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hw6-sol

# hw6-sol - 2.24 FIJI IlEJtﬂtleL‘l mnrmllvnc-n lot the...

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Unformatted text preview: 2.24 FIJI IlEJtﬂtleL‘l] mnrmllvnc-n. lot the middle Imtltr be "0:10 B. KC'L at "(My B yields {i 5L 20 -1 L". + ' — I} 11L .5 kn II I, — . rm 1', = '2 111.54.. The voltage [Imp r'I fulltm'ri jlllnlmllamly mi 1': =3 k9 x1}. = 16 V. To ﬁnd 11.1. mutt.- that hy' Clhm's. Ian-.1 f'rj — TA _ .. 2km —|J IIL-‘ln. {1.} when! r'” 111515: I10 mh'vd by noting that 1'” = 13 + r', = 23 ‘5" NFL thirs is an appliL-atiou 0F KVL}. Suhrilitutiug inlu {1]. 1w haw that r31 = 10 V. {1 L1"! in I: f? . g (p D~—" th‘lg—j ___ ___ _ UL) ‘_—.__ I J x“:- H-J I I a“? ‘— l ‘J I 2—. ‘6' I? k I“ C I? l r 4 _ X = F" F‘ﬂtlll (jh 5* 115:“. l '6 _.. an- g. P II II a _l 2' '1' i:- .la. 3.11 Parta The mesh equation follows [iilt‘t'tijif he itnqiet-timl ms [3.1+ Rn —I?:I.z :RAI —| [ iii—l = [ ray—l i Ruthie-+1?” RA+RT+REJ L Partb The nod-1| eqlmtiun Folhm'ﬁ by inﬁpec-tinn after grounding Hit.- ithttJIn node. There is 1:10 good lt‘étﬁﬂll m ground an}; nther nude ms doing rm would [IIII]_'i-' make the problem 11mm diHicult. henc-e. nther £11151.‘.'I.'.‘T.'.‘i- Iu1de1' different lt‘flifl'lirllll'li.‘ nudes will not rec-[tire full L-I'edit. The center nude Ziﬁ denoted nude .4 and the (enter—right nude is denoted node B. The nudal equation is I l 1 l 4-__ ".I + "1.1 + "e I _]lr' I :| |: PA :| =|: “.1 :| . 4'.- _i'?-:' He' + 1E1: + m: I.” It]: Parts d and 3 A5 Hanna stmlentis nun: nut knuu' of :n' have net-ere; tn synﬂmlic- waivers. and ms plum-Inn" ﬁhIH intn NATL-U3 dues. nut mntl‘ihnte .‘iignificunth' tEJ understanding c-iu-nitﬁ. these parts ﬁlmuld not be graded. though Iieeere this tn the a 'aui:.~|"ﬁ diﬁu‘etiun. For reference. the MATLAB 5c-11'lit and eulﬁu'el'ﬁ are ﬁttpﬂl'éltlifljl' attached. 31'; I“; ( # i" A) if? ' rig/'3 ‘ ‘ ’3 'x l D d) / y n “10 — v: r ’ 13 3» A + ‘23— V4 ‘. K A + m ,. a VB \ -2"‘ 3 V8 ‘ 2Vrs=C> 17§= 7V3 Vm=’?5v LHB + 1k.) VEZJ’VX 7 IX I3 Sam; (’urrenF mej-h (GSWLVOF \ic“\/E, 3 ~Z‘f— f”? T ' —’ 3 \y m“ ELI: 73kg 3.46 : ‘ :J-wémAJ ,. J (:3 2° (9 5 f a CD 0 <3 A U 2 3.4? This problem is made signiﬁcanlllv [easier hfu' apllh'lug sourw tmnsfornutinus. Consldvr tht‘ hm output natives as tlu1 nodes which ('[JIIIJE‘L'I the fuse. Transfunujug HIL- think. we haw. {1m +1}’1Dk Fuse 9“ 10f: ohms From this new circuit: the current through the fuse is. g 1" s: .T5 111A. ‘=m[;_.+1jkﬂ+1nukn — After scum: calculations we have that N E 5.3 in order to satisfy the current constraint across the fuse. Thus. the rmximmu munb-er of loads is N = .5. 4.5? The transducer is described by r-rn = {LISP — 1n1'l.-r and it has an i11ten1al resistance of RTE = EDDIE-i. The desired output characteristic: on the other hand, is for a pressure of P = 20 111111 Hg to produce co = 1 V and for a pressure of P = m mm Hg to produce 1'9 = ID V. Hence: the desired L'Tn to 1'0 equation is 1-0 — in _ a L'.— — 9.25 1113. _ El 1113.: % 1'0 = +1031}. One possible realization of this circuit is as follows. Note that the designed circuit is possible because there is no 1.1.1:1certaint}r in the transducer resistance. This may.r be seen as a practical flaw in the proposed solution. 300 300 EUDk 1UE|U 1|:UD v i sum: 1 V ["1_-_:111'-- l: lu':=~:h'..ulc----seu-- in E.‘ hi! 1.‘.'i'.]] u] -\".- 41‘s lLiiléLH' In. |{-'-']:i:'.1n:u:i1]1;.: '.E'_-'-1-'-.u'--1||.1LL_'-']-::~n;l-l-'-II--|s1,¥.1n. mm:- -d' WILL-f: Jul-'4' :L' -t ='\""I'-11-‘~i11:-C 1'12" “i-Illl" ‘5'"J'-'-I'-n.-.- ::|' 'II']—.-".::'.[I1-lI-:kh. ...
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• Spring '10
• VladoLubarda
• Trigraph, The Fuse, nther nude ms, Hit.- ithttJIn node, nod-1| eqlmtiun Folhm'ﬁ, diHicult. henc-e. nther

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