Exam1Solutions[1]

Exam1Solutions[1] - (f w . . C/ W! ‘ , e (25) 1. Consider...

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Unformatted text preview: (f w . . C/ W! ‘ , e (25) 1. Consider the 2-winding device. Negiect leakage. Provide numerical answers and indicate units inside {O i] [ ] — use scientific notation for large/smali numbers (e. g. 3 x 10”6 ). Must show work for credit. ft; is i All dimenswns meters 0.0% L a! A); far il=5A no = 43: x 10"7 H/m w.“ . 9““ “Mqu w as} M = 3.595%,43 {wa‘fl .5: {VJ an???“ ( {Ha/w x a; , a m} 3,3 A F a a WWW L21 =Mi 25:? 1 A? g 2 if: I” t, M . Initials L“ "W V E: / (25) 2. Negiect resistances. Answers may be 0 or 00. Assume steady state. [0 Lil] r 1/ " ‘1‘“ ( 939(2) t X : 60 Q n “ii” h “'wa‘Q” W W7} (1) ll / ZN}: 5)???” ft?) [IX/‘5; (b) X” Z *—"© 9 Lin WM x (gaflgfi r”: X ~ #3 Q d W A? P (C) Xmi ~ 3206 Q L W? i :5": t a» “"i / (d) [2 if (59/ A 3%??? @m—li (6)120) = ME 02: coswéto “$10) A Wng 5% g fig” Initials NS = 20 N,. = 10 mm = 2000 1-1“ Neglect leakage mg} I him . H {831’ it???) 5:3. / j l L w 0C“! (a) Lsr(er) : L+ 0‘! COS( 1 91.4“ /80°) H (b) L <0 > = L53- +WO We + W O) H ME" (53190: g; '2 33 r r —.——~ 20510 w M j (0)111‘1-(90 = a05+ 0 cos( w- er+ w“ o) 1-1 “- j‘gtwk ,M g/ 4;?“ (53205:? f 0;“ c: / (05(ammfl 15 M ‘5: Ma (ts/f5 (6)1fis = 1A, i,. = 2A and 8r 2 30°, 0/ V v. “I r l ‘ | I k (ff) 50 C91”) :2.- w 5% gig-r3 'fm/ EOE) ":Lf‘ 4‘0v2 5/1”? ‘ 03' :2: 01%;? $2510 753/96 0:4: 0. / Te(1A,2A,30°) = 0?? / N-m W Lia‘s III , I I Wow“ (25) 4. [0 U1] L88,max : 6 mH L = 4 mI--I 83, min f“ m (I) Z of: ’W j; (1011849).) 2 5 + { Icos($§,_+2?5>°) 1111-1 . I . I It I 9 s ‘ I_ (b) If 1,3 — 5A sketch (111 figure above) loeauon of N and S . (Slimmm > (c) If is = 2A and the rotor is free to rotate, what is the equilibrium position closest to that shown above. or: 4§ ° (d) If is = 2A, rs : O, and 8,. = 40t, the applied voltage must be (fl » 1 “ w? d ,1 " -\ . 4’1» 1 f a? 115,551” § - Z. gag-S 5105:151;1~+2"?0")€€Q) x? I f ..« w.» w: ‘M‘f E f mm * _‘ . P” - j S t“ “E ,7 d’, .. ,5 0 $14” 1;)": ‘a I m 1,. [/5051‘1’1 253(1- +270” USU) = 0 + 561:} cos(§§2t+_§2_°) 1W? ( 1 j W t , N ,1 l": / {a W ) /(_~;.r;.;> (1.53:3 3 (e) IfiS = 2A,and 0,. a 0°, 4&5 War 2’ 5"". ‘\_ | ‘ N m I .N‘ W % 3; 17;" 1:? I1 1251/1565 3,31," “1"” {j 71)” I) T42 A,0°) = 1M N—m. 6» " 1’ x ’ (25) [0 i] 1. Initiais Consider the 2-winding device. Negiect leakage. Provide numerical answers and indicate units inside [ ] — use scientific notation for iarge/smali numbers (eg. 3 x 10_6 ). Must Show work for credit. 0.0431 All dimensions meters ll Hes. Ii 5A 0 Q. W. m ‘ ‘ W 651303 W i O‘L.) r 1 (m :7” “3””: £32 M ED Awe ( 0 é) m [viii a . 1M1“ v M) f 5:) «- 3R / ")1: 3K}; Cber if {OQL}; “if! j “W Ki?“ fir J" E“ \‘i ‘)Z~f h} Chi/J“: “’7 4: C J (25) 2. Neglect resistances. Answers may be 0 or 00. Assume steady state. [0 i,ii] an“ m 10, 000 H“1 31” = 00 (a) Xml =Wfl 9 (13X!1 : tJED Q (c)X}2 = O Q ~' * {3.20. ((1)12 : A A ((3)1203) = fl (TM coswfim Q30) A Initiais Initials (:25) 3. -------- 4- [Oi,ii'] NS 3 10 N,. = 20 mm 2 100011“ Neglect leakage "NS"? ‘96:} 2‘ C9“ 6 O M” W' 0 WW" MA “'“““‘““"“"'""“"‘ gr... N‘I (a) Lsswr) + cos( e,.+ ) H mm {0m} (1,; Mr (:10) M b L,.G. = , "W _ m0 H ::. ; g ( ) H( I) 0% +meflCOS(_—Or + ) “LEW fix?! or i go MM", icfl’x‘jw (C) LS,.(9,.) = + Uri-00M E 8F+E8©0) H :5 TOOK) “WE/K ' ‘ A ‘ Z g "2 ’ (d) Wf(iS,L,.,9,) — 2, L233 1m, «2‘ L«[r L{ "+0 Lw U” {K ‘ .5 "a, u" a 1m)” ; t :22“ mm W a ’r ““ 1/0 4) h»- w% (2 <1 vii??? ‘ > > f “2" Wu (,5?) {lg} 6W3 (6)1fis =1A,i,= 2A and 9,. = 30°, . ‘ fl g . Cm 54) U“ “T” 3W} 6. if» M M g "2, ir'“ by W / ‘ 3 (EM Te(lA,2A,30°): 051» N—m film/i Initials (25) 4‘ [o Lii} L88, = 61111-1 r L = 4 mH 33, min (3) 1283(6)?) : _§_ 1‘ 003(é8?‘+wfl0) mH (b) If is = 5A sketch (in figure above) location ()st and SS. 6 30W 5:... L»; (J k w i (c) If is = 5A and the rotor is free to rotate, what is the equilibrium position closest to that shown above. e‘mfir’fi o P (d)1fis = 5A, r n S Us O , and 9,. = 40$ , the applied voltage must be i ‘ \ “"1 5L w « (L a i cl \ :_: 5:1 52‘: ‘- if“ M i L». J J '3 .. s I {b r W 1M3)” Q, j {:5 030): O + 00 cos(€i§t+__©_°) EMU (e) If is = 5A,and 0,. 3 0°, “"“H‘ W!“ ‘ 2- (3k {MAL-3g W. f.“ ,. * l - v v i l 5; ’7 (‘n (bl Age {is} g” Qt} ‘ Z“ i, v Te“ A, 0") = MN—m. -- ' ‘ e i ...
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This note was uploaded on 11/29/2010 for the course ECE 321 taught by Professor Staff during the Spring '08 term at Purdue.

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Exam1Solutions[1] - (f w . . C/ W! ‘ , e (25) 1. Consider...

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