EEE202-Hmwk3-Solution - PEI-36 “1 wish 1 solve the...

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Unformatted text preview: PEI-36 “1% wish 1:: solve the circuit in Figure P2—36 by using thenude method. Let 2,“) be the nude pmfial at line indicated node. (:1) Express is) innerms nfsafi) and 11,3). (b) Write uKCLequafinn at line supemmie Ibuteucircles Ihe dqudent voltage snurce. 'Ihis equafinn should involve mfly fine wfinhles 2.“) and u,(t}. (c) Find 3,,(1). Fig. 14-36. Circuit for Problem PIE-35. ”/767 Wan“; [45‘ “/29 U549 ijCIQ/ army/7513’ “7/23 «gcfifb’e ~74”; Frog/gm; /<CL 694’ 7/19/9 Six/aarwcaxr; Vie/“[5 L): _ .va W» (2) ,2 _,0_ i 42 ’A .0, 12?? Cans+rwin4~ aging-Lion 1‘35 Vb+ 3,5 ‘I’ bi? J [SGML ,L“ : VS" V“? / SO Wax can suésfiéthu'ILe 42.12 #6 x exmfession 23/1479 147:9 ¢o#7.3"7irofn7‘* erOfiKOW—A ’ . 'w~%)_( Vb+3(“—_§;T*‘7 “V? . vb=-%VQ'-%\é (I) Niovv 55 UIQ5°4r+w+e (I) {07LO (Q) / VV/H‘CZ) yin/€25 stwé—gvo:é\/Ofi3\'é 7 L? TVS : :2 V4 7 P2-37 {a}Defineand]abelanappmpriatesetofnode potentials for the circuit in Figure P267. (13) Write a set of KCL equations that can be solved fior the node ponenfials final: you defined in part (:1), using the node polentials as variables. (e) Solve your equations to express nit) in terms of u,(1). AK) 0411325 v" flock?) 0/1 «VI/V :33 10 “My”; real/[3,3 331/33 Fig. Pl-3T.E5IenirfioerblemPZ-31 4""0 modes 423% KCL KCL 63% 37303333,: \/ _ /3"\/CL 37-f— nodE V . . /'_- /- ‘/ V5—\//(_ \/-\/, __\_/L \/ V ‘ \2 m- +-_V :13; + 31;; “ :3 132 33.31 .331 7 - 1_ , _. 1/- 3 V1 = ‘43 31) -v(/+3+3 3 1/, — 3 14 // , , _ , "’ Z— \/ + \/l = 3L V3 \ / - L 1/.— ~£— '1/ 9\ \/~1—f§ 33/331331 :—\/,3 v L1 3333 H v (3- 3333) Vin—“~— 1/ (—31—?) W7 3?): vc (“’3 3) m v-=('3)(§)1/S:~—1/WWJ P289 Solve fur 1:6) in terms of LG) in the circuit in Figure P2439. fajc) H7345 49th 5Pecf~péc9 2) CAL 11+"; "H443 +0 L151” mesh mméwfs, 776 @437; M85; £36311“: +0 Fig. Pz-a9. CircuitmembhmPZ—BQ. :1 an i‘mmaclievLe Cons-iraz‘nvL «274144123» of I/ '5 is .. [7.1. 54:23 I’M KVL. QV‘L'I'CJl’Icé mfda’ It? MES“L/ f :2 (’1le +1 (I1) + 2,5 *= O gun‘- we. movie ‘7”7afiL i t In; / $0 eve. cow #84}.th fill/:13 above, gkfa/‘e‘sriorx “7L0 'SIQ~—Q.I/'r~ QIl‘r‘S" Qj;?_L/=O .-__. - 3‘.” f_ g ., we +2- <5», (.5 MGM/J f’erb‘penr‘m KVL armour/Cf Pixfiirf mes}; I *\ »-~;2,z+3’_1:5,+¢2l:3,: ”f" QJC’ ._ 54—1—3 NH ‘5 ”‘1” s h P242 Salve fm: mm in the circuit in FigurePZ—d-E. Z. dole /, La @7on- ),44 951:] 42/2 a/y :1}: 7800 at. ffifc 6 fly V71, 2:: am 29 1/ " O 4 r ‘ C? ‘ 5,3) 0 9 23.1) W 67, n +a #7 7i / I 122 on a“ Fig- P342. [fimuitfiermblfimPZ-d-Z. nga /j/ “I 11> TA ,3 [e z)“{“'+ mail) [WMMQCIIQfi/‘fiff 7/8/61; I]: is 'Ue 5" KW, 4+ +28 Wage/4648?» ”755/, j 3(IQ—r,>+g{rl>+1(I 11): (j ”31’,+4IQM_’T‘3:O , “3115*“611‘Is “0 (1) KW mL 7%. “7144 Mesa >212: (CI; if-fs )+«>7(~3)"“»4«¢ IO "BI ISL:2 +2 I:z =0 3 3:3 +42:O g I :~—37;I“3 (:2) Sués +f+u+g (2;) m+o (I) v 3”,: ~=é'.:_2 IIB zér «[‘g’jI s 3,5 _ 1" P244 (:0 Which method, the mesh method er the node methnd, will mull: in fewer equafinns In solve in order to find my) in the circuit in Figure P244? (b) Compute ”(1), using the meltmd that yen selected in (a). Fig. Pie-4:1. [3mm far Emblem P244. 1.7g yé’JU Clo/74 COL/ML} cowysvtrm‘n‘yl— £7’UQ‘F/bfi_£' fig” Mask aide/7 91‘s F5 éfi‘ch’r" 3/469 gag/y €3,762 161/4; fleecfs 4-0 {'96 Per-“pvrmza/ W lie/“tidf 7/31/53 KCZJ‘ W€>£J(c{ have. 412 (7c? uj'cc/ 2:70 ”OJ-é/ Carma/ff} i3 Vfi’ff, Co /1 5" #Fa I‘M ‘/ e3; (,1 Q‘7LIK0P75’ Lil—’45], I;:’is £2 /<VL £24; (Lea/“I'VEK‘I/ewlkcjcl/e Mask ‘ ‘ .LQQQ (1‘3 ~13) + 5652 (I3) + £00.52. (JE‘PIQ) :0 (/00 + 50 + 92059 3‘73 2 7001, - 3003; 35‘ @133 ~= /’[email protected] 4:51-- 2630-sz 5.0 I __ 1’60 u “(206' . 3 _ "7 15,” ‘7 .453 \ v' x 4 V“ p 12630 x : V: 503:9 “ '7 1.55] - quésfl I V/ P245 Findu(t)inflnecircuitinFigureP2-46by any methudthatyuu chmse. A432 ([6,, or- 0442.550 ai’TCIIy-Si} V9061 [J Frog/{ace vLWo aljgg ”ii/C eyuaho/U gawk”, éuvL flea/u/ qulc! ifI/I&flC4/;G7L/:BI/V frat/fag “i475 walha/ c'mste/x A/f” Cl [ Am a7/ 5‘55 . .2 c? \/, _ V,- A V] , \/{ _ _ ._ ,L / v”, — :2 v, Q ‘22 ~ ' , V _ V, _ V Ka. 0* Va 3 ”L533“"[email protected]«Ls=fi§i I ”9““, 413' [(WOM/ V; SO [:1ij éfli‘M/fnw‘I/‘g V3 3 3V3: 6 \r/f — a ‘2 r ~21. .3. , _ + , P260 The circuit in Figure P260 represents a medal of an amplifier that is driving a. loudspeakfi, which is modelled by the 1653 resistor. The input to the amplifier is the signal em = Dzmflflflk). (.5) Compute the value of the meistance R such that 13(1) = 21) magnum). (b) The poem: in the signal x0) = A gnaw!) is 1121'2. Compute the mtin 0f the power :in the signal 1;“) deliveredm thelnudspeakerm the pewerin theinput 515m ”4” Fig. 132-61}. Birmit far Problem 112.31. (at) M77115: UOZ‘ILaj'cg VOL, cal-VJ die jaw/(cl é/ L/O/‘LOJ‘C’ gob/322),?) , w —€3CJ/<.19—_( > SOC) O06 VQL’ _ V5 $002.21—a CJCtQ 01263560001 300‘ 300 7:12, Ake‘vb/Je/ Uo/jlgje CI!U¢SEOV’ +70 70/ch (/ejfi'Afi) V(«+.‘>n Iéf’ __ 4.2,; ’ 300000 5,, V" (SO V045) A-t-ié—Q “(/bcfi“? SOC/goggo5flOOOf " ' ”Va w'o 277’— “ AAAA 3'00 (Doc) 2 O 495—” »--—iC’C)O:t :(30 ){é’figw‘ 00 QOO>C0£ {/6663 £3 /‘6 +/ Ii) Bf :2 '\ /<rv‘\ r.mn\ x /._-m ,m-\ )oC/ uvcz) f€+ié: (/QJ(§(::(: 7'0‘6)(J6j (2419(3—+— “'00 goo) (Qéf>(::§? «ié r 7078é -41 (19):” 7.4»er mum» V -C) :2 dog (£3001) o 5 2 C4352)”; ., 53g A: 0,22 cm; 4149. Fewer '2 = a ” ”£32; Gui-[(74% Pom/er“ “ V : ‘26) COS([€0CJO 1,531 (3(3):; . S0 .A‘ :5 2C5 and VAL) é- [Don/Jew i5 ‘ 3» Z: V :"2 2 {:90 9'7” / IQO+IO (:24? aunL/pu‘l’ 1L0 ihlpu+ Pcvlér pWT »: ff =2er 10: 000 _1OU PM 553—3 (033“ P3-1 0 A switch is a device that acts like an open circuit when it is open and a short circuit when it is closed. Find the equivalent resistance Req at the indicated terminals in the circuit of Figure P3—10. All resistances are measured in ohms. Hint: Redraw the circuits for the two indicated M cases. ...___\_ Mort). 4.;ch 1.2 /,,,,+ Z (a) Find the equivalent resistance when the switch is open. (b) Find the equivalent resistance when the switch is closed. (CZ) . @EC?I"I :2 7—77 a [7/5 <14 4‘57‘7‘6 ,‘3‘ £746?“ View #713 fwd 2.12 _ ‘ i‘cyt‘yilox: or? Er; parq/é/ i Q//;2~>3L+2L‘= 1J2 ”3 =(Q (2 7 5 //’9+€ War-7L 4217410374 7%“? éflSWc’fS (2/6 #6 $0t'7’7é‘2 +{4’37 are Sac fibw ewe/4&2? ca/fg’ergyni— Lease NJ"- p344 1,, .. V (a) Determine 11(1) (in terms of v3 (1)) for the circuit in 1'10) 19 19 19 152 Figure P3-14. (b) Determine i2 (1) (in terms of vs (1)). e “Wk/>24 WM... max curr (2.4+ d/‘w‘s ,0 W ’46“? Fig. P3—14. Circuit for Problem 133—14. 3 3 3 3 3 3 i 2 3 . H , ‘ . \ ,1 ,, 3 l <62) 7—23 “IQ/Id .L / / “f“ke $3“ fakes+ Way 214131774 19:? 7‘1. 1Lrfl<[ a”. Key - “Pg y" “7-14 8 8-7 “Hf e 4;,ch 137‘. 49,4 ‘13 14/9 r12” wn‘fi Jus'f' '73-}, 5, 2" ($7514 £7de Pansifcpn +0 éejfifi, 3 j x \\\\\ w// 14/675122 AOVJ‘ 7%};- ,qqe 0w; 9420+ 7%.5 ne‘FWor‘é Aer: Age/1 3 “3/3’ E Wedacad +0 soma‘f-é/‘nj .1 1 ’6‘- x/ef’y sl‘mf/Or. 3 J W6“ we can «5238/- ;H‘mul' 344/5 waeJJ cayswfifflue +0 ocean 9 ,4 3 (b) I” $341550?" “pastnxow +0 7419 £150ch /e‘+/_5‘ MchE 3 50w (2% 7%.: m‘ 19% horlc/ +L7a+ er‘rérT‘IL 13‘ din/{Jed {\fl‘l'o 1:1 affip 3 , .11.. 3 C __ .9 2 :L j. WEI-'2 ‘3‘). : 53 921-(9‘47) 1 17 .._ __ . ___ 3 3 , _ r . /AI\$ loqrflern rip bah/W73 Wcufla’ Ea: fefd’fi‘ed' fiyyfiree fl . 3 +1344 25- such «Mac-:1 . LL?- .. .. z .1. a; ) «Lag-=(z 4, “g x, 8613/3 P345 Find the equivalent resistance of the two—terminal network in Figure P3-45. A11 resistances are in ohms. ...
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