30 )u(t) (b) x(t) = 2tu(t) 4 t (c) x(t) = 5u (d) x(t) = 5e 3 t 2 u(t) d (t) dt j 355 356 j Network Theory SOLUTION (a) Let us first find the Laplace transform of cos(2t + 30 )u(t) L fcos(2t + 30 )u(t)g = L fcos 30 cos 2tu(t) = cos 30 Lfcos 2tu(t)g = cos 30 = sin 30 sin 2tu(t)g sin 30 Lfsin 2tu(t)g 2 sin 30 s2 + 4 s s2 +4 2 sin 30 s2 + 4 s cos 30 The Laplace transform of x(t) is now found by using differentiation in s domain property. L d2 L fcos(2t + 30 )u(t)g 2 ds d2 s cos 30 2 sin 30 = 2 ds s2 + 4 t2 cos(2t + 30 ) = 3 2p 3 s 17 d2 6 2 7 = 26 ds 4 s2 + 4 5 2p 3 3 s 17 d d 6 6 2 7 = ds ds 4 s2 + 4 5 d d = ds ds d = ds " p " p 3 ( 2s) = 2 (s2 + 4)2 (b) x(t) = 2tu(t) 4 8 F 12 3s s2 + 4 1 3 2 s +4 2 p = ! 3 s 2 ! p 3 s 1 2 2 (s2 + 4)2 6s2 + 3 s 2 2s F ! p 1 p (s2 + 4)3 # 1 s2 + 4 1 3 2 + (s2 + 4)2 2s 4L 3 s 2 (s2 + 4)3 d (t) dt Lfx(t)g = X (s) = 2Lftu(t)g # 2 p 8s2 3s2 d (t) dt ! 1 Laplace Transform We know that whenever a function f (t) is discontinuous at the origin, we have L 357 j d f (t) dt = sF (s) f (0 ). Applying this relation to the second term on the right side of the above equation, we get X (s) = 2 = = 1 4[s 1 s2 2 4[s s2 2 s2 (0 )] 0] 4s t (c) x(t) = 5u 3 Using scaling property, s Lff (at)g = a1 F a Lfx(t)g = X (s) = 5 11=3 Lfu(t)g we get; 1 1 =5 1=3 s = s ! 5 s t (d) x(t) = 5e 2 u(t) We know the Laplace transform pair: Lfe at u(t)g = 1 s+a Lfx(t)g = X (s) = 5L Hence; =5 R.P n e 1 s+ 1 2 1 t 2 = 5.4 Find the Laplace transform of the following functions: (a) x(t) = t cos at (b) x(t) = (c) x(t) = 1 sin at sinh(at) 2a2 sin2 !t t o u(t) 10 2s + 1 s 1 3 s ! s 1 3 358 j Network Theory SOLUTION (a) x(t) = t cos at d F (s) ds f (t) = cos at s F (s) = 2 s + a2 Lftf (t)g = We know that Let ) s d ds s2 + a2 Lft cos atg = Lftf (t)g = Hence = x(t) = (b) s2 a2 (s2 + a2 )2 1 sin at sinh at 2a2 1 2a2 1 = 2 4a = 1 at e sin at 2 eat sin at 1 e 2 e at at sin at sin at We know the shifting in s domain property: Lfes t f (t)g = F (s)js!(s 0 s0 ) Applying this property along with linearity property, we get Lfx(t)g = X (s) 1 Lfeat sin atg 4a2 1 a = 2 4a s2 + a2 s!s Lfe = 1 = 2 4a (s = (c) x(t) = 1 t [(s at sin atg 2 2 s +a a a a a)2 + a2 s !s+a a (s + a)2 + a2 s a)2 + a2 ] [(s + a)2 + a2 ] sin2 !t We know that Lff (t)g = F (s) = Z1 f (t)e 0 st dt Laplace Transform Z1 Hence; Z1 F (s)ds = s Z1 f (t) s 0 Z1 f (t) = e 0 Z1 = 0 =L st e f (t) e t f (t) t 1 t s 1 j!t 2 e j2 ej 2!t 2 + e j 2!t = 4 1 1 1 1 + F (s) = 4 s j 2! 2 s 1 j!t = e j2 Hence; Hence, X (s) = L =L 1 2 1 f (t) !1 = lim x 1 ln 4 = R.P Zx x s F (s)ds = lim = 1 1 4 s + j 2! sin !t t t Z1 = dt dt f (t) = sin2 !t In the present case; 359 dsdt st st j ln(x !1 f (x)dx s j 2! ) x2 + 4! 2 x2 s2 + 4ω 1 ln 4 s2 ln(s 2 x !1 + j 2! ) 1 ln 4 2 ln x + 2 ln s + ln(x + j 2! ) 4 2 2 s + 4! ln(s + j 2! ) s2 5.5 Consider the pulse shown in Fig. R.P. 5.5, where f (t) = e2t for 0 < t < T . Find F (s) for the pulse. 360 j Network Theory Figure R.P. 5.5 SOLUTION The discrete pulse f (t) could be imagined as the product of signal x(t) and g (t) as shown in Fig. R.P. 5.5(a) and (b) respectively. That is; f (t) = x(t)g (t) = e2t [u(t) 2t = e u(t) 2t = e u(t) 2t = e u(t) Hence; u(t T )] 2t e u(t e e 2T 2(t T ) = 1 s 1 e u(t e T (s 2) 2 s e T (s (s 2) T) u(t T) e2T e T s Lff (t)g = F (s) = s 1 2 = T) 2(t T +T ) 2) s 2 2 Figure R.P. 5.5(a) Alternate method: Lff (t)g = F (s) ZT = 0 = 1 Z1 f (t)e 0 e2t e st dt e T (s (s 2) 2) st dt Laplace Transform R.P 5.6 Find the Laplace transform of f (t) shown in Fig. R.P. 5.6. Figure R.P. 5.6 SOLUTION f (t) is a discrete pulse and can be expressed mathematically as: f (t) = x(t)g (t) ) = sin t[u(t) u(t = sin tu(t) sin tu(t = sin tu(t) sin[ (t 1 + 1)]u(t 1) f (t) = sin tu(t) sin[ (t 1)] cos (t 1) cos[ (t 1)] 1)] sin u(t = sin tu(t) + sin[ (t Hence; F (s) = Lff (t)g = = R.P 1) 1) 1)]u(t 1) 1s e + 2 s2 + 2 s + 2 π [1 + e s ] s2 + π 2 Figure R.P. 5.6(a) 5.7 Determine the Laplace transform of f (t) shown in Fig. R.P. 5.7. Figure R.P. 5.7 j 361 362 j Network Theory SOLUTION We can write, f (t) = x(t)g (t) 5 t [u(t) u(t 2)] 2 5 5 = tu(t) tu(t 2) 2 2 5 5 = tu(t) (t 2 + 2)u(t 2) 2 2 5 5 = tu(t) (t 2)u(t 2) 5u(t 2) 2 2 5 1 5 1 5 2s e 2s e Hence; Lff (t)g = F (s) = 2 2 2 s 2 s s 5 = 2 [1 e 2s 2se 2s ] 2s = R.P Figure R.P. 5.7(a) 5.8 Find the Laplace transform of f (t) shown in Fig. R.P. 5.8. Figure R.P. 5.8 SOLUTION The equation of a straight line is y = mx + c, where m = slope of the line and c = intercept on y -axis. Hence, f (t) = When f (t) = 5 t+5 3 2, let us find t. Laplace Transform j 363 5 t+5 3 ) t = 4:2 Seconds Mathematically, That is; 2= f (t) = x(t)g (t) 5 = t + 5 [u(t) u(t 4:2)] 3 5 5 = tu(t) + tu(t 4:2) + 5u(t) 5u(t 4:2) 3 3 5 5 = tu(t) + (t 4:2 + 4:2)u(t 4:2) 3 3 +5u(t) 5u(t 4:2) 5 5 = tu(t) + (t 4:2)u(t 4:2) + 7u(t 4:2) 3 3 +5u(t) 5u(t 4:2) 5 5 = tu(t) + (t 4:2)u(t 4:2) + 2u(t 4:2) + 5u(t) 3 3 Hence; F (s) = Lff (t)g 5 5 2 = 2 + 2 e 4:2s + e 3s 3s s 4 : 2 s + 6se 5 + 5e = 3s2 R.P Figure R.P. 5.8(a) 4:2s + 4:2s 5.9 5 s + 15s If f (0 ) = 3 and 15u(t) 4 (t) = 8f (t) + 6f 0 (t), find f (t) (hint: by taking the Laplace transform of the differential equation, solving for F (s) and by inverting, find f (t)). SOLUTION Given, 15u(t) 4 (t) = 8f (t) + 6f 0 (t) Taking Laplace transform on both the sides, we get 15 s ) Therefore; 15 s 4 = 8F (s) + 6[sF (s) 4 = 8F (s) + 6sF (s) + 18 F (s)(6s + 8) = ) f (0 )] 18 + 4s 15 s 18 15 4s + (6s + 8) s(6s + 8) K K2 22s + 15 = 1+ = 4 4 s s+ 6s s + 3 3 F (s) = 364 j Network Theory The constants K1 and K2 are found using the theory of partial fractions. 22s + 15 K1 = = 1:875 4 6 s+ 3 s=0 22s + 15 K2 = 4 = 5:542 6s s= 1:875 5:542 4 s+ 3 Hence; F (s) = Taking the inverse, we get f (t) = 1.875 R.P s h 3 5.542e 4 t 3 i u(t) 5.10 Find the inverse Laplace transform of the following functions: s+1 (a) F (s) = 2 s + 4s + 13 (b) F (s) = s2 3e s + 2s + 17 SOLUTION F (s) = (a) s+1 (s + 2)2 + 9 (s + 2) 1 (s + 2)2 + 9 s+2 = (s + 2)2 + 32 s+2 = (s + 2)2 + 32 = 1 (s + 2)2 + 32 3 1 3 (s + 2)2 + 32 The determination of the Laplace inverse makes use of the following two Laplace transform pairs: Lfe bt a sin atg = (s + b)2 + a2 Lfe bt cos atg = (s +sb+)2 b+ a2 Hence; f (t) = L =e (b) F (s) = 1 fF (s)g 2t cos 3t 3e s s2 + 2s + 17 1 e 3 2t sin 3t Laplace Transform Let F (s) = e where X (s) = ) Since we get; s j 365 X (s) 3 3 = + 2s + 17 (s + 1)2 + 42 3 4 = 4 (s + 1)2 + 42 3 x(t) = e t sin 4t 4 F (s) = e s X (s) s2 f (t) = x(t 1) 3 (t 1) Therefore; f (t) = e sin[4(t 1)]; t>1 4 3 f (t) = e (t 1) sin[4(t 1)]u(t 1) 4 Laplace transform method for solving a set of differential equations: 1. Identify the circuit variables such as inductor currents and capacitor voltages. 2. Obtain the differential equations describing the circuit and keep a watch on the initial conditions of the circuit variables. 3. Obtain the Laplace transform of the various differential equations. 4. Using Cramer’s rule or a similar technique, solve for one or more of the unknown variables, obtaining the solution in s domain. 5. Find the inverse transform of the unknown variables and thus obtain the solution in the time domain. R.P 5.11 Referring to the RL circuit of Fig. R.P. 5.11, (a) write a differential equation for the inductor current iL (t). (b) Find IL (s), the Laplace transform of iL (t). (c) Solve for iL (t). Figure R.P. 5.11 366 j Network Theory SOLUTION (a) Applying KVL clockwise, we get 10iL (t) + 5 diL dt 5u(t 2) = 0 (b) Taking Laplace transform of the above equation, we get 5 10IL (s) + 5[sIL (s) iL (0 )] = e 2s s 5 ) IL (s) = s e 2s e 2s =e 2s = + 5iL (0 ) 5s + 10 + 5 10 s(s + 2) 1 IL (s) = e 2 Hence; 5 10 3 s K1 K2 + + s s+2 s(s + 2) 1 1 K1 = = s + 2 s=0 2 1 1 K2 = = s s= 2 2 where 3s 2s 1 s 5 10 3 1 + s+2 (s + 2) (c) Taking Inverse Laplace transform, we get 1 u(t) e 2t u(t) t!t 2 + 5 10 3 e 2t u(t) iL (t) = 2 1 = u(t 2) e 2t u(t 2) + 5 10 3 e 2t u(t) 2 R.P 5.12 Obtain a single integrodifferential equation in terms of iC for the circuit of Fig. R.P. 5.12. Take the Laplace transform, solve for IC (s), and then find iC (t) by making use of inverse transform. Figure R.P. 5.12 Laplace Transform j 367 SOLUTION Applying KVL clockwise to the right-mesh, we get Z1 iC dt + 4[iC 4u(t) + iC + 10 0:5 (t)] = 0 0 Taking Laplace transform, we get 1 10IC (s) 4 + IC (s) + + 4IC (s) s s 2=0 2s 4 5s + 10 1.6 = 0.4 s+2 ) IC (s) = Taking inverse Laplace transform, we get iC (t) = 0.4δ(t) R.P 1.6e 2t u(t) Amps. 5.13 Refer the circuit shown in Fig. R.P. 5.13. Find i(0) and i(1) using initial and final value theorems. Figure R.P. 5.13 SOLUTION Applying KVL we get i+2 di = 10 dt Taking Laplace transform, on both the sides, we get I (s) + 2[sI (s) ) ) i(0 )] = I (s) + 2[sI (s) 1] = I (s)[1 + 2s] = 10 s 10 s 10 s +2 368 j Network Theory ) 10 2 + s(1 + 2s) 1 + 2s 10 + 2s = s(1 + 2s) 5+s = 1 s s+ 2 I (s) = According to initial value theorem, i(0) = lim sI (s) s !1 (s + 5) = lim s s!1 1 s s+ 2 5 1+ s =1 = lim 1 s!1 1+ 2s We know from fundamentals for an inductor, i(0+ ) = i(0 ) = i(0). Hence, i(0) found using initial value theorem verifies the initial value of i(t) given in the problem. From final value theorem, i(1) = lim sI (s) s !0 5 s(s + 5) = = 10 A = lim s !0 1 1=2 s s+ 2 R.P 5.14 Find i(t) and vC (t) for the circuit shown in Fig. R.P. 5.14 when vC (0) = 10 V and i(0) = 0 A. The input source is vi = 15u(t) V. Choose R so that the roots of the characteristic equation are real. Figure R.P. 5.14 Laplace Transform j 369 SOLUTION Applying KVL clockwise, we get L di + vC + Ri = vi (t) dt (5.15) The differential equation describing the variable vC is C dvC =i dt (5.16) The Laplace transform of equation (5.15) is L[sI (s) i(0) + VC (s) + RI (s) = Vi (s)] (5.17) The Laplace transform of equation (5.16) is C [sVC (s) vC (0) = I (s)] (5.18) Noting that i(0) = 0, substituting for C and L and rearranging equation (5.17) and (5.18), we get, [R + s]I (s) + VC (s) = Vi (s) = I (s) + 1 sVC (s) = 5 2 15 s (5.19) (5.20) Putting equations (5.19) and (5.20) in matrix form, we get 2 4 R+s 1 3" # 2 15 3 1 I (s) =4 s 5 1 5 VC (s) s 5 2 (5.21) Solving for I (s) using Cramer’s rule, we get I (s) = 5 s2 + Rs + 2 The inverse Laplace transform of I (s) will depend on the value of R. The equation s2 + Rs + 2 = 0 is defined as the characteristic equation. For the roots of this equation to be real, it is essential that b2 4ac 0 . R2 This means that; ) The condition b2 4120 p R2 2 4ac 0 is with respect to algebraic equaion ax2 + bx + c = 0. 370 j Network Theory Let us choose the value of R as 3 I (s) = Then ) where Hence; I (s) = K1 = 5 5 = + 3s + 2 (s + 1)(s + 2) s2 K2 K1 + s+1 s+2 5 s + 2 s = 1 =5 5 = K2 = s + 1 s = 2 5 5 I (s) = s+1 s+2 5 Taking inverse Laplace transform, we get i(t) = 5e t u(t) 5e 2t u(t) Please note that t = 0 gives i(0) = 0 and t = 1 gives i(1) = 0. Solving the matrix equation (5.21) for VC (s), using Cramer’s rule, we get VC (s) = 10s2 + 10Rs + 30 s(s2 + Rs + 2) Substituting the value of R, we get VC (s) = 10(s2 + 3s + 3) s(s + 1)(s + 2) Using partial fraction expansion, we can write, VC (s) = where, K1 = 15, K2 = Hence; K1 K2 K3 + + s s+1 s+2 10, K3 = 5 VC (s) = 15 s 10 5 + s+1 s+2 Taking inverse Laplace transform, we get vC (t) = 15u(t) 10e t u(t) + 5e Verification: Putting t = 0, we get vC (0) = 15 10 + 5 = 10 V vC (1) = 15 0 + 0 = 15 V This checks the validity of results obtained. 2t u(t) Laplace Transform R.P j 371 5.15 For the circuit shown in Fig. R.P. 5.15, the steady state is reached with the 100 V source. At t = 0, 1 switch K is opened. What is the current through the inductor at t = seconds ? 2 Figure R.P. 5.15 SOLUTION At t = 0 , the circuit is as shown in Fig. 5.15(a). i2 (0+ ) = i2 (0 ) = 2:5 A Figure R.P. 5.15(a) Figure R.P. 5.15(b) For t 0+ , the circuit diagram is as shown in Fig. 5.15(b). Applying KVL clockwise to the circuit, we get 80i(t) + 4 di =0 dt 372 j Network Theory Taking Laplace transform, we get 80I (s) + 4[sI (s) ) i(0 )] = 0 80I (s) + 4sI (s) = 4 2:5 ) [20 + s]I (s) = 2:5 2:5 I (s) = s + 20 Taking inverse Laplace transform, we get, i(t) = 2:5e At t = 0:5 sec, we get i(0.5) = 2.5e R.P 10 20t = 1.135 10 4 A 5.16 Refer the circuit shown in Fig. R.P. 5.16. Find: (a) vo (t) for t 0 (b) io (t) for t 0 (c) Does your solution for io (t) make sense when t = 0? Explain. Figure R.P. 5.16 SOLUTION Figure R.P. 5.16(a) Laplace Transform j 373 KCL at node A: (for t 0) Idc = 1 L Zt vo dt + 0 vo dvo +C R dt Taking Laplace transform, Idc Vo (s) Vo (s) = + + sCVo (s) s sL R Idc C Vo (s) = 1 1 s+ s2 + RC LC Hence; Substituting the values of Idc , R, L, and C , we find that Vo (s) = 120; 000 120; 000 = s2 + 10; 000s + 16 106 (s + 2000)(s + 8000) Using partial fractions, we get Vo (s) = where K1 = 20, and K2 = Hence; K1 K2 + s + 2000 s + 8000 20 Vo (s) = 20 20 s + 2000 s + 8000 Taking inverse Laplace transform, we get vo (t) = 20e (b) io (t) = C Hence 2000t u(t) 20e 8000t u(t) dvo dt Io (s) = C [sVo (s) vo (0)] For t 0 , since the switch was in closed state, the circuit was not activated by the source. This means that vo (0) = vo (0 ) = vo (0+ ) = 0 and iL (0+ ) = iL (0 ) = 0: Then; Io (s) = CsVo (s) 25 10 9 s 120; 000 s2 + 10; 000s + 16 106 3 10 3 s = (s + 2000)(s + 8000) = = K1 s + 2000 + K2 s + 8000 374 Network Theory j We find that, K1 = 10 3, Therefore; and K2 = 4 10 Io (s) = 3 10 3 4 10 3 + s + 2000 s + 8000 Taking inverse Laplace transform, we get io (t) = 4e (c) io (0+ ) = 4 8000t u(t) e 2000t u(t)mA 1 = 3mA Yes. The initial inductor current is zero by hypothesis iL (0+ ) = IL (0 ) = 0 . Also, the initial resistor current is zero because vo (0+ ) = vo (0 ) = 0. Thus at t = 0+ , the source current appears in the capacitor. R.P 5.17 Refer the circuit shown in Fig. R.P. 5.17. The circuit parameters are R = 10kΩ, L = 800 mH and C = 100nF, if Vdc = 70V, find: (a) vo (t) for t 0 (b) io (t) for t 0 (c) Use initial and final value theorems to check the inital and final values of current and voltage. Figure R.P. 5.17 SOLUTION At t = 0 , switch is open and at t = 0+ , the switch is closed. Since at t = 0 , the circuit is not energized by dc source, io (0 ) = 0 and vo (0 ) = 0. Then by the hypothesis, that the current in an inductor and voltage across a capacitor cannot change instantaneously, io (0+ ) = io (0 ) = 0 and vo (0+ ) = vo (0 ) = 0 Laplace Transform The KCL equation when the switch is closed (for t 0+ ) is given by vo 1 dvo + + C dt R L Zt (vo Vdc )d = 0 0 Zt dvo vo 1 C + + dt R L ) ) C vo d = 0 Zt dvo vo 1 + + dt R L vo d = 0 Zt 1 Vdc d L 1 L 0 Vdc t Laplace transform of the above equation gives C [sVo (s) vo (0)] + Vo (s) 1 Vo (s) 1 Vdc + = R L s L s2 Since vo (0) is same as vo (0 ), we get CsVo (s) + Vo (s) 1 Vo (s) Vdc + = R L s Ls2 ) Vdc LC Vo (s) = s s2 + 1 RC s+ 1 LC Substituting the values of Vdc , R, L, and C , we get 875 106 s [s2 + 1000s + 1250 104 ] 875 106 = s(s s1 )(s s2 ) Vo (s) = where s1 ; s 2 = = Hence; Vo (s) = 500 p 25 104 1250 104 500 j 3500 s(s + 500) 875 106 j 3500)(s + 500 + j 3500) Using partial fractions, we get Vo (s) = We find that 875 106 = 70 125 105 875 106 K2 = ( 500 + j 3500)(j 7000) K1 = K1 K2 K2 + + s s + 500 j 3500 s + 500 + j 3500 j 375 376 j Network Theory p = 5 50 /171:87 p p 70 5 50 /171:87 5 50 / 171:87 Vo (s) = + + s s + 500 j 3500 s + 500 + j 3500 Taking inverse Laplace transform, we get h p vo (t) = 70 + 5 50 /171:87 e (500 j 3500)t p + 5 50 / 171:87 e (500+j 3500)t i u(t) The inverse of Vo (s) can be expressed in a better form by following the technique described below: Let us consider a transformed function C + jd C jd + s + a j! s + a + j! m/ m / + = s + a j! s + a + j! F (s) = m= where p c2 + d2 and = tan 1 d c The inverse transform of F (s) is given by f (t) = 2me at cos(!t + )u(t) (For the proof see R.P. 5.19) In the present context, p m = 5 50; = 171:87 ! = 3500 and a = 500 L This means that, 1 ) p 5 50 /171:87 5 50 / 171:87 + s + 500 j 3500 s + 500 + j 3500 ( p p = 2 5 50e 500t cos (3500t + 171:87 ) p = 10 50e 500t cos (3500t + 171:87 ) Hence; F vo (t) = 70 + 10 50e 500t cos (3500t + 171.87 ) u(t) Laplace Transform (b) vo dvo +C R dt io (t) = Taking Laplace transforms on both the sides, we get Vo (s) + C sVo (s) R Vo (s) Io (s) = + CsVo (s) R Io (s) = ) ) Io (s) = CVo (s) s + = Vdc L 2 vo (0 ) 1 RC s+ 6 6 4 s s2 + 1 3 1 RC RC s+ 1 7 7 5 LC Substituting the values of Vdc , R, L, and C , we get Io (s) = = 87:5(s + 1000) s(s + 500 j 3500)(s + 500 + j 3500) K1 K2 K2 + + s s + 500 j 3500 s + 500 + j 3500 We find that, 87:5 1000 = 7mA 1250 104 87:5(500 + j 3500) K2 = ( 500 + j 3500)(j 7000) = 12:5 / 106:26 mA 7 12:5 / 106:26 12:5 /106:26 + Io (s) = + s s + 500 j 3500 s + 500 + j 3500 K1 = The inverse Laplace transform yields, h io (t) = 7 + 12:5 / 106:26 e = 7 + 25e (c) s s2 + 1 RC s+ 1 + 12:5 /106:26 e 106.26 ) u(t)mA 500t cos(3500t Vdc LC Vo (s) = (500 j 3500)t (500+j 3500)t i u(t) LC Vdc LC = 70V LC The same result may be obtained by putting t = 1 in the expression for vo (t). From Final Value theorem: vo (1) = lim vo (t) = lim sVo (s) = t !1 s !0 j 377 378 j Network Theory From initial value theorem : vo (0) = lim sVo (s) s !1 !1 6 ss2 + 6s = lim s Vdc LC 1 RC s+ 1 LC =0 This verifies our beginning analysis that vo (0+ ) = vo (0 ) = 0. The same result may be obtained by putting t = 0 in the expression for vo (t). 1 s+ Vdc RC We know that, Io (s) = 1 1 L 2 s s + s+ RC From final value theorem : Io (1) = lim sIo (s) s !0 = lim 6 s s !0 Vdc L s2 + 1 s+ 6s LC RC 1 RC s+ 1 LC 1 Vdc RC L 1 LC Vdc 70 = = 7 mA = R 10 103 = The same result may be obtained by putting t = 1 in the expression for io (t). From initial value theorem : io (0) = lim sIo (s) s !1 2 6 Vdc = lim s 6 s!1 4 L s+ s s2 + =0 1 3 RC 1 RC s+ 1 7 7 5 LC This agrees with our initial analysis that the initial current through the inductor is zero. The same result can be obtained by putting t = 0 in the expression for io (t). Laplace Transform R.P j 379 5.18 Apply the initial and final value theorems to each of the functions given below: s2 + 5s + 10 s2 + 5s + 10 (a) F (s) = (b) F (s) = s+6 5(s2 + 6s + 8) SOLUTION Since in F (s) referred in (a) and (b) are improper 1 fractions, the corresponding time domain counterparts, f (t) contain impulses. Thus, neither the initial value theorem nor the final value theorems may be applied to these transformed functions. R.P 5.19 c + jd c jd + s + a j! s + a + j! Find the inverse Laplace transform of F (s) = SOLUTION Expressing c + jd and c jd in the exponenetial from, we get, F (s) = m= where mej me j + s + a j! s + a + j! p f (t) = L Hence; c2 1 + d2 (a j! )t = me e = 2me d c fF (s)g j = me and = tan 1 at j ( +!t) e " at e (a+j! )t e j ( +!t) u(t) + me j u(t) + me at ej (+!t) + e j ( +!t) 2 # u(t) u(t) u(t) = 2me at cos(θ + ωt)u(t) R.P 5.20 Find the initial and final values of f (t) when F (s) = s2 60 2s + 1 SOLUTION Initial value theorem f (0) = lim sF (s) s !1 = lim s s !1 s2 60 =0 2s + 1 1 If the degree of the numerator polynomial is greater than or equal to the degree of the denominator polynomial, the fraction is said to be improper. 380 Network Theory j Final value theorem: The poles of F (s) are given by finding the roots of the denominator polynomial. That is, s2 2s + 1 = 0 ) (s 1)2 = 0 ) s = 1; 1 Since both the poles of F (s) lie to the right of the s plane, final value theorem cannot be used to find f (1). R.P 5.21 Find i(t) for the circuit of Fig.R.P. 5.21, when i1 (t) = 7e i(1). 6t A for t 0 and i(0) = 0. Also find Figure R.P. 5.21 SOLUTION Refer Fig. R.P. 5.21(a). KCL at node v1 : v1 5 + i = 7e 6t di dt di 1 3i + 4 + i = 7e 6t Hence; 5 dt 4 di 8 ) + i = 7e 6t 5 dt 5 35 di + 2i = e 6t ) dt 4 Also; v1 = 3i + 4 Figure R.P. 5.21(a) Taking Laplace transform of the differential equation, we get [sI (s) ) 35 1 4 s+6 35 1 I (s) = 4 (s + 2)(s + 6) i(0)] + 2I (s) = Laplace Transform j 381 Using partial fraction expansion, we get I (s) = 35 35 and K2 = 16 16 1 35 1 35 I (s) = 16 s + 2 16 s + 6 35 2t i(t) = e 6t u(t) e 16 K1 = and find that Hence; 5.7 K1 K2 + s+2 s+6 Circuit element models and initial conditions In the analysis of a circuit, the Laplace transform can be carried one step further by transforming the circuit itself rather than the differential equation. Earlier we have seen how to represent a circuit in time domain by differential equations and then use Laplace transform to transform the differential equations into algebraic equations. In this section, we will see how we can represent a circuit in s domain using the Laplace transform and then analyze it using algebraic equations. 5.7.1 Resistor The voltage-current relationship for a resistor R is given by Ohm’s law: v (t) = i(t)R (5.22) Taking Laplace transform on both the sides, we get V (s) = I (s)R (5.23) Fig. 5.19 (a) shows the representation of a resistor in time domain and Fig. 5.19(b) in frequency domain using Laplace transform. Figure 5.19(a) Resistor represented in Figure 5.19(b) Resistor represented in the time domain frequency domain using Laplace transform The impedance of an element is defined as Z (s) = V (s) I (s) 382 j Network Theory provided all initial conditions are zero. Please note that the impedance is a concept defined only in frequency domain and not in time domain. In the case of a resistor, there is no initial condition to be set to zero. Comparision of equations (5.22) and (5.23) reveals that, resistor R has same representation in both time and frequency domains. 5.7.2 Capacitor For a capacitor with capacitance C , the time-domain voltage-current relationship is v (t) = 1 C Zt i( )d + v (0) (5.24a) 0 The s domain characterization is obtained by taking the Laplace transform of the above equation. That is, V (s) = 1 Cs I (s) + v (0) s (5.24b) To find the impedance of a capacitor, set the initial condition v (0) to zero. Then from equation V (s) 1 (5.24b), we get Z (s) = = as the impedance of the capacitor. With the help of equation I (s) Cs (5.24b), we can draw the frequency domain representation of a Capacitor and the same is shown in Fig. 5.20(b). This equivalent circuit is drawn so that the KVL equation represented by equation (5.24 b) is satisfied. Performing source transformation on the equivalent s domain circuit for a capacitor which is shown in Fig. 5.20(b), we get an alternate frequency domain representation as shown in Fig. 5.20(c). Figure 5.20(a) A capacitor represented in time domain (b) A capacitor represented in the frequency domain (c) Alternate frequency domain representation for a capacitor Laplace Transform j 383 5.7.3 Inductor For an inductor with inductance L, the time domain voltage-current relation is v (t) = L di(t) dt (5.25) The Laplace transform of equation (5.25) yields, V (s) = LsI (s) Li(0) (5.26) To find impedance of an inductor, set the initial condition i(0) to zero. Then from equation (5.26), we get Z (s) = V (s) = Ls I (s) (5.27) which represents the impedance of the inductor. Equation (5.26) is used to get the frequency domain representation of an inductor and the same is shown in Fig. 5.21(b). The series connection of elements corresponds to sum of the voltages in equation (5.26). Converting the voltage source in Fig.5.21(b) into an equivalent current source, we get an alternate representation for the inductor in frequency domain which is as shown in Fig. 5.21(c). To find the frequency domain representation of a circuit, we replace the time domain representation of each element in the circuit by its frequency domain representation. Figure 5.21(a) An inductor represented in time domain (b) An inductor represented in the frequency domain (c) An alternate frequency domain representation To find the complete response of a circuit, we first get its frequency domain representation. Next, using KV L or KCL, we find the variables of interest in s doamin. Finally, we use the inverse Laplace transform to represent the variables of interest in time domain. 384 j Network Theory EXAMPLE 5.16 Determine the voltage vC (t) and the current iC (t) for t 0 for the circuit shown in Fig. 5.22. C C Figure 5.22 SOLUTION We shall analyze this circuit using nodal technique. Hence we represent the capacitor in frequency domain by a parallel circuit since it is easier to account for current sources than voltage sources while handling nodal equations. The symbol for switch indicates that at t = 0 it is closed and at t = 0+ , it is open. The circuit at t = 0 is shown in Fig. 5.23(a). Let us assume that at t = 0 , the circuit is in steady state. Under steady state condition, capacitor acts as on open circuit as shown in Fig. 5.23(a). 26 12 4 = = 6+3 9 3 4 vC (0 ) = 3 = 4V 3 vC (0) = vC (0+ ) = vC (0 ) = 4V i1 (0 ) = Hence; Fig. 5.23(b) represents the frequency domain representation of the circuit shown in Fig. 5.22. Figure 5.23(a) KCL at top node: VC (s) 2 s + VC (s) = 2 + 2 s 6 2 ) VC (s) = 2 s s+ 3 3 Figure 5.23(b) Laplace Transform Inverse Laplace transform yields h vC (t) = 6 Also; IC (s) = VC (s) iC (t) = 2 e 3 2 t 3 2 t 3 i 2= 2 s EXAMPLE 2e j 385 u(t)V 2 3 s+ 2 3 u(t)A 5.17 Determine the current iL (t) for t 0 for the circuit shown in Fig. 5.24. Figure 5.24 SOLUTION At t = 0 , switch is closed and at t = 0+ , it is open. Let us assume that at t = 0 , the circuit is in steady state. In steady state, capacitor is open and inductor is short. The equivalent circuit at t = 0 is as shown in Fig. 5.25(a). 12 = 1A 8+4 vC (0 ) = 1 8 = 8V iL (0 ) = Therefore; iL (0) = iL (0+ ) = iL (0 ) = 1A vC (0) = vC (0+ ) = vC (0 ) = 8V Figure 5.25(a) For t 0+ , the circuit in frequency domain is as shown in Fig. 5.25(b). We will use KVL to find iL (t). Hence, we use series circuits to represent both the capacitor and inductor in the frequency domain. These series circuits contain voltage sources rather than current sources. It is easier to account for voltage sources than current sources when writing mesh equations. This justifies the selection of series representation for both the capacitor and inductor. 386 Network Theory j Applying KVL clockwise to the right mesh, we get 8 s + 20 s IL (s) + 4sIL (s) 4+8IL (s) = 0 ) 8 ) IL (s) = ) s +4= 20 s + 8 + 4s IL (s) 2+s (s + 1) + 1 = s2 + 2s + 5 (s + 1)2 + 4 2 s+1 1 + IL (s) = (s + 1)2 + 22 2 (s + 1)2 22 Figure 5.25(b) We know the Laplace transform pairs: s+a (s + a)2 + b2 L e at sin bt = (s + ab)2 + b2 L and e at cos bt = iL (t) = e Hence; EXAMPLE t cos 2t + 1 t e sin 2t u(t)A 2 5.18 Find vo (t) of the circuit shown in Fig. 5.26. Figure 5.26 SOLUTION The unit step function u(t) is defined as follows: u(t) = 1; t 0+ 0; t 0 Figure 5.27(a) Laplace Transform j 387 Since the circuit has two independent sources with u(t) associated with them, the circuit is not energized for t 0 . Hence the initial current through the inductor is zero. That is, iL (0 ) = 0. Since current through an inductor cannot change instantaneously, iL (0) = iL (0+ ) = iL (0 ) = 0 vC (0) = vC (0+ ) = vC (0 ) = 0 Also; The equivalent circuit for t 0+ in frequency domain is as shown in Fig. 5.27(b). Figure 5.27(b) KCL at supernode: V1 (s) 1+ 2 ) 1 + s3 V2 (s) 2 V2 (s) + = s 2 s 2 1 7 1 1 6 = + V1 (s) 4 + V2 (s) 15 s 1+ ) s s s 2 2+s 2 + V2 (s) = V1 (s) s+1 2s s The constraint equation: Applying KVL to the path comprising of current source ! voltage source ! inductor, V1 (s) we get, 1 + V2 (s) = 0 s+2 V2 (s) ) V1 (s) 1 s+2 1 V2 (s) = s+2 V1 (s) = 388 j Network Theory Putting the above two equations in matrix form, we get 2 2 3 3 2+s 2 V1 (s) 6 2s 7 5=6 54 4 V2 (s) 1 s 6 s+1 4 1 2 3 7 7 1 5 s s+2 Solving for V2 (s) and then applying the principle of voltage division, we get 2(3s2 + 6s + 4) 1 V2 (s) = 2 2(s + 2)(3s2 + 3s + 2) 4 s2 + 2s + 3 Vo (s) = (s + 2)(s + 0:5 j 0:646)(s + 0:5 + j 0:646) Vo (s) = ) Using partial fractions, we can write K1 K2 K2 + + s+2 s + 0:5 j 0:646 s + 0:5 + j 0:646 K1 = 0:5 Vo (s) = We find that Hence; We know that, L 1 K2 = 0:316 / 37:76 0:5 0:316 / 37:76 0:316 /37:76 Vo (s) = + + s+2 s + 0:5 j 0:646 s + 0:5 + j 0:646 1 =e s+a L at u(t) 1 = 2me Hence, EXAMPLE vo (t) = 0.5e 2t u(t) m / m/ + s + a j! s + a + j! at cos(!t + )u(t) + 0.632e 0:5t cos [0.646t 37.76 ] u(t) 5.19 For the network shown in Fig. 5.28, find vo (t), t > 0, using mesh equations. Figure 5.28 Laplace Transform SOLUTION The step function u(t) is defined as follows. u(t) = 1; t 0+ 0; t 0 Since the circuit is not energized for t 0 , there are no initial conditions in the circuit. For t 0+ , the frequency domain equivalent circuit is shown in Fig. 5.29(b). Figure 5.29(a) Figure 5.29(b) 2 By inspection, we find that I1 (s) = s KVL clockwise for mesh 2: 4 s + 1 [I2 (s) 4 ) s I1 (s)] + 2I2 (s) + 1 [I2 (s) I3 (s)] = 0 I1 (s) + I2 (s) [1 + 2 + 1] I3 (s) = 0 Substituting the value of I1 (s), we get 4 s ) + 4I2 (s) 4I2 (s) I3 (s) = I3 (s) = 2 s 6 s KVL clockwise for mesh 3: 1 [I3 (s) I2 (s)] + sI3 (s) + 1I3 (s) = 0 ) I2 (s) + I3 (s) [s + 2] = 0 Putting the KVL equations for mesh 2 and mesh 3 in matrix form, we get 2 4 4 1 1 s+2 32 54 I2 (s) I3 (s) 2 6 3 5=4 s 5 3 0 j 389 390 j Network Theory Solving for I3 (s), using Cramer’s rule, we get I3 (s) = 1:5 7 s s+ 4 Vo (s) = I3 (s) 1 = ) 1:5 s s+ 7 4 Using partial fractions, we can write Vo (s) = 4 6 6 K1 = ; and K2 = 72 37 We find that, Hence; EXAMPLE K1 K2 + 7 s s+ Vo (s) = 6 61 4 7 s vo (t) = 6h 1 7 7 1 75 s+ 4 e 7 t 4 i u(t) 5.20 Use mesh analysis to find vo (t), t > 0 in the network shown in Fig. 5.30. Figure 5.30 SOLUTION The circuit is not energized for t 0 because the independent current source is associated with u(t). This means that there are no initial conditions in the circuit. The frequency domain circuit for t 0+ is shown in Fig. 5.31. By inspection we find that: I1 (s) = 4 s ; I2 (s) = Ix (s) = I3 (s) 4 s Ix (s) 2 ) 2I2 (s) = I3 (s) 4 s ) 1 I3 (s) I2 (s) = 2 4 s Laplace Transform j 391 KVL for mesh 3 gives s [I3 (s) I2 (s)] + 1 I3 (s) 4 s +1 I3 (s) = 0 1 4 s I3 (s) I3 (s) 2 s 4 + I3 (s) = 0 + I3 (s) ) s 4(s 2) s(s + 4) Vo (s) = 1[I3 (s)] 4(s 2) = s(s + 4) ) I3 (s) = and Figure 5.31 By partial fractions, we can write K1 K2 + s s+4 K1 = 2; K2 = 6 Vo (s) = We find that Hence; Vo (s) = 2 6 s s+4 6e 4t u(t) Taking inverse Laplace transform, we get vo (t) = 2u(t) EXAMPLE 5.21 Using the principle of superposition, find vo (t) for t > 0. Refer the circuit shown in Fig. 5.32. Figure 5.32 SOLUTION Since both the independent sources are associated with u(t), which is zero for t 0 , the circuit will not have any initial conditions. The frequency domain circuit for t 0+ is shown in Fig. 5.33(a). 392 j Network Theory Figure 5.33(a) As a first step, let us find the contribution to Vo (s) due to voltage source alone. This needs the deactivation of the current source. Referring to Fig. 5.33(b), we find that 4 I (s) = ) s s+1+ 2 +1 s Vo1 (s) = I (s)[1] = s2 4 + 2s + 2 Figure 5.33(b) Next let us find the contribution to the output due to current source alone. Refer to Fig. 5.33(c). Using the principle of current division, 2 I1 (s) = ) s s s+1+ 2 s Vo2 (s) = 1 [I1 (s)] = +1 s2 2s + 2s + 2 Figure 5.33(c) Finally adding the two contributions, we get Vo (s) = Vo1 (s) + Vo2 (s) = = We find that, Hence; s2 4 2s 2s + 4 + 2 = 2 + 2s + 2 s + 2s + 2 s + 2s + 2 K1 + K1 s + 1 j1 s + 1 + j1 p K1 = 2 / 45 p p 2 / 45 2 /+45 Vo (s) = + s + 1 j1 s + 1 + j1 Laplace Transform We know that: m / m/ + s + a jb s + a + jb = 2me at 393 cos(bt + )u(t) F vo (t) = 2 2e t cos(t + 45 )u(t) Hence; EXAMPLE L 1 j 5.22 (a) Convert the circuit in Fig. 5.34 to an appropriate s domain representation. (b) Find the Thevein equivalent seen by 1Ω resistor. (c) Analyze the simplified circuit to find an expression for i(t). Figure 5.34 SOLUTION (a) Since the independent current source has u(t) in it, the circuit is not activated for t 0 . In otherwords, all the initial conditions are zero. Fig.5.35 (a) shows the s domain equivalent circuit for t 0+ . Figure 5.35(a) (b) Sine we are interested in the current in 1Ω using the Thevenin theorem, remove the 1Ω resistor from the circuit shown in Fig. 5.35(a). The resulting circuit thus obtained is shown in Fig. 5.35(b). 394 j Network Theory Zt (s) is found by deactivating the independent current source. Zt (s) = (5 + 0:001s)jj 500 s 2500 + 0:5s = Ω 0:001s2 + 5s + 500 Referring to Fig. 5.35 (b), Vt (s) = = 3 s [Zt (s)] Figure 5.35(b) 106 + 1500s 7:5 Volts 2 s(s + 5000s + 5 105 ) The Thevenin equivalent circuit along with 1Ω resistor is shown in Fig. 5.35 (c). I (s) = Vt (s) Zt (s) + 1 7:5 106 + 1500s s(s2 + 5500s + 3 106 ) 7:5 106 + 1500s = s(s + 4886)(s + 614) = Figure 5.35(c) Using partial fractions, we get I (s) = 2:5 s + 0:008 s + 4886 2:508 s + 614 Taking inverse Laplace transforms, we get i(t) = 2.5 + 0.008e 4886t 2.508e 614t u(t)A Check: i(0) = 2:5 + 0:008 2:508 = 0 i(1) = 2:5: and These could be verified by evaluating i(t) at t = 0 and t = 1 using the concepts explained in Chapter 4. EXAMPLE 5.23 Refer the RLC circuit shown in Fig. 5.36. Find the complete response for v (t) if t 0+ . Take v (0) = 2V. Laplace Transform j 395 Figure 5.36 SOLUTION Since we wish to analyze the circuit given in Fig. 5.36 using KVL, we shall represent L and C in frequency domain using series circuits to accomodate the initial conditions. Accordingly, we get the frequency domain circuit shown in Fig. 5.36 (a). Applying KVL clockwiseto the circuit shown in Fig. 5.36 (a), we get 2s 2 9 I (s) + = 0 + 6+s+ 2 s + 16 s s 32 ) I (s) = 2 (s + 6s + 9)(s2 + 16) 2 9 + Hence; V (s) = I (s) s = 2 s + s 288 s(s + 3)2 (s2 + 16) Using partial fraction, we get K4 K2 K3 K4 K1 + + + + V (s) = + 2 s s s+3 (s + 3) s j4 s + j4 2 Solving for K1 , K2 , K3 , and K4 , we get Figure 5.36(a) 288 K1 = = 2 2 2 (s + 3) (s + 16) s=0 288 d = 2:2 K2 = ds s(s2 + 16) s= 3 288 K3 = s(s2 + 16) s= 3 = 3:84 288 = 0:36 / 106:2 K4 = 2 s(s + 3) (s + j 4) s=j 4 396 j Network Theory Therefore V (s) = 2 2 s s + 0:36 / 106:2 2:2 3:84 0:36 /106:2 + + + 2 s+3 (s + 3) s j4 s + j4 Taking inverse Laplace Transform we get, v(t) = 2.2e 3t + 3.84te 3t 106.2 ) + 0.72 cos(4t Verification: Putting t = 0 in the above equation v (0) = 2:2 + 0 + 0:72 cos( 106:2 ) = 2:2 0:2 = 2V (The same quantity is given in the problem) 5.8 Waveform synthesis The three important singularity functions explained in section 5.3 are very useful as building blocks in constructing other waveforms. In this section, we illustrate the concept of waveform synthesis with a number of exmaples, and also determine expressions for these waveforms. EXAMPLE 5.24 Express the voltage pulse shown in Fig.5.37 in terms of unit step function and then find V (s). dv (t) . Also find L dt Figure 5.37 SOLUTION The pulse shown in Fig. 5.37 is the gate function. This function may be regarded as a step function that switches on at t = 2 secs and switches off at t = 4 secs. Laplace Transform j 397 Figure 5.37(a) Referring to Figs. 5.37 and 5.37 (a), we can write v (t) = v1 (t) + v2 (t) ) Hence; v (t) = 5u(t V (s) = = 5 s e 2) 5u(t 5 4s 2s 5 e s s 2s 4) e e 4s Figure 5.37(b) Taking the derivative of v (t), we get dv (t) = 5 [ (t dt Fig. 5.37(b) shows the graph of (t 2) 4)] dv (t) . dt We can obtain Fig. 5.37(b) directly from Fig. 5.36 by observing that at t = 2 seconds, there is a sudden rise of 5V leading to 5 (t 2). Similarly, at t = 4 seconds, a sudden fall of 5V leading to 5 (t 4). We know the Laplace trasnform pair Lf (t Hence; L a)g = e dv(t) dt =e as as =5 e Lf (t)g 2s e 4s 398 j Network Theory EXAMPLE 5.25 Express the current pulse in Fig.5.38 in terms of the unit step. Find: (i) L fi(t)g (ii) L R i(t)dt . Figure 5.38 SOLUTION Figure 5.39(a) Figure 5.39(b) Referring to Figs. 5.39 (a) and (b), using the principle of synthesis, we can write i(t) = i1 (t) + i2 (t) + i3 (t) = 5u(t) 10u(t 2) + 5u(t 4) Laplace Transform j 399 The Laplace transform of the above equation yields I (s) = = 5 10 s s 5 1 s 5 1 = Zs Let f (t) = then; f (t) = Z 2s e 2e e 5 + e s 2s +e 4s 4s 2s 2 i(t)dt [5u(t) = 5r (t) 10u(t 10r (t 2) + 5u(t 2) + 5r (t 4)]dt 4) = f1 (t) + f2 (t) + f3 (t) The function f1 (t) is a ramp of slope = 5 as shown in Fig. 5.39 (c). To this, if we add a ramp of slope = 10, the effect of this addition is, we get a ramp of slope = 5 10 = 5 for t 2 secs till we encounter the next ramp. At t = 4 seconds, if we add a ramp with a slope of +5, the net slope beyond t = 4 seconds is 5 + 5 = 0. Thus figure f (t) is drawn as shown in Fig. 5.39 (d). Figure 5.39(c) L ff (t)g = F (s) = L f5r (t) = = 5 10 s2 s2 5 1 s2 10r (t e 2s 2e + 2s 2) + 5r (t 5 e 4s +e 4s s2 4)g 400 j Network Theory Figure 5.39(d) EXAMPLE 5.26 Express the sawtooth function in terms of singularity functions. Then find Lfv (t)g. Figure 5.40 SOLUTION There are three methods to solve this problem. Method 1: The function v1 (t) is a ramp function of slope = +5. This slope +5 should continue till t = 1 second. Hence at t = 1 second, a ramp of slope t = 5 is added to v1 (t). The graph of v1 (t) + v2 (t) is shown in Fig. 5.41(a). Next, to v1 (t) + v2 (t), a step of 5V is added at t = 1 second. Hence; v (t) = v1 (t) + v2 (t) + v3 (t) = 5r (t) 5r (t V (s) = Lff (t)g = = 5 1 2 s e s 5u(t 1) 5 5 s2 s2 se s e s 1) 5 s e s Laplace Transform Figure 5.41(a) Figure 5.41(b) Method 2: This method involves graphical manipulation. Figure 5.41(c) j 401 402 j Network Theory The equation of a straight line passing through the origin is y = mx, where m = slope of the line. This allows us to write v1 (t) = 5t. From Fig. 5.41(c), we can write v (t) = v1 (t)v2 (t) Hence; = 5t [u(t) u(t 1)] = 5tu(t) 5tu(t 1) = 5tu(t) 5(t 1 + 1)u(t = 5tu(t) 5(t 1)u(t = 5r (t) 5r (t 5 V (s) = 2 1 e s s 1) se 1) 5u(t s 1) 5u(t 1) 1) Method 3: Figure 5.41(d) This method also involves graphical manipulation. We observe from Fig. 5.41(d) that v (t) is a multiplication of a ramp function and a unit step function. Thus; v (t) = v1 (t)v2 (t) = 5r (t) [u( t + 1)] Figure 5.41(e) Laplace Transform From Fig. 5.41(e), we can write v2 (t) = v3 (t) + v4 (t) ) v2 (t) = 1 u(t 1) ) u( t + 1) = 1 u(t 1) Hence; v (t) = 5r (t) [1 = 5r (t) We know that, r (t) = tu(t) Hence; v (t) = 5r (t) Please note that, u(t)u(t u(t 1)] 5r (t)u(t 1) 5tu(t)u(t 1) = 5r (t) 5(t 1 + 1)u(t)u(t = 5r (t) 5(t 1)u(t)u(t 1) = u(t 1) 1) 5u(t)u(t 1) 1) [Refer Fig. 5.41(f)] Figure 5.41(f) Thus; v (t) = 5r (t) = 5r (t) 5r (t 5 V (s) = 2 1 e s s Hence; EXAMPLE 5(t 1)u(t 1) 1) 5u(t se s 5u(t 1) 1) 5.27 Given the signal 8 < 3; 2; x(t) = : 2t 4; t<0 0<t<1 t>1 Express x(t) in terms of singularity functions. Also find Lfx(t)g. SOLUTION The signal x(t) may be viewed as follows: (i) in the interval, t < 0, x(t) may be regarded as 3u( t) (ii) in the interval, 0 < t < 1, x(t) may be viewed as 2[u(t) (iii) for t > 1, x(t) may be viewed as (2t 4)u(t 1) u(t 1)] and j 403 404 j Network Theory Thus; x(t) = 3u( t) ) x(t) = 3 [1 2 [u(t) u(t)] u(t 1)] + (2t 2u(t) + 2u(t 4)u(t 1) 1) + 2tu(t 1) 1) =3 5u(t) 2u(t 1) + 2(t 1 + 1)u(t =3 5u(t) 2u(t 1) + 2(t 1)u(t =3 5u(t) + 2r(t 4u(t 1) + 2u(t 1) 1) 1) Lfx(t)g cannot be found because x(t) contains a constant 3 for 1 < t < 0 (a noncausal signal). EXAMPLE 5.28 Express f (t) in terms of singularity functions and then find F (s). Figure 5.42 SOLUTION To find f (t) for 0 < t < 2 : Equation of the straight line 1 is y x y1 y2 = x1 x2 y1 x1 Here, y is f (t) and x is t. Hence, ) f (t) 3 3 3 = t 0 2 0 2f (t) 6 = 6t ) f (t) = 3 3t Figure 5.43 Laplace Transform To find f (t) for 2 < t < 3: Here, f (t) + 3 0+3 = t 2 3 2 ) f (t) + 3 = 3t 6 ) f (t) = 3t 9 8 < 3 Hence 3t; 0 < t < 2 3t 9; 2 < t < 3 f (t) = : 0; otherwise The above equation may also be written as : f (t) = [3 3t] [u (t) = 3u (t) 3u (t 3tu (t ) 3) u (t 2) 9] [u (t 9u (t 2) + 9u (t 2) 3tu (t) + 6tu (t = 3u (t) 12u (t 2) 3tu (t) + 6 (t = 3u (t) 12u (t +12u (t f (t) = 3u (t) 2) 3) + 9u (t 3 (t 3) u (t 3tu (t) + 6 (t = 3 s 3) + 9u (t 3) 2) 3) 3) 2) u (t 2) u (t 9u (t 2) F (s) = L ff (t)g Hence; 2) 3tu (t 2) 2 + 2) u (t 3tu (t) + 6 (t 2) 3)] 3) 12u (t 3 + 3) u (t u (t 2) 2) + 3tu (t 3tu (t) + 3tu (t f (t) = 3u (t) 3 (t EXAMPLE 2)] + [3t 6 3 + 2e 2 s s 2s 2) 3) + 9u (t 3 (t 3 e s2 3) u (t 3) 3) 3s 5.29 Express the function f (t) shown in Fig. 5.44 using singularity functions and then find F (s). Figure 5.44 j 405 406 j Network Theory SOLUTION Equation of the straight line shown in Fig. 5.45(a) is f1 (t) + 1 2+1 = t 1 2 1 f1 (t) + 1 = t + 1 ) ) f1 (t) = t The above equation is for the values t lying between 1 and 2. This could be expressed, by writing f (t) = f1 (t) g (t) Figure 5.45(a) Figure 5.45(b) ) f (t) = Hence; t [u (t u (t 1) = (t 1 + 1) u (t = (t 1) u (t = r (t 1) 1) u (t 2)] 1) + (t u (t 1) + (t 1) + r (t F (s) = L ff (t)g = 2 + 2) u (t 1 e s s2 2) 2) u (t 2) + 2u (t 1 s 1 e + 2e s s 2) + 2u (t 2) 2s + 2 e s 2s 2) Laplace Transform EXAMPLE 5.30 Find the Laplace transform of the function f (t) shown in Fig. 5.46. Figure 5.46 SOLUTION Method 1: Figure 5.47(a) We can write; f (t) = fA (t) + fB (t) = sin tu (t) + sin (t Hence; F (s) = L ff (t)g = = s2 s2 + 2 π 1+e s 2 +π 1) u (t + s2 + 2 1) e s j 407 408 j Network Theory Method 2 : Figure 5.47(b) Graphically, we can manipulate f (t) as f (t) = fC (t) g (t) = sin t [u (t) u (t 1)] = sin tu (t) sin tu (t = sin tu (t) sin t (t 1 + 1) [u (t 1)] = sin tu (t) sin ( (t 1) + ) u (t 1) = sin tu (t) + sin (t F (s) = L ff (t)g = Hence; = EXAMPLE s2 s2 + 2 π 1+e s 2 +π 1) 1) u (t + s s2 + 2 5.31 Find the Laplace transform of the signal x(t) shown in Fig. 5.48. Figure 5.48 1) e s Laplace Transform j 409 SOLUTION Figure 5.49 Mathematically, we can write x(t) as x (t) = xA (t) = sin (t 1) u (t = s2 π e s + π2 sin (t 1) L fx (t)g = X (s) = s2 + 2 e Hence; EXAMPLE xB (t) e s s2 + 2 3) u (t e 3) 3s 3s 5.32 Refer the waveform shown in Fig. 5.50. The equation for the waveform is sin t from 0 to ; sin t s 1 coth : from to 2 . Show that the Lapalce transform of this waveform is F (s) = 2 s +1 2 Figure 5.50 410 j Network Theory SOLUTION f (t) is a periodic waveform with a period T = seconds. Let f1 (t) be the waveform f (t) described over only one period. The Laplace transform of f (t) and f1 (t) are related as F1 (s) 1 e sT Let us now proceed to find F1 (s). From Fig. 5.51 (b), we can write F (s) = f1 (t) = fA (t) + fB (t) = sin tu (t) + sin (t ) u (t 1 1 ) F1 (s) = 2 + 2 e s s +1 s +1 (1 + e s ) = s2 + 1 F1 (s) F1 (s) Hence; F (s) = = sT 1 e 1 e s s (1 + e ) ) F (s) = 2 (s + 1) (1 e s ) , es=2 + e s=2 = 1 e, 2 s2 + 1 , es=2 s=2 e, s ) ) Figure 5.51(a) s=2 e s=2 2 1 cosh 2 s2 + 1 sinh s 2 πs 1 = 2 coth s +1 2 F (s) = EXAMPLE Figure 5.51(b) 5.33 Find the Laplace transform of the pulse shown in Fig. 5.52. Figure 5.52 Laplace Transform j 411 SOLUTION We can describe Fig. 5.52 mathematically as 0<t<2 Vo ; Vo t + 3Vo ; 2 < t < 3 f (t) = The expression for f (t) for 2 < t < 3 is obtianed as follows : Equation of a straight line between two points is given by y1 x1 y2 y1 = x1 x2 y x In the present context, y = f (t); x = t; (x1 ; y1 ) = (2; Vo ) and (x2 ; y2 ) = (3; 0) 0 Vo f (t) Vo = t 2 3 2 f (t) = Vo t + 3Vo Hence; ) The time domain expression for f (t) between t = 0 and 3 could be written using graphical manipulation as f (t) = Vo [u (t) u (t 2)] + [ Vo t + 3Vo ] [u (t u (t 2) 3)] The first term on the right-side of the above equation defines f (t) for 0 < t < 2 and the second term on the right-side defines f (t) for 2 < t < 3. f (t) = Vo u (t) Vo u (t 2) Vo tu (t = Vo u (t) Vo u (t 2) Vo (t +Vo (t = Vo u (t) 3 + 3) u (t Vo u (t +3Vo u (t = Vo u (t) ) f (t) = Vo u (t) 2) u (t 2) 2) 3Vo u (t 3) 2) 3Vo u (t 2Vo u (t 2) 3Vo u (t 2) + Vo (t 2) + Vo r (t 2) 3) + 3Vo u (t 3) 2) + Vo (t 3) u (t 3) 3) 3) u (t 3) 3) F (s) = L ff (t)g Hence; = EXAMPLE Vo (t 2) 2) u (t Vo r (t 2 + 2) u (t 3) + 3Vo u (t 3) + 3Vo u (t Vo (t 2) + Vo tu (t Vo s Vo e s2 2s + Vo e s2 3s 5.34 Consider a staircase waveform which extends to infinity and at t = nt0 jumps to the value n + 1, being a superposition of unit step functions. Determine the Laplace transform of this waveform. 412 Network Theory j SOLUTION We can write, F (s) = L ff (t)g = 1 1+e = Let e t0 s s =x then F (s) = t0 s 2t0 ) + 1 + e t0 s + e 2t0 s + t0 ) + u (t f (t) = u (t) + u (t 1 s +e 1 s s 2t0 s + 1 1 + x + x2 + s From Binomial theorem, we have (1 Hence; EXAMPLE 1 = 1 + x + x2 + 1 F (s) = s (1 x) 1 = s (1 e t0 s ) x) Figure 5.53 5.35 (a) Find the Laplace transform of the staircase waveform shown in Fig. 5.54. (b) If this voltage were applied to an RL series circuit with R = 1Ω and L = 1H , find the current i(t). Figure 5.54 SOLUTION (a) We can express mathematically, the voltage waveform shown in Fig. 5.54 as, v (t) = 8 1; 1 < t < 2 > > > > < 2; 2 < t < 3 3; 3 < t < 4 > > 4; 4 < t < 5 > > : 0; elsewhere Laplace Transform v (t) = [u (t or 1) +3 [u (t = u (t u (t 2)] + 2 [u (t u (t 3) 1) + u (t u (t 2) 4)] + 4 [u (t 2) + u (t 3) + u (t 4) 413 3)] u (t 4) j 5)] 4u (t 5) Taking the Laplace transform, we get V (s) = 1 s e +e s 2s 3s +e +e 4s 4e 5s (b) Assuming all initial conditions to be zero, the time domian circuit shown in Fig. 5.55 gets transformed to a circuit as shown in Fig. 5.56. Figure 5.55 Time Domain Circuit Figure 5.56 Frequency . Domain Circuit From Fig. 5.56, we can write I (s) = V (s) s+1 1 1 1 1 e s+ e 2s + e 3s + e s (s + 1) s (s + 1) s (s + 1) s(s + 1) 1 1 1 1 1 1 s 2s + e + e e ) I (s) = s s+1 s s+1 s s+1 1 1 1 1 + e 4s 4 e 5s s s+1 s s+1 ) I (s) = 4s 4 e s (s + 1) 5s 3s Taking the inverse Laplace transform, we get i (t) = u (t) i (t) = e + 1 t !t 1 + u (t) e u (t) t!t 2 + u (t) t e u (t) t!t 4 4 u (t) e t u (t) t!t 5 + u (t) 1 e t u (t) (t 1) e (t t u (t 4) 1) + 1 u (t 4) e 4 1 (t 2) e u (t (t 5) 2) + 1 u (t 5) e t u (t) e (t 3) t !t 3 u (t 3) 414 j Network Theory EXAMPLE 5.36 A voltage pulse of 10 V magnitude and 5 sec duration is applied to the RC network shown in Fig. 5.57. Find the current i(t) if R = 10Ω and C = 0:05F . Figure 5.57 SOLUTION Figure 5.58(a) Mathematically, we can express v (t) as follows : v (t) = v1 (t) v2 (t) = 10u(t) Hence; V (s) = L fv (t)g = 10 1 10u(t e s t0 s t0 ) Assuming all initial conditions to be zero, the Laplace transformed network is as shown in Fig. 5.58(b). I (s) = = V (s) R+ 1 Cs e 10 1 s R+ t0 s 1 Cs Figure 5.58(b) Laplace Transform 10Cs 1 e t0 s s (RCs + 1) 1 10 = 1 e t0 s 1 R s+ I (s) = 2 RC = 10 6 4 R 1 s+ Taking inverse Laplace transform yields i (t) = = 10 R 10 R e t RC e t RC t u (t) u (t) i (t) = e 0:510 EXAMPLE 6 10 e R 10 R t RC u (t) t !t 10 0:5 t0 s 7 5 e RC t0 u (t (t 510 e 1 s+ RC (t t0 ) RC e u (t) 1 1 3 t0 ) 6 6 ) u t 5 10 6 5.37 Find the Laplace transform of the waveform shown in Fig. 5.59. Figure 5.59 SOLUTION v (t) = or v (t) = 3t [u (t) 3t; 0 < t < 1 2; 1 < t < 2 u (t 1)] + 2 [u (t = 3tu (t) 3tu (t 1) + 2u (t = 3tu (t) 3 (t 1 + 1) u (t = 3tu (t) 3 (t 1) u (t 1) 1) 1) u (t 2)] 2u (t 2) 1) + 2u (t 3u (t 1) 2u (t 1) + 2u (t 1) 2) 2u (t 2) j 415 416 Network Theory j v (t) = 3tu (t) = 3r (t) 3r (t V (s) = L fv (t)g Hence; = 5.9 3 (t 3 s2 3 e s s2 1) u (t 1) u (t 1) u (t 1 s e s 2u (t 1) 2 e s 2u (t 1) 2) 2) 2s The System function The system function or transfer function of a linear time-invariant system is defined as the ratio of Laplace transform of the output to Laplace transform of the input under the assumption that all initial conditions are zero. Hence, for relaxed LTI system, the response Y (s) to an input X (s) is H (s) X (s), where H (s) is the system function. The system function H (s) may be found in several ways: 1. For a system defined by a linear differential equation, by taking Laplace transform of the Y (s) . differential equation and then finding the ratio X (s) 2. From the Laplace transform of impulse response h(t). 3. From the s domain model of a physical system like an electrical system. EXAMPLE 5.38 The output y (t) of an LTI system is found to be e 3t u(t) when the input x(t) is 0:5u(t). (a) Find the impulse response h(t) of the system. (b) Find the output when the input is x(t) = e t u(t). SOLUTION (a) Taking Laplace transforms of x(t) and y (t), we get 1 0:5 ; X (s) = s+3 s Y (s) 2s H (s) = = X (s) s+3 2 (s + 3) 6 6 H (s) = =2 (s + 3) s+3 Y (s) = Hence ) Taking inverse Laplace transform, we get h (t) = 2δ (t) 6e 3t u (t) Laplace Transform j 417 x (t) = e t u (t) (b) ) Thus; 1 s+1 Y (s) = X (s) H (s) 2s = (s + 1) (s + 3) X (s) = = K2 K1 + s+1 s+3 2s K1 = s + 3 s = where Therefore; 1 = 1 2s =3 K2 = s + 1 s = 3 3 1 + Y (s) = s+1 s+3 Taking inverse Laplace transform of Y (s), we get y (t) = y (t) = or EXAMPLE e e t + 3e t + 3e 3t ; t0 3t ; u(t) 5.39 Determine the output v (t) for the circuit shown in Fig. 5.60. Figure 5.60 SOLUTION The transformed network of Fig. 5.60 with the assumption that all initial conditions are zero is shown in Fig. 5.61(a). 418 j Network Theory V (s) = = 1 I (s) s2 3 1 6 Vs (s) 7 4 15 s 1+ s 1 V (s) = H (s) = Vs (s) s+1 ) Figure 5.61(a) The inverse Laplace transform of H (s) is called the impulse response of the circuit and is denoted by h(t). h (t) = e t u (t) I method : From Convolution theorem, we have, v (t) = h (t) vs (t) Z1 h ( ) vs (t = ) d 0 Z1 e = 0 = 2e t u ( ) 2e (t ) u (t ) d Z1 u ( ) u (t ) d 0 Let us compute the product u ( ) u (t u ( ) = u (t ) = Hence; u ( ) u (t ) = ) for different values of 1; < 0 0; > 0 1; t 0; t > 0 or < t < 0 or > t 1; 0 < < t; t > 0 0; otherwise Figure 5.61(b) Laplace Transform Therefore; v (t) = 2e t Zt t d = 2te j 419 ; t0 0 = 2te t u (t) II method : In the frequency domain, convolution operation is transformed into a multiplicative operation. That is; V (s) = H (s) Vs (s) 1 2 (s + 1) (s + 1) 2 = (s + 1)2 = Inverse Laplace transform yields, v(t) = 2te t u(t)Volts Reinforcement problems R.P 5.22 (a) Find H (s) = Vo (s) for the circuit shown in Fig. R.P. 5.22. (b) Determine vo (t) when the Vi (s) intital current in the inductor is zero. Figure R.P.5.22 SOLUTION The Laplace transformed network with all initial conditions set to zero is shown in Fig. R.P. 5.22(a). Vo (s) = I (s) 150 + 2 10 = ) H (s) = 3 s Vi (s) 150 + 2 10 100 + 3 10 Vo (s) 1.5 = Vi (s) 2.5 3s 3s + 150 + 2 10 105 + 2s 105 + 5s 3s 420 j Network Theory (b) Vo (s) = H (s) Vi (s) = 1:5 105 + 2s 100 2:5 105 + 5s s 40 s + 0:75 105 = s [s + 0:5 105 ] = K1 K2 + s s + 0:5 105 Figure R.P. 5.22(a) = 60 s=0 5 40 s + 0:75 10 K2 = = 20 s 5 40 s + 0:75 105 K1 = [s + 0:5 105 ] where s= 0:5 10 Hence; Vo (s) = 60 s 20 s + 0:5 105 Taking inverse Laplace transform, we get vo (t) = 60 R.P 20e 0:5 105t u (t) Volts 5.23 Refer the circuit shown in Fig. R.P. 5.23. The switch closes at t = 0. Determine the voltage v (t) after the switch closes. Figure R.P. 5.23 SOLUTION The switch is open at t = 0 and closed at t = 0+ . Let us assume that at t = 0 , the circuit is in steady state. The circuit at t = 0 is shown in Fig. R.P. 5.23(a). Laplace Transform j 421 Figure R.P. 5.23(a) Referring to Fig. R.P. 5.23(a), we get 8 = 2A 2+2 v 0 =0 i 0 = From switching principles, we know that the current through an inductor and the voltage across a capacitor cannot change instantaneously. Therefore, + i (0) = i 0+ = i 0 v (0) = v 0 and =v 0 = 2A = 0V We shall solve this probelm using nodal technique. Hence, in the frequency domain, we will use the parallel models for the capacitor and inductor because the parallel models contain current sources rather than voltage sources. The frequency domain circuit is shown in Fig. R.P. 5.23(b). Figure R.P.5.23(b) KCL at node V (s) : V (s) 2 ) ) 8 s + V (s) + V (s) + 2 = 0 1 s s s 1 1 + +s = 2 s s + 2 + 2s2 = V (s) 2s V (s) 4 2 s s 2 s 422 j Network Theory ) V (s) = = = 2s2 4 +s+2 s2 2 + 0:5s + 1 s2 2 + 0:5s + (0:25)2 = 2 (s + 0:25) + (0:96824)2 = 2 0:96824 0:96824 (s + 0:25)2 + (0:96824)2 2 = 2:066 We know that, L 1 0:96824 (s + 0:25)2 + (0:96824)2 a 2 (s + b) + Hence, v (t) = 2.066e R.P (0:25)2 + 1 a2 =e bt sin at u (t) 0:25t sin (0.96824t) u (t) Volts 5.24 Find the impulse response of the circuit shown in Fig. R.P. 5.24. Figure R.P. 5.24 SOLUTION The frequency domain representation of the circuit is shown in Fig. R.P. 5.24(a) by assuming that all initial conditions to be zero. Laplace Transform Figure R.P. 5.24(a) KCL at node a: Va (s) ) ) ) Vb (s) Va (s) 1 + Vg (s) + =0 1 2 2 2s Va (s) Vb (s) Va (s) Vb (s) + + 2sVa (s) = 0 2 2 1 1 1 1 =0 + + 2s Vb (s) + Va (s) 2 2 2 2 Va (s) [1 + 2s] Vb (s) = 0 KCL at node b: Vb (s) Vi (s) s ) Va (s) ) ) + Vb (s) Va (s) 2 =0 1 1 Vi (s) 1 + Vb (s) = + 2 s 2 s Va (s) Vi (s) (2 + s) + Vb (s) = 2 2s s sVa (s) + (2 + s) Vb (s) = 2Vi (s) Putting the above nodal equations in matrix form, we get 1 + 2s s 1 2+s Va (s) Vb (s) = 0 2Vi (s) Solving, we get 2Vi (s) 2 + s + 4s + 2s2 s Va (s) 1 = Vi (s) (s + 1)2 vi (t) = (t) ) Vi (s) = 1 Va (s) = ) Given j 423 424 j Network Theory Va (s) Hence; 1 (s + 1)2 1 Va (s) = (s + 1)2 1 ) = Taking inverse Laplace transform, we get va (t) = h (t) = te t u (t) R.P 5.25 Find the convolution of h (t) = t and f (t) = e H (s) F (s). SOLUTION h (t) f (t) = L /t for t > 0, using the inverse transform of 1 fH (s) F (s)g 1 H (s) = L fh (t)g = 2 where s F (s) = L ff (t)g = Hence; H (s) F (s) = = 1 s+ 1 + ) s2 (s K3 K2 K1 + 2 + s s s+ Solving the partial fractions yields 1 K1 = Hence; H (s) F (s) = ) 2 1 = R.P 1 s 1 + 1 1 ; 1 s2 1 K3 = + 2 1 2 1 s+ fH (s) F (s)g 1 = K2 = 2 h (t) f (t) = L ; 2 u (t) + 1 tu (t) + 1 2 e t u (t) 1 t 1 + + 2 e t u (t) α2 α α 5.26 Consider a pulse of amplitude 5V for a duration of 4 seconds with its starting point t = 0. Find the convolution of this pulse with itself and draw the convolution x (t) x (t) versus time. Laplace Transform j Figure R.P. 5.26(a) SOLUTION x (t) = 5u (t) ) Let 5 5u (t 4) 5 e 4s s s y (t) = x (t) x (t) X (s) = Taking Laplace transform, we get Figure R.P. 5.26(b) Y (s) = X (s) X (s) = 25 50 s2 s2 4s e + 25 s2 e 8s Taking inverse Laplace transform, we get Hence; R.P y (t) = 25tu (t) 50 (t y(t) = 25r(t) 50r(t 5.27 Show that L Ktr (r 1 e at 1)! = 4) u (t 4) + 25 (t 4) + 25r(t K (s + a)r SOLUTION Let f (t) = 1 then F (s) = Thus; We know that; 1 s dn F (s) ( 1)n n! = dsn sn 1 n F (s) L ftn f (t)g = ( 1)n d ds n With f (t) = 1, we get L ft n n g = ( 1) = n! sn+1 ( 1)n n! sn+1 8) 8) u (t 8) 425 426 j Network Theory Putting n = r 1, we get L K R.P (r tr 1 L 1 e at L 1)! 1 e at and Therefore; tr tr = (r 1)! sr = (r 1)! (s + a)r = K (s + a)r 5.28 Tests conducted on a certain network revealed that the current was i(t) = 2e t + 4e 3t when a unit step voltage was suddenly applied to the input terminals of the network at t = 0. What voltage must be applied to get an output current of i(t) = 2e t if the network remains unchanged? SOLUTION Given, Hence; i(t) = 2e t + 4e 3t ; I (s) = V (s) = and t 0 when v (t) = u(t) 2 4 + s+1 s+3 1 s Laplace transform of the output System function = H (s) = Laplace transform of the input I (s) ) H (s) = V (s) 2s (s 1) = (s + 1) (s + 3) We have to find v (t) when i(t) = 2e t . First we will find V (s) when I (s) = Hence; 2 I (s) using the relation H (s) = . s+1 V (s) V (s) = I (s) H (s) 2 s+1 = 2s (s 1) (s + 1) (s + 3) (s + 3) = s (s 1) = K2 K1 + s s 1 Laplace Transform Using partial fractions, we find that K1 = Hence; V (s) = v (t) = ) R.P 3 and K2 = 4 3 s 4 + s 1 3u (t) + 4et u (t) Volts 5.29 Find the Laplace transform of the periodic waveform shown in Fig. R.P. 5.29. Figure R.P. 5.29 SOLUTION The Laplace transform of a periodic waveform is found using the relation F (s) = 1 F1 (s) e sT where F1 (s) = L ff1 (t)g = Laplace transform of f (t) over 0 < t < T . Where T = fundamental period of f (t). Figure R.P. 5.29(a) Referring to Fig. R.P. 5.29(a) we can write: f1 (t) = ) f1 (t) = 1 a t [u (t) + 1 a 0<t<a 1; a < t < 3a 1 a tu (t 3a < t < 4a a)] + [u (t t + 4 [u (t 1 + tu (t a t ; a > > 1 > > > : a t + 4; u (t 1 = tu (t) a 8 > > > > > < 3a) a) u (t u (t 4a)] a) u (t a) + u (t 4a) + 4u (t 3a) 4u (t 3a)] 3a) 4a) 1 a tu (t 3a) j 427 428 j Network Theory ) f1 (t) = 1 a 1 tu (t) 1 a a (t + 4u (t 3a) 1 a a 1 a + a) u (t 3a + 3a) u (t 1 = tu (t) (t 3a) + 4u (t (t a) + u (t 4a) a) u (t a) 1 a a 1 = r (t) Hence; a 1 a 1 (t a) u (t r (t 1 a) a = 1 1 as2 as2 1 as2 1 e as2 e e 3as 3as +e 4a) u (t 4a) u (t 3a) u (t 1 as 3a) a) (t 3a) + 1 a 3a) 4a) + 4u (t (t 4a) u (t 4a) a as e (t 1 a 1 3a) + r (t r (t a a F1 (s) = L ff1 (t)g = a) + u (t 3a) + 1 a) u (t 4a + 4a) u (t u (t (t 3a) u (t 3a) 3u (t a +4u (t 3a) 4u (t 4a) 1 = tu (t) (t a) + 4as 1 as2 e 4as Alternate method for finding F1 (s): From Figs. R.P. 5.29(b), (c), (d), we can write f1 (t) = fA (t) + fB (t) + fC (t) + fD (t) 1 = tu (t) a + 1 a (t 1 a (t a) u (t 4a) u (t Figure R.P. 5.29(b) 4a) a) 1 a (t 3a) u (t 3a) 4a) 4a) Laplace Transform Figure R.P. 5.29(c) Figure R.P. 5.29(d) Hence; F1 (s) = L ff1 (t)g = Finally; 1 as2 e as F (s) = L ff (t)g = where T = 4a 1 as2 1 1 F (s) = as2 1 1 as2 F1 (s) e sT e as e 3as + e (1 e 4as ) 4as e 3as + 1 as2 e 4as j 429 430 j Network Theory R.P 5.30 Find the Laplace transform of the function f (t) shown in Fig. R.P. 5.30. Figure R.P. 5.30 SOLUTION Let f (t) = x(t) + u(t), where x(t) is a periodic triangular wave and is as shown in Fig. R.P. 5.30(a). Figure R.P.5.30(a) Figure R.P.5.30(b) Let x1 (t) be x(t) within its first period as shown in Fig. R.P.5.30(b). Referring to Fig. R.P. 5.30(b), we can write x1 (t) = ) x1 (t) = 2t [u (t) u (t = 2tu (t) 2tu (t = 2tu (t) 2 (t 2 (t = 2tu (t) 2 (t 4 1)] + (4 1) u (t x1 (t) = 2tu (t) 4 (t ) x1 (t) = 2r (t) 4r (t 1) 1) u (t 1) + 4u (t 1) + 2 (t 2u (t 2)] 2) 2tu (t 1) 4u (t 2) u (t 1) + 2tu (t 2) 2) 1) + 4u (t 1) + 2 (t 1) + 2 (t 2) u (t 2 + 2) u (t 2u (t 1) 1) + 2r (t 1) 4u (t 1) 1 + 1) u (t 1) u (t ) 2t) [u (t 1) + 4u (t 1 + 1) u (t 2 (t 2t; 0<t<1 2t; 1 < t < 2 2) u (t 2) 1) 4u (t 2) + 4u (t 2) 2) 2) Laplace Transform Hence; X1 (s) = L fx1 (t)g = = = 2 4 s2 s2 2 s e s2 1 2e s2 1 e 2 + s 2 s2 e +e 2s 2s s 2 Since x(t) is periodic, X (s) = L fx (t)g = X1 (s) 1 e sT where T = 2 seconds 2 Hence; We know that; 2 (1 e s ) s2 (1 e 2s ) f (t) = x(t) + u(t) X (s) = Applying linearity property, F (s) = X (s) + U (s) 2 1 = s (1 R.P 2 e s 1 + 2 s e ) s 5.31 Find f (t) using convolution integral for the function, F (s) = 4s (s + 1) (s2 + 4) SOLUTION Let where F (s) = F1 (s) F2 (s) 4 F1 (s) = s+1 s F2 (s) = 2 s +4 f (t) = L Z1 1 ) f1 (t) = 4e t u (t) ) f2 (t) = cos 2tu (t) [F1 (s) F2 (s)] f1 () f2 (t = 0 ) d j 431 432 j Network Theory u ( ) u (t We know that 1; 0 < < t; t > 0 0; otherwise ) = Zt Hence; (t ) cos 24e f (t) = 0 t = 4e Zt d e cos 2 d 0 Using the standard integral formula Z eax cos bx dx = eax a2 + b2 f (t) = 4e we get t (a cos bx + b sin bx) t e 1+4 (cos 2 + 2 sin 2) =0 4 = e t et (cos 2t + 2 sin 2t 1) 5 4 8 4 t = cos 2t + sin 2t e ; t0 5 5 5 8 4 cos 2t + sin 2t f (t) = 5 5 R.P 4 e 5 t u (t) 5.32 If h(t) = 2e 3t u(t) (t). Find y (t) = h(t) x(t) by (a) using convolution in and x(t) = u(t) the time-domain (b) Finding H (s) and X (s) and then obtaining L 1 [H (s)X (s)] SOLUTION 3t Given h(t) = 2e and x(t) = u(t) (a) u(t) (t) y (t) = x (t) h (t) Z1 x () h (t = ) d 0 Z1 ju () = ()j 2e 3(t ) u (t ) d 0 Z1 2e = 0 3(t ) Z1 u (t ) u () d e 2 0 3(t ) u (t ) () d Laplace Transform j 433 1; 0 < < t; t > 0 0; otherwise The second integral on the right-hand side is evaluated using the sifting property for an impulse function. We know that, u (t ) u () = Zt Hence; 2e y (t) = 0 ) y (t) = 2e = 3t 2 1 3 3t 3 e d e3 t 3 e 2e 0 3t 3(t ) 2e 3t 2e 3t u (t )j=0 u (t) u (t) Since t > 0, we associate u(t) in the first component on the right hand side of y (t). 2 1 e 3t u (t) 2e 3 2 8 3t u (t) e = 3 3 Then; y (t) = 3t u (t) (b) Verification : 1 2 ; X (s) = s+3 s Y (s) = X (s) H (s) 2 (1 s) = s (s + 3) H (s) = ) = 1 K2 K1 + s s+3 Using partial fractions, we find that K1 = Hence; ) R.P 2 8 ; K2 = 3 3 8 1 2 1 Y (s) = 3 s 3 s+3 2 8 3t y (t) = u (t) e u (t) 3 3 2 8 3t u (t) e = 3 3 5.33 When an impulse (t) V is applied to a certain network, the ouput voltage is vo (t) = 4u(t) 4u(t 2) V. Find and sketch vo (t) if the imput voltage is 2u (t 1) V. 434 j Network Theory SOLUTION When vi (t) = (t), it is given that vo (t) = 4u(t) 4u(t 2). From this data, we can find the transfer function H (s) as follows: L fvo (t)g L fvi (t)g L f4u (t) 4u (t = L f (t)g H (s) = = 4 1 e s 2s 2)g The transfer function H (s) can be used to find vo (t) when vi (t) = 2u(t 1) V. This procedure is as follows: Vo (s) Vi (s) Vo (s) = Vi (s) H (s) H (s) ) 2 = e s = 8 s2 s e s 4 4 s s 8 s2 e e 2s 3s Taking inverse Laplace transform, we get vo (t) = 8 (t = 8r (t 1) u (t 1) 1) 8r (t 8 (t 3) u (t 3) 3) The corresponding wave form for vo (t) is sketched in Fig. R.P. 5.33 Figure R.P. 5.33 R.P 5.34 Refer the two circuits shown in Fig. R.P. 5.34(a) and (b). Given that v1 (t) = sin 103 t and v2 (t) = e 1000t for t 0 and c = 1 F. (a) Show that it is possible to have i1 (t) = i2 (t) for all t 0. (b) Determine the required values of R and L for the condition in part (a) to hold good. Laplace Transform Figure R.P. 5. 34 (a) j 435 Figure R.P.5.34 (b) SOLUTION Referring Fig. R.P. 5.34(a) we can write in Laplace domain V1 (s) I1 (s) = R+ 1 Cs Similarly, referring Fig. R.P. 5.34(b), we can write in Laplace domain I2 (s) = V2 (s) sL + 1 Cs i1 (t) = i2 (t) means that I1 (s) = I2 (s) 103 V1 (s) = L sin 103 t = Also; V2 (s) = L e 1000t = 2 s2 + (103 ) 1 s + 103 Hence, the condition I1 (s) = I2 (s) gives, 103 s2 + 106 ) R s+ 103 106 R 1 R+ 106 = 1 s + 103 s 1 sL + + 106 ) s 1 = (s2 106 L (s + 103 ) s2 + 106 L If the above equation is satisfied, then it is possible to have i1 (t) = i2 (t). For this to happen, it is required that R 103 = L; 106 R = 103 The above conditions give L = 1H and R = 103 Ω and 106 = 106 L 436 j Network Theory R.P 5.35 For the circuit shown in Fig. R.P. 5.35 has zero initial conditions. At t = 0, the switch K is p closed. Find p the value of R such that the response v (t) = 0:5 sin 2t volts. Take the excitation as i(t) = te 2t A. Figure R.P.5.35 SOLUTION p Given i(t) = te 2t Taking Laplace transform of i(t) gives I (s) = (s + 1 p 2)2 p Laplace transform of the response v (t) = 0:5 sin 2t is " p # 2 1 V (s) = 2 2 s +2 Hence; Z (s) = V (s) I (s) p 1 (s + 2)2 =p 2 s2 + 2 (5.29) For the circuit shown in Fig. R.P. 5.35 we can write Z (s) = R + s 2 ) 1 + 1 s 2s Z (s) = R + 2 s +2 (5.30) Laplace Transform j 437 Equating equations 5.29 and 5.30, we get p 2 1 s+ 2 2s p =R+ 2 2 s +2 2 s +2 p 2 1 s2 + 2 R 2s = p s + 2 2 2 2 s = p + 2s + p 2 2 ) Equating the like powers of s, we get 1 2 R= p Ω Exercise Problems E.P 5.1 Find the Laplace transform of the following functions : (a) f1 (t) = sin(!t + ) (b) f2 (t) = sin2 t (c) f3 (t) = 1 [sinh(at) 2a3 Ans: F1 (s) = E.P sin(at)] s sin θ + ω cos θ 2 , F2 (s) = , F3 (s) = 2 2 2 2 s +ω s (s + 4) (s 1 a2 ) (s2 + a2 ) 5.2 In the network shown in Fig. E.P. 5.2, the switch K is moved from position a to position b at t = 0, a steady state having previously been established at position a. Solve for i(t), using the Laplace transformation method. Figure E.P. 5.2 Ans: i (t) = Va e RA RA +RB L u (t) 438 j E.P Network Theory 5.3 Find i1 (t) and i2 (t) for t > 0 for the circuit shown in Fig. E.P. 5.3 using Laplace transform. Figure E.P. 5.3 h Ans : i1 (t) = 2.4e h i2 (t) = 1.2e E.P i + 0.6e 6510 t + 3 u (t) mA i 6510 t 1.2e 510 t u (t) mA 5 105 t 5 5 5 5.4 Using Laplace transform technique, find i(t) when i1 = 0:1e 5.4 when b = 105 . Assume steady state conditions at t = 0 . A for the circuit shown in E.P. Figure E.P. 5.4 1 27 Ans: i (t) = e bt + e 30 40 E.P bt 17 e 24 6bt 10bt u (t) A 5.5 The current source shown in Fig. E.P. 5.5 is i(t) = tu(t) A. Find vo (t) when the initial value of vo is zero. Ans: vo (t) = t 10 3 Figure E.P. 5.5 1 e 103 t mV, t 0 Laplace Transform E.P j 439 5.6 Find the inverse Laplace transform of the following F (s): 8s 3 4s2 (a) F (s) = 2 (b) F (s) = s + 4s + 13 (s + 3)2 2 t Ans: (a) f (t) = 10.2e cos (3t + 38.3 ) u (t) (b) f (t) = 4e E.P 3t 24te 3t 3t + 18t2 e u (t) 5.7 Using convolution integral, find f (t) if F (s) = Ans: f (t) = 2 1 E.P e 5t 10 s (s + 5) u (t) 5.8 Refer the network shown in Fig. E.P. 5.8. Assume the network is in steady state for t < 0. Determine the current i(t) for t > 0. Figure E.P. 5.8 Ans: E.P i (t) = 4.22e t cos (3t 18.4 ) u (t) A 5.9 Find vo (t) in the circuit shown in Fig. E.P. 5.9. Figure E.P. 5.9 Ans: vo (t) = 4 8.93e 3:73t + 4.93e 0:27t u (t) V 440 j Network Theory E.P 5.10 Find vo (t) for t > 0. Refer the circuit shown in Fig. E.P. 5.10. Figure E.P. 5.10 4 Ans: vo (t) = + 2.55e 3 E.P 1 t 3 cos F 17t + 10.1 u (t) 5.11 For the circuit shown in E.P. 5.11. Vo (s) Find: (a) H (s) = Vi (s) (c) Step response (b) h (t) (d) The response when vi (t) = 8 cos 2t V Figure E.P. 5.11 2 s+4 (b) h (t) = 2e 4t u (t) Ans: (a) (c) H (s) = vo (t) = 0.5 1 (d) vo (t) = 1.5 e E.P e 4t 4t u (t) V + cos 2t + 0.5 sin 2t u (t) V 5.12 Refer the circuit shown in Fig. E.P. 5.12. The switch is closed at t = 0 Find : (a) i1 (t) and (b) i2 (t) Laplace Transform Figure E.P. 5.12 Ans: (a) i1 (t) = 3.33 E.P 1.67e 6:34t 1.67e (b) i2 (t) = 3.33 + 1.22e 6:34t 4.55e u (t) 23:66t 23:66t u (t) 5.13 Find the Laplace transform of the waveform shown in Fig. E.P. 5.13. Ans: E.P F (s) = A 1 e s s2 Figure E.P. 5.13 A e s 2s 5.14 Find the Laplace transform of the periodic waveform shown in Fig. E.P. 5.14. Ans: 1 F (s) = 2 s Figure E.P. 5.14 1 e s2 2s 2 e s 2s 1 1 e 2s j 441 442 j E.P Network Theory 5.15 Find the Laplace transform of the waveform shown in Fig. E.P. 5.15 Ans: F (s) = E.P Figure E.P. 5.15 1 2 2 2 + + 2e s 2 s s s 1 e s 5.16 Obtain the Laplace transform of the f (t) shown in Fig. E.P. 5.16. Ans: F (s) = E.P 1 5 s Figure E.P. 5.16 3e s + 3e 3s 5e 4s 5.17 Obtain the Laplace transform of the unit impulses shown in Fig. E.P. 5.17 Figure E.P. 5.17 Ans: X(s) = 1 1 e s Laplace Transform E.P j 443 5.18 Refer the circuit shown in Fig. E.P. 5.18. Let i(0) = 1A; vo (0) = 2V and vs (t) = 4e Find vo (t) for t > 0. 2t u(t)V. Figure E.P. 5.18 Ans: E.P vo (t) = 2 + 4.33e 0:5t + 1.33e 2t u (t) volts 5.19 Find i(t) in the circuit shown in Fig. E.P. 5.19. Assume that the circuit is initially relaxed. Figure E.P. 5.19 Ans: E.P i (t) = 0.5 0.5e 4t te 4t u (t) 5.20 Refer the circuit shown in Fig. E.P. 5.20. Assume zero initial conditions. Use convolution theorem to find i(t). Figure E.P. 5.20 Ans: i (t) = t e 2 5t (t 2) 2 e 5(t 2) 444 j E.P Network Theory 5.21 There is no energy stored in the circuit shown in Fig. E.P. 5.21 at the time when the switch is opened. Show that sI (s) g V2 (s) = R1 1 2 C1 s + s+ L1 L 1 C1 Figure E.P. 5.21 E.P 5.22 Refer the circuit shown in Fig. E.P. 5.22. If is (t) = 6u(t)mA, find v2 (t). Figure E.P. 5.22 Ans: v2 (t) = 10e E.P 4000t cos (3000t 90 ) u (t) V 5.23 Find Vo (s) and vo (t) in the circuit shown in Fig. E.P. 5.23 if the initial energy is zero and the switch is closed at t = 0 Figure E.P. 5.23 Ans: vo (t) = 30 60e 5000t + 30e 10000t u (t) Laplace Transform E.P 5.24 The initial energy in the circuit in Fig. E.P. 5.24 is zero. (a) Find Vo (s). (b) Use the initial and final value theorems to find vo (0+ ) and vo (1). (c) Do the values obtained in part (b) agree with known circuit behaviour? Explain. (d) Find vo (t). Figure E.P. 5.24 Ans: (a) Vo (s) = + 4200 21 2 s (s + 8s + 25) 103 s (b) vo 0+ = 0, vo () = 168V (c) YES (d) vo (t) = 168 + 7225.95e 4t cos (3t + 91.33 ) u (t) V E.P 5.25 Find the initial and final value of H (s) = Ans: 1, 2 E.P s3 + 25 + 6 s(s + 1)2 (s + 3) 5.26 Verify final value theorem and initial value theorem for the function, f (t) = 2 + e 3t cos 2t E.P 5.27 Using the convolution theorem, find the Laplace inverse of the following functions: 1 s 1 (i) F (s) = (iii) F (s) = (ii) F (s) = 2 s(s + 1) (s a) (s + 1)(s + 2) Ans: (i) f (t) = 1 e t (ii) f (t) = teat (iii) f (t) = e t + 2e 2t 2e t j 445 446 j E.P Network Theory 5.28 In the circuit shown in Fig. E.P. 5.28, find the voltage across the resistance vR (t) using convolution integral. Given that vg (t) = e 2t and RC = 1 second. Figure E.P. 5.28 Ans: vR (t) = 2e E.P 2t e t, t0 5.29 Find the inverse Laplace transform of the following functions: 3s s2 + 3 1 (i) 2 (iii) (ii) (s + 1)(s2 + 4) (s + 1)(s + 2)2 (s2 + 2s + 2)(s + 2) Ans: (i) cos t cos 2t (ii) e t e (iii) E.P 7 e 2 2t 2t (1 + t) 2.5e t cos t + 0.5e t sin t 5.30 In the circuit shown in Fig. E.P. 5.30, switch K is open for a long time so that steady state is reached and at t = 0, switch is closed. Determine the current i(t) in 10 ohm resistor. Figure E.P.5.30 Ans: Current in each 10 Ω resistor = 2u(t) e 5t Laplace Transform E.P j 447 5.31 Synthesize the wave form shown in Fig. E.P. 5.31 using ramp function and obtain the Laplace transform of f (t). Figure E.P.5.31 Ans: E.P F (s) = 5 [1 s2 2e s + e 2s ] 5.32 Find the Laplace transform of the voltage wave form as shown in Fig. E.P. 5.32. Figure E.P.5.32 Ans: E.P V (s) = 2 [1 s2 3e s + 5e 1:5s 6e 2s + 6e 3s ] 5.33 Find the Laplace transform of the perodic wave forms shown in Figs. E.P. 5.33(a) and (b). Figure E.P.5.33(a) 448 j Ans: Network Theory (i) F (s) = (ii) F (s) = E.P 1 1 1 e 1 e 1 4s s2 2 2s s Figure E.P.5.33(b) 2 1 e s + 2 e 2s 2 s s 4e s 2 2s + e s s 1 e s 2s 2 + 2e s 3s 1 e s2 4s 5.34 For the circuit shown in Fig. E.P. 5.34, find the current transients in both the loops using Laplace transformation method. Figure E.P.5.34 12 7 2 i2 (t) = + 7 Ans: i1 (t) = E.P 5 2t e 5t Ampere; t 0 e 7 5 7t e 5t Ampere; t 0 e 7 5.35 Find the Laplace transform of the saw tooth wave as shown in Fig. E.P. 5.35. Figure E.P. 5.35 Laplace Transform Ans: F (s) = E.P 5.36 V (1 j 449 e T s T se T s ) T s2 (1 e T s ) For the circuit shown in Fig. E.P. 5.36 switch K is closed at t = 0. Determine the current i(t) for t 0. Figure E.P. 5.36 Ans: E.P i(t) = 0.357e 2t 5 e j 25t 25 + j2 5 25 j2 e j 25t 5.37 For the circuit shown in Fig. E.P. 5.37, determine the source current when the switch K is closed at t = 0. Assume zero initial conditions. Figure E.P. 5.37 Ans: i(t) = 2.57e t 0.57e 0:3t Amperes; t 0 Chapter 6 6.1 Resonance Introduction A.C Circuits made up of resistors, inductors and capacitors are said to be resonant circuits when the current drawn from the supply is in phase with the impressed sinusoidal voltage. Then 1. the resultant reactance or susceptance is zero. 2. the circuit behaves as a resistive circuit. 3. the power factor is unity. A second order series resonant circuit consists of R, L and C in series. At resonance, voltages across C and L are equal and opposite and these voltages are many times greater than the applied voltage. They may present a dangerous shock hazard. A second order parallel resonant circuit consists of R; L and C in parallel. At resonance, currents in L and C are circulating currents and they are considerably larger than the input current. Unless proper consideration is given to the magnitude of these currents, they may become very large enough to destroy the circuit elements. Resonance is the phenomenon which finds its applications in communication circuits: The ability of a radio or Television receiver (1) to select a particular frequency or a narrow band of frequencies transmitted by broad casting stations or (2) to suppress a band of frequencies from other broad casting stations, is based on resonance. Thus resonance is desired in tuned circuits, design of filters, signal processing and control engineering. But it is to be avoided in other circuits. It is to be noted that if R = 0 in a series RLC circuit, the circuits acts as a short circuit at resonance and if R = in parallel RLC circuit, the circuit acts as an open circuit at resonance. 1 452 6.2 | Network Analysis Transfer Functions As ω is varied to achieve resonance, electrical quantities are expressed as functions of ω, normally denoted by F (jω) and are called transfer functions. Accordingly the following notations are used. V (jω) = Impedance function I(jω) I(jω) Y (jω) = = Admittance function V (jω) V2 (jω) G(jω) = = Voltage ratio transfer function V1 (jω) I2 (jω) α(jω) = = Current ratio transfer function I1 (jω) Z(jω) = If we put jω = s then the above quantities will be Z(s), Y (s), G(s), α(s) respectively. These are treated later in this book. 6.3 Series Resonance Fig. 6.1 represents a series resonant circuit. Resonance can be achieved by 1. varying frequency ω 2. varying the inductance L 3. varying the capacitance C Figure 6.1 Series Resonant Circuit The current in the circuit is I= E E = R + j(XL − XC ) R + jX At resonance, X is zero. If ω0 is the frequency at which resonance occurs, then 1 1 = resonant frequency. or ω0 = √ ω0 L = ω0 C LC V The current at resonance is Im = = maximum current. R The phasor diagram for this condition is shown in Fig. 6.2. The variation of current with frequency is shown in Fig. 6.3. 2 Figure 6.2 Figure 6.3 Resonance | 453 It is observed that there are two frequencies, one above and the other below the resonant frequency, ω0 at which current is same. 1 and |Z| with ω. Fig. 6.4 represents the variations of XL = ωL; XC = ωC 1 we see that any constant product of L and C give a particular From the equation ω0 = √ LC L resonant frequency even if the ratio is different. The frequency of a constant frequency source C can also be a resonant frequency for a number of L and C combinations. Fig. 6.5 shows how the L sharpness of tuning is affected by different ratios, but the product LC remaining constant. C 0 0 Figure 6.4 Figure 6.5 L For larger ratio, current varies more abruptly in the region of ω0 . Many applications call for C narrow band that pass the signal at one frequency and tend to reject signals at other frequencies. 6.4 Bandwidth, Quality Factor and Half Power Frequencies At resonance I = Im and the power dissipated is 2 Pm = Im R watts. Im When the current is I = √ power dissipated is 2 Pm I2 R = m watts. 2 2 From ω − I characteristic shown in Fig. 6.3, it is observed that there are two frequencies Im ω1 and ω2 at which the current is I = √ . As at these frequencies the power is only one half of 2 that at ω0 , these are called half power frequencies or cut off frequencies. The ratio, current at half power frequencies 1 Im =√ =√ Maximum current 2Im 2 1 When expressed in dB it is 20 log √ = −3dB. 2 454 Network Analysis j Therefore !1 and !2 are also called pIm2 = pE2R , the p2R = jR + j(XL XC )j As 3 dB frequencies. magnitude of the impedance at half-power frequencies is Therefore, the resultant reactance, X = XL XC = R. The frequency range between half - power frequencies is passband or band width. BW = !2 !1 = B: !2 !1 , and it is referred to as R The sharpness of tuning depends on the ratio , a small ratio indicating a high degree of L selectivity. The quality factor of a circuit can be expressed in terms of R and L of the inductor. Quality factor = Q = !0 L R Writing !0 = 2f0 and multiplying numerator and denominator by 2 2 1 1 Q = 2f0 12 ILIm2 R = 2 12ILI2mRT 2 m = 2 Selectivity is the reciprocal of Q. m Maximum energy stored total energy lost in a period Q = !R0 L and !0 L = ! 1C ; As 0 and since !0 = 6.5 2 1 pLC , we have Q = ! 1CR 0 rL Q= R C 1 Expressions for ω1 and ω2 , and Bandwidth At half power frequencies !1 and !2 , E I = pE = 2 2R fR + (XL XC )2g jXL XC j = R i:e:; !L !C1 = R 1 2 ∴ At ! = !2 , Simplifying, !22 LC !2 CR R = !2 L ! 1C 2 1=0 1 2 I , we get, 2 m Resonance Solving, we get !2 = p RC + R C 2 2LC 2 + 4LC R+ = 2L s R 2 2L + 1 LC j 455 (6.1) Note that only + sign is taken before the square root. This is done to ensure that !2 is always positive. At ! = !1 , R = 1 !1 L ) Solving; !1 C ! LC + !1 CR p1 = 0 2 2 !1 = RC + 2RLCC + 4LC s 2 R + 1 R + = 2L 2L LC 2 1 (6.2) While determining !1 , only positive value is considered. Subtracting equation(6.1) from equation (6.2), we get Since Q = and therefore !2 !1 = RL = Band width. !0 L , Band width is expressed as R B = !2 !1 = RL = !Q0 : Q = ! !0 ! = !B0 2 1 Multiplying equations (6.1) and (6.2), we get 2 1 !1 !2 = 4RL2 + LC !0 = p !1 !2 or R2 = 1 4L2 LC = !02 The resonance frequency is the geometric mean of half power frequencies. R << 1 ; in which case Q 5 Normally 2L LC 1 R+ 1 Then; !1 2RL + LC and !2 2L LC p ' = ∴ r R 2L + !0 and ' !2 = 2RL + !0 p !0 = !1 +2 !2 = Arithmetic mean of !1 and !2 R = !0 , Equations for ! and ! as given by equations (6.1) and (6.2) can be expressed 1 2 L Q in terms of Q as s 2 ! 0 !2 = 2Q + 2!Q0 + !02 Since 456 j Network Analysis = !0 Similarly !1 = !0 2 s 3 4 21Q + 1 + 21Q 5 2 s 3 4 21Q + 1 + 21Q 5 2 2 R << p 1 and then Q > 5. 2L LC Consequently !1 and !2 can be approximated as r1 R R !1 ' 2L + LC = 2L + !0 = B2 + !0 r1 R R !2 ' 2L + LC = + 2L + !0 = B2 + !0 so that ! = !1 + !2 : Normally, 0 6.6 2 Frequency Response of Voltage across L and C As frequency is varied, both the voltages across L and C increase with frequency upto !0 and they are equal at !0 : But their maximum values do not occur at !0 : Vc reaches its maximum at ! < !0 and VL reaches its maximum at ! > !0 . This can be verified by calculating the frequency at which each occurs. 6.7 Expression for ω at which VL is Maximum Current in the circuit shown in Figure 6.1 is I=q Voltage across L is VL = !LI = q Squaring VL2 = This is maximum when R2 + E !L dVL 2 = 0 d! 1 !C 2 E!L R2 + !L E 2 ! 2 L2 R2 + !L !C1 2 1 !C 2 Resonance "( That is, EL 2 2 or ) 2 + 1 2 = LC 1 1 R2 C 2L 1 Let this frequency be !L . Then; !L2 = !02 1 !L = !0 That is, !L 6.8 2! ! 2 !L L + !2C !C !C !C 1 2 1 1 2 R + !L !C = !L !C !L + !C 1 L 2 L#2 1 2 R2 + !2#L# + 2 2 2 = !# 2 !C C ! C2 2 2 2 2 R ! C + 1 2! LC = 1 !2 (2LC R2 C 2 ) = 2 !2 = 2LC 2 R2 C 2 R 2 1 # =0 1 1 2Q2 s 1 1 > !0 . 1 2Q2 : Expression for ω at which VC is Maximum E !C R2 + !2 L 2 VC2 = 2 2 2 E ! C R + !L q VC = Now This is maximum when That is, " ( E 2 !2 C2 2 R 2 + d 2 d! (VC ) = 0: !L !C 1 !L !C 1 L !2C 1 2 = + 2! ( !L !C 1 1 1 !C !C 2 2 R2 + !L !C 1 !L + !C 1 )# 2 =0 j 457 458 j Network Analysis 1# L 1# 2 2 R 2 + ! 2 L2 + ! # 2 = # 2C 2 2 C ! C2 ! L L 2! 2 L2 + R2 = 2 C L 2 1 R2 !2 = 2 C2L2R = LC 2L2 2 1 R C = !2 1 = 1 0 LC 2 L Let this frequency be !C 1 2Q2 r !C = !0 1 2Q1 2 i:e:; ! C < !0 Variations of VC and VL as functions of ! are shown in Fig. 6.6. We know that Figure 6.6 VC = r E n !C R 2 2 Consider ! 2 C 2 R2 + (! 2 LC !C C R 2 2 2 + (! 2 LC 1)2 !2 C 2 2 2 2 E + (! 2 LC Q2 1 2 LC = !0 = 2 2 0 and 2Q2 2 2 + !0 CR = Q1 2 g (6.3) 2 4Q4 = Q2 1 1 2 0 2Q4 + 4Q4 LC 2Q2 2 0 2 2 0 2 1 1)2 1)2 and at ! = !C . Then equation(6.3) becomes 1 + (!C LC 1) = ! 1 CR + ! 1 2Q 1 11 1 1 ! 1 2Q = Q 1 2Q + ! 1 1 1 1 1 1 2 2 since 2 o = p fR ! C = Q2 1 2 1 1 4Q2 Substituting the above expression in the denominator of equation (6.3), we get Vcm = q EQ 1 1 6.9 4Q2 Selectivity with Variable L In a series resonant circuit connected to a constant voltage, with a constant frequency, when L is varied to achieve resonance, the following conditions prevail: Resonance j 459 XC is constant and I = q E when L = 0. R2 + XC2 V at X = X 2. With increase in L; XL increases and Im = R L C 3. With further increase in L; I proceeds to fall. All these conditions are depicted in Fig. 6.7 VC max occurs at !0 but VL max occurs at a point beyond !0 . L at which VL becomes a maximum is obtained in terms of 1. other constants. EXL fR2 + (XL XC )2g 2 2 VL2 = R2 + (EX XL X )2 L C 2 dVL = 0. This is maximum when dXL VL = R Therefore; 2 + (XL 1 2 Figure 6.7 XC )2 2XL = XL2 f2(XL XC )g R2 + XL2 + XC2 2XL XC = XL2 XL = R X+ XC C 2 Therefore; Let the corresponding value of L is Lm . Then; and L0 = value of L at !0 such that XL XC 2 Lm = C (R2 + XC2 ) !0 L = ! 1C : 0 6.10 Selectivity with Variable C In a series resonant circuit connected to a constant voltage, constant frequency supply, if varied to achive resonance, the following conditions prevail: 1. XL is constant. 2. XC varies as inversely as C when C = 0, 3. C is I = 0. V , I = Im = . when !C = !L R with further increase in C; I starts decreasing as shown in Fig. 6.8, where Cm is the value of capacitance at maximum voltage across C and C0 is the value of the capacitance at !0 . 1 460 j Network Analysis C at which VC becomes maximum can be determined in terms of other circuit constants as follows. EXC R2 + (XL XC )2 2 2 VC2 = R2 + (EX XC X )2 L C VC = p Figure 6.8 dVC2 = 0 dXC R2 + (XL XC )2 2XC X 2 f2(XL XC )( Then; C R2 + XL2 +XC2 2XL XC = XL XC +XC2 2 2 XC = R X+ XL For maximim VC ; g 1) = 0 L Let the correrponding value of C be Cm . Cm = R2 +L X 2 : Then; L 6.11 Transfer Functions 6.11.1 Voltage ratio transfer function of a series resonant circuit and frequency response For the circuit shown in Fig. 6.9, we can write j!) = R H (j!) = VV0((j! ) R + j !L s !CR n! L 1 ! o 1 = 1 + j !L R = 1+j = = 1 1 ! R! h 0 0 1 + jQ !!0 !!0 CR !0 ! i 1 1+Q 2 h! !0 Figure 6.9 0 1 !C !0 ! i 2 , 1 2 tan 1 Q ! !!0 0 ! Resonance Let be a measure of the deviation in ! from !0 . It is defined as = ! ! !0 = !! 0 0 1 ! !0 = ( + 1) 1 = ( + 1)2 !0 ! +1 +1 For small deviations from !0 ; << 1: Then, ! !0 !0 ! ' 2 1 Then Then; H (j!) = 1 + j12Q = p 1 1 + 4Q2 2 = tan 1 2 + 2 +1 2Q The amplitfude and phase response curves are as shown in Fig. 6.10. Figure 6.10 (a) and (b): Amplitude and Phase response of a series resonance circuit 6.11.2 Impedance function The Impedance as a function of j! is given by Z (j!) = R + j !L !C !L 1 =R 1+j R! !CR ! 0 = R 1 + jQ !0 ! s ! !0 ! ! 0 2 , 1 2 tan Q =R 1+Q !0 ! !0 ! 1 For small deviations from !0 , we can write Z (j!) ' R[1 + j 2Q ] = R p1 + 4Q tan 2 2 1 2Q j 461 462 j 6.12 Network Analysis Parallel Resonance The dual of a series resonant circuit is often considered as a parallel resonant circuit and it is as shown in Fig. 6.11. The phasor diagram for resonance is shown in Fig. 6.12. The admittance as seen by the current source is Y (j!) = YR + YL + YC 1 1 = + j !C R !L = G + jB Figure 6.11 Parallel Resonance Circuit Figure 6.12 Phasor Diagram If the resonance occurs at !0 ; then the susceptance B is zero. That is, !0 C = !1L 0 or !0 = p 1 rad= sec : LC At resonance, and IC 0 = IL0 = j!0 CRI ILC = IC 0 + IL0 = 0 The quality factor, as in the case of series resonant circuit is defined as Maximum energy stored Q = 2 Energy dissipated in a period CV 2 1 2 R T = 2f0 CR = !0 CR: !0 C = !1L ; 0 R Q = ! L: 0 1 = 2 2 V 2 m m Since Resonance On either side of !0 there are two frequencies at which the voltage is same. At resonance, the voltage is maximum and is given by Vm = IR and is evident from the response curve as shown in Fig. 6.13. At this frequency, p pm V2 = m watts. The frequencies at which the R 1 times the maximum voltage are called half voltage is 2 power frequencies or cut off frequencies, since at these frequencies, = p Vp m 2 p = R2 At any ! , V = m = half of the maximum power. 2R Y = 1 + j !C 2 At !1 and !2 ; R jY j = p 1 = 2R s 1 2 R Squaring, Therefore; At ! = !2 , Hence; Figure 6.13 1 2R2 1 !C !L = = 1 R2 + + 1 !L !C !L 1 !C !L 1 2 2 1 R !2 C !1L = R1 2 2 !2 LC 1 = !R2 L !22 LCR R !2 L = 0 p 2 + 4LCR2 !2 = L + L2LCR Note that only positive sign is used before the square root to ensure that !2 is positive. Thus; Similarly; So that, bandwidth !2 = 2RC + 1 !1 = 1 2RC s + 1 2 s2RC 1 2RC 1 B = !2 !1 = RC + 1 LC 2 + 1 LC j 463 464 j Network Analysis !1 !2 = 1 2 2RC 1 = = !02 and Thus; !0 = p 1 LC As LC + 1 2RC = and 2 rC p R Q = LC = R 0 2 L !2 = B2 + s B +! s2 !1 = B2 + B = !Q0 , !2 = !0 !1 = !0 and 6.13 2RC Q = !0 RC = !RL B and Using 1 LC ! 0 = p !1 !2 L Since 1 2 B 2 2 2 0 + !02 2 s 3 4 21Q + 1 + 21Q 5 2 s 3 4 21Q + 1 + 21Q 5 2 2 Transfer Function and Frequency Response The transfer function for a parallel RLC circuit shown in Fig. 6.14. is H (j! ), the current ratio transfer function. ) 1 = H (j!) = II0 ((j! j!) RY (j!) 1 = = 1 1 R R + j !C 1 1+j 1 !! CR 1 = 1 + jR !C !L !L 1! ! = !R 1 1 0 0 1 + jQ !0 !0 !0 !L As in the case of series resonance, here also let 0 ! = ! ! !0 = !! 0 0 1 Figure 6.14 Parallel RLC Circuit Resonance j 465 then, ! !0 = 2 + 2 !0 ! + 1 For << 1, for small deviations from !0 ! !0 ' 2 !0 ! Therefore, H (j!) = 1 + j12Q 6.14 Resonance in a Two Branch RL RC Parallel Circuit Consider the two branch parallel circuit shown in Fig. 6.15. Let E be the voltage across each of the parallel circuit shown in the figure. The vector diagram at resonance is shown in Figure 6.1. Figure 6.15 Two branch Parallel Circuit Figure 6.16 The admittance of the circuit is Y (j! ) = GL For resonance, jBL + GC + jBC BL = BC If this occurs at ! = !0 , then 1 !0 L !C = RL2 + !02 L2 RC2 + ! 1C !0 C = 2 2 2 RC !0 C + 1 L(1 + !02 C 2 RC2 ) = C (RL2 + !02 L2 ) !02 (LC 2 RC2 L2 C ) = RL2 C L 0 2 0 2 466 | Network Analysis ω02 = 2 C −1 RL 2 − L2 C LC 2 RC 2 − L 2 C −L 1 RL 1 RL C = 2 C −L 2 − L LC (RC LC (RC C 2 − L RL 1 C ω0 = √ 2 − L LC RC C = Therefore, This is the expression for resonant frequency. It is to be noted that 1. resonance is not possible for certain combination of circuit elements unlike in a series circuit where resonance is always possible. 2. resonance is also possible by varying of RL or RC . Consider the case where 2 RC < or L 2 < RL C L 2 < RC C In both these cases, the quantity under radical is negative and therefore resonance is not possible. The admittance at resonance of the above parallel circuit is " ! RC RL S Y0 = 2 + X 2 + R2 + X 2 RL L0 C C0 2 RL < where XL0 and XC0 are the inductive and capacitive reactances respectively at resonance. L If RL = RC = C 1 then ω0 = √ LC as in R, L, C series circuit. If RL = RC = L C which means 2 2 RL = RC = R2 = Then, L = XL XC . C XC XL − 2 2 + XL RC + XC2 1 1 = − =0 XL + XC XL + XC BL − BC = 2 RL Resonance j 467 In this case, the circuit acts as a pure resistive circuit irrespective of frequency. That is, the circuit is resonant for all frequencies. In this case the circuit admittance is Y RL + RC RL + XL2 RC2 + XC2 R2 + X 2 + R2 + X 2 L C =R R4 + R2 (XL2 + XC2 ) + XL2 XC2 2R2 + XL2 + XC2 =R 4 2R + R2 (XL2 + XC2 ) = 2 2R2 + X 2 + X 2 R L C = 2 R 2R2 + XL2 + XC2 rC 1 = R= L rL Z =R= C or 6.14.1 Resonance by varying inductance If resonance is achieved by varying only L in the circuit shown in Figure 6.15 but with constant current constant frequency source, then the condition for resonance is BL = BC ) R2 X+LX 2 = R2 X+CX 2 = XZ 2C where ZC2 = RC2 + XC2 L L C C C 2 2 2 Then; XL XC XL ZCq+ XC RL = 0 ZC2 ZC4 4XC2 RL2 Solving, forXL we get XL = 2XC q 4 2 2 L C 2 Therefore; L = 2 ZC ZC 4XC RL since XL XC = C The following conditions arise: 1. If ZC4 > 4XC2 RL2 ; L has two values for the circuit to resonate. 2. For ZC4 = 4XC2 3. RL2 , L = 12 CZC2 for reasonance. 2 , No value of L makes the circuit to resonate. For ZC4 < 4XC2 RL 468 j 6.14.2 Network Analysis Resonance by varying capacitance As in the previous case, we have at resonance„ BL = BC XC XL 2 2 2 RC2 + XC2 = ZL2 ; where ZL = RL + XL ) Simplifying we get, XC2 XL XC ZL2q+ RC2 XL = 0 ZL2 ZL4 4XL2 RC2 XC = 2XL 2L q C= ZL2 ZL4 4XL2 RC2 Therefore; The following conditions arise: 1. For ZL4 2. 3. 6.14.3 > 4XL2 RC2 , there are two values for C to resonante. 2 , resonance occurs at C = 2L . For Z44 = 4XL2 RC ZL2 2 , no value of C makes the circuit to resonate. For ZL4 < 4XL2 RC Resonance by varying RL or RC It is often possible to adjust a two branch parallel combination to resonate by varying either RL or RC . This is because, when the supply is of constant current and, constant frequency, these resistors control inphase and quadratare components of the currents in the two parallel paths. From the condition BL = BC , we get XL = XC RL + XL2 RC2 + XC2 XL R 2 + X X X 2 RL2 = X L C C L rCXL RL = X RC2 + XL XC XL2 (6.4) C This equation gives the value of RL for resonance when all other quantities are constant and 2 the term under radical is positive. Similarly if only RC is variable, keeping all other quantities constant, the value of resonanace is given by r XC RC = X RL2 + XL XC XC2 L provided the term under radical is positive. RC for Resonance 6.15 | 469 Practical Parallel and Series Resonant Circuits A practical resonant parallel circuit contains an inductive coil of resistance R and inductance L in parallel with a capacitor C as shown in Fig. 6.17. It is called a tank circuit because it stores energy in the magnetic field of the coil and in the electric field of the capacitor. Note that resistance RC of the capacitor is negligibly small. Condition for parallel resonance is shown by the phasor diagram of Fig. 6.18. IC = IL sin φ That is, ⇒ BC = BL ωL = ωC. R2 + ω 2 L2 Let the value of ω which satisfy this condition be ω0 . Then, R2 + ω02 L2 = L C 1 L 1 = − R2 C L2 LC 1 R2 C ω0 = √ 1− L LC ω02 = Figure 6.17 1− R2 C L (6.5) (6.6) Figure 6.18 Admittance of the circuit shown in figure 6.17 is 1 + jωC R + jωL R jωL = 2 − + jωC R + ω 2 L2 R2 + ω 2 L2 Y (jω) = At ω = ω0 , Y (jω) is purely real. Hence, Y (jω0 ) = R2 R + ω02 L2 (6.7) 470 | Network Analysis Substituting for ω0 in equation (6.7), Y (jω0 ) = R2 R R RC = L = 2 2 L + ω0 L C (6.8) L , which is called the dynamic resistance of CR R2 C > 1, there is the circuit. This is greater than R if there is resonance. However, note that if L no resonance. Fig. 6.19 shows a practical series resonant circuit. The input impedance as a function of ω is and the circuit is a pure resistive with R0 = Z(jω) = jωL + G2 ωC G −j 2 2 2 +ω C G + ω2C 2 Condition for resonance is ωC ωL = 2 G + ω 2 C2 1 C 1 1 ω2 = = − G2 − 2 2 2 L C LC C R L 1 1− = LC CR2 L 1 1− ω=√ CR2 LC Impedance at resonance is Z0 = G2 Figure 6.19 G L G = C = 2 + ωC CR L L The circuit at resonance is a purely resistive, and Z0 = R0 = . However, note that here CR L also resonance is not possible for > 1. CR2 In both the circuits, shown in Figs 6.18 and 6.19, resonance is achieved by varying either C or L until the input impedance or admittance is real and this process is called tuning. For this reason these circuits are called tuned circuits. Series circuits EXAMPLE 6.1 Two coils, one of R1 = 0.51 Ω, L1 = 32 mH, the other of R2 = 1.3 Ω and L2 = 15 mH and two capacitors of 25 μF and 62 μF are all in series with a resistance of 0.24 Ω. Determine the following for this circuit (i) Resonance frequency (ii) Q of each coil Resonance (iii) Q of the circuit (iv) Cut off frequencies (v) Power dissipated at resonance if E = 10 V. SOLUTION From the given values, we find that Rs = 0.51 + 1.3 + 0.24 = 2.05Ω Ls = 32 + 15 = 47 mH 25 × 62 Cs = μF = 17.816 μF 87 (i) Resonant frequency: ω0 = √ 1 Ls Cs 1 47 × × 17.816 × 10−6 = 1092.8 rad/ sec =√ (ii) Q of coils: For Coil 1, For Coil 2, (iii) Q of the circuit: 10−3 ω0 L1 R1 1092.8 × 32 × 10−3 = = 68.57 0.51 ω0 L2 Q2 = R2 1092.8 × 15 × 10−3 = = 12.6 1.3 Q1 = ω0 Ls Rs 1092.8 × 47 × 10−3 = = 25 2.05 Q= (iv) Cut off frequencies: Band width is, 1092.8 ω0 = = 43.72 B= Q 25 Considering Q > 5, the cut off frequencies, B = 1092.8 ± 21.856 ω2,1 = ω0 ± 2 Therefore, ω2 = 1115 rad/ sec and ω1 = 1071 rad/ sec . | 471 472 j Network Analysis (v) Power dissipated at resonance: Given E = 10 V We know that at resonance, only the resistance portion will come in to effect. Therefore P = ER 2 EXAMPLE = 102 = 48:78 W 2:05 6.2 For the circuit shown in Fig. 6.20, find the out put voltages at (i) (ii) (iii) ! = !0 ! = !1 ! = !2 when vs (t) = 800 cos !t mV. Figure 6.20 SOLUTION For the circuit, using the values given, we can find that resonant frequency !0 = 1 pLC 1 =p 312 10 1:25 10 3 Quality factor: Band width: As Q > 5, Hence; and Q = !R0 L 1:6 106 312 10 = 62:5 103 12 = 1:6 3 =8 B = !Q0 1:6 106 = = 0:2 106 rad= sec 8 !2;1 = !0 B2 = (1:6 0:1)106 rad= sec !2 = 1:7 106 rad= sec !1 = 1:5 106 rad= sec 10 6 rad= sec Resonance | 473 (i) Output voltage at ω0 : Using the relationship of transfer function, we get 50Im 62.5Im = 0.8 0◦ H(jω)|ω=ω0 = Since the current is maximum at resonance and is same in both resistors, vo (t) = 0.8 × 800 cos(1.6 × 106 t) mV = 640 cos(1.6 × 106 t) mV At ω1 and ω2 , Zin = √ 2Rs ±45◦ . Therefore, Rout 50 =√ 45◦ Zin 2 × 62.5 = 0.5657 45◦ H(jω)|ω=ω1 = H(jω)|ω=ω2 = 0.5657 45◦ and (ii) Out put voltage at ω = ω1 vo (t) = 0.5657 × 800 cos(1.6 × 106 t + 45◦ ) mv = 452.55 cos(1.6 × 106 t + 45◦ ) mV (iii) Out put voltage at ω = ω2 vo (t) = 452.55 cos(1.6 × 106 t − 45◦ ) mV EXAMPLE 6.3 In a series circuit R = 6 Ω, ω0 = 4.1 × 106 rad/sec, band width = 105 rad/sec. Compute L, C, half power frequencies and Q. SOLUTION We know that Quality factor, Q= Also, Q= Therefore, L= and Hence, Q= C= = ω0 4.1 × 106 = 41 = B 105 ω0 L R QR 41 × 6 = = 60 μH ω0 4.1 × 106 1 ω0 CR 1 ω0 QR 1 = 991.5 pF 6 4.1 × 10 × 41 × 6 | Network Analysis 474 As Q > 5, B 105 = 4.1 × 106 ± 2 2 6 ω2 = 4.15 × 10 rad/ sec ω2,1 = ω0 ± That is, ω1 = 4.05 × 106 rad/ sec and EXAMPLE 6.4 In a series resonant circuit, the current is maximum when C = 500 pF and frequency is 1 MHz. If C is changed to 600 pF, the current decreases by 50%. Find the resistance, inductance and quality factor. SOLUTION Case 1 Given, C = 500 pF I = Im f = 1 × 106 Hz ω0 = 2π × 106 rad/ sec ⇒ We know that ω0 = √ 1 LC Therefore, Inductance, 1012 1 = (2π × 106 )2 × 500 ω02 C = 0.0507 mH L= Case 2 When C = 600 pF, Im E = ⇒ |Z| = 2R 2 2R √ R2 + X 2 = 2R ⇒ X = 3R I= X = XL − XC = 2π × 106 × 0.0507 × 10−3 − = 318.56 − 265.26 √ = 53.3 Ω = 3R 1012 2π × 106 × 600 Resonance Therefore resistance, j 475 p:3 = 30:77Ω R = 53 3 Quality factor, EXAMPLE :56 = 10:35 Q = !R0 L = 318 30:77 6.5 In a series circuit with R = 50 Ω, L = 0.05 H and C = 20 F, frequency is varied till the voltage across C is maximum. If the applied voltage is 100 V, find the maximum voltage across the capacitor and the frequency at which it occurs. Repeat the problem for R = 10 Ω. SOLUTION Case 1 Given R = 50 Ω, We know that L = 0.05 H, C = 20 F !0 = p 1 = p 10 = 103 rad/sec 0:05 20 LC 3 0:05 10 L ! Q = R0 = 20 = 1 3 Using the given value of E = 100 V in the relationship VCm = q QE 1 1 we get 4Q2 VCm = q100 1 = 115:5 V r !C = ! 1 r1 1 1 4 and the corresponding frequency at this voltage is 0 = 103 2 2Q2 = 707 rad= sec Case 2 When R = 10 Ω, 0:05 = 5 10 5 100 = 502:5 V VCm = q 1 r 1 3 Q = 10 1 4 25 !C = 103 1 50 = 990 rad= sec 476 j Network Analysis EXAMPLE 6.6 (i) A series resonant circuit is tuned to 1 MHz. The quality factor of the coil is 100. What is the ratio of current at a frequency 20 kHz below resonance to the maximum current? (ii) Find the frequency above resonance when the current is reduced to 90% of the maximum current. SOLUTION (i) Let !a be the frequency 20 kHz below the resonance, Ia be the current and Za be the impedance at this frequency. Then !a !0 20 103 = 980 kHz 103 980 40:408 10 3 = 2 = Now the ratio of current, !a = 106 !0 = 980 !a 103 Ia = R = 1 Im Za 1 + j (2)Q 1 1 j 100(40:408 1 = 1 j 4:0408 = 0:2402 /76 = 10 (ii) Let !b be the frequency at which Ib = 0:9Im Then or where Then; or and We know that Hence Ib I = 1 + j1(2)Q = 0:9 m p1 + x 1 0:9 x = (2)100 1 1 + x2 = = 1:2346 0:81 x2 = 0:2346 2 = x = 0:4843 0:4843 200 0 0:4843 !b = 1 + 200 !0 = 1:00242 MHz = !!b 1= 3) Resonance EXAMPLE 6.7 For the circuit shown in Fig. 6.21, obtain the values of ω0 and vC at ω0 . Figure 6.21 SOLUTION For the series circuit, ω0 = √ 1 LC = 1 4× 1 4 × 10−6 = 103 rad/sec At this ω0 , I = Im . Therefore, V1 = 125Im and the circuit equation is 1.5 = V1 + (Im − 0 · 105V1 )10 + jVL − jVC Since VL = VC , the above equation can be modified as 1.5 = 125Im + 10Im − 1.05 × 125Im 1.5 A 3.75 1.5 4 × 106 Vc = × 3.75 103 = 1600 V Hence, Im = and EXAMPLE 6.8 For the circuit shown in Fig. 6.22(a), obtain Zin and then find ω0 and Q. Figure 6.22(a) | 477 478 j Network Analysis SOLUTION Taking I as the input current, we get VR = 10I and the controlled current source, 0:3VR = 0:3 = 3I 10I The input impedance can be obtained using the standard formula V voltage = Zin (j!) = Applied Input current I (6.6) For futher analysis, the circuit is redrawn as shown in Fig. 6.22(b). It may be noted that the controlled current source is transformed to its equivalent voltage source. Figure 6.22(b) Referring Fig. 6.22(b), the circuit equation may be obtained as V = 10 + j 10 3 9 ! j3010! 30! Substituting equation (6.7) in equation (6.6), we get Zin = 10 + j 10 3 ! 4 For resonance, Zin should be purely real. This gives 10 Rearranging, 3 10 9 10 Ω 9 30! ! = 4 3010 ! 9 !2 = 304 10103 = 0:133 1012 9 j3 I (6.7) Resonance Solving we get j 479 p0:133 10 = 365 10 rad= sec ! = !0 = Quality factor 12 3 Q = !R0 L = 365 = 36:5 10 10 3 3 10 Parallel circuits EXAMPLE 6.9 For the circuit shown in Fig. 6.23(a), find !0 , Q, BW and half power frequencies and the out put voltage V at !0 . Figure 6.23(a) SOLUTION Transforming the voltage source into current source, the circuit in Fig. 6.28(a) can be redrawn as in Fig. 6.23(b). Then; !0 = p 1 LC p400109 100 = 5 10 rad= sec = 6 Q = !0 CR 10 100 10 100 10 ! B = = 5 10 = 10 rad= sec =5 6 12 6 0 Q 50 5 3 = 50 Figure 6.23(b) 480 j Network Analysis As Q > 10, !2;1 = B2 + !0 =5 10 102 6 5 !2 = 5:05 M rad= sec Hence; Output voltage, and V = I 80 kΩ 10 3 80 103 j 5 106 400 10 = 0:04 90 V = EXAMPLE !1 = 4:95 M rad= sec 6 6.10 In a parallel RLC circuit, C = 50 F. Determine BW, Q, R and L for the following cases. (i) (ii) !0 = 100; !2 = 120 !0 = 100; !1 = 80 SOLUTION (i) !0 = 100; !2 = 120 We know that Rearraging we get !0 = p !1 !2 2 !1 = !!0 2 1002 = = 83:33 rad=sec 120 Band width B = !2 !1 = 120 Quality factor, Q = !B0 = We know that 83:33 = 36:67 rad= sec 100 = 2:73 36:67 Q = !RL = !0 RC 0 (6.8) Resonance Rearraging equation (6.8), j 481 R = !QC 0 2:73 106 = 546 Ω = 100 1 L = !2C Similarly 50 0 = (ii) 106 =2H 1002 50 !0 = 100; !1 = 80: Solving the same way as in case (i), we get 2 = 125 !2 = 100 80 BW = B = 125 80 = 45 rad= sec Q = 100 = 2:22 45 EXAMPLE 6.11 In the circuit shown in Fig. 6.24(a), vs (t) = 100 cos !t volts. Find resonance frequency, quality factor and obtain i1 ; i2 ; i3 . What is the average power loss in 10 kΩ. What is the maximum stored energy in the inductors? 10k 40k Figure 6.24(a) SOLUTION The circuit in Fig. 6.24(a) is redrawn by replacing its voltage source by equivalent current source as shown in Fig. 6.24(b). Resonance frequency, !0 = p 1 LC = p50 10 1 1:25 10 3 = 4000 rad= sec Quality factor, 6 Figure 6.24(b) Q = !0 C Req = 4000 = 40 1:25 10 8 10 6 3 482 j Network Analysis At resonance, the current source will branch into resistors only. Hence, v(t) jj = (10kΩ 40kΩ) vs (t) 10000 = 80 cos 4000t volts i1 (t) lags v(t) by 90 . Therefore, i1 (t) = 50 10803 4000 sin 40000t = 400 sin 4000t mA i2 (t) = 40 801000 cos 4000t = 2 cos 4000t mA i3 (t) = i1 (t) = 400 sin 4000t mA Average power in 10 kΩ: p 802 2 Pav = 10 103 = 0:32 W Maximum stored energy in the inductance: E = 12 LIm2 1 50 10 = 2 = 4 mJ EXAMPLE 6.12 For the network shown in Fig. 6.25(a), obtain frequency and quality factor. 3 (400 10 3 2 ) Yin and then use it to determine the resonance Figure 6.25(a) Figure 6.25(b) SOLUTION Considering V as the input voltage and I as the input current, it can be found that 10kΩ IR = V ) 10 4 IR = V Resonance j 483 The circuit in Fig. 6.25(a) is redrawn by replacing the controlled voltage source in to its equivalent current source by taking s = j! and is shown in Fig. 6.25(b). Referring Fig. 6.25(b), I 10V 1 1 = V sC + + R1 sL 11 sL ) I = V sC + R + sL Input admittance, with s is being replaced by j! is 103 Yin = VI = 1014 + j!1 10 8 j 11 ! 4:4 j 2500 = 10 4 + j! 10 8 ! At resonance, Yin should be purely real. This enforces that 10 Therefore; ! = 2500 ! p !0 = 108 2500 = 500 K rad= sec 8 Quality factor: Q = !0 RC = 500 10 10 10 3 4 8 = 50 EXAMPLE 6.13 In a parallel RLC circuit, cut off frequencies are 103 and 118 rad/sec. 10 Ω. Find R, L and C . jZ j at ! = 105 rad/sec is SOLUTION Given !1 = 103 rad= sec !2 = 118 rad= sec Therefore Resonant frequency, B = 118 103 = 15 rad= sec !0 = p p !1 !2 = 118 103 = 110:245 rad= sec 484 | Network Analysis Quality factor ω0 B 110.245 = 7.35 = 15 Q= Admittance, Since we get Note that, Therefore, 1 1 Y = + j ωC − R ωL R 1 1 + j ωCR − = R ωL ω0 ωCR Rω0 1 1+j − = R ω0 ωω0 L R Q = ω0 RC = , ω0 L ω ω0 1 1 + jQ − Y = R ω0 ω ω ω0 105 110.245 − = − = −0.0975 ω0 ω 110.245 105 1 (1 + j7.35(−0.0975)) R 1 = (1 − j0.7168) R 1.23 1 1 + (0.7168)2 = ⇒ |Y | = R R Y = It is given that |Z| = 10 and therefore |Y | = get 1 1 = 1.23 10 R 1 . Putting this value of Y in equation (6.9), we 10 ⇒ R = 12.3Ω From the relationship Q = ω0 CR, we get Therefore, (6.12) ω0 CR = 7.35 7.35 1 C= × 12.3 110.245 = 5.42 μF Resonance Inductance, j 485 L = !21C 0 110: = 15:18 mH = EXAMPLE 2452 1 5:42 10 3 6.14 For the circuit shown in Fig. 6.26(a), find !0 , V1 at !0 , and V1 at a frequency 15 k rad/sec above !0 . Figure 6.26(a) SOLUTION Changing voltage source of Fig. 6.26(a) into its equivalent current source, the circuit is redrawn as shown in Fig. 6.26(b). Referring Fig. 6.26(b), k !0 = 1 pLC 1 =p 100 10 10 10 = 10 rad=sec 6 9 Figure 6.26(b) 6 Voltage across the inductor at !0 is, V1 = j 106 3 10 5 10 9 3 = j 15 V Quality factor, Q = !0 CR = 106 10 10 5 10 9 = 50 Given !a = !0 + 15 k rad/sec 10 + 10 = 1:015 10 rad=sec = 15 Now; !a !0 = 1:015 !0 !a 3 6 6 1 = 0:03 1:015 3 486 | Network Analysis Using this relation in the equation, ωa ω0 1 1 + jQ Y = − R ω0 ωa 1 Y = (1 + j50 × 0.03) 5000 = 3.6 × 10−4 56.31◦ we get The corresponding value of V1 is V1 = I Y −1 = jωa × 3 × 10−9 × Y −1 = j 1.015 × 106 × 3 × 10−9 3.6 × 10−4 56.31◦ = 8.444 33.69◦ V EXAMPLE 6.15 A parallel RLC circuit has a quality factor of 100 at unity power factor and operates at 1 kHz and dissipates 1 Watt when driven by 1 A at 1 kHz. Find Bandwidth and the numerical values of R, L and C. SOLUTION Given f = 1 kHz, P = 1 W, I = 1 A, Q = 100, cos φ = 1 B= 103 × 2π ω0 = = 20π rad/sec Q 100 P = I 2R Therefore R=1Ω R L= ω0 Q 1 20π × 100 = 159 μH 1 C= 2 ω0 L = = 10 (20π)2 159 = 16.9 μF Resonance EXAMPLE 6.16 For the circuit shown in Fig. 6.27, determine resonance frequency and the input impedance. Figure 6.27 SOLUTION Equation for resonance frequency is # ! " $ 2 − L $ 1 RL C % ωL = 2 − L LC RC C = 1 0.1 × 10−3 = 98.47 rad/sec We know that X L = ω0 L = 98.47 × 0.1 and = 9.847 Ω 1 XC = ω0 C 1 = 98.47 × 10−3 = 10.16 Ω 22 − 100 1 − 100 | 487 488 | Network Analysis Admittance Y at resonance is purely real and is given by Y = G1 + G2 + G3 1 2 1 = 2 2 + 5 + 3 2 + (0.1ω0 ) 1 + 10 ω0 2 1 1 + + 2 + 9.847 5 1 + 10.162 = 0.23 S Y = 22 and the input impedance, Z= EXAMPLE 1 = 4.35 Ω Y 6.17 The impedance of a parallel RLC circuit as a function of ω is depicted in the diagram shown in Fig. 6.28. Determine R, L and C of the circuit. What are the new values of ω0 and bandwidth if C is increased by 4 times? 0.4 Figure 6.28 SOLUTION It can be seen from the figure that ω0 = 10 rad/sec B = 0.4 rad/sec R = 10 Ω Then Quality factor Q= 10 ω0 = = 25 BW 0.4 L= 10 R = = 0.04 H ω0 Q 10 × 25 We know that Resonance j 489 As Q = !0 CR, C = 10 25 10 = 0:25 F If C is increased by 4 times, the new value of C is 1 Farad. Therefore, !0 = p 1 LC = p01:04 = 5 and the corresponding bandwith 1 B = RC EXAMPLE = 0:1 6.18 In a two branch RL RC parallel resonant circuit, L = 0:4 H and C = 40 F. Obtain resonant frequency for the following values of RL and RC . (i) (ii) (iii) (iv) (v) RL = 120; RC = 80 RL = RC = 80 RL = 80; RC = 0 RL = RC = 100 RL = RC = 120 SOLUTION As RL and RC are given separately, we can use the following formula to calculate the resonant frequency. !0 v u 1 u t =p LC RL2 RC2 L C L C Let us compute the following values LC = 0:4 40 10 = 16 10 6 6 1 pLC = 250 (i) RL = 120; RC = 80 Using equation (6.10), L = 104 C r 120 !0 = 250 2 502 104 104 ! (6.10) 490 | Network Analysis As the result is an imaginary number resonance is not possible in this case. (ii) RL = RC = 80 802 − 104 802 − 104 = 250 rad/sec ω0 = 250 (iii) RL = 80; RC = 0 802 − 104 −104 = 150 rad/sec ω0 = 250 (iv) RL = RC = 100 1002 − 104 1002 − 104 As the result is indeterminate, the circuit resonates at all frequencies. ω0 = 250 (v) RL = RC = 120 1202 − 104 1202 − 104 = 250 rad/sec ω0 = 250 EXAMPLE 6.19 The following information is given in connection with a two branch parallel circuit: RL = 10 Ω, RC = 20 Ω, XC = 40 Ω, E = 120 V and frequency = 60 Hz. What are the values of L for resonance and what currents are drawn from the supply under this condition? SOLUTION As the frequency is constant, the condition for resonance is XC XL = 2 2 + XL RC + XC2 XL 40 1 = 2 = 2 2 2 20 + 40 50 10 + XL 2 RL ⇒ XL2 − 50XL + 100 = 0 ⇒ Solving we get XL = 47.913 Ω or 2.087 Ω Then the corresponding values of inductances are L= XL = 0.127 H ω or 5.536 mH Resonance | 491 The supply current is I = EG = E (GL + GC ) 10 + 0.02 = 1.7 A for XL = 47.913 Ω I = 120 102 + 47.9132 10 + 0.02 = 12.7 A for XL = 2.087 Ω I = 120 102 + 2.0872 Thus, or Exercise Problems E.P 6.1 Refer the circuit shown in Fig. E.P. 6.1, where Ri is the source resistance (a) Determine the transfer function of the circuit. (b) Sketch the magnitude plot with Ri = 0 and Ri = 0. Figure E.P. 6.1 Ans: E.P R Vo (s) L s = H(s) = R +R Vi (s) i s2 + s+ L 1 LC 6.2 For the circuit shown in Fig. E.P. 6.2, calculate the following: (b) Q, (c) fc1 , (d) fc2 and (e) B (a) f0 , Figure E.P. 6.2 Ans: (a) 254.65 kHz (b) 8 (c) 239.23 kHz (d) 271.06 kHz (e) 31.83 kHz 492 j E.P Network Analysis 6.3 Refer the circuit shown in Fig. E.P. 6.3, find the output voltage, when (a) ! = !0 (b) ! = !1 , and (c) ! = !c2 . Figure E.P. 6.3 Ans: (a) 640 cos(1.6 106 t)mV (b) 452.55 cos(1.5 106 t + 45 )mV (c) 452.55 cos(1.7 106 t 45 )mV E.P 6.4 Refer the circuit shown in Fig. E.P. 6.4. Calculate Zi (s) and then find (a) !0 and (b) Q. Figure E.P. 6.4 Ans: (a) 364.69 krad/sec, (b) 36 E.P 6.5 Refer the circuit shown in Fig. E.P. 6.5. Show that at resonance, Figure E.P. 6.5 jVoj max = GQjV j s 1 1 4Q2 . Resonance E.P 6.6 Refer the circuit given in Fig. E.P. 6.6, calculate !0 ; Q and jVoj j 493 max Figure E.P. 6.6 Ans: 3513.64 rad/sec, 26.35, 316 volts. E.P 6.7 A parallel network, which is driven by a variable frequency of 4 A current source has the following values: R = 1 kΩ , L = 10 mH, C = 100 F. Find the band width of the network, the half power frequenies and the voltage across the network at half-power frequencies. Ans: 10 rad/sec, 995 rad/sec, 10005 rad/sec E.P 6.8 For the circuit shown in Fig. E.P. 6.8, determine the expression for the magnitude response, 1 . versus ! and Zin at !0 = pLC Ans: (a) jZin j = E.P 6.9 qR ( jZinj Figure E.P. 6.8 !2 RLC )2 +(!L)2 , 1+(!RC )2 (b) jZin j = G 1 C L (1+R2 C L) A coil under test may be represented by the model of L in series with R. The coil is connected in series with a variable capacitor. A voltage source v (t) = 10 cos 1000 t volts is connected to the coil. The capacitor is varied and it is found that the current is maximum when C = 10F. Also, when C = 12:5F, the current is 0.707 of the maximum value. Find Q of the coil at ! = 1000 rad/sec. Ans: 5 494 j E.P Network Analysis 6.10 A fresher in the devices lab for sake of curiosity sets up a series RLC network as shown in Fig. E.P.6.10. The capacitor can withstand very high voltages. Is it safe to touch the capacitor at resonance? Find the voltage across the capacitor. Figure E.P. 6.10 Ans: Not safe, jVc jmax = 1600 V Chapter 7 7.1 Two Port Networks Introduction A pair of terminals through which a current may enter or leave a network is known as a port. A port is an access to the network and consists of a pair of terminals; the current entering one terminal leaves through the other terminal so that the net current entering the port equals zero. There are several reasons why we should study two-ports and the parameters that describe them. For example, most circuits have two ports. We may apply an input signal in one port and obtain an output signal from the other port. The parameters of a two-port network completely describes its behaviour in terms of the voltage and current at each port. Thus, knowing the parameters of a two port network permits us to describe its operation when it is connected into a larger network. Two-port networks are also important in modeling electronic devices and system components. For example, in electronics, two-port networks are employed to model transistors and Op-amps. Other examples of electrical components modeled by two-ports are transformers and transmission lines. Four popular types of two-ports parameters are examined here: impedance, admittance, hybrid, and transmission. We show the usefulness of each set of parameters, demonstrate how they are related to each other. Fig. 7.1 represents a two-port network. A four terminal network is called a two-port network when the current entering one terFigure 7.1 A two-port network minal of a pair exits the other terminal in the pair. For example, I1 enters terminal a and exits terminal b of the input terminal pair a-b. We assume that there are no independent sources or nonzero initial conditions within the linear two-port network. 496 7.2 j Network Theory Admittance parameters The network shown in Fig. 7.2 is assumed to be linear and contains no independent sources. Hence, principle of superposition can be applied to determine the current I1 , which can be written as the sum of two components, one due to V1 and the other due to V2 . Using this principle, we can write I1 = y11 V1 + y12 V2 Figure 7.2 A linear two-port network where y11 and y12 are the constants of proportionality with units of Siemens. In a similar way, we can write I2 = y21 V1 + y22 V2 Hence, the two equations that describe the two-port network are I1 = y11 V1 + y12 V2 (7.1) I2 = y21 V1 + y22 V2 (7.2) Putting the above equations in matrix form, we get I1 I2 = y11 y12 y21 y22 Here the constants of proportionality y11 ; y12 ; y21 and y22 are called y parameters for a network. If these parameters y11 , y12 , y21 and y22 are known, then the input/output operation of the two-port is completely defined. From equations (7.1) and (7.2), we can determine y parameters. We obtain y11 and y21 by connecting a current source I1 to port 1 and short-circuiting port 2 as shown in Fig. 7.3, finding V1 and I2 , and then calculating, y11 I1 = V1 V2 =0 V1 V2 Figure 7.3 Determination of y11 and y12 y21 I2 = V1 V2 =0 Since y11 is the admittance at the input measured in siemens with the output short-circuited, it is called short-circuit input admittance. Similarly, y21 is called the short-circuit transfer admittance. Two Port Networks j 497 Similarly, we obtain y12 and y22 by connecting a current source I2 to port 2 and shortcircuiting port 1 as in Fig. 7.4, finding I1 and V2 , and then calculating, y12 I1 = V2 V1 =0 y12 is called the short-circuit transfer admittance and y22 is called the shortcircuit output admittance. Collectively the y parameters are referred to as short-circuit admittance parameters. Please note that y12 = y21 only when there are no dependent sources or Op-amps within the two-port network. EXAMPLE y22 I2 = V2 V1 =0 Figure 7.4 Determination of y12 and y22 7.1 Determine the admittance parameters of the T network shown in Fig. 7.5. Figure 7.5 SOLUTION To find y11 and y21 , we have to short the output terminals and connect a current source I1 to the input terminals. The circuit so obtained is shown in Fig. 7.6(a). V1 V1 = 22 5 4+ 2 + 2 I1 1 y11 = = S V1 V2 =0 5 I1 = Hence; Using the principle of current division, I1 2 I1 I2 = = 2 + 2 2 1 V1 ) I2 = 2 5 I2 1 S Hence; y21 = = V1 V2 =0 10 Figure 7.6(a) 498 j Network Theory To find y12 and y22 , we have to short-circuit the input terminals and connect a current source I2 to the output terminals. The circuit so obtained is shown in Fig. 7.6(b). V2 I2 = 42 2+ 4+2 V2 = 4 2+ 3 3V2 = 10 I2 3 Hence; y22 = = S V2 V1 =0 10 Employing the principle of current division, we have Figure 7.6(b) I2 2 2+4 2I2 I1 = 6 1 3V2 I1 = 3 10 I1 1 y12 = S = V2 V1 =0 10 I1 = ) ) Hence; It may be noted that, y12 = y21 . Thus, in matrix form we have I = YV 3 2 1 I1 5 4 5=6 4 1 I2 10 2 ) EXAMPLE 3 1 32 V1 10 7 4 5 5 3 V2 10 7.2 Find the y parameters of the two-port network shown in Fig. 7.7. Then determine the current in a 4Ω load, that is connected to the output port when a 2A source is applied at the input port. Figure 7.7 Two Port Networks j 499 SOLUTION To find y11 and y21 , short-circuit the output terminals and connect a current source I1 to the input terminals. The resulting circuit diagram is shown in Fig. 7.8(a). V1 V1 = 12 1Ωjj2Ω 1+2 3 I1 = V1 2 I1 3 y11 = = S V1 V2 =0 2 I1 = ) Hence; Using the principle of current division, I2 = ) I2 = ) I2 = Hence; y21 = I1 1 1+2 1 I1 3 1 3 V1 3 2 I2 1 = S V1 2 Figure 7.8(a) To find y12 and y22 , short the input terminals and connect a current source I2 to the output terminals. The resulting circuit diagram is shown in Fig. 7.8(b). I2 = = V2 2Ωjj3Ω V2 5V2 = 23 6 2+3 I2 5 y22 = = S V2 V1 =0 6 Figure 7.8(b) Employing the current division principle, I1 = ) I2 3 2+3 3 I1 = I2 5 500 j Network Theory ) 3 5V2 I1 = 5 6 1 I1 = V2 2 I1 1 y12 = S = V2 V1 =0 2 ) Hence; Therefore, the equations that describe the two-port network are 3 1 I1 = V1 V2 2 2 1 5 I2 = V1 + V2 2 6 (7.3) (7.4) Putting the above equations (7.3) and (7.4) in matrix form, we get 2 3 6 2 4 1 2 3 2 3 1 32 I1 V1 2 74 5=4 5 5 5 V2 I2 6 Referring to Fig. 7.8(c), we find that I1 = 2A and V2 = 4I2 Substituting I1 = 2A in equation (7.3), we get 3 2 = V1 4 Multiplying equation (7.4) by Figure 7.8(c) 1 V2 2 (7.5) 4, we get 20 V2 6 20 V2 = 2V1 V2 6 20 0 = 2V1 + 1 V2 6 1 13 0= V1 + V2 2 12 4I2 = 2V1 ) ) ) Putting equations (7.5) and (7.6) in matrix form, we get 2 3 6 2 4 1 2 3 2 3 1 32 2 V1 2 74 5=4 5 5 13 V2 0 12 (7.6) Two Port Networks j 501 It may be noted that the above equations are simply the nodal equations for the circuit shown in Fig. 7.8(c). Solving these equations, we get 3 V2 = V 2 1 3 I2 = V2 = A 4 8 and hence, EXAMPLE 7.3 Refer the network shown in the Fig. 7.9 containing a current-controlled current source. For this network, find the y parameters. Figure 7.9 SOLUTION To find y11 and y21 short the output terminals and connect a current source I1 to the input terminals. The resulting circuit diagram is as shown in Fig. 7.10(a) and it is further reduced to Fig. 7.10(b). Figure 7.10(a) V1 22 2+2 I1 = V1 I1 y11 = = 1S V1 V2 =0 I1 = ) Hence; Figure 7.10(b) 502 j Network Theory Applying KCL at node A gives (Referring to Fig. 7.10(a)). ) ) ) Hence; I3 + I2 = 3I1 V1 + I2 = 3I1 2 V1 + I2 = 3V1 2 5V1 = I2 2 I2 5 y21 = = S V1 2 To find y22 and y12 , short the input terminals and connect a current source I2 at the output terminals. The resulting circuit diagram is shown in Fig. 7.10(c) and further reduced to Fig. 7.10(d). Figure 7.10(c) I2 = ) Hence; I01 = V2 2 V2 2 I1 1 = S y12 = V2 2 I1 = Applying KCL at node B gives V2 V2 + + 3I1 2 2 V2 I1 = 2 V2 V2 V2 I2 = + 3 2 2 2 I2 y22 = = 0:5S V2 V1 =0 I2 = But Hence; ) Figure 7.10(d) Two Port Networks j 503 Short-cut method: Referring to Fig. 7.9, we have KCL at node V1 : V1 V1 V2 + 2 2 = V1 0:5V2 I1 = Comparing with I1 = y11 V1 + y12 V2 we get y11 = 1S and y12 = 0:5S KCL at node V2 : V2 V2 V1 + 2 2 V2 V2 V1 = 3 [V1 0:5V2 ] + + 2 2 5 I2 = V1 0:5V2 2 I2 = 3I1 + ) Comparing with I2 = y21 V1 + y22 V2 we get y21 = 2:5S and y22 = EXAMPLE 0:5S 7.4 Find the y parameters for the two-port network shown in Fig. 7.11. Figure 7.11 SOLUTION To find y11 and y21 short-circuit the output terminals as shown in Fig. 7.12(a). Also connect a current source I1 to the input terminals. 504 j Network Theory Figure 7.12(a) KCL at node V1 : V1 V1 Va + 1 1 2 2Va = I1 I1 = ) 3V1 (7.7) KCL at node Va : V1 Va 0 + 2V1 = 0 + 1 1 2 2Va 2V1 + Va + 2V1 = 0 Va ) ) Va = 0 (7.8) Making use of equation (7.8) in (7.7), we get 3V1 = I1 I1 y11 = = 3S V1 V2 =0 Hence; Since Va = 0, I2 = 0, ) y21 I2 = = 0S V1 V2 =0 To find y22 and y12 short-circuit the input terminals and connect a current source I2 to the output terminals. The resulting circuit diagram is shown in Fig. 7.12(b). Two Port Networks j 505 Figure 7.12(b) KCL at node V2 : V2 V2 Va = I2 + 1 1 2 3V2 Va = I2 ) (7.9) KCL at node Va : Va V2 1 ) or + Va 0 +0=0 1 2 3Va V2 = 0 1 Va = V2 3 (7.10) Substituting equation (7.10) in (7.9), we get 3V2 ) Hence; We have; Also; ) 1 V2 = I2 3 8 V2 = I2 3 8 I2 = S y22 = V2 3 1 Va = V2 3 I1 + I3 = 0 I1 = = I3 Va = 1 2 (7.11) 2Va (7.12) 506 j Network Theory Making use of equation (7.12) in (7.11), we get I1 1 = V2 2 3 I1 2 y12 = S = V2 V1 =0 3 Hence; EXAMPLE 7.5 Find the y parameters for the resistive network shown in Fig. 7.13. Figure 7.13 SOLUTION Converting the voltage source into an equivalent current source, we get the circuit diagram shown in Fig. 7.14(a). To find y11 and y21 , the output terminals of Fig. 7.14(a) are shorted and connect a current source I1 to the input terminals. This results in a circuit diagram as shown in Fig. 7.14(b). Figure 7.14(a) KCL at node V1 : Figure 7.14(b) V1 V1 V2 + = I1 + 3V1 2 1 Two Port Networks j 507 Since V2 = 0, we get ) Hence; KCL at node V2 : Since V2 = 0, we get ) Hence V1 + V1 = I1 + 3V1 2 3 I1 = V1 2 I1 3 y11 = S = V1 V2 =0 2 V2 V1 V2 + 3V1 + = I2 2 1 0 + 3V1 V1 = I2 I2 = 2V1 I1 = 2S y21 = V2 To find y21 and y22 , the input terminals of Fig. 7.14(a) are shorted and connect a current source I2 to the output terminals. This results in a circuit diagram as shown in Fig. 7.14(c). Figure 7.14(c) Figure 7.14(d) 508 j Network Theory KCL at node V2 : V2 V2 0 + = I2 2 1 3 V2 = I2 2 3 I2 = S y22 = V2 2 ) Hence; KCL at node V1 : I1 = V1 V1 V2 + =0 2 1 Since V1 = 0, we get I1 = V2 I1 = y12 = V2 Hence; EXAMPLE 1S 7.6 The network of Fig. 7.15 contains both a dependent current source and a dependent voltage source. Find the y parameters. Figure 7.15 SOLUTION While finding y parameters, we make use of KCL equations. Hence, it is preferable to have current sources rather than voltage sources. This prompts us to convert the dependent voltage source into an equivalent current source and results in a circuit diagram as shown in Fig. 7.16(a). To find y11 and y21 , refer the circuit diagram as shown in Fig. 7.16(b). KCL at node V1 : V1 V1 V2 + + 2V1 = 2V2 + I1 1 1 Two Port Networks Figure 7.16(a) Figure 7.16(b) Since V2 = 0, we get ) Hence; KCL at node V2 : V1 + V1 + 2V1 = I1 4V1 = I1 I1 y11 = = 4S V1 V2 V2 V1 + = 2V1 + I2 1 1 Since V2 = 0, we get ) Hence; V1 = 2V1 + I2 3V1 = I2 I2 y21 = = V1 V2 =0 3S j 509 510 j Network Theory To find y22 and y12 , refer the circuit diagram shown in Fig. 7.16(c). KCL at node V1 : V1 V1 V2 + + 2V1 = 2V2 + I1 1 1 Since V1 = 0, we get V2 = 2V2 + I1 ) 3V2 = I1 I1 y12 = = V2 V1 =0 Hence; 3S Figure 7.16(c) KCL at node V2 : Since V1 = 0, we get ) V2 V2 V1 + = 2V1 + I2 1 1 V2 + V2 = 0 + I2 2V2 = I2 I2 y22 = = 2S V2 V1 =0 Hence; 7.3 Impedance parameters Let us assume the two port network shown in Fig. 7.17 is a linear network that contains no independent sources. Then using superposition theorem, we can write the input and output voltages as the sum of two components, one due to I1 and other due to I2 : V1 = z11 I1 + z12 I2 V2 = z21 I1 + z22 I2 Figure 7.17 Two Port Networks j 511 Putting the above equations in matrix from, we get V1 V2 = z11 z12 z21 z22 I1 I2 The z parameters are defined as follows: z11 V1 = I1 I2 =0 z12 V1 = I2 I1 =0 z21 V2 = I1 I2 =0 z22 V2 = I2 I1 =0 In the preceeding equations, letting I1 or I2 = 0 is equivalent to open-circuiting the input or output port. Hence, the z parameters are called open-circuit impedance parameters. z11 is defined as the open-circuit input impedance, z22 is called the open-circuit output impedance, and z12 and z21 are called the open-circuit transfer impedances. If z12 = z21 , the network is said to be reciprocal network. Also, if all the z-parameter are identical, then it is called a symmetrical network. EXAMPLE 7.7 Refer the circuit shown in Fig. 7.18. Find the z parameters of this circuit. Then compute the current in a 4Ω load if a 24 0 V source is connected at the input port. Figure 7.18 SOLUTION To find z11 and z21 , the output terminals are open circuited. Also connect a voltage source V1 to the input terminals. This gives a circuit diagram as shown in Fig. 7.19(a). Figure 7.19(a) 512 j Network Theory Applying KVL to the left-mesh, we get ) 12I1 + 6I1 = V1 V1 = 18I1 V1 z11 = = 18Ω I1 I2 =0 Hence; Applying KVL to the right-mesh, we get ) Hence; V2 + 3 0 + 6I1 = 0 z21 V2 = 6I1 V2 = 6Ω = I1 To find z22 and z12 , the input terminals are open circuited. Also connect a voltage source V2 to the output terminals. This results in a network as shown in Fig. 7.19(b). Figure 7.19(b) Applying KVL to the left-mesh, we get V1 = 12 0 + 6I2 ) V1 = 6I2 V1 z12 = = 6Ω I2 I1 =0 Hence; Applying KVL to the right-mesh, we get ) Hence; V2 + 3I2 + 6I2 = 0 V2 = 9I2 V2 z22 = = 9Ω I2 I1 =0 The equations for the two-port network are, therefore V1 = 18I1 + 6I2 (7.13) V2 = 6I1 + 9I2 (7.14) Two Port Networks j 513 The terminal voltages for the network shown in Fig. 7.19(c) are V1 = 24 0 (7.15) V2 = (7.16) 4I2 Figure 7.19(c) Combining equations (7.15) and (7.16) with equations (7.13) and (7.14) yields 24 0 = 18I1 + 6I2 0 = 6I1 + 13I2 Solving, we get EXAMPLE I2 = 0:73 0 A 7.8 Determine the z parameters for the two port network shown in Fig. 7.20. Figure 7.20 SOLUTION To find z11 and z21 , the output terminals are open-circuited and a voltage source is connected to the input terminals. The resulting circuit is shown in Fig. 7.21(a). Figure 7.21(a) 514 j Network Theory By inspection, we find that I3 = V1 Applying KVL to mesh 1, we get R1 (I1 I3 ) = V1 R1 I1 R1 I3 = V1 R1 I1 R1 V1 = V1 (1 + R1 ) V1 = R1 I1 V1 R1 z11 = = I1 I2 =0 1 + R1 ) ) ) Hence; Applying KVL to the path V1 ! R2 ! R3 ! V2, we get V1 + R2 I3 R3 I2 + V2 = 0 Since I2 = 0 and I3 = V1 , we get ) V1 + R2 V1 0 + V2 = 0 R2 ) R 1 I1 = (1 R2 ) 1 + R1 V2 R1 (1 R2 ) z21 = = I1 I2 =0 1 + R1 V2 = V1 (1 Hence; The circuit used for finding z12 and z22 is shown in Fig. 7.21(b). Figure 7.21(b) By inspection, we find that I2 ) ) I3 = V1 and V1 = I3 R1 I2 I3 = (I3 R1 ) I3 (1 + R1 ) = I2 Two Port Networks Applying KVL to the path R3 ) Hence; 515 ! R2 ! R1 ! V2, we get R3 I2 + (R2 + R1 ) I3 ) j V2 = 0 I2 R3 I2 + (R2 + R1 ) = V2 1 + R1 R2 + R1 = V2 I2 R 3 + 1 + R1 V2 z22 = I2 I1 =0 R2 + R1 Ω = R3 + 1 + R1 Applying KCL at node a, we get V1 + I3 = I2 V1 V1 + = I2 R1 1 = I2 V1 + R1 V1 1 z12 = = 1 I2 I1 =0 + R1 R1 = 1 + R1 ) ) ) EXAMPLE 7.9 Construct a circuit that realizes the following z parameters. z= SOLUTION Comparing z with = z11 z12 z21 z22 12 4 4 8 , we get z11 = 12Ω; z12 = z21 = 4Ω; z22 = 8Ω Let us consider a T network as shown in Fig. 7.22(a). Our objective is to fit in the values of R1 ; R2 and R3 for the given z. Applying KVL to the input loop, we get V1 = R1 I1 + R3 (I1 + I2 ) = (R1 + R3 ) I1 + R3 I2 516 j Network Theory Comparing the preceeding equation with V1 = z11 I1 + z12 I2 we get z11 = R1 + R3 = 12Ω ) z12 = R3 = 4Ω R1 = 12 Figure 7.22(a) R3 = 8Ω Applying KVL to the output loop, we get V2 = R2 I2 + R3 (I1 + I2 ) ) V2 = R3 I1 + (R2 + R3 ) I2 Comparing the immediate preceeding equation with V2 = z21 I1 + z22 I2 we get z21 = R3 = 4Ω z22 = R2 + R3 = 8Ω ) R2 = 8 Figure 7.22(b) R3 = 4Ω Hence, the network to meet the given z parameter set is shown in Fig. 7.22(b). EXAMPLE If z = 7.10 40 10 20 30 Ω for the two-port network, calculate the average power delivered to 50Ω resistor. Figure 7.23 Two Port Networks j 517 SOLUTION We are given z11 = 40Ω z12 = 10Ω z21 = 20Ω z22 = 30Ω Since z12 6= z21 , this is not a reciprocal network. Hence, it cannot be represented only by passive elements. We shall draw a network to satisfy the following two KVL equations: V1 = 40I1 + 10I2 V2 = 20I1 + 30I2 One possible way of representing a network that is non-reciprocal is as shown in Fig. 7.24(a). Figure 7.24(a) Now connecting the source and the load to the two-port network, we get the network as shown in Fig. 7.24(b). Figure 7.24(b) KVL for mesh 1: ) 60I1 + 10I2 = 100 6I1 + I2 = 10 KVL for mesh 2: ) ) 80I2 + 20I1 = 0 4I2 + I1 = 0 I1 = 4I2 518 j Network Theory Solving the above mesh equations, we get ) ) 24I2 + I2 = 10 23I2 = 10 10 I2 = 23 = j I2 j 2 R L Power supplied to the load 100 50 (23)2 = 9:45 W = EXAMPLE 7.11 Refer the network shown in Fig. 7.25. Find the z parameters for the network. Take = 4 3 Figure 7.25 SOLUTION To find z11 and z21 , open-circuit the output terminals as shown in Fig. 7.26(a). Also connect a voltage source V1 to the input terminals. Figure 7.26(a) Applying KVL around the path V1 ! 4Ω ! 2Ω ! 3Ω, we get 4I1 + 5I3 = V1 Also; V2 V2 = 3I3 ; so I3 = 3 (7.17) (7.18) Two Port Networks j 519 KCL at node b leads to I1 V2 I3 = 0 (7.19) Substituting equation (7.18) in (7.19), we get Hence; V2 1 V2 = + I 1 = V2 + 3 3 4 1 = V2 + 3 3 V2 3 z21 = = Ω I1 I2 =0 5 Substituting equation (7.18) in (7.17), we get V2 3 5 3 = 4I1 + I1 3 5 V1 = 5Ω z11 = I1 V1 = 4I1 + 5 Therefore; Since V2 3 = I1 5 To obtain z22 and z12 , open-circuit the input terminals as shown in Fig. 7.26(b). Also, connect a voltage source V2 to the output terminals. Figure 7.26(b) KVL for the mesh on the left: V1 + 5I4 3I2 = 0 (7.20) V2 + 3I4 3I2 = 0 (7.21) KVL for the mesh on the right: Also; I 4 = V2 (7.22) 520 j Network Theory Substituting equation (7.22) in (7.21), we get ) Hence; V2 + 3V2 3I2 = 0 V2 (1 + 3) = 3I2 V2 3 z22 = = I2 I1 =0 1 + 3 3 3 = Ω = 4 5 1+3 3 Substituting equation (7.22) in (7.20), we get V1 + 5V2 = 3I2 3 Substituting V2 = I2 , we get 5 3 V1 + 5 5 I2 = 3I2 V1 Hence; z12 = I2 I1 =0 = 3 3 4 =3 3 = 3 Finally, in the matrix form, we can write 2 Please note that z12 EXAMPLE z11 z12 3 2 1Ω 3 5 1 5=4 5 3 5 z=4 z21 z22 3 5 6 z21, since a dependent source is present in the circuit. = 7.12 Find the Thevenin equivalent circuit with respect to port 2 of the circuit in Fig. 7.27 in terms of z parameters. Figure 7.27 Two Port Networks j 521 SOLUTION The two port network is defined by V1 = z11 I1 + z12 I2 ; V2 = z21 I1 + z22 I2 ; here, V1 = Vg Zg I1 and V2 = IL ZL = I2 ZL To find Thevenin equivalent circuit as seen from the output terminals, we have to remove the load resistance RL . The resulting circuit diagram is shown in Fig. 7.28(a). Vt = V2 jI2 =0 = z21 I1 Figure 7.28(a) (7.23) With I2 = 0, we get V1 = z11 I1 Vg I1 Zg V1 I1 = = z11 z11 ) Solving for I1 , we get I1 = Vg z11 + Zg (7.24) Substituting equation (7.24) into equation (7.23), we get Vt = z21 Vg z11 + Zg To find Zt , let us deactivate all the independent sources and then connect a voltage source V2 across the output terminals as shown in Fig. 7.28(b). V2 Zt = ; where V2 = z21 I1 +z22 I2 I2 Vg =0 We know that V1 = z11 I1 + z12 I2 Substituting, V1 = I1 Zg in the preceeding equation, we get Solving; I1 Zg = z11 I1 + z12 I2 z12 I2 I1 = z11 + Zg Figure 7.28(b) 522 j Network Theory We know that, V2 = z21 I1 + Z22 I2 z12 I2 + z22 I2 = z21 z11 + Zg V2 z21 z12 Zt = = z22 I2 z11 + Zg Thus; The Thevenin equivalent circuit with respect to the output terminals along with load impedance ZL is as shown in Fig. 7.28(c). EXAMPLE Figure 7.28(c) 7.13 (a) Find the z parameters for the two-port network shown in Fig. 7.29. (b) Find V2 (t) for t > 0 where vg (t) = 50u(t)V. Figure 7.29 SOLUTION The Laplace transformed network with all initial conditions set to zero is as shown in Fig. 7.30(a). Figure 7.30(a) Two Port Networks j 523 (a) To find z11 and z21 , open-circuit the output terminals and then connect a voltage source V1 across the input terminals as shown in Fig. 7.30(b). Applying KVL to the left mesh, we get 1 V1 = s + s I1 V1 z11 = I1 I2 =0 Hence; =s+ 1 1 s = s2 + 1 s Also; V2 = I1 Hence; V2 1 z21 = = I1 I2 =0 s s To find z21 and z22 , open-circuit the input terminals and then connect a voltage source V2 across the output terminals as shown in Fig. 7.30(c). Figure 7.30(b) Figure 7.30(c) Applying KVL to the right mesh, we get V2 = s + s I2 V2 s2 + 1 z22 = = I2 I1 =0 s 1 V1 = I2 ) Also; s V1 1 z12 = = I2 I1 =0 s ) Summarizing, 1 2 z=4 z11 z12 z21 z22 3 2 s2 + 1 6 s 5=6 4 1 s 1 3 7 s 7 s2 + 1 5 s 524 j Network Theory (b) Figure 7.30(d) Refer the two port network shown in Fig. 7.30(d). V1 = Vg ) ) I1 Zg = z11 I1 + z12 I2 Vg = (z11 + Zg ) I1 + z12 I2 V2 Vg = (z11 + Zg ) I1 + z12 ZL V2 = z21 I1 + z22 I2 V2 V2 = z21 I1 z22 ZL z22 1 1+ V2 I1 = z21 ZL and ) ) (7.25) (7.26) Substituting equation (7.26) in equation (7.25) and simplifying, we get V2 Vg = z21 zL (ZL + z22 ) (z11 + Zg ) (7.27) z12 z21 Substituting for ZL , Zg and z-parameters, we get V2 (s) = 2 Vg (s) s +1 = ) V2 (s) = Vg (s) = Hence; V2 (s) = 1 s2 s +1 +1 +1 s s s 2 (s + s + 1)2 1 1 s3 + 2s2 + 3s + 2 1 (s + 1) (s2 + s + 2) Vg (s) (s + 1) (s2 + s + 2) 1 s2 (7.28) Two Port Networks The equation s2 + s + 2 = 0 gives s1;2 = 1 2 j V2 (s) = This means that, (s + 1) s + ) Hence; V2 (s) = = 525 p 7 2 Vg (s) p! p! 1 7 7 s+ +j j 2 2 2 1 2 vg (t) = 50u(t) 50 Vg (s) = s Given j 50 1 s(s + 1) s + 2 K1 K2 + + s s+1 p 7 j 2 ! p 1 7 s+ +j 2 2 ! K3 K3 p + p 7 7 1 1 s+ s+ +j j 2 2 2 2 By performing partial fraction expansion, we get K1 = 25; K2 = 25; K3 = 9:45 /90 9:45 / 90 25 25 9:45 /90 p p + Hence; V2 (s) = + s s+1 7 7 1 1 s+ s+ +j j 2 2 2 2 Taking inverse Laplace transform of the above equation, we get V2 (t) = 25 25e t + 18:9e 0:5t cos(1:32t + 90 ) u(t)V Verification: V2 (0) = 25 25 + 18:9 cos 90 = 0 V2 (1) = 25 + 0 + 0 = 25V Please note that at t = 1, the circuit diagram of Fig. (7.29) looks as shown in Fig. 7.30(e). I(1) = 50 = 25A 2 Hence; V2 (1) = VC (1) = 25V Figure 7.30(e) } 526 j Network Theory EXAMPLE 7.14 The following measurements were made on a resistive two-port network: Measurement 1: With port 2 open and 100V applied to port 1, the port 1 current is 1.125A and port 2 voltage is 104V. Measurement 2: With port 1 open and 50V applied to port 2, the port 2 current is 0.3A, and the port 1 voltage is 30 V. Find the maximum power that can be delivered by this two-port network to a resistive load at port 2 when port 1 is driven by an ideal voltage source of 100 Vdc. SOLUTION V1 100 z11 = = 88:89Ω = I1 I2 =0 1:125 V2 104 z21 = = 92:44Ω = I1 I2 =0 1:125 z12 = V1 30 = 100Ω = I2 I1 =0 0:3 V2 50 z22 = = 166:67Ω = I2 I1 =0 0:3 We know from the previous example 7.12 that, Zt = z22 z12 z21 z11 + Zg = 166:67 92:44 100 88:89 + 0 = 166:67 103:99 = 62:68Ω For maximum power transfer, ZL = Zt = 62:68Ω (For resistive load) Vt = = z21 Vg z11 + Zg 92:44 100 88:89 + 0 = 104 V Two Port Networks The Thevenin equivalent circuit with respect to the output terminals with load resistance is as shown in Fig. 7.31. Pmax = I2t RL 2 104 62:68 = 62:68 2 = 43:14 W EXAMPLE Figure 7.31 7.15 Refer the network shown in Fig. 7.32(a). Find the impedance parameters of the network. Figure 7.32(a) SOLUTION Figure 7.32(b) j 527 528 j Network Theory Referring to Fig. 7.32(b), we can write V3 = 2 (I1 + I2 ) KVL for mesh 1: 2I1 + 2I2 + 2 (I1 + I2 ) = V1 ) 4I1 + 4I2 = V1 KVL for mesh 2: ) ) ) 2 (I2 2V3 ) + 2 (I1 + I2 ) = V2 4 2 (I1 + I2 ) + 2 (I1 + I2 ) = V2 2I2 6 (I1 + I2 ) = V2 2I2 6I1 z11 = 4I2 = V2 V1 4I1 + 4I2 = = 4Ω I1 I2 =0 I1 I2 =0 V2 = z21 = I1 I2 =0 6I1 4I2 = I1 I2 =0 6Ω V1 4I1 + 4I2 = = 4Ω z12 = I2 I1 =0 I2 I1 =0 V2 = z22 = I2 I1 =0 EXAMPLE 6I1 4I2 = I2 I1 =0 4Ω 7.16 Is it possible to find z parameters for any two port network ? Explain. SOLUTION It should be noted that for some two-port networks, the z parameters do not exist because they cannot be described by the equations: V1 = I1 z11 + I2 z12 V2 = I1 z21 + I2 z22 As an example, let us consider an ideal transformer as shown in Fig. 7.33. (7.29) Two Port Networks j 529 Figure 7.33 The defining equations for the two-port network shown in Fig. 7.33 are: V1 = 1 V2 I1 = n n I2 It is not possible to express the voltages in terms of the currents, and viceversa. Thus, the ideal transformer has no z parameters and no y parameters. 7.4 z and y parameters by matrix partitioning For z parameters, the mesh equations are V1 = z11 I1 + z12 I2 + + z1n In V2 = z21 I1 + z22 I2 + + z2n In 0 = 0 = zn1 I1 + zn2 I2 + + znn In By matrix partitioning, the above equations can be written as z11 z12 z1n z21 z22 z2n z31 z32 z3n zn1 zn2 znn 530 j Network Theory The above equation can be simplified as (exact analysis not required) V1 V2 = M N Q 1 P I1 I2 M NQ 1 P gives z parameters. Similarly for y parameters, ) M NQ EXAMPLE 1P I1 I2 = M N Q 1 P V1 V2 gives y parameters. 7.17 Find y and z parameters for the resistive network shown in Fig. 7.34(a). Verify the result by using Y Δ transformation. 3 I1 I3 Figure 7.34(a) SOLUTION For the loops indicated, the equations in matrix form, 1 2 I2 Two Port Networks Then; ) V1 V2 = y=z 3 0 0 0:5 1 3:5 1:8571 0:2857 0:2857 0:4285 1 = 0:6 0:4 2 0:5 2 0:5 I1 I2 j 531 = [z] 0:4 2:5 Verification Figure 7.34(b) Refer Fig 7.34(b), converting T of 1, 1’, 2 into equation, 11+12+12 =5 1 Z2 = 5 5 Z3 = 2 1 5 2 = 5 Z2 = 5:5 11 Therefore, 3 2 13 y11 = ; y12 = y21 = ; y22 = 5 5 5 The values with transformed circuit is shown in Fig 7.34(c). Z1 = EXAMPLE Figure 7.34(c) 7.18 Find y and z parameters for the network shown in Fig.7.35 which contains a current controlled source. Figure 7.35 532 j Network Theory SOLUTION At node 1, 0:5V2 = I1 1:5V1 At node 2, 0:5V1 + V2 = I2 3I1 In matrix form, 1:5 0:5 0:5 1 ) V1 V2 V1 V2 = = = Therefore; [z] = [y] = [z] 7.5 1 0 3 1 1:5 0:5 0:5 1 0:4 0:4 3:2 1:2 0:4 0:4 3:2 1:2 1 I1 I2 = 1:5 4 1 I1 I2 0:5 0:5 1 0 3 1 I1 I2 Hybrid parameters The z and y parameters of a two-port network do not always exist. Hence, we define a third set of parameters known as hybrid parameters. In the pair of equations that define these parameters, V1 and I2 are the dependent variables. Hence, the two-port equations in terms of the hybrid parameters are or in matrix form, V1 = h11 I1 + h12 V2 (7.30) I2 = h21 I1 + h22 V2 (7.31) V1 I2 h11 h12 h21 h22 I1 V2 These parameters are particularly important in transistor circuit analysis. These parameters are obtained via the following equations: h11 V1 = I1 V2 =0 h12 V1 = V2 I1 =0 h21 I2 = I1 V2 =0 h22 I2 = V2 I1 =0 The parameters h11 ; h12 ; h21 and h22 represent the short-circuit input impedance, the opencircuit reverse voltage gain, the short-circuit forward current gain, and the open-circuit output admittance respectively. Because of this mix of parameters, they are called hybrid parameters. Two Port Networks EXAMPLE j 533 7.19 Refer the network shown in Fig. 7.36(a). For this network, determine the h parameters. Figure 7.36(a) SOLUTION To find h11 and h21 short-circuit the output terminals so that V2 = 0. Also connect a current source I1 to the input port as in Fig. 7.36(b). Figure 7.36(b) Applying KCL at node x: I1 + ) ) Vx Vx 0 + + I1 = 0 RB RC I1 [ Vx 1 1 + RB RC (1 )I1 RB RC Vx = RB + RC 1] = 534 j Network Theory V1 h11 = I1 V2 =0 Hence; Vx + I1 RA = I1 V2 =0 (1 )I1 RB RC + R A I1 = (RB + RC ) I1 (1 )RB RC + RA = RB + RC KCL at node y: ) ) Hence; I1 + I2 + I3 = 0 Vx 0 =0 I1 + I2 + RC 1 (1 )I1 RB RC =0 I1 + I2 + RC RB + RC I2 h21 = I1 V2 =0 (1 )RB = RB + RC (RC + RB ) = RB + RC To find h22 and h12 open-circuit the input port so that I1 = 0. Also, connect a voltage source V2 between the output terminals as shown in Fig. 7.36(c). Figure 7.36(c) KCL at node y: V1 V1 V2 + + I1 = 0 RB RC Two Port Networks j 535 Since I1 = 0, we get V1 ) V1 V2 =0 RC RC 1 V2 1 = + V1 RB RC RC V1 RB h12 = = V2 I1 =0 RB + RC RB + Applying KVL to the output mesh, we get V2 + RC (I1 + I2 ) + RB I2 = 0 Since I1 = 0, we get ) EXAMPLE R C I2 + R B I2 = V2 I2 1 h22 = = V2 I1 =0 RC + RB 7.20 Find the hybrid parameters for the two-port network shown in Fig. 7.37(a). Figure 7.37(a) SOLUTION To find h11 and h21 , short-circuit the output port and connect a current source I1 to the input port as shown in Fig. 7.37(b). Figure 7.37(b) 536 j Network Theory Referring to Fig. 7.37(b), we find that V1 = I1 [2Ω + (8Ωjj4Ω)] = I1 4:67 V1 h11 = = 4:67Ω I1 V2 =0 Hence; By using the principle of current division, we find that 2 I1 8 = I1 8+ 3 4 I2 2 h21 = = I1 V2 =0 3 I2 = Hence; To obtain h12 and h22 , open-circuit the input port and connect a voltage source V2 to the output port as in Fig. 7.37(c). Using the principle of voltage division, 8 2 V2 = V2 8+4 3 2 V1 = h12 = V2 3 V2 = (8 + 4)I2 V1 = Hence; Also; ) EXAMPLE = 12I2 I2 1 h22 = = S V2 I1 =0 12 Figure 7.37(c) 7.21 Determine the h parameters of the circuit shown in Fig. 7.38(a). Figure 7.38(a) Two Port Networks j 537 SOLUTION Performing Δ to Y transformation, the network shown in Fig. 7.38(a) takes the form as shown in 1 Fig. 7.38(b). Please note that since all the resistors are of same value, RY = RΔ . 3 Figure 7.38(b) To find h11 and h21 , short-circuit the output port and connect a current source I1 to the input port as in Fig. 7.38(c). V1 = I1 [4Ω + (4Ωjj4Ω)] Hence; = 6I1 V1 h11 = = 6Ω I1 V2 =0 Using the principle of current division, I1 4 4+4 I1 I2 = 2 I2 1 h21 = = I1 V2 =0 2 I2 = ) Hence; To find h12 and h22 , open-circuit the input port and connect a voltage source V2 to the output port as shown in Fig. 7.38(d). Using the principle of voltage division, we get V2 4 4 +4 V1 1 h12 = = V2 I1 =0 2 V1 = ) Also; ) V2 = [4 + 4] I2 = 8I2 I2 1 h22 = = S V2 I1 =0 8 Figure 7.38(c) Figure 7.38(d) 538 j Network Theory EXAMPLE 7.22 Determine the Thevenin equivalent circuit at the output of the circuit in Fig. 7.39(a). Figure 7.39(a) SOLUTION To find Zt , deactivate the voltage source Vg and apply a 1 V voltage source at the output port, as shown in Fig. 7.39(b). Figure 7.39(b) The two-port circuit is described using h parameters by the following equations: V1 = h11 I1 + h12 V2 (7.32) I2 = h21 I1 + h22 V2 (7.33) But V2 = 1 V and V1 = I1 Zg Substituting these in equations (7.32) and (7.33), we get ) I1 Zg = h11 I1 + h12 h12 I1 = Zg + h11 I2 = h21 I1 + h22 (7.34) (7.35) Two Port Networks j 539 Substituting equation (7.34) into equation (7.35), we get h21 h12 h11 + Zg h11 h22 h21 h12 + h22 Zg = h11 + Zg h11 + Zg V2 1 Zt = = = I2 I2 h11 h22 h12 h21 + h22 Zg I2 = h22 Therefore; To get Vt , we find open circuit voltage V2 with I2 = 0. To find Vt , refer the Fig. 7.39(c). Figure 7.39(c) At the input port, we can write Vg + I1 Zg + V1 = 0 V1 = Vg I1 Zg ) (7.36) Substituting equation (7.36) into equation (7.32), we get Vg I1 Zg = h11 I1 + h12 V2 Vg = (h11 + Zg ) I1 + h12 V2 ) (7.37) and substituting I2 = 0 in equation (7.33), we get 0 = h21 I1 + h22 V2 h22 I1 = V2 h21 ) (7.38) Finally substituting equation (7.38) in (7.37), we get Vg = (h11 + Zg ) ) V2 = Vt = h12 h21 h22 V2 + h12 V2 h21 Vg h21 h11 h22 Zg h22 Hence, the Thevenin equivalent circuit as seen from the output terminals is as shown in Fig. 7.39(d). Figure 7.39(d) 540 j Network Theory EXAMPLE 7.23 Find the input impedance of the network shown in Fig. 7.40. Figure 7.40 SOLUTION For the two-port network, we can write V1 = h11 I1 + h12 V2 (7.39) I2 = h21 I1 + h22 V2 (7.40) But V2 = IL ZL = where ZL = 75 kΩ I2 Z L (7.41) Substituting the value of V2 in equation (7.40), we get ) I2 = h21 I1 h22 I2 ZL h21 I1 I2 = 1 + ZL h22 (7.42) Substituting equation (7.42) in equation (7.41), we get V2 = ZL h21 I1 1 + ZL h22 Substituting equation (7.43) in equation (7.39), we get V1 = h11 I1 Hence Zin = V1 I1 h12 ZL h21 I1 1 + ZL h22 ZL h12 h21 1 + ZL h22 75 103 10 5 200 = 3 103 1 + 75 103 10 6 = 2:86kΩ = h11 (7.43) Two Port Networks EXAMPLE 7.24 Find the voltage gain, V2 Vg for the network shown in Fig. 7.41. S Figure 7.41 SOLUTION For the two-port network we can write, V1 = h11 I1 + h12 V2 ; I2 = h21 I1 + h22 V2 ; Hence; ) ) Also; ) here V1 = Vg Zg I1 here V2 = ZL I2 Vg Zg I1 = h11 I1 + h12 V2 Vg = (h11 + Zg ) I1 + h12 V2 Vg h12 V2 I1 = h11 + Zg V2 I2 = = h21 I1 + h22 V2 ZL Vg h12 V2 V2 + h22 V2 = h21 ZL h11 + Zg From the above equation, we find that V2 Vg = h21 ZL (h11 Zg ) (1 + h22 ZL ) h12 h21 ZL 100 50 103 = (2 103 + 1 103 ) (1 + 10 5 50 103 ) (10 = 1250 4 100 50 103) j 541 542 j Network Theory EXAMPLE 7.25 The following dc measurements were done on the resistive network shown in Fig. 7.42(a). Measurement 1 V1 = 20 V I1 = 0:8 A V2 = 0 V I2 = 0:4 A Measurement 2 V1 = 35 V I1 = 1 A V2 = 15 V I2 = 0 A Find the value of Ro for maximum power transfer. Figure 7.42(a) SOLUTION For the two-port network shown in Fig. 7.41, we can write: V1 = h11 I1 + h12 V2 I2 = h21 I1 + h22 V2 From measurement 1: V1 20 h11 = = 25Ω = I1 V2 =0 0:8 I2 0:4 = = h21 = I1 V2 =0 0:8 0:5 From measurement 2: ) ) Then; ) V1 = h11 I1 + h12 V2 35 = 25 1 + h12 15 10 h12 = = 0:67 15 I2 = h21 I1 + h22 V2 0 = h21 1 + h22 15 h21 0:5 h22 = = = 0:033 S 15 15 Two Port Networks | 543 For example (7.22), Vt = = = Zt = = = Vg h21 h12 h21 − h11 h22 − Zg h22 50 × (−0.5) 0.67 × (−0.5) − 25 × 0.033 − 20 × 0.033 −25 = 13.74 Volts −1.82 h11 + Zg h11 h22 − h12 h21 + h22 Zg 25 + 20 25 × 0.033 − 0.67 × (−0.5) + 0.033 × 20 45 = 24.72 Ω 1.82 For maximum power transfer, ZL = Zt = 24.72 Ω (Please note that, ZL is purely resistive). The Thevenin equivalent circuit as seen from the output terminals along with ZL is shown in Fig. 7.42(b). Pmax = It2 × 24.72 2 13.74 = × 24.72 24.72 + 24.72 = (13.74)2 = 1.9 Watts 4 × 24.72 Figure 7.42(b) EXAMPLE 7.26 Determine the hybird parameters for the network shown in Fig. 7.43. C Figure 7.43 544 j Network Theory SOLUTION To find h11 and h21 , short-circuit the output terminals so that V2 = 0. Also connect a current source I1 to the input port as shown in Fig. 7.44(a). Figure 7.44(a) Applying KVL to the mesh on the right side, we get ) ) ) Hence; 1 [I1 + I2 ] = 0 R2 [I1 + I2 ] + j!C 1 I1 + R 2 + I2 = 0 R2 + j!C j!C [ + j!R2 C ] I1 = I2 = [1 + j!CR2 ] I2 [ + j!R2 C ] I1 1 + j!R2 C I2 h21 = I1 V2 =0 = + j!CR2 1 + j!R2 C Applying KVL to the mesh on the left side, we get V1 = R1 I1 + R2 [I1 + I2 ] Hence; = [R1 + R2 ] I1 + R2 I2 R2 ( + j!CR2 ) I1 = R1 + R2 1 + j!R2 C V1 h11 = I1 V2 =0 Two Port Networks j 545 R2 ( + j!R2 C ) 1 + j!R2 C R1 + R2 (1 ) + j!R1 R2 C = 1 + j!R2 C = R1 + R2 To find h22 and h12 , open-circuit the input terminals so that I1 = 0. Also connect a voltage source V2 to the output port as shown in Fig. 7.44(b). The dependent current source is open, because I1 = 0. V1 = I2 R2 V2 = R2 1 R2 + Hence; j!C V1 h12 = V2 I1 =0 j!CR2 1 + j!CR2 V2 j!C V2 I2 = = 1 1 + j!CR2 R2 + j!C I2 j!C h22 = = V2 I1 =0 1 + j!CR2 = ) 7.6 Figure 7.44(b) Transmission parameters The transmission parameters are defined by the equations: V1 = AV2 BI2 I1 = CV2 DI2 Figure 7.45 Terminal variables used to define the ABCD Parameters 546 j Network Theory Putting the above equations in matrix form we get V1 I1 = A B C D V2 I2 Please note that in computing the transmission parameters, I2 is used rather than I2 , because the current is considered to be leaving the network as shown in Fig. 7.45. These parameters are very useful in the analysis of circuits in cascade like transmission lines and cables. For this reason they are called Transmission Parameters. They are also known as ABCD parameters. The parameters are determined via the following equations: V1 A= V2 I2 =0 V1 B= I2 V2 =0 I1 C= V2 I2 =0 I1 D= I2 V2 =0 A, B, C and D represent the open-circuit voltage ratio, the negative short-circuit transfer impedance, the open-circuit transfer admittance, and the negative short-circuit current ratio, respectively. When the two-port network does not contain dependent sources, the following relation holds good. AD BC = 1 EXAMPLE 7.27 Determine the transmission parameters in the s domain for the network shown in Fig. 7.46. Figure 7.46 SOLUTION The s domain equivalent circuit with the assumption that all the initial conditions are zero is shown in Fig. 7.47(a). Figure 7.47(a) Two Port Networks j 547 To find the parameters A and C, open-circuit the output port and connect a voltage source V1 at the input port. The same is shown in Fig. 7.47(b). V1 I1 = 1+ 1 1 s = sV1 s+1 Then V2 = I1 Also; 1 sV1 V1 V2 = = ss+ 1 s+1 V1 A= =s+1 V2 I2 =0 1 V2 = I1 ) ) ) s Figure 7.47(b) s I1 C= =s V2 I2 =0 To find the parameters B and D, short-circuit the output port and connect a voltage source V1 to the input port as shown in Fig. 7.47(c). Figure 7.47(c) The total impedance as seen by the source V1 is 1 Z=1+ s 1 s 1 +1 1 s+2 = s+1 s+1 V1 V1 (s + 1) I1 = = Z (s + 2) =1+ (7.44) Using the principle of current division, we have Hence; 1 I1 I1 s I2 = = 1 s+1 +1 s I1 D= =s+1 I2 V2 =0 From equation (7.44) and (7.45), we can write V1 (s + 1) (s + 2) V1 B= =s+2 I2 V2 =0 I2 (s + 1) = Hence; (7.45) 548 j Network Theory Verification We know that for a two port network without any dependent sources, AD (s + 1) (s + 1) EXAMPLE BC = 1 s (s + 2) = 1 7.28 Determine the ABCD parameters for the two port network shown in Fig. 7.48. Figure 7.48 SOLUTION To find the parameters A and C, open-circuit the output port as shown in Fig. 7.49(a) and connect a voltage source V1 to the input port. Applying KVL to the output mesh, we get V2 + mI1 + 0 RC + I1 RA = 0 ) V2 = I1 (m + RA ) I1 1 Hence; C = = V2 I2 =0 m + RA Applying KVL to the input mesh, we get Hence; V1 = I1 (RA + RB ) V1 RA + RB A= = V2 I2 =0 m + RA Figure 7.49(a) To find the parameters B and D, short-ciruit the output port and connect a voltage source V1 to the input port as shown in Fig. 7.49(b). Two Port Networks Figure 7.49(b) Applying KVL to the right-mesh, we get mI1 + RC I2 + RB (I1 + I2 ) = 0 ) ) Hence; (m + RB ) I1 = (RC + RB ) I2 (RC + RB ) I2 I1 = (m + RB ) I1 (RC + RB ) D= = I2 V2 =0 (m + RB ) Applying KVL to the left-mesh, we get V1 + RA I1 + RB (I1 + I2 ) = 0 ) Hence; EXAMPLE V1 = (RA + RB ) I1 + RB I2 (RC + RB ) = (RA + RB ) I2 + R B I2 (m + RB ) R C R A + R C R B + R B R A mR B I2 = m + RB V1 B= I2 V2 =0 R C R A + R C R B + R B R A mR B = m + RB 7.29 The following direct-current measurements were done on a two port network: Port 1 open V1 = 1 mV V2 = 10 V I2 = 200 A Port 1 Short-circuited I1 = 0:5 A I2 = 80 A V2 = 5 V Calculate the transmission parameters for the two port network. j 549 550 j Network Theory SOLUTION For the two port network, we can write V1 = AV2 BI2 I1 = CV2 DI2 From I1 = 0 (port 1 open): 1 10 From V1 = 0 (Port 1 short): 3 = A 10 0=A5 B 200 10 B 80 10 6 6 Solving simultaneously yields, From I1 = 0: From V1 = 0: 4 10 4 ; B = 25Ω 0 = C 10 D 200 10 6 0:5 10 6 = C 5 D 80 10 6 A= Solving simultaneously yields, C= In summary, 5 10 S; D= A= 4 10 B= 25Ω C= D= EXAMPLE 7 5 10 0:025 4 7 S 0:025 7.30 Find the transmission parameters for the network shown in Fig. 7.50. Figure 7.50 SOLUTION To find the parameters A and C, open the output port and connect a voltage source V1 to the input port as shown in Fig. 7.51(a). Two Port Networks Figure 7.51(a) Applying KVL to the input loop, we get V1 = 1:5 103 I1 + 10 3 V2 Also KCL at node a gives V2 =0 40 103 V2 ) I1 = = 6:25 10 6 V2 160 103 Substituting the value of I1 in the preceeding loop equation, we get 40I1 + ) V1 = 1:5 103 V1 = 9:375 10 Also; 3 6 V2 + 10 V2 + 10 3 8:375 10 V2 V1 = 8:375 10 A= V2 I2 =0 = Hence; 6:25 10 I1 C= = V2 I2 =0 V2 3 6:25 10 3 6 To find the parameters B and D, refer the circuit shown in Fig. 7.51(b). Figure 7.51(b) 3 V2 j 551 552 j Network Theory Applying KCL at node b, we find ) 40I1 + 0 = I2 I2 = 40I1 I1 1 D= = I2 V2 =0 40 Hence; Applying KVL to the input loop, we get V1 = 1:5 103 I1 ) V1 = 1:5 103 V1 B= I2 V2 =0 1:5 103 = 40 = 37:5Ω Hence; EXAMPLE I2 40 7.31 Find the Thevenin equivalent circuit as seen from the output port using the transmission parameters for the network shown in Fig. 7.52. Figure 7.52 SOLUTION For the two-port network, we can write V1 = AV2 BI2 (7.46) I1 = CV2 DI2 (7.47) Two Port Networks j 553 Figure 7.53(a) Refer the network shown in Fig. 7.53(a) to find Vt . At the input port, Vg Also; I1 Zg = V1 (7.48) I2 = 0 (7.49) Making use of equations (7.48) and (7.49) in equations (7.46) and (7.47) we get, Vg and I1 Zg = AV2 (7.50) I1 = CV2 (7.51) Making use of equation (7.51) in (7.50), we get Vg ) CV2 Zg = AV2 V2 = Vt = Vg A + CZg To find Rt , deactivate the voltage source Vg and then connect a voltage source V2 = 1 V at the output port. The resulting circuit diagram is shown in Fig. 7.53(b). Figure 7.53(b) 554 j Network Theory Referring Fig. 7.53(b), we can write V1 = I1 Z g Substituting the value of V1 in equation (7.46), we get I1 Zg = AV2 BI2 B A V2 + I2 I1 = ) Zg Zg (7.52) Equating equations (7.47) and (7.52) results Hence DI2 = A V2 + B I2 Zg Zg B A = D+ I2 V2 C + Zg Zg B D+ Zg V2 = Zt = A I2 C+ Zg B + DZg = A + CZg CV2 Figure 7.54 Hence, the Thevenin equivalent circuit as seen from the output port is as shown in Fig. 7.54. EXAMPLE 7.32 For the network shown in Fig. 7.55(a), find RL for maximum power transfer and the maximum power transferred. Figure 7.55(a) Two Port Networks j 555 SOLUTION From the previous example 7.31, B + DZg 20 + 3 20 20 + 60 = = = 6.67Ω A + CZg 4 + 0:4 20 4+8 Vg 100 100 Vt = = = = 8.33V A + CZg 4 + 0:4 20 12 Zt = For maximum power transfer, RL = Zt = 6:67Ω (purely resistive) Hence, the Thevenin equivalent circuit as seen from output terminals along with RL is as shown in Fig. 7.55(b). 8:33 = 0:624A 6:67 + 6:67 (PL )max = It2 6:67 It = = (0:624)2 6:67 Figure 7.55(b) = 2.6Watts EXAMPLE 7.33 Refer the bridge circuit shown in Fig. 7.56. Find the transmission parameters. Figure 7.56 SOLUTION Performing Δ to Y transformation, as shown in Fig. 7.57(a) the network reduces to the form as shown in Fig. 7.57(b). Please note that, when all resistors are of equal value, 1 RY = RΔ 3 556 j Network Theory Figure 7.57(a) Figure 7.57(b) To find the parameters A and D, open the output port and connect a voltage source V1 at the input port as shown is Fig. 7.57(c). Applying KVL to the input loop we get I1 + 4I1 = V1 ) Also; Also; Hence; ) V1 = 5I1 V2 I1 = 4 ) I1 1 C= = S V2 I2 =0 4 5 V1 = 5I1 = V2 4 V1 5 A= = V2 I2 =0 4 To find the parameters B and D, refer the circuit shown in Fig. 7.57(d). Figure 7.57(c) Two Port Networks j 557 I1 4 4 = I1 4 + 1 5 I1 5 D= = I2 v2 =0 4 I2 = Hence Applying KVL to the input loop, we get V1 + 1 I1 + 4 (I1 + I2 ) = 0 Substituting I1 = Figure 7.57(d) 5 I2 in the preceeding equation, we get 4 V1 ) ) Hence; 5 5 I2 + 4 I2 + I2 = 0 4 4 5 V1 I2 5I2 + 4I2 = 0 4 4V1 = 9I2 V1 9 B= = Ω I2 v2 =0 4 Verification: For a two port network which does not contain any dependent sources, we have AD 5 4 7.7 54 1 4 BC = 1 25 94 = 16 9 =1 16 Relations between two-port parameters If all the two-port parameters for a network exist, it is possible to relate one set of parameters to another, since these parameters interrelate the variables V1 ; I1 ; V2 and I2 : To begin with let us first derive the relation between the z parameters and y parameters. The matrix equation for the z parameters is ) V1 V2 = z11 z12 z21 z22 I1 I2 V = zI Similarly, the equation for y parameters is ) I1 I2 y11 y12 = y21 y22 I = yV (7.53) V1 V2 (7.54) 558 j Network Theory Substituting equation (7.54) into equation (7.53), we get V = zyV Hence; z=y 1 = Δy = y11 y22 where adj(y) Δy y21 y12 This means that we can obtain z matrix by inverting y matrix. It is quite possible that a two-port network has a y matrix or a z matrix, but not both. Next let us proceed to find z parameters in terms of ABCD parameters. The ABCD parameters of a two-port network are defined by V1 = AV2 ) ) BI2 I1 = CV2 DI2 1 V2 = (I1+ DI2 ) C 1 D V2 = I1 + I2 C C I1 DI2 BI2 + V1 = A C C AI1 AD = + B I2 C C (7.55) (7.56) Comparing equations (7.56) and (7.55) with V1 = z11 I1 + z12 I2 V2 = z21 I1 + z22 I2 respectively, we find that AD BC 1 D A z12 = z21 = z22 = C C C C Next, let us derive the relation between hybrid parameters and z parameters. z11 = V1 = z11 I1 + z12 I2 (7.57) V2 = z21 I1 + z22 I2 (7.58) From equation (7.58), we can write I2 = z21 V2 I1 + z22 z22 Substituting this value of I2 in equation (7.57), we get (7.59a) z21 I1 V2 + V1 = z11 I1 + z12 z22 z22 z12 V2 z11 z22 z12 z21 I1 + = z22 z22 (7.59b) Two Port Networks j 559 Comparing equations (7.59b) and (7.59a) with V1 = h11 I1 + h12 V2 I2 = h21 I1 + h22 V2 Δz z11 z22 z12 z21 = h11 = z22 Z22 z12 z21 h12 = h21 = z22 z22 Δz = z11 z22 z12 z21 we get, where h22 = 1 z22 Finally, let us derive the relationship between y parameters and ABCD parameters. I1 = y11 V1 + y12 V2 (7.60) I2 = y21 V1 + y22 V2 (7.61) From equation (7.61), we can write I2 y22 V2 y21 y21 y22 1 = V2 + I2 y21 y21 V1 = (7.62) Substituting equation (7.62) in equation (7.60), we get y11 y22 y11 V2 + y12 V2 + I2 y21 y21 y11 Δy V2 + I2 = y21 y21 I1 = (7.63) Comparing equations (7.62) and (7.63) with the following equations, V1 = AV2 we get where y22 A= y21 BI2 I1 = CV2 DI2 1 Δy B= C= y21 y21 Δy = y11 y22 y12 y21 D= y11 y21 Table 7.1 lists all the conversion formulae that relate one set of two-port parameters to another. Please note that Δz, Δy, Δh, and ΔT, refer to the determinants of the matrices for z, y, hybrid, and ABCD parameters respectively. 560 j Network Theory Table 7.1 Parameter relationships z 2 4 z z11 z12 3 5 z21 z22 y 2 z 22 6 Δz 4 z21 Δz 2 h 3 z11 6 z21 6 4 1 z21 T Δz z21 7 7 z22 5 z21 Δz 6 z22 6 4 z21 z22 Δz = z11 z22 EXAMPLE y22 6 Δy 6 4 y21 Δy z12 3 Δz 7 5 z11 Δz 2 3 z12 z22 7 7 1 5 z22 y 2 2 4 3 y12 Δy 7 7 y11 5 Δy y11 y12 y22 6 y21 6 4 Δy y21 2 1 6 y11 6 4 y21 y11 z12 z21 ; Δy = y11 y22 1 A 6 C 6 4 1 C 2 3 5 y21 y22 2 T 2 3 ΔT B 7 7 A 5 B 2 3 y21 7 7 y11 5 y21 3 ΔT C 7 7 D 5 C D 6 B 6 4 1 B 3 y12 y11 7 7 Δy 5 y11 3 4 A B B 6 D 6 4 1 D 5 3 Determine the y parameters for a two-port network if the z parameters are z= SOLUTION 10 5 5 9 Δz = 10 9 5 5 = 65 z22 9 y11 = = S Δz 65 z12 5 y12 = = S Δz 65 z21 5 y21 = = S Δz 65 z11 10 y22 = = S Δz 65 1 6 h11 6 4 h21 h11 2 4 3 h12 h22 7 7 1 5 h22 3 h22 h11 7 7 Δh 5 h11 3 h11 h21 7 7 1 5 h21 h11 h12 3 5 h21 h22 h12 h21 ; ΔT = AD 7.34 2 Δh 6 h21 6 4 h22 h21 ΔT D 7 7 C 5 D y12 y21 ; Δh = h11 h22 Δh 6 h22 6 4 h21 h22 2 C D 2 h 2 BC Two Port Networks EXAMPLE j 561 7.35 Following are the hybrid parameters for a network: h11 h12 h21 h22 = 5 2 3 6 Determine the y parameters for the network. Δh = 5 6 3 2 = 24 1 1 y11 = = S h11 5 h22 6 = S y12 = h11 5 h21 3 = S y21 = h11 5 Δh 24 = S y22 = h11 5 SOLUTION Reinforcement problems R.P 7.1 The network of Fig. R.P. 7.1 contains both a dependent current source and a dependent voltage source. Determine y and z parameters. Figure R.P. 7.1 SOLUTION From the figure, the node equations are Iab = I2 V2 2 At node a, I1 = V1 2V2 I2 V2 2 562 j Network Theory At node b, V1 = V2 I2 2V1 V2 2 Simplifying, the nodal equations, we get 3 V2 2 3 3V1 + V2 2 I1 + I2 = V1 I2 = In matrix form, ) 1 1 0 1 I1 I2 I1 I2 2 6 =4 3 = 2 y=4 Therefore; Z= and R.P 1 1 1 0 1 4 3 3 3 2 7 V1 5 V2 3 2 2 3 3 1 1 6 2 7 V1 4 5 V2 3 3 2 3 3 3 5 2 2 3 14 2 3 3 3 3 5=4 4 7.2 Compute y parameters for the network shown in Fig. R.P. 7.2. Figure R.P. 7.2 2 0:5 1 3 1 4 5 3 Two Port Networks j 563 SOLUTION The circuit shall be transformed into s-domain and then we shall use the matrix partitioning method to solve the problem. From Fig 7.2, Node equations in matrix form, I I R.P s s s P V Q V s V V V M N V1 I1 = P Q N 1 M V2 I2 1 2 s+3 s 2 1 = s s+2 5 1 8 2 4 2 39 > > < = s+3 s 6 5 5 7 = 4 5 s s+2 > 2 1 > : ; 5 5 s + 2:2 (s + 0:4) y= (s + 0:4) s + 1:8 V V1 V2 V1 V2 7.3 Determine for the circuit shown in Fig. R.P. 7.3(a): (a) Y1 , Y2 , Y3 and gm in terms of y parameters. (b) Repeat the problem if the current source is connected across Y3 with the arrow pointing to the left. Figure R.P. 7.3(a) SOLUTION (a) Refering Fig. R.P. 7.3(a), the node equations are: At node 1 I1 = Y1 V1 + (V1 V2 )Y3 = V1 (Y1 + Y3 ) Y3 Y2 (7.64) At node 2 I2 = gm V1 + V2 Y2 + (V2 = (gm V1 )Y3 Y3 )V1 + (Y2 + Y3 )V2 (7.65) 564 j Network Theory Then from equations (7.64) and (7.65), y11 = Y1 + Y3 ; y12 = y21 = gm y22 Y3 ; Y3 = Y2 + Y3 Solving, Y3 = y12 ; Y1 = y11 + y12 Y2 = y22 + y12 ; gm = y21 y12 (b) Making the changes as given in the problem, we get the circuit shown in Fig R.P. 7.3(b). Figure R.P. 7.3(b) Node equations : At node 1 I1 = Y1 V1 + (V1 = (Y1 + Y3 V2 )Y3 gm )V1 gm V1 Y3 V2 (7.66) At node 2, I2 = V2 Y2 + (V2 = (gm V1 )Y3 + gm V1 Y3 )V1 + V2 (Y2 + Y3 ) (7.67) From equations (7.66) and (7.67), y11 = Y1 + Y3 y21 = gm gm ; Y3 ; Y3 y22 = Y2 + Y3 y12 = Solving, Y3 = y12 ; Y2 = y22 y12 gm = y21 y12 Y1 = y11 Y3 + gm = y11 + y12 + y21 y12 = y11 y21 Two Port Networks R.P j 565 7.4 Complete the table given as part of Fig. R.P. 7.4. Also find the values for y parameters. Table Sl.no 1 2 3 4 5 Figure R.P. 7.4 V1 50 100 200 SOLUTION From article 7.2, I1 I2 = y11 y12 y21 y22 V1 V2 Substituting the values from rows 1 and 2, 1 7 27 24 Post multiplying by [V] = y11 y12 y21 y22 For row 4: For row 5: I1 I2 V1 V2 V1 V2 50 100 100 50 1, = = For row 3: y11 y12 y21 y22 = = = 1 7 27 24 0:1 0:14 0:1 0:14 0:06 0: 2 0:1 0:14 0:06 0: 2 0:1 0:14 0:06 0: 2 0:06 0: 2 1 1 200 0 1 50 100 100 50 20 0 10 30 = 20 28 = = 140:84 98:59 133:8 56:338 V2 100 50 0 I1 1 7 I2 27 24 20 10 0 30 566 j Network Theory R.P 7.5 Find the condition on a and b for reciprocity for the network shown in Fig. R.P. 7.5. Figure R.P. 7.5 SOLUTION The loop equations are aV2 = 3(I1 + I3 ) V2 = (I2 I3 ) bI1 I3 = V2 (V1 aV2 ) = (1 + a)V2 V1 V1 (7.68) (7.69) (7.70) Substituting equation (7.70) in equations (7.68) and (7.69), aV2 = 3I1 + 3(1 + a)V2 3V1 4V1 = (3 + 4a)V2 + 3I1 V2 = I2 [(1 + a) V2 V1 ] bI1 V1 + (2 + a) V2 = bI1 + I2 V1 ) ) Putting equations (7.71) and (7.72) in matrix form and solving V1 V2 = 4 1 (3 + 4a) 2+a 1 1 2 + a 3 + 4a 1 4 Δ 1 M 3 + 4a = N Δ 3 4b = For reciprocity, 3 + 4a = 3 Therefore; a= b 4b 3 0 b 1 3 0 b 1 I1 I2 I1 I2 I1 I2 (7.71) (7.72) Two Port Networks R.P 7.6 For what value of a is the circuit reciprocal? Also find h parameters. Figure R.P. 7.6 SOLUTION The node equations are 0:5V1 V1 I1 = V2 V2 = (I1 + I2 )2 + aI1 h= = 1 Δ 2 6 0:5 0 0 2 1 2 0 0 0:5 2 =4 2+a 2 2 0:5 For reciprocity, ) ) Therefore h21 2+a 2= 2 4 = 2 + a; a = 2 2 2 h= 2 0:5 h12 = 3 7 5 1 2+a 1 1 1 2+a 1 1 (Δ = 1) j 567 568 j Network Theory R.P 7.7 Find y12 and y21 for the network shown in Fig. R.P. 7.7 for n = 10. What is the value of n for the network to be reciprocal? Figure R.P. 7.7 SOLUTION Equations for I1 and I2 are 0:01V2 5 = 0:2V1 0:002V2 V2 V2 0:01V2 I2 = + n I1 + 20 50 Substituting the value of I1 from equation (7.73) in equation (7.74), we get V2 V2 0:01V2 I2 = n(0:2V1 0:002V2 ) + + 20 20 Simplifying the above equation with n = 10, I1 = V1 I2 = 2V1 + 0:0498V2 From equation (7.73), y12 = 0:002 and from equation (7.75), y21 = 0:2n For reciprocity y12 = y21 Hence; R.P ) 0:002 = 0:2n n = 0:01 7.8 Find T parameters (ABCD) for the two-port network shown in Fig. R.P. 7.8. Figure R.P. 7.8 (7.73) (7.74) (7.75) (7.76) Two Port Networks j 569 SOLUTION Network equations are V1 1:5V1 10I1 = V2 V2 I1 1:5V1 + I2 25 (7.77) V2 =0 20 (7.78) Simplifying, 2:5V1 10I1 = V2 0:06V1 + I1 = 0:09V2 I2 In matrix form, Therefore 2:5 0:06 T= R.P 2:5 0:06 10 1 10 1 V1 I1 1 = 1 0 0:09 1 1 0 0:09 1 = V2 I2 0:613 3:23 0:053 0:806 7.9 (a) Find T parameters for the active two port network shown in Fig. R.P. 7.9. (b) Find new T parameters if a 20 Ω resistor is connected across the output. Figure R.P. 7.9 SOLUTION (a) With Therefore; Vx = 10I1 , 0:08Vx = 0:8I1 V1 10I1 = V2 I1 + I2 = V2 V1 0:8I1 = 5(I2 0:08Vx ) 5I2 + 4I1 10I1 50 (7.79) (7.80) 570 j Network Theory Simplying the equations (7.79) and (7.80),we get V1 14I1 = V2 V1 + 20I1 = and Therefore, 1 1 T= 14 20 1 1 5 0 50 5I2 50I2 = 3:33 133:33 0:167 9:17 (b) Treating 20 Ω across the output as a second T network for which # " 1 0 T= 1 1 20 Then new T-parameters, T= R.P 3:33 133:33 0:167 9:17 " 1 0 # 10 133:33 = 1 0:625 9:17 1 20 7.10 Obtain z parameters for the network shown in Fig. R.P. 7.10. Figure R.P. 7.10 SOLUTION At node 1, V1 = (I1 = 10I1 At node 2, V2 = I2 ) 0:3V2 )10 + V2 (7.81) 2V2 V2 6 8 V2 = V1 + 10I2 3 10 + V1 = 10I2 + V1 5 V2 3 (7.82) Two Port Networks Putting in matrix form, 1 1 Therefore, z= R.P 1 1 2 −8 3 2 −8 3 V1 V2 −1 = 10 0 0 −10 10 0 0 −10 = I1 I2 5.71 −4.286 2.143 2.143 7.11 Obtain z and y parameters for the network shown in Fig. R.P. 7.11. Figure R.P. 7.11 SOLUTION For the meshes indicated, the equations in matrix form is | 571 572 j Network Theory By matrix partitioning, 2 s+2 2 6 s s z=4 s+4 2 s 2s 2 s+2 2 6 s s =4 s+4 2 s 2s 2 s+2 2 6 s s =6 4 2 s+4 s 2s 2 2 s + 10s + 8 6 s(3s + 4) 6 =6 6 s2 + 6s + 8 4 s(3s + 4) R.P 3 7 5 3 7 5 2 3 1 2s 4 1 5 3s + 4 2 2 2s 6 4 3s + 4 3 2 7 7 5 4 2s 6 3s + 4 6 s 3s + 4 3 s2 + 6s + 8 s(3s + 4) 7 7 7 2 s + 8s + 8 7 5 s(3s + 4) 1 1 2 1 1 2 1 3 2 7 5 1 4 3 s 3s + 4 7 7 8s 5 3s + 4 7.12 Find z and y parameters at ! = 108 rad/sec for the transistor high frequency equivalent circuit shown in Fig. R.P. 7.12. Figure R.P. 7.12 SOLUTION In the circuit, Vx = V1 . Therefore the node equations are 5 + j 6 10 j 10 4 V2 I2 = j 10 4 V1 + 0:01V1 + 10 4 (1 + j )V2 I1 = (10 4 )V1 Two Port Networks Simplifying the above equations, I1 = 10−4 [(0.1 + j6)V1 − j1V2 ] Therefore, I2 = 10−4 [(100 + j1)V1 + (1 + j)V2 ] ⎡ ⎤ ⎡ ⎤ 0.1 + j6 −j1 V 1 I1 ⎦ × 10−4 ⎣ ⎦ =⎣ I2 100 − j1 1 + j1 V2 ωC1 = 108 × 5 × 10−12 = 5 × 10−4 ωC2 = 108 × 10−12 = 10−4 Δ = 10−8 [(0.1 + j6)(1 + j) + (100 − j1)(j1)] = 10−8 × 106.213 /92.64◦ ⎡ Therefore, y=⎣ 6 /89◦ 100 /−0.6◦ ⎡ z = y−1 = ⎣ Then, ⎡ =⎣ R.P √ 2 /45◦ √ 2 /45◦ j1 100 /−180.6◦ 6 /89◦ √ 2 /45◦ j1 100 /−180.6◦ 6 /89◦ ⎡ =⎣ −j1 ⎤ ⎦ × 10−4 ⎤ ⎦ × 10−4 ÷ Δ ⎤ ⎦× 10−4 10−8 × 106.213 /92.64◦ 133.15 /−47.64◦ 94.16 /−2.64◦ 94.16 /86.8◦ 565 /−31.6◦ ⎤ ⎦ 7.13 Obtain TA , TB , TC for the network shown in Fig. R.P. 7.13 and obtain overall T. A B Figure R.P. 7.13 C | 573 574 j Network Theory SOLUTION Using the equation for T-parameters for a T -network Z1 + Z3 ; A= Z3 We have for A For B; For C; Overall T : P B= Z1 Z3 ; Z3 C= 1 ; Z3 2 7 D= 2 3 Z2 + Z3 Z3 7 6 5 5 1 1 5 2 9 54 3 6 6 6 7 TB = 4 5 1 10 6 6 # " 1 0 TC = 1 1 7 TA = 4 T = [TA ][TB ][TC ] = 4:709 15:93 0:962 3:46 This dervation is left as an exercise to the reader. Verification: Using T Δ transformation, that is changing T (3, 4, 6 ) of Fig. R.P. 7.13, in to Δ, Zxz = 13:5 Zxy = 9 Zyz = 18 Figure R.P.7.13(a) Putting the values in the circuit of Fig. R.P. 7.13, we get Figure R.P.7.13(b) Two Port Networks Reducing, the above circuit, we get the circuit shown in Fig. R.P. 7.13c. Figure R.P. 7.13(c) Converting the circuit into T, we get the circuit shown in Fig. R.P. 7.13(d). Now from Fig. R.P. 7.13(d), 3:8564 + 1:0396 = 4:709 1:0396 1:0396(3:8564 + 2:5644) + 3:8564 2:5644 B= 1:0396 = 15:93 Ω 1 1 C= = = 0:962 Zp 1:0396 2:5644 + 1:0396 D= = 3:46 1:0396 A= Figure R.P. 7.13(d) Exercise Problems E.P 7.1 Find the y parameters for the network shown in Fig. E.P. 7.1. Figure E.P. 7.1 Ans: y11 = α + RA + RB 1 ; y12 = ; y21 = RA RB RB (α + RA ) 1 ; y22 = : RA RB RB j 575 576 j E.P Network Theory 7.2 Find the z parameters for the network shown in Fig. E.P. 7.2. Figure E.P. 7.2 Ans: E.P z11 = 13 Ω; 7 z12 = 2 Ω; 7 z21 = 2 Ω; 7 z22 = 3 Ω: 7 7.3 Find the h parameters for the network shown in Fig. E.P. 7.3. Figure E.P. 7.3 Ans: h12 E.P sC) R) R* + R) + (1 sC) R* + 1 sC) R* ; = sC) R* + 1 h11 = m)R* ; h22 7.4 Find the y parameters for the network shown in Fig. E.P. 7.4. Figure E.P. 7.4 Ans: y11 = y22 = 7 S; 15 y12 = y21 = 2 S. 15 sC) R* + m . sC) R* + 1 sC) = . sC) R* + 1 h21 = Two Port Networks E.P j 577 7.5 Find the y parameters for the network shown in Fig. E.P. 7.5. Give the result in s domain. Figure E.P. 7.5 Ans: E.P y11 = y22 = 2s(2s + 1) ; 4s + 1 y12 = y21 = 4s2 . 4s + 1 7.6 Obtain the y parameters for the network shown in Fig. E.P. 7.6. Figure E.P. 7.6 Ans: E.P y11 = 0:625 S; y12 = 0:125 S; y21 = 0:375 S; y22 = 0:125 S: 7.7 Find the z parameters for the two-port network shown in Fig. E.P. 7.7. Keep the result in s domain. Figure E.P. 7.7 Ans: z11 = 2s + 1 ; s z12 = z21 = 2; z22 = 2s + 2 . s 578 j E.P Network Theory 7.8 Find the h parameters for the network shown in Fig. E.P. 7.8. Keep the result in s domain. Figure E.P. 7.8 Ans: h11 = E.P 5s + 4 ; 2(s + 2) h12 = s+4 ; 2(s + 2) h21 = (s + 4) ; 2(s + 2) h22 = s : 2(s + 2) 7.9 Find the transmission parameters for the network shown in Fig. E.P. 7.9. Keep the result in s domain. Figure E.P. 7.9 Ans: A = E.P 3s + 4 ; s+4 B= 2s + 4 ; s+4 C= 4s ; s+4 D= 3s + 4 : s+4 7.10 For the same network described in Fig. E.P. 7.9, find the h parameters using the defining equations. Then verify the result obtained using conversion formulas. 2s + 4 s+4 (s + 4) 4s ; h12 = ; h21 = ; h22 = : Ans: h11 = 3s + 4 3s + 4 3s + 4 3s + 4 Two Port Networks E.P j 579 7.11 Select the values of RA , RB , and RC in the circuit shown in Fig. E.P. 7.11 so that A =1, B = 34 Ω, C = 20 mS and D = 1.4. I1 I2 V1 V2 Figure E.P. 7.11 Ans: E.P R) = 10Ω; R* = 20Ω; R+ = 50Ω: 7.12 Find the s domain expression for the h parameters of the circuit in E.P. 7.12. Figure E.P. 7.12 Ans: E.P h11 = 1 s C s2 + 1 LC ; h12 = h21 1 LC ; = 1 2 s + LC h22 2 Cs + LC = 1 s2 + LC s2 7.13 Find the y parameters for the network shown in Fig. E.P. 7.13. Figure E.P. 7.13 Ans: y11 = 0:04S; y12 = 0:04S; y21 = 0:04S; y22 = 0:03S: : 580 j E.P Network Theory 7.14 Find the two-port parameters h12 and y12 for the network shown in Fig. E.P. 7.14. Figure E.P. 7.14 Ans: h12 = 1:2; E.P y12 = 0:24S: 7.15 Find the ABCD parameters for the 4Ω resistor of Fig. E.P. 7.15. Also show that the ABCD parameters for a single 16Ω resistor can be obtained by (ABCD)4 : Figure E.P. 7.15 Ans: Verify your answer using the relation between the parameters. E.P 7.16 For the T -network shown in Fig. E.P. 7.16. show that AD Figure E.P. 7.16 BC =1. Two Port Networks E.P j 581 7.17 Find y21 for the network shown in Fig. E.P. 7.17. Ans: E.P y21 Figure E.P. 7.17 s = : 4s + 1 7.18 Determine the y-parameters for the network shown in Fig. E.P. 7.18. Figure E.P. 7.18 Ans: E.P y11 = s3 s2 + + 2s + 1 ; s(s2 + 2) y12 = y21 = 1 ; 2 s(s + 2) y22 = s3 + s2 + 2s + 1 : s(s2 + 2) 7.19 Obtain the h-parameters for the network shown in Fig. 7.19. Figure E.P. 7.19 Ans: E.P h11 = 30 Ω; 11 h21 = 1 ; 11 h12 = 1 ; 11 h22 = 4 11 7.20 The following equations are written for a two-port network. Find the transmission parameters for the network. (Hint: use relation between y and T parameters). I1 = 0:05V1 0:4V2 I2 = 0:4V1 + 0:1V2 582 j Network Theory E.P 7.21 Find the network shown in figure, determine the z and y parameters. Figure E.P. 7.21 Ans: y11 = 4 ; z11 = 1Ω; E.P y22 = 3; 4 z22 = Ω; 3 y12 = y21 = 3, z12 = z21 = 1Ω: 7.22 Determine the z, y and Transmission parameters of the network shown in Fig. 7.22. Figure E.P. 7.22 3 , 55 = 20Ω, Ans: y11 = z11 4 , 55 = 15Ω y22 = z12 = z21 z22 A = 55Ω E.P 1 , 55 = 5Ω y12 = y21 = B = 55Ω C = 0.2, D = 3. 7.23 For the network shown in Fig. E.P. 7.23 determine z parameters. Figure E.P. 7.23 Ans: z11 = 2s(5s + 1) 2s 2s3 + 5s2 + 3s + 5 = z = = , z , z 12 21 22 2s2 + 5s + 1 2s2 + 5s + 1 2s2 + 5s + 1 The unit and S are same Two Port Networks E.P 7.24 Determine the y parameters of the two-port network shown in Fig. E.P. 7.24. Figure E.P. 7.24 Ans: y11 = 1 , 4 y21 = 1 , 4 y12 = 5 , 4 y22 = 4 . 3 j 583