dv i= C. dt 1. Assuming the passive sign convention, (a) i = 0 (dc) (b) (220)( 9)(16.2) t t Engineering Circuit Analysis 8 th Edition

Transcription

1 . Assuming he passive sign convenion, dv i= C. d (a) i = (dc) 9 9 (b) (22)( 9)(6.2) i= e = 32.8e A 9 3 (c) i (22 = )(8 )(.) sin. = 7.6sin. pa 9 (d) i= (22 )(9)(.8) cos.8 = 58.4 cos.8 na

2 2. (a) C = 3 pf, assume passive sign convenion. v =.5 V for - 5 s i C dv = C = A for d i c (A).5 5 (s).5 (b) vc = 4cosπ dv i C ( ) d c 2 c = = (3 ) 4 π sinπ =.633sin π na for s i c (na) (s) -.633

3 3. (a) C = µf, assume passive sign convenion. For > 4 s, v = V herefore i C = For all oher imes, v= (a) C = µf, assume passive sign convenion. dv ic = C = = d 2 i c (µa) 6 6 ( ).5 A for 4 s.5 4 (s).5 (b) i C = (7.5 ) = 8.75 pa 2 i c (pa) 2 for 4 s 4 (s) 8.75

4 4. (a) ε =.35ε ( ) ( ) A= 2.5 = A C= ε = = 6 d m pf (b) ε = 3.5ε 2 6 A C= ε = = 6 d pf (c) ε = ε ; d = 3.5µ m 2 6 A C= ε = = 6 d pf 6 2 (d) ε ε ; A 2A 5 = m ; d µ m = = = 2 6 A C= ε = = 6 d pf

5 5. Gadolinium is a meallic (conducing) elemen A A= ( )(75 ) = 7.5 m. C= ε. d (a) C = 9.64 pf (b) C = 3.32 nf (c) C = 3.32 pf (d) C = 3.28 pf

6 6. One possible soluion. We require a nf capacior consruced from µm hick gold foil ha fis enirely wihin he volume of a sandard AAA baery wih he available dielecric having permiiviy 3.ε. A sandard AAA baery has approximaely a lengh of 4.45cm and he circular diameer of.5cm. 3.ε A C 2 4 d = = A= Am 9 If we selec he area of he plaes as 4 cm 2, he gap spacing beween he plaes is d =.97µm.

7 7. <Design> One possible soluion: We recall ha for a parallel-plae (air separaed) capacior, C = εε A d where e is he relaive permiiviy of he spacer maerial, A = he plae area, d = he plae spacing, and he remaining erm is he free-space permiiviy. We propose o vary he capaciance value by sliding one conducor pas he oher. We being wih a fla piece of glass m by m and 2 mm hick (none of hese dimensions are criical). Coa he cener of he glass wih a. mm hick layer of gold (maerial and hickness no criical) having dimensions mm by mm. Take a piece of plasic (e = 3). mm hick. Coa he op side wih a. mm hick layer of gold measuring 336 mm by 336 mm. We expec he meal o siffen he plasic sufficienly so sliding will work. Place he plasic on he meal-coaed glass slide, boh meal layers facing upwards. Make elecrical conac off o he side and neglec is conribuion o capaciance (assume negligibly hin wire). When he op meal layer is compleely over he boom meal layer, (he smaller area is he A in our equaion), we have C = (3)(8.854e-2)(336e-3)(336e-3)/(.e-3) = 3 nf (o hree significan figures). We hen slide he op meal layer along glass unil only onehird is lengh overlaps he boom meal layer. This reduces he effecive area of he capacior by one-hird, and hence he capaciance becomes nf.

8 8. <Design> One possible soluion: In he firs prining, he capaciance range was saed as 5 nf o nf, bu his should be changed o 5 pf o pf o obain more realisic areas. We begin by noing ha here is no requiremen for linear correlaion beween knob posiion and capaciance, alhough ha migh be desirable. Take wo discs of plasic, each approximaely 3 mm hick (so ha hey are siff) and radius 85 mm. Coa one side of each disc wih gold (hickness unimporan) bu over only half he area (i.e. from o 8 degrees). Moun he discs wih he meal layers facing each oher, separaed by a mm hick plasic spacer of very small diameer using a hin nonconducing rod as an axis. Then, wih C = εε A d where e is he relaive permiiviy of he spacer maerial, A = he plae area, d = he plae spacing, when he wo meal halves are aligned over one anoher he capaciance is pf. Roaing one disc by 9 degrees reduces he effecive (overlap) area by one-half, and hence he new capaciance is 5 pf.

9 9. C = (.8)( )( 4 4 ), W in cm W 4 2 (2)(.8)(8.854 ) W = 9 8 (.57 V A ) (.62 )( ) V A W(cm) C - V ff -5 V ff - V af

10 . (a) Graph (b) Graph

11 (c) Graph

12 . C = 33 mf. (a) Graph dv i= C so d v= id C. 5 v (V) (s) (b) v(.3 s) = half-way beween and V = V v(.6 s) = V v(. s) = = 2.2 V

13 Wc = Cvc = = 44.8J Wc = Cvc = = 7.52 pj Wc = Cvc = =.59µ J (a) ( ) 2 (b) ( ) 2 (c) ( ) 2

14 3. C = 37 pf 2 V, < vc ( ) = 2 2 e V, C v C. 2. Energy = [ ] 2 (a) = : nj (b) =.2 s: nj (c) =.5 s:.335 nj (d) = s: 8.7 pj

15 4. (a) Afer being conneced o DC source for a very long ime, he capacior ac as open circui. Therefore, curren hrough he circui is, V.2 i= = = mA R P4 Ω = i R= ( ) 4= 4.984mW v C = =.4258V PSpice Verificaion: (b) Afer being conneced o DC source for a very long ime, he capacior ac as open circui. Therefore, curren hrough he circui is, i= A P v 4Ω C = W =.2V

16 5. (a) no curren hrough 5 ohm resisor so 7 vc = = = (4.5 )(3) 3.65 V so vc 3.65 nv (b) v C = volage across curren source so (4.5)[+ 7 8] = nv

17 6. One possible design soluion: The general equaion o calculae inducance is given by: 2 µ N A L= l µ π H 7 where = 4 / N = Number of urns A = Area of cross-secion = π r, m 2 2 l= Lengh of he wire, m r = ouside radius of he coil(form + wire), m m l d If we ake a plasic form and wound 29 AWG copper wire (diameer =.286mm) round i, we can consruc a 3nH inducor. For ha if we choose N = 22 urns, r = 6 mm, we can find he lengh of he wire µ N π r 4π 22 π (3.43 ) l= = = 57.87m 9 L 3

18 7. L = 75 mh. di v= L. d Volage can change insananeously bu curren canno (or volage becomes infinie). From = - o, i = 2( + ) so slope = +2. From = 2 o 2.5, slope = -2. (a) v (mv) (s) -5 (b) =, 2.9 s, 3. s v() = ; v(2.9) = ; v(3.) = ;

19 8. Given, L = 7nH il = A for µ s 5 i = 2 A for 5 µ s L dil vl = L = 7 = 6.8 fv d 5 v L (fv) (µs)

20 9. di di 3 v= L = (4.2 ) d d (a) v = (b) 75.6 cos 6 mv (c) π sin(π 9 ) V (d) -54.6e - pv (e) (4.2 ) ( 4 ) + = (4.2 )[ 4 + ] e e e = 4.2( 4 ) e mv

21 2. (a) i = 8 ma, v = V L (b) i = 8 ma, v = V L (c) i = 8 A, v = V L L L L (d) dil 2 il = 4 e A, vl = L = 8 4e = 32e pv d di il = 3 + e A, vl = L = 8 ( e ) = 8e ( ) pv d L 2 (e) ( )

22 2. is = ma, vs = 2. V (a) vl = ; il = is = ma (b) 4 kω resisor is irrelevan here. vl = ; il = is = ma (c) vl = ; il = vs/4.7x 3 = µa

23 22. (a) Rise ime, fall imes: 2ms, 2ms (b) Rise ime, fall imes: ms, 5ms

24 (c) Rise ime, fall imes: ns, 2ns

25 dil 23. vl = L. d Beween -2 and s, slope = (2 )/(-2 ) = - Beween and 3 s, slope = -/2 For > 3 s, slope = -2/-3 = 2/3 (a) v(-) = ()(-) = - V (b) v() = ()(-) = - V (c) v(.5) = ()(-/2) = -.5 V (d) v(2.5) = ()(-/2) = -.5 V (e) v(4) = ()(2/3) =.67 V (f) v(5) =.67 V

26 5 24. (a) il = 5 d il () A 3 L + = = 6 cos2π il d A 3 L 6 2π (b) = sin2π = = 22.69( cos2π )

27 25. v = 4.3, 5 ms and il (.) = µ A L 4.3 il= 4.3 d il(.) (.) L + = (a) -.65 ma (b) -.65 ma (c) ma

28 2 26. (a) E= LiL = J (b) ( ) 2 6 E= LiL =.5 = 5 J (c) ( ) 2 7 E Li.5 = L = 2 = 2 J (d) ( ) E= LiL =.5 5sin 6 =.25sin 6 J 2

29 27. L = 33 mh, = ms. w =.5Li 2 (a) 88.5 mj (b).595 nj

30 28. (a) Afer being conneced o DC source for a very long ime, he inducors ac as shor circuis. The oal curren in he given circui hen, i= =.563mA Therefore, curren ix = = µ A (b) Afer being conneced o DC source for a very long ime, he capacior ac as open circui and inducor ac as shor circuis. Therefore, curren ix ( ) = = 2.857A 3 5

31 29. (a) 5 + V / 27 + V / 3 + ( V ) / 2= Solving, V = 4.95 V By volage division, Vx = (2/27)(4.95) = 8.64 V (b) 5 + V / 27 + V / 2 + ( V ) / 2= Solving, V = V By volage division, Vx = (2/27)(36.85) = 6.38 V (c) 5 + V / 27 + ( V ) / 2= V = 58.2 V By volage division, Vx = (2/27)(58.2) = V (d) Same as case (b): Vx = 6.38 V

32 3. (a) Taking inducor as he load, he hevenin equivalen resisance is given by, Req ( ) 3 = + 47 = 8.245kΩ The hevenin volage is given by, VT 3 47 = 4 = 3.298V ( ) (b) Afer being conneced o DC source for a very long ime, he inducor acs as a shor circui. 4 il = = 4µ A 3 kω 47kΩ ( ) L P = i R=.4 =.6mW P = W WL = LiL =.5 5 (.4 ) = 4nJ 2

33 3. C eq = + + = mf

34 32. 7 Lequiv = L L + + = L L 2L L

35 33. Creae a series srand of 5 H inducors (Leq = 5 H). Place five such srands in parallel. (/5 + /5 + /5 + /5 + /5) - =.25 H

36 34. Saring from he righmos end, we have, a series combinaion of 2F, 2F and 2F, for 2 which he equivalen capaciance is, Ceq= = F This is in parallel wih he series combinaion of 8F and 5F. Therefore, 2 52 Ceq2 = + = F Now, his is in series wih 4F and F which yields he new capaciance as, 2 Ceq3 = = F This combinaion is in parallel wih 5F and he final equivalen capaciance is, Ceq2 = = 3.35F

37 35. (/7 + /22) = 9.3 F 5 + (/2 + /) - = F Thus, C eq = (/9.3 + /5.923) - = 3.62 F

38 36. Towards he righmos end (b erminal) of he given circui, 2H is in series wih H, he combinaion is in parallel wih 5H. Therefore, he equivalen inducance is given by, 65 Leq = = H + 8 (2+ ) 5 On he lef side (a erminal) 2H is in parallel wih H, and he combinaion is in series wih 7H and he whole combinaion is in parallel wih 4H. Therefore, he equivalen 548 inducance on his side is given by, Leq 2 = = H The oal equivalen inducance seen from he a b erminal hen becomes, Leq = Leq+ Leq 2 = + = 6.638H 8 8

39 37. The circui can be simplified as:

40 38. (a) For each elemen as Ω resisor, he equivalen resisance is given as, R eq = + ( R+ R+ R) ( R R ) R ( R + R + R ) =.379R=.379Ω (b) For each elemen as H inducor, he equivalen inducance is given as, L eq = + ( L+ L+ L) ( L L ) L ( L + L + L ) =.379L=.379H (c) For each elemen as F capacior, he equivalen capaciance is given as, C eq = + C + C + C C+ C + C + C+ C+ C ( ) ( ) ( ) ( ) =.8787C = F

41 39. /Leq = / + ½ + / + /7 + ½ + ¼ Thus, Leq = ph

42 4. The circui can be simplified as:

43 4.

44 42. (a) Lequiv = + 3H + = (b) For he given nework having 3sages, we can wrie, L equiv = + + = For he general nework of his ype having N sages, we can wrie, L equiv = N N = = N 2 3 N 2 3 N

45 (2 pf) + (2 pf) = pf 43. Far righ: Which is in parlle wih 2 pf, for a combined value of 3 pf. 6 Since (3 pf) + (2 pf) = pf and his is in parallel wih 2 pf, 5 we obain a oal of 6/5 pf. This equivalen value is in sries wih 2 pf, hence he final value is.23 pf.

46 44. For he righmos end, nh is in series wih nh, he combinaion is in parallel wih nh. 2 Leq = = nh This combinaion is in series wih nh and he new combinaion is in parallel wih nh. 5 Leq 2 = = nh Now, his combinaion is in series wih nh. Hence he ne equivalen inducance is given by, 5 Leq = + =.625 nh 8

47 45. (a) ( ) d v v dv v v = C + C + d d R dv v v dv C+ C2 + = C [] d R R d 2 3 or ( ) v v v is= + + v2 v5 d il ( ) R R L or 2 2 Nex, ( ) v v = L ( ) + v d i is vsd R R L L [2] (b) vs+ ( i i ) d + ( i i ) d = C C 2 L 2 C 2 ( i i ) d R( i i ) R( i i ) + + = 2 2 L 2 dil L + R( i i ) + ( i i ) d = d 2 L L C s

48 46. (a) v v dv v v 2 d v v v d ' 2= C S 6 C C L 5 = C L 3 8 L (b) v = 2 ( ) ' 2 S i + i i d + L ( ) 2 L 6 i i ' 2 i 8 3 d + = + L di d L

49 47. is() = 6 ma herefore i2 = i3 i = 4 ma (a) is = vd i () vd il() di s d (b) (c) = v+ v v = e ( ) 5 V 2 ( ) = () ma 6 + = + i vd i e 2 2( ) = 2() ma 4 + = + i vd i e

50 dv i( ) = C eq = + 4 ( 8) e = 6.4e ma d S 48. (a) ( ) (b) v ( ) = v () + i( ) d ' C ( 6.4 ) ' = + e d = e V (c) v2( ) = v2() + i( ) d ' C ( 6.4 ) ' = + e d = e + V

51 49. I is assumed ha all he sources in he given circui have been conneced for a very long ime. We can replace he capacior wih an open circui and he inducor wih a shor circui and apply principle of superposiion. By superposiion, we ge, v v C L = = 9.6V = = V Pspice verificaion:

52 5. Le us assign he node volages as V, V 2, V 3 and V 4 wih he boom node as he reference. Supernode: v v2 e 3 ( v3 v4) d 5 5,4 Supernode: 2 + ' = and.8v +.2v v = v v2 6 dv2 v2 4e Node 2: = + 5 d 2

53 5. (a) Assuming an ideal op amp, dv C f ic = C f f d dvs v C = d R f ou dv Thus, vou = RC f d s (b) Assuming a finie A, dv v Av C = d R f s d d v ou v dv ou s C A f = d R dvs vou RC f = v ou d A A Solving, vou = RC A f dvs d

54 52. (a) A=, R =,and R = ; i R = kω, C f = 5 µ F, vs = 2sin 54 mv v ( ) = v d ' v () ou s cf RC f 3 6 o 3 = 2sin 54 d ' =.747 cos 54+ Cµ V 5 C being a consan value (b) A= 5, R = MΩ, and R = 3 Ω ; i R = kω, C f = 5 µ F, vs = 2sin 54 mv o Wriing he nodal equaions we ge, vd + vs vd d + + C ( vd + vou) =...() R R d d C vd v d i ou d ( + ) + =...(2) ou v Av R On solving hese wo equaions, we ge, v = 2 v 6 v V 4 9 d ou s ou o v = d d cos 54 ' 4 sin 54 ' = sin cos 54 C V C being a consan value

55 53. i = v d + i ( ). Assuming an ideal op amp, L s L L L Thus, v vsd + il ( ) = R f ou R v v d R i f ou = s f L L + ( )

56 54. (a) Assuming an ideal op amp, dvc f vou i = C ; i = ; v = v f f f d R C f R C ou f dvc f vou i = C ; i = ; v = v f f f d R C f R C ou f dv v v ou ou s C f + = d R f R dvou vou vs + = d R C R C f f f (b) Equaion 7 is: v = v d v ou s C f RC f () In case of pracical inegraor circui, when R f is very large, he soluion of he equaion obained in (a) is equaion 7 iself. In case of ideal inegraor under DC condiions, he capacior acs as an open circui and herefore op-amp gain is very high. As we know ha op-amp has inpu offse volage, he inpu offse volage ges amplified and appears as an error volage a he oupu. Due o such error volage he op-amp can ge easily sauraed. So in order o reduce he effec of such error volage, usually a very large resisor is added in parallel wih he feedback capacior in case of pracical inegraor circui. Thereby he DC gain is limied o R f /R.

57 55. One possible soluion: We wan, V = C/s v ou dv = RC d s 3 V vou = = RC s RC= If we arbiarily selec R= MΩ, C= µ F

58 6. (a, b) (c) Original circui, nodal: ( 2 ) ( is v v d i ) = 3 v v2 d i v2d i v2d i2 ( ) + ( ) ( ) ( ) = New circui, nodal: = v v s ( ) d v v dv d d =

59 6. (a,b) (c) Original circui, mesh: 2+ i + 4 d ( i i2 ) = d 7i2+ 4 d ( i2 i ) = d New, mesh: Nodal: v = - 2 V v2 v v2 = + + v 2d + i ( ) i2 ( ) i = -2 A 7 4 = + i2 i+ i2d + v2 ( ) 7 4

60 62. (a, b) Original circui, mesh: v ( i i ) = s 2 ( i ) ( i ) i i d i d = d i 2 3 ( 2) 2 3 i 6 3d vc ( ) = d New circui, mesh: i = i s

61 63. (a)-(b) (c) Original circui, mesh: di 2i + + 6( i i2 ) = d di i i i i d di3 8i3+ 3 2( i2 i3 ) = d 2 + 2( 2 3) 6( 2) = Original circui, nodal: v = V v v2 2 ( ) 3 ( ) ( ) ( v v ) v v d i ( ) = v v v d + i ( ) v v d i ( ) = ( v v ) v ( ) 2 v v d i ( ) = v5d i3 ( ) = 8 3

62 New circui, mesh: i = A i i2 2 ( ) 3 ( ) ( ) ( i i ) i i d v ( ) = i i i d + v ( ) i i d v ( ) = ( i i ) i ( ) 2 i i d v ( ) = i5d v3( ) = 8 3 New circui, nodal: dv 2v+ + 6( v v2) = d dv v v v v d dv3 8v3+ 3 2( v2 v3 ) = d ( 2 3) 6( 2) =

63 64. (a)

64 65. (a) DC. Thus, i L = 7/8x 3 = 87.5 µa (b) P diss = (il)2(8x 3 ) = 62.5 µa

65 66. (a) For DC condiions, he capacior acs as an open circui. Therefore, he oal curren I 7 I = = 55.56µ A ( ) ( ) ( ) in he circui is: P = I R= = µ W 8k 46k P = I R= = 4.99µ W (b) Volage across he capacior: (c) Energy sored in he capacior: (d) Pspice Verificaion 2 3 vc = I 46 = 2.556V wc = Cvc = = µ J 2

66 67. (a) By curren division, i 44 L = = Thus, v x = (44(i L ) = V ma. (b) = =. 2 2 wl Li 32.7 µ J 2 2 wc = Ci = 2.85 µ J

67 WL = LiL = 6 = 3 nj (a) 3 3 ( ) 2 (b) L R i ( ) = e = e ma L i ( ms) = ma L i (3 ns) = e ma=.369ma L i (26 ns) = e ma=.362ma L i (5 ns) = e ma=.26ma L (c) Fracion of energy a seleced imes: A, = 3ns, (.369 W ) 2 L = LiL = =.48 nj = 3.6% of he iniial energy sored. 2 2 A, = 5ns, (.26 W ) 2 L = LiL = =.399 pj =.46% of he iniial energy sored. 2 2

68 69. (a) w = = µ 2 6 ( )( )(8) 45 J (b) no - he resisor will slowly dissipae he energy sored in he capacior (c) Transien required. (d) Fracion of energy a seleced imes: RC =.46 s = 46 ms So vc(46 ms) = 9e- = 3.3 V Hence w = 54.8 mj or 3.5% of iniial amoun sored. vc(2.3 s) =.66 V Hence w = 8.39 nj =.45% of iniial amoun sored.

69 7. (a) When we analyze he given circui wih he DC condiions for a long ime, he circui can be redrawn as below. Applying node volage analysis, we see ha v = v2. Therefore, v c = 4 V Energy sored in he µf capacior is: Wc = CvC = 4 = 8 µ J 2 2 Curren, i L = A. Energy sored in he 2mH inducor is: W L = J (b) Pspice Verificaion:

70 7. dv vou = R f C = d s π cos(2π ) = cos(2π ) If we plo, v ou as a funcion of ime over 5ms, our graph looks as below. V V(V) (s) Bu, since he oupu for he op-amp is ±5V, we can say ha in his case, all he volage above 5 V will be clipped. This can be easily seen in he oupu in he Pspice. (b) Pspice-verificaion

71

72 72. (a) v L dv s ou = = cos(2 ) R d = π cos(2 ) π V If we plo, v ou as a funcion of ime over ms, our graph looks as below. V(V) (s) (b) The oupu volage graph is differen in Pspice simulaion.

73

74 73. a) R vou = v d = d L 3 f 3 s 5sin(4π ) ' = cos(4π ) 5 V V(V) (s) (b) Pspice verificaion In his case, he op-amp ges sauraed around ±5V.

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76 74. (a) dvou vou vs + = d R C R C f f f vou = cos(2 π) mv V(mV) (ms) (b) Pspice verificaion:

77