Electric Circuits, 9th Edition P36 docx

326 Sinusoidal Steady-State Analysis ^ASSESSMENT PROBLEMS Objective 3—Know how to use circuit analysis techniques to solve a circuit in the frequency domain 9.7 A 20 fl resistor is connected in parallel with a 5 mH inductor. This parallel combination is connected in series with a 5 ft resistor and a 25 ^iF capacitor. a) Calculate the impedance of this inter- connection if the frequency is 2 krad/s. b) Repeat (a) for a frequency of 8 krad/s. c) At what finite frequency does the imped- ance of the interconnection become purely resistive? d) What is the impedance at the frequency found in (c)? NOTE: Also try Chapter Problems 9.28, 9.29, and 9.32. Answer: (a) 9 - /12 fl; (b) 21 + /3 O; (c) 4 krad/s; (d) 15 ft, 9.8 The interconnection described in Assessment Problem 9.7 is connected across the terminals of a voltage source that is generating v = 150 cos 4000f V. What is the maximum amplitude of the current in the 5 mH inductor? Answer: 7.07 A. Figure 9.20 • The delta-to-wye transformation. Delta-to-Wye Transformations The A-to-Y transformation that we discussed in Section 3.7 with regard to resistive circuits also applies to impedances. Figure 9.20 defines the A-connected impedances along with the Y-equivalent circuit. The Y impedances as functions of the A impedances are (9.51) Z 3 z, = z d z. ZbZ c + z b + z; Z C Z, A + z b + z»z b z/ z, + z h + z c (9.52) (9.53) The A-to-Y transformation also may be reversed; that is, we can start with the Y structure and replace it with an equivalent A structure. The A impedances as functions of the Y impedances are Z, = Z\Z 2 ZiZ 2 Z)Z 2 + Z 2 Zi Z\ + Z 2 Z 3 z 2 + z 2 z 3 + Z 3 Zi + z 3 z x + Zj,Z x z, (9.54) (9.55) (9.56) The process used to derive Eqs. 9.51-9.53 or Eqs. 9.54-9.56 is the same as that used to derive the corresponding equations for pure resistive cir- cuits. In fact, comparing Eqs. 3.44-3.46 with Eqs. 9.51-9.53, and Eqs. 3.47-3.49 with Eqs. 9.54-9.56, reveals that the symbol Z has replaced 9.6 Series, Parallel, and Delta-to-Wye Simplifications 327 the symbol R. You may want to review Problem 3.62 concerning the deri- vation of the A-to-Y transformation. Example 9.8 illustrates the usefulness of the A-to-Y transformation in phasor circuit analysis. Example 9.8 Using a Delta-to-Wye Transform in the Frequency Domain Use a A-to-Y impedance transformation to find I () , Ii, I 2 ,13,14,15, V|, and V 2 in the circuit in Fig. 9.21. C_) 12(M£ Figure 9.21 A The circuit for Example 9.8. Solution First note that the circuit is not amenable to series or parallel simplification as it now stands. A A-to-Y impedance transformation allows us to solve for all the branch currents without resorting to either the node-voltage or the mesh-current method. If we replace either the upper delta (abc) or the lower delta (bed) with its Y equivalent, we can further simplify the resulting circuit by series-parallel com- binations. In deciding which delta to replace, the sum of the impedances around each delta is worth checking because this quantity forms the denomi- nator for the equivalent Y impedances. The sum around the lower delta is 30 + /40, so we choose to eliminate it from the circuit. The Y impedance con- necting to terminal b is (21) + /60)(10) Zl = 30 + /40 = 12 + > 4a ' the Y impedance connecting to terminal c is 10(-/20) and the Y impedance connecting to terminal d is (20 + ;60)(-y20) Inserting the Y-equivalent impedances into the cir- cuit, we get the circuit shown in Fig 9.22, which we can now simplify by series-parallel reductions. The impedence of the abn branch is Z ahn = 12 + ./4 - /4 = 12 ft, and the impedance of the acn branch is Z acn = 63.2 + /2.4 - /2.4 - 3.2 = 60 ft. 4ft -/2.4 ft Figure 9.22 • The circuit shown in Fig. 9.21, with the lower delta replaced by its equivalent wye. Note that the abn branch is in parallel with the acn branch. Therefore we may replace these two branches with a single branch having an impedance of •^an (60)(12) 72 10 ft. Combining this 10 ft resistor with the impedance between n and d reduces the circuit shown in Fig. 9.22 to the one shown in Fig. 9.23. From the latter circuit. 120/0° h = 7Z ~. =4/53.13° 1 18 - /24 l 2.4 + /3.2 A. Z 3 = 30 + /40 = 8 - /24 ft. Once we know I () , we can work back through the equivalent circuits to find the branch currents in the original circuit. We begin by noting that I 0 is the current in the branch nd of Fig. 9.22. Therefore V nd = (8 - /24)1« = 96 - /32 V. 328 Sinusoidal Steady-State Analysis We may now calculate the voltage V an because * T an ' T nd and both V and V nd are known. Thus V an = 120 - 96 + /32 = 24 + /32 V. We now compute the branch currents I abn and I acn : 24 + /32 .8 A Iabn = J^ = 2 + J ~ A, 24 + /32 _ _4_ _8_ 60 " ~ 10 + ; 15 A " In terms of the branch currents defined in Fig. 9.21, Ii = Iabn = 2 + /3 A ' i_ •_§_ 10 +7 15 h = Iacn = T7T + 777 A « We check the calculations of Ii and I? by noting that I, + I 2 = 2.4 + /3.2 = I 0 . •V a 120/Q!/ + V i8n -/2412 Figure 9.23 • A simplified version of the circuit shown in Fig. 9.22. To find the branch currents I 3 , I 4 , and I 5 , we must first calculate the voltages V! and V 2 . Refering to Fig. 9.21, we note that 328 V t = 120/CT - (-/4)1, = — + /8 V, V 2 = 120/0° - (63.2 + /2.4)I 2 = 96 - j~ V. We now calculate the branch currents I 3 ,1 4 , and I 5 ; V L - V 2 4 . 12.8 13 = -^- = 3 +i— A, l4 = 2^T7rS = 3" 7l - 6A ' ^-g**"- We check the calculations by noting that I 4 + Is = 3 + jg - /1-6 + /4.8 = 2.4 + /3.2 = I 0 , 4 2 12.8 M , „ .8 w I3 + I 4 = 3 + 3 + ;— - ;1.6 = 2 + 7- = l h 4 4 12.8 8 26 •/ASSESSMENT PROBLEM Objective 3—Know how to use circuit analysis techniques to solve a circuit in the frequency domain 9.9 Use a A -to-Y transformation to find the current I in the circuit shown. 1 ^ I Answer: I = 4 /28.07° A. 136zD!/+^ 1411 /40 fl ^ -/15 H ion NOTE: Also try Chapter Problem 9.37. 9.7 Source Transformations and Thevenin-Norton Equivalent Circuits 329 9 J Source Transformations and Thevenin-Norton Equivalent Circuits The source transformations introduced in Section 4.9 and the Thevenin- Norton equivalent circuits discussed in Section 4.10 are analytical tech- niques that also can be applied to frequency-domain circuits. We prove the validity of these techniques by following the same process used in Sections 4.9 and 4.10, except that we substitute impedance (Z) for resist- ance (/?). Figure 9.24 shows a source-transformation equivalent circuit with the nomenclature of the frequency domain. Figure 9.25 illustrates the frequency-domain version of a Thevenin equivalent circuit. Figure 9.26 shows the frequency-domain equivalent of a Norton equivalent circuit. The techniques for finding the Thevenin equivalent voltage and impedance are identical to those used for resistive circuits, except that the frequency-domain equivalent circuit involves the manipulation of complex quantities. The same holds for finding the Norton equivalent current and impedance. Example 9.9 demonstrates the application of the source-transformation equivalent circuit to frequency-domain analysis. Example 9.10 illustrates the details of finding a Thevenin equivalent circuit in the frequency domain. Z s v,/z. Figure 9.24 A A source transformation in the frequency domain. Frequency-domain linear circuit; may contain both independent and dependent sources. •" Figure 9.25 • The frequency-domain version of a Thevenin equivalent circuit. Frequency-domain linear circuit; may contain both independent and dependent , sources. * b Figure 9.26 • The frequency-domain version of a Norton equivalent circuit. Example 9.9 Performing Source Transformations in the Frequency Domain Use the concept of source transformation to find the phasor voltage V 0 in the circuit shown in Fig. 9.27. in y'3 0 0.2 a ;'°- 6a I "WV rv-w>—^_/yy\, ry\~r\ ^ /^ + \40ZQ! Figure 9.27 A The circuit for Example 9.9. Solution We can replace the series combination of the voltage source (40/0°) and the impedance of 1 +/3Q with the parallel combination of a current source and the 1 + /3 ft impedance. The source current is I = TTp = To (1 - ;3) /12 A. Thus we can modify the circuit shown in Fig. 9.27 to the one shown in Fig. 9.28. Note that the polarity reference of the 40 V source determines the refer- ence direction for I. Next, we combine the two parallel branches into a single impedance, Z = (1 + /3)(9 - /3) 10 = 1.8 + /2.4 a, 330 Sinusoidal Steady-State Analysis which is in parallel with the current source of 4 - y*12 A. Another source transformation con- verts this parallel combination to a series combina- tion consisting of a voltage source in series with the impedance of 1.8 + /2.4 ft. The voltage of the volt- age source is V = (4 - /12)(1.8 + /2.4) = 36 - /12 V. Using this source transformation, we redraw the circuit as Fig. 9.29. Note the polarity of the voltage source. We added the current Io to the circuit to expedite the solution for VQ. 0.2 0, /0.6 H >VvV '^YYV Figure 9.28 • The first step in reducing the circuit shown in Fig. 9.27. 1.8 n /2.4 n o.2 o /o.6 n l /\/W orw-\ /y^ OTTTL c Figure 9.29 • The second step in reducing the circuit shown in Fig. 9.27. Also note that we have reduced the circuit to a simple series circuit. We calculate the current Io by dividing the voltage of the source by the total series impedance: = 36 - /12 = 12(3 - /1) 0 12 - /16 ~ 4(3 - /4) 39 + /27 25 1.56 +/1.08 A. We now obtain the value of V 0 by multiplying I 0 by the impedance 10 - /19: V fl = (1.56 + /1.08)(10 - /19) = 36.12 - /18.84 V. Example 9.10 Finding a Thevenin Equivalent in the Frequency Domain Find the Thevenin equivalent circuit with respect to terminals a,b for the circuit shown in Fig. 9.30. 12 H -/40 a no a 120ffi(^ v J 6Qn ^ 10Vj Figure 9.30 • The circuit for Example 9.10. Solution We first determine the Thevenin equivalent voltage. This voltage is the open-circuit voltage appearing at terminals a,b. We choose the reference for the Thevenin voltage as positive at terminal a. We can make two source transformations relative to the 120 V, 12 CI, and 60 CI circuit elements to simplify this portion of the circuit. At the same time, these transfor- mations must preserve the identity of the controlling voltage Vj because of the dependent voltage source. We determine the two source transformations by first replacing the series combination of the 120 V source and 12 CI resistor with a 10 A current source in parallel with 12 CI. Next, we replace the parallel combination of the 12 and 60 ft resistors with a single 10 ft resistor. Finally, we replace the 10 A source in parallel with 10 ft with a 100 V source in series with 10 ft. Figure 9.31 shows the resulting circuit. We added the current I to Fig. 9.31 to aid fur- ther discussion. Note that once we know the current 9.7 Source Transformations and Thevenin-Norton Equivalent Circuits 331 I, we can compute the Thevenin voltage. We find I by summing the voltages around the closed path in the circuit shown in Fig. 9.31. Hence 100 = 101 - /401 + 1201 + 10V t = (130 - /40)1 + 10V V . -/40 ft 100/0!' V I 120 ft 'WV- 10 V v -•a + v- ni Figure 9.32 A A circuit for calculating the Thevenin equivalent impedance. Figure 9.31 • A simplified version of the circuit shown in Fig. 9.30. We relate the controlling voltage V v to the current I by noting from Fig. 9.31 that 'Hi en, V v = 100 - 101. = 18/-126.87° A. 30 - /40 we now calculate V x : V v = 100 - 180/-126.87° = 208 + /144 V. Finally, we note from Fig. 9.31 that V-rh = 10¼ + 1201 = 2080 + /1440 + 120(18)/-126.87° = 784 - /288 = 835.22/-20.17° V. To obtain the Thevenin impedance, we may use any of the techniques previously used to find the Thevenin resistance. We illustrate the test- source method in this example. Recall that in using this method, we deactivate all independent sources from the circuit and then apply either a test voltage source or a test current source to the terminals of interest. The ratio of the voltage to the current at the source is the Thevenin imped- ance. Figure 9.32 shows the result of applying this technique to the circuit shown in Fig. 9.30. Note that we chose a test voltage source \ T . Also note that we deactivated the independent voltage source with an appropriate short-circuit and pre- served the identity of V v . The branch currents I a and I b have been added to the circuit to simplify the calculation of I T . By straightforward applications of Kirchhoff s circuit laws, you should be able to verify the following relationships: Ih = 10 - /40 v r - iov A 120 _ -V 7 (9 + /4) ' 120(1 - /4) * IT = la + lb ^ V, = 10I a , V r f 9 + /4 10 - /40 V 12 VH3-/4) ' 12(10-/40) 1 Z-n, = T 1 = 91.2 -/38.411. If Figure 9.33 depicts the Thevenin equivalent circuit. 784-/288/ + V 91.2 ft /yyV— -/38.4X2 + Figure 9.33 • The Thevenin equivalent for the circuit shown in Fig. 9.30. 332 Sinusoidal Steady-State Analysis /ASSESS MEN Objective 3—Know how to use circuit analysis techniques to solve a circuit in the frequency domain 9.10 Find the steady-state expression for v 0 {t) in the circuit shown by using the technique of source transformations. The sinusoidal voltage sources are v x = 240 cos (4000/ + 53.13°) V, th = 96 sin 4000/ V. 20 H Answer: 48 cos (4000/ + 36.87°) V NOTE: Also try Chapter Problems 9.45, 9.46, and 9.49. 9.11 Find the Thevenin equivalent with respect to terminals a,b in the circuit shown. Answer: V Th = V ab = 10/45^ V; Z Th = 5 - /5 H. 9.8 The Node-Voltage Method In Sections 4.2-4.4, we introduced the basic concepts of the node-voltage method of circuit analysis. The same concepts apply when we use the node-voltage method to analyze frequency-domain circuits. Example 9.11 illustrates the solution of such a circuit by the node-voltage technique. Assessment Problem 9.12 and many of the Chapter Problems give you an opportunity to use the node-voltage method to solve for steady-state sinu- soidal responses. Example 9.11 Using the Node-Voltage Method in the Frequency Domain Use the node-voltage method to find the branch currents I a , I b , and I c in the circuit shown in Fig. 9.34. Figure 9.34 • The circuit for Example 9.11. Solution We can describe the circuit in terms of two node volt- ages because it contains three essential nodes. Four branches terminate at the essential node that stretches across the bottom of Fig. 9.34, so we use it as the refer- ence node. The remaining two essential nodes are labeled 1 and 2, and the appropriate node voltages are designated V] and V 2 . Figure 9.35 reflects the choice of reference node and the terminal labels. 10.6/0! A 1 lft /2 ft 2 5 a f WV rwy~\—« ,yvV- Mj 100 Figure 9.35 A The circuit shown in Fig. 9.34, with the node voltages defined. Summing the currents away from node 1 yields Vi Vi - V 7 -10.6 + 777 + -r ~ = °- 10 1 + j2 Multiplying by 1 + /2 and collecting the coeffi- cients of V] and V 2 generates the expression V^l.l + /0.2) - V 2 = 10.6 + /21.2. 9.9 The Mesh-Current Method 333 Summing the currents away from node 2 gives V2-V1 , V 2 , V 2 ~ 201 + 1 = U. 1 + /2 -/5 5 The controlling current I, is I, 1 + /2 Substituting this expression for l x into the node 2 equation, multiplying by 1 + /2, and collecting coefficients of V, and V 2 produces the equation -5V t + (4.8 + /0.6) V 2 = 0. The solutions for V^ and V 2 are V! = 68.40 -/16.80 V, V 2 = 68 - /26 V. Hence the branch currents are I a = ^ = 6.84-/1.68 A, Vi - V 2 1 +/2 = 3.76 + /1.68 A, V 2 - 20I X I b = -*— = -1.44 - /11.92 A, V, I c = -^+ = 5.2 + /13.6 A. -;5 To check our work, we note that I a + I, = 6.84 - /1.68 + 3.76 + /1.68 = 10.6 A, I v = I b + I c = -1.44 - /11.92 + 5.2 + /13.6 = 3.76 + /1.68 A. ^ASSESSMENT PROBLEM Objective 3—Know how to use circuit analysis techniques to solve a circuit in the frequency domain 9.12 Use the node-voltage method to find the steady- state expression for v(t) in the circuit shown. The sinusoidal sources are i s = 10 cos cot A and v s = 100 sin cot V, where to - 50 krad/s. Answer: v(t) = 31.62 cos (50,000r - 71.57°) V. NOTE: Also try Chapter Problems 9.55 and 9.59. 20 a 9.9 The Mesh-Current Method We can also use the mesh-current method to analyze frequency-domain cir- cuits. The procedures used in frequency-domain applications are the same as those used in analyzing resistive circuits. In Sections 4.5-4.7, we intro- duced the basic techniques of the mesh-current method; we demonstrate the extension of this method to frequency-domain circuits in Example 9.12. Example 9.12 Using the Mesh-Current Method in the Frequency Domain Use the mesh-current method to find the voltages Vj, V 2 ,and V 3 in the circuit shown in Fig. 9.36 on the next page. Solution The circuit has two meshes and a dependent volt- age source, so we must write two mesh-current equations and a constraint equation. The reference direction for the mesh currents I] and I 2 is clock- wise, as shown in Fig. 9.37. Once we know ^ and I 2 , we can easily find the unknown voltages. Summing the voltages around mesh 1 gives 150 = (1 +/2)^ + (12-/16)(1, -I 2 ), or 150 = (13 - /14)¾ - (12 - /16)I 2 . Summing the voltages around mesh 2 generates the equation 0 = (12 - /16)(¾ - 10 + (1+ /3)¾ + 391,. Figure 9.37 reveals that the controlling current I, is the difference between I, and I 2 ; that is, the con- straint is Ir = It - h 334 Sinusoidal Steady-State Analysis + V, - + V, I VvV L_rY-w-\ m /y^ L_TY>nr\_ ui yzn+T m /3 a 120¾ O'T v = -;'16ft: 39 I A Figure 9.36 • The circuit for Example 9.12. 1 fl /2 A 1 (1 /3 £1 i <wv o"v>-> • »yy^ orw>_ Solving for I] and I 2 yields I, = -26- /52 A, I 2 = -24 - /58 A, I v = -2 + /6 A. The three voltages are Vj = (1 + /2)1 1 = 78 -/104 V, V 2 = (12 - /16)1, = 72 + /104 V, V 3 = (1 + /3)I 2 = 150 - /130 V. Also 391 v = -78 + /234 V. Figure 9.37 • Mesh currents used to solve the circuit shown in Fig. 9.36. Substituting this constraint into the mesh 2 equa- tion and simplifying the resulting expression gives 0 = (27 + /16)¾ - (26 + /13)¾. We check these calculations by summing the volt- ages around closed paths: -150 + Vj + V 2 = -150 + 78 - /104 + 72 + /104 = 0, -V 2 + V 3 + 39¾ = -72 - /104 + 150 - /130 - 78 + /234 = 0, -150 + V! + V 3 + 39¾. = -150 + 78 - /104 + 150 - /130 - 78 + /234 = 0. ^ASSESSMENT PROBLEM Objective 3—Know how to use circuit analysis techniques to solve a circuit in the frequency domain 9.13 Use the mesh-current method to find the pha- sor current I in the circuit shown. Answer: I = 29 + /2 = 29.07/3.95° A. NOTE: Also try Chapter Problems 9.60 and 9.64. /^+^33.8/0° 9.10 The Transformer A transformer is a device that is based on magnetic coupling. Transformers are used in both communication and power circuits. In communication cir- cuits, the transformer is used to match impedances and eliminate dc signals from portions of the system. In power circuits, transformers are used to estab- lish ac voltage levels that facilitate the transmission, distribution, and con- sumption of electrical power. A knowledge of the sinusoidal steady-state 9.10 The Transformer 335 behavior of the transformer is required in the analysis of both communication and power systems. In this section, we will discuss the sinusoidal steady-state behavior of the linear transformer, which is found primarily in communica- tion circuits. In Section 9.11, we will deal with the ideal transformer, which is used to model the ferromagnetic transformer found in power systems. Before starting we make a useful observation. When analyzing circuits containing mutual inductance use the meshor loop-current method for writ- ing circuit equations. The node-voltage method is cumbersome to use when mutual inductance in involved. This is because the currents in the various coils cannot be written by inspection as functions of the node voltages. The Analysis of a Linear Transformer Circuit A simple transformer is formed when two coils are wound on a single core to ensure magnetic coupling. Figure 9.38 shows the frequency-domain cir- cuit model of a system that uses a transformer to connect a load to a source. In discussing this circuit, we refer to the transformer winding con- nected to the source as the primary winding and the winding connected to the load as the secondary winding. Based on this terminology, the trans- former circuit parameters are jR] = the resistance of the primary winding, R 2 = the resistance of the secondary winding, Lj = the self-inductance of the primary winding, L 2 - the self-inductance of the secondary winding. M = the mutual inductance. The internal voltage of the sinusoidal source is V v , and the internal impedance of the source is Z s . The impedance Z L represents the load con- nected to the secondary winding of the transformer. The phasor currents Ij and I 2 represent the primary and secondary currents of the transformer, respectively. Analysis of the circuit in Fig. 9.38 consists of finding I, and I 2 as func- tions of the circuit parameters V v , Z s , R h L t , L 2 , R 2 , M, Z L , and OJ. We are also interested in finding the impedance seen looking into the transformer from the terminals a,b.To find l\ and I 2 , we first write the two mesh-cur- rent equations that describe the circuit: Source Transformer Load Figure 9.38 A The frequency domain circuit model for a transformer used to connect a load to a source. V v = (Z v + Ri + jo)L l )l l - ja>Ml 2 , (9.57) 0 = -;©Afl| + (R 2 + joiL 2 4- Z L )I 2 . (9.58) To facilitate the algebraic manipulation of Eqs. 9.57 and 9.58, we let Z,, = Z, + R x + jtoL u (9.59) Z 22 = R 2 + jcoL 2 + ZJL, (9.60) where Z (1 is the total self-impedance of the mesh containing the primary winding of the transformer, and Z 22 is the total self-impedance of the mesh containing the secondary winding. Based on the notation introduced in Eqs. 9.59 and 9.60, the solutions for Ij and I 2 from Eqs. 9.57 and 9.58 are I. = Z ]] Z 22 4- o)~M V ~> * 17 x ' (9.61) JOjM Z U Z 22 + w'M io)M -V = - 1, (9.62) '22 . portions of the system. In power circuits, transformers are used to estab- lish ac voltage levels that facilitate the transmission, distribution, and con- sumption of electrical power. A knowledge. finding the Thevenin equivalent voltage and impedance are identical to those used for resistive circuits, except that the frequency-domain equivalent circuit involves the manipulation of complex