Sinusoidal Steady State Power Calculations

Transcription

1 10 Sinusoidal Steady State Power Calculations Assessment Problems AP 10.1 [a] V = 100/ 45 V, Therefore I = 20/15 A P = 1 (100)(20)cos[ 45 (15)] = 500W, 2 A B Q = 1000sin 60 = VAR, B A [b] V = 100/ 45, I = 20/165 P = 1000cos( 210 ) = W, Q = 1000sin( 210 ) = 500 VAR, B A A B [c] V = 100/ 45, I = 20/ 105 P = 1000cos(60 ) = 500W, A B Q = 1000sin(60 ) = VAR, A B [d] V = 100/0, I = 20/120 P = 1000cos( 120 ) = 500W, B A Q = 1000sin( 120 ) = VAR, B A AP 10.2 pf = cos(θ v θ i ) = cos[15 (75)] = cos( 60 ) = 0.5 leading rf = sin(θ v θ i ) = sin( 60 ) =

2 10 2 CHAPTER 10. Sinusoidal Steady State Power Calculations AP 10.3 From Ex. 9.4 I eff = I ρ 3 = A ( ) P = IeffR 2 = (5000) = 54W 3 AP 10.4 [a] Z = (39 + j26) ( j52) = 48 j20 = 52/ Ω Therefore I l = 250/0 48 j j4 = 4.85/18.08 A (rms) V L = ZI l = (52/ )(4.85/18.08 ) = / 4.54 V (rms) I L = V L 39 + j26 = 5.38/ A (rms) [b] S L = V L I L = (252.20/ 4.54 )(5.38/ ) = 1357/33.69 = ( j752.73)va P L = W; Q L = VAR [c] P l = I l 2 1 = (4.85) 2 1 = 23.52W; Q l = I l 2 4 = VAR [d] S g (delivering) = 250I l = ( j376.36)va Therefore the source is delivering W and absorbing magnetizing VAR. [e] Q cap = V L 2 52 = (252.20)2 = VAR 52 Therefore the capacitor is delivering magnetizing VAR. Check: = VAR and = W AP 10.5 Series circuit derivation: S = 250I = (40,000 j30,000) Therefore I = 160 j120 = 200/ A (rms) I = 200/36.87 A (rms) Z = V I = /36.87 = 1.25/ = (1 j0.75)ω Therefore R = 1Ω, X C = 0.75Ω

3 Problems 10 3 Parallel circuit derivation P = (250)2 (250)2 ; therefore R = R 40,000 = Ω Q = (250)2 X C ; therefore X C = (250)2 30,000 = 2.083Ω AP 10.6 S 1 = 15,000(0.6) + j15,000(0.8) = j12,000va S 2 = 6000(0.8) j6000(0.6) = 4800 j3600va S T = S 1 + S 2 = 13,800 + j8400va S T = 200I ; therefore I = 69 + j42 I = 69 j42a V s = ji = j = j69 = /15.91 V (rms) AP 10.7 [a] The phasor domain equivalent circuit and the Thévenin equivalent are shown below: Phasor domain equivalent circuit: Thévenin equivalent: V Th = 3 j j40 = 48 j24 = 53.67/ V Z Th = 4 + j18 + j j40 = 20 + j10 = 22.36/26.57 Ω For maximum power transfer, Z L = (20 j10)ω

4 10 4 CHAPTER 10. Sinusoidal Steady State Power Calculations AP / [b] I = = 1.34/ A 40 ( ) Therefore P = 20 = 17.96W 2 [c] R L = Z Th = 22.36Ω 53.67/ [d] I = = 1.23/ A j10 ( ) Therefore P = (22.36) = 17W 2 Mesh current equations: 660 = (34 + j50)i 1 + j100(i 1 I 2 ) + j40i 1 + j40(i 1 I 2 ) AP 10.9 [a] 0 = j100(i 2 I 1 ) j40i I 2 Solving, I 2 = 3.5/0 A;.. P = 1 2 (3.5)2 (100) = W 248 = j400i 1 j500i (I 1 I 2 ) 0 = 375(I 2 I 1 ) + j1000i 2 j500i I 2 Solving, I 1 = 0.80 j0.62a; I 2 = 0.4 j0.3 = 0.5/ P = 1 (0.25)(400) = 50W 2

5 Problems 10 5 [b] I 1 I 2 = 0.4 j0.32a P 375 = 1 2 I 1 I 2 2 (375) = 49.20W [c] P g = 1 (248)(0.8) = 99.20W 2 Pabs = = 99.20W (checks) AP [a] V Th = 210V; V 2 = 1 4 V 1; I 1 = 1 4 I 2 Short circuit equations: 840 = 80I 1 20I 2 + V 1 0 = 20(I 2 I 1 ) V 2.. I 2 = 14A; R Th = = 15Ω [b] P max = ( ) = 735W 30 AP [a] V Th = 4(146/0 ) = 584/0 V (rms) [b] P = V 2 = 4V 1 ; I 1 = 4I 2 Short circuit equations: 146/0 = 80I 1 20I 2 + V 1 0 = 20(I 2 I 1 ) V 2.. I 2 = 146/365 = 0.40A; R Th = = 1460Ω ( ) = 58.40W 2920

6 10 6 CHAPTER 10. Sinusoidal Steady State Power Calculations Problems P 10.1 [a] P = 1 2 (250)(4)cos( ) = 500cos 75 = W (abs) Q = 500sin 75 = VAR (abs) [b] P = 1 2 (18)(5)cos( ) = 45cos(105 ) = 11.65W (del) Q = 45sin(105 ) = VAR (abs) [c] P = 1 2 (150)(2)cos( 65 50) = 150cos( 115 ) = 63.39W (del) Q = 150sin( 115 ) = VAR (del) [d] P = 1 2 (80)(10)cos( ) = 400cos( 50 ) = W (abs) Q = 400sin( 50 ) = VAR (del) P 10.2 [a] coffee maker = 1200 W frying pan = 1196 W microwave oven = 1450W toaster = 1146W P = 4992W P 10.3 Therefore I eff = = 41.6A The breaker will not trip. [b] P = = 6482W; I eff = = 54.02A The breaker will trip because the current is greater than 20A. p = P + P cos2ωt Q sin 2ωt; dp dt = 2ωP sin 2ωt 2ωQ cos 2ωt dp dt = 0 when 2ωP sin 2ωt = 2ωQ cos 2ωt or tan2ωt = Q P cos 2ωt = P P 2 + Q 2; sin2ωt = Q P 2 + Q 2

7 Problems 10 7 Let θ = tan 1 ( Q/P), then p is maximum when 2ωt = θ and p is minimum when 2ωt = (θ + π). Therefore p max = P + P and p min = P P P P 2 + Q Q( Q) 2 P 2 P + Q = P Q 2 2 P P 2 + Q Q Q 2 P 2 P + Q = P 2 + Q 2 2 P 10.4 [a] P = 1 (240) = 60W 1 ωc Q = 1 2 p max = P + = (5000)(5) = 360Ω (240) 2 ( 360) = 80VAR P 2 + Q 2 = 60 + (60) 2 + (80) 2 = 160W (del) [b] p min = = 40W (abs) [c] P = 60W from (a) [d] Q = 80VAR from (a) [e] generates, because Q < 0 [f] pf = cos(θ v θ i ) I = j360 = j0.67 = 0.83/53.13 A.. pf = cos( ) = 0.6 leading P 10.5 [g] rf = sin( ) = 0.8 I g = 4/0 ma; 1 jωc = j1250ω; jωl = j500ω Z eq = [ j1250 ( j500)] = 1500 j500ω P g = 1 2 I 2 Re{Z eq } = 1 2 (0.004)2 (1500) = 12mW The source delivers 12 mw of power to the circuit.

8 10 8 CHAPTER 10. Sinusoidal Steady State Power Calculations P 10.6 jωl = j20,000( ) = j10ω; 1 jωc = 10 6 j20,000(1.25) = j40ω 6 + V o j10 + V o 30(V o /j10) 30 j40 [. 1. V o j j3 ] = 6 30 j40.. V o = 100/ V = 0.. I = V o j10 = 10/36.87 A I o = 6/0 I = 6 8 j6 = 2 j6 = 6.32/ A P 30Ω = 1 2 I o 2 30 = 600W P 10.7 Z f = j10,000 20,000 = 4000 j8000ω Z i = 2000 j2000ω.. Z f Z i = 4000 j j2000 = 3 j1 V o = Z f Z i V g ; V g = 1/0 V V o = (3 j1)(1) = 3 + j1 = 3.16/ V P = 1 Vm 2 2 R = 1 (10) = = 5mW

9 Problems 10 9 P 10.8 [a] From the solution to Problem 9.63 we have: I a = 67.5 j7.5ma and I b = j22.5ma S j18v = 1 ( j18)( j0.0075) = j607.5 mva 2 P j18v = 67.5 mw; and Q j18v = mvar S 12V = 1 (12)( j0.0225) = 135 j135 mva 2 P 12V = 135 mw; and Q 12V = 135 mvar P j400ω = 0W; and Q j400ω = 1 2 (400) I a 2 = mvar P 400Ω = 1 2 (400) I b 2 = 202.5mW; and Q 400Ω = 0 mvar P j100ω = 0W; and Q j100ω = 1 2 (100) I a I b 2 = 450 mvar [b] P gen = = 202.5mW = P abs [c] Q gen = = mVAR Qabs = = mVAR (check) P 10.9 [a] From the solution to Problem 9.64 we have I = j2.5a; and I a = j2.5a S 75V = 1 (75)(0.5 j2.5) = j93.75 VA 2

10 10 10 CHAPTER 10. Sinusoidal Steady State Power Calculations P 75V = 18.75W; and Q 75V = VAR S dep.source = 1 2 (100I )(I a I ) = 12.5 j62.5 VA P dep.source = 12.5W; and Q dep.source = 62.5 VAR P j50ω = 0W; and Q j50ω = 1 2 (50) I 2 = VAR P j20ω = 0W; and Q j20ω = 1 2 (20) I a 2 = 62.5 VAR P 10Ω = 1 2 (10) I a 2 = 31.25W; and Q 10Ω = 0 VAR P j550ω = 0, W; and Q j550ω = 1 2 (550) I I a 2 = VAR [b] P dev = = 31.25W = P abs [c] Q dev = = 225VAR Qabs = = 225VAR = Q dev P [a] line loss = = 2000W line loss = I g I g 2 = 80 I g = 80 A I g 2 R L = R L = 75Ω I g 2 X L = X L = 100Ω Thus, Z = (100) 2 + (X l 100) 2 I g = ,000 + (X l 100) 2

11 Problems ,000 + (X l 100) 2 = = 12,500 Solving, (X l 100) = ±50. Thus, X l = 150Ω or X l = 50Ω [b] If X l = 150Ω: I g = j50 = 8 j4a S g = 1000I g = 8000 j4000va Thus, the voltage source is delivering 8 kw and 4 kvars. Q j150 = I g 2 X l = 80(150) = 12 kvars Therefore the line reactance is absorbing 12 kvars. Q j100 = I g 2 X L = 80( 100) = 8kvars Therefore the load reactance is generating 8 kvars. Qgen = 12kvars = Q abs If X l = 50Ω: I g = j50 = 8 + j4a S g = 1000I g = j4000va Thus, the voltage source is delivering 8 kw and absorbing 4 kvars. Q j50 = I g 2 (50) = 80(50) = 4kvars Therefore the line reactance is absorbing 4 kvars. The load continues to deliver 8 kvars. Qgen = 8kvars = Q abs P [a] I eff = 40/115 = 0.35A [b] I eff = 130/115 = 1.13A P i(t) = 250t 0 t 80ms i(t) = t 80ms t 100ms { }. I rms = (250) t 2 dt + ( t) 2 dt (250) 2 t 2 dt = (250) 2 t =

12 10 12 CHAPTER 10. Sinusoidal Steady State Power Calculations ( t) 2 = t t dt = t dt = 10 5 t = t 2 dt = t = I rms = 10{(32/3) (488/3)} = 11.55A P P = I 2 rmsr.. R = 1280 (11.55) 2 = 9.6Ω P [a] A = 40 2 (5) + ( 40) 2 )(5) = 16,000 mean value = A 20 = 800 V rms = 800 = V(rms) [b] P = V 2 rms R = = 20W [c] R = V 2 rms P = = 80kΩ P [a] Area under one cycle of v 2 g: A = (100)( ) + 400( ) + 400( ) + 100( ) = 1000( ) Mean value of v 2 g : M.V. = A = 1000( ) = V rms = 250 = 15.81V (rms) [b] P = V 2 rms R = = 62.5W

13 Problems P W dc = V dc 2 to+t R T; W s = t o v 2 s R dt.. V 2 dc R T = to+t t o v 2 s R dt P [a] V 2 dc = 1 T V dc = 1 T to+t t o to+t t o v 2 s dt v 2 s dt = V rms = V eff Z eq = 50 (25 j25) = 20 j10ω.. V o = 0.05Z eq = 1 j0.5 V(rms) I 1 = V o 25 j25 = 30 + j10 ma(rms) S g = V g I g = (1 j0.5)(0.05) = 50 + j25mva [b] Source is delivering 50 mw. [c] Source is absorbing 25 mvar. [d] Q cap = I 1 2 (75) = 75 mvar P 50Ω = V o 2 = 25mW 50 P 25Ω = I 1 2 (25) = 25mW Q ind = I 1 2 (50) = 50 mvar P middle branch = 25mW; Q middle branch = 0 P right branch = 25mW; Q right branch = = 25 mvar

14 10 14 CHAPTER 10. Sinusoidal Steady State Power Calculations [e] P del = 50mW Pdiss = = 50mW = P del [f] Q abs = = 75mVAR = Q dev P jωl = j25ω; 1 jωc = j75ω I o = j j25 = j1.2a P = 1 2 I o 2 (50) = 1 (7.2)(50) = 180W 2 Q = 1 2 I o 2 (25) = 90VAR S = P + jq = j90va S = VA P [a] Let V L = V m /0 : S L = 2500(0.8 + j0.6) = j1500va I l = 2000 V m + j 1500 V m ; I l = 2000 V m j /θ = V m + ( 2000 j 1500 ) (1 + j2) V m V m 250V m /θ = V 2 m + (2000 j1500)(1 + j2) = V 2 m j2500 V m

15 Problems V m cos θ = Vm ; 250V m sinθ = 2500 (250) 2 Vm 2 = (Vm ) ,500Vm 2 = Vm ,000Vm or Vm 4 52,500Vm = 0 Solving, Vm 2 = 26,250 ± 25,647.86; V m = V and V m = 24.54V [b] If V m = V: sinθ = 2500 (227.81)(250) = 0.044;.. θ = 2.52 If V m = V: sinθ = 2500 (24.54)(250) = ;.. θ = 24.05

16 10 16 CHAPTER 10. Sinusoidal Steady State Power Calculations P [a] 1 jωc = j40ω; jωl = j80ω Z eq = 40 j40 + j = 80 + j60ω I g = 40/0 = 0.32 j0.24a 80 + j60 S g = 1 2 V gi g = 1 40( j0.24) = 6.4 j4.8va 2 P = 6.4W (del); Q = 4.8VAR(del) [b] I 1 = S = S g = 8VA j40 40 j40 I g = 0.04 j0.28a P 40Ω = 1 2 I 1 2 (40) = 1.6W P 60Ω = 1 2 I g 2 (60) = 4.8W Pdiss = = 6.4W = P dev [c] I j40ω = I g I 1 = j0.04a Q j40ω = 1 2 I j40ω 2 ( 40) = 1.6VAR(del) Q j80ω = 1 2 I g 2 (80) = 6.4VAR(abs) Qabs = = 4.8VAR = Q dev

17 Problems P S T = 40,800 + j30,600va S 1 = 20,000(0.96 j0.28) = 19,200 j5600va S 2 = S T S 1 = 21,600 + j36,200 = 42,154.48/ VA rf = sin( ) = pf = cos( ) = lagging P [a] S 1 = 10,000 j4000va S 2 = V L 2 Z 2 S 1 + S 2 = 16 + j4kva = (1000)2 60 j80 = 6 + j8kva 1000I L = 16,000 + j4000;.. IL = 16 j4a(rms) V g = V L + I L (0.5 + j0.05) = (16 j4)(0.5 + j0.05) = j1.2 = / Vrms [b] T = 1 f = 1 50 = 20ms = t 20 ms ;.. t = 3.79µs [c] V L leads V g by or 3.79µs P [a] I 1 = 6000 j = 40 j20a (rms)

18 Pdev = 13, ,490 = 27,720W = P dis Qdel = = 12,000VAR = Q abs CHAPTER 10. Sinusoidal Steady State Power Calculations I 2 = I 3 = j ,000 j = 50 + j30a (rms) = 40 j30a (rms) I g1 = I 1 + I 3 = 80 j50a (rms) I n = I 1 I 2 = 10 j50a (rms) I g2 = I 2 + I 3 = 90 + j0a V g1 = 0.1I g I n = 156 j15v(rms) V g2 = 0.2I n I g2 = j10v(rms) S g1 = [(156 j15)(80 + j50)] = [13,230 + j6600] VA S g2 = [(161 + j10)(90 + j0)] = [14,490 + j900] VA Note: Both sources are delivering average power and magnetizing VAR to the circuit. [b] P 0.1 = I g1 2 (0.1) = 890W P 0.2 = I n 2 (0.2) = 520W P 0.1 = I g2 2 (0.1) = 810W Pdis = ,000 = 27,720W Qabs = = 12,000VAR P [a] 120I 1 = 4800 j2400;.. I 1 = 40 + j20a(rms)

19 Problems I 2 = j3600;.. I 2 = 40 j30a(rms) I 3 = j6 = 10 j40a(rms) I g1 = I 1 + I 3 = 50 j20a S g1 = 120(50 + j20) = 6000 j2400va Thus the V g1 source is delivering 6 kw and 2.4 kvars. I g2 = I 2 + I 3 = 50 j70a(rms) S g2 = 120(50 + j70) = 6000 j8400va Thus the V g2 source is delivering 6 kw and 8.4 kvars. [b] P gen = = 12kW Pabs = (240)2 = 12kW = P gen 24 Qdel = = 13.2kVAR Qabs = (240)2 6 P S 1 = = j0va.. I 1 = = 34.51A S 2 = = j0va.. I 2 = !0 = A S 3 = = j0va.. I 3 = = 24.4A I g1 = I 1 + I 3 = 58.91A I g2 = I 2 + I 3 = A = 13.2kVAR = Q del The breaker protecting the upper service conductor will trip because its feeder current exceeds 50 A. The breaker protecting the lower service conductor will not trip because its feeder current is less than 50 A. Thus, service to Loads 1 and 3 will be interrupted.

20 10 20 CHAPTER 10. Sinusoidal Steady State Power Calculations P I 1 = 6000 j8000 I 1 = 24 j32;.. I1 = 24 + j32a(rms) 250I 2 = j3000 I 2 = 36 + j12;.. I 2 = 36 j12a(rms) I 3 = 250/0 25 = 10 + j0a; I 4 = 250/0 j5 = 0 + j50a I g = I 1 + I 2 + I 3 + I 4 = 70 + j70a V g = (70 + j70)(j0.1) = j7 = 243.1/1.65 V (rms) P [a] From Problem 9.78, Z ab = 20 + j27.25 I 1 = so j = 75 = 0.9 j1.2a 30 + j40 I 2 = jωm Z 22 I 1 = j54 (0.9 j1.2) = j0ma j120 V L = (60 + j20)( j0) = j8.1 V L = 25.61V [b] P g (ideal) = 75(0.9) = 67.5W P g (practical) = 67.5 I 1 2 (10) = 45W P L = I 2 2 (60) = W % delivered = (100) = 21.87% 45

21 Problems P [a] S 1 = 3 + j0kva; S 2 = 4 j3kva; S 3 = 5 j6kva S T = S 1 + S 2 + S 3 = 12 j9kva 300I = (12 j9) 10 3 ;.. I = 40 + j30a Z = j30 = 4.8 j3.6ω = 6/ Ω [b] pf = cos( ) = 0.8 leading P [a] From the solution to Problem we have [b] I L = I L = 40 + j30a (rms).. V s = 300/0 + (40 + j30)(0.2 + j0.05) = j8 = 306.6/1.5 V (rms) 16,000 P l = (2500)(0.2) = 500W [c] P s = 12, = 12.5kW [d] η = 12 (100) = 96% 12.5 P [a] Z 1 = j70 = 250/16.26 Ω Q l = (2500)(0.05) = 125VAR Q s = = 8.875kVAR pf = cos(16.26 ) = 0.96 lagging rf = sin(16.26 ) = 0.28 Z 2 = 160 j120 = 200/ Ω pf = cos( ) = 0.8 leading rf = sin( ) = 0.6 Z 3 = 30 j40 = 50/ Ω pf = cos( ) = 0.6 leading rf = sin( ) = 0.8 [b] Y = Y 1 + Y 2 + Y 3 Y 1 = 1 250/16.26 ; Y 2 = Y = j17.88ms 1 200/ ; Y 3 = 1 50/ 53.13

22 10 22 CHAPTER 10. Sinusoidal Steady State Power Calculations Z = 1 Y = 37.44/ Ω pf = cos( ) = 0.74 leading rf = sin( ) = 0.67 P [a] I = 270/ j48 = 2.7 j3.6 = 4.5/ A(rms) P [b] Y L = [c] Z L = P = (4.5) 2 (6) = 121.5W j40.. X C = = 12 j16 ms 1 = 62.5Ω = 83.33Ω [d] I = 270/ j8 = 3.01/ 5.12 A P = (3.01) 2 (6) = 54.37W [e] % = (100) = 44.75% Thus the power loss after the capacitor is added is 44.75% of the power loss before the capacitor is added. I L = 120,000 j90, = 25 j18.75a (rms) I C = 4800 jx C = j 4800 X C = ji C I l = 25 j ji C = 25 + j(i C 18.75)

23 Problems V s = (10 + j5)[25 + j(i C 18.75)] = ( I C ) + j( I C ) V s 2 = ( I C ) 2 + ( I C ) 2 = (4800) I 2 C 52,687.5I C + 3,422, = 0 I C = A(rms) and I C = A(rms) *Select the smaller value of I C to minimize the magnitude of I l... X C = = C = 1 (59.838)(120π) = 44.43µF P [a] S L = 20,000( j0.53) = 17,000 + j10,535.65va 125I L = (17,000 + j10,535.65);.. I L = 136 j84.29a(rms) I L = j84.29a(rms) V s = (136 j84.29)( j0.08) = j10.04 = /4.31 V(rms) V s = V(rms) [b] P l = I l 2 (0.01) = (160) 2 (0.01) = 256W [c] (125)2 X C = 10,535.65; X C = Ω 1 ωc = ; C = 1 ( )(120π) = µF [d] I l = j0a(rms) V s = ( j0.08) = j10.88 = /4.92 V(rms) V s = V(rms) [e] P l = (136) 2 (0.01) = W

24 10 24 CHAPTER 10. Sinusoidal Steady State Power Calculations P [a] S o = original load = j 1600 (0.6) = j1200kva 0.8 S f = final load = j 1920 (0.28) = j560kva Q added = = 640kVAR [b] deliver [c] S a = added load = 320 j640 = / kva [d] I L [e] I L = pf = cos( 63.43) = leading = ( j1200) = j500a I L = j500 = / A(rms) I L = A(rms) ( j560) = j I L = 800 j = / A(rms) I L = A(rms) P [a] P before = P after = (833.33) 2 (0.25) = kW P [a] [b] V s (before) = ( j500)( j0.1) = j58.33 = / 1.28 V(rms) V s (before) = V(rms) V s (after) = (800 j233.33)( j0.1) = j21.67 = /0.47 V(rms) V s (after) = V(rms) 50 = j10(i 1 I 2 ) + j10(i 3 I 2 ) + 10(I 1 I 3 )

25 Problems = 10I 2 + j20(i 2 I 3 ) + j10(i 2 I 1 ) + j10(i 2 I 1 ) + j10(i 2 I 3 ) 0 = j10i (I 3 I 1 ) + j20(i 3 I 2 ) + j10(i 1 I 2 ) Solving, I 1 = j0.5a(rms); I 2 = 3 + j2a(rms); I 3 = 2 + j2a(rms) I a = I 1 = j0.5a I c = I 2 = 3 + j2a I e = I 1 I 3 = 3.5 j1.5a I b = I 1 I 2 = 2.5 j1.5a I d = I 3 I 2 = 1A I f = I 3 = 2 + j2a [b] V a = 50V V c = 10I c = 30 + j20v V e = 10I e = 35 j15v V b = j10i b + j10i d = 15 + j15v V d = j20i d + j10i b = 15 + j5v V f = j10i f = 20 j20v S a = 50I a = j25va S b = V b I b = 15 + j60va S c = V c I c = j0va S d = V d I d = 15 j5va S e = V e I e = 145 j0va S f = V f I f = 0 j80va [c] P dev = = 290W Pabs = = 290W Note that the total power absorbed by the coupled coils is zero: = 0 = P b + P d

26 10 26 CHAPTER 10. Sinusoidal Steady State Power Calculations [d] Q dev = = 85VAR Qabs = = 85VAR Q absorbed by the coupled coils is Qb + Q d = 55 P [a] 272/0 = 2I g + j10i g + j14(i g I 2 ) j6i 2 +j14i g j8i 2 + j20(i g I 2 ) 0 = j20(i 2 I g ) j14i g + j8i 2 + j4i 2 +j8(i 2 I g ) j6i g + 8I 2 Solving, I g = 20 j4a(rms); I 2 = 24/0 A(rms) P 8Ω = (24) 2 (8) = 4608W [b] P g (developed) = (272)(20) = 5440W [c] Z ab = V g I g 2 = j4 2 = j2.62 = 11.38/13.28 Ω [d] P 2Ω = I g 2 (2) = 832W Pdiss = = 5440W = P dev P [a] 300 = 60I 1 + V (I 1 I 2 )

27 Problems = 20(I 2 I 1 ) + V I 2 V 2 = 1 4 V 1; I 2 = 4I 1 Solving, V 1 = 260V (rms); I 1 = 0.25A (rms); V 2 = 65V (rms) I 2 = 1.0A (rms) P [a] V 5A = V (I 1 I 2 ) = 285V (rms).. P = (285)(5) = 1425W Thus 1425 W is delivered by the current source to the circuit. [b] I 20Ω = I 1 I 2 = 1.25A(rms).. P 20Ω = (1.25) 2 (20) = 31.25W (30 j40)i 1 + V 1 + V 2 = 250 (5 + j10)i 2 V 2 = 0 V = V I 1 = 300(I 1 I 2 ) Solving, V 1 = j300 V(rms); I 1 = 5 + j0 A(rms); V 2 = 50 j100 V(rms) I 2 = 10 + j0 A(rms) P 30Ω = (5) 2 (30) = 750W; and P 5Ω = (10) 2 (5) = 500W [b] P g = 250(5/0 ) = 1250W Pabs = = 1250W = P dev

28 10 28 CHAPTER 10. Sinusoidal Steady State Power Calculations P [a] 25a a2 2 = 500 I 25 = a 1 I; P 25 = a 2 1I 2 (25) I 4 = a 2 I; P 4 = a 2 2I 2 (4) P 4 = 4P 25 ; a 2 2I 2 4 = 100a 2 1I a 2 1 = 4a a a2 1 = 500; a 1 = 2 25(4) + 4a 2 2 = 500; a 2 = 10 [b] I = 2000/ = 2/0 A(rms) I 25 = a 1 I = 4A P 25Ω = (16)(25) = 400W [c] I 4 = a 2 I = 10(2) = 20A(rms) V 4 = (20)(4) = 80/0 V(rms) P [a] Z Th = j (4000)( j4000) 4000 j Z L = Z Th = 2000 j2000ω = j2000ω [b] V Th = 10/0 (4000) 4000 j4000 = 5 + j5 = 5 2/45 V I = 5 2/ I rms = 1.25mA = /45 ma P load = ( ) 2 (2000) = 3.125mW

29 Problems [c] The closest resistor values from Appendix H are 1.8kΩ and 2.2kΩ. Find the capacitor value: 1 = 2000 so C = 62.5nF 8000C The closest capacitor value is 47 nf. Try R = 1.8kΩ: I = 5/ j j = 1.3/54.85 ma(rms) = j1.06 ma(rms) P load = (0.0013) 2 (1800) = 3.03mW (instead of mw) Try R = 2.2kΩ: I = 5/ j j = 1.176/53.92 ma(rms) = j ma(rms) P load = ( ) 2 (2200) = 3.04mW (instead of mw) Therefore, use the 2.2 kω resistor to give a load impedance of Z L = 2200 j ω. P [a] From Problem 9.75, Z Th = 85 + j85ω and V Th = j850v. Thus, for maximum power transfer, Z L = Z Th = 85 j85ω: I 2 = j = 5 + j5a 425/0 = (5 + j5)i 1 j20(5 + j5)

30 10 30 CHAPTER 10. Sinusoidal Steady State Power Calculations.. I 1 = j j5 = 42.5 j22.5a S g (del) = 425( j22.5) = 18, j9562.5va P g = 18,062.5W [b] P loss = I 1 2 (5) + I 2 2 (45) = 11, = 13,812.5W % loss in transformer = 13,812.5 (100) = 76.47% 18,062.5 P [a] 156 j Z Th.. Z Th = j j500 = j j Z L = j300ω = 400 j300ω [b] I = 300/0 800/0 = 0.375/0 A(rms) P = (0.375) 2 (400) = 56.25W [c] Let R = 180Ω + 220Ω = 400Ω 2π(500)L = 300 so L = π = 95.5mH Use 9 series-connected 10 mh inductors to get 90 mh. Use 2 parallel-connected 10 mh inductors to get 5 mh. Use 2 parallel-connected 1 mh inductors to get 0.5 mh. Combine the 90 mh, the 5 mh, and the 0.5 mh in series to get 95.5 mh. P [a] Open circuit voltage: V 1 = 5Iφ = I φ 25 + j10 (25 + j10)i φ = 100 5Iφ I φ = j10 = 3 j A

31 Problems V Th = j3 1 + j3 (5I φ) = 15V Short circuit current: V 2 = 5I φ = 100 5I φ 25 + j10 I φ = 3 j1a I sc = 5I φ 1 = 15 j5a Z Th = 15 = j0.3ω 15 j5 Z L = Z Th = 0.9 j0.3ω I L = = 8.33A(rms) P = I L 2 (0.9) = 62.5W [b] V L = (0.9 j0.3)(8.33) = 7.5 j2.5v(rms) I 1 = V L j3 = j2.5a(rms)

32 10 32 CHAPTER 10. Sinusoidal Steady State Power Calculations I 2 = I 1 + I L = 7.5 j2.5a(rms) 5I φ = I 2 + V L.. Iφ = 3 j1a I d.s. = I φ I 2 = j1.5a S g = 100(3 + j1) = 300 j100va S d.s. = 5(3 j1)( 4.5 j1.5) = 75 + j0va P dev = = 375W % developed = 62.5 (100) = 16.67% 375 Checks: P 25Ω = (10)(25) = 250W P 1Ω = (67.5)(1) = 67.5W P 0.9Ω = 62.5W Pabs = = 375 = P dev Q j10 = (10)(10) = 100VAR Q j3 = (6.94)(3) = 20.82VAR Q j0.3 = (69.4)( 0.3) = 20.82VAR Q source = 100VAR Q = = 0 P Z L = Z L /θ = Z L cos θ + j Z L sin θ Thus I = V Th (R Th + Z L cos θ) 2 + (X Th + Z L sinθ) 2 Therefore P = 0.5 V Th 2 Z L cosθ (R Th + Z L cos θ) 2 + (X Th + Z L sin θ) 2 Let D = demoninator in the expression for P, then dp d Z L = (0.5 V Th 2 cos θ)(d 1 Z L dd/d Z L ) D 2 dd d Z L = 2(R Th + Z L cos θ)cos θ + 2(X Th + Z L sin θ)sinθ

33 Problems ( ) dp dd d Z L = 0 when D = Z L d Z L Substituting the expressions for D and (dd/d Z L ) into this equation gives us the relationship R 2 Th + X2 Th = Z L 2 or Z Th = Z L. P [a] Z Th = 200 j R = Z Th = 400Ω [b] V Th = (j200)(500 + j300) j500 j j300 + j200 (300/0 ) = 60 + j60v(rms) = 240 j320 = 400/ Ω I = 60 + j j320 = j112.5 ma(rms) = /71.57 ma(rms) P = ( ) 2 (400) = 5.625W [c] Pick the 390Ω resistor from Appendix H for the closest match: I = 60 + j j320 = /71.93 ma(rms) P = ( ) 2 (390) = 5.624W P [a] Open circuit voltage: V φ V φ j5 0.1V φ = 0.. V φ = 40 + j80v(rms) V Th = V φ + 0.1V φ ( j5) = V φ (1 j0.5) = 80 + j60v(rms) Short circuit current:

34 10 34 CHAPTER 10. Sinusoidal Steady State Power Calculations [b] I sc = 0.1V φ + V φ j5 = (0.1 + j0.2)v φ V φ V φ j5 + V φ j5 = 0.. V φ = 100V(rms) I sc = (0.1 + j0.2)(100) = 10 + j20a(rms) Z Th = V Th I sc =.. R o = Z Th = 4.47Ω 80 + j j20 = 4 j2ω I = 80 + j j2 = j8.82a (rms) P = (11.49) 2 ( 20) = W [c] I = 80 + j60 8 = 10 + j7.5a (rms) P = ( )(4) = 625W

35 Problems [d] V φ V φ j5 + V o (25 + j50) j5 V φ = 50 + j25v (rms) 0.1V φ = 5 + j2.5v (rms) 5 + j2.5 + I C = 10 + j7.5 I C = 5 + j5a (rms) = 0 I L = V φ j5 = 5 j10a (rms) I R = I C + I L = 10 j5a (rms) I g = I R + 0.1V φ = 15 j2.5a (rms) S g = 100I g = 1500 j250va 100 = 5(5 + j2.5) + V cs j50.. V cs = 50 j62.5v (rms) S cs = (50 j62.5)(5 j2.5) = j437.5va Thus, Pdev = 1500 % delivered to R o = 625 (100) = 41.67% 1500 P [a] First find the Thévenin equivalent: jωl = j3000ω Z Th = ,000 + j3000 = j3000ω V Th = 12,000 (180) = 120V ,000

36 10 36 CHAPTER 10. Sinusoidal Steady State Power Calculations j ωc = j1000ω I = j2000 = 18 j6ma P = 1 2 I 2 (2000) = 360mW [b] Set C o = 0.1µF so j/ωc = j2000ω Set R o as close as possible to R o = ( ) 2 = Ω [c] I =.. R o = 4000Ω j1000 = j1.85ma P = 1 2 I 2 (4000) = 443.1mW Yes; 443.1mW > 360mW [d] I = = 15mA P = 1 2 (0.015)2 (4000) = 450mW [e] R o = 4000Ω; [f] Yes; C o = 66.67nF 450mW > 443.1mW P [a] Set C o = 0.1µF, so j/ωc = j2000ω; also set R o = Ω P [a] I = [b] Yes; [c] Yes; j1000 = j1.79ma P = 1 2 I 2 (4123.1) = mW mW > 360mW mW < 450mW 1 ωc = 100Ω; C = 1 (60)(200π) = 26.53µF

37 Problems [b] V swo = (40)( j10) = j400 = /5.64 V(rms) V sw = (40 j40)( j10) = j350 = /4.50 V(rms) ( ) % increase = (100) = 9.68% [c] P lwo = (40 2) 2 (1.25) = 4000W P [a] P lw = 40 2 (1.25) = 2000W ( ) 4000 % increase = (100) = 100% 5120 = 30I 1 + j100i 1 + j40(i 1 I 2 ) + j64(i 1 I 2 ) + j40i 1 0 = 160I 2 + j64(i 2 I 1 ) j40i 1 Solving, I 1 = 8 j20a(rms); I 2 = 13/0 A(rms).. V o = (13)(160) = 2080V(rms) [b] P = (13) 2 (160) = 27,040W [c] S g = (5120)(8 + j20) = 40,960 j102,400va.. P g = 40,960W % delivered = 27,040 (100) = 66% 40,960 P [a] Open circuit voltage: 5120 = 30I 1 + j100i 1 + j40i 1 + j64i 1 + j40i 1

38 10 38 CHAPTER 10. Sinusoidal Steady State Power Calculations.. I 1 = j244 = 2.54 j20.67a(rms) V Th = j64i 1 + j40i 1 = j104i 1 = j264.32v Short circuit current: 5120 = 30I 1 + j100i 1 + j40(i 1 I sc ) + j64(i 1 I sc ) + j40i 1 0 = j64(i sc I 1 ) j40i 1 Solving, I sc = j95.63a Z Th = V Th j = I sc j95.63 = j20.33ω I L = j P L = (201.67) 2 (5.37) = 218.4kW = j24.61 = /7.01 A [b] Z o I 2 + j64(i 2 I 1 ) j40i 1 = 0 so I 1 = Z o + j64 I 2 j104.. I 1 = 5.37 j j64 ( j24.61) = 85.32/0 A(rms) j104 P dev = (5120)(85.32) = 436.8kW [c] Begin by choosing the capacitor value from Appendix H that is closest to the required reactive impedance, assuming the frequency of the source is 60 Hz: = 1 2π(60)C so C = 1 2π(60)(20.33) = µF

39 Problems Choose the capacitor value closest to this capacitance from Appendix H, which is 100µF. Then, 1 X L = 2π(60)( ) = Ω Now set R L as close as possible to R L = ( ) 2 = 8.2Ω R 2 Th + (X L + X Th ) 2 : The closest single resistor value from Appendix H is 10Ω. The resulting real power developed by the source is calculated below, using the Thévenin equivalent circuit: I = j j j = 130.7/28.96 P = j (130.7) = kw P Open circuit voltage: (instead of kw) I 1 = 30/ j30 = 0.4 j0.8a V Th = j30i 1 j18i 1 = j12i 1 = j4.8 = 10.73/26.57 Short circuit current: 30/0 = 15I 1 + j30(i 1 I sc ) + j18i sc 0 = j15i sc j18(i sc I 1 ) + j30(i sc I 1 ) j18i sc Solving, I sc = 1.95/ A

40 10 40 CHAPTER 10. Sinusoidal Steady State Power Calculations Z Th = j4.8 = j5.16ω 1.95/ I 2 = j = 2.795/26.57 A 30/0 = 15I 1 + j30(i 1 I 2 ) + j18i 2.. I 1 = 30 + j12i j30 = 30 + j12(2.795/26.57 ) 15 + j30 = 1A Z g = 30/0 1 = 30 + j0 = 30/0 Ω P [a] Open circuit: V Th = 120 (j10) = 36 + j48v 16 + j12 Short circuit: (16 + j12)i 1 j10i sc = 120 j10i 1 + (11 + j23)i sc = 0 Solving, I sc = 2.4A Z Th = 36 + j = 15 + j20ω

41 Problems Z L = ZTh = 15 j20ω I L = V Th 36 + j48 = Z Th + Z L 30 P L = I L 2 (15) = 60W = j1.6a(rms) [b] I 1 = Z 22I 2 jωm = 26 + j3 j10 (1.2 + j1.6) = 5.23/ A (rms) P transformer = (120)(5.23)cos( ) (5.23) 2 (4) = 432.8W % delivered = 60 (100) = 13.86% P [a] jωl 1 = jωl 2 = j(400)( ) = j250ω jωm = j(400)( ) = j125ω 400 = (125 + j250)i g j125i L [b] k = 0 = j125i g + (375 + j250)i L Solving, I g = 0.8 j1.2a; I L = 0.4A Thus, i g = 1.44cos(400t )A i L = 0.4cos 400t A M L1 L 2 = = 0.5 [c] When t = 1.25π ms: 400t = (400)(1.25π) 10 3 = 0.5π rad = 90 i g (1.25π ms) = 1.44cos( ) = 1.2A i L (1.25π ms) = 0.4cos(90 ) = 0A w = 1 2 L 1i L 2i 2 2 Mi 1i 2 = 1 2 (0.625)(1.2) = 450mJ

42 10 42 CHAPTER 10. Sinusoidal Steady State Power Calculations When t = 2.5π ms: 400t = (400)(2.5π) 10 3 = π = 180 i g (2.5π ms) = 1.44cos( ) = 0.8A i L (2.5π ms) = 0.4cos(180) = 0.4A w = 1 2 (0.625)(0.8) (0.625)( 0.4)2 (0.3125)( 0.8)( 0.4) = 150mJ [d] From (a), I L = 0.4 A,.. P = 1 2 (0.4)2 (375) = 30W [e] Open circuit: [f] V Th = Short circuit: 400 (j125) = j80v j = (125 + j250)i 1 j125i sc 0 = j125i 1 + j250i sc Solving, I sc = j Z Th = V Th I sc =.. R L = Ω j j = 25 + j200ω I = j j200 = 0.592/ A P = 1 2 (0.592)2 (201.56) = 35.3W [g] Z L = ZTh = 25 j200ω j80 [h] I = = 3.58/ P = 1 2 (3.58)2 (25) = 160W

43 Problems P [a] 54 = I 1 + j2(i 1 I 2 ) + j3i 2 0 = 7I 2 + j2(i 2 I 1 ) j3i 2 + j8i 2 + j3(i 1 I 2 ) Solving, I 1 = 12 j21a (rms); I 2 = 3A (rms) P [a] V o = 7I 2 = 21/0 V(rms) [b] P = I 2 2 (7) = 63W [c] P g = (54)(12) = 648W % delivered = 63 (100) = 9.72% = I 1 + j2(i 1 I 2 ) + j4ki 2 0 = 7I 2 + j2(i 2 I 1 ) j4ki 2 + j8i 2 + j4k(i 1 I 2 ) Place the equations in standard form: 54 = (1 + j2)i 1 + j(4k 2)I 2 0 = j(4k 2)I 1 + [7 + j(10 8k)]I 2 I 1 = 54 I 2j(4k 2) (1 + j2) Substituting, I 2 = j54(4k 2) [7 + j(10 8k)](1 + j2) (4k 2) For V o = 0,I 2 = 0, so if 4k 2 = 0, then k = 0.5.

44 10 44 CHAPTER 10. Sinusoidal Steady State Power Calculations P [a] [b] When I 2 = 0 I 1 = 54 = 10.8 j21.6a(rms) 1 + j2 P g = (54)(10.8) = 583.2W Check: P loss = I 1 2 (1) = 583.2W [b] Open circuit: V Th = j3i 1 + j2i 1 = ji 1 I 1 = 54 = 10.8 j j2 V Th = 21.6 j10.8v Short circuit: 54 = I 1 + j2(i 1 I sc ) + j3i sc 0 = j2(i sc I 1 ) j3i sc + j8i sc + j3(i 1 I sc ) Solving, I sc = j5.82 Z Th = V Th 21.6 j10.8 = I sc j5.82 = j3.6 = 3.6/86.86 Ω.. R L = Z Th = 3.6Ω I = 21.6 j j3.6 = 4.614/163.1

45 Problems P = I 2 (3.6) = 76.6W, which is greater than when R L = 7Ω ( P [a] Z ab = 50 j400 = 1 N ) 2 1 Z L N 2 [b].. Z L = 1 (1 6) 2(50 j400) = 2 j16ω I 1 = = 240/0 ma N 1 I 1 = N 2 I 2 I 2 = 6I 1 = 1.44/0 A I L = I 1 + I 2 = 1.68/0 A V L = (2 j16)i L = j26.88 = 27.1/97.13 V(rms) P [a] Replace the circuit to the left of the primary winding with a Thévenin equivalent: V Th = (15)(20 j10) = 60 + j120v Z Th = j10 = 6 + j8ω Transfer the secondary impedance to the primary side: Z p = 1 25 (100 jx C) = 4 j X C 25 Ω

46 10 46 CHAPTER 10. Sinusoidal Steady State Power Calculations P [b] I = Now maximize I by setting (X C /25) = 8Ω:.. C = 60 + j ( ) = 0.25µF = 6 + j12a P = I 2 (4) = 720W [c] R o 25 = 6Ω;.. Ro = 150Ω 60 + j120 [d] I = = 5 + j10a 12 P = I 2 (6) = 750W V a 1 = V o 50 ; 1I a = 50I o.. V a I a = V o/50 50I o = V o/i o 50 2 = = 2Ω V b 25 = V a 1 ; 25I b = 1I a.. V b /25 25I b = V b/i b 25 2 = V a I a.. V b I b = 25 2V a I a = 25 2 (2) = 1250Ω Thus I b = [145/( )] = 100mA (rms); since the ideal transformers are lossless, P 5kΩ = P 1250Ω, and the power delivered to the 1250Ω resistor is (0.1) 2 (1250) or 12.5W.

47 Problems P [a] V b I b = 252 (5000) a 2 = 200Ω; therefore a 2 = 15,625, a = 125 P [a] [b] I b = = 362.5mA; P = (0.3625)2 (200) = W For maximum power transfer, Z ab = 90kΩ ( Z ab = 1 N ) 2 1 Z L N 2.. ( 1 N ) 2 1 = 90,000 N = N 1 N 2 = ±15; N 1 N 2 = = 16 ( ) [b] P = I i 2 (90,000) = (90,000) = 90mW 180,000 ( ) 180 [c] V 1 = R i I i = (90,000) = 90V 180,000 [d] V g = ( )(100,000 80,000) = 100V P g (del) = ( )(100) = 225mW % delivered = 90 (100) = 40% 225 P [a] Z Th = j Z ab = 1700Ω ( ) (40 j30) = j1020 = 1700/36.87 Ω 50

48 10 48 CHAPTER 10. Sinusoidal Steady State Power Calculations Z ab = Z L (1 + N 1 /N 2 ) 2 (1 + N 1 /N 2 ) 2 = 6800/1700 = 4.. N 1 /N 2 = 1 or N 2 = N 1 = 1000 turns [b] V Th = 255/ j30 (j200) = 1020/53.13 V [c] I L = 1020/ j1020 = 0.316/34.7 A(rms) Since the transformer is ideal, P 6800 = P P = I 2 (1700) = 170W 255/0 = (40 + j30)i 1 j200( j0.18).. I 1 = 4.13 j1.80a(rms) P gen = (255)(4.13) = 1053W P diss = = 883W % dissipated = 883 (100) = 83.85% 1053 P [a] Open circuit voltage: 40/0 = 4(I 1 + I 3 ) + 12I 3 + V Th

49 Problems I 1 4 = I 3; I 1 = 4I 3 Solving, V Th = 40/0 V Short circuit current: 40/0 = 4I 1 + 4I 3 + I 1 + V 1 4V 1 = 16(I 1 /4) = 4I 1 ;.. V 1 = I /0 = 6I 1 + 4I 3 Also, 40/0 = 4(I 1 + I 3 ) + 12I 3 Solving, I 1 = 6A; I 3 = 1A; I sc = I 1 /4 + I 3 = 2.5A R Th = V Th I sc = = 16Ω I = 40/0 32 = 1.25/0 A(rms) P = (1.25) 2 (16) = 25W [b]

50 Pabs = = 230W = P dev CHAPTER 10. Sinusoidal Steady State Power Calculations 40 = 4(I 1 + I 3 ) + 12I V 1 = 4I (I 1 /4 + I 3 );.. V 1 = 2I 1 + 4I 3 40 = 4I 1 + 4I 3 + I 1 + V 1.. I 1 = 6A; I 3 = 0.25A; I 1 + I 3 = 5.75/0 A P 40V (developed) = 40(5.75) = 230W.. % delivered = 25 (100) = 10.87% 230 [c] P RL = 25W; P 16Ω = (1.5) 2 (16) = 36W P 4Ω = (5.75) 2 (4) = W; P 1Ω = (6) 2 (1) = 36W P 12Ω = ( 0.25) 2 (12) = 0.75W P [a] Open circuit voltage: 500 = 100I 1 + V 1 V 2 = 400I 2 V 1 1 = V V 2 = 2V 1 I 1 = 2I 2 Substitute and solve: 2V 1 = 400I 1 /2 = 200I 1.. V1 = 100I 1

51 Problems = 100I I 1.. I 1 = 500/200 = 2.5A.. I 2 = 1 2 I 1 = 1.25A V 1 = 100(2.5) = 250V; V 2 = 2V 1 = 500V V Th = 20I 1 + V 1 V I 2 = 150V(rms)

52 10 52 CHAPTER 10. Sinusoidal Steady State Power Calculations Short circuit current: 500 = 80(I sc + I 1 ) + 360(I sc + 0.5I 1 ) 2V 1 = 40 I (I sc + 0.5I 1 ) 500 = 80(I 1 + I sc ) + 20I 1 + V 1 Solving, I sc = 1.47A R Th = V Th I sc = = 102Ω P = = 55.15W [b] 500 = 80[I 1 (75/102)] [I 2 (75/102)] , I 1 = 3.456A = 80I I 2

53 Problems P source = (500)[3.456 (75/102)] = W % delivered = (100) = 4.05% [c] P 80Ω = 80(I 1 + I L ) 2 = W P 20Ω = 20I 2 1 = W P 40Ω = 40I 2 2 = W P 102Ω = 102I 2 L = 55.15W P 360Ω = 360(I 2 + I L ) 2 = W Pabs = = W = P dev P [a] [b] [c] [d] 30[5(44.28) + 19(15.77)] = kwh [5(44.28) + 19(8.9)] = kwh [5(44.28) + 19(4.42)] = 9.16 kwh [5(44.28) + 19(0)] = 6.64 kwh 1000 Note that this is about 40 % of the amount of total power consumed in part (a). 30[0.2(1433) (3.08)] P [a] = 10.8 kwh 1000 [b] The standby power consumed in one month by the microwave oven when in the ready state is 30[23.8(3.08)] 1000 = 2.2 kwh This is (2.2/10.8) 100 = 20.4% of the total power consumed by the microwave in one month. Since it is not practical to unplug the microwave when you are not using it, this is the cost associated with having a microwave oven. P jωl 1 = j(2π)(60)(0.25) = j94.25ω I = j94.25 = 1.27/ A(rms) P = R 1 I 2 = 5(1.27) 2 = 8.06W

54 10 54 CHAPTER 10. Sinusoidal Steady State Power Calculations P jωl 1 = j(2π)(60)(0.25) = j94.25ω I = j94.25 = 1.27/ A(rms) P = R 1 I 2 = 0.05(1.27) 2 = 81.1mW Note that while the current supplied by the voltage source is virtually identical to that calculated in Problem 10.69, the much smaller value of transformer resistance results in a much smaller value of real power consumed by the transformer. P An ideal transformer has no resistance, so consumes no real power. This is one of the important characteristics of ideal transformers.