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ELE203-MT1(2009-10)

# ELE203-MT1(2009-10) - HACETTEPE UNIVERSITY DEPARTMENT OF...

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Unformatted text preview: HACETTEPE UNIVERSITY DEPARTMENT OF ELECTRICAL & ELECTRONICS ENGINEERING ELE 203 CIRCUIT THEORY- I, MIDTERMJ. 24 November 2009 I. Show all of your steps in detail and present your work clearly. 2. Write your name on all the papers you turn in. 3. Use the back side of your paper if necessary. Ql. Consider the circuit given in Fig.1, 2) Use the node voltage method to determine Vb V2, V3, and V}, b) Construct a proper tree. Choose 1 ﬂ mistor (associated by (Assign all currents and voltages on the tree and co~tree) and to ﬁnd tree-branch voltages, d) Find the node voltages shown in the circuit (Vb V2, V3, and V‘) in terms of tree-branch voltages of the topology. Verify your results with those found in part (a). VI) certainly as a tree branch use the topological approach 02. For the circuit shown in Fig.2, 3) Find the Norton Equivalent of the circuit seen by the load, RL, a,b. (Use mesh current method for your analysis), b) Find the value of RL so that the power delivered to RI, is maximum and what is the value of this power? c) Find the ratio of the power delivered to R1, to the total power generated in I connected to the terminals he circuit. Q3. 3) Find To by using superposition for the circuit shown in F ig.3. b) Calculate the power on the dependent current source (2:5). Is the power generated or dissipated? Q4 Consider the following circuit given in Fig.4 in which all OPAMPS are ideal. 2) Express the voltage 1:, as a function of the input voltages v; and v2, b) Find the value of v. if v; =lV and v2=2V. c) Find the value of v, if v; =0V and V2=2V. IOU (9.1.41) ”3016 agcTuaJ-Vmﬁ, (”4‘ ”H 12r- + ~9va m3 (Va—33>) -— (m 1—. o . 4J7, 2) klxj;v {1) \lH——V3+éﬂ4ﬁ LO ‘2) " 7 \IV'VZL-év Vl‘2/ VH:_ v2 N3 {A A“) V2 -\}g 1H1 [VﬁQmLHVHQH :2) 3v2_+\13:-2H/ m wamwg— La‘JH. —é\h1 +6V3 :12? \x m, wkgvs 4AA. _—: AQJV. 46‘2’r Q)“; + ‘0 («ax/2’20 ——l DUI—1 =‘le '7 "’2. C”3\J2"'2L‘\\ + EVL‘ : '—\2_/ -_S_U,L :Lfo -—~:.:_£> VQ,:‘89V J V :JLW/ U‘:-Q\/ U35VZ”L’N\-— I9: ,; 2£a+2I4+2(Iu-3) +3 (In +I.)-O .3) QI\$+3II=~6 a» 8135+I,=-.9. ('51 ﬂop I,: g;-3+2(I,-3)+3(L*-Q)+4-E=o 7'31; rgl: 3?? 13433-‘3 3V\ _.Q_\JQ_4- HU‘B 5: ‘éOA— _:——-—’—A _— - .__-—-——"" .—-'——— 'LNL 4\O\Js 1" 60 :=—\ my, __ Lr \1?> éo U3; \LyFFé \l 1) :54, Var-1L3 V \I\=—_2€5?7 v .ﬂ. 9233 Q ##— {—- ’3\h .-—Q.‘J-2__J( 8V3. ‘-'=- O 92AM : , vl=i 43V \I\=—-3_é, V \l\_.\/?_ O-‘v'l. '— +-‘ :3 —L .éD’O- :ﬂé- {I ‘2’ g ‘3 Lt :51 .. (2w) \‘1. -——- —(Q.A€ﬂyiﬁ-3€VB ._ .m. a? \M ”inadeo‘ ﬂ1+ﬂﬂoé V‘:—\$';\f02 -)\/;=-LL'( of J“; I W. WW :_8l_(_2_0V;+9_0va) b) u;=lv,\5_=.9_v / iv-s _ ~1ov5%.=-§V5 “W [I ”-5034’3) / 1 —1m/ 5 vofw 6 40" ”f "533/ gm“ .__ «10V 4 Vo=§Vé 10V '/ 0) ‘4-‘0", ”fl" 4w .4 vow-av; 10V '/ ﬂ ...
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