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EEL 311C Test 3D

# EEL 311C Test 3D - Tﬁst 3_r" 1"...

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Unformatted text preview: Tﬁst #_ 3 '_r :" _.. . 1--"'_ I-N'oteAllbold—facelettersrepresenteempiexnumhers ' - (3135110111101 ”‘5 I .' 51:3? {Zn-IIihefeﬂowiﬂg3111:1111ﬁﬁdw ‘31'1eh_t_h_3tI'I{he-31111111131133 "fat, is pn'Jrelz-rr real. 5 - ' [311311313 that the 1111pedei1eeIef-3Teai'ste'1' in series with 3 'eep3"eite1 e311 never be - equivejeﬁt to that {if 3_1’eeie_t{11' 3111- 311311131 Wiﬂ1 311 fndﬁeter. ' 3-H...- 23:. FE} ‘3’? . .. .' :- . '. £9”.— R+JwL FL.- 4', . :1 . Iee L "F" +"J 11:31:? L_ L 'I-I 111+“ 13"” 11.433”?— Ima‘ﬁq “my; 5" . . . .. I g f I - - ' ' ' “33%“3'1 9‘ I ' I .‘fﬂNTINUED ------ " ' ' an Era) = (3 + 45mm) v .5. Anita? r“: 4? +3 . _ :‘j'-.;""3{iﬂ. Find Eh: EMS; aalcm cf the E‘cﬂcaiing'. wavcfcmis: :-__.H_i_i'iE:- _Iiji evaluating withcut integraticii kiiuwmg' er '- KW? far a sinusciﬂ, {_r'chcaIciE + 31.; 'Y:1_:iu may. alsa a::d .tc- Lia: ifiéizis2 EaP= {1.5 whcr: 4;} Elcnc'tcs _Eim: __av:ragc - IE1; EIE1 =1551nmE| EL. Th: vc'rtical this rcprcscnt aﬁciilii_E: valu: I {I Earn/we, ' (15:, “Pkg/P- 53:. S'Swmi: Ms " 1AM” ” E a 43 (Law-vi: E15 GELWE? ' Té 1, i3?- 4.1137V 55 § in 1': a H Id1'WhaE is physical signiﬁcance cf reactive pcwcr? "Li: is lame. Pom. "EiEVEEMPJIZE’Ef +5: MAE fawn THE-i. Amﬂtre @Ezmaﬁ :ﬁuesiiéﬁ'ii 2_i 3. - 4-. Camderihﬁ fulluﬁingi-téiﬁ'mitin_hquepégfdﬁmaﬂﬂ'}: : II I -_- ‘2': I' - I -._.—. ...'..\_,._.___“__I___ i 5 _;Ifa_z)1 Fwd-_ﬁﬁseen- hy'tﬁﬂéi'éﬁé 153m. éi'i're the ansiiréf'iﬁ'gﬁl‘artfar'in. ' ' I" r _-r_ —_. _- - .-‘. . ..-. 3' ""-' _..— u._ _.d _h - _ "'-—._._ . ....__ N _ " "'fizgjmmﬁgg ‘1 -.‘ 15 .. Jib) Fiﬁd_Zg'Th-5Eéﬂ lay-Rn. Giire the answelr'in'rectan iﬂai‘ a . 7— (‘3 + J 1+ I) V r . inn -:. Q1: +39 IL 2: ‘3 I?! GALE-{3:6 ~51 5 (a) What is the vain-3 of matching Ru. for maximum average power transfer? . W RE? 2‘31““:— 3601440- 5 ' {d}'ﬂFi1‘t-dthe maximum ayarage power transfer‘redtu the Inad. \$71., E '2 C‘éﬁﬂ '- '3 w I -' maﬁa—61%“ V111 33 R6 E) = H 6665166 m . n. 3-: 1' _ .. 1 (I. ﬁzxg ') EFF-I1! “‘ “*T’tz'w Queeuen He; 3 . -; The foilewmg 1111111111 was analyzed yieldmg V1] = 5 66:5?‘1’11113, 1'2" ' .. ' 12:11 Uaiii'gieriiy SKVL,'KC L, and am Law; ﬁnd the 11115 1111131115 IL, 11:, In, and 111. 7 Give all the'anSWei'a in' ' [er _I 15- {b} Find eem’plea pewer_ absorbed 111 each. element using the space previded far ' each. Give all the were in EDIE! £01111. Indepeﬂdentaemee: . ...,_ m - '5' = -— Va 1:" = 4.526;, Le?§)(é1/—3 3 ﬂ. - ' 1.1a Si ___.._._( iEe—ji'é)VF’r =11.GL1 (1135’ - . .- .1. ~ “11., _ 0111 1:51:11: _-Depenc_ienteeuree: 5' _,_ '31-‘13, '19 2 JO] 6, NIH .— CGH TIHUED ..-_. 3 (c) #11111- the priiiciple 11f“ paiiseniﬁtipn: 11f simplex 11111111311. W n. ' m 31_-_1.—5L+ safﬁs‘geirgs. = (— 1-13+J.1.sj+ Jars + 16 -1-J \$1?— *J-V-Iii ”:1; CI . 5 I [:1] Using ﬂie vaiues 11f .11112151115 alreatiy given. find the- equivalent impedance 51:11:11 _ I131 the 1ndep¢ndent source Give: 11r- a11sw'1'11'i11 1rasweilasrectanruiari‘orm. .11.. " ' . ' . #5 3.11"». = 5-“1‘ ._1msz.11=f_ ﬁg: [5| ifB-DF II L1. 4‘3? 211.. won— - I‘_ {13} Wt 15 the power timing 511-1111 I13: tiie Hidepentient 5111;111:111“? bf ‘ CasC'Lig)” 0%?(imﬂvgl) [5 ...
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EEL 311C Test 3D - Tﬁst 3_r" 1"...

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