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ECE2010_HW5 - ECE201 1 HOMEWORK 5 A09" Name 07">...

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Unformatted text preview: ECE201 1 HOMEWORK # 5 A09 \" Name : __ 07" > ECE Box Number: This assignment is due Friday, October 9, bring to class. Make a photocopy for your refer— ence. Problem 1 (10 pts) — Complex algebra Use your calculator to find the rectangular complex representation for each of the following polar form quantities I («5 151300 : 1321‘; +113: =11.M+575 (53271100 1 7.5.37 + 1231' (a ”LOZ‘7g: : term - H.221- .n/ - H on 7m” 6 ' ramq+375 m Ia (JV = CHE am; “1.13 «42,733 Problem 2 (10 pts) ~ Complex algebra Use your calculator to find the polar representation for each of the following rectangular complex quantities. Express your answer in polar form with angle in degrees, for example: 31300, I? O (A) 6' — j 6 3 gfirvofl ’5 [avg (-33.6? m [rmoq' : 533553.04“ (l r5; 3 + 3'9 (03 Q7 2 3 17.2112: (d3 0 17 +5 ;, 17,72 M cl 3 as 2 16 432‘ Problem 3 (10 pts) v Complex algebra Find the value of each of the following complex quantities. Express your final answer each time in polar form7 for example: 3130‘”. (Q? (2+323é5—5AA : 23,00. Lil-{£0 (5) (81.1950ch ; ”2%" (:03 151;? ”all équo+5“L-_‘Q°+e+j4 : .muumgéfl 1: £49?) 0 ‘ . ‘3 f0 0 e\ “Wat L” .. 3.6QLL."21 4 [S 0 ° 4 3 1% W Problem 4 (10 pts) ~ Impedance Computations Find the value of the total impedance of each of the circuits given below. Express your final answer each time in polar form as in the last problem. M 3 H (C?\ r/qu‘rW/‘me ~1— l *— 5]: ca: :5 r/g El 2 ’35 o" = MAME" Problem 5 (60 piss)? Solving AC circuits Solve for the unknowns shown in each of the following circuits. In each case there are two parts to the answer: The phasor answer and the time waveform that the phasor represents. SHOW BOTH. 0c (3 7‘5 H t 3‘“ 2w M 1’" U ‘ 3? /§ 20 cocfloot +60% L 7‘” SJ‘MHQ 20—0“ (U; *5 3m 01.9!0’9 rgfimg 10 X59? 60° m w o 2,; Le .—. mar-aw 2.0 +£1.39 I a ,x A. . fl .. o VR 2; £49“: {flzoMé‘gfiq 333.43") a; I189£33wz i—s V 1 l air/V1930 ‘ , i “L3,... wwww o r - ° :3 «Ll-z VL ”i 1“. 2L :: (3503(é.§4&‘{ (fill) 8011; 4L :. 2334/2d , x _ 2R ELL; (12 ff.) 1 0.354?! w‘éwi hggvé/fo)‘: N v‘( :2 17.39 ,2 Q4130 “HRH“ “E It) : nee, “gédfli 4;» 3143‘ ‘ 5»: IC 2 £44,345”? {2657" ‘ - W Colt) : H.H7€«D§{§'04: +2é,§70\ \——__\ O Ia : 2.2» 42:33. . M (RH) -; 2~2¥€¢o§(5‘o£—53,qg°) v V 810‘“ til-313‘ “EM “5* gfl‘fmi (370 t w 15331 3‘9 (2’ ’ w m »' 9 “MM " V2, .. :32 a ":3?‘ = Q.i>272‘,,’.5__§7 3 a ‘ ’ 33.3% a a!) w. V1- . ¥ 6: - “-3 “V a V Ia”,z;§fi2¢??gfi:fl33 1:1": “£33; :2.H‘3i:§§;§3 2 f}? ,1? M ‘ , __l VJ W Ufl’é) *‘ 7,0 “who-£3 ; ‘ «24:13 ° .. » K2 3* {’1 L——— g 411m _,, 9133759§ 39% - ‘51 3%”? \_,___,3__________~ an): M _ 9 ‘nr -%§,%“ . V , a J “’ 2 “1“, LL”) “ Zunwbot— Sam‘s) ’ aH‘I. 3 - /V_‘ “ W UTH) £735.36 evffélx‘t“ #533363 : I *2 58’6” , VIP L” ’ PM” 1711+) ’ am mam: as» urcfl \W- § (A (H ; ‘Hm «:5 (new: 4r Hrs”) r y a; 5930 . 3:41 “v “KW (1'13”? : 31.2wr(:uu-t+w" ...
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