Sinusoidal Steady State Power Calculations: Assessment Problems, Exercises of Electrical Circuit Analysis

Download Sinusoidal Steady State Power Calculations: Assessment Problems and more Exercises Electrical Circuit Analysis in PDF only on Docsity! 10 Sinusoidal Steady State Power Calculations Assessment Problems AP 10.1 [a] V = 100/− 45◦ V, I = 20/15◦ A Therefore P = 1 2 (100)(20) cos[−45 − (15)] = 500 W, A → B Q = 1000 sin −60◦ = −866.03 VAR, B → A [b] V = 100/− 45◦, I = 20/165◦ P = 1000 cos(−210◦) = −866.03 W, B → A Q = 1000 sin(−210◦) = 500 VAR, A → B [c] V = 100/− 45◦, I = 20/− 105◦ P = 1000 cos(60◦) = 500 W, A → B Q = 1000 sin(60◦) = 866.03 VAR, A → B [d] V = 100/0◦, I = 20/120◦ P = 1000 cos(−120◦) = −500 W, B → A Q = 1000 sin(−120◦) = −866.03 VAR, B → A AP 10.2 pf = cos(θv − θi) = cos[15 − (75)] = cos(−60◦) = 0.5 leading rf = sin(θv − θi) = sin(−60◦) = −0.866 10–1 docsity.com 10–2 CHAPTER 10. Sinusoidal Steady State Power Calculations AP 10.3 From Ex. 9.4 Ieff = Iρ√ 3 = 0.18√ 3 A P = I2effR = (0.0324 3 ) (5000) = 54 W AP 10.4 [a] Z = (39 + j26)‖(−j52) = 48 − j20 = 52/− 22.62◦ Ω Therefore I = 250/0◦ 48 − j20 + 1 + j4 = 4.85/18.08 ◦ A(rms) VL = ZI = (52/− 22.62◦)(4.85/18.08◦) = 252.20/− 4.54◦ V(rms) IL = VL 39 + j26 = 5.38/− 38.23◦ A(rms) [b] SL = VLI∗L = (252.20/− 4.54◦)(5.38/+ 38.23◦) = 1357/33.69◦ = (1129.09 + j752.73) VA PL = 1129.09 W; QL = 752.73 VAR [c] P = |I|21 = (4.85)2 · 1 = 23.52 W; Q = |I|24 = 94.09 VAR [d] Sg(delivering) = 250I∗ = (1152.62 − j376.36) VA Therefore the source is delivering 1152.62 W and absorbing 376.36 magnetizing VAR. [e] Qcap = |VL|2 −52 = (252.20)2 −52 = −1223.18 VAR Therefore the capacitor is delivering 1223.18 magnetizing VAR. Check: 94.09 + 752.73 + 376.36 = 1223.18 VAR and 1129.09 + 23.52 = 1152.62 W AP 10.5 Series circuit derivation: S = 250I∗ = (40,000 − j30,000) Therefore I∗ = 160 − j120 = 200/− 36.87◦ A(rms) I = 200/36.87◦ A(rms) Z = V I = 250 200/36.87◦ = 1.25/− 36.87◦ = (1 − j0.75) Ω Therefore R = 1 Ω, XC = −0.75 Ω docsity.com Problems 10–5 [b] I1 − I2 = 0.4 − j0.32 A P375 = 1 2 |I1 − I2|2(375) = 49.20 W [c] Pg = 1 2 (248)(0.8) = 99.20 W ∑ Pabs = 50 + 49.2 = 99.20 W (checks) AP 10.10 [a] VTh = 210/0◦ V; V2 = 14V1; I1 = 1 4I2 Short circuit equations: 840 = 80I1 − 20I2 + V1 0 = 20(I2 − I1) − V2 ·. . I2 = 14 A; RTh = 21014 = 15 Ω [b] Pmax = (210 30 )2 15 = 735 W AP 10.11 [a] VTh = −4(146/0◦) = −584/0◦ V(rms) = 584/180◦ V(rms) V2 = 4V1; I1 = −4I2 Short circuit equations: 146/0◦ = 80I1 − 20I2 + V1 0 = 20(I2 − I1) + V2 ·. . I2 = −146/365 = −0.40 A; RTh = −584−0.4 = 1460 Ω [b] P = (−584 2920 )2 1460 = 58.40 W docsity.com 10–6 CHAPTER 10. Sinusoidal Steady State Power Calculations Problems P 10.1 [a] P = 1 2 (100)(10) cos(50 − 15) = 500 cos 35◦ = 409.58 W (abs) Q = 500 sin 35◦ = 286.79 VAR (abs) [b] P = 1 2 (40)(20) cos(−15 − 60) = 400 cos(−75◦) = 103.53 W (abs) Q = 400 sin(−75◦) = −386.37 VAR (del) [c] P = 1 2 (400)(10) cos(30 − 150) = 2000 cos(−120◦) = −1000 W (del) Q = 2000 sin(−120◦) = −1732.05 VAR (del) [d] P = 1 2 (200)(5) cos(160 − 40) = 500 cos(120◦) = −250 W (del) Q = 500 sin(120◦) = 433.01 VAR (abs) P 10.2 p = P + P cos 2ωt − Q sin 2ωt; dp dt = −2ωP sin 2ωt − 2ωQ cos 2ωt dp dt = 0 when − 2ωP sin 2ωt = 2ωQ cos 2ωt or tan 2ωt = −Q P cos 2ωt = P√ P 2 + Q2 ; sin 2ωt = − Q√ P 2 + Q2 Let θ = tan−1(−Q/P ), then p is maximum when 2ωt = θ and p is minimum when 2ωt = (θ + π). Therefore pmax = P + P · P√ P 2 + Q2 − Q(−Q)√ P 2 + Q2 = P + √ P 2 + Q2 and pmin = P − P · P√ P 2 + Q2 − Q · Q√ P 2 + Q2 = P − √ P 2 + Q2 docsity.com Problems 10–7 P 10.3 [a] hair dryer = 600 W vacuum = 630 W sun lamp = 279 W air conditioner = 860 W television = 240 W ∑ P = 2609 W Therefore Ieff = 2609 120 = 21.74 A Yes, the breaker will trip. [b] ∑ P = 2609 − 909 = 1700 W; Ieff = 1700120 = 14.17 A Yes, the breaker will not trip if the current is reduced to 14.17 A. P 10.4 [a] Ieff = 40/115 ∼= 0.35 A; [b] Ieff = 130/115 ∼= 1.13 A P 10.5 Wdc = V 2dc R T ; Ws = ∫ to+T to v2s R dt ·. . V 2 dc R T = ∫ to+T to v2s R dt V 2dc = 1 T ∫ to+T to v2s dt Vdc = √ 1 T ∫ to+T to v2s dt = Vrms = Veff P 10.6 [a] Area under one cycle of v2g : A = (52)(2)(30 × 10−6) + 22(2)(37.5 × 10−6) = 1800 × 10−6 Mean value of v2g : M.V. = A 200 × 10−6 = 1800 × 10−6 200 × 10−6 = 9 ·. . Vrms = √ 9 = 3 V(rms) [b] P = V 2rms R = 32 2.25 = 4 W P 10.7 i(t) = 200t 0 ≤ t ≤ 75 ms i(t) = 60 − 600t 75 ms ≤ t ≤ 100 ms Irms = √ 1 0.1 {∫ 0.075 0 (200)2t2 dt + ∫ 0.1 0.075 (60 − 600t)2 dt } = √ 10(5.625) + 10(1.875) = √ 75 = 8.66 A(rms) docsity.com 10–10 CHAPTER 10. Sinusoidal Steady State Power Calculations |Ig|2RL = 2500 ·. . RL = 10 Ω |Ig|2XL = −5000 ·. . XL = −20 Ω Thus, |Z| = √ (30)2 + (X − 20)2 |Ig| = 500√ 900 + (X − 20)2 ·. . 900 + (X − 20)2 = 25 × 10 4 250 = 1000 Solving, (X − 20) = ±10. Thus, X = 10 Ω or X = 30 Ω [b] If X = 30 Ω: Ig = 500 30 + j10 = 15 − j5 A Sg = −500I∗g = −7500 − j2500 VA Thus, the voltage source is delivering 7500 W and 2500 magnetizing vars. Qj30 = |Ig|2X = 250(30) = 7500 VAR Therefore the line reactance is absorbing 7500 magnetizing vars. Q−j20 = |Ig|2XL = 250(−20) = −5000 VAR Therefore the load reactance is generating 5000 magnetizing vars.∑ Qgen = 7500 VAR = ∑ Qabs If X = 10 Ω: Ig = 500 30 − j10 = 15 + j5 A Sg = −500I∗g = −7500 + j2500 VA Thus, the voltage source is delivering 7500 W and absorbing 2500 magnetizing vars. Qj10 = |Ig|2(10) = 250(10) = 2500 VAR Therefore the line reactance is absorbing 2500 magnetizing vars. The load continues to generate 5000 magnetizing vars.∑ Qgen = 5000 VAR = ∑ Qabs docsity.com Problems 10–11 P 10.13 Zf = −j10,000‖20,000 = 4000 − j8000 Ω Zi = 2000 − j2000 Ω ·. . Zf Zi = 4000 − j8000 2000 − j2000 = 3 − j1 Vo = −Zf Zi Vg; Vg = 1/0◦ V Vo = (3 − j1)(1) = 3 − j1 = 3.16/− 18.43◦ V P = 1 2 V 2m R = 1 2 (10) 1000 = 5 × 10−3 = 5 mW P 10.14 [a] P = 1 2 (240)2 480 = 60 W − 1 ωC = −9 × 106 (5000)(5) = −360 Ω Q = 1 2 (240)2 (−360) = −80 VAR pmax = P + √ P 2 + Q2 = 60 + √ (60)2 + (80)2 = 160 W(del) [b] pmin = 60 − √ 602 + 802 = −40 W(abs) [c] P = 60 W from (a) [d] Q = −80 VAR from (a) [e] generate, because Q < 0 [f] pf = cos(θv − θi) I = 240 480 + 240 −j360 = 0.5 + j0.67 = 0.83/53.13 ◦ A ·. . pf = cos(0 − 53.13◦) = 0.6 leading [g] rf = sin(−53.13◦) = −0.8 docsity.com 10–12 CHAPTER 10. Sinusoidal Steady State Power Calculations P 10.15 [a] The mesh equations are: (10 − j20)I1 + (j20)I2 = 170 (j20)I1 + (12 − j4)I2 = 0 Solving, I1 = 4 + j1 A; I2 = 3.5 − j5.5 A S = −VgI∗1 = −(170)(4 − j1) = −680 + j170 VA [b] Source is delivering 680 W. [c] Source is absorbing 170 magnetizing VAR. [d] P10Ω = ( √ 17)2(10) = 170 W P12Ω = ( √ 42.5)2(12) = 510 W (I1 − I2) = 0.5 + j6.5 A Q−j20Ω = ( √ 42.5)2(20) = −850 VAR |I1 − I2| = √ 42.5 Qj16Ω = ( √ 42.5)2(16) = 680 VAR [e] ∑ Pdel = 680 W ∑ Pdiss = 170 + 510 = 680 W ·. . ∑Pdel = ∑Pdiss = 680 W [f] ∑ Qabs = 170 + 680 = 850 VAR ∑ Qdev = 850 VAR ·. . ∑ mag VAR dev = ∑ mag VAR abs = 850 docsity.com Problems 10–15 P 10.20 ST = 4500 − j 45000.96 (0.28) = 4500 − j1312.5 VA S1 = 2700 0.8 (0.8 + j0.6) = 2700 + j2025 VA S2 = ST − S1 = 1800 − j3337.5 = 3791.95/− 61.66◦ VA pf = cos(−61.66◦) = 0.4747 leading P 10.21 2400I∗1 = 60,000 + j40,000 I∗1 = 25 + j16.67; ·. . I1 = 25 − j16.67 A(rms) 2400I∗2 = 20,000 − j10,000 I∗2 = 8.33 − j4, 167; ·. . I2 = 8.33 + j4.167 A(rms) I3 = 2400/0◦ 144 = 16.67 + j0 A; I4 = 2400/0◦ j96 = 0 − j25 A Ig = I1 + I2 + I3 + I4 = 50 − j37.5 A Vg = 2400 + (j4)(50 − j37.5) = 2550 + j200 = 2557.83/4.48◦ V(rms) P 10.22 [a] S1 = 60,000 − j70,000 VA S2 = |VL|2 Z∗2 = (2500)2 24 − j7 = 240,000 + j70,000 VA S1 + S2 = 300,000 VA 2500I∗L = 300,000; ·. . IL = 120/0◦ A(rms) Vg = VL + IL(0.1 + j1) = 2500 + (120)(0.1 + j1) = 2512 + j120 = 2514.86/2.735◦ V(rms) docsity.com 10–16 CHAPTER 10. Sinusoidal Steady State Power Calculations [b] T = 1 f = 1 60 = 16.67 ms 2.735◦ 360◦ = t 16.67 ms ; ·. . t = 126.62 µs [c] VL lags Vg by 2.735◦ or 126.62 µs P 10.23 [a] From the solution to Problem 9.56 we have: Vo = j80 = 80/90◦ V Sg = −12VoI ∗ g = − 1 2 (j80)(10 − j10) = −400 − j400 VA Therefore, the independent current source is delivering 400 W and 400 magnetizing vars. I1 = Vo 5 = j16 A P5Ω = 1 2 (16)2(5) = 640 W Therefore, the 8 Ω resistor is absorbing 640 W. I∆ = Vo −j8 = −10 A Qcap = 1 2 (10)2(−8) = −400 VAR Therefore, the −j8 Ω capacitor is developing 400 magnetizing vars. 2.4I∆ = −24 V I2 = Vo − 2.4I∆ j4 = j80 + 24 j4 = 20 − j6 A = 20.88/− 16.7◦ A docsity.com Problems 10–17 Qj4 = 1 2 |I2|2(4) = 872 VAR Therefore, the j4 Ω inductor is absorbing 872 magnetizing vars. Sd.s. = 12(2.4I∆)I ∗ 2 = 1 2(−24)(20 + j6) = −240 − j72 VA Thus the dependent source is delivering 240 W and 72 magnetizing vars. [b] ∑ Pgen = 400 + 240 = 640 W = ∑ Pabs [c] ∑ Qgen = 400 + 400 + 72 = 872 VAR = ∑ Qabs P 10.24 [a] From the solution to Problem 9.58 we have Ia = −j10 A; Ib = −20 + j10 A; Io = 20 − j20 A S100V = −12(100)I ∗ a = −50(j10) = −j500 VA Thus, the 100 V source is developing 500 magnetizing vars. Sj100V = −12(j100)I∗b = −j50(−20 − j10) = −500 + j1000 VA Thus, the j100 V source is developing 500 W and absorbing 1000 magnetizing vars. P10Ω = 1 2 |Ia|2(10) = 500 W Thus the 10 Ω resistor is absorbing 500 W. Q−j10Ω = 1 2 |Ib|2(−10) = −2500 VAR Thus the −j10 Ω capacitor is developing 2500 magnetizing vars. Qj5Ω = 1 2 |Io|2(5) = 2000 VAR Thus the j5 Ω inductor is absorbing 2000 magnetizing vars. [b] ∑ Pdev = 500 W = ∑ Pabs docsity.com 10–20 CHAPTER 10. Sinusoidal Steady State Power Calculations I2 = 5000 − j2000 125 = 40 − j16 A (rms) I3 = 10,000 + j0 250 = 40 + j0 A (rms) ·. . Ig1 = I1 + I3 = 72 − j8 A (rms) In = I1 − I2 = −8 + j8 A (rms) Ig2 = I2 + I3 = 80 − j16 A(rms) Vg1 = 0.05Ig1 + 125 + 0.15In = 127.4 + j0.8 V(rms) Vg2 = −0.15In + 125 + 0.05Ig2 = 130.2 − j2 V(rms) Sg1 = [(127.4 + j0.8)(72 + j8)] = [9166.4 + j1076.8] VA Sg2 = [(130.2 − j2)(80 + j16)] = [10,448 + j1923.2] VA Note: Both sources are delivering average power and magnetizing VAR to the circuit. [b] P0.05 = |Ig1|2(0.05) = 262.4 W P0.15 = |In|2(0.15) = 19.2 W P0.05 = |Ig2|2(0.05) = 332.8 W∑ Pdis = 262.4 + 19.2 + 332.8 + 4000 + 5000 + 10,000 = 19,614.4 W ∑ Pdev = 9166.4 + 10,448 = 19,614.4 W = ∑ Pdis ∑ Qabs = 1000 + 2000 = 3000 VAR ∑ Qdel = 1076.8 + 1923.2 = 3000 VAR = ∑ Qabs P 10.29 [a] Let VL = Vm/0◦: SL = 600(0.8 + j0.6) = 480 + j360 VA I∗ = 480 Vm + j 360 Vm ; I = 480 Vm − j 360 Vm docsity.com Problems 10–21 120/θ = Vm + (480 Vm − j 360 Vm ) (1 + j2) 120Vm/θ = V 2m + (480 − j360)(1 + j2) = V 2m + 1200 + j600 120Vm cos θ = V 2m + 1200; 120Vm sin θ = 600 (120)2V 2m = (V 2 m + 1200) 2 + 6002 14,400V 2m = V 4 m + 2400V 2 m + 18 × 105 or V 4m − 12,000V 2m + 18 × 105 = 0 Solving, Vm = 108.85 V and Vm = 12.326 V If Vm = 108.85 V: sin θ = 600 (108.85)(120) = 0.045935; ·. . θ = 2.63◦ If Vm = 12.326 V: sin θ = 600 (12.326)(120) = 0.405647; ·. . θ = 23.93◦ [b] P 10.30 [a] SL = 20,000(0.85 + j0.53) = 17,000 + j10,535.65 VA 125I∗L = (17,000 + j10,535.65); I ∗ L = 136 + j84.29 A(rms) ·. . IL = 136 − j84.29 A(rms) Vs = 125 + (136 − j84.29)(0.01 + j0.08) = 133.10 + j10.04 = 133.48/4.31◦ V(rms) |Vs| = 133.48 V(rms) [b] P = |I|2(0.01) = (160)2(0.01) = 256 W docsity.com 10–22 CHAPTER 10. Sinusoidal Steady State Power Calculations [c] (125)2 XC = −10,535.65; XC = −1.483 Ω − 1 ωC = −1.48; C = 1 (1.48)(120π) = 1788.59 µF [d] I = 136 + j0 A(rms) Vs = 125 + 136(0.01 + j0.08) = 126.36 + j10.88 = 126.83/4.92◦ V(rms) |Vs| = 126.83 V(rms) [e] P = (136)2(0.01) = 184.96 W P 10.31 IL = 153,600 − j115,200 4800 = 32 − j24 A(rms) IC = 4800 −jXC = j 4800 XC = jIC I = 32 − j24 + jIC = 32 + j(IC − 24) Vs = 4800 + (2 + j10)[32 + j(IC − 24)] = (5104 − 10IC) + j(272 + 2IC) |Vs|2 = (5104 − 10IC)2 + (272 + 2IC)2 = (4800)2 ·. . 104I2C − 100,992IC + 3,084,800 = 0 Solving, IC = 31.57 A(rms); IC = 939.51 A(rms) *Select the smaller value of IC to minimize the magnitude of I. ·. . XC = − 480031.57 = −152.04 ·. . C = 1 (152.04)(120π) = 17.45 µF docsity.com Problems 10–25 Short circuit current: V2 = 5Iφ = 100 − 5Iφ 25 + j10 Iφ = 3 − j1 A Isc = 5Iφ 1 = 15 − j5 A ZTh = 15 15 − j5 = 0.9 + j0.3 Ω ZL = Z∗Th = 0.9 − j0.3 Ω IL = 15 1.8 = 8.33 A(rms) P = |IL|2(0.9) = 62.5 W [b] VL = (0.9 − j0.3)(8.33) = 7.5 − j2.5 V(rms) I1 = VL j3 = −0.833 − j2.5 A(rms) docsity.com 10–26 CHAPTER 10. Sinusoidal Steady State Power Calculations I2 = I1 + IL = 7.5 − j2.5 A(rms) 5Iφ = I2 + VL ·. . Iφ = 3 − j1 A Id.s. = Iφ − I2 = −4.5 + j1.5 A Sg = −100(3 + j1) = −300 − j100 VA Sd.s. = 5(3 − j1)(−4.5 − j1.5) = −75 + j0 VA Pdev = 300 + 75 = 375 W % developed = 62.5 375 (100) = 16.67% Checks: P25Ω = (10)(25) = 250 W P1Ω = (62.5)(1) = 62.5 W P0.9Ω = 62.5 W∑ Pabs = 250 + 62.5 + 62.5 = 375 W = ∑ Pdev Qj10 = (10)(10) = 100 VAR Qj3 = (6.94)(3) = 20.83 VAR Q−j0.3 = (69.4)(−0.3) = −20.83 VAR Qsource = −100 VAR∑ Q = 100 + 20.83 − 20.83 − 100 = 0 P 10.37 [a] Open circuit voltage: Vφ − 100 5 + Vφ j5 − 0.1Vφ = 0 ·. . Vφ = 40 + j80 V(rms) docsity.com Problems 10–27 VTh = Vφ + 0.1Vφ(−j5) = Vφ(1 − j0.5) = 80 + j60 V(rms) Short circuit current: Isc = 0.1Vφ + Vφ −j5 = (0.1 + j0.2)Vφ Vφ − 100 5 + Vφ j5 + Vφ −j5 = 0 ·. . Vφ = 100 V(rms) Isc = (0.1 + j0.2)(100) = 10 + j20 A(rms) ZTh = VTh Isc = 80 + j60 10 + j20 = 4 − j2 Ω ·. . Ro = |ZTh| = 4.47 Ω [b] I = 80 + j60 4 + √ 20 − j2 = 7.36 + j8.82 A(rms) P = (11.49)2( √ 20) = 590.17 W docsity.com 10–30 CHAPTER 10. Sinusoidal Steady State Power Calculations P 10.39 [a] Set Co = 0.1 µF, so −j/ωC = −j2000 Ω; also set Ro = 4123.1 Ω I = 120 8123.1 + j1000 = 14.55 − j1.79 mA P = 1 2 |I|2(4123.1) = 443.18 mW [b] Yes; 443.18 mW > 360 mW [c] Yes; 443.18 mW < 450 mW P 10.40 [a] 1 ωC = 100 Ω; C = 1 (100)(120π) = 26.53 µF [b] Iwo = 13,800 300 + 13,800 j100 = 46 − j138 A(rms) Vswo = 13,800 + (46 − j138)(1.5 + j12) = 15,525 + j345 = 15,528.83/1.27◦ V(rms) Iw = 13,800 300 = 46 A(rms) Vsw = 13,800 + 46(1.5 + j12) = 13,869 + j552 = 13,879.98/2.28◦ V(rms) % increase = ( 15,528.82 13,879.98 − 1 ) (100) = 11.88% [c] Pwo = |46 − j138|21.5 = 31.74 kW Pw = 462(1.5) = 3174 W % increase = (31,740 3174 − 1 ) (100) = 900% P 10.41 [a] So = original load = 1600 + j 1600 0.8 (0.6) = 1600 + j1200 kVA Sf = final load = 1920 + j 1920 0.96 (0.28) = 1920 + j560 kVA ·. . Qadded = 560 − 1200 = −640 kVAR [b] deliver [c] Sa = added load = 320 − j640 = 715.54/− 63.43◦ kVA pf = cos(−63.43) = 0.4472 leading docsity.com Problems 10–31 [d] I∗L = (1600 + j1200) × 103 2400 = 666.67 + j500 A IL = 666.67 − j500 = 833.33/− 36.87◦ A(rms) |IL| = 833.33 A(rms) [e] I∗L = (1920 + j560) × 103 2400 = 800 + j233.33 IL = 800 − j233.33 = 833.33/− 16.26◦ A(rms) |IL| = 833.33 A(rms) P 10.42 [a] Pbefore = Pafter = (833.33)2(0.05) = 34,722.22 W [b] Vs(before) = 2400 + (666.67 − j500)(0.05 + j0.4) = 2633.33 + j241.67 = 2644.4/5.24◦ V(rms) |Vs(before)| = 2644.4 V(rms) Vs(after) = 2400 + (800 + j233.33)(0.05 + j0.4) = 2346.67 + j331.67 = 2369.99/8.04◦ V(rms) |Vs(after)| = 2369.99 V(rms) P 10.43 [a] 180 = 3I1 + j4I1 + j3(I2 − I1) + j9(I1 − I2) − j3I1 0 = 9I2 + j9(I2 − I1) + j3I1 Solving, I1 = 18 − j18 A(rms); I2 = 12/0◦ A(rms) ·. . Vo = (12)(9) = 108/0◦ V(rms) [b] P = (12)2(9) = 1296 W [c] Sg = −(180)(18 + j18) = −3240 − j3240 VA ·. . Pg = −3240 W % delivered = 1296 3240 (100) = 40% docsity.com 10–32 CHAPTER 10. Sinusoidal Steady State Power Calculations P 10.44 [a] Open circuit voltage: 180 = 3I1 + j4I1 − j3I1 + j9I1 − j3I1 ·. . I1 = 1803 + j7 = 9.31 − j21.72 A(rms) VTh = j9I1 − j3I1 = j6I1 = 130.34 + j55.86 V = 141.81/23.20◦ V(rms) Short circuit current: 180 = 3I1 + j4I1 + j3(Isc − I1) + j9(I1 − Isc) − j3I1 0 = j9(Isc − I1) + j3I1 Solving, Isc = 20 − j20 A I1 = 30 − j20 A ZTh = VTh Isc = 130.34 + j55.86 20 − j20 = 1.86 + j4.66 Ω IL = 130.34 + j55.86 3.72 = 35 + j15 = 38.08/23.20◦ A docsity.com Problems 10–35 [b] When I2 = 0 I1 = 54 1 + j2 = 10.8 − j21.6 A(rms) Pg = (54)(10.8) = 583.2 W Check: Ploss = |I1|2(1) = 583.2 W P 10.48 [a] From Problem 9.67, ZTh = 85 + j85 Ω and VTh = 850 + j850 V. Thus, for maximum power transfer, ZL = Z∗Th = 85 − j85 Ω: I2 = 850 + j850 170 = 5 + j5 A 425/0◦ = (5 + j5)I1 − j20(5 + j5) ·. . I1 = 325 + j1005 + j5 = 42.5 − j22.5 A Sg(del) = 425(42.5 + j22.5) = 18,062.5 + j9562.5 VA Pg = 18,062.5 W [b] Ploss = |I1|2(5) + |I2|2(45) = 11,562.5 + 2250 = 13,812.5 W % loss in transformer = 18,062.5 − 13,812.5 18,062.5 (100) = 23.53% docsity.com 10–36 CHAPTER 10. Sinusoidal Steady State Power Calculations P 10.49 [a] From Problem 9.70, Zab = 100 + j136.26 so I1 = 50 100 + j13.74 + 100 + 136.26 = 50 200 + j150 = 160 − j120 mA I2 = jωM Z22 I1 = j270 800 + j600 (0.16 − j0.12) = 51.84 + j15.12 mA VL = (300 + j100)(51.84 + j15.12)103 = 14.04 + j9.72 V |VL| = 17.08 V [b] Pg(ideal) = 50(0.16) = 8 W Pg(practical) = 8 − |I1|2(100) = 4 W PL = |I2|2(300) = 874.8 mW % delivered = 0.8748 4 (100) = 21.87% P 10.50 [a] Open circuit: VTh = 120 16 + j12 (j10) = 36 + j48 V Short circuit: (16 + j12)I1 − j10Isc = 120 −j10I1 + (11 + j23)Isc = 0 Solving, Isc = 2.4/0◦ A ZTh = 36 + j48 2.4 = 15 + j20 Ω ·. . ZL = Z∗Th = 15 − j20 Ω IL = VTh ZTh + ZL = 36 + j48 30 = 1.2 + j1.6 A(rms) = 2.0/53.13◦ A(rms) PL = |IL|2(15) = 60 W docsity.com Problems 10–37 [b] I1 = Z22I2 jωM = 26 + j3 j10 (1.2 + j1.6) = 5.23/− 30.29◦ A)rms) Ptransformer = (120)(5.23) cos(−30.29◦) − (5.23)2(4) = 432.8 W % delivered = 60 432.8 (100) = 13.86% P 10.51 [a] jωL1 = j(10,000)(1 × 10−3) = j10 Ω jωL2 = j(10,000)(1 × 10−3) = j10 Ω jωM = j10 Ω 200 = (5 + j10)Ig + j5IL 0 = j5Ig + (15 + j10)IL Solving, Ig = 10 − j15 A; IL = −5 A Thus, ig = 18.03 cos(10,000t − 56.31◦) A iL = 5 cos(10,000t − 180◦) A [b] k = M√ L1L2 = 0.5√ 1 = 0.5 [c] When t = 50π µs: 10,000t = (10,000)(50π) × 10−6 = 0.5π rad = 90◦ ig(50π µs) = 18.03 cos(90◦ − 56.31◦) = 15 A iL(50π µs) = 5 cos(90◦ − 180◦) = 0 A w = 1 2 L1i 2 1 + 1 2 L2i 2 2 + Mi1i2 = 1 2 (10−3)(15)2 + 0 + 0 = 112.5 mJ When t = 100π µs: 10,000t = (104)(100π) × 10−6 = π = 180◦ ig(100π µs) = 18.03 cos(180 − 56.31◦) = −10 A iL(100π µs) = 5 cos(180 − 180◦) = 5 A w = 1 2 (10−3)(10)2 + 1 2 (10−3)(5)2 + 0.5 × 10−3(−10)(5) = 37.5 mJ docsity.com 10–40 CHAPTER 10. Sinusoidal Steady State Power Calculations Sc = VcI∗c = 31.25 + j0 VA Sd = VdI∗d = −6.25 + j25 VA Se = VeI∗e = 0 − j101.5625 VA Sf = VfI∗f = 31.25 VA [c] ∑ Pdev = 62.5 W∑ Pabs = 6.25 + 31.25 − 6.25 + 31.25 = 62.5 W Note that the total power absorbed by the coupled coils is zero: 6.25 − 6.25 = 0 = Pb + Pd [d] ∑ Qdev = 101.5625 VAR The capacitor is developing magnetizing vars.∑ Qabs = 75 + 1.5625 + 25 = 101.5625 VAR∑ Q absorbed by the coupled coils is Qb + Qd = 26.5625 VAR P 10.53 Open circuit voltage: I1 = 10/0◦ 1 + j2 = 2 − j4 A VTh = j2I1 + j1.2I1 = j3.2I1 = 12.8 + j6.4 = 14.31/26.57◦ V Short circuit current: 10/0◦ = (1 + j2)I1 − j3.2Isc docsity.com Problems 10–41 0 = −j3.2I1 + j5.4Isc Solving, Isc = 5.89/− 5.92◦ A ZTh = 14.31/26.57◦ 5.89/− 5.92◦ = 2.43/32.49 ◦ = 2.048 + j1.304 Ω ·. . I2 = 14.31/26.57 ◦ 4.096 = 3.49/26.57◦ A 10/0◦ = (1 + j2)I1 − j3.2I2 ·. . I1 = 10 + j3.2I21 + j2 = 10 + j3.2(3.49/26.57◦) 1 + j2 = 5/0◦ A Zg = 10/0◦ 5/0◦ = 2 + j0 = 2/0◦ Ω P 10.54 [a] 272/0◦ = 2Ig + j10Ig + j14(Ig − I2) − j6I2 +j14Ig − j8I2 + j20(Ig − I2) 0 = j20(I2 − Ig) − j14Ig + j8I2 + j4I2 +j8(I2 − Ig) − j6Ig + 8I2 docsity.com 10–42 CHAPTER 10. Sinusoidal Steady State Power Calculations Solving, Ig = 20 − j4 A(rms); I2 = 24/0◦ A(rms) P8Ω = (24)2(8) = 4608 W [b] Pg(developed) = (272)(20) = 5440 W [c] Zab = Vg Ig − 2 = 272 20 − j4 − 2 = 11.08 + j2.62 = 11.38/13.28 ◦ Ω [d] P2Ω = |Ig|2(2) = 832 W∑ Pdiss = 832 + 4608 = 5440 W = ∑ Pdev P 10.55 [a] 300 = 60I1 + V1 + 20(I1 − I2) 0 = 20(I2 − I1) + V2 + 40I2 V2 = 1 4 V1; I2 = −4I1 Solving, V1 = 260 V(rms); V2 = 65 V(rms) I1 = 0.25 A(rms); I2 = −1.0 A(rms) V5A = V1 + 20(I1 − I2) = 285 V(rms) ·. . P = −(285)(5) = −1425 W Thus 1425 W is delivered by the current source to the circuit. [b] I20Ω = I1 − I2 = 1.25 A(rms) ·. . P20Ω = (1.25)2(20) = 31.25 W docsity.com Problems 10–45 Pg(del) = (2.25 × 10−3)(100) = 225 mW % delivered = 90 225 (100) = 40% P 10.60 [a] Zab = ( 1 + N1 N2 )2 (1 − j2) = 25 − j50 Ω ·. . I1 = 100/0 ◦ 15 + j50 + 25 − j50 = 2.5/0 ◦ A I2 = N1 N2 I1 = 10/0◦ A ·. . IL = I1 + I2 = 12.5/0◦ A(rms) P1Ω = (12.5)2(1) = 156.25 W P15Ω = (2.5)2(15) = 93.75 W [b] Pg = −100(2.5/0◦) = −250 W∑ Pabs = 156.25 + 93.75 = 250 W = ∑ Pdev P 10.61 [a] 25a21 + 4a 2 2 = 500 I25 = a1I; P25 = a21I 2(25) I4 = a2I; P4 = a22I 2(4) P4 = 4P25; a22I 24 = 100a21I 2 ·. . 100a21 = 4a22 25a21 + 100a 2 1 = 500; a1 = 2 25(4) + 4a22 = 500; a2 = 10 [b] I = 2000/0◦ 500 + 500 = 2/0◦ A(rms) I25 = a1I = 4 A P25Ω = (16)(25) = 400 W [c] I4 = a2I = 10(2) = 20 A(rms) V4 = (20)(4) = 80/0◦ V(rms) docsity.com 10–46 CHAPTER 10. Sinusoidal Steady State Power Calculations P 10.62 [a] Open circuit voltage: 500 = 100I1 + V1 V2 = 400I2 V1 1 = V2 2 ·. . V2 = 2V1 I1 = 2I2 Substitute and solve: 2V1 = 400I1/2 = 200I1 ·. . V1 = 100I1 500 = 100I1 + 100I1 ·. . I1 = 500/200 = 2.5 A ·. . I2 = 12I1 = 1.25 A V1 = 100(2.5) = 250 V; V2 = 2V1 = 500 V VTh = 20I1 + V1 − V2 + 40I2 = −150 V(rms) Short circuit current: 500 = 80(Isc + I1) + 360(Isc + 0.5I1) 2V1 = 40 I1 2 + 360(Isc + 0.5I1) 500 = 80(I1 + Isc) + 20I1 + V1 docsity.com Problems 10–47 Solving, Isc = −1.47 A; I1 = 4.41 A; V1 = 176.47 V RTh = VTh Isc = −150 −1.47 = 102 Ω P = 752 102 = 55.15 W [b] 500 = 80[I1 − (75/102)] − 75 + 360[I2 − (75/102)] 575 + 6000 102 + 27,000 102 = 80I1 + 180I1 ·. . I1 = 3.456 A Psource = (500)[3.456 − (75/102)] = 1360.29 W % delivered = 55.15 1360.29 (100) = 4.05% [c] P80Ω = 80 ( I1 − 75102 )2 = 592.13 W P20Ω = 20I21 = 238.86 W P40Ω = 40I22 = 119.43 W P102Ω = 752 102 = 55.15 W P360Ω = 360 ( I2 − 75102 )2 = 354.73 W ∑ Pabs = 592.13 + 238.86 + 119.43 + 55.15 + 354.73 = 1360.3 W = ∑ Pdev docsity.com 10–50 CHAPTER 10. Sinusoidal Steady State Power Calculations Transfer the secondary impedance to the primary side: Zp = 1 25 (100 + jXC) = 4 + j XC 25 Ω Now maximize I by setting (XC/25) = −8 Ω: ·. . C = 1 200(20 × 103) = 0.25 µF [b] I = 60 + j120 10 = 6 + j12 A P = |I|2(4) = 720 W [c] Ro 25 = 6 Ω; ·. . Ro = 150 Ω [d] I = 60 + j120 12 = 5 + j10 A P = |I|2(6) = 750 W P 10.65 [a] Zab = 50 − j400 = ( 1 − N1 N2 )2 ZL = ( 1 − 2800 700 )2 ZL = 9ZL ·. . ZL = 19(50 − j400) = 5.556 − j44.444 Ω [b] I1 = 24 100 = 240/0◦ mA docsity.com Problems 10–51 N1I1 = −N2I2 I2 = −4I1 = 960/180◦ mA IL = I1 + I2 = 720/180◦ mA(rms) VL = (5.556 − j44.444)IL = −4 + j32 = 32.25/97.13◦ V(rms) P 10.66 [a] Begin with the MEDIUM setting, as shown in Fig. 10.31, as it involves only the resistor R2. Then, Pmed = 500 W = V 2 R2 = 1202 R2 Thus, R2 = 1202 500 = 28.8 Ω [b] Now move to the LOW setting, as shown in Fig. 10.30, which involves the resistors R1 and R2 connected in series: Plow = V 2 R1 + R2 = V 2 R1 + 28.8 = 250 W Thus, R1 = 1202 250 − 28.8 = 28.8 Ω [c] Note that the HIGH setting has R1 and R2 in parallel: Phigh = V 2 R1‖R2 = 1202 28.8‖28.8 = 1000 W If the HIGH setting has required power other than 1000 W, this problem sould not have been solved. In other words, the HIGH power setting was chosen in such a way that it would be satisfied once the two resistor values were calculated to satisfy the LOW and MEDIUM power settings. docsity.com 10–52 CHAPTER 10. Sinusoidal Steady State Power Calculations P 10.67 [a] PL = V 2 R1 + R2 ; R1 + R2 = V 2 PL PM = V 2 R2 ; R2 = V 2 PM PH = V 2(R1 + R2) R1R2 R1 + R2 = V 2 PL ; R1 = V 2 PL − V 2 PM PH = V 2V 2/PL( V 2 PL − V 2 PM ) ( V 2 PM ) = PMPLPM PL(PM − PL) PH = P 2M PM − PL [b] PH = (750)2 (750 − 250) = 1125 W P 10.68 First solve the expression derived in P10.67 for PM as a function of PL and PH. Thus PM − PL = P 2 M PH or P 2M PH − PM + PL = 0 P 2M − PMPH + PLPH = 0 ·. . PM = PH2 ± √( PH 2 )2 − PLPH = PH 2 ± PH √ 1 4 − ( PL PH ) For the specified values of PL and PH PM = 500 ± 1000 √ 0.25 − 0.24 = 500 ± 100 ·. . PM1 = 600 W; PM2 = 400 W Note in this case we design for two medium power ratings If PM1 = 600 W R2 = (120)2 600 = 24 Ω docsity.com