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Spring05-ECT-Exam2-solution1

# Spring05-ECT-Exam2-solution1 - Exam 2 9‘3 06%omlx‘0n W...

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Unformatted text preview: Exam 2 - 9‘3 06 %omlx‘0n W Name: ID # : Problem (1) ( 15 point) For the OP. Amp circuits shown below, find the output voltage V0 Assume that the OP—AMPs are ideal. - "Rom oomﬁmo" VF ‘5 I m I kCL “0&2“ m) w. 4° _\’n-° = 0 £2, + a —.:— V“ LR\‘*R23 -: VOLR\1 3“ 22 WW? Wynn \I :KR‘;\VZL)VI\ 3% “°'“'V+TW . Rem mnnedﬁm 0 \$21 Yn:0 kcL mob. m) m+G—l9‘ V51=° R l: .0 dW__:5_ 3 at Mayo“ ' el'Vs g , N c “-9 C?:° , No: —- RC dl’ g UCL“ (A “x “ Cn:o ., N5 = 4%) Wmc] A ..‘h.. \< ’Vo\ SH!“ gm M02 WWWWE E “’02 3%«661; E i: 5 A‘U’cﬁ Name: Problem (2) |D#: ( 10 point) For the element shown, the waveforms for the element voltage and element current are found to be: v(t) = 5 cos (500t) V i(t) = 10 sin (50m) A + v(t) _ —l:]— «’5’ 0 Sketch the two waveforms, on the same graph, and show which one in leading . - Determine the phasors representing the waveforms of the voltage and the current. For the circuit shown ﬁnd the equivalent capacitance, . Cab between terminals (a) and (b) . Cac between terminals (a) and (c) a 25 nF b SOnF 25 nF C . Deﬁne the element type ( inductor, capacitor, resistor) and ﬁnd its value. ‘ vbl‘taﬁe legals cow-mall =3) \\I’\§’(L§€i’ét}€‘ . ,l \l Lil; C , (25“.?) (2%} Gt M ’23“? 423ml: ., gonf‘: @254? W Name: ID # 3 Problem (3) ( 15 point) The switch in the circuit shown has been open for a longtime before closing at t = 0. 6k§2 12kg ZkQ a) Find vc(0+). “(£65 =Mg(d) : [3.26m][q.z\s] = 2‘8-8\ V b) Find vc(oo). (1mg) 125+6‘5 Mom) : 3" c) Find the time constant T for l‘ 2 0. Re} : 61\$ lﬂ2 K : 033‘”) = Lus t: ch 18 a = (MBNOQM 253mm“: t; a 4 é d) Find capacitor voltage v60) , t2 0 . - 25 ﬁg mm: NCCDO) +[wc(0*)-«vcccn)] e . -24 8' New: 24+ [288’ 3 e) Sketch v60) for t 2 0. Name: ID # : Problem (4) ( 20 point) The OP.AMP circuit shown below is ideal. Assume that no energy stored in the inductor at t = 0. Lilo l a O ’ ’ ” Citei ho ’ ’ ’" ’ ””””””””””” " 01060 a} i=0+ mdudor 4'7 ~ . ‘ e W or“ w N?(6\:° (1* t = no 8*w SW“ WW mam)?“ 'P 5km a) For t 2 0, derive the differential equation for vp (t) . N? (on) = VA KCL. main L?) W A. R r W WWWWW w ’ E \— 4. demogjw wgmmm b) Find expression for VP (1‘) in terms of the circuit parameters( VS R and L). - H 1,3” _‘ Mth: pru) + (”FLO ) _ WAG”) 8 NW"): Vs+ [O—VS) e L {:30 ‘c) Find the output voltage v00) in terms of( st R, R1 ,R2 and L). KcL neck CM Uri—Us + U“ - °c c \/\.I\.r\/\/\/\"I R a R‘ {.M .. B. e E u 99‘2” N , E ’Rﬂeq ]§_ Vé ’ V5 e 3 E “°= KT] “ ° 3”“ “i i; . “ WWW WWWWW .\S 3 Nos-x d) Find v00) for(VS = 2 V, R=100£2, R1=1k£2,R2 = IOkQ and L=0.01H), “if“ z t\ :J is = ‘00 3 \03000 ’c \ 00‘ — \O 3000 J ~\6 \$N¢\$ +6 wed): n [2 _. 2 e e) Find the time required for the OP. AMP to saturate. a a? 22 22 {3‘ \03000 M it + — [(7)000 t, émfmimmiiums Name: ID # : Problem (5) ( 20 point) For the circuit shown , the switch was closed for long time and it is open at t = 0. (LU? 3: {116) _ (V 'Vc... ”I. ~\- R00 300 = NL + mum) ~00 Name: ID #: b) The differential equation describing the circuit for I 2 0+ is : diL +(5x106)iL=0 """""""""""""""""""""""""""""""" Wr ite’therreiatedwharacterist'rceguation’f WWWW WWWWW WWWW ’ WW WW W'WWWWWW WWWWW \$2+ 2000\$ +Q5 M0) = o c) Find the roots s1 and S2 of the characteristic equation . g : -i000+ ' 2000 \$3-01: iota-W; ‘ 3 z s : JON-32000 i 6 G 2 S\ : -\000: \O - 6".‘0 2 d) Write the solution form in terms of constants and deﬁne its type ( overdamped, critically damped , underdamped). .iooob {Lou}: [was zooo’c + 82. sin moot} e Linwoolamved e) Calculate the constants and write the solution for i L (t) . .t. . Rom some“ {mm at he t=° 1‘29. : -Qoooxm + 290082. \00m -Ug/ = — 03(63 + L2)“ )82 0:05 \. , _L _, l WW”W“W*WW":T(3§0‘FW§ Semen 9. \$11 ct): {2 cos 1000’“ +059“- moot] e wwwé' Name: ID # : Problem (6) (20 point) For the circuit shown below, the switch was open for long time and closed at t = 0 a) At t = 0— (switch is open ) find : vc(0_) and iL (0*) Nci6320 (Ltdico b) At I = 0+ (switch is closed) ﬁnd : vC(0+) and i L (0*) NQL0* } : NC L6) : D (LL01): l'dO- ) : 0 c) At t = 00 (switch is closed ) find : vc(00) and iL (0°) Nclm): 0 (LL00) : 5%)- d) Derive the differential equation describing the circuit in terms of ~15» t 2 0+ Vs: W; + UK lie-t Link“: V9: vc .mc £1.32; + ewe/Angeli. —_— V5’ 9 Vc=VL . {c-H‘l— = A‘R e) Derive the differential equation describing the circuit in terms of i L . MIAWMWAKWWYAMM» ...
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