356 Sinusoidal Steady-State Analysis 9.67 The op amp in the circuit seen in Fig. P9.67 is ideal. PSPICE Find the steady-state expression for v (> (t) when - =2cosl(/Y V Figure P9.67 too kn 40 kO 9.68 The op amp in the circuit in Fig. P9.68 is ideal. MULTISIM a ) ^ n< ^ ^ e stea dy-state expression for v 0 (t). b) How large can the amplitude of v g be before the amplifier saturates? Figure P9.68 v g = 25 cos 50,000* V 9.69 The sinusoidal voltage source in the circuit shown in PSPICE pig P9.69 is generating the voltage v„ = 4 cos 200r V. MULTISIM 11,-1 i If the op amp is ideal, what is the steady-state expres- sion for v 0 (t)1 Figure P9.69 10 kO 20 kQ 20 kH -f VW <b :250 nF 33 ka 9.70 The 250 nF capacitor in the circuit seen in Fig. P9.69 PSPICE is replaced with a variable capacitor. The capacitor MULTISIM - s ac jj USTec ] U xitil the output voltage leads the input voltage by 135°. a) Find the value of C in microfarads. b) Write the steady-state expression for v () (t) when C has the value found in (a). 9.71 The operational amplifier in the circuit shown in PSPICE Fig. P9.71 is ideal. The voltage of the ideal sinu- MULTISIM • , 1 . ir» i r\f\* t / soida 1 source is v g = 30 cos 10°t V. a) How small can C a be before the steady-state output voltage no longer has a pure sinusoidal waveform? b) For the value of C 0 found in (a), write the steady-state expression for v a . Figure P9.71 10 nF loo a loo a 9.72 a) Find the input impedance Z ab for the circuit in Fig. P9.72. Express Z ab as a function of Z and K where K = (R 2 /R\). b) If Z is a pure capacitive element, what is the capacitance seen looking into the terminals a,b? Figure P9.72 9.73 For the circuit in Fig. P9.73 suppose v t = 20 cos(2000f - 36.87°) V v 2 = 10cos(5000/ + 16.26°) V a) What circuit analysis technique must be used to find the steady-state expression for v v (t)1 b) Find the steady-state expression for v C) (t). Problems 357 Figure P9.73 lmH /YYYV 100 |xF If 10 a 9.74 For the circuit in Fig. P9.61, suppose v. d = 5 cos 80,000/ V v b = -2.5 cos 320,000/ V. b) Find the coefficient of coupling. c) Find the energy stored in the magnetically cou- pled coils at t = 1007T /xs and t = 200-7T ^s. Figure P9.77 30 a a) What circuit analysis technique must be used to find the steady-state expression for j„(r)? b) Find the steady-state expression for / 0 (/)? 9.78 For the circuit in Fig. P9.78, find the Thevenin equivalent with respect to the terminals c,d. Section 9.10 9.75 A series combination of a 300 O resistor and a 100 mH inductor is connected to a sinusoidal volt- age source by a linear transformer. The source is operating at a frequency of 1 krad/s. At this fre- quency, the internal impedance of the source is 100 + /13.74 CI. The rms voltage at the terminals of the source is 50 V when it is not loaded. The param- eters of the linear transformer are R\ = 41.68 O, L { = 180 mH, R 2 = 500 ft, L 2 = 500 mH, and M = 270 mH. a) What is the value of the impedance reflected into the primary? b) What is the value of the impedance seen from the terminals of the practical source? 9.76 The sinusoidal voltage source in the circuit seen in PSPICE Fig. P9.76 is operating at a frequency of 200 krad/s. LTISIM ^ e coe ff} c j etl t 0 f coupling is adjusted until the peak amplitude of i x is maximum. a) What is the value of kl b) What is the peak amplitude of /j if v g = 560 cos(2 X 10¾ V ? Figure P9.76 150 n so a loo a 2oo a •—vw—i 12.5 nF 9.77 a) Find the steady-state expressions for the cur- rents ig and i L in the circuit in Fig. P9.77 when PSPICE MU LTISIM v g = 70 cos 5000/ V. Figure P9.78 425/0° 45 a -WV- V (rms) 9.79 The value of k in the circuit in Fig. P9.79 is adjusted so that Z ab is purely resistive when a> = 4 krad/s. Find Z ab . Figure P9.79 a»- 20 a ^VW- 12.5 mH !8mH 5 a -WW 12.5 /JLF Section 9.11 9.80 At first glance, it may appear from Eq. 9.69 that an inductive load could make the reactance seen look- ing into the primary terminals (i.e., X ah ) look capac- itive. Intuitively, we know this is impossible. Show that X) b can never be negative if X L is an inductive reactance. 9.81 a) Show that the impedance seen looking into the terminals a,b in the circuit in Fig. P9.81 on the next page is given by the expression <ab 358 Sinusoidal Steady-State Analysis b) Show that if the polarity terminals of either one of the coils is reversed, -ab Figure P9.81 a« Z, /V, A'\ Z, 9.82 a) Show that the impedance seen looking into the terminals a,b in the circuit in Fig. P9.82 is given by the expression Zab - Z L 1 + N-, b) Show that if the polarity terminal of either one of the coils is reversed that 'ab 1 -^ Figure P9.82 /V, Z a b" b* AT,: infinity. The amplitude and phase angle of the source voltage are held constant as R x varies. Figure P9.84 >\ = V,n cos U) ' ^wv 9.83 Find the impedance Z ab in the circuit in Fig. P9.83 if Z L = 80/60'H. R, 9.85 The parameters in the circuit shown in Fig. 9.53 are /?, = 0.1 il,o)L x = 0.8 ft,fl 2 = 24 il,(oL 2 = 32 ft, and V L = 240 + /0 V. a) Calculate the phasor voltage V s . b) Connect a capacitor in parallel with the inductor, hold V L constant, and adjust the capacitor until the magnitude of I is a minimum. What is the capacitive reactance? What is the value of V v ? c) Find the value of the capacitive reactance that keeps the magnitude of I as small as possible and that at the same time makes lYvl = |V/J = 240 V. 9.86 a) For the circuit shown in Fig. P9.86, compute V v and V/. b) Construct a phasor diagram showing the rela- tionship between V s , V/, and the load voltage of 240/0° V. c) Repeat parts (a) and (b), given that the load voltage remains constant at 240 /0° V, when a capacitive reactance of -5 Cl is connected across the load terminals. Figure P9.86 + Vj_ + 0.1 Q~ /0.8 Q + v, 240/0° vis a Figure P9.83 a« 1/6 n -pft;f; b«- 8:1 Ideal • 10:1 Ideal Z Section 9.12 9.84 Show by using a phasor diagram what happens to PSPICE the magnitude and phase angle of the voltage v„ in MULTISIM the circuit in Fig p9 84 as R ^ js yaricd from zero tQ Sections 9.1-9.12 9.87 You may have the opportunity as an engineering graduate to serve as an expert witness in lawsuits involving either personal injury or property damage. As an example of the type of problem on which you may be asked to give an opinion, consider the follow- ing event. At the end of a day of fieldwork, a farmer returns to his farmstead, checks his hog confinement building, and finds to his dismay that the hogs are Problems 359 dead. The problem is traced to a blown fuse that caused a 240 V fan motor to stop. The loss of ventila- tion led to the suffocation of the livestock. The inter- rupted fuse is located in the main switch that connects the farmstead to the electrical service. Before the insurance company settles the claim, it wants to know if the electric circuit supplying the farmstead functioned properly. The lawyers for the insurance company are puzzled because the farmer's wife, who was in the house on the day of the accident convalescing from minor surgery, was able to watch TV during the afternoon. Furthermore, when she went to the kitchen to start preparing the evening meal, the electric clock indicated the correct time. The lawyers have hired you to explain (1) why the electric clock in the kitchen and the television set in the living room continued to operate after the fuse in the main switch blew and (2) why the second fuse in the main switch didn't blow after the fan motor stalled. After ascertaining the loads on the three-wire distribu- tion circuit prior to the interruption of fuse A, you are able to construct the circuit model shown in Fig. P9.87. The impedances of the line conductors and the neutral conductor are assumed negligible. a) Calculate the branch currents I t , I 2 , I3, I 4 , I5, and I 6 prior to the interruption of fuse A. b) Calculate the branch currents after the interrup- tion of fuse A. Assume the stalled fan motor behaves as a short circuit. c) Explain why the clock and television set were not affected by the momentary short circuit that interrupted fuse A. d) Assume the fan motor is equipped with a ther- mal cutout designed to interrupt the motor cir- cuit if the motor current becomes excessive. Would you expect the thermal cutout to oper- ate? Explain. e) Explain why fuse B is not interrupted when the fan motor stalls. Figure P9.87 Fuse A (100 A) 120. V 'FQ Momentary ' short circuit X interrupts fuse A 120 V FQ -*\fi- 9.88 a) Calculate the branch currents I]-I<s in the cir- pRAcncAL cuit in Fie. 9.58. PERSPECTIVE 0 b) Find the primary current I p . 9.89 Suppose the 40 ft resistance in the distribution cir- pRAcncAL cuit in Fie. 9.58 is replaced bv a 20 ft resistance. PERSPECTIVE r a) Recalculate the branch current in the 2 (1 resistor, I 2 . b) Recalculate the primary current, I p . c) On the basis of your answers, is it desirable to have the resistance of the two 120 V loads be equal? 9.90 A residential wiring circuit is shown in Fig. P9.90. In PRACTICAL this model, the resistor Ri, is used to model a 250 V PERSPECTIVE appliance (such as an electric range), and the resis- tors R] and R 2 are used to model 125 V appliances (such as a lamp, toaster, and iron). The branches carrying ^ and I 2 are modeling what electricians refer to as the hot conductors in the circuit, and the branch carrying \„ is modeling the neutral conduc- tor. Our purpose in analyzing the circuit is to show the importance of the neutral conductor in the sat- isfactory operation of the circuit. You are to choose the method for analyzing the circuit. a) Show that l n is zero if R^ = R 2 . b) Show that V! = V 2 if Ri = R 2 . c) Open the neutral branch and calculate Vi and V 2 if R } = 40 ft, R 2 = 400 ft, and R 3 = 8 ft. d) Close the neutral branch and repeat (c). e) On the basis of your calculations, explain why the neutral conductor is never fused in such a manner that it could open while the hot conduc- tors are energized. Figure P9.90 + • 14./0° kV- -VAr- • + 0.02x2 /0.02 n 125/0° V V, £ /?, -A<W Ideal • + 0.03X1 125/0° V /0.03 (1 mm «- — L + K,fV 3 0.02 a /0.02 a —-WV 1-^ 0 ^ •- 9.91 a) Find the primary current I p for (c) and (d) in P™ nw„r PRACTICAL Problem 9.90. Km moan PERSPECTIVE Fuse B( 100 A) b) Do your answers make sense in terms of known circuit behavior? CHAPTER * i _\ CHAPTER CONTENTS 10.1 Instantaneous Power p. 362 10.2 Average and Reactive Power p. 363 10.3 The rms Value and Power Calculations p. 368 10.4 Complex Power p. 370 10.5 Power Calculations p. 371 10.6 Maximum Power Transfer p. 375 ^CHAPTER OBJECTIVES 1 Understand the following ac power concepts, their relationships to one another, and how to calculate them in a circuit: Instantaneous power; Average (real) power; Reactive power; Complex power; and Power factor. Understand the condition for maximum real power delivered to a load in an ac circuit and be able to calculate the load impedance required to deliver maximum real power to the load. Be able to calculate all forms of ac power in ac circuits with linear transformers and in ac circuits with ideal transformers. 360 Sinusoidal Steady-State Power Calculations Power engineering has evolved into one of the important sub- disciplines within electrical engineering. The range of problems dealing with the delivery of energy to do work is considerable, from determining the power rating within which an appliance operates safely and efficiently, to designing the vast array of gen- erators, transformers, and wires that provide electric energy to household and industrial consumers. Nearly all electric energy is supplied in the form of sinusoidal voltages and currents. Thus, after our Chapter 9 discussion of sinusoidal circuits, this is the logical place to consider sinusoidal steady-state power calculations. We are primarily interested in the average power delivered to or supplied from a pair of termi- nals as a result of sinusoidal voltages and currents. Other meas- ures, such as reactive power, complex power, and apparent power, will also be presented. The concept of the rms value of a sinusoid, briefly introduced in Chapter 9, is particularly pertinent to power calculations. We begin and end this chapter with two concepts that should be very familiar to you from previous chapters: the basic equa- tion for power (Section 10.1) and maximum power transfer (Section 10.6). In between, we discuss the general processes for analyzing power, which will be familiar from your studies in Chapters 1 and 4, although some additional mathematical tech- niques are required here to deal with sinusoidal, rather than dc, signals. w -_.:. Practical Perspective Heating Appliances In Chapter 9 we calculated the steady-state voltages and cur- rents in electric circuits driven by sinusoidal sources. In this chapter we consider power in such circuits. The techniques we develop are useful for analyzing many of the electrical devices we encounter daily, because sinusoidal sources are the pre- dominant means of providing electric power in our homes, schools, and businesses. One common class of electrical devices is heaters, which transform electric energy into thermal energy. Examples include electric stoves and ovens, toasters, irons, electric water heaters, space heaters, electric clothes dryers, and hair dryers. One of the critical design concerns in a heater is power con- sumption. Power is important for two reasons: The more power a heater uses, the more it costs to operate, and the more heat it can produce. Many electric heaters have different power settings corre- sponding to the amount of heat the device supplies. You may wonder just how these settings result in different amounts of heat output. The Practical Perspective example at the end of this chapter examines the design of a handheld hair dryer with three operating settings (see the accompanying figure). You will see how the design provides for three different power levels, which correspond to three different levels of heat output. Heater tube Fan and motor Hot air 361 362 Sinusoidal Steady-State Power Calculations 10.1 Instantaneous Power l —*" + V Figure 10.1 A The black box representation of a circuit used for calculating power. We begin our investigation of sinusoidal power calculations with the familiar circuit in Fig. 10.1. Here, v and /' are steady-state sinusoidal signals. Using the passive sign convention, the power at any instant of time is VI. (10.1) This is instantaneous power. Remember that if the reference direction of the current is in the direction of the voltage rise, Eq. 10.1 must be written with a minus sign. Instantaneous power is measured in watts when the voltage is in volts and the current is in amperes. First, we write expressions for v and i; v = V„, cos (cot + 0j,), i — I„, cos {ait -I- 0,), (10.2) (10.3) where 0,, is the voltage phase angle, and 0-, is the current phase angle. We are operating in the sinusoidal steady state, so we may choose any convenient reference for zero time. Engineers designing systems that transfer large blocks of power have found it convenient to use a zero time corresponding to the instant the current is passing through a positive max- imum. This reference system requires a shift of both the voltage and cur- rent by 0, Thus Eqs. 10.2 and 10.3 become v = V m cos (ait + 0,, - 0,), i = 1,,, cos cot. (10.4) (10.5) When we substitute Eqs. 10.4 and 10.5 into Eq. 10.1, the expression for the instantaneous power becomes p = V m I m cos {cot + 0 V - 0j) cos cot. (10.6) We could use Eq. 10.6 directly to find the average power; however, by sim- ply applying a couple of trigonometric identities, we can put Eq. 10.6 into a much more informative form. We begin with the trigonometric identity 1 1 cos a cos /3 = — cos (a /3) +-cos(a + /3) to expand Eq. 10.6; letting a = cot + 0,, — 0, and fS = cot gives p = —— cos (6 V - 0,) + —r— cos {loot + 0„ - 0,-). (10.7) Now use the trigonometric identity cos (a + /3) = cos a cos /3 — sin a sin (3 See entry 8 in Appendix F. to expand the second term on the right-hand side of Eq. 10.7, which gives y 171*111 / n n\ , r 111*111 //, ., \ * p = —— cos (6 V - 6-) H — cos (0,, - 0,) cos 2(ot V I - -^- sin (0 V - 6j) sin 2u>t. (10.8) Figure 10.2 depicts a representative relationship among v, i, and p, based on the assumptions 0.,, - 60° and 6-, = 0°. You can see that the fre- quency of the instantaneous power is twice the frequency of the voltage or current. This observation also follows directly from the second two terms on the right-hand side of Eq. 10.8. Therefore, the instantaneous power goes through two complete cycles for every cycle of either the voltage or the current. Also note that the instantaneous power may be negative for a portion of each cycle, even if the network between the terminals is passive. In a completely passive network, negative power implies that energy stored in the inductors or capacitors is now being extracted. The fact that the instantaneous power varies with time in the sinusoidal steady-state operation of a circuit explains why some motor-driven appliances (such as refrigerators) experience vibration and require resilient motor mountings to prevent excessive vibration. We are now ready to use Eq. 10.8 to find the average power at the ter- minals of the circuit represented by Fig. 10.1 and, at the same time, intro- duce the concept of reactive power. Figure 10.2 • Instantaneous power, voltage, and current versus vt for steady-state sinusoidal operation. 10.2 Average and Reactive Power We begin by noting that Eq. 10.8 has three terms, which we can rewrite as follows: p = P + Pcos2wt - Qs'm2a)t, (10.9) 364 Sinusoidal Steady-State Power Calculations where Average (real) power • P = V I cos (0 V - e t ), (10.10) Reactive power • Q = V I •sinfo-0,-)- (10.11) P is called the average power, and Q is called the reactive power. Average power is sometimes called real power, because it describes the power in a circuit that is transformed from electric to nonelectric energy. Although the two terms are interchangeable, we primarily use the term average power in this text. It is easy to see why P is called the average power. The average power associated with sinusoidal signals is the average of the instantaneous power over one period, or, in equation form, >-i h+T pdt. (10.12) where T is the period of the sinusoidal function. The limits on Eq. 10.12 imply that we can initiate the integration process at any convenient time t {) but that we must terminate the integration exactly one period later. (We could integrate over nT periods, where n is an integer, provided we multi- ply the integral by \fnT.) We could find the average power by substituting Eq. 10.9 directly into Eq. 10.12 and then performing the integration. But note that the average value of/? is given by the first term on the right-hand side of Eq. 10.9, because the integral of both cos 2cot and sin 2eot over one period is zero. Thus the average power is given in Eq. 10.10. We can develop a better understanding of all the terms in Eq. 10.9 and the relationships among them by examining the power in circuits that are purely resistive, purely inductive, or purely capacitive. 0.01 0.015 Time (s) 0.025 Figure 10.3 • Instantaneous real power and average power for a purely resistive circuit. Power for Purely Resistive Circuits If the circuit between the terminals is purely resistive, the voltage and cur- rent are in phase, which means that $ v = 0,. Equation 10.9 then reduces to p = P + P cos 2oot. (10.13) The instantaneous power expressed in Eq. 10.13 is referred to as the instantaneous real power. Figure 10.3 shows a graph of Eq. 10.13 for a representative purely resistive circuit, assuming co = 377 rad/s. By defini- tion, the average power, P, is the average of/; over one period. Thus it is easy to see just by looking at the graph that P = 1 for this circuit. Note from Eq. 10.13 that the instantaneous real power can never be negative, which is also shown in Fig. 10.3. In other words, power cannot be extracted from a purely resistive network. Rather, all the electric energy is dissi- pated in the form of thermal energy. Power for Purely Inductive Circuits If the circuit between the terminals is purely inductive, the voltage and current are out of phase by precisely 90°. In particular, the current lags the voltage by 90° (that is, 6-, = $ v -90'); therefore 6,, - 6-, = +90°. The expression for the instantaneous power then reduces to -Q sin 2(ot. (10.14) 10,2 Average and Reactive Power 365 In a purely inductive circuit, the average power is zero. Therefore no transformation of energy from electric to nonelectric form takes place. The instantaneous power at the terminals in a purely inductive circuit is continually exchanged between the circuit and the source driving the cir- cuit, at a frequency of 2co. In other words, when p is positive, energy is being stored in the magnetic fields associated with the inductive elements, and when p is negative, energy is being extracted from the magnetic fields. A measure of the power associated with purely inductive circuits is the reactive power Q.The name reactive power comes from the character- ization of an inductor as a reactive element; its impedance is purely reac- tive. Note that average power P and reactive power Q carry the same dimension.To distinguish between average and reactive power, we use the units watt (W) for average power and var (volt-amp reactive, or VAR) for reactive power. Figure 10.4 plots the instantaneous power for a represen- tative purely inductive circuit, assuming u> = 311 rad/s and Q = 1 VAR. Power for Purely Capacitive Circuits If the circuit between the terminals is purely capacitive, the voltage and current are precisely 90° out of phase. In this case, the current leads the voltage by 90° (that is, B t = 6 V + 90°); thus, 0 V - 0,- = -90°. The expres- sion for the instantaneous power then becomes p = —Qsm2(ot. (10.15) Again, the average power is zero, so there is no transformation of energy from electric to nonelectric form. In a purely capacitive circuit, the power is continually exchanged between the source driving the circuit and the electric field associated with the capacitive elements. Figure 10.5 plots the instantaneous power for a representative purely capacitive circuit, assum- ing (o = 377 rad/s and Q = -1 VAR. Note that the decision to use the current as the reference leads to Q being positive for inductors (that is, $ v — 0, : = 90° and negative for capac- itors (that is, 6 V - 0, = -90°. Power engineers recognize this difference in the algebraic sign of Q by saying that inductors demand (or absorb) mag- netizing vars, and capacitors furnish (or deliver) magnetizing vars.We say more about this convention later. o Q, Q (VAR) § 0 0.005 0.01 0.015 0.02 0.025 | Time (s) Figure 10.4 • Instantaneous real power, average power, and reactive power for a purely inductive circuit. a 0 0.005 0.01 0.015 0.02 0.025 Time (s) Figure 10.5 • Instantaneous real power and average power for a purely capacitive circuit. The Power Factor The angle 6 V - 0, plays a role in the computation of both average and reactive power and is referred to as the power factor angle. The cosine of this angle is called the power factor, abbreviated pf, and the sine of this angle is called the reactive factor, abbreviated rf. Thus pf = cos (0,, - 0,), (10.16) -4 Power factor rf = sin (0,, - 0/). (10.17) Knowing the value of the power factor does not tell you the value of the power factor angle, because cos (0,, - 0 ( ) = cos (0, - 0„). To completely describe this angle, we use the descriptive phrases lagging power factor and leading power factor. Lagging power factor implies that current lags volt- age—hence an inductive load. Leading power factor implies that current leads voltage—hence a capacitive load. Both the power factor and the reac- tive factor are convenient quantities to use in describing electrical loads. Example 10.1 illustrates the interpretation of P and Q on the basis of a numerical calculation. . providing electric power in our homes, schools, and businesses. One common class of electrical devices is heaters, which transform electric energy into thermal energy. Examples include electric. transformation of energy from electric to nonelectric form. In a purely capacitive circuit, the power is continually exchanged between the source driving the circuit and the electric field associated. (such as an electric range), and the resis- tors R] and R 2 are used to model 125 V appliances (such as a lamp, toaster, and iron). The branches carrying ^ and I 2 are modeling what electricians