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APPENDIX A Exercises EA.1 Given Z = − j and Z = + j 6, we have: Z + Z = 10 + j Z − Z = −6 − j Z Z = 16 − j 24 + j 12 − j 18 = 34 − j 12 − j − j 16 − j 12 − j 24 + j 18 Z1 / Z2 = × = = −0.02 − j 0.36 8+ j6 8− j6 100 EA.2 Z = 15∠45 o = 15 cos( 45 o ) + j 15 sin( 45 o ) = 10.6 + j 10.6 Z = 10∠ − 150 o = 10 cos( −150 o ) + j 10 sin( −150 o ) = −8.66 − j Z = 5∠90 o = cos(90 o ) + j sin(90 o ) = j EA.3 Notice that Z1 lies in the first quadrant of the complex plane Z = + j = 32 + ∠ arctan( / 3) = 5∠53.13o Notice that Z2 lies on the negative imaginary axis Z = − j 10 = 10∠ − 90 o Notice that Z3 lies in the third quadrant of the complex plane Z = −5 − j = 52 + 52 ∠(180 o + arctan( −5 / − 5)) = 7.07 ∠225 o = 7.07 ∠ − 135 o EA.4 Notice that Z1 lies in the first quadrant of the complex plane Z = 10 + j 10 = 10 + 10 ∠ arctan(10 / 10) = 14.14∠45 o = 14.14 exp( j 45 o ) Notice that Z2 lies in the second quadrant of the complex plane Z = −10 + j 10 = 10 + 10 ∠(180 o + arctan( −10 / 10)) = 14.14∠135 o = 14.14 exp( j 135 o ) 685 EA.5 Z 1Z = (10∠30 o )(20∠135 o ) = (10 × 20)∠(30 o + 135 o ) = 200∠(165 o ) Z / Z = (10∠30 o ) /(20∠135 o ) = (10 / 20)∠(30 o − 135 o ) = 0.5∠( −105 o ) Z − Z = (10∠30 o ) − (20∠135 o ) = (8.66 + j 5) − ( −14.14 + j 14.14) = 22.8 − j 9.14 = 24.6∠ − 21.8o Z + Z = (10∠30 o ) + (20∠135 o ) = (8.66 + j 5) + ( −14.14 + j 14.14) = −5.48 + j 19.14 = 19.9∠106o Problems PA.1 Given Z = + j and Z = − j 3, we have: Z1 + Z2 = + j Z − Z = −2 + j Z Z = − j + j 12 − j = 17 + j Z1 / Z2 = PA.2 + j + j − + j 18 = = 0.04 + j 0.72 × − j3 + j3 25 Given that Z = − j and Z = + j 3, we have: Z1 + Z2 = + j Z − Z = −1 − j Z1 Z2 = + j − j − j = − j Z1 / Z2 = − j2 − j3 − − j = = −0.3077 − j 0.5385 × + j3 − j3 13 686 PA.3 Given that Z = 10 + j and Z = 20 − j 20, we have: Z + Z = 30 − j 15 Z − Z = −10 + j 25 Z Z = 200 − j 200 + j 100 − j 100 = 300 − j 100 Z1 / Z2 = PA.4 PA.5 PA.6 10 + j 20 + j 20 100 + j 300 = = 0.125 + j 0.375 × 20 − j 20 20 + j 20 800 (a) Z a = − j = 7.071∠ − 45o = 7.071 exp(− j 45o ) (b) Z b = −10 + j = 11.18∠153.43o = 11.18 exp(j 153.43o ) (c) Z c = −3 − j = 5∠ − 126.87 o = exp(− j 126.87 o ) (d) Z d = − j 12 = 12∠ − 90 o = 12 exp(− j 90 o ) (a) Z a = 5∠45o = exp(j 45o ) = 3.536 + j 3.536 (b) Z b = 10∠120 o = 10 exp(j 120 o ) = −5 + j 8.660 (c) Z c = 15∠ − 90 o = 15 exp(− j 90 o ) = − j 15 (d) Z d = −10∠60 o = 10 exp(− j 120 o ) = −5 − j 8.660 (a) Z a = 5e j 30 = 5∠30 o = 4.330 + j 2.5 (b) Z b = 10e − j 45 = 10∠ − 45o = 7.071 − j 7.071 (c) Z c = 100e j 135 = 100∠135o = −70.71 + j 70.71 o o o 687 PA.7 (d) Z d = 6e j 90 = 6∠90 o = j (a) Z a = + j + 10∠30 o = 13.66 + j 10 (b) Z b = 5∠45o − j 10 = 3.536 − j 6.464 (c) Zc = 10∠45 o 10∠45 o 2∠ − 8.13o = 1.980 − j 0.283 = + j4 5∠53.13o (d) Zd = 15 = 3∠ − 90 o = − j o 5∠90 o 688 APPENDIX C PC.1 Because the capacitor voltage is zero at t = 0, the charge on the capacitor is zero at t = Then using Equation 3.5 in the text, we have t q(t ) = ∫ i (t )dx + 0 t = ∫ 3dx = 3t For t = µs, we have q(3) = × × 10 −6 = µC PC.2 Refer to Figure PC.2 in the book Combining the 10-Ω resistance and the 20-Ω resistance we obtain a resistance of 6.667 Ω, which is in series with the 5-Ω resistance Thus, the total resistance seen by the 15-V source is + 6.667 = 11.667 Ω The source current is 15/11.667 = 1.286 A The current divides between the 10-Ω resistance and the 20-Ω resistance Using Equation 2.27, the current through the 10-Ω resistance is i 10 = 20 × 1.286 = 0.8572 A 20 + 10 Finally, the power dissipated in the 10-Ω resistance is P10 = 10i 10 = 7.346 W PC.3 The equivalent capacitance of the two capacitors in series is given by Ceq = = µF / C1 + / C The charge supplied by the source is q = CeqV = 200 × × 10 −6 = 800 µC 689 PC.4 The input power to the motor is the output power divided by efficiency × 746 = 1865 W 0.80 η However the input power is also given by Pin = Pout = Pin = Vrms I rms cos(θ ) in which cos(θ ) is the power factor Solving for the current, we have I rms = Pin 1865 = = 11.30 A Vrms cos(θ ) 220 × 0.75 PC.5 j = 30 + j 40 − j 80 = 30 − j 40 = 50∠ − 53.1o ωC Thus the impedance magnitude is 50 Ω PC.6 We have Z = R + jωL − Apparent power = Vrms I rms Also, the power factor is cos(θ ) = 0.6 from which we find that θ = 53.13 o (We selected the positive angle because the power factor is stated to be lagging.) Then we have Q = Vrms I rms sin(θ ) = ( Apparent power) × sin(θ ) = 2000 × 0.8 = 1600 VAR PC.7 For practical purposes, the capacitor is totally discharged after twenty time constants and all of the initial energy stored in the capacitor has been delivered to the resistor The initial stored energy is W = 21 CV = 21 × 150 × 10 −6 × 100 = 0.75 J PC.8 ω = 2πf = 120π Z = R + jωL − j = 50 + j 56.55 − j106.10 = 50 − j 49.55 = 70.39∠ − 44.74 o ωC 690 I rms = PC.9 Vrms 110 = = 1.563 A 70.39 Z See Example 4.2 in the book In this case, we have K = K = VS R = A and τ = L R = 0.5 s Then the current is given by i (t ) = − exp( −t / τ ) = − exp(−2t ) PC.10 We have VBC = −VCB = −50 V and VAB = VAC −VBC = 200 − ( −50) = 250 The energy needed to move the charge from point B to point A is W = QVAB = 0.2(250) = 50 J 691 CONTENTS Chapter Chapter 24 Chapter 84 Chapter 121 Chapter 174 Chapter 221 Chapter 329 Chapter 286 Chapter 347 Chapter 10 359 Chapter 11 407 Chapter 12 458 Chapter 13 487 Chapter 14 520 Chapter 15 572 Chapter 16 608 Chapter 17 646 Appendix A 685 Appendix C 689 Complete solutions to the in-chapter exercises, answers to the end-ofchapter problems marked by an asterisk *, and complete solutions to the Practice Tests are available to students at www.pearsonhighered.com/hambley CHAPTER Exercises E1.1 Charge = Current × Time = (2 A) × (10 s) = 20 C E1.2 i (t ) = E1.3 Because i2 has a positive value, positive charge moves in the same direction as the reference Thus, positive charge moves downward in element C dq (t ) d = (0.01sin(200t) = 0.01 × 200cos(200t ) = 2cos(200t ) A dt dt Because i3 has a negative value, positive charge moves in the opposite direction to the reference Thus positive charge moves upward in element E E1.4 Energy = Charge × Voltage = (2 C) × (20 V) = 40 J Because vab is positive, the positive terminal is a and the negative terminal is b Thus the charge moves from the negative terminal to the positive terminal, and energy is removed from the circuit element E1.5 E1.6 iab enters terminal a Furthermore, vab is positive at terminal a Thus the current enters the positive reference, and we have the passive reference configuration (a) pa (t ) = v a (t )ia (t ) = 20t 10 10 20t w a = ∫ pa (t )dt = ∫ 20t dt = 0 10 = 20t = 6667 J (b) Notice that the references are opposite to the passive sign convention Thus we have: pb (t ) = −v b (t )ib (t ) = 20t − 200 10 10 0 w b = ∫ pb (t )dt = ∫ (20t − 200)dt = 10t − 200t 10 = −1000 J E1.7 (a) Sum of currents leaving = Sum of currents entering ia = + = A (b) = + + ib ib = -2 A ⇒ (c) = + ic + + ⇒ ic = -8 A E1.8 Elements A and B are in series Also, elements E, F, and G are in series E1.9 Go clockwise around the loop consisting of elements A, B, and C: -3 - +vc = ⇒ vc = V Then go clockwise around the loop composed of elements C, D and E: - vc - (-10) + ve = ⇒ ve = -2 V E1.10 Elements E and F are in parallel; elements A and B are in series E1.11 The resistance of a wire is given by R = substituting values, we have: = 1.12 × 10 −6 × L π (1.6 × 10 − )2 / ρL A Using A = πd / and ⇒ L = 17.2 m E1.12 P =V R ⇒ R =V / P = 144 Ω E1.13 P =V R ⇒ V = PR = 0.25 × 1000 = 15.8 V ⇒ I = V / R = 120 / 144 = 0.833 A I = V / R = 15.8 / 1000 = 15.8 mA E1.14 Using KCL at the top node of the circuit, we have i1 = i2 Then, using KVL going clockwise, we have -v1 - v2 = 0; but v1 = 25 V, so we have v2 = -25 V Next we have i1 = i2 = v2/R = -1 A Finally, we have PR = v 2i2 = ( −25) × ( −1) = 25 W and Ps = v 1i1 = (25) × ( −1) = −25 W E1.15 At the top node we have iR = is = 2A By Ohm’s law we have vR = RiR = 80 V By KVL we have vs = vR = 80 V Then ps = -vsis = -160 W (the minus sign is due to the fact that the references for vs and is are opposite to the passive sign configuration) Also we have PR = v R iR = 160 W P17.39 (a) (b) (c) (d) (e) (f) P17.40 Output power remains constant Mechanical speed remains constant Output torque remains constant Armature current increases in magnitude and its phase leads the source voltage Power factor decreases and becomes leading Reactive power is produced by the machine Torque angle decreases Synchronous speed for the machine is: 2π 60 ω ωs = = = 125.7 rad s (P 2) ns = 1200 rpm Tdev = Pdev = 29.68 Nm ωs According to Equation 17.37, we have: Tdev = KBr Btotal sin δ We define Kt = KBr Btotal Then from the initial operating conditions, we find: Kt = Tdev sin δ = 29.68 = 340.6 sin(5o ) Now when the torque doubles, we have T × 29.68 sin δ = dev = Kt 340.6 which yields δ = 10.04 o The pullout torque occurs for δ = 90 o Thus we have: Tmax = Kt sin(90 o ) = 340.6 newton meters Pdev, max = Tmax ωs = 42.82 kW or 57.39 hp P17.41* ωs = 2π 60 ω = = 75.40 rad s (P 2) 10 Tdev, rated = Pdev, rated ωs = ns = 720 rpm 100 × 746 = 989.4 newton meters 75.40 670 According to Equation 17.37, we have: Tdev = KBr Btotal sin δ We define Kt = KBr Btotal Then from the rated operating conditions, we find: Kt = Tdev sin δ = 989.4 = 2893 sin(20 o ) The pullout torque occurs for δ = 90 o Thus we have: Tmax = Kt sin(90 o ) = 2893 newton meters The torque speed characteristic is: P17.42 Refer to Figure 17.22 for the phasor diagrams with constant developed power and variable field current The phasor diagram for the initial operating conditions is: Xs Ia1 Er Va 240 = 248.47 cos δ1 cos(15o ) = E r sin δ1 = 64.31 = 1.2 × E r = 298.16 Er = = 671 The phasor diagram for the second operation condition is: Notice that the vertical component of jX s Ia is the same in both diagrams Now we can write: (Va + X s I a tan θ2 )2 + (X s I a )2 = (E r )2 (240 + 64.31 tan θ2 )2 + (64.31)2 = (298.16)2 Solving, we find θ2 = 38.51o and the power factor is cos θ2 = 78.24% leading Finally, we have: X I 64.31 sin δ = s a = Er 298.16 which yields: δ2 = 12.46o P17.43 The developed power is 50 × 746 = 37300 W Neglecting losses, this is the input power On a per phase basis, we have Pin1 = 37300 = Va I a cos θ1 = 240I a cos o This yields I a = 51.81 A As a phasor Ia = 51.81∠0 o The phasor diagram is: Va 240 = 248.47 cos δ1 cos 15o X s I a = Va tan δ1 = 64.31 Er = = 672 Xs = X s I a 64.31 = = 1.241 Ω 51.81 Ia The phasor diagram with the load removed is: Notice that Er is constant in magnitude Also with zero developed power, the torque angle is zero We have X s Ia = Er − Va = 248.47∠0 − 240∠0 = 8.47 8.47 = 6.821∠90 o Ia = j 1.241 Thus, θ2 = 90 o , and the power factor is zero P17.44* For zero developed power, the torque angle is zero The phasor diagram is: Er = Va − jX s Ia = 480 − j 5(15∠ − 90 o ) = 405∠0 o To achieve zero armature current, we must have Er = Va = 480∠0 o The magnitude of Er is proportional to the field current Thus we have: E 480 If = If r = = 5.93 A Er 405 673 P17.45 (a) ωs = Tdev = (b) 2π 60 ω = = 125.66 rad s (P 2) Pdev ωs = ns = 1200 rpm 50 × 746 = 296.8 newton meters 125.66 Pin = Pdev = 50 × 746 = 3I aVa cos θ = 3I a 240(0.9) Thus, I a = 57.56 θ = cos − (0.9) = 25.84 o Thus, Ia = 57.56∠25.84 o Er = Va − jX s Ia = 240 − j 0.5Ia = 240 − j 0.5(51.80 + j 25.21) Er = 252.6 − j 25.90 = 253.9∠ − 5.85 o Thus, the torque angle is δ1 = 5.85o (c) To double the power, we must double the torque According to Equation 17.37, the developed torque is proportional to sin δ Thus, sin δ2 =2 sin δ1 which yields the new torque angle δ2 = 11.77 o Er remains constant in magnitude, thus we have Er = 253.9∠ − 11.77 o V − Er 240 − 253.9∠ − 11.77 o = = 105.0∠9.38o Ia = a jX s j The power factor is cos(9.38o ) = 98.66% leading P17.46 Because the developed power includes the losses, we have: Pin = Pdev = 100 × 746 = 74600 W = 3I a 1Va cos θ1 Solving, we have Ia1 = Pin 3Va cos θ1 = 74600 = 121.9 A 3(240 )0.85 θ1 = cos − (0.85 ) = 31.79o Thus, Ia = 121.9∠ − 31.79o 674 Er = Va = jX s Ia = 207.9 − j 51.81 = 214.3∠ − 13.99o The phasor diagram is: For 100% power factor, the phasor diagram becomes: Notice that (refer to Figure 17.22) the imaginary (i.e., vertical) component of Er is the same in both diagrams Thus we have: X s Ia = Er sin(13.99o ) = 51.81 Er = (Va )2 + (X s I a )2 = 2402 + (51.81) = 245.5 The magnitude of Er is proportional to the field current, so we have: E 245.5 If = If r = 10 × = 11.46 A 214.3 Er P17.47* (a) The speed of a synchronous machine is related to the frequency of the armature voltages by Equation 17.14: 120f ns = P The frequency of the voltage applied to the 12-pole motor is 60 Hz and the speed is 600 rpm Thus, the generator is driven at 600 rpm and the voltages induced in its armature have a frequency of: ns Pgen 600 × 10 fgen = = = 50 Hz 120 120 675 Thus, the two machines can be used to convert 60-Hz power into 50-Hz power (b) 1000 rpm is not the synchronous speed for any 60-Hz motor Thus, we will try to convert 60-Hz into some other frequency for which 1000 rpm is a synchronous speed The block diagram of this system is: The speeds of motor and the generator are the same, so we have: 120 × 60 120f2 ns = = Pm Pg The frequencies are the same for the generator and motor 2, so we can write: n P 1000Pm f2 = s m2 = 120 120 Substituting and rearranging, we have: 72 Pm 1Pm = 10 Pg Of course, the number of poles on each machine must be an even integer One solution is: Pg = 10 Pm = 12 Pm = and for which f2 = 50 Hz Another solution is: Pg = 10 Pm = and Pm = 12 for which f2 = 100 Hz P17.48 Referring to the V-curves shown in Figure 17.23, we see that unity power factor occurs when the armature current is minimum Thus, we would adjust the rheostat in the field circuit to obtain minimum armature 676 current The setting required would change with load P17.49 The machine has copper-losses in the armature windings and rotational losses (We not consider the power that must be supplied to the field circuit.) With no load and current adjusted for minimum armature current, the machine operates with unity power factor For a delta connection, the phase current is the line current divided by The input power is: Pin, no -load = 3(480 )9.5 = 13.68 kW The copper loss is: Pcopper, no -load = 3(0.05 )(9.5)2 = 13.5 W Thus, the rotational loss is: Prot = Pin, no -load − Pcopper, no -load = 13.68 kW We assume that the rotational loss is independent of the load At fullload, we have Pout + Prot + 3(I a )2 Ra = 3Va I a cos(θ ) = Pin 200(746) + 13680 + 0.15(I a )2 = 3(480 )I a (0.9 ) Solving for Ia , we have I a = 127.6 or 8510 The appropriate root is I a = 127.6 A Thus, we have: Pin = 3Va I a cos(θ ) = 165.4 kW η= P17.50* (a) Pout × 100% = 90.22% Pin Pout 746 = 932.5 W η 0.8 P 932.5 = 76.2% power factor = cosθ = in = VI 120(10.2) Pin = = Of course, the power factor is lagging for an induction motor (b) Z = (V I )∠ cos −1 (power factor ) = 120 10.2 ∠ cos −1 (0.762) = 11.76∠40.36 o Ω (c) Since the motor runs just under 1800 rpm, evidently we have a four-pole motor 677 P17.51 At full load, the slip is s full = (3600 − 3500 ) 3600 = 2.778% At no load, the slip is s no = (3600 − 3595 ) 3600 = 0.1388% Thus, we can write 0.02778K − K = 0.5 0.001388K − K = Solving, we find K = 18.95 and K = 0.02630 Now, for 0.2 hp out 18.95s − 0.02630 = 0.2 Finally, the speed for 0.2 hp out is n = 3600(1 − s ) = 3557 rpm P17.52 = −53.13o Thus the phase of Ia needs to be θm + 90 o = 36.87 o The phase angle of Im is θm = − arctan The imaginary part of Za is 9− = 12 tan θa = −9 ωC Of course, ω = 2π 60 , and we find C = 147.4 µF P17.53* Under full load, the current of the motor is × 746 I full = = 10.36 A 0.8(0.75)240 We estimate the starting current as I start = × I full = 62.2 A Neglecting other loads that might be connected, the equivalent circuit is 678 During starting, the voltage across the motor is Vmotor = 240 − (0.2 + j 0.2) × 62.2∠θ start The phase angle of the starting current is unknown However, we can anticipate that the motor appears inductive and θ start is negative In the worst case, we would have θ start = − tan − (0.2 0.2) = −45o Then, voltage drops from 240 V to 240 - (0.2)2 + (0.2)2 × 62.2 = 222.4 V when the motor starts The percentage drop in voltage is 7.33% This would cause a noticeable momentary dimming of the lights in the farmhouse P17.54 To reverse the direction of rotation, reverse the connections to either the main winding or the starting winding (but not both) P17.55 A universal motor would be better than an induction motor in a portable vacuum cleaner because the universal motor gives a higher power to weight ratio An induction motor would be the better choice for a fan in a heating system because the life of a universal motor is relatively short compared to that of an induction motor The induction motor would be better for a refrigerator compressor, again because of longer service life For a variable speed hand-held drill, we should choose the universal motor because it gives higher power for a given weight P17.56 A sketch of a stepper motor cross section with stator poles and rotor poles is: The rotation is 15° per step 679 P17.57 Many sites dealing with stepper motors can be found on the internet but the URLs change often P17.58 Compared to conventional dc motors, the advantages of brushless dc motors are long service life with little maintance, freedom from radio interference, ability to operate in explosive environments, and capability for very high speeds Practice Test T17.1 (a) The magnetic field set up in the air gap of a four-pole three-phase induction motor consists of four magnetic poles spaced 90° from one another in alternating order (i.e., north-south-north-south) The field points from the stator toward the rotor under the north poles and in the opposite direction under the south poles The poles rotate with time at synchronous speed around the axis of the motor (b) The air gap flux density of a two-pole machine is given by Equation 17.12 in the book: Bgap = Bm cos(ωt − θ ) in which Bm is the peak field intensity, ω is the angular frequency of the three-phase source, and θ is angular displacement around the air gap This describes a field having two poles: a north pole corresponding to ωt − θ = and a south pole corresponding to ωt − θ = π The location of either pole moves around the gap at an angular speed of ωs = ω For a four pole machine, the field has four poles rotating at an angular speed of ωs = ω/2 and is given by Bgap = Bm cos(ωt − θ / 2) in which Bm is the peak field intensity, ω is the angular frequency of the three-phase source, and θ denotes angular position around the gap T17.2 Five of the most important characteristics for an induction motor are: Nearly unity power factor High starting torque Close to 100% efficiency Low starting current High pull-out torque T17.3 An eight-pole 60-Hz machine has a synchronous speed of ns = 900 rpm, 680 and the slip is: n − nm 900 − 864 s = s = = 0.04 ns 900 Because the machine is wye connected, the phase voltage is 240 / = 138.6 V (We assume zero phase for this voltage.)The per phase equivalent circuit is: Then, we have Z s = + j + j 40(0.5 + 12 + j 0.8) j 40 + 0.5 + 12 + j 0.8 = 11.48 + j 6.149 Ω = 13.03∠28.17 o Ω power factor = cos(28.17 o ) = 88.16% lagging Vs 138.6∠0 o = 10.64∠ − 28.17 o A rms o Z s 13.03∠28.17 Pin = 3I sVs cos θ = 3.898 kW Is = = For a wye-connected motor, the phase current and line current are the same Thus, the line current magnitude is 10.64 A rms Next, we compute Vx and Ir′ j 40(0.5 + 12 + j 0.8) Vx = Is j 40 + 0.5 + 12 + j 0.8 = 123.83 − j 16.244 = 124.9∠ − 7.473o Ir′ = Vx j 0.8 + 0.5 + 12 = 9.971∠ − 11.14 o The copper losses in the stator and rotor are: 681 Ps = 3Rs I s2 = 3(0.5)(10.64 ) and = 169.7 W Pr = 3Rr′(I r′ )2 = 3(0.5)(9.971) = 149.1 W Finally, the developed power is: 1−s Pdev = × Rr′(I r′ )2 s = 3(12)(9.971) = 3.579 kW Pout = Pdev − Prot = 3.429 kW The output torque is: Tout = Pout = 37.90 newton meters ωm The efficiency is: η= T17.4 Pout × 100% = 87.97% Pin At 60 Hz, synchronous speed for an eight-pole machine is: 120f 120(60 ) ns = = = 900 rpm P The slip is given by: n − nm 900 − 850 s = s = = 5.56% ns 900 The frequency of the rotor currents is the slip frequency From Equation 17.17, we have ωslip = sω For frequencies in the Hz, this becomes: fslip = sf = 0.05555 × 60 = 3.333 Hz In the normal range of operation, slip is approximately proportional to output power and torque Thus at 80% of full power, we estimate that s = 0.8 × 0.05555 = 0.04444 This corresponds to a speed of 860 rpm T17.5 The stator of a six-pole synchronous motor contains a set of windings 682 (collectively known as the armature) that are energized by a three-phase ac source These windings produce six magnetic poles spaced 60° from one another in alternating order (i.e., north-south-north-south-northsouth) The field points from the stator toward the rotor under the north stator poles and in the opposite direction under the south stator poles The poles rotate with time at synchronous speed (1200 rpm) around the axis of the motor The rotor contains windings that carry dc currents and set up six north and south magnetic poles evenly spaced around the rotor When driving a load, the rotor spins at synchronous speed with the north poles of the rotor lagging slightly behind and attracted by the south poles of the stator (In some cases, the rotor may be composed of permanent magnets.) T17.6 Figure 17.22 in the book shows typical phasor diagrams with constant developed power and variable field current The phasor diagram for the initial operating conditions is: Notice that because the initial power factor is unity, we have θ1 = 0° and Ia1 is in phase with Va Also, notice that jXsIa1 is at right angles to Ia1 Now, we can calculate the magnitudes of Er1 and of XsIa1 Va 440 Er = = = 446.79 V cos δ cos(10 o ) X s I a = E r sin δ = 77.58 V Then, the field current is reduced until the magnitude of Er2 is 75% of its initial value E r = 0.75 × E r = 335.09 V The phasor diagram for the second operating condition is: 683 Because the torque and power are constant, the vertical component of jX s Ia is the same in both diagrams as illustrated in Figure 17.22 in the book Thus, we have: sin δ = Xs Ia1 77.58 = Er 335.09 which yields: δ = 13.39 o (Another solution to the equation is δ = 166.61 o , but this does not correspond to a stable operating point.) Now, we can write: (Va − X s I a tan θ )2 + (X s I a )2 = (E r )2 (440 − 77.58 tan θ )2 + (77.58)2 = (335.09)2 Solving, we find θ = 55.76°, and the power factor is cos θ = 56.25% lagging 684 [...]... source is delivering energy (c) The circuit has three nodes, one on each of the top corners and one along the bottom of the circuit (d) First, we must find the voltage vI across the current source We choose the reference shown: Then, going around the circuit counterclockwise, we have − v I +Vs + v R = 0 , which yields v I =Vs + v R = 10 − 6 = 4 V Next, the power for the current source is PI = I sv