Sinusoidal Steady State Analysis-Circuit And Network Analysis-Solution Manual, Exercises of Electrical Circuit Analysis

Download Sinusoidal Steady State Analysis-Circuit And Network Analysis-Solution Manual and more Exercises Electrical Circuit Analysis in PDF only on Docsity! 9 Sinusoidal Steady State Analysis Assessment Problems AP 9.1 [a] V = 170/−40◦ V [b] 10 sin(1000t + 20◦) = 10 cos(1000t − 70◦) ·. . I = 10/−70◦ A [c] I = 5/36.87◦ + 10/−53.13◦ = 4 + j3 + 6 − j8 = 10 − j5 = 11.18/−26.57◦ A [d] sin(20,000πt + 30◦) = cos(20,000πt − 60◦) Thus, V = 300/45◦ − 100/−60◦ = 212.13 + j212.13 − (50 − j86.60) = 162.13 + j298.73 = 339.90/61.51◦ mV AP 9.2 [a] v = 18.6 cos(ωt − 54◦) V [b] I = 20/45◦ − 50/− 30◦ = 14.14 + j14.14 − 43.3 + j25 = −29.16 + j39.14 = 48.81/126.68◦ Therefore i = 48.81 cos(ωt + 126.68◦) mA [c] V = 20 + j80 − 30/15◦ = 20 + j80 − 28.98 − j7.76 = −8.98 + j72.24 = 72.79/97.08◦ v = 72.79 cos(ωt + 97.08◦) V AP 9.3 [a] ωL = (104)(20 × 10−3) = 200 Ω [b] ZL = jωL = j200 Ω 9–1 docsity.com 9–2 CHAPTER 9. Sinusoidal Steady State Analysis [c] VL = IZL = (10/30◦)(200/90◦) × 10−3 = 2/120◦ V [d] vL = 2 cos(10,000t + 120◦) V AP 9.4 [a] XC = −1 ωC = −1 4000(5 × 10−6) = −50 Ω [b] ZC = jXC = −j50 Ω [c] I = V ZC = 30/25◦ 50/−90◦ = 0.6/115 ◦ A [d] i = 0.6 cos(4000t + 115◦) A AP 9.5 I1 = 100/25◦ = 90.63 + j42.26 I2 = 100/145◦ = −81.92 + j57.36 I3 = 100/−95◦ = −8.72 − j99.62 I4 = −(I1 + I2 + I3) = (0 + j0) A, therefore i4 = 0 A AP 9.6 [a] I = 125/−60◦ |Z|/θz = 125 |Z| /(−60 − θZ) ◦ But −60 − θZ = −105◦ ·. . θZ = 45◦ Z = 90 + j160 + jXC ·. . XC = −70 Ω; XC = − 1 ωC = −70 ·. . C = 1 (70)(5000) = 2.86 µF [b] I = Vs Z = 125/−60◦ (90 + j90) = 0.982/−105◦A; ·. . |I| = 0.982 A AP 9.7 [a] ω = 2000 rad/s ωL = 10 Ω, −1 ωC = −20 Ω Zxy = 20‖j10 + 5 + j20 = 20(j10)(20 + j10) + 5 − j20 = 4 + j8 + 5 − j20 = (9 − j12) Ω docsity.com Problems 9–5 AP 9.11 Use the lower node as the reference node. Let V1 = node voltage across the 20 Ω resistor and VTh = node voltage across the capacitor. Writing the node voltage equations gives us V1 20 − 2/45◦ + V1 − 10Ix j10 = 0 and VTh = −j10 10 − j10(10Ix) We also have Ix = V1 20 Solving these equations for VTh gives VTh = 10/45◦V. To find the Thévenin impedance, we remove the independent current source and apply a test voltage source at the terminals a, b. Thus It follows from the circuit that 10Ix = (20 + j10)Ix Therefore Ix = 0 and IT = VT −j10 + VT 10 ZTh = VT IT , therefore ZTh = (5 − j5) Ω AP 9.12 The phasor domain circuit is as shown in the following diagram: docsity.com 9–6 CHAPTER 9. Sinusoidal Steady State Analysis The node voltage equation is −10 + V 5 + V −j(20/9) + V j5 + V − 100/−90◦ 20 = 0 Therefore V = 10 − j30 = 31.62/−71.57◦ Therefore v = 31.62 cos(50,000t − 71.57◦) V AP 9.13 Let Ia, Ib, and Ic be the three clockwise mesh currents going from left to right. Summing the voltages around meshes a and b gives 33.8 = (1 + j2)Ia + (3 − j5)(Ia − Ib) and 0 = (3 − j5)(Ib − Ia) + 2(Ib − Ic). But Vx = −j5(Ia − Ib), therefore Ic = −0.75[−j5(Ia − Ib)]. Solving for I = Ia = 29 + j2 = 29.07/3.95◦ A. AP 9.14 [a] M = 0.4 √ 0.0625 = 0.1 H, ωM = 80 Ω Z22 = 40 + j800(0.125) + 360 + j800(0.25) = (400 + j300) Ω Therefore |Z22| = 500 Ω, Z∗22 = (400 − j300) Ω Zr = ( 80 500 )2 (400 − j300) = (10.24 − j7.68) Ω [b] I1 = 245.20 184 + 100 + j400 + Zr = 0.50/− 53.13◦ A i1 = 0.5 cos(800t − 53.13◦) A [c] I2 = ( jωM Z22 ) I1 = j80 500/36.87◦ (0.5/− 53.13◦) = 0.08/0◦ A i2 = 80 cos 800t mA docsity.com Problems 9–7 AP 9.15 I1 = Vs Z1 + Z2/a2 = 25 × 103/0◦ 1500 + j6000 + (25)2(4 − j14.4) = 4 + j3 = 5/36.87◦ A V1 = Vs − Z1I1 = 25,000/0◦ − (4 + j3)(1500 + j6000) = 37,000 − j28,500 V2 = − 125V1 = −1480 + j1140 = 1868.15/142.39 ◦ V I2 = V2 Z2 = 1868.15/142.39◦ 4 − j14.4 = 125/− 143.13 ◦ A Also, I2 = −25I1 docsity.com 9–10 CHAPTER 9. Sinusoidal Steady State Analysis P 9.6 [a] T 2 = 8 + 2 = 10 ms; T = 20 ms f = 1 T = 1 20 × 10−3 = 50Hz [b] v = Vm sin(ωt + θ) ω = 2πf = 100π rad/s 100π(−2 × 10−3) + θ = 0; ·. . θ = π 5 rad = 36◦ v = Vm sin[100πt + 36◦] 80.9 = Vm sin 36◦; Vm = 137.64 V v = 137.64 sin[100πt + 36◦] = 137.64 cos[100πt − 54◦] V P 9.7 u = ∫ to+T to V 2m cos 2(ωt + φ) dt = V 2m ∫ to+T to 1 2 + 1 2 cos(2ωt + 2φ) dt = V 2m 2 {∫ to+T to dt + ∫ to+T to cos(2ωt + 2φ) dt } = V 2m 2 { T + 1 2ω [ sin(2ωt + 2φ) |to+Tto ]} = V 2m 2 { T + 1 2ω [sin(2ωto + 4π + 2φ) − sin(2ωto + 2φ)] } = V 2m ( T 2 ) + 1 2ω (0) = V 2m ( T 2 ) P 9.8 Vm = √ 2Vrms = √ 2(120) = 169.71 V P 9.9 [a] The numerical values of the terms in Eq. 9.8 are Vm = 20, R/L = 1066.67, ωL = 60 √ R2 + ω2L2 = 100 φ = 25◦, θ = tan−1 60/80, θ = 36.87◦ Substitute these values into Equation 9.9: i = [ −195.72e−1066.67t + 200 cos(800t − 11.87◦) ] mA, t ≥ 0 [b] Transient component = −195.72e−1066.67t mA Steady-state component = 200 cos(800t − 11.87◦) mA [c] By direct substitution into Eq 9.9 in part (a), i(1.875 ms) = 28.39 mA [d] 200 mA, 800 rad/s, −11.87◦ docsity.com Problems 9–11 [e] The current lags the voltage by 36.87◦. P 9.10 [a] From Eq. 9.9 we have L di dt = VmR cos(φ − θ)√ R2 + ω2L2 e−(R/L)t − ωLVm sin(ωt + φ − θ)√ R2 + ω2L2 Ri = −VmR cos(φ − θ)e−(R/L)t√ R2 + ω2L2 + VmR cos(ωt + φ − θ)√ R2 + ω2L2 L di dt + Ri = Vm [ R cos(ωt + φ − θ) − ωL sin(ωt + φ − θ)√ R2 + ω2L2 ] But R√ R2 + ω2L2 = cos θ and ωL√ R2 + ω2L2 = sin θ Therefore the right-hand side reduces to Vm cos(ωt + φ) At t = 0, Eq. 9.9 reduces to i(0) = −Vm cos(φ − θ)√ R2 − ω2L2 + Vm cos(φ − θ)√ R2 + ω2L2 = 0 [b] iss = Vm√ R2 + ω2L2 cos(ωt + φ − θ) Therefore L diss dt = −ωLVm√ R2 + ω2L2 sin(ωt + φ − θ) and Riss = VmR√ R2 + ω2L2 cos(ωt + φ − θ) L diss dt + Riss = Vm [ R cos(ωt + φ − θ) − ωL sin(ωt + φ − θ)√ R2 + ω2L2 ] = Vm cos(ωt + φ) P 9.11 [a] Y = 50/60◦ + 100/− 30◦ = 111.8/− 3.43◦ y = 111.8 cos(500t − 3.43◦) [b] Y = 200/50◦ − 100/60◦ = 102.99/40.29◦ y = 102.99 cos(377t + 40.29◦) [c] Y = 80/30◦ − 100/− 225◦ + 50/− 90◦ = 161.59/− 29.96◦ y = 161.59 cos(100t − 29.96◦) docsity.com 9–12 CHAPTER 9. Sinusoidal Steady State Analysis [d] Y = 250/0◦ + 250/120◦ + 250/− 120◦ = 0 y = 0 P 9.12 [a] 1000Hz [b] θv = 0◦ [c] I = 200/0◦ jωL = 200 ωL /− 90◦ = 25/− 90◦; θi = −90◦ [d] 200 ωL = 25; ωL = 200 25 = 8 Ω [e] L = 8 2π(1000) = 1.27 mH [f] ZL = jωL = j8 Ω P 9.13 [a] ω = 2πf = 314,159.27 rad/s [b] I = V ZC = 10 × 10−3/0◦ 1/jωC = jωC(10 × 10−3)/0◦ = 10 × 10−3ωC/90◦ ·. . θi = 90◦ [c] 628.32 × 10−6 = 10 × 10−3 ωC 1 ωC = 10 × 10−3 628.32 × 10−6 = 15.92 Ω, ·. . XC = −15.92 Ω [d] C = 1 15.92(ω) = 1 (15.92)(100π × 103) C = 0.2 µF [e] Zc = j (−1 ωC ) = −j15.92 Ω P 9.14 [a] jωL = j(2 × 104)(300 × 10−6) = j6 Ω 1 jωC = −j 1 (2 × 104)(5 × 10−6) = −j10 Ω; Ig = 922/30 ◦ A docsity.com Problems 9–15 P 9.20 1 jωC = 1 (1 × 10−6)(50 × 103) = −j20 Ω jωL = j50 × 103(1.2 × 10−3) = j60 Ω Vg = 40/0◦ V Ze = −j20 + 30‖j60 = 24 − j8 Ω Ig = 40/0◦ 24 − j8 = 1.5 + j0.5 mA Vo = (30‖j60)Ig = 30(j60)30 + j60(1.5 + j0.5) = 30 + j30 = 42.43/45 ◦ V vo = 42.43 cos(50,000t + 45◦) V P 9.21 [a] Z1 = R1 − j 1 ωC1 Z2 = R2/jωC2 R2 + (1/jωC2) = R2 1 + jωR2C2 = R2 − jωR22C2 1 + ω2R22C22 Z1 = Z2 when R1 = R2 1 + ω2R22C22 and 1 ωC1 = ωR22C2 1 + ω2R22C22 or C1 = 1 + ω2R22C 2 2 ω2R22C2 [b] R1 = 1000 1 + (40 × 103)2(1000)2(50 × 10−9)2 = 200 Ω C1 = 1 + (40 × 103)2(1000)2(50 × 10−9)2 (40 × 103)2(1000)2(50 × 10−9) = 62.5 nF docsity.com 9–16 CHAPTER 9. Sinusoidal Steady State Analysis P 9.22 [a] Y2 = 1 R2 + jωC2 Y1 = 1 R1 + (1/jωC1) = jωC1 1 + jωR1C1 = ω2R1C 2 1 + jωC1 1 + ω2R21C21 Therefore Y1 = Y2 when R2 = 1 + ω2R21C 2 1 ω2R1C21 and C2 = C1 1 + ω2R21C21 [b] R2 = 1 + (50 × 103)2(1000)2(40 × 10−9)2 (50 × 103)2(1000)(40 × 10−9)2 = 1250 Ω C2 = 40 × 10−9 1 + (50 × 103)2(1000)2(40 × 10−9)2 = 8 nF P 9.23 [a] Z1 = R1 + jωL1 Z2 = R2(jωL2) R2 + jωL2 = ω2L22R2 + jωL2R 2 2 R2 + ω2L22 Z1 = Z2 when R1 = ω2L22R2 R22 + ω2L22 and L1 = R22L2 R22 + ω2L22 [b] R1 = (4000)2(1.25)2(5000) 50002 + 40002(1.25)2 = 2500 Ω L1 = (5000)2(1.25) 50002 + 40002(1.25)2 = 625 mH P 9.24 [a] Y2 = 1 R2 − j ωL2 Y1 = 1 R1 + jωL1 = R1 − jωL1 R21 + ω2L21 Therefore Y2 = Y1 when R2 = R21 + ω 2L21 R1 and L2 = R21 + ω 2L21 ω2L1 [b] R2 = 80002 + 10002(4)2 8000 = 10 kΩ L2 = 80002 + 10002(4)2 10002(4) = 20 H docsity.com Problems 9–17 P 9.25 Vg = 500/30◦ V; Ig = 0.1/83.13◦ mA Z = Vg Ig = 5000/− 53.13◦ Ω = 3000 − j4000 Ω z = 3000 + j ( ω − 32 × 10 3 ω ) ω − 32 × 10 3 ω = −4000 ω2 + 4000ω − 32 × 103 = 0 ω = 7.984 rad/s P 9.26 [a] Zeq = 50,000 3 + −j20 × 106 ω ‖(1200 + j0.2ω) = 50,000 3 + −j20 × 106 ω (1200 + j0.2ω) 1200 + j[0.2ω − 20×106 ω ] = 50,000 3 + −j20×106 ω (1200 + j0.2ω) [ 1200 − j ( 0.2ω − 20×106 ω )] 12002 + ( 0.2ω − 20×106 ω )2 Im(Zeq) = −20 × 10 6 ω (1200)2 − 20 × 10 6 ω [ 0.2ω ( 0.2ω − 20 × 10 6 ω )] = 0 −20 × 106(1200)2 − 20 × 106 [ 0.2ω ( 0.2ω − 20 × 10 6 ω )] = 0 −(1200)2 = 0.2ω ( 0.2ω − 20 × 10 6 ω ) 0.22ω2 − 0.2(20 × 106) + 12002 = 0 ω2 = 64 × 106 ·. . ω = 8000 rad/s ·. . f = 1273.24 Hz [b] Zeq = 50,000 3 + −j2500‖(1200 + j1600) = 50,000 3 + (−j2500)(1200 + j1600) 1200 − j900 = 20,000 Ω Ig = 30/0◦ 20,000 = 1.5/0◦ mA ig(t) = 1.5 cos 8000t mA docsity.com 9–20 CHAPTER 9. Sinusoidal Steady State Analysis [b] When L = 8 H: ZT = 400 + 500 × 106(8)2 20002 + 5002(8)2 = 2000 Ω Ig = 200/0◦ 2000 = 100/0◦ mA ig = 100 cos 500t mA When L = 2 H: ZT = 400 + 500 × 106(2)2 20002 + 500(2)2 = 800 Ω Ig = 200/0◦ 800 = 250/0◦ mA ig = 250 cos 500t mA P 9.31 [a] Y1 = 11 2500 × 103 = 4.4 × 10 −6 S Y2 = 1 14,000 + j5ω = 14,000 196 × 106 + 25ω2 − j 5ω 196 × 106 + 25ω2 Y3 = jω2 × 10−9 YT = Y1 + Y2 + Y3 For ig and vo to be in phase the j component of YT must be zero; thus, ω2 × 10−9 = 5ω 196 × 106 + 25ω2 or 25ω2 + 196 × 106 = 5 2 × 10−9 ·. . 25ω2 = 2304 × 106 ·. . ω = 9600 rad/s [b] YT = 4.4 × 10−6 + 14,000196 × 106 + 25(9600)2 = 10 × 10 −6 S ·. . ZT = 100 kΩ Vo = (0.25 × 10−3/0◦)(100 × 103) = 25/0◦ V vo = 25 cos 9600t V docsity.com Problems 9–21 P 9.32 [a] Zg = 500 − j 10 6 ω + 103(j0.5ω) 103 + j0.5ω = 500 − j 10 6 ω + 500jω(1000 − j0.5ω) 106 + 0.25ω2 = 500 − j 10 6 ω + 250ω2 106 + 0.25ω2 + j 5 × 105ω 106 + 0.25ω2 ·. . If Zg is purely real, 10 6 ω = 5 × 105ω 106 + 0.25ω2 2(106 + 0.25ω2) = ω2 ·. . 4 × 106 = ω2 ·. . ω = 2000 rad/s [b] When ω = 2000 rad/s Zg = 500 − j500 + (j1000‖1000) = 1000 Ω ·. . Ig = 20/0 ◦ 1000 = 20/0◦ mA Vo = Vg − IgZ1 Z1 = 500 − j500 Ω Vo = 20/0◦ − (0.02/0◦)(500 − j500) = 10 + j10 = 14.14/45◦ V vo = 14.14 cos(2000t + 45◦) V P 9.33 Zab = 1 − j8 + (2 + j4)‖(10 − j20) + (40‖j20) = 1 − j8 + 3 + j4 + 8 + j16 = 12 + j12 Ω = 16.971/45◦ Ω P 9.34 First find the admittance of the parallel branches Yp = 1 2 − j6 + 1 12 + j4 + 1 2 + 1 j0.5 = 0.625 − j1.875 S Zp = 1 Yp = 1 0.625 − j1.875 = 0.16 + j0.48 Ω Zab = −j4.48 + 0.16 + j0.48 + 2.84 = 3 − j4 Ω Yab = 1 Zab = 1 3 − j4 = 120 + j160 mS = 200/53.13◦ mS docsity.com 9–22 CHAPTER 9. Sinusoidal Steady State Analysis P 9.35 Simplify the top triangle using series and parallel combinations: (1 + j1)‖(1 − j1) = 1 Ω Convert the lower left delta to a wye: Z1 = (j1)(1) 1 + j1 − j1 = j1 Ω Z2 = (−j1)(1) 1 + j1 − j1 = −j1 Ω Z3 = (j1)(−j1) 1 + j1 − j1 = 1 Ω Convert the lower right delta to a wye: Z4 = (−j1)(1) 1 + j1 − j1 = −j1 Ω Z5 = (−j1)(j1) 1 + j1 − j1 = 1 Ω Z6 = (j1)(1) 1 + j1 − j1 = j1 Ω The resulting circuit is shown below: Simplify the middle portion of the circuit by making series and parallel combinations: (1 + j1 − j1)‖(1 + 1) = 1‖2 = 2/3 Ω Zab = −j1 + 2/3 + j1 = 2/3 Ω docsity.com Problems 9–25 [b] ib = 2.5 cos(800t + 90◦) A ic = 10 cos(800t + 36.87◦) A vg = 358.47 cos(800t + 67.01◦) V P 9.41 [a] jωL = j(1000)(100) × 10−3 = j100 Ω 1 jωC = −j 10 6 (1000)(10) = −j100 Ω Using voltage division, Vab = (100 + j100)‖(−j100) j100 + (100 + j100)‖(−j100)(247.49/45 ◦) = 350/0◦ VTh = Vab = 350/0◦ V [b] Remove the voltage source and combine impedances in parallel to find ZTh = Zab: Yab = 1 j100 + 1 100 + j100 + 1 −j100 = 5 − j5 mS ZTh = Zab = 1 Yab = 100 + j100 Ω [c] docsity.com 9–26 CHAPTER 9. Sinusoidal Steady State Analysis P 9.42 Using voltage division: VTh = 36 36 + j60 − j48(240) = 216 − j72 = 227.68/− 18.43 ◦ V Remove the source and combine impedances in series and in parallel: ZTh = 36‖(j60 − j48) = 3.6 + j10.8 Ω P 9.43 Open circuit voltage: V2 10 + 88Iφ + V2 − 15V2 −j50 = 0 Iφ = 5 − (V2/5) 200 Solving, V2 = −66 + j88 = 110/126.87◦ V = VTh Find the Thévenin equivalent impedance using a test source: IT = VT 10 + 88Iφ + 0.8Vt −j50 Iφ = −VT /5 200 IT = VT ( 1 10 − 881/5 200 + 0.8 −j50 ) docsity.com Problems 9–27 ·. . VT IT = 30 − j40 = ZTh IN = VTh ZTh = −66 + j88 30 − j40 = −2.2 + j0 A = 2.2/180 ◦ A The Norton equivalent circuit: P 9.44 Short circuit current Iβ = −6Iβ 2 2Iβ = −6Iβ; ·. . Iβ = 0 I1 = 0; ·. . Isc = 10/−45◦ A = IN The Norton impedance is the same as the Thévenin impedance. Find it using a test source VT = 6Iβ + 2Iβ = 8Iβ, Iβ = j1 2 + j1 IT docsity.com 9–30 CHAPTER 9. Sinusoidal Steady State Analysis V1 20 + j10 + j3V1 50 − j100 + V1 50 − j100 = 250 20 + j10 V1 = 500 − j250 V; Vo = 300 − j400 V = VTh = 500/− 53.13◦ V Short circuit current: Isc = 250/0◦ 70 + j10 = 3.5 − j0.5 A ZTh = VTh Isc = 300 − j400 3.5 − j0.5 = 100 − j100 Ω The Thévenin equivalent circuit: P 9.49 Open circuit voltage: (9 + j4)Ia − Ib = −60/0◦ docsity.com Problems 9–31 −Ia + (9 − j4)Ib = 60/0◦ Solving, Ia = −5 + j2.5 A; Ib = 5 + j2.5 A VTh = 4Ia + (4 − j4)Ib = 10/0◦ V Short circuit current: (9 + j4)Ia − 1Ib − 4Isc = −60 −1Ia + (9 − j4)Ib − (4 − j4)Isc = 60 −4Ia − (4 − j4)Ib + (8 − j4)Isc = 0 Solving, Isc = 2.07/0◦ ZTh = VTh Isc = 10/0◦ 2.07/0◦ = 4.83 Ω docsity.com 9–32 CHAPTER 9. Sinusoidal Steady State Analysis Alternate calculation for ZTh: ∑ Z = 4 + 1 + 4 − j4 = 9 − j4 Z1 = 4 9 − j4 Z2 = 4 − j4 9 − j4 Z3 = 16 − j16 9 − j4 Za = 4 + j4 + 4 9 − j4 = 56 + j20 9 − j4 Zb = 4 + 4 − j4 9 − j4 = 40 − j20 9 − j4 Za‖Zb = 2640 − j320864 − j384 Z3 + Za‖Zb = 16 − j169 − j4 + 2640 − j320 864 − j384 = 4176 − j1856 864 − j384 = 4.83 Ω docsity.com Problems 9–35 P 9.53 From the solution to Problem 9.52 the phasor-domain circuit is Making two source transformations yields Ig1 = 16 − j12 j2 = −6 − j8 A Ig2 = −14 − j48 −j5 = 9.6 − j2.8 A Y = 1 j2 + 1 10 + 1 −j5 = (0.1 − j0.3) S Z = 1 Y = 1 + j3 Ω Ie = Ig1 + Ig2 = 3.6 − j10.8 A Hence the circuit reduces to Vo = ZIe = (1 + j3)(3.6 − j10.8) = 36/0◦ V ·. . vo(t) = 36 cos 2000t V docsity.com 9–36 CHAPTER 9. Sinusoidal Steady State Analysis P 9.54 The circuit with the mesh currents identified is shown below: The mesh current equations are: −20/− 36.87◦ + j2I1 + 10(I1 − I2) = 0 50/− 106.26◦ + 10(I2 − I1) − j5I2 = 0 In standard form: I1(10 + j2) + I2(−10) = 20/− 36.87◦ I1(−10) + I2(10 − j5) = −50/− 106.26◦ = 50/73.74◦ Solving on a calculator yields: I1 = −6 + j10A; I2 = −9.6 + j10A Thus, Vo = 10(I1 − I2) = 36V and vo(t) = 36 cos 2000tV P 9.55 From the solution to Problem 9.52 the phasor-domain circuit with the right-hand source removed is V′o = 10‖ − j5 j2 + 10‖ − j5(16 − j12) = 18 − j26 V docsity.com Problems 9–37 With the left hand source removed V′′o = 10‖j2 −j5 + 10‖j2(−14 − j48) = 18 + j26 V Vo = V′o + V ′′ o = 18 − j26 + 18 + j26 = 36 V vo(t) = 36 cos 2000t V P 9.56 Write a KCL equation at the top node: Vo −j8 + Vo − 2.4I∆ j4 + Vo 5 − (10 + j10) = 0 The constraint equation is: I∆ = Vo −j8 Solving, Vo = j80 = 80/90◦ V P 9.57 Write node voltage equations: Left Node: V1 40 + V1 − Vo/8 j20 = 0.025/0◦ Right Node: Vo 50 + Vo j25 + 16Io = 0 docsity.com 9–40 CHAPTER 9. Sinusoidal Steady State Analysis P 9.61 jωL = j5000(60 × 10−3) = j300 Ω 1 jωC = −j (5000)(2 × 10−6) = −j100 Ω −400/0◦ + (50 + j300)Ia − 50Ib − 150(Ia − Ib) = 0 (150 − j100)Ib − 50Ia + 150(Ia − Ib) = 0 Solving, Ia = −0.8 − j1.6 A; Ib = −1.6 + j0.8 A Vo = 100Ib = −160 + j80 = 178.89/153.43◦ V vo = 178.89 cos(5000t + 153.43◦) V P 9.62 10/0◦ = (1 − j1)I1 − 1I2 + j1I3 −5/0◦ = −1I1 + (1 + j1)I2 − j1I3 docsity.com Problems 9–41 1 = j1I1 − j1I2 + I3 Solving, I1 = 11 + j10 A; I2 = 11 + j5 A; I3 = 6 A Ia = I3 − 1 = 5 A = 5/0◦ A Ib = I1 − I3 = 5 + j10 A = 11.18/63.43◦ A Ic = I2 − I3 = 5 + j5 A = 7.07/45◦ A Id = I1 − I2 = j5 A = 5/90◦ A P 9.63 Va − (100 − j50) 20 + Va j5 + Va − (140 + j30) 12 + j16 = 0 Solving, Va = 40 + j30 V IZ + (30 + j20) − 140 + j30−j10 + (40 + j30) − (140 + j30) 12 + j16 = 0 Solving, IZ = −30 − j10 A Z = (100 − j50) − (140 + j30) −30 − j10 = 2 + j2 Ω docsity.com 9–42 CHAPTER 9. Sinusoidal Steady State Analysis P 9.64 [a] 1 jωC = −j50 Ω jωL = j120 Ω Ze = 100‖ − j50 = 20 − j40 Ω Ig = 2/0◦ Vg = IgZe = 2(20 − j40) = 40 − j80 V Vo = j120 80 + j80 (40 − j80) = 90 − j30 = 94.87/− 18.43◦ V vo = 94.87 cos(16 × 105t − 18.43◦) V [b] ω = 2πf = 16 × 105; f = 8 × 10 5 π T = 1 f = π 8 × 105 = 1.25π µs ·. . 18.43 360 (1.25π µs) = 201.09 ns ·. . vo lags ig by 201.09 ns P 9.65 jωL = j106(10 × 10−6) = j10 Ω 1 jωC = −j (106)(0.1 × 10−6) = −j10 Ω Va = 50/− 90◦ = −j50 V Vb = 25/90◦ = j25 V (10 − j10)I1 + j10I2 − 10I3 = −j50 docsity.com Problems 9–45 ωM = (200 × 103)(2k × 10−3) = 400k Zr = [ 400k 500 ]2 (300 − j400) = k2(192 − j256) Ω Zin = 200 + j200 + 192k2 − j256k2 |Zin| = [(200 + 192k2)2 + (200 − 256k2)2] 12 d|Zin| dk = 1 2 [(200 + 192k2)2 + (200 − 256k2)2]− 12 × [2(200 + 192k2)384k + 2(200 − 256k2)(−512k)] d|Zin| dk = 0 when 768k(200 + 192k2) − 1024k(200 − 256k2) = 0 ·. . k2 = 0.125; ·. . k = √ 0.125 = 0.3536 [b] Zin (min) = 200 + 192(0.125) + j[200 − 0.125(256)] = 224 + j168 = 280/36.87◦ Ω I1 (max) = 560/0◦ 224 + j168 = 2/− 36.87◦ A ·. . i1 (peak) = 2 A Note — You can test that the k value obtained from setting d|Zin|/dk = 0 leads to a minimum by noting 0 ≤ k ≤ 1. If k = 1, Zin = 392 − j56 = 395.98/− 8.13◦ Ω Thus, |Zin|k=1 > |Zin|k=√0.125 If k = 0, Zin = 200 + j200 = 282.84/45◦ Ω Thus, |Zin|k=0 > |Zin|k=√0.125 P 9.69 jωL1 = j50 Ω jωL2 = j32 Ω docsity.com 9–46 CHAPTER 9. Sinusoidal Steady State Analysis 1 jωC = −j20 Ω jωM = j(4 × 103)k √ (12.5)(8) × 10−3 = j40k Ω Z22 = 5 + j32 − j20 = 5 + j12 Ω Z∗22 = 5 − j12 Ω Zr = [ 40k |5 + j12| ]2 (5 − j12) = 47.337k2 − j113.609k2 Zab = 20 + j50 + 47.337k2 − j113.609k2 = (20 + 47.337k2) + j(50 − 113.609k2) Zab is resistive when 50 − 113.609k2 = 0 or k2 = 0.44 so k = 0.66 ·. . Zab = 20 + (47.337)(0.44) = 40.83 Ω P 9.70 [a] jωLL = j100 Ω jωL2 = j500 Ω Z22 = 300 + 500 + j100 + j500 = 800 + j600 Ω Z∗22 = 800 − j600 Ω ωM = 270 Ω Zr = ( 270 1000 )2 [800 − j600] = 58.32 − j43.74 Ω [b] Zab = R1 + jωL1 + Zr = 41.68 + j180 + 58.32 − j43.74 = 100 + j136.26 Ω P 9.71 ZL = V3 I3 = 80/60◦ Ω docsity.com Problems 9–47 V2 10 = V3 1 ; 10I2 = 1I3 V1 8 = −V2 1 ; 8I1 = −1I2 Zab = V1 I1 Substituting, Zab = V1 I1 = −8V2 −I2/8 = 82V2 I2 = 82(10V3) I3/10 = (8)2(10)2V3 I3 = (8)2(10)2ZL = 512, 000/60◦ Ω P 9.72 In Eq. 9.69 replace ω2M2 with k2ω2L1L2 and then write Xab as Xab = ωL1 − k 2ω2L1L2(ωL2 + ωLL) R222 + (ωL2 + ωLL)2 = ωL1 { 1 − k 2ωL2(ωL2 + ωLL) R222 + (ωL2 + ωLL)2 } For Xab to be negative requires R222 + (ωL2 + ωLL) 2 < k2ωL2(ωL2 + ωLL) or R222 + (ωL2 + ωLL) 2 − k2ωL2(ωL2 + ωLL) < 0 which reduces to R222 + ω 2L22(1 − k2) + ωL2ωLL(2 − k2) + ω2L2L < 0 But k ≤ 1 hence it is impossible to satisfy the inequality. Therefore Xab can never be negative if XL is an inductive reactance. docsity.com 9–50 CHAPTER 9. Sinusoidal Steady State Analysis P 9.75 [a] I = 240 24 + 240 j32 = (10 − j7.5) A Vs = 240/0◦ + (0.1 + j0.8)(10 − j7.5) = 247 + j7.25 = 247.11/1.68◦ V [b] Use the capacitor to eliminate the j component of I, therefore Ic = j7.5 A, Zc = 240 j7.5 = −j32 Ω Vs = 240 + (0.1 + j0.8)10 = 241 + j8 = 241.13/1.90◦ V [c] Let Ic denote the magnitude of the current in the capacitor branch. Then I = (10 − j7.5 + jIc) = 10 + j(Ic − 7.5) A Vs = 240/α = 240 + (0.1 + j0.8)[10 + j(Ic − 7.5)] = (247 − 0.8Ic) + j(7.25 + 0.1Ic) It follows that 240 cos α = (247 − 0.8Ic) and 240 sin α = (7.25 + 0.1Ic) Now square each term and then add to generate the quadratic equation I2c − 605.77Ic + 5325.48 = 0; Ic = 302.88 ± 293.96 Therefore Ic = 8.92 A (smallest value) and Zc = 240/j8.92 = −j26.90 Ω. P 9.76 The phasor domain equivalent circuit is Vo = Vm/0◦ 2 − IRx; I = Vm/0 ◦ Rx − jXC As Rx varies from 0 to ∞, the amplitude of vo remains constant and its phase angle decreases from 0◦ to −180◦, as shown in the following phasor diagram: docsity.com Problems 9–51 P 9.77 [a] I = 120 7.5 + 120 j12 = 16 − j10 A V = (0.15 + j6)(16 − j10) = 62.4 + j94.5 = 113.24/56.56◦ V Vs = 120/0◦ + V = 205.43/27.39◦ V [b] [c] I = 120 2.5 + 120 j4 = 48 − j30 A V = (0.15 + j6)(48 − j30) = 339.73/56.56◦ V Vs = 120 + V = 418.02/42.7◦ V The amplitude of Vs must be increased from 205.43 V to 418.02 V (more than doubled) to maintain the load voltage at 120 V. docsity.com 9–52 CHAPTER 9. Sinusoidal Steady State Analysis [d] I = 120 2.5 + 120 j4 + 120 −j2 = 48 + j30 A V = (0.15 + j6)(48 + j30) = 339.73/120.57◦ V Vs = 120 + V = 297.23/100.23◦ V The amplitude of Vs must be increased from 205.43 V to 297.23 V to maintain the load voltage at 120 V. P 9.78 Vg = 4/0◦ V; 1 jωC = −j20 kΩ Let Va = voltage across the capacitor, positive at upper terminal Then: Va − 4/0◦ 20,000 + Va −j20,000 + Va 20,000 = 0; ·. . Va = (1.6 − j0.8) V 0 − Va 20,000 + 0 − Vo 10,000 = 0; Vo = −Va2 ·. . Vo = −0.8 + j0.4 = 0.89/153.43◦ V vo = 0.89 cos(200t + 153.43◦) V P 9.79 [a] Va − 4/0◦ 20,000 + jωCoVa + Va 20,000 = 0 Va = 4 2 + j20,000ωCo Vo = −Va2 (see solution to Prob. 9.78) docsity.com Problems 9–55 [b] Vo = 6(1 − j1) 1 + j1 = −j6 V vo = 6 cos(105t − 90◦) V P 9.83 [a] Because the op-amps are ideal Iin = Io, thus Zab = Vab Iin = Vab Io ; Io = Vab − Vo Z Vo1 = Vab; Vo2 = − ( R2 R1 ) Vo1 = −KVo1 = −KVab Vo = Vo2 = −KVab ·. . Io = Vab − (−KVab) Z = (1 + K)Vab Z ·. . Zab = Vab(1 + K)Vab Z = Z (1 + K) [b] Z = 1 jωC ; Zab = 1 jωC(1 + K) ; ·. . Cab = C(1 + K) P 9.84 [a] I1 = 120 24 + 240 8.4 + j6.3 = 23.29 − j13.71 = 27.02/−30.5◦ A I2 = 120 12 − 120 24 = 5/0◦ A I3 = 120 12 + 240 8.4 + j6 = 28.29 − j13.71 = 31.44/−25.87◦ A I4 = 120 24 = 5/0◦ A; I5 = 120 12 = 10/0◦ A I6 = 240 8.4 + j6.3 = 18.29 − j13.71 = 22.86/−36.87◦ A docsity.com 9–56 CHAPTER 9. Sinusoidal Steady State Analysis [b] I1 = 0 I3 = 15 A I5 = 10 A I2 = 10 + 5 = 15 A I4 = −5 A I6 = 5 A [c] The clock and television set were fed from the uninterrupted side of the circuit, that is, the 12 Ω load includes the clock and the TV set. [d] No, the motor current drops to 5 A, well below its normal running value of 22.86 A. [e] After fuse A opens, the current in fuse B is only 15 A. P 9.85 [a] The circuit is redrawn, with mesh currents identified: The mesh current equations are: 120/0◦ = 23Ia − 2Ib − 20Ic 120/0◦ = −2Ia + 43Ib − 40Ic 0 = −20Ia − 40Ib + 70Ic Solving, Ia = 24/0◦ A Ib = 21.96/0◦ A Ic = 19.40/0◦ A The branch currents are: I1 = Ia = 24/0◦ A I2 = Ia − Ib = 2.04/0◦ A I3 = Ib = 21.96/0◦ A I4 = Ic = 19.40/0◦ A I5 = Ia − Ic = 4.6/0◦ A I6 = Ib − Ic = 2.55/0◦ A docsity.com Problems 9–57 [b] Let N1 be the number of turns on the primary winding; because the secondary winding is center-tapped, let 2N2 be the total turns on the secondary. From Fig. 9.58, 13,200 N1 = 240 2N2 or N2 N1 = 1 110 The ampere turn balance requires N1Ip = N2I1 + N2I3 Therefore, Ip = N2 N1 (I1 + I3) = 1 110 (24 + 21.96) = 0.42/0◦ A Check voltages — V4 = 10I4 = 194/0◦ V V5 = 20I5 = 92/0◦ V V6 = 40I6 = 102/0◦ V All of these voltages are low for a reasonable distribution circuit. P 9.86 [a] The three mesh current equations are 120/0◦ = 23Ia − 2Ib − 20Ic 120/0◦ = −2Ia + 23Ib − 20Ic 0 = −20Ia − 20Ib + 50Ic Solving, Ia = 24/0◦ A; Ib = 24/0◦ A; Ic = 19.2/0◦ A ·. . I2 = Ia − Ib = 0 A [b] Ip = N2 N1 (I1 + I3) = N2 N1 (Ia + Ib = 1 110 (24 + 24) = 0.436/0◦ A docsity.com