Sinusoidal Steady State Analysis

Sinusoidal Steady State Analysis 9 Assessment Problems AP 9. [a] V = 70/ 40 V [b] 0 sin(000t +20 ) = 0 cos(000t 70 ).. I = 0/ 70 A [c] I =5/36.87 + 0/ 53.3 =4+j3+6 j8 =0 j5 =.8/ 26.57 A [d] sin(20,000πt +30 ) = cos(20,000πt 60 ) Thus, V = 300/45 00/ 60 = 22.3 + j22.3 (50 j86.60) = 62.3 + j298.73 = 339.90/6.5 mv AP 9.2 [a] v =8.6 cos(ωt 54 ) V [b] I = 20/45 50/ 30 =4.4 + j4.4 43.3+j25 = 29.6 + j39.4=48.8/26.68 Therefore i =48.8 cos(ωt + 26.68 ) ma [c] V =20+j80 30/5 =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 = (0 4 )(20 0 3 ) = 200 Ω [b] Z L = jωl = j200 Ω 9

9 2 CHAPTER 9. Sinusoidal Steady State Analysis [c] V L = IZ L = (0/30 )(200/90 ) 0 3 = 2/20 V [d] v L = 2 cos(0,000t + 20 ) V AP 9.4 [a] X C = ωc = [b] Z C = jx C = j50 Ω 4000(5 0 6 ) = 50 Ω [c] I = V Z C = 30/25 50/ 90 =0.6/5 A [d] i =0.6 cos(4000t + 5 ) A AP 9.5 I = 00/25 =90.63 + j42.26 I 2 = 00/45 = 8.92 + j57.36 I 3 = 00/ 95 = 8.72 j99.62 I 4 = (I + I 2 + I 3 )=(0+j0) A, therefore i 4 =0A AP 9.6 [a] I = 25/ 60 Z /θ z = 25 Z /( 60 θ Z) But 60 θ Z = 05.. θz =45 Z =90+j60 + jx C.. X C = 70 Ω; X C = ωc = 70.. C= =2.86 µf (70)(5000) [b] I = V s Z = 25/ 60 (90 + j90) =0.982/ 05 A;.. I =0.982 A AP 9.7 [a] ω = 2000 rad/s ωl =0Ω, ωc = 20 Ω Z xy =20 j0+5+j20 = 20(j0) +5 j20 (20 + j0) =4+j8+5 j20=(9 j2) Ω

Problems 9 3 [b] ωl =40Ω, ωc = 5Ω Z xy =5 j5+20 j40=5 j5+ =5 j5+6+j8 = (2 + j3) Ω [ ] ( ) 20(jωL) [c] Z xy = + 5 j06 20 + jωl 25ω [ ] (20)(j40) 20 + j40 = 20ω2 L 2 400 + ω 2 L + j400ωl j06 +5 2 400 + ω 2 L2 25ω The impedance will be purely resistive when the j terms cancel, i.e., 400ωL 400 + ω 2 L = 06 2 25ω Solving for ω yields ω = 4000 rad/s. [d] Z xy = 20ω2 L 2 400 + ω 2 L +5=0+5=5Ω 2 AP 9.8 The frequency 4000 rad/s was found to give Z xy =5Ωin Assessment Problem 9.7. Thus, V = 50/0, I s = V = 50/0 Z xy 5 Using current division, I L = = 0/0 A 20 20 + j20 (0) = 5 j5 =7.07/ 45 A i L =7.07 cos(4000t 45 ) A, I m =7.07 A AP 9.9 After replacing the delta made up of the 50 Ω, 40 Ω, and 0 Ω resistors with its equivalent wye, the circuit becomes

9 4 CHAPTER 9. Sinusoidal Steady State Analysis The circuit is further simplified by combining the parallel branches, (20 + j40) (5 j5) = (2 j6) Ω Therefore I = 36/0 4+2 j6+4 =4/28.07 A AP 9.0 V = 240/53.3 = 44 + j92 V V 2 = 96/ 90 = j96 V jωl = j(4000)(5 0 3 )=j60 Ω jωc = j 6 0 6 (4000)(25) = j60 Ω Perform source transformations: V j60 V 2 20 = 44 + j92 j60 = j 96 20 = j4.8 A =3.2 j2.4 A Combine the parallel impedances: Y = j60 + 30 + j60 + 20 = j5 j60 = 2 Z = Y =2Ω V o = 2(3.2+j2.4)=38.4+j28.8 V = 48/36.87 V v o = 48 cos(4000t +36.87 ) V

Problems 9 5 AP 9. Use the lower node as the reference node. Let V = node voltage across the 20 Ω resistor and V Th = node voltage across the capacitor. Writing the node voltage equations gives us V 20 2/45 + V 0I x j0 We also have I x = V 20 =0 and V Th = j0 0 j0 (0I x) Solving these equations for V Th gives V Th = 0/45 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 0I x = (20 + j0)i x Therefore I x =0 and I T = V T j0 + V T 0 Z Th = V T I T, therefore Z Th =(5 j5) Ω AP 9.2 The phasor domain circuit is as shown in the following diagram:

9 6 CHAPTER 9. Sinusoidal Steady State Analysis The node voltage equation is 0 + V 5 + V j(20/9) + V j5 + V 00/ 90 20 =0 Therefore Therefore V =0 j30=3.62/ 7.57 v =3.62 cos(50,000t 7.57 ) V AP 9.3 Let I a, I b, and I c be the three clockwise mesh currents going from left to right. Summing the voltages around meshes a and b gives 33.8 =(+j2)i a +(3 j5)(i a I b ) and 0=(3 j5)(i b I a )+2(I b I c ). But V x = j5(i a I b ), therefore I c = 0.75[ j5(i a I b )]. Solving for I = I a =29+j2 =29.07/3.95 A. AP 9.4 [a] M =0.4 0.0625 = 0. H, ωm =80Ω [b] I = Z 22 =40+j800(0.25) + 360 + j800(0.25) = (400 + j300) Ω Therefore Z 22 = 500 Ω, Z22 = (400 j300) Ω ( ) 80 2 Z r = (400 j300) = (0.24 j7.68) Ω 500 245.20 84 + 00 + j400 + Z r =0.50/ 53.3 A i =0.5 cos(800t 53.3 ) A ( ) jωm j80 [c] I 2 = I = 500/36.87 (0.5/ 53.3 )=0.08/0 A Z 22 i 2 = 80 cos 800t ma

Problems 9 7 AP 9.5 I = V s Z + Z 2 /a 2 = =4+j3 =5/36.87 A 25 0 3 /0 500 + j6000 + (25) 2 (4 j4.4) V = V s Z I =25,000/0 (4 + j3)(500 + j6000) = 37,000 j28,500 V 2 = 25 V = 480 + j40 = 868.5/42.39 V I 2 = V 2 = 868.5/42.39 Z 2 4 j4.4 = 25/ 43.3 A Also, I 2 = 25I

9 8 CHAPTER 9. Sinusoidal Steady State Analysis Problems P 9. [a] ω =2πf = 3769.9 rad/s, f = ω 2π = 600 Hz P 9.2 V rms = [b] T =/f =.67 ms [c] V m =0V [d] v(0) = 0 cos( 53.3 )=6V [e] φ = 53.3 ; φ = 53.3 (2π) = 0.9273 rad 360 [f] V =0when 3769.9t 53.3 =90. Now resolve the units: (3769.9 rad/s)t = 43.3 =2.498 rad, t = 662.64 µs (80 /π) [g] (dv/dt) =( 0)3769.9 sin(3769.9t 53.3 ) T/2 0 (dv/dt) =0 when 3769.9t 53.3 =0 or 3769.9t = 53.3 57.3 /rad Therefore t = 245.97 µs T T/2 0 V 2 m sin 2 2π T tdt ( ) 2π Vm 2 sin 2 tdt= V m 2 T 2 T/2 0 =0.9273 rad ( cos 4π ) T t dt = V mt 2 4 Therefore V rms = VmT 2 T 4 = V m 2 P 9.3 [a] 40 V [b] 2πf = 00π; f =50Hz [c] ω = 00π = 34.59 rad/s [d] θ(rad) = 2π 360 (60 )= π =.05 rad 3 [e] θ =60 [f] T = f = 50 =20ms [g] v = 40 when 00πt + π 3 = π;.. t=6.67 ms

Problems 9 9 [ ( [h] v = 40 cos 00π t 0.0 ) + π ] 3 3 = 40 cos[00πt (π/3)+(π/3)] = 40 cos 00πt V [i] 00π(t t o )+(π/3) = 00πt (π/2).. 00πt o = 5π 6 ; t o =8.33 ms [j] 00π(t + t o )+(π/3) = 00πt +2π.. 00πt o = 5π 3 ; t o =6.67 ms 6.67 ms to the left P 9.4 [a] Left as φ becomes more positive [b] Left P 9.5 [a] By hypothesis v = 80 cos(ωt + θ) dv = 80ω sin(ωt + θ) dt.. 80ω =80,000; ω = 000 rad/s [b] f = ω 2π = 59.55 Hz; T = f =6.28 ms 2π/3 6.28 = 0.3333,.. θ= 90 ( 0.3333)(360) = 30.. v= 80 cos(000t +30 ) V

9 0 CHAPTER 9. Sinusoidal Steady State Analysis P 9.6 [a] T 2 =8+2=0ms; T =20ms P 9.7 u = f = T = =50Hz 20 0 3 [b] v = V m sin(ωt + θ) ω =2πf = 00π rad/s 00π( 2 0 3 )+θ =0;.. θ= π 5 v = V m sin[00πt +36 ] 80.9 =V m sin 36 ; V m = 37.64 V rad =36 v = 37.64 sin[00πt +36 ] = 37.64 cos[00πt 54 ] V to+t = V 2 m = V 2 m 2 = V 2 m 2 = V 2 m 2 = V 2 m V 2 m cos 2 (ωt + φ) dt t o to+t t o 2 + cos(2ωt +2φ) dt 2 { } to+t to+t t o dt + cos(2ωt +2φ) dt { T + 2ω { T + ( T 2 ) t o [ ] } sin(2ωt +2φ) t o+t t o } 2ω [sin(2ωt o +4π +2φ) sin(2ωt o +2φ)] + ( ) T 2ω (0) = V m 2 2 P 9.8 P 9.9 V m = 2V rms = 2(20) = 69.7 V [a] The numerical values of the terms in Eq. 9.8 are V m =20, R/L = 066.67, ωl =60 R2 + ω 2 L 2 = 00 φ =25, θ = tan 60/80, θ =36.87 Substitute these values into Equation 9.9: i = [ 95.72e 066.67t + 200 cos(800t.87 ) ] ma, t 0 [b] Transient component = 95.72e 066.67t ma Steady-state component = 200 cos(800t.87 ) ma [c] By direct substitution into Eq 9.9 in part (a), i(.875 ms) =28.39 ma [d] 200 ma, 800 rad/s,.87

Problems 9 [e] The current lags the voltage by 36.87. P 9.0 [a] From Eq. 9.9 we have L di dt = V mr cos(φ θ) e (R/L)t ωlv m sin(ωt + φ θ) R2 + ω 2 L 2 R2 + ω 2 L 2 Ri = V mr cos(φ θ)e (R/L)t R2 + ω 2 L 2 + V mr cos(ωt + φ θ) R2 + ω 2 L 2 L di [ ] R cos(ωt + φ θ) ωl sin(ωt + φ θ) dt + Ri = V m R2 + ω 2 L 2 But R R2 + ω 2 L = cos θ and ωl 2 R2 + ω 2 L = sin θ 2 Therefore the right-hand side reduces to V m cos(ωt + φ) At t =0, Eq. 9.9 reduces to i(0) = V m cos(φ θ) R2 ω 2 L 2 + V m cos(φ θ) R2 + ω 2 L 2 =0 V m [b] i ss = cos(ωt + φ θ) R2 + ω 2 L2 Therefore L di ss dt and = ωlv m sin(ωt + φ θ) R2 + ω 2 L2 V m R Ri ss = cos(ωt + φ θ) R2 + ω 2 L2 L di [ ] ss R cos(ωt + φ θ) ωl sin(ωt + φ θ) dt + Ri ss = V m R2 + ω 2 L 2 = V m cos(ωt + φ) P 9. [a] Y = 50/60 + 00/ 30 =.8/ 3.43 y =.8 cos(500t 3.43 ) [b] Y = 200/50 00/60 = 02.99/40.29 y = 02.99 cos(377t +40.29 ) [c] Y = 80/30 00/ 225 + 50/ 90 = 6.59/ 29.96 y = 6.59 cos(00t 29.96 )

9 2 CHAPTER 9. Sinusoidal Steady State Analysis [d] Y = 250/0 + 250/20 + 250/ 20 =0 y =0 P 9.2 [a] 000Hz [b] θ v =0 [c] I = 200/0 jωl = 200 ωl / 90 = 25/ 90 ; [d] 200 200 = 25; ωl = ωl 25 =8Ω 8 [e] L = =.27 mh 2π(000) [f] Z L = jωl = j8ω θ i = 90 P 9.3 [a] ω = 2πf = 34,59.27 rad/s [b] I = V = 0 0 3 /0 = jωc(0 0 3 )/0 =0 0 3 ωc/90 Z C /jωc.. θ i =90 [c] 628.32 0 6 =0 0 3 ωc P 9.4 [d] C = ωc = 0 0 3 628.32 0 =5.92 Ω,.. XC = 5.92 Ω 6 5.92(ω) = (5.92)(00π 0 3 ) C =0.2 µf [e] Z c = j ( ) = j5.92 Ω ωc [a] jωl = j(2 0 4 )(300 0 6 )=j6ω jωc = j (2 0 4 )(5 0 6 ) = j0 Ω; I g = 922/30 A

Problems 9 3 [b] V o = 922/30 Z e Z e = Y e ; Y e = 0 + j 0 + 8+j6 Y e =0.8 + j0.04 S Z e = 0.8 + j0.04 =5.42/ 2.53 Ω V o = (922/30 )(5.42/ 2.53 ) = 5000.25/7.47 V [c] v o = 5000.25 cos(2 0 4 t +7.47 ) V P 9.5 [a] Z L = j(8000)(5 0 3 )=j40 Ω Z C = j (8000)(.25 0 6 ) = j00 Ω P 9.6 600/20 [b] I = 40 + j40 j00 =8.32/76.3 A [c] i =8.32 cos(8000t +76.3 ) A Z =4+j(50)(0.24) j (50)(0.0025) =4+j4 =5.66/45 Ω I o = V Z = 0./ 90 5.66/45 =7.68/ 35 ma i o (t) =7.68 cos(50t 35 ) ma P 9.7 [a] Y = 3+j4 + 6 j2 + j4 =0.2 j0.6+0.04 + j0.03 + j0.25 =0.6 + j0.2 = 200/36.87 ms [b] G = 60 ms [c] B = 20 ms

9 4 CHAPTER 9. Sinusoidal Steady State Analysis [d] I =8/0 A, V = I Y = 8 0.2/36.87 = 40/ 36.87 V I C = V Z C = 40/ 36.87 4/ 90 = 0/53.3 A i C = 0 cos(ωt +53.3 ) A, I m =0A P 9.8 Z L = j(2000)(60 0 3 )=j20 Ω Z C = j (2000)(2.5 0 6 ) = j40 Ω Construct the phasor domain equivalent circuit: Using current division: I = (20 j40) (0.5)=0.25 j0.25 A 20 j40+40+j20 V o = j20i =30+j30=42.43/45 V v o =42.43 cos(2000t +45 ) V P 9.9 [a] V g = 300/78 ; I g =6/33.. Z= V g I g = 300/78 6/33 = 50/45 Ω [b] i g lags v g by 45 : 2πf = 5000π; f = 2500 Hz; T =/f = 400 µs.. i g lags v g by 45 (400 µs) =50µs 360

Problems 9 5 P 9.20 jωc = ( 0 6 )(50 0 3 ) = j20 Ω jωl = j50 0 3 (.2 0 3 )=j60 Ω V g = 40/0 V Z e = j20+30 j60=24 j8ω I g = 40/0 24 j8 =.5+j0.5 ma V o = (30 j60)i g = 30(j60) 30 + j60 (.5+j0.5)=30+j30=42.43/45 V v o =42.43 cos(50,000t +45 ) V P 9.2 [a] Z = R j ωc Z 2 = R 2 /jωc 2 R 2 +(/jωc 2 ) = R 2 +jωr 2 C 2 = R 2 jωr 2 2C 2 +ω 2 R 2 2C 2 2 Z = Z 2 when R = R 2 +ω 2 R 2 2C 2 2 and [b] R = ωc = ωr2 2C 2 +ω 2 R 2 2C 2 2 or C = +ω2 R 2 2C 2 2 ω 2 R 2 2C 2 000 + (40 0 3 ) 2 (000) 2 (50 0 9 ) 2 = 200 Ω C = + (40 03 ) 2 (000) 2 (50 0 9 ) 2 (40 0 3 ) 2 (000) 2 (50 0 9 ) =62.5 nf

9 6 CHAPTER 9. Sinusoidal Steady State Analysis P 9.22 [a] Y 2 = R 2 + jωc 2 Y = R +(/jωc ) = jωc +jωr C = ω2 R C 2 + jωc +ω 2 R 2 C 2 Therefore Y = Y 2 when R 2 = +ω2 R 2 C 2 ω 2 R C 2 and C 2 = C +ω 2 R 2 C 2 [b] R 2 = + (50 03 ) 2 (000) 2 (40 0 9 ) 2 (50 0 3 ) 2 (000)(40 0 9 ) 2 = 250 Ω C 2 = P 9.23 [a] Z = R + jωl 40 0 9 + (50 0 3 ) 2 (000) 2 (40 0 9 ) 2 =8nF Z 2 = R 2(jωL 2 ) R 2 + jωl 2 = ω2 L 2 2R 2 + jωl 2 R 2 2 R 2 + ω 2 L 2 2 Z = Z 2 when R = ω2 L 2 2R 2 R 2 2 + ω 2 L 2 2 and L = R2 2L 2 R 2 2 + ω 2 L 2 2 [b] R = (4000)2 (.25) 2 (5000) 5000 2 + 4000 2 (.25) 2 = 2500 Ω L = P 9.24 [a] Y 2 = R 2 j ωl 2 Y = (5000) 2 (.25) = 625 mh 5000 2 + 4000 2 (.25) 2 R + jωl = R jωl R 2 + ω 2 L 2 Therefore Y 2 = Y when R 2 = R2 + ω 2 L 2 R and L 2 = R2 + ω 2 L 2 ω 2 L [b] R 2 = 80002 + 000 2 (4) 2 8000 L 2 = 80002 + 000 2 (4) 2 000 2 (4) =0kΩ =20H

Problems 9 7 P 9.25 V g = 500/30 V; I g =0./83.3 ma Z = V g I g z = 3000 + j = 5000/ 53.3 Ω = 3000 j4000 Ω ( ω ) 32 03 ω ω 32 03 ω = 4000 ω 2 + 4000ω 32 0 3 =0 ω =7.984 rad/s P 9.26 [a] Z eq = 50,000 + 3 = 50,000 3 = 50,000 3 Im(Z eq )= j20 06 (200 + j0.2ω) ω + + j20 06 ω j20 0 6 (200 + j0.2ω) 200 + j[0.2ω 20 06 ω ] (200 + j0.2ω) [ 200 j ( )] 0.2ω 20 06 ω ω 200 2 + ( 0.2ω 20 06 ω [ ( 20 06 (200) 2 20 06 ω ω ( 20 0 6 (200) 2 20 0 6 [0.2ω (200) 2 =0.2ω ( 0.2ω ) 20 06 ω 0.2ω 0.2ω ) 2 0.2ω )] 20 06 =0 ω )] 20 06 =0 ω 0.2 2 ω 2 0.2(20 0 6 ) + 200 2 =0 ω 2 =64 0 6.. ω= 8000 rad/s.. f= 273.24 Hz [b] Z eq = 50,000 + j2500 (200 + j600) 3 = 50,000 3 + ( j2500)(200 + j600) 200 j900 =20,000 Ω I g = 30/0 20,000 =.5/0 ma i g (t) =.5 cos 8000t ma

9 8 CHAPTER 9. Sinusoidal Steady State Analysis P 9.27 [a] Find the equivalent impedance seen by the source, as a function of L, and set the imaginary part of the equivalent impedance to 0, solving for L: Z C = j (500)(2 0 6 ) = j000 Ω Z eq = j000 + j500l 2000 = j000 + 2000(j500L) 2000 + j500l = j000 + Im(Z eq )= 000 + 2000(j500L)(2000 j500l) 2000 2 + (500L) 2 20002 (500L) 2000 2 + (500L) 2 =0.. 2000 2 (500L) 2000 2 + (500L) 2 = 000 [b] I g =.. 500 2 L 2 2 20002 L + 2000 2 =0 Solving the quadratic equation, L =4H 00/0 j000 + j2000 2000 = 00/0 000 =0./0 A i g (t) =0. cos 500t A P 9.28 [a] jωl + R ( j/ωc)=jωl + jr/ωc R j/ωc = jωl + jr ωcr j = jωl + Im(Z ab )=ωl.. L= jr(ωcr + j) ω 2 C 2 R 2 + CR 2 ω 2 C 2 R 2 + ωcr 2 ω 2 C 2 R 2 + =0.. ω 2 C 2 R 2 += CR2 L.. ω 2 = (CR2 /L) = C 2 R 2 ω = 300 krad/s (25 0 9 )(00) 2 60 0 6 = 900 (25 0 9 ) 2 08 (00) 2

Problems 9 9 [b] Z ab (300 0 3 )=j48 + (00)( j33.33) 00 j33.33 =64Ω P 9.29 jωl = j00 0 3 (0.6 0 3 )=j60 Ω jωc = j (00 0 3 )(0.4 0 6 ) = j25 Ω V T = j25i T +5I 30I I = j60 30 + j60 I T j60 V T = j25i T +25 30 + j60 I T V T I T = Z ab =20 j5 = 25/ 36.87 Ω P 9.30 0 6 [a] Z = 400 j 500(2.5) = 400 j800 Ω Z 2 = 2000 j500l = j0 6 L 2000 + j500l Z T = Z + Z 2 = 400 j800 + j0 6 L 2000 + j500l = 400 + 500 06 L 2 2000 2 + 500 2 L 2 j800 + j 2 0 9 L 2000 2 + 500 2 L 2 Z T is resistive when 2 0 9 L 2000 2 + 500 2 L = 800 2 or 5002 L 2 25 0 5 L + 2000 2 =0 Solving, L =8Hand L 2 =2H.

9 20 CHAPTER 9. Sinusoidal Steady State Analysis P 9.3 [a] Y = [b] When L =8H: Z T = 400 + 500 06 (8) 2 2000 2 + 500 2 (8) 2 = 2000 Ω I g = 200/0 2000 = 00/0 ma i g = 00 cos 500t ma When L =2H: Z T = 400 + 500 06 (2) 2 2000 2 + 500(2) 2 = 800 Ω I g = 200/0 = 250/0 ma 800 i g = 250 cos 500t ma Y 2 = 2500 0 3 =4.4 0 6 S = 4,000 + j5ω 4,000 96 0 6 +25ω 2 j 5ω 96 0 6 +25ω 2 Y 3 = jω2 0 9 Y T = Y + Y 2 + Y 3 For i g and v o to be in phase the j component of Y T must be zero; thus, ω2 0 9 = or 25ω 2 + 96 0 6 = 5ω 96 0 6 +25ω 2 5 2 0 9.. 25ω 2 = 2304 0 6.. ω= 9600 rad/s [b] Y T =4.4 0 6 4,000 + 96 0 6 + 25(9600) =0 2 0 6 S.. Z T = 00 kω V o =(0.25 0 3 /0 )(00 0 3 ) = 25/0 V v o = 25 cos 9600t V

Problems 9 2 P 9.32 [a] Z g = 500 j 06 ω + 03 (j0.5ω) 0 3 + j0.5ω = 500 j 06 ω + 500jω(000 j0.5ω) 0 6 +0.25ω 2 = 500 j 06 ω + 250ω 2 0 6 +0.25ω 2 + j 5 0 5 ω 0 6 +0.25ω 2.. If Z g is purely real, 0 6 ω = 5 05 ω 0 6 +0.25ω 2 2(0 6 +0.25ω 2 )=ω 2.. 4 0 6 = ω 2.. ω= 2000 rad/s [b] When ω = 2000 rad/s Z g = 500 j500+(j000 000) = 000 Ω.. I g = 20/0 000 = 20/0 ma V o = V g I g Z Z = 500 j500 Ω V o = 20/0 (0.02/0 )(500 j500) = 0 + j0=4.4/45 V v o =4.4 cos(2000t +45 ) V P 9.33 Z ab = j8+(2+j4) (0 j20) + (40 j20) = j8+3+j4+8+j6=2+j2ω=6.97/45 Ω P 9.34 First find the admittance of the parallel branches Y p = 2 j6 + 2 + j4 + 2 + j0.5 Z p = Y p = 0.625 j.875 =0.625 j.875 S =0.6 + j0.48 Ω Z ab = j4.48+0.6 + j0.48+2.84=3 j4ω Y ab = Z ab = 3 j4 = 200/53.3 ms = 20 + j60 ms

9 22 CHAPTER 9. Sinusoidal Steady State Analysis P 9.35 Simplify the top triangle using series and parallel combinations: ( + j) ( j)=ω Convert the lower left delta to a wye: Z = Z 2 = (j)() +j j = jω ( j)() +j j = jω Z 3 = (j)( j) +j j =Ω Convert the lower right delta to a wye: Z 4 = ( j)() +j j = jω Z 5 = ( j)(j) +j j =Ω Z 6 = (j)() +j j = jω The resulting circuit is shown below: Simplify the middle portion of the circuit by making series and parallel combinations: ( + j j) (+)= 2 =2/3Ω Z ab = j+2/3+j =2/3Ω

Problems 9 23 P 9.36 V o = V g Z o Z T = 500 j000 300 + j600 + 500 j000 (00/0 ) =.8/ 00.3 V v o =.8 cos(8000t 00.3 ) V P 9.37 jωc = j400 Ω jωl = j200 Ω Let Z = 200 j400 Ω; Z 2 = 600 + j200 Ω I g = 400/0 ma I o = Z 2 Z + Z 2 I g = 600 + j200 800 + j800 (0.4/0 ) = 450 + j50 ma = 474.34/8.43 ma i o = 474.34 cos(20,000t +8.43 ) ma P 9.38 V = j5( j2)=0v 25+0+(4 j3)i =0.. I = 5 4 j3 =2.4+j.8 A I 2 = I j5 =(2.4+j.8) j5 =2.4 j3.2 A

9 24 CHAPTER 9. Sinusoidal Steady State Analysis V Z = j5i 2 +(4 j3)i = j5(2.4 j3.2)+(4 j3)(2.4+j.8) = j2 V 25+(+j3)I 3 +( j2)=0.. I3 =6.2 j6.6 A I Z = I 3 I 2 =(6.2 j6.6) (2.4 j3.2)=3.8 j3.4 A Z = V Z I Z = j2 3.8 j3.4 =.42 j.88 Ω P 9.39 I s =3/0 ma jωc = j0.4ω jωl = j0.4ω After source transformation we have V o = j0.4 j0.4 5 28 + j0.4 j0.4 5 (66 0 3 )=0mV v o = 0 cos 200t mv P 9.40 [a] V a = (50 + j50)(2/0 ) = 00 + j300 V 00 + j300 I b = 20 j40 = j2.5 A =2.5/90 A I c =2/0 + j2.5+6+j3.5 =8+j6A = 0/36.87 A V g =5I c + V a = 5(8 + j6) + 00 + j300 = 40 + j330 V = 358.47/67.0 V

Problems 9 25 [b] i b =2.5 cos(800t +90 ) A i c = 0 cos(800t +36.87 ) A v g = 358.47 cos(800t +67.0 ) V P 9.4 [a] jωl = j(000)(00) 0 3 = j00 Ω jωc = j 0 6 (000)(0) = j00 Ω Using voltage division, V ab = (00 + j00) ( j00) j00 + (00 + j00) ( j00) (247.49/45 ) = 350/0 V Th = V ab = 350/0 V [b] Remove the voltage source and combine impedances in parallel to find Z Th = Z ab : Y ab = j00 + 00 + j00 + =5 j5ms j00 [c] Z Th = Z ab = Y ab = 00 + j00 Ω

9 26 CHAPTER 9. Sinusoidal Steady State Analysis P 9.42 Using voltage division: V Th = 36 36 + j60 j48 (240) = 26 j72 = 227.68/ 8.43 V Remove the source and combine impedances in series and in parallel: Z Th =36 (j60 j48)=3.6+j0.8ω P 9.43 Open circuit voltage: V 2 0 +88I φ + V 2 V 5 2 j50 I φ = 5 (V 2/5) 200 Solving, =0 V 2 = 66 + j88 = 0/26.87 V = V Th Find the Thévenin equivalent impedance using a test source: I T = V T 0 +88I φ + 0.8V t j50 I φ = V T /5 200 ( I T = V T 0 88/5 200 + 0.8 ) j50

Problems 9 27.. V T I T =30 j40 = Z Th I N = V Th Z Th = 66 + j88 30 j40 = 2.2+j0 A =2.2/80 A The Norton equivalent circuit: P 9.44 Short circuit current I β = 6I β 2 2I β = 6I β ;.. Iβ =0 I =0;.. Isc = 0/ 45 A = I N The Norton impedance is the same as the Thévenin impedance. Find it using a test source V T =6I β +2I β =8I β, I β = j 2+j I T

9 28 CHAPTER 9. Sinusoidal Steady State Analysis Z Th = V T I T = 8I β = j8 [(2 + j)/j]i β 2+j =.6+j3.2Ω P 9.45 Using current division: I N = I sc = 50 80 + j60 (4)=.6 j.2 =2/ 36.87 A Z N = j00 (80 + j60) = 00 j50 Ω The Norton equivalent circuit: P 9.46 ω = 2π(200/π) = 400 rad/s Z c = j 400(0 6 ) = j2500 Ω V T = (0,000 j2500)i T + 00(200)I T Z Th = V T I T =30 j2.5 kω P 9.47 I N = 5 j5 Z N +( j3) ma, Z N in kω

Problems 9 29 I N = 8 j3.5 Z N +4.5 j6 ma, Z N in kω 5 j5 Z N + j3 = 8 j3.5 Z N +(4.5 j6) 23 j.5 Z N =3.5 j3.. ZN =4+j3 kω I N = 5 j5 4+j3 + j3 = j6 ma =6/ 90 ma P 9.48 Open circuit voltage: V 250 20 + j0 0.03V V o + 50 j00 =0.. V o = j00 50 j00 V

9 30 CHAPTER 9. Sinusoidal Steady State Analysis V 20 + j0 + j3v 50 j00 + V 50 j00 = 250 20 + j0 V = 500 j250 V; V o = 300 j400 V = V Th = 500/ 53.3 V Short circuit current: I sc = 250/0 70 + j0 =3.5 j0.5 A Z Th = V Th I sc = 300 j400 3.5 j0.5 = 00 j00 Ω The Thévenin equivalent circuit: P 9.49 Open circuit voltage: (9 + j4)i a I b = 60/0

Problems 9 3 I a +(9 j4)i b = 60/0 Solving, I a = 5+j2.5 A; I b =5+j2.5 A V Th =4I a +(4 j4)i b = 0/0 V Short circuit current: (9 + j4)i a I b 4I sc = 60 I a +(9 j4)i b (4 j4)i sc =60 4I a (4 j4)i b +(8 j4)i sc =0 Solving, I sc =2.07/0 Z Th = V Th I sc = 0/0 2.07/0 =4.83 Ω

9 32 CHAPTER 9. Sinusoidal Steady State Analysis Alternate calculation for Z Th : Z =4++4 j4 =9 j4 Z = 4 9 j4 Z 2 = 4 j4 9 j4 Z 3 = 6 j6 9 j4 Z a =4+j4+ 4 9 j4 Z b =4+ 4 j4 9 j4 Z a Z b = Z 3 + Z a Z b = 2640 j320 864 j384 = 56 + j20 9 j4 = 40 j20 9 j4 6 j6 9 j4 + 2640 j320 864 j384 = 476 j856 864 j384 =4.83 Ω

Problems 9 33 P 9.50 [a] I T = V T 000 + V T αv T j000 I T = ( α) V T 000 j000 = j +α j000.. Z Th = V T I T = j000 α +j Z Th is real when α =. [b] Z Th = 000 Ω [c] Z Th = 500 j500 = j000 α +j = 000 000(α ) + j (α ) 2 + (α ) 2 + Equate the real parts: 000 (α ) 2 + = 500.. (α ) 2 +=2.. (α ) 2 = so α =0 Check the imaginary parts: (α )000 (α ) 2 = 500 + α= Thus, α =0. 000 000(α ) [d] Z Th = + j (α ) 2 + (α ) 2 + For Im(Z Th ) > 0, α must be greater than. SoZ Th is inductive for <α 0.

9 34 CHAPTER 9. Sinusoidal Steady State Analysis P 9.5 V 240 j0 + V 50 + V 30 + j0 =0 Solving for V yields V = 98.63/ 24.44 V V o = 30 30 + j0 (V ) = 88.43/ 42.88 V P 9.52 jωl = j(2000)( 0 3 )=j2ω jωc = j 0 6 (2000)(00) = j5ω V g = 20/ 36.87 =6 j2 V V g2 = 50/ 06.26 = 4 j48 V V o (6 j2) j2 Solving, V o = 36/0 V v o (t) = 36 cos 2000t V + V o 0 + V o ( 4 j48) j5 =0

Problems 9 35 P 9.53 From the solution to Problem 9.52 the phasor-domain circuit is Making two source transformations yields I g = I g2 = 6 j2 j2 4 j48 j5 Y = j2 + 0 + j5 Z = Y =+j3ω = 6 j8 A =9.6 j2.8 A =(0. j0.3) S I e = I g + I g2 =3.6 j0.8 A Hence the circuit reduces to V o = ZI e =(+j3)(3.6 j0.8) = 36/0 V.. v o (t) = 36 cos 2000t V

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 + j2i + 0(I I 2 )=0 50/ 06.26 + 0(I 2 I ) j5i 2 =0 In standard form: I (0 + j2) + I 2 ( 0) = 20/ 36.87 I ( 0) + I 2 (0 j5) = 50/ 06.26 = 50/73.74 Solving on a calculator yields: I = 6+j0A; I 2 = 9.6+j0A Thus, V o = 0(I I 2 )=36V and v o (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 = 0 j5 (6 j2)=8 j26 V j2+0 j5

Problems 9 37 With the left hand source removed V o = 0 j2 ( 4 j48)=8+j26 V j5+0 j2 V o = V o + V o =8 j26+8+j26=36v v o (t) = 36 cos 2000t V P 9.56 Write a KCL equation at the top node: V o j8 + V o 2.4I + V o (0 + j0)=0 j4 5 The constraint equation is: I = V o j8 Solving, V o = j80 = 80/90 V P 9.57 Write node voltage equations: Left Node: V 40 + V V o /8 =0.025/0 j20 Right Node: V o 50 + V o j25 +6I o =0

9 38 CHAPTER 9. Sinusoidal Steady State Analysis The constraint equation is I o = V V o /8 j20 Solution: V o =(4+j4)=5.66/45 V V =(0.8+j0.6)=.0/36.87 V I o =(5 j5)=5.8/ 7.57 ma P 9.58 (0 + j5)i a j5i b = 00/0 j5i a j5i b = j00 Solving, I a = j0 A; I b = 20 + j0 A I o = I a I b =20 j20=28.28/ 45 A i o (t) =28.28 cos(50,000t 45 ) A P 9.59 (2 j2)i a 2I g 5( j8)=0

Problems 9 39 2I a + (2 + j4)i g + j20 5(j4)=0 Solving, I g =4 j2 =4.47/ 26.57 A P 9.60 Set up the frequency domain circuit to use the node voltage method: At V : 5/0 + V V 2 j8 + V 20/90 j4 =0 At V 2 : V 2 V j8 + V 2 j4 + V 2 20/90 2 =0 In standard form: ( V j8 + ) ( + V 2 ) =5/0 + 20/90 j4 j8 j4 ( V ) ( + V 2 j8 j8 + j4 + ) = 20/90 2 2 Solving on a calculator: V = 8 3 + j 4 3 V V 2 = 8+j4 V Thus V 0 = V 20/90 = 8 3 j 56 3 =8.86/ 98.3 V

9 40 CHAPTER 9. Sinusoidal Steady State Analysis P 9.6 jωl = j5000(60 0 3 )=j300 Ω jωc = j (5000)(2 0 6 ) = j00 Ω 400/0 + (50 + j300)i a 50I b 50(I a I b )=0 (50 j00)i b 50I a + 50(I a I b )=0 Solving, I a = 0.8 j.6 A; I b =.6+j0.8 A V o = 00I b = 60 + j80 = 78.89/53.43 V v o = 78.89 cos(5000t + 53.43 ) V P 9.62 0/0 =( j)i I 2 + ji 3 5/0 = I +(+j)i 2 ji 3

Problems 9 4 =ji ji 2 + I 3 Solving, I =+j0 A; I 2 =+j5a; I 3 =6A I a = I 3 =5A =5/0 A I b = I I 3 =5+j0 A =.8/63.43 A I c = I 2 I 3 =5+j5A =7.07/45 A I d = I I 2 = j5 A =5/90 A P 9.63 V a (00 j50) 20 Solving, V a =40+j30 V + V a j5 + V a (40 + j30) 2 + j6 =0 I Z + (30 + j20) 40 + j30 j0 + (40 + j30) (40 + j30) 2 + j6 =0 Solving, I Z = 30 j0 A Z = (00 j50) (40 + j30) 30 j0 =2+j2Ω

9 42 CHAPTER 9. Sinusoidal Steady State Analysis P 9.64 [a] jωc = j50 Ω jωl = j20 Ω Z e = 00 j50=20 j40 Ω I g =2/0 V g = I g Z e = 2(20 j40)=40 j80 V V o = j20 80 + j80 (40 j80)=90 j30=94.87/ 8.43 V v o =94.87 cos(6 0 5 t 8.43 ) V [b] ω =2πf =6 0 5 ; f = 8 05 π T = f = π 8 0 =.25πµs 5.. 8.43 (.25πµs) = 20.09 ns 360.. v o lags i g by 20.09 ns P 9.65 jωl = j0 6 (0 0 6 )=j0 Ω jωc = j (0 6 )(0. 0 6 ) = j0 Ω V a = 50/ 90 = j50 V V b = 25/90 = j25 V (0 j0)i + j0i 2 0I 3 = j50

Problems 9 43 j0i +0I 2 0I 3 =0 0I 0I 2 +20I 3 = j25 Solving, I =0.5 j.5 A; I 3 = +j0.5 A I 2 = 2.5 A I a = I = 0.5+j.5 =.58/08.43 A I b = I 3 = j0.5 =.2/ 26.57 A i a =.58 cos(0 6 t + 08.43 ) A i b =.2 cos(0 6 t 26.57 ) A P 9.66 [a] jωl = j(5000)(2 0 3 )=j0 Ω jωl 2 = j(5000)(8 0 3 )=j40 Ω jωm = j0 Ω 70 = (0 + j0)i g + j0i L 0=j0I g + (30 + j40)i L Solving, I g =4 j3 A; I L = A [b] k = i g = 5 cos(5000t 36.87 ) A i L = cos(5000t 80 ) A M L L 2 = 2 6 =0.5

9 44 CHAPTER 9. Sinusoidal Steady State Analysis [c] When t = 00πµs, 5000t = (5000)(00π) 0 6 =0.5π = π/2 rad =90 i g (00πµs) = 5 cos(53.3 )=3A i L (00πµs) = cos( 90 )=0A w = 2 L i 2 + 2 L 2i 2 2 + Mi i 2 = 2 (2 0 3 )(9)+0+0=9mJ When t = 200πµs, 5000t = π rad = 80 i g (200πµs) = 5 cos(80 36.87 )= 4A i L (200πµs) = cos(80 80 )=A w = 2 (2 0 3 )(6) + 2 (8 0 3 )() + 2 0 3 ( 4)() = 2 mj P 9.67 Remove the voltage source to find the equivalent impedance: Z Th =45+j25 + Using voltage division: V Th = V cd = j20i = j20 ( ) 2 20 (5 j5)=85+j85 Ω 5+j5 ( ) 425 = 850 + j850 V = 202./45 V 5+j5 P 9.68 [a] jωl = j(200 0 3 )(0 3 )=j200 Ω jωl 2 = j(200 0 3 )(4 0 3 )=j800 Ω jωc = j (200 0 3 )(2.5 0 9 ) = j400 Ω.. Z 22 = 00 + 200 + j800 j400 = 300 + j400 Ω.. Z 22 = 300 j400 Ω M = k L L 2 =2k 0 3

Problems 9 45 ωm = (200 0 3 )(2k 0 3 ) = 400k [ ] 2 400k Z r = (300 j400) = k 2 (92 j256) Ω 500 Z in = 200 + j200 + 92k 2 j256k 2 Z in = [(200 + 92k 2 ) 2 + (200 256k 2 ) 2 ] 2 d Z in dk = 2 [(200 + 92k2 ) 2 + (200 256k 2 ) 2 ] 2 [2(200 + 92k 2 )384k + 2(200 256k 2 )( 52k)] d Z in =0when dk 768k(200 + 92k 2 ) 024k(200 256k 2 )=0.. k 2 =0.25;.. k= 0.25=0.3536 [b] Z in (min) = 200 + 92(0.25) + j[200 0.25(256)] = 224 + j68 = 280/36.87 Ω I (max) = 560/0 224 + j68 =2/ 36.87 A.. i (peak) =2A Note You can test that the k value obtained from setting d Z in /dk =0 leads to a minimum by noting 0 k. Ifk =, Z in = 392 j56 = 395.98/ 8.3 Ω Thus, Z in k= > Z in k= 0.25 If k =0, Z in = 200 + j200 = 282.84/45 Ω Thus, Z in k=0 > Z in k= 0.25 P 9.69 jωl = j50 Ω jωl 2 = j32 Ω

9 46 CHAPTER 9. Sinusoidal Steady State Analysis jωc = j20 Ω jωm = j(4 0 3 )k (2.5)(8) 0 3 = j40k Ω Z 22 =5+j32 j20=5+j2 Ω Z 22 =5 j2 Ω [ ] 2 40k Z r = (5 j2)=47.337k 2 j3.609k 2 5+j2 Z ab =20+j50+47.337k 2 j3.609k 2 =(20+47.337k 2 )+j(50 3.609k 2 ) Z ab is resistive when 50 3.609k 2 =0 or k 2 =0.44 so k =0.66.. Z ab = 20 + (47.337)(0.44)=40.83 Ω P 9.70 [a] jωl L = j00 Ω jωl 2 = j500 Ω Z 22 = 300 + 500 + j00 + j500 = 800 + j600 Ω Z 22 = 800 j600 Ω ωm = 270 Ω ( ) 270 2 Z r = [800 j600] = 58.32 j43.74 Ω 000 [b] Z ab = R + jωl + Z r =4.68 + j80+58.32 j43.74 = 00 + j36.26 Ω P 9.7 Z L = V 3 I 3 = 80/60 Ω

Problems 9 47 V 2 0 = V 3 ; 0I 2 =I 3 V 8 = V 2 ; 8I = I 2 Z ab = V I Substituting, Z ab = V I = 8V 2 I 2 /8 = 82 V 2 I 2 = 82 (0V 3 ) I 3 /0 = (8)2 (0) 2 V 3 I 3 = (8) 2 (0) 2 Z L = 52, 000/60 Ω P 9.72 In Eq. 9.69 replace ω 2 M 2 with k 2 ω 2 L L 2 and then write X ab as X ab = ωl k2 ω 2 L L 2 (ωl 2 + ωl L ) R22 2 +(ωl 2 + ωl L ) 2 { } = ωl k2 ωl 2 (ωl 2 + ωl L ) R22 2 +(ωl 2 + ωl L ) 2 For X ab to be negative requires R 2 22 +(ωl 2 + ωl L ) 2 <k 2 ωl 2 (ωl 2 + ωl L ) or R 2 22 +(ωl 2 + ωl L ) 2 k 2 ωl 2 (ωl 2 + ωl L ) < 0 which reduces to R 2 22 + ω 2 L 2 2( k 2 )+ωl 2 ωl L (2 k 2 )+ω 2 L 2 L < 0 But k hence it is impossible to satisfy the inequality. Therefore X ab can never be negative if X L is an inductive reactance.

9 48 CHAPTER 9. Sinusoidal Steady State Analysis P 9.73 [a] Z ab = V ab I + I 2 = V 2 I + I 2 = N I = N 2 I 2, I 2 = N N 2 I V V 2 = N N 2, V = N N 2 V 2 V + V 2 = Z L I = ( ) N + V 2 N 2 V 2 ( + N /N 2 )I Z ab = I Z L (N /N 2 + )( + N /N 2 )I.. Z ab = Z L [+(N /N 2 )] 2 Q.E.D. [b] Assume dot on the N 2 coil is moved to the lower terminal. Then V = N N 2 V 2 and I 2 = N N 2 I As before Z ab = V 2 I + I 2 and V + V 2 = Z L I.. Z ab = Z ab = V 2 ( N /N 2 )I = Z L [ (N /N 2 )] 2 Q.E.D. Z L I [ (N /N 2 )] 2 I

Problems 9 49 P 9.74 [a] Z ab = V ab I = V + V 2 I V N = V 2 N 2, V 2 = N 2 N V N I = N 2 I 2, I 2 = N N 2 I ( V 2 =(I + I 2 )Z L = I + N ) Z L N 2 V + V 2 = ( ) ( N + V 2 = + N ) 2 Z L I N 2 N 2.. Z ab = ( + N /N 2 ) 2 Z L I I ( Z ab = + N ) 2 Z L Q.E.D. N 2 [b] Assume dot on N 2 is moved to the lower terminal, then V = V 2, V = N V 2 N N 2 N 2 N I = N 2 I 2, I 2 = N N 2 I As in part [a] V 2 =(I 2 + I )Z L and Z ab = V + V 2 I Z ab = ( N /N 2 )V 2 I = ( N /N 2 )( N /N 2 )Z L I I Z ab =[ (N /N 2 )] 2 Z L Q.E.D.

9 50 CHAPTER 9. Sinusoidal Steady State Analysis P 9.75 [a] I = 240 24 + 240 j32 = (0 j7.5) A V s = 240/0 +(0.+j0.8)(0 j7.5) = 247 + j7.25 = 247./.68 V [b] Use the capacitor to eliminate the j component of I, therefore I c = j7.5 A, Z c = 240 j7.5 = j32 Ω V s =240+(0.+j0.8)0 = 24 + j8 = 24.3/.90 V [c] Let I c denote the magnitude of the current in the capacitor branch. Then I = (0 j7.5+ji c )=0+j(I c 7.5) A V s = 240/α =240+(0.+j0.8)[0 + j(i c 7.5)] = (247 0.8I c )+j(7.25+0.i c ) It follows that 240 cos α = (247 0.8I c ) and 240 sin α =(7.25+0.I c ) Now square each term and then add to generate the quadratic equation I 2 c 605.77I c + 5325.48=0; I c = 302.88 ± 293.96 Therefore I c =8.92 A (smallest value) and Z c = 240/j8.92 = j26.90 Ω. P 9.76 The phasor domain equivalent circuit is V o = V m/0 2 IR x ; I = V m/0 R x jx C As R x varies from 0 to, the amplitude of v o remains constant and its phase angle decreases from 0 to 80, as shown in the following phasor diagram:

Problems 9 5 P 9.77 [a] [b] I l = 20 7.5 + 20 =6 j0 A j2 V l =(0.5 + j6)(6 j0)=62.4+j94.5 = 3.24/56.56 V V s = 20/0 + V l = 205.43/27.39 V [c] I l = 20 2.5 + 20 =48 j30 A j4 V l =(0.5 + j6)(48 j30) = 339.73/56.56 V V s = 20 + V l = 48.02/42.7 V The amplitude of V s must be increased from 205.43 Vto48.02 V (more than doubled) to maintain the load voltage at 20 V.

9 52 CHAPTER 9. Sinusoidal Steady State Analysis [d] I l = 20 2.5 + 20 j4 + 20 j2 =48+j30 A V l =(0.5 + j6)(48 + j30) = 339.73/20.57 V V s = 20 + V l = 297.23/00.23 V The amplitude of V s must be increased from 205.43 Vto297.23 V to maintain the load voltage at 20 V. P 9.78 V g =4/0 V; = j20 kω jωc Let V a = voltage across the capacitor, positive at upper terminal Then: V a 4/0 20,000 + V a j20,000 + V a 20,000 =0;.. Va =(.6 j0.8) V 0 V a 20,000 + 0 V o 0,000 =0; V o = V a 2.. V o = 0.8+j0.4 =0.89/53.43 V v o =0.89 cos(200t + 53.43 ) V P 9.79 [a] V a 4/0 20,000 V a = + jωc o V a + V a 20,000 =0 4 2+j20,000ωC o V o = V a 2 (see solution to Prob. 9.78)

Problems 9 53 V o = 2 2+j4 0 6 C o =.. denominator angle =45 2/80 2+j4 0 6 C o so 4 0 6 C o =2.. Co =0.5 µf [b] V o = 2/80 2+j2 =0.707/35 V v o =0.707 cos(200t + 35 ) V P 9.80 jωc = j0 kω jωc 2 = j00 kω V a 2 5000 + V a j0,000 + V a 20,000 + V a V o 00,000 =0 20V a 40 + j0v a +5V a + V a V o =0.. (26 + j0)v a V o =40 0 V a 20,000 + 0 V o j00,000 =0 j5v a V o =0 Solving, V o =.43 + j7.42=7.55/79. V v o (t) =7.55 cos(0 6 t +79. ) V

9 54 CHAPTER 9. Sinusoidal Steady State Analysis P 9.8 [a] V g = 25/0 V V p = 20 00 V g =5/0 ; V n = V p =5/0 V 5 80,000 + 5 V o =0 Z p Z p = j80,000 40,000=32,000 j6,000 Ω V o = 5Z p 80,000 +5=7 j =7.07/ 8.3 V v o =7.07 cos(50,000t 8.3 ) V [b] V p =0.2V m /0 ; V n = V p =0.2V m /0 P 9.82 [a] 0.2V m 80,000 + 0.2V m V o 32,000 j6,000 =0.. V o =0.2V m +.. 0.2V m (.4 j0.2) 0.. V m 35.36 V jωc = j20 Ω V n 20 + V n V o =0 j20 V o j20 = V n 20 + V n j20 V o = jv n + V n =( j)v n V p = V g(/jωc o ) 5+(/jωC o ) = V g =6/0 V V p = 6/0 +j5 0 5 C o = V n 32,000 j6,000 V m (0.2)=0.2V m (.4 j0.2) 80,000 V g +j(5)(0 5 )C o.. V o = V o = ( j)6/0 +j5 0 5 C o 2(6) +25 0 0 C 2 o =6 Solving, C o =2µF

Problems 9 55 [b] V o = 6( j) +j = j6 V v o = 6 cos(0 5 t 90 ) V P 9.83 [a] Because the op-amps are ideal I in = I o, thus Z ab = V ab I in V o = V ab ; V o = V o2 = KV ab = V ab ; I o = V ab V o I o Z V o2 =.. I o = V ab ( KV ab ) Z.. Z ab = ( R2 V ab ( + K)V ab Z = [b] Z = jωc ; Z ab = R ) V o = KV o = KV ab = ( + K)V ab Z Z ( + K) jωc( + K) ;.. Cab = C( + K) P 9.84 [a] I = 20 24 + 240 8.4+j6.3 =23.29 j3.7=27.02/ 30.5 A I 2 = 20 2 20 24 =5/0 A I 3 = 20 2 + 240 8.4+j6 =28.29 j3.7=3.44/ 25.87 A I 4 = 20 24 =5/0 A; I 5 = 20 2 = 0/0 A I 6 = 240 8.4+j6.3 =8.29 j3.7=22.86/ 36.87 A

9 56 CHAPTER 9. Sinusoidal Steady State Analysis [b] I =0 I 3 =5A I 5 =0A I 2 =0+5=5A I 4 = 5 A I 6 =5A [c] The clock and television set were fed from the uninterrupted side of the circuit, that is, the 2 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 5 A. P 9.85 [a] The circuit is redrawn, with mesh currents identified: The mesh current equations are: 20/0 =23I a 2I b 20I c 20/0 = 2I a +43I b 40I c 0= 20I a 40I b +70I c Solving, I a = 24/0 A I b =2.96/0 A I c =9.40/0 A The branch currents are: I = I a = 24/0 A I 2 = I a I b =2.04/0 A I 3 = I b =2.96/0 A I 4 = I c =9.40/0 A I 5 = I a I c =4.6/0 A I 6 = I b I c =2.55/0 A

Problems 9 57 [b] Let N be the number of turns on the primary winding; because the secondary winding is center-tapped, let 2N 2 be the total turns on the secondary. From Fig. 9.58, 3,200 = 240 N 2 or = N 2N 2 N 0 The ampere turn balance requires N I p = N 2 I + N 2 I 3 Therefore, I p = N 2 (I + I 3 )= N 0 (24+2.96)=0.42/0 A Check voltages V 4 = 0I 4 = 94/0 V V 5 = 20I 5 = 92/0 V V 6 = 40I 6 = 02/0 V All of these voltages are low for a reasonable distribution circuit. P 9.86 [a] The three mesh current equations are 20/0 =23I a 2I b 20I c 20/0 = 2I a +23I b 20I c 0= 20I a 20I b +50I c Solving, I a = 24/0 A; I b = 24/0 A; I c =9.2/0 A.. I 2 = I a I b =0A [b] I p = N 2 N (I + I 3 )= N 2 N (I a + I b = 0 (24 + 24) = 0.436/0 A

9 58 CHAPTER 9. Sinusoidal Steady State Analysis [c] Check voltages V 4 = 0I 4 = 0I c = 92/0 V V 5 = 20I 5 = 20(I a I c ) = 96/0 V V 6 = 40I 6 = 20(I b I c ) = 96/0 V Where the two loads are equal, the current in the neutral conductor (I 2 ) is zero, and the voltages V 5 and V 6 are equal. The voltages V 4, V 5, and V 6 are too low for a reasonable dirtribution circuit. P 9.87 [a] 25=(R +0.05 + j0.05)i (0.03 + j0.03)i 2 RI 3 25 = (0.03 + j0.03)i +(R +0.05 + j0.05)i 2 RI 3 Subtracting the above two equations gives 0=(R +0.08 + j0.08)i (R +0.08 + j0.08)i 2.. I = I 2 so I n = I I 2 =0A [b] V = R(I I 3 ); V 2 = R(I 2 I 3 ) [c] Since I = I 2 (from part [a]) V = V 2

Problems 9 59 250 = (660.04 + j0.04)i a 660I b 0= 660I a + 670I b Solving, I a =25.28/ 0.23 =25.28 j0.0 A I b =24.90/ 0.23 =24.90 j0.0 A I = I a I b =0.377 j0.0053 A V =60I =22.63 j0.095 = 22.64/ 0.23 V V 2 = 600I = 226.3 j0.95 = 226.4/ 0.23 V [d] 25 = (60.05 + j0.05)i (0.03 + j0.03)i 2 60I 3 25 = (0.03 + j0.03)i + (600.05 + j0.05)i 2 600I 3 0= 60I 600I 2 + 670I 3 Solving, I =26.97/ 0.24 =26.97 j0.3 A I 2 =25.0/ 0.24 =25, 0 j0.04 A I 3 =24.90/ 0.24 =24.90 j0.04 A V = 60(I I 3 ) = 24.4/ 0.27 V V 2 = 600(I 2 I 3 ) = 24.6/ 0.20 V [e] Because an open neutral can result in severely unbalanced voltages across the 25 V loads.

9 60 CHAPTER 9. Sinusoidal Steady State Analysis P 9.88 [a] Let N = primary winding turns and 2N 2 = secondary winding turns. Then 4,000 = 250 ; N 2.. = N 2N 2 N 2 = a In part c), I p =2aI a.. I p = 2N 2I a N = 56 I a = (25.28 j0.0) 56 I p = 45.4 j.8 ma = 45.4/ 0.23 ma In part d), I p N = I N 2 + I 2 N 2.. I p = N 2 N (I + I 2 ) = (26.97 j0.+25.0 j0.0) 2 = (52.07 j0.22) 2 I p = 464.9 j.9 ma = 464.9/ 0.24 ma [b] Yes, because the neutral conductor carries non-zero current whenever the load is not balanced.