20102ee110_1_HW3_2010_solution

20102ee110_1_HW3_2010_solution - émdtd F‘P‘CobU/ms...

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Unformatted text preview: émdtd F‘P‘CobU/ms 01> AS in any true design problem, there is more than one posaibk: solution. Mcflcl answers foIiow: (a) Y = 1 1/4 S at (3.2% 1 racks. Lionstmm this; using a 1 S conductance in pm‘aiiei with an inductance L web that 1-5011?! v 4, 01' L 2&50 111E. W (b) Y x 200 R13 (purely tea} at m = i radxs). This can be constructed using a 200 £118 cmuluctmme (R = 5 Q), in parailei wish 2m inductor L and capacitmf C such that mC ----- Eme 3 0. Arbitrarily sciecting L = 1 H we find that C = I F. Om: goiutian therefore is: a 5 £2 resistor in parallel with a 1 F capacitor in parallel with 21‘ i H inductor. (2} Y = 7180" HS 2 G +113 at mm 100 radffi. G = 8.6%"; = 7COSSOO = 1.21% (an 822.7 LO. 3% res£3tcr)‘ B = KIMY} = 79111800 = 5.894‘6. We may realize £115: suscepmnce by placing a capacitor C in parallel with the resistor such thatij = jéfigg‘or C = 68.94 “E fib‘. [m l ‘ I «La— Wimm 444444444444444444 amwaw On: 5011mm: therefore: 18 an 822.7 In?! I‘fiSISCOl‘ m pau‘alls} With a 68.94 mi: (d) The simpiest solution is :1 single comiuctzmce Gr =1 200 1118 {a 5 i2 resiswr).‘ 51L,» C 3 ‘ C .1. Q % “BU; Ha ?€‘EM€ ‘WB nodd chgfi VH9?) M +k€ ngwfivw ENE???“ H16 4WD Aefenéesn'f Sue/WM, VF" : 3595([037Q’ *3) --~+> SAVE C : 5mm: -~> ___.-,..i... 3%“ SOs/Ml? W? ”5,152, ..... L ‘- ,2‘m H '7) iii ”3 ?L 5), ago/1% KC}, mm +27%: §f0} (1/2) W ““4“?” + {35:32 SHE/6L "§'3“/.L “'1 W 1 [05’ ”13/04; V“); B); MSG/"g ML 021‘: +Le 575% m.) m CVJ~W V‘) V; W V3. “'" "7“” W J— w® 392%ng 3 We £qu M $2M; Amizéw (fi‘qvggrfi‘yg back «419, «H46 +aMe éaz/Ma‘m) my :— TM? w [179% HEBRJMV CD Pmb‘em DJ GW‘P‘AKCUJZ ‘95-“ko sis EXAVWPE. 0 “me, best boat? h; OMAWR 0 act amvwscan (S h, 0 EVAUAOLVE 0M etewxeyd‘ M‘ (A hwe 3‘) 58a 861‘ A refmemg 05: M GAVH‘M’ m“ ““4 +b “Move “\‘bwwck 444;. Sound. . ’ [ASH/j SuMcsi'V'Wn and ‘06) deh‘ts loo 939(1xlosé ”32') V (.3: 1x (o‘F rod/S . 8:3 avamonh‘vtj each 00M?ona~f FesPed h) V: . we now Rm; Hag reabwired Correcfibn Ruhr ’F’O oxclm‘xwde GM UMNL. (Le) V; : (QR) . ‘Z/ ’ Th9“ msfinj graphical} Maluleie. (D WED” ”(Mum‘s Law @ node 2.. I! :2 VI.‘ “i“! ”I. Va! 7" ’5‘ ’ 2 4“?er “I" (=2) KCL @ “0&8 2. ofdds Kev? we 3d“. (flow. (2.)) 92. /.A ?~% -— [.0 051535- {I (25> Considering the a): 2x104 rad/ 3 source first, we make the following replacements: 100 cos (2x104; + 3") v ~> 100 43" V 33 “I? 1» 711.515 :2 H2 pH 7.71224 (2 92 “F ——> 10.5435 52 l‘h‘cu (vl’ 100 z 3“)! 47x103 + V,’- (—13.515) + (W WVI')!‘ (56x103 V4.48) = 0 m (‘3' w V1 ')z’ (56x103 +1448) + V37" {7105435)RO {2} Solving. we find that + V}: /(47 “0%) "3' = 3.223 4 -87° 111V and V3' 1 31.28 4 477° 11V Thus, 1'1'0) —" 3.223 cos (2x104! » 87°) 111V and 1:31!) =1 31,28 cosc2x1047~ 177°) 11V fl . . y ‘ 5 . (. ousulermg the effects of the a.) = 2x10 rack 3 source next, 100 005 (2x105: - 3°) v «a. 100 A3" V 33 “I: ‘——> 10.1515 Q 122 “H ””224 Q 92 up -> 70.05435 :2 Then __ V'7(47Xl3’)+V;"f 510.1515 + (Vfi' —V3")/' (56x103 +j44.8) = O [3] (5.3" — V1")! (56in3 +j44.8') + m" — 100 4 3")147x103 + V3"/(-j0.05435) = 0 [4] Soh'ing, we find that 71” = 312.8 K 177° pV and VB" = 115.7 A -930 “V Thus. NW) : 312,8 605(2x105f “W 177°) 13V ar1d123"(r): H17 cos(2><105I- 93°) LN Adding, we find mm : 3.223x10"3c03(2X104i~87°)+312.8)(10'12 005(2x105r+177°}V and V20) x3__1.23mf<10’9003(2><104rw177°)+ 115.7x10‘”ms<2x105r~—93°) V ...
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