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ECEN214 Exam 3 Spring 06

# ECEN214 Exam 3 Spring 06 - EE 214 Exam#3 Spring 2006...

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Unformatted text preview: EE 214 Exam #3 Spring 2006 Instructions: This is a closed book, closed notes test. Show all work on all problems. Forfull creditmyou mostuse the method speciﬁed, Place answer in the space p‘Fo'vided. i. ﬂ .4 F”: V v; ‘ \ i m If 5g 1 € {V ID # 1)Fincl Thevenin’s equivalent circuit at terminals a-b for the circuit shown. Express the voltage in phasor form with rms volts. (20 points) - i a «V7 .._ g Zth“ ; f; \$7; '1’ 1/“ “AH—~— .. . .__._..... - a; - r: " T .- ... ,' u Var" :2 «z: s w W i; x - w. - u ‘ “ J “" J’ ,4« “ —- V E ;_C} y r "I ~"‘" 9/ w .c-i_.J J. ' “ . w a» .w—M W ~ l _ we ,_ --r’ d c v ' L, /r‘=~. f ”’ «see a AWN“ - H“ -: . (W x W i "3 u * V b t "‘7‘ U" 5; ';,\._~70" M A “7’01...” 'F M , a [ﬁx-f. /--. ‘2'. “(r v W k {D MMMWWWA.WM 2; ,2 f?” v u c..— - w W u . ~w~~wwrm , v i y j a “it «12,: I/ I». p7“ I is]; I) W. k ‘ ,,,. K V? a} V *" Q Li ,a i” W e ,,,.. 6) i/ r "'1 : V3 I f, } a“ i . .J. . {(2 U C,’ _ j ‘3’} ‘m M :1 3’ s.__...._.___ -. ._.,. .. If: 4k -/"j\ ’ .2 . W I ~ ‘ \' 7 A. J l x‘ d] «F: L < (7/1 w<_> ff I J x. A." \-"\V at!» f v'". x j, A. "’ , i, “A: "‘ “L “ 3 WT 6" J r m x’ a 37 l F’s“ *‘ c.» , .3 <5, 9 ’ wt“) 5. “UV * .'\ 7 ‘l . i , .- if r - i * I (3/ L N»— "v ’4 L’vﬁ/M 319— i/ ’ ¥ '77” w... ‘9 M/ WT W Q I 3 l g a u to . f v l , “'7 ~~ v - *“‘ i i w t - - f . , ,— ‘f 4; ’ » In. «w—ﬁ \3 ._.- - 1 (J I N. 4 ‘2 i J H v (‘7’v 'J“ :r" a... '7 "‘ - ’ ' ’1 ’ / v" . " 4.1"» 4 .7 if; Raf/y “1’ r L/ W2; M- I, w J U ‘ . . k . . 7 I. . «an ‘ Q r' ; , x. ,v" 1 u .1. f7 ” at 5‘V M k “'0 a“ 73") Lt" - ‘51? y ’ ....-~ .3 : _W—-— s- 5 ’ w "M MM 1 v "~ I A ' -~ , w / ,’ ;.,'v . ,- -' ,r v; r . I V1.12,» ‘( i . A.» V ’x 3 “M 2) A 6 brake horse power motor has an efficiency of 87% and a 0.75 power factor lagging. It requires a voltage of 220 volts and is served from a conductor with a 2 ohm resistance“ Find the'VOltage, VS, required to provide 220 volts at the motor at full load. (20 points) 212. Vs 2—. 4W.:.._. ” / m I / r H “L m. .. 1 L“ é U x’ imp «Kim u ,3 A} I "D ' V. _ n ﬂ, M {.4 dz! 7,» we, r» ._ , y, , f , i’ [gig f “f; C» e _M~ ﬂ ‘jM 3 w; l” 53‘? W /V . [1/ /{ \ M j 1...») -- J j“ / r” 1:, u l i M / at”? a. W“ A I 1 L” g ‘N‘i A jié‘i’zéﬁ A 5? .7 4" ’mi {Liz -« If ” 33/): 9’ l ' .‘d z a} 5 g I “f‘ L i ) aM/l' : {MV\£ Q {J p J" «f w #— i!‘ v / ii if" 1 W3 1%,. \$9; 'J K ~ L, k —.. M a a“ / .r; M a a I; to w I «1;, if“ “:33 w 3 iv; 4.; “>3” /%f 7/ u if] M fir/gr! a y, ___ ‘‘‘‘‘‘‘ IMWMW”M_T.;:.‘W'~W~_NMu t... hm” y’ "N W q" a: ,r’ L’ aﬂm '''' “"7 i: “y i '2; [2/1- I; a? :r __‘> if 555 f “7" if a; W L........._._..- ‘ fr .- v “‘1 9 i’ i, .L. ‘7 M )M, w; - g9 if a writ/«L dbzﬁ / 3 W K ‘ l, 7‘" y»; z. l.)- ijj; — “h 1m r L... “’1': / N31 I" . “K h 5‘" «‘1 \$3311; if f «a \ I}, {a 1/ «WW-w “‘ ‘“ ’ Lemme “:1 9 / 7 2 , A c, m - ,5 ~ .- A w r ' v: x A e «i / r" f or J a 5 ; ’ ”~ r" 1‘ «flied 4‘: {vi- 2’ wf’ ‘“ ’ i" I (CM 1 3) The following loads are served at 480 volts: Load 1 = 10,000 volt amps, 0.7 power factor lagging Load 2 = 5,000 watts, 0.9 power factor leading Load 3 = 6,000 watts, 1.0 power factor Find the magnitude of the current, 11 L], serving the combined loads. (20 points) “7("1: ' .x [1 L; = £3 “Wm WE WW” . r N . .v r "g 5'» zu/ I EM; 6N1? , . ., M. :53 ,x "a 3 if x -------------- w—m—mmw____~_h__ﬂ_ 3:“) ,. ,\$053819tiwﬁﬁiﬁﬂY'Z’mtr‘q—g-w-~w~w~fi M g7 U m; ﬂag/J ‘ m} 74? f? ; 11" Vilma : (3.” 1,3,3 ‘“ 2 ‘r 2 2;- M a?” 6 g W" /’ *Mw “’ Viv/C if 'V ' ., i/ :2 “‘1? If r ~ L 5& 1 f6 "T"? a,» ? ~> " w x g x J: z: :2 5y rigs. f/wyéa’ flin ﬁg 4) Use the nodal voltage method to find the current, I,in rms amps in the following circuit. '(20 points) yo ¢ ’5' w ,z‘ ,2: ll , Irms == ’W ’ 72 '4 MMWM ....... mm 3: "T" ..... My... WWWWW W»; 1J1 i» F ‘ u, ~ N ‘x ‘1: ,r —- I ’ ‘- 9 I, f' “"37 c1 V”; :: joj‘, ’3” éii 37% '5’ {’2} { __.__ “‘1‘”! M v - ‘ wL * «Nomi WWWWWWW M 3: {a 7/ ii ,jwﬁ: I :1 52! /m V%‘3 5) The switch in the circuit has been closed for a long time and is opened at t=0. Is the circuit under damped, critically damped, or over damped? (circle one) (5 points) Find the voltage Vo(t) for t__>_0. (15 points) Circuit damping: over damped under damped critically damped Vo(t) :2 ‘ r! .i . W hot) l {1! 1““‘M'Vwi » n “A "3: _ N— l' tray 3 “ft; ) L, £2": 2’ do i i” " ' ‘ M v 2 ,1 . _ V “r in: 1‘54 {x t ‘V’v L, "V w M [£2 w gawk; :N Jib/J “is” u F Jr.” —- WM ...... N ~— iigwﬁw M3 ’ ., j 4 M L a m_3§ (in, 3*: r" ! ._.. 4 i ‘ V *‘Hkigtdi ’ Vii“ } "f “is 6 (L w. I“; ~ ;: ﬂ f “' L [(\$154 “' 6‘" v ‘5!“ i ( J4, : “AL” .3 W WWWWWWW N : (f/ 37:)(‘0‘ ,7. Nap—M uuuuuuuuuuuu wwwmu‘j’ .‘T .g L3 '1 C) «1‘ iii" ‘ “H y‘“ L .1 {a M x , I \H' h ‘1’! * f 4.4,. yr 7/ r ‘r (a L ’7 73/1 la w W J r434 “J I A): w (My M _M_\> 5:]? V ﬁ/ééﬂ wiggle/717W”? a “Um”, \“ I”) ,. .i l‘ i: ’7‘ llv" 1 fr «fwd, '1 ‘1‘ / W’ka M (a l» «AU —-— barrio we: i f’” w“ ’ L J ’ "7', Q5051”! / A, I “(if i1 M,- r n Jul 1 T‘ .1 i ff“ M (.1147 ﬁsélL/cLA w “ W3 1/”; twat“? i v I W ~—-~- ‘ L, “w ... i x ; 7 AZ LL5b)- f1": “:2 [KEV I: ”” cabiar M.“ A ._': 7 rm». f a m a... ‘7 ,u. ' 3, 7! 5.0:) f "" C “” M I kﬂjw~ﬂ y , ’ A W, :1 ‘ 5:7 »~ A ﬁgﬁ W7“ (:65 (£06 “"- ‘ 'h ’I'If/ ’ #7 w. n.- («~ /{«"~ “"6 w ,1 Lay), .>[email protected] .5 ...
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