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Lec 32 - 0 EQLLCLHM:5 each lim‘sofimewl‘ Name{0 r C 0...

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Unformatted text preview: 0) EQLLCLHM :5? each lim‘sofimewl‘ Name {0] r C} 0 35'. 201 QUIZ E Spring EDGE (apnea hook quiz War]: on the quiz page. Yeu may use a calculator hul: net share it with semeene else.] The voltage pulse in the figure is applied to the ideal ihtegratihg amplifier shown. Assume that vh{fij = D. a. Determine volit} for the time interval from 40 me to EU me. h- Find the value of Va at 60 me; vu{ED me}. big-“UN! 5n nF J mm H 1 _r=u VII '- 1! [emcee = eeee Liam's: b§{e=.e U 1: Wattle—emf .ecfix-t-Q = -E§ebt U J eateeems, e [Jailbwi‘fi 2 Flex! Maeeeea «is maze = {Ls—arse} x! i: we? —e:a (Ia—zeflda Hit: .04 a X=t = #15:” ‘3‘ t + £333?" -115 K:,OL[- L zeal} : H15Qt+e15efi +3e—Lc— ID @A the) = (matinee—Ha) U: matemws Unh- Ugifoeas): ‘iZJSV Esme :ELRC) EE 2G1 QUIZ E Spring 2003 {Open beak gui:. Hark an the quiz page. You may use a calculate: but not she:e it with sumsune else.] The veitage pulse in the figure is sppliefl tn the iéeal integrating amplifies shown. Assume that vgifi} = D. a. Determine VD{E} fer the time interval from 40 ms tn 6% ms. b. Find the value of V; at EU ms, vu{ED msJ. ”‘22 my a 2:: In 6- a we {ms} '1‘“ [-1 11-25% I'I-F ‘ s Get; E Lie EC} ('13 n3 {355;t;é=.uibwfi. [JE&EE ='—"L{U t 558%: “emf (-J-hcg-K +c) = East ‘4') e. at -'_—_ Lit-cums. o b‘eELLDws‘i = sv. Hos—tfimms U‘gfiA = (ALA-mt} V t. EJ3351= -GEI1 [—Lz,tflgxhdbg~ 4-E5 .bH i. K =15, == Eigcffi kk‘_£3 _‘ UDICEIEFE: k 1_EE “”34 L iii-1.0L!» = ems—amtz' “14+?4-s [4A that?) = C—Emfl-tf' +{Dnot —a‘) U} Meeteéews (L11 Unseen = 1m L: :3 '31 ‘-L~ {filmm- 105A“? \JZS} "‘3‘ m :1”? \} L3; 5-" "‘ EVA A We :2- 5— .2: 2%: ""1r.:2.<.x~fl$ ‘EJ 2.53., 'nr‘w __L___ g; 56:25.5 Qajw 'I I}; W i? Q. 34 0:... 3575.: A ,-..;.- m 17 x f w .1; 3 “’- ;= I fifli}: a: {I‘VI-é‘xt‘LdafiLg ____ .1339? 341-10 \a E: 2.91 ’3‘ r :35 2’2 Mallard EE ED] Spring EUHS MW lflflfl - Humewsrk: M.S.32 Page 1 0f 2 % Homework: M.8.32 % Ynu may submit this Human-'07}: a. tutal of 11'] times and receive- full credii. View numglarive Bumpigjnu Ems This is similar in problem 6.23 in our Lexthnck m quesnpn 1': adapted. with pamissinn {mm HIE publishe'n fiam pmblsm 3.32 of am' rmbaak. Eiecm‘c Cir-miss, Missal: and Rigid; 5:}. edition. Addison-Wash}. I 995: if 13' intended'far use by registered madam: at Iowa Starr.- University and may m: be mgdfar 91.1m- purposes wiriism the expressed permission of the publisher. Consider Lhe schematic shown bsinw. C = 1125 microfarsds [:1th VA = 143 volts RA = 14 kiln-shrug RR = 4 kiln—ohms RC = [115 kiln-ohms I A = 1.1 amptrcs 1. Find the voltage acm'ssflm capanilm'. VD at t = G‘. i : volts. 2. Find the Nepal" fie Haney, alpha. for Ihis circuit. I - radiansfsscund. 3. Find the resonant radian fretiuency. omega sub zero. for fliis circuiL I _ i rsdisnsfsssnnd. 4, Find the. fins] value of 111: current fl'Imugh LIL-final.- |__ _ __ _ ___!amperes. lLfi} has the form shown balls-w: 111+ (ti-alpha“; * (Elcosflwdfl + Egsm{wdt)] ampsres. 5. Find 131: I _ amperes. httpstffscennniinn.cs.isstsns.ndm’EEEfl1533113Inwf1flfwebqniz.cgi?Sessi0nTD=krunrnan_1113121... 4141212103 Mallard-EB 201 Spring 2W3 MWF lflflfl - Haméwfirk: H.332 Page 2 of 2 .mnrr: 'flrfl£9— _ _ _ .Cciurs: Hamid Copyright ® 1993- Eflifawa SIM: Uniwm‘gr htrps-flcqa-bnfincm.iastaia.edu!EE2fl.1 spflflmwfl flfwehquiz.cgi ?Sfissinnm=kmampa]_l USU... 4M-flr2UU3. Figure 6311 A Circuit used to describe the step response of a parallel RLC circuit. “--..M*7—~‘~"=:"‘ raw ‘ ’1} mi N - u W“ wwwwwwww . ........................ i ......................... Z‘ a ‘ oi LL. +_L_c?-LLL #4,, m I (fizz 12C. 51:} LC. LC. 1 Rd \Jtflrm: LLQQ :17 M 1 1% “ ~ "m a... a 1m ; mutt { "Ki: . LL53?) = 1—5: +31%: waficyt—i—Bza «ELM x . ,.__.. . i “EL—a: i “$65: LL69 --— 3...; +55% +Dza wart Cl). (“Rio-Q = H7553 = €Qmfi __ _.[_..r5... OE: by L406: 1- “LL—LR. (3.41:1.- LAM: __l A) bi fizlmmgg —, ___ Amt .. e) LL69 = Lfififa CM'I‘IOE wie- lcifl: 75m, 1+ = —-l A .5=_l+13,’ B12154: “83':- [LL59 = L— 1 +61. (i-5m150t+Z-0335$m750‘t.)] D. M9 = (a M = 2.0335 A 7‘30 amt A “hard: 0'1me :0 (EBWQJJLLLLJ‘) U“ (0+3 = 40 V db El: “10V FE . 6 (0+5: pi = ‘LEE :_ _|.58’auo% C- (Al-L “[58 =1: m" = 750?); -. Imofiiéd Tb; 2: ~20933>3 V do mEmem Lou, EOE Emtzo we; mfim 93E“. Ami: E :3. cow 0% am: 33 GNP oar _ _ q _ _ _ _ _ a com n E UEEmu EEEG _ H H r; EmQEmEmpo _ _ a E8 9. w: tqumEmbcD _ _ .,_ _ _ _ _ _ __ _ aw _ _ _ _ _ fl _ _ _ _ om ow. om EEV d ...
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