Midterm & Solution

Midterm & Solution - EE 110 NAME: CIRCUIT ANALYSIS II A. N....

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Unformatted text preview: EE 110 NAME: CIRCUIT ANALYSIS II A. N. Willson. Jr. May 8.2008 MID-TERM EXAMINATION Do all work in this examination gacket. There are three Questions. Each counts 10 goints. Good luck! 1. The switch has been in position a for 21 1009 200 Q 6009 long time. At 1‘: O. the switch moves instanlancously 10 position b. a.) (I point) At fr- O+, whnl is the initial value of Va? 13.) (2 points) What is the initial value of due/d! ? c.) (2 points) For I“) 0. find the roots of the system’s characteristic equation. d.) (1 point) For t> O, the sub-system is (check one box): Ci undamped; El underdnmped; El critically damped; [:1 overdamped. e.) (4 points) Obtain the numerical expression for Va“) for I; 0. 300 V ."‘ “J? C '7 "i n" I’M ....|.. .. v- I #WFM-x/VWH I " - I“ rid—1‘3"": l ' " ,— .;'; ‘3 “H (3) Fm L“! 7i.)- . ' . 1 ' ‘ 1W5 /' firm 2‘ ‘3“.‘4’: .Hr—H- r. WW» " {of} “fir-2U V} '57 wax- x (L: ) ~ { (0 T) '— ‘W‘ 5 REM FW— ------------- --- . ‘i ._ i _ ’21 C Li + (‘5 )U'b' S 4;?!)4E/‘9C + 1.3 (-1 '-'—’ f _ _ - '1" .‘ In)” a) (ARV/h j I fit"- + (MSW L “D {F g ———— ,- w ,:.~——- _- * my: — . x; .I .. K ’21:; K: '1’ “f. ) ' A "- did fl "MW; iv 1 a 4’ w a - 4 " -»+ ..) C- ‘t ,, _ “(C 4.“ A; '_ QIX‘KLI) - ______ “3 j v\:'____',,___ _ . 1 y; d A f i : Qtl? ‘ (1‘ I] {-(J (b ) (Muir aim, flL- (Li-i) a: ,3 C L? T. ,5 '1; w. (la-3301‘ “an? Adi (“1 xi” _ M____,.§...{ r“ / ’ {Bib :— wiifiu V/M Jig}: ...... ....w-...«._..w_Ww Few-*- " i KC) L i _.1. J (I )r') be 1 s as ‘ u, / ‘ Cm L i ‘ ‘ M . vL. 1.. b I“ l_ I {I .l I" L :‘i'fji 4 \Cbi/i A \l { . ~ ‘3 "1'1 '3 U i M» ' k I (Vii-L ("A n /_l Mk3." k («A-I'vfi 5i ~25 “ML-19 ..... .. I I i: 8..) (war-Lew: c1-) ._ $53“ .- 80} (w W (a :3; (E)! 6 Mg 61'} ,_(_ jrfiivx (yd- \ 17.7 (‘2) L -: law"; Ul -. MW 39+ T37 \Aftlsf) :: 36‘99 midi +2506. ‘5";‘ 6“} MW - J1 _ __ ': I NJ . R , i EE 11G NAMEWWELL 2A. (5 points) Determine the current supplied by the source VS. H1=12§z 82:29 33 = 0.552 Idea! Idem [NW .2»; f‘ .4? T;va {{5 arm a; 1 ‘ “1...; n.1,, 1. __ CW .2 { i {Q at” l) :5 i -- I; gkld't': §A$L1+3LXZAQ {I ."2 t 1 (“3“ 3'31 ““(i “ram-«2.412 ~51: ‘5) "TI-m5- {I | i...1_:%*.:‘ J... r ._._ 1‘ _ .. 1.2:.» , I __ _ O 3:? mflgfllemfij .52 - {Er} + 0 W93“‘\ CAth-rfivxih'd;\n‘ii{fh .' r; (-56] fuj‘iugv‘laiq‘ W 2- __ [35- I) (“L-54f; ~32.) 3 L744 # [4653?) EE 110 NAME: a.) {1 point) Find the average power dissipated in the line for this sourcei'liner‘load circuit. in.) (2 points) Find the capacitive reactance that, when connected in parallel with the load. will make the load lock purely resistive. c.) [1 point) What is the equivalent impedance of the load in (b)? d.) (l point}l Find the average power dissipated in the line when the capacitive reactance is connected across the load. e.) (1 point) Express the power loss in (d) as a percentage ol'thc power loss found in (a). Source —9[-e‘———- Line —-—9~1«é—~ Load 38.. (four points) The op amp in this circuit is ideal. Find the steady—state expression for v0“). . '3 n -_ I! - l fl : J —' 5‘ -- :2 #J m *1. it! ill YL. 11 ; i-dfi‘c “5'23 m l xc I_{¢‘._le..3 Mo; 7 A "24 . a. ..-.,.NM_.VMW I d ‘ w c - :- 2} 3 IE: :. “Law-WM r \3‘" 53:3 "r r. 463?}; .3: '2.hjz:faq ,a -~>'k C MT“. 1L ' "»">">r'-i="3 ""'""““" ‘/”: “"° Kor'wjfi :‘lEZes ‘ y -- .wr‘ “ . . _ .A \ n .v . .. .W J AWN ‘— r J. 3. g d A Unto fun» the}: {iv-5., all—av Hm: c 144:1 _ U)? CZ : Lie-(ton) -;(Cas‘§*fli) “la-“wing. in 3- ?: a WWW“? 1;“, an“, “NJ/4. ‘ .._..u_._#fim WM: .... .. I ........... __I:....:_ ___________ m M W (I _ t5 : 2.340 J. m?- vr r: v.3 g: sic-,5} 4.. oh “1 o; r 5 4.; ‘v. .r U1. 1"— ; - F + J—--—---. _“”" [,1 (5i); Us)“: 8P _N______. :5 of Gov M WM. M- ______..._._._._..—~---~~-~-'f~'-"_ ’ " “I q I U -- . T5 LTZ-l’? > 2 '.....‘"_*? 9;“ m_(._‘?if_;1i;5 :3; .) 3.1-4 ...
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This note was uploaded on 08/12/2008 for the course EE 110 taught by Professor Gupta during the Spring '08 term at UCLA.

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Midterm & Solution - EE 110 NAME: CIRCUIT ANALYSIS II A. N....

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