NodeAnalysisSolutions

NodeAnalysisSolutions - EE 1305 February 15, 2007 4A Unit...

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Unformatted text preview: EE 1305 February 15, 2007 4A Unit 2: Node Analysis Problems Solutions 120 A 200 + 6V 1. Use Node Anaiysis to determine the eiectric potential at node A. B I1 =— VB - V3 : ®V”Vfi (Clara‘s Law) I; '— 4 fl (Cum/44:1? Saw/m) I3 : Var-O 2 iii (@éwihys Law.) 20% (MOS) QV~Vfl _+ 271/5}: VA We Mmjr Safve (bv‘yfin 's'ZJL 290$; Comma/l ®ewom.‘flq+gv ,5 ' 120521,. ['0 (EV—‘VfiD-E— 32QJL avg @0V‘” IQVH-i» 4g'gv 3- 6V“, 5% v” = :6 Va W: 33.%\/ ) 20 A 30 + + 3V 6V 2. Use Node Analysis to determine the electric potential at node A. 'l“ “t”. w 3v© ._ , V8: ) Var—6V I: : Va -\/A f; (akin: Law) 25'?— 2.5L ,) - Va "Vfil .. (DVflvfi (@éwmk Law; ml— in 43:2,. 35'2". I3 : Vfi-ov 3 VA {@MWE Law) [52, EI’L I: + :2 3' 3\/— Vi”? Jr (,JVWVfi‘ S VA comma“ berm/m.”qu “5) 2151-" “g. m o _ 3 (fix/:4) + 2. (Wave) é VA am $2M veg) . :: (oqu + v8 “Visa; .- V35} (Commam fléflflmi'rffifl {2:99'4'5) 25—7-— "3 3-2, in {IL 3 (2%ng 1‘2 (ngvya) .: Q‘s/ya; 6V—«5vfi +2Vg~ 2V9 : eve ..n.—“f'"e9"f b _, V g __ (IETFu y) , m W 2 am wawas!’ i _ _ m # .TH‘V“ V3 ' 5W8 . wavfi am) 3.515 J T" ) Law! ,_. 2.5L. _ , I; +4.9 _—_- 1‘5- CKCL\ Vfi “VB \/6 M Comm-am dgvffilmifi «vwa “ - l "“ -""""""““ {:3 gm 1. x 1 chfl“vfi>+ 3(éV“\/g>= éva QVH "' H VB 1'. V {2d €?ua‘fimcx> Sch/6’. Slmwg‘f‘aflfiflfififi Qfiyum+t’@‘15 gr VA await/{3 —-H\/;q +1Vgrnbv. (I) 2% ~IIVB;MI€V (2) ’22 V9: + 4Vr2>rnIZV 2?.“ Vfi —— FEEV3.$”IQ%'V WWHMM 417%; = «- amx/ V5: l.7‘f\/ __..—-———-—-—- a PE +2VB :wévv “Hm +2.07W3mméo‘k/ HM Va-+3.5<g\/:—~Qv .mu Vfizz-Ci.5"$ \/ via: 87/ m/ ...
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NodeAnalysisSolutions - EE 1305 February 15, 2007 4A Unit...

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