Electric Circuits, 9th Edition P35 pptx

316 Sinusoidal Steady-State Analysis Thus the phasor representation is the sum of the phasors of the individual terms. We discuss the development of Eq. 9.24 in Section 9.5. Before applying the phasor transform to circuit analysis, we illus- trate its usefulness in solving a problem with which you are already familiar: adding sinusoidal functions via trigonometric identities. Example 9.5 shows how the phasor transform greatly simplifies this type of problem. Example 9.5 If yi = 20 cos express y = Adding Cosines Using Phasors , (cot - 30°) and y 2 = 40 cos (oot + 60°), y { + y 2 as a single sinusoidal function. a) Solve by using trigonometric identities. b) Solve by using the phasor concept. Solution a) First we expand both y 1 and y 2 , using the cosine of the sum of two angles, to get >'! = 20 cos cot cos 30° + 20 sin w/ sin 30°; y 2 = 40 cos a>r cos 60° - 40 sin cot sin 60°. Adding y\ and y 2 , we obtain y = (20 cos 30 + 40 cos 60) cos cot + (20 sin 30 - 40 sin 60) sin cot — 37.32 cos cot - 24.64 sin cot. To combine these two terms we treat the co-efficients of the cosine and sine as sides of a right triangle (Fig. 9.6) and then multiply and divide the right-hand side by the hypotenuse. Our expression for y becomes ^/37.32 24.64 . \ y = 44.72 _ A „„ cos cot AAmn sm (at \ 44.72 44.72 ) = 44.72( cos 33.43° cos cot - sin 33.43° sin cot). Again, we invoke the identity involving the cosine of the sum of two angles and write y = 44.72 cos (cot + 33.43°). 44.72// /33.43° / * 37.32 24.64 Figure 9.6 • A right triangle used in the solution for y. b) We can solve the problem by using phasors as follows: Because y = yi + y 2 , then, from Eq. 9.24, Y = Y, + Y 2 = 20/-30° + 40/60° = (17.32 - /10) + (20 + /34.64) = 37.32 + /24.64 = 44.72/33.43°. Once we know the phasor Y, we can write the corresponding trigonometric function for y by taking the inverse phasor transform: y = ^-^44.726^ 3343 } = ^ {44.72^ 3343 ^} = 44.72 cos (cot + 33.43°). The superiority of the phasor approach for adding sinusoidal functions should be apparent. Note that it requires the ability to move back and forth between the polar and rectangular forms of complex numbers. 9.4 The Passive Circuit Elements in the Frequency Domain 317 /"ASSESSMENT PROBLEMS Objective 1—Understand phasor concepts and be able to perform a phasor transform and an inverse phasor transform 9.1 Find the phasor transform of each trigonomet- ric function: a) v = 170cos(377r - 40°) V. b) i = 10 sin (1000/ 1 + 20°) A. c) i = [5 cos (art + 36.87°) + 10cos(art -53.13°)] A. d) v = [300cos(20,00077t + 45°) - 100 sin(20,0007rr + 30°)] mV. Answer: (a) 170/-40° V; (b) 10/-70° A; NOTE: Also try Chapter Problem 9.11. (c) 11.18/-26.57° A; (d) 339.90/61.51° mV, 9.2 Find the time-domain expression correspon- ding to each phasor: a) V = 18.6/-54° V. b) I = (20/45^ - 50/-30°) mA. c) V = (20 + /80 - 30/15°) V. Answer: (a) 18.6 cos (cot - 54°) V; (b) 48.81 cos (cot + 126.68°) mA; (c) 72.79 cos (cot + 97.08°) V. 9.4 The Passive Circuit Elements in the Frequency Domain The systematic application of the phasor transform in circuit analysis requires two steps. First, we must establish the relationship between the phasor current and the phasor voltage at the terminals of the passive cir- cuit elements. Second, we must develop the phasor-domain version of Kirchhoff s laws, which we discuss in Section 9.5. In this section, we estab- lish the relationship between the phasor current and voltage at the termi- nals of the resistor, inductor, and capacitor. We begin with the resistor and use the passive sign convention in all the derivations. The V-I Relationship for a Resistor From Ohm's law, if the current in a resistor varies sinusoidally with time — that is, if i = I m cos (cot + 0,) —the voltage at the terminals of the resistor, as shown in Fig. 9.7, is v = R[I m cos (cot + Of)] = RI m [ cos (cot + Si)], (9.25) where /,„ is the maximum amplitude of the current in amperes and 0,- is the phase angle of the current. The phasor transform of this voltage is Figure 9.7 • A resistive element carrying a sinusoidal current. V = RI m e^ = Rln/Oi. (9.26) But I m /0j is the phasor representation of the sinusoidal current, so we can write Eq. 9.26 as V = Rl, (9.27) -4 Relationship between phasor voltage and phasor current for a resistor 318 Sinusoidal Steady-State Analysis Figure 9.8 • The frequency-domain equivalent circuit of a resistor. V,l V / i \ 1 I -r/4 0 \ i/ \ T//2 / \if V fi \\ T\ 1 / 37/2/ V I '\ IT Figure 9.9 A A plot showing that the voltage and cur- rent at the terminals of a resistor are in phase. which states that the phasor voltage at the terminals of a resistor is simply the resistance times the phasor current. Figure 9.8 shows the circuit dia- gram for a resistor in the frequency domain. Equations 9.25 and 9.27 both contain another important piece of information—namely, that at the terminals of a resistor, there is no phase shift between the current and voltage. Figure 9.9 depicts this phase rela- tionship, where the phase angle of both the voltage and the current wave- forms is 60°. The signals are said to be in phase because they both reach corresponding values on their respective curves at the same time (for example, they are at their positive maxima at the same instant). 2T The V-I Relationship for an Inductor We derive the relationship between the phasor current and phasor voltage at the terminals of an inductor by assuming a sinusoidal current and using Ldi/dt to establish the corresponding voltage. Thus, for /' = /,„ cos (cot + 0,-), the expression for the voltage is v — L— = -coLI in sin (cot + fy). at (9.28) We now rewrite Eq. 9.28 using the cosine function: v = -coLI m cos (cot + 0, - 90°). (9.29) The phasor representation of the voltage given by Eq. 9.29 is V = -coLlJ^-^ Relationship between phasor voltage and • phasor current for an inductor = -<oLI,J e <e-J 90 ° = jtoLlJ 6 ' (9.30) = jcoLl. Note that in deriving Eq. 9.30 we used the identity -^ = cos90° - /sin90° = -j. Figure 9.10 • The frequency-domain equivalent circuit for an inductor. Equation 9.30 states that the phasor voltage at the terminals of an inductor equals jtoL times the phasor current. Figure 9.10 shows the frequency- domain equivalent circuit for the inductor. It is important to note that the relationship between phasor voltage and phasor current for an inductor applies as well for the mutual inductance in one coil due to current flowing in another mutually coupled coil. That is, the phasor voltage at the termi- nals of one coil in a mutually coupled pair of coils equals jcoM times the phasor current in the other coil. 9.4 The Passive Circuit Elements in the Frequency Domain 319 We can rewrite Eq. 9.30 as V = (0)/,/90-)/,,,/0,- = a>LI m /(0j + 90)°, (9.31) which indicates that the voltage and current are out of phase by exactly 90°. In particular, the voltage leads the current by 90°, or, equivalently, the current lags behind the voltage by 90°. Figure 9.11 illustrates this concept of voltage leading current or current lagging voltage. For example, the volt- age reaches its negative peak exactly 90° before the current reaches its negative peak. The same observation can be made with respect to the zero-going-positive crossing or the positive peak. We can also express the phase shift in seconds. A phase shift of 90° corresponds to one-fourth of a period; hence the voltage leads the current by T/4, or^y second. The V-I Relationship for a Capacitor We obtain the relationship between the phasor current and phasor voltage at the terminals of a capacitor from the derivation of Eq. 9.30. In other words, if we note that for a capacitor that v. i Figure 9.11 • A plot showing the phase relationship between the current and voltage at the terminals of an inductor (0,- = 60"). i = C Mv df and assume that v = V,,, cos (to/ + (9,,), then I = jwC\. (9.32) Now if we solve Eq. 9.32 for the voltage as a function of the current, we get V = j W C (9.33) A Relationship between phasor voltage and phasor current for a capacitor Equation 9.33 demonstrates that the equivalent circuit for the capacitor in the phasor domain is as shown in Fig. 9.12. The voltage across the terminals of a capacitor lags behind the current by exactly 90°. We can easily show this relationship by rewriting Eq. 9.33 as 1 v = ^lz*Lh„M. 1/jioC + V I Figure 9.12 • The frequency domain equivalent circuit of a capacitor. coC m - 9oy (9.34) 320 Sinusoidal Steady-State Analysis Figure 9.13 • A plot showing the phase relationship between the current and voltage at the terminals of a capacitor (0, = 60°). The alternative way to express the phase relationship contained in Eq. 9.34 is to say that the current leads the voltage by 90°. Figure 9.13 shows the phase relationship between the current and voltage at the ter- minals of a capacitor. Impedance and Reactance We conclude this discussion of passive circuit elements in the frequency domain with an important observation. When we compare Eqs. 9.27,9.30, and 9.33, we note that they are all of the form Definition of impedance • V = ZI, (9.35) TABLE 9.1 Impedance and Reactance Values Circuit Element Resistor Inductor Capacitor Impedance R j(oL K-l/wQ Reactance coL -1/taC where Z represents the impedance of the circuit element. Solving for Z in Eq. 9.35, you can see that impedance is the ratio of a circuit element's volt- age phasor to its current phasor. Thus the impedance of a resistor is R, the impedance of an inductor is jcoL, the impedance of mutual inductance is jcoM, and the impedance of a capacitor is 1/ytuC. In all cases, impedance is measured in ohms. Note that, although impedance is a complex number, it is not a phasor. Remember, a phasor is a complex number that shows up as the coefficient of e j<ot . Thus, although all phasors are complex numbers, not all complex numbers are phasors. Impedance in the frequency domain is the quantity analogous to resistance, inductance, and capacitance in the time domain. The imaginary part of the impedance is called reactance. The values of impedance and reactance for each of the component values are summarized in Table 9.1. And finally, a reminder. If the reference direction for the current in a passive circuit element is in the direction of the voltage rise across the ele- ment, you must insert a minus sign into the equation that relates the volt- age to the current. ^/ASSESSMENT PROBLEMS Objective 2—Be able to transform a circuit with a sinusoidal source into the frequency domain using phasor concepts 9.3 The current in the 20 mH inductor is 10 cos (10,000? + 30°) mA. Calculate (a) the inductive reactance; (b) the impedance of the inductor; (c) the phasor voltage V; and (d) the steady-state expression for v(t). 20 mH v - » i 9.4 The voltage across the terminals of the 5 /iF capacitor is 30 cos (4000r + 25°) V. Calculate (a) the capacitive reactance; (b) the impedance of the capacitor; (c) the phasor current I; and (d) the steady-state expression for i(t). 5/xF —1(— Answer: (a) 200 ft; (b)/200 O; (c) 2/120° V; (d) 2 cos (10,000? + 120°) V. Answer: (a) -50 O; (b) -/50 O; (c) 0.6/115° A; (d) 0.6 cos (4000/ + 115°) A. NOTE: Also try Chapter Problems 9.13 and 9.14. 9.5 Kirchhoffs Laws in the Frequency Domain 321 9.5 Kirchhoffs Laws in the Frequency Domain We pointed out in Section 9.3, with reference to Eqs. 9.23 and 9.24, that the phasor transform is useful in circuit analysis because it applies to the sum of sinusoidal functions. We illustrated this usefulness in Example 9.5. We now formalize this observation by developing Kirchhoffs laws in the fre- quency domain. Kirchhoffs Voltage Law in the Frequency Domain We begin by assuming that v x — v n represent voltages around a closed path in a circuit. We also assume that the circuit is operating in a sinusoidal steady state. Thus Kirchhoffs voltage law requires that v, + v 2 + ••• + v„ = 0, (9.36) which in the sinusoidal steady state becomes complex V my cos (mt + 6]) + V im cos (a>t + 0 2 ) + • • • + V;„ w cos (tat + $„) = 0. (9.37) We now use Euler's identity to write Eq. 9.37 as ^{V llh e j0 ^ wl } + M{V mi e>°ieJ°>'} + ••• + »{V W- C / V*} (9.38) which we rewrite as %t{V m e^e>** + V tlu e> (h ei 10 ' + ••• + K„,/V w } = 0. (9.39) Factoring the term e? 0 * from each term yields *t{{V m /* + V,n/ h + ••• + Vn^y 0 "} = 0, or &{(Vi + V 2 + ••• + V„)6>'} = 0. (9.40) But e Ju)t * 0, so Yi + V 2 + • • • + \ a = 0, (9.41) < KVL in the frequency domain which is the statement of Kirchhoffs voltage law as it applies to phasor voltages. In other words, Eq. 9.36 applies to a set of sinusoidal voltages in the time domain, and Eq. 9.41 is the equivalent statement in the fre- quency domain. Kirchhoffs Current Law in the Frequency Domain A similar derivation applies to a set of sinusoidal currents. Thus if ij + / 2 + • • • + /„ = 0, (9.42) 322 Sinusoidal Steady-State Analysis then KCL in the frequency domain • Ii + I? + + 1. (9.43) where l h I 2 , • • •, I„ are the phasor representations of the individual cur- rents ii, / 2 , • • •, i n . Equations 9.35, 9.41, and 9.43 form the basis for circuit analysis in the frequency domain. Note that Eq. 9.35 has the same algebraic form as Ohm's law, and that Eqs. 9.41 and 9.43 state Kirchhoff s laws for phasor quantities. Therefore you may use all the techniques developed for analyzing resistive circuits to find phasor currents and voltages. You need learn no new analytic techniques; the basic circuit analysis and simplification tools covered in Chapters 2-4 can all be used to analyze circuits in the frequency domain. Phasor circuit analysis consists of two fundamental tasks: (1) You must be able to construct the frequency-domain model of a circuit; and (2) you must be able to manipulate complex numbers and/or quantities algebraically. We illustrate these aspects of phasor analysis in the discussion that follows, beginning with series, parallel, and delta-to-wye simplifications. /"ASSESSMENT PROBLEM Objective 3—Know how to use circuit analysis techniques to solve a circuit in the frequency domain 9.5 Four branches terminate at a common node. The reference direction of each branch current (*j, i 2 , /3, and / 4 ) is toward the node. If NOTE: Also try Chapter Problem 9.15. i x = 100 cos (G>* + 25°) A, i 2 = 100cos(o)f + 145°) A, and 13 = 100 cos (eat - 95°) A, find/ 4 . Answer: i 4 = 0. 9.6 Series, Parallel, and Delta-to-Wye Simplifications The rules for combining impedances in series or parallel and for making delta-to-wye transformations are the same as those for resistors. The only difference is that combining impedances involves the algebraic manipula- tion of complex numbers. + m Z, z 2 I z„ Figure 9.14 • Impedances in series. Combining Impedances in Series and Parallel Impedances in series can be combined into a single impedance by simply adding the individual impedances.The circuit shown in Fig. 9.14 defines the problem in general terms. The impedances Zj, Z 2 , • • •, Z n are connected in series between terminals a,b. When impedances are in series, they carry the same phasor current I. From Eq. 9.35, the voltage drop across each imped- ance is Z]l, Z 2 I, • • •, Z,J, and from Kirchhoff s voltage law, ah Z,I + Z 2 I + • • • + Z n l (Z t + Z 2 + • • • + Z„)I. (9.44) The equivalent impedance between terminals a,b is V ab Z ab = -y = Z { + Z 2 + Example 9.6 illustrates a numerical application of Eq. 9.45. + Z, (9.45) 9.6 Series, Parallel, and Delta-to-Wye Simplifications 323 Example 9.6 Combining Impedances in Series A 90 ft resistor, a 32 mH inductor, and a 5 /xF capacitor are connected in series across the termi- nals of a sinusoidal voltage source, as shown in Fig. 9.15. The steady-state expression for the source voltage v s is 750 cos (5000/ + 30°) V. a) Construct the frequency-domain equivalent circuit. b) Calculate the steady-state current / by the phasor method. The phasor transform of v s is V, = 750 /30° V. Figure 9.16 illustrates the frequency-domain equivalent circuit of the circuit shown in Fig. 9.15. b) We compute the phasor current simply by divid- ing the voltage of the voltage source by the equiv- alent impedance between the terminals a,b. From Eq. 9.45, 90 fi 32 mH 5/iF Figure 9.15 • The circuit for Example 9.6. Solution a) From the expression for v s , we have ay = 5000 rad/s. Therefore the impedance of the 32 mH inductor is Z L = ju>L = /(5000)(32 X 10~ 3 ) = /160 ft, and the impedance of the capacitor is Zab = 90 + /160- /40 = 90 + /120 = 150/53.13° ft. Thus 750/30° We may now write the steady-state expression for / directly: i = 5cos(5000r - 23.13°) A. 750/30°/ V -/40 n -1 10 6 Figure 9.16 A The frequency-domain equivalent circuit of the circuit shown in Fig. 9.15. ^ASSESSMENT PROBLEM Objective 3—Know how to use circuit analysis techniques to solve a circuit in the frequency domain 9.6 Using the values of resistance and inductance in the circuit of Fig. 9.15, let Y s = 125 /-60° V and a) = 5000 rad/s. Find a) the value of capacitance that yields a steady-state output current i with a phase angle of -105°. b) the magnitude of the steady-state output current f. Answer: (a) 2.86 fiF; (b) 0.982 A. NOTE: Also try Chapter Problem 9.24. 324 Sinusoidal Steady-State Analysis Impedances connected in parallel may be reduced to a single equiva- lent impedance by the reciprocal relationship 1 Z„b 1 1 + + Z v Z 2 + (9.46) Zah V I, j Z : Figure 9.17 A Impedances in parallel. Figure 9.17 depicts the parallel connection of impedances. Note that when impedances are in parallel, they have the same voltage across their termi- nals. We derive Eq. 9.46 directly from Fig. 9.17 by simply combining Kirchhoffs current law with the phasor-domain version of Ohm's law, that is, Eq. 9.35. From Fig. 9.17, I = Ii + h + + l ir or V V V V — = — + — + • • • + —. Z a b Z[ Z 2 Z n (9.47) Canceling the common voltage term out of Eq. 9.47 reveals Eq. 9.46. From Eq. 9.46, for the special case of just two impedances in parallel, Z ab = Z\Z 1^2 Zi + Z, (9.48) We can also express Eq. 9.46 in terms of admittance, defined as the recip- rocal of impedance and denoted F.Thus Y = — = G + jB (Siemens). (9.49) Admittance is, of course, a complex number, whose real part, G, is called conductance and whose imaginary part, B, is called susceptance. Like admittance, conductance and susceptance are measured in Siemens (S). Using Eq. 9.49 in Eq. 9.46, we get Y ab Yi +Y? + + Y„. (9.50) The admittance of each of the ideal passive circuit elements also is worth noting and is summarized in Table 9.2. Example 9.7 illustrates the application of Eqs. 9.49 and 9.50 to a spe- cific circuit. TABLE 9.2 Admittance and Susceptance Values Circuit Element Admittance (Y) Resistor G (conductance) Inductor j(-l/wL) Capacitor j<oC Susceptance -\/u)L (oC 9.6 Series, Parallel, and Delta-to-Wye Simplifications 325 Combining Impedances in Series and in Parallel The sinusoidal current source in the circuit shown in Fig. 9.18 produces the current i s = 8 cos 200,000f A. a) Construct the frequency-domain equivalent circuit. b) Find the steady-state expressions for v, i h i 2 , and / 3 . Solution a) The phasor transform of the current source is 8 /0°; the resistors transform directly to the fre- quency domain as 10 and 6 ft; the 40 /xH inductor has an impedance of /8 fl at the given frequency of 200,000 rad/s; and at this fre- quency the 1 /xF capacitor has an impedance of —/'5 ft. Figure 9.19 shows the frequency-domain equivalent circuit and symbols representing the phasor transforms of the unknowns. b) The circuit shown in Fig. 9.19 indicates that we can easily obtain the voltage across the current source once we know the equivalent impedance of the three parallel branches. Moreover, once we know V, we can calculate the three phasor currents l h I 2 , and I3 by using Eq. 9.35. To find the equivalent impedance of the three branches, we first find the equivalent admittance simply by adding the admittances of each branch. The admittance of the first branch is y ' = To = 01 s ' the admittance of the second branch is Yi = 1 /8 = 0.06 - /0.08 S, 6 + /8 100 and the admittance of the third branch is y 3 = -L = ,0.2S. The admittance of the three branches is Y = Y t + Y 2 + Y 3 = 0.16 + /0.12 = 0.2/36.87° S. The impedance at the current source is Z = — = 5/-36.87° a. xQijv ion 1/xF Figure 9.18 • The circuit for Example 9.7. -/5 n Figure 9.19 • The frequency-domain equivalent circuit. The Voltage V is V = ZI = 40/-36.87° V. Hence 40/-36.87 Ii = I 2 = 10 40/-36.87' 6 + /8 = 4/-36.87° = 3.2 - /2.4 A, - 4/-90° = -/4 A, and 40/-36.87° h = / , = = 8/53.13° = 4.8 + /6.4 A. 5/-90 L J We check the computations at this point by veri- fying that Ii + I 2 + I3 = I Specifically, 3.2 - /2.4 - /4 + 4.8 + /6.4 = 8 + /0. The corresponding steady-state time-domain expressions are v = 40cos(200,000r - 36.87°) V, i t = 4 cos (200,000/ - 36.87°) A, / 2 = 4cos(200,000r - 90°) A, /3 = 8cos (200,000/ + 53.13") A.