# HW5sols - P 4.?1 [a] First ﬁnd the ThéTEllill Equivalent...

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Unformatted text preview: P 4.?1 [a] First ﬁnd the ThéTEllill Equivalent with reapEct to £1.13 using E1 51112129551011 0f sﬂurce transformationa. a 11:0 101113 '5} 41:9 25mm 455:?! b 5 Q a 50V 45kg 1:1 4 5kg! 4 51:51 + 54V 35.5kﬂ “ECU: 5-1, H m .1 1.1%; = 3183.0} = :11 3 ‘1 51.3 — 5-1~ [h] l371321111211 = { Hi :I X lﬂﬂ = —5'§7E 1 D' x P 4.73 The nude voltage Equations ELIE: f."_ — f':_ i.‘=_ — ['3 _ U 2000 mam] 5mm _ L“: — r'- if”: f“: — r.'- r:- '_ ' + _—' + 5—5 + SE] ' = u aunt] on. [Jun mum] ELLEIUU ['3 — fig ‘_ f'g _ 3U f':_ = U 1L1.r.Jr.Jr.J -1U.f_]f_] mum In at mldﬂrd £01111: 1 _I .1 F _: 1.| (ﬁx 1 r: + + +r~a — +r.=--. .-=— ' 2mm] 20;qu .3an SUUUE 5-“ 2000 i. ’_ :i __ 3U H. 1. * :i * E *7. ’_ 1' _U 50m ELLUUU 1 5mm 50.000 10.0er 5( 2mm.th _ 3U ) a 1 1 1 i':_ _ "' ['2 _' "' i'.'3 _' ‘- = 2f]. [I10 ( EU. UUU .- .UUU -1EI. [MU 1 Solving: r.'- - Y: r); = —'_E ‘5": #3 = 23E] ‘5' T-En-L = #3 = 28D 1' II to H19 1119511 current Equatiﬂns are: —-1El -— EDGE}; -- EUJIJUHL —1'g} = D 500012 - Eﬂﬂﬂﬂﬁg — 1511:?"- ELLEIUU-ir'g — 1'1} = Cl BREED-:15: — 3'2} -- IRENE-:15: — 301;} = D H13 canstraint Equation is: L1. 2 1'1 — 12 Put 1119313 Equﬂ‘tiﬁng '111 atandard form: 1";{221000} u— 12-1—2111110} -* ESE-{ﬂ} -- 1'53]:- = if] iyﬂ-EUﬂﬂU} m 12-{1.5._111111j:- -- EEC-::—SD._EE]U} -— iii-:0} = ﬂ 111:0}m12155111111111}mix-{130111110}Lid-j—SUUHDD} = U 1-4—1}4211:1411-1113-+1111j1 = u Sﬂlx'il1g1 ii = 13.5 11111: 1'; = 12.95 11111: is: = 1-1 11111: ii = Fin-1011.11 280 R 1 = = ED 11?? T' 11.1114 13. 14W 211111? P 1.?5 [3.] Find the Thex'euin equivalent with respect to the terminals of the altimeter. This is. 11103-12 easily done by ﬁrat ﬁnding the Thex'enill with respect te the terminals {11' the 4.3 D. register. Tllevenin TﬁltﬂgE‘: llete 1'5, 1': zero. a 29 24 a 160 JUL] 25 EU 2 Solving. rm. = 2L] ‘5'. Short—Circuit current: 251-5: 9 241-! 165.? h 3£=12+2i5m 3;¢=—12.'L ED u Hrl1:—_i2 =_.|_C'.'3 [SENSE 43!} 20v b 3' 3.13;} EA 20V m 3’ ED Rfﬂtai = K = Rumm- = 3.33 — 3.13 = £1.25? [1]] Actual cnrrcnt: 313\$} El] Iactna'. = = _r‘ g 95 mar = >< mm = —EEFE 5.33 P 4.77 i—"Th = ll since circnit 120111111115 no inclcpcnrlcnt annrcE's. HT — r.':_ 1'2 La: In standard form: 1 1 1 IX I If I: — T Li * E; E) ‘“ “T L‘I) * 35-1-1; _ U 1 1 1 n] r * r1 *J‘A[-F)I =1 Sﬂlr'in 5.31 r.':_ = 2 ‘57: [T = E ‘5": [1.5 .-‘L Li. H 51'! F 4.78 1311 = E] since there are n0 indEpcndent murcee'. in the circuit. Thus we need 0111? ﬁnd Hm. iazii—Eiilz—Eﬂi; ii = mumA = HT 3'1» = — 201330 = mat-T 'gl =1;’n.us = 12.551 1'1" cm = 12.55:: 12.5“ P 4.79 [a] Seh'ingf r' = 7.03125 T I _ lﬂJIJUI’JH 11”“ _ 12.500” rim = 1- — in = -1313 r 125} = 7.625 1r en = j-{121mm|2nmﬂj:- + 2500; = .5 kn I?” = H“ = .3 kn [h] A43: 5% 5m 4.3?5‘u’ 5RD mm = {-1315 x 10—43:}2-EUUU} = 9.5mm- [c] ﬁle register CIGEE‘EI to 5 k5? frern Appendix H 1135 El. 1redne ef -1.T k5}. L'se Teltnge dix-'i.'-_'.i+2+n to ﬁnd the 1.‘dltELge drep hernia. this lead resisterf and use the voltage to ﬁnd the pm‘rer delivered te it: REID if, r = Ti-J. 75‘? = —2.12 1-" 1' k irﬂﬂ a. aﬂﬂﬂ‘ ' 6—2.1??? _ pin-k = # = 955.12 In.“ {HUD The percent error hetn'eer1 the Inexinnnn power and the pmrer delivered to the heat resister t'reln Appendix H is W: errer =1) —U.1 9-37 9% {1 nu :- P 4.83 We begin by ﬁnding the Tllei'e11i11 equivalent with reapeet to R3. After Il'iELking EL CDIIIJiE of source trenefernietiene the circuit SiIl'LpiiﬁEE to , 15m — 30:; u = T: 1-?“ = 2ml + 3m; = em; = mm 3: Using the teat—murce methed trJ ﬁnd the Thex'enin IEEEtﬂIIIZE’ give: 1 i _ FT HT — 30-i:—r.rq~,r"30::- _ 30 2:1 .51“ i T 1 1 5.1 e m» 3U 1U an E LT 1c: _ _ . FE-rh = _— = ? = (.05? :T .. Thus our problem i5 reduced te analyzing the Circuit 51101111 helmr. T551 130 1m —e., =2.“n 2235 A 1522.. + 55.25 3 15422 250“ = 533-21525. -55 25 5:. = 12.5 5: 12155.25 — 55.25 =12.5 5:10 a. = 22.55“: 22.. = 2.552 P r13? [51] Find the Thex'enin equivalent with respect to the tEl’l‘ﬂil‘lﬂiE. of BL. r{iipen circuit voltage: The Ineeh current equatione. are: —2in*3{5._—52}-2n{5nay-25 = n 252“'1ii2—i5i'“3i52—hi' = D 1553“153_2n-;23—2lj-m~1-{ig—iz} = n The dependent 5ource constraint equation i5: i3 = 52 — i1 Place these equations in standard form: iii-:3 -- 20 an - biz—3} «- i3'i—2EI} -~ i333} = Edi] 5553:1522 -- .2 - 3} - .ia'i:—r1:i' n 23m} = [J n-g—2n} +i-1-i—4} -- igﬂ a EU - -- ig-ilﬂ} = U in:— 1} “- :‘E-ﬁl} m ia-{D} m in-i:-1} = '21 Solving i: = 99.5 A: 1'; = 7’8 2'1: 1'3 = 100.8 A: ﬁsh 2 Bil-ii; — i3} = —2r11-" i3 = 21.5 A Shﬂrt—Circuit current“. The 1119511 current Equations are: —En’lD--3{:é-.—1'g}-Eiu_ = [J Eigma-gig—iajms-jsg—a} = 11 The 1219119111219111 30111129 4201151131111 Equatiﬂn 1.5: 1'3 = 1'; — 1'1 Place 111E159 Equations 111 standard 101111: 113:3 - 2} m 12-1—3} - 13-31} - 131:0} = 2:10 1.1—3} — 121:“ A d a 3} 43-511} m 131:0} = 11 Ill-:0} m iii—d1} w 135:4 -1}~13{lﬂ} = D 1'._{—1}mJig-:1}wig-::U::--—13{—1} = [J Solving 1'1: 92 3'1; 1'; = T333 1‘1: 1'3 2 95 1‘1: 1'3 218.5? A . . . VII-l —2-’l 12:11-13=—-11‘1: HT|._=, =—=Eﬂ 15c —-1 ED + 24” 12v an EL 2 Fin 2 5E? ...
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## This note was uploaded on 07/10/2011 for the course EEL 3113c taught by Professor Staff during the Spring '11 term at University of Florida.

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HW5sols - P 4.?1 [a] First ﬁnd the ThéTEllill Equivalent...

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