ENGR210 Lecture17

ENGR210 Lecture17 - Name : CWRU e-mail: ENGR 210....

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Unformatted text preview: Name : CWRU e-mail: ENGR 210. Introduction to Circuits and Instruments (4) Quiz No. 5 10/3/08 Problem 1 (10 points) Using source transformations determine V0 in the following circuit. V0: \ .. I? = 'ZmA 4,11 3M 61¢- 4». :fizk W _L J— r. ‘* 32* {WW 2“ _.)\MA + ._ .. 3rd: 2"»: 2L VD- LR—QmJ‘VZ“) Name : CWRU e-mail: Problem 2 (10 points) rush: . \0. . Hm 2L‘-'Z\3+1L"‘I°I.:O iGL‘-io‘\,_- 2i; =0 sopeamski , t + “(co-HOLoz'IOI'rZIS ~1Il+8Ig=O (a) Write the mesh equations (in standard form) to determine i1, i2 and i3. PLEASE WRITE ALL COEFFICIENTS IN DECIMAL - NO FRACTIONS! I1+_ _ "IO i2+ '2' ' -iz. i1+ +IO M +10 is: 60 ____—_:__i1+_:I_i2 +___+_I__I3 =______O L5 \.73 it: \.4‘\- I3: 6.53 (b) Assume that i1=+2Amps, i2 = +2 Amps, and i3=+8Amps. What is the SIGNED power associated with the dependent current source. Power = 2A é— (,0 ’53 I” P=—Uw=_(eo>(c)=—3bo (no . ‘ ‘ warts . tram 45w (90 Name : SOL-U770 U 5’ CWRU e-mail: ENGR 210. Introduction to Circuits and Instruments (4) Quiz No. 5 10/3/08 Problem 1 (10 points) Using source transformations determine V0 in the following circuit. V0: 2 UO'fi—s ' — + ML _1 av ~ in: : ZMA 'U'kk)(z"“)‘|7— 41 in i" iZL trumsi-‘orm currewi‘munLL Was-Com UOng... Sou/1Q). . combine (ml-lay. souncw , Name : $ORW‘OU a" ' CWRU e-mail: Problem 2 (10 points) —>i0 ~ maskli ‘ . ‘ => «at, _ ICC; ~2i3 = 0 Supermes I’M ‘ \ ‘ . __ -¢o+ IOLZ— \OL,+Z I3-2 L.+ 313-0 =3 —{2Ll+ IO Lz+iot3= 9° 3L0:I3—l2 (-0 49 8Q 3L\=(3"lz ’;_>—3LI—I‘Z+ 13:0 (a) Write the mesh equations (in standard form) to determine i1, i2 and i3. PLEASE WRITE ALL COEFFICIENTS IN DECIMAL - NO FRACTIONS! __+\(f_i1+_’£__i2+ _2 i3: 0 “I2. i1+ +|O i2+ +10 i3: (90 '3 i1+ _| i2+ i3— 0 pdmg 50mm”: is 1.73A. i2=i++A i3: MBA (b) Assume that i1=+2Amps, i2 = +2 Amps, and i3=+8Amps. What is the SIGNED power associated with the dependent current source. Power: m 300 wafig' e...“ = 1' 90 uth s - _ +- / i a; =3(L>:¢ Amps. O — PTAQ’AL:\<(00‘)((0)2-3Q0wafist (90 I C.— 32 > - : cw + E on 388:3 9QO N, on 3-88:3 » DO 4m-ocw0_..3 “:1 6 17.. t0 4 R100 "Vccf (0;: + Va; "Vggé ’= + Vck _!&_‘:4q)'-fl)—4 Vcc A‘ P “+7: 1? ivxfiuanSk-n’r ,1, ,J,,,., .J _J 1 - ‘77,, , 7 - ; .1 j i J 5. I V 1 ‘ ; F .31 z. 1 fl, 3 . ‘ f, ‘ ,, . L‘ j i -, ‘ ‘ ,4; Book,‘ iOP’iRmPS , ,, 7 ‘_ ‘ I I ‘ ‘ 7 1 x ; i _; _ 1 i. ‘ 3 r' \ ‘ 1 i ‘ ‘ ‘ ‘ ‘ 3 ‘ ‘ ‘ ‘ j ‘ ? ‘ ‘ T 1 u ' " - “~ ' H -‘fi "W’" 4‘ - W‘ ‘ ‘ ‘ s j : I I ‘_ ‘ : 4b, 4 A) ’44 1 ._.oPermMmeejr~i _,—, 72%; meeamqwz? I'wapiérfiéhfi ; v i I a i I l m,_alctm.m&mcdiL-R4m;héjn_s ; _L,,, ‘ J; 5 f ‘ § .‘ i i 1‘ ‘z i ‘ ‘ 1 1 3 ‘ § 1 ‘ ; ‘ ‘ WWV hwmwwww , m, fl * f, 9,1. L,L,,3i-§s§)b‘rn¢ghmg_4-47,,V L w i I 7 ‘ |_ ‘ ‘ ‘ I I ‘ T, w, 37,, 1 . .irjfl V. . _,, I; 1 ‘ ,i3‘7;,,,ji;77 W 1,4. —— Slam . , ,,_ 5,37,, ‘ _., ., ;_4‘ i ‘ I | * ‘ | ; ‘ ‘ E 1 ' Ag;,gwgwgflyflgnJ j- §cclLLn3 j .1 ., Q 3, 14%;; ‘ ;,, __J_4,Jigii_fi _‘:J,_;,_:,,A J‘MLH;W:7\_3M:: ahafufiiwwg 7, * g-..‘ i .j , V,‘ pf”,_ Jfifiii‘, 1 ~;%7_,,7fi44777lm~1d:¥£r¢nh¢hq’p4 ._ h L _, Lg, {4,74, . E 3 ‘ I ' ‘ i 5 » 1 ‘ ‘ ‘ ‘ ‘ *7 I iii—71777.7",m‘ 777V 7 ‘ _ .7, 1,7 ._ ,VWJ:VV,,,,;-__,E..__, \ . 77—? A 1 ‘ 1 ‘ w 1 3 ‘ ‘ 1 73W??? f 1 e_~w*f‘— 73“; 7' u j '7 7 ‘ ‘7 i 7’ j a : ' ~ ’ ‘ " ' 7‘ ’ "‘ ' ' "' ' " ‘ ’ ‘ "" '* ‘ H” ‘ ,,,! l J | i 1 ' i 1 1 a _ 'w ! ’ 4 i a t E E7 E f ...
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ENGR210 Lecture17 - Name : CWRU e-mail: ENGR 210....

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