Unformatted text preview: 2.24 FIJI IlEJtﬂtleL‘l] mnrmllvncn. 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 appliLatiou 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“: HJ 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 itnqiettiml ms [3.1+ Rn —I?:I.z :RAI — [ iii—l = [ ray—l
i Ruthie+1?” RA+RT+REJ L Partb The nod1 eqlmtiun Folhm'ﬁ by inﬁpectinn 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. hence. nther £11151.‘.'I.'.‘T.'.‘i Iu1de1' different lt‘flifl'lirllll'li.‘ nudes will not rec[tire full LI'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 netere; tn synﬂmlic waivers. and ms plumInn" ﬁhIH intn NATLU3 dues. nut mntl‘ihnte .‘iignificunth' tEJ understanding ciunitﬁ. these parts ﬁlmuld not be graded. though Iieeere this
tn the a 'aui:.~"ﬁ diﬁu‘etiun. For reference. the MATLAB 5c11'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 mejh (GSWLVOF \ic“\/E, 3 ~Z‘f— f”? T ' —’ 3 \y m“
ELI: 73kg 3.46 : ‘ :JwémAJ ,. 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 munber of loads is N = .5. 4.5? The transducer is described by rrn = {LISP — 1n1'l.r and it has an i11ten1al resistance of RTE = EDDIEi. 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
10 — 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 1UEU 1:UD v i
sum: 1 V ["1__:111' l: lu':=~:h'..ulcseu in E.‘ hi! 1.‘.'i'.]] u] \". 41‘s lLiiléLH' In. {'']:i:'.1n:u:i1]1;.: '.E'_'1'.u'1.1LL_'']::~n;ll'IIs1,¥.1n. mm: d' WILLf: Jul'4' :L' t ='\""I'11‘~i11:C 1'12" “iIlll" ‘5'"J''I'n.. ::'
'II']—.".::'.[I1lI:kh. ...
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 Spring '10
 VladoLubarda
 Trigraph, The Fuse, nther nude ms, Hit. ithttJIn node, nod1 eqlmtiun Folhm'ﬁ, diHicult. hence. nther

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