First and Second Order Transient Circuits Solved Problems, Exercises of Electronic Circuits Design

Download First and Second Order Transient Circuits Solved Problems and more Exercises Electronic Circuits Design in PDF only on Docsity! Chapter Seven: First- and Second-Order Transient Circuits  516 Irwin, Basic Engineering Circuit Analysis, 8/E 7. Use the differential equation approach to find w-(1) for ¢ > O in the circuit in Fig. P7.2 and plot the response i including the time interval just prior to closing the switch. 2 kL 4kG ica I es . + “> ev(*) = 100 pF & 2kG r=0 7 fe Figure P7.2 SOLUTION: y iyh eth io) = @ (byrtds . ay Birdy ha ay Ht fry € 20: Veg Mel 4 Bab Ve Wh 2 ake iy Lgl ley { a. a \ + é ® Re Rate / qrdds kyeo taf Fa (Rasta)? 2 LHS LB its ee 3 we t (a) Chapter Seven: First- and Second-Order Transient Circuits 517 7.3 Use the differential equation approach to find uc(7) for ¢ > O in the circutt in Fig. P7.3, ©% f= 0 is Z vc(f) FR 100 pF 12Vv Figure P72  Chapter Seven: First- and Second-Order Transient Circuits 7.8 Use the ditferential equation approach to find vc(t) for ¢ > 0 inthe circuit in Fig. P7.5 and plot the response including the time interval just prior to opening the switch, © 2) A 6 kO 12V vet + 7 iy, 7 yy ~ gS t2kO yp S12 ko BK S 200 LF aR wld x f= | | oF > Figure P7.5 SOLUTIONS “wio-)> tions 2 be Rye Pep alem Atos Ear kardy ter 4 >0y ve, 6 Mog y de, HE x0 Ruy ce Bq c Vth ehe om de . Be <o de” OE tv = BAC K;z0 Velotl= 42 kp bk, lén=4V | te C8) a & | . & 7 Chapter Seven: First- and Second-Order Transient Circuits 521 “i Use the differential equation approach.to find (1) for ? > 0 in the circuit in Fig. P7.7 and plot the response including the time interval just prior to closing the switch. 12kO 200 He 6 ko oe it Sih * yc ee a p= L BV +) S12 kO 2 S 6kKO —} . Vey <! ight) Figure P7.7 SOLUTION: UctoTi= Vets GPa 1 py Ty Ct) = Ue foe bo, Caen ap ke We fertys BV te lense a.25ma Tee ty ima) rr eecemaedis ff St §22 Irwin, Basic Engineering Circuit Analysis, 8/E ?.& Use the differential equation approach to find o,{7) for f > 0 in the circuit in Fig. P7.8 and plot the response including the time interval just prior to closing the switch. ee i p Bae y+ dy > 7 eth Res 2, Heys Lesh. t 2 3 y Uy Cot)* Uy lov = S4V = CU Ah a0 Bed 524 Irwin, Basic Engineering Circuit Analysis, 8/E 7.1G Use the differential equation approach to find i,(1) for t > O in the circuit in Fig. P7.10 and plot the response including the time interval just prior to opening the switch, © 50 pF r=0 PEs = z ete SK Sp AKO St, goa 12 ko fot Fa gh Figure P7.70 SOLUTION: = - * = &y = 6k rie OT to roma bs iy Las rt.) z = b> Ly Wty 3M : PPE 4 “: ‘ Veloys tele = BFA _ av ZB, Cor)e Velo) tm 2a rls Bp drt 0, veg ti karte, wo fF laze ie t lo Be gst TES wields, dUe y Ve oO at Cy read : -th: a, 7 where tr Kyprkpe TE ct Eten) yet, Ts Cl hme]: ass K ps0 7 1, Cot)s ~ Ue cot) Ry ia = -0.Smh =k +h Ko ~osmAa ly bfx ~ oS 2 mal ad ‘ i  ( r és 4 Zed, ——_s L aoe By J i ee [a aon a ape Chapter Seven: Firsi- and Second-Order Transient Circuits 525 7.17 Use the differential equation approach to find 7,(1) for ¢ > O inthe circuit in Fig. P7.11 and plot the response including the time interval just prior to-opening the switch. i=0 2k 2kO Cy C) ; “ys fem é s 12V < a ye BK S @, 3kO 4k 2 iol) Figure P7.71 SOLUTION: ; a ' te . re 9 > rm et ~ Pas ect ze ile a 2 + = fb yp 2,3 Be way bye en gs Fas Oya) 2 dhe 1 ane nne bea WG Yor By Abe = sn. Be id a oo sty TCA Chs BYSg Kye Kea = Joa tl Kev Bhp av Fy tO hy oy EO SD bb th la Ze BO | Ghapter Seven: First- and Second-Order Transient Circuits 527 7.13 Use the differential equation approach to find v,(1) for ¢ > OQ in the circuit in Fig. P7.13 and. plot the response including the time interval just prior to opening the switch. Figure P7.13 SOLUTION: _ Up fobs Ue fo) Re hz rihy  3 Chapter Seven: First- and Second-Order Transient Circuits 529 ak 7.49 Use the differential equation approach to find (4) for t => 0 in the network in Fig. P7.15. Figure P7.15 SOLUTION: tee”: & fo7}e C4, = ea = 6 fat} Plorist BT) = oA = ky tky EL bdt Pp Ryo 202? Uy [B\ ven at at” UE 4 ot C2, thw => teh ods eed RS  530 trwin, Basic Engineering Circuit Analysis, 8/E 7.76 Use the differential equation approach to find i(7) for ¢ > O in the circuit in Fig. P7.16 and plot the response including the time interval just prior to switch movement. Figure P7.46é SOLUTION: ~ beon bbe) = lima = ier) feet bs -l, = ~lamA tre Ldt ‘ Pee G a dt fay yt. Gt bt ¢ & + Gt oO Pek the VE oy take ls G0 zB . Lime) be Utot}~% = -Iem& wt a ~)0%t io i lt) + lhe mA | 5s | Op remem ¥ f(y) : ° t - 3 q = a / -s i a" Chapter Seven: First- and Second-Order Transient Circuits 531 ?.V% To the circuit in Fig. 7.17. find 7,(¢) fort > 0 using the differential equation approach. 40 r= 60 t C} sel Aga a Z £ Beg g, ¥ - = é sav) p, 2120 ui ig(d) tLe) Figure P7177 SOLUTION: ~ ~ en OT tat = Lt. fey = Et Pe. Hl 8S. yds ZA a ty iot'S bd 2 Ug (Bre) Fee ye dt eB) At A , te, tee Le + + & ok Ow Chapter Seven: First- and Second-Order Transient Circuits 533 7.1% In the network in Fig. 7.19, find i,(7) for ¢ > 0 using the differential equation approach. © 60 r=0 —_ af he , a IQ) fh # 2HE t 408 Or “120 Se L Ree Figure P7.19 SOLUTION: é wp > 534 Irwin, Basic Engineering Circuit Analysis, 8/E 7.2 Use the differential equation approach to find i(r) for : > Q in the circuit. in Fig. P7.20 and plot the response mcluding the time interval just prior to switch movement, PS¥ 3kO f= 0 O % ev) » 2K Chama “O.2mH ~~ i(f) Figure P7.20 SOLUTION: trot bios Be 2m = Ry fone Zak 4s Ryd thd gt to yo . tft oe be : Ce Vee O.! pes Kye #4, = mA Kee Uiot\-G 26 . ci) Cr — a ‘ ile = mA | wo 4 i —— 536 Irwin, Basic Engineering Circuit Analysis, 8/E Y.2e Use the differential equation approach to find i,(¢) for > ( in the circuit in Fig. P7.22 and plot the response including the time interval just, prior Lo opening the switch, 2H r=0 A XE / 12¢ i & 20 Figure P7.22 SOLUTION: sig tot 2a > oA Lares i Pgs Padiite 47,602 i ree Kae Ty lot kya -& A va CRY Chapter Seven: First- and Second-Order Transient Circuits 537 7.23 Using the differential equation approach. find i,(t} for f > O in the circuit in Fig. P7.23 and plot the response including the time interval just prior to opening the switch, ei 8 » S120 bes etry 12V ©) ° g(t) Figure P7.23 SOLUTION: By snadee Beanpr trad | ie lO bre iods Gas) Ye. , . “4 pz. esol: ty (othe -G tet a- Z A C7  ve £ OF) 1 I Chapter Seven: First- and Second-Order Transient Circuits Ym Use the differential equation approach to find i(t) for t > Qin the circuit in Fig. P7.25 and plot the response inchiding the time interval just prior to opening the switch. 5 kO \ T=0 2kO 1kO fy 5 “fh z ge i) 3 | bo - : Sy . (Hs mA ikX Se Soko © 1 kA | 1mH Figure P?7.25 SOLUTION: bso thos Sma trot: Plots blo) som A pee: be Ul ven & Vie kde [+ em ak tr u_ Bt Pa at dC fake ye ef = -~ LFo bee 2 + fee Ales rte] de (tar @Q\b Rae 3m vs Eyt hye Ys i (bg tla) _ gus Geo ka Ulor-& = SeA Rae  irwin, Basic Engineering Circuit Analysis, 8/E 7.28 Use the step-by-step technique to find 7,() for ¢ > 0 in the network in Fig. P7.28. ee Sar SP 2KO AKO PP Ad a a, io] wv 200 WF 9 BE 6KO Figure P7.28 SOLUTION” = ve) = rue SO Ai ak By am bage deinsaom > Fe D> 12 (Ban) ty pe £ a, bo7 = an a ae Akio = ev iy Lots lO AL, = ZS A Ky thy § fo) de by hole Bef am A ak, po Ryree ToC Reg Regs Ry My phe Vor O2b7 Ss Chapter Seven: First- and Second-Order Transient Circuits 543 my Y.ae Use the step-by-step method to find »,{7) fort > 0 in the network in Fig. P7.29, 12 mA Gd g 26 kO BK S Volt) Figure P7.29 SOLUTION wuss w+ ewe tea PPT wt Vy, tole Teno? CER) 2 ev = we © an “ay os) ag — teot — Pd i b Wy (oO = BE e lgtks, ama) BF sw te Alor) L. __t dt 4 sae — rT Wy aden kK, Me (oe) .e Uy le) qd me 4 oa ~ 4 544 Irwin, Basic Engineering Circuit Analysis, 8/E 7.00 Use the step-by-step method to find i,(¢) fort > O in the circuit in Fig. P7.30. 4kQ 4ko) mn ‘ p 200 we 5 i I “ t t 0 4 Pa | + aay ©) r=a 3 ako BoD | ig(t) _ Oo Figure P7.30 SOLUTION: 7 faye ve “EE thik +h € : ey a ao WY Mae Sa Vertg = 2ths nw BV E eB Peoda = Bytes zwv(Ce Bev. ° F4 bus o* / N's ige SF kyo Paks 2be Sarees Poe Bi z4v 4) . 7 mF a Lop tee pig ha Ime kyr Pr thy ae = hes pope Ts Reg eye Be eye 4he a Sp, ky Re tts, i — ta OFS  % poy —e faF GP Seng = Pye Bets 5 38k " __ “Rex e+ Chapter Seven: First- and Second-Order Transient Circuits 547 7.03 Find v(t) for: > 0 m the network in Fig. P7.33 using the step-by-step method. ©&5 2kO x fs AAS y O Fi SS 50 pF 24V ee S DKA Sennen + Figure P7.32 SOLUTION: “4g @)e Kat et tao dL ‘ (0 g0 % los Z24y g ¥ +h Ve t)eav cae Ly  Zb= ty Co + teste) ~ (8g +24 ” t (hyrhetly) = (re i i | 2 = Uz (esr) v2¥e 0 ey Uy by = ty Ceyeey Sein 4, =-f2 mA Vs Coty? ty ho i yy, lot) s FF Vo kth, ° Ww Vy bO=o sk, Win e43e — “| | fo  a2 oa 4 2 ° “ we 550 Irwin, Basic Engineering Circuit Analysis, 8/E 7.36 Use the step-by-step technique to find v,(¢) for > 0 in the circuit in Fig. P7.36, % oO 50 pF volt} ¢ O A Bs 6KO > Figure P7.36 SOLUTION: 4, ct). Kae, ot a Ye 5 : Cee Ve lots 12h. ey : a ate, : RF / ' ae en Fue GYR = Shee : Ae CO bP na kyle 2 Ay Chapter Seven: Firsi- and Second-Order Transient Circuits 551 P.27 Find i,() for ¢ > 0 in the network in Fig. P7.37 using the step-by-step method. ©© ‘ 12¥ AA Cr ie 2kO 4 n 200 pF f=0 BS 2kO ee 2k @ 2 4ko 5 | il ‘Figure P7.37 eR _ @ aigco ety = 1ehy y= yrdye aif L Repnity 2 “ ] aie Gs BV a5 BY ee a a oa ine akte les fore . 3. 88 mar Bt, “Pe tRa Chapter Seven: First- and Second-Order Transient Circuits 553 7.39 Use the step-by-step method to find i,(2) fort > 0 in the circuit in Fig. P7.39. S 6a ; 2H @) 24V i | yee ys Figure P7.39 aD AA oo’ bod: 4g. +h 554 trwin, Basic Engineering Circuit Analysis, 8/E 7.40 Find i,(¢) for ¢ > 0 in the network in Fig. P7.40 using the step-by-step method. © zg <= 60 22S 62 Figure P7.40 SOLUTION: yc). wake TE ge ‘ LI te Ke UAE 2 3 Tey Hh - sa Tt tbo) : , LL Vd ye his}: hx 2 da Ryrks teot % Ltage Ay — ame r ae aL io! fn Ge Fe” | Le ame L, fo 2-4 2 GIG Cytol o= Ky Chapter Seven: First- and Second-Order Transient Circuits 55 7A Find i,(1) for r > 0 in the network in Fig. P7.41 using the step-by-step method. "= Figure 7.41 SOLUTION: Las Belge ~ tA Le TFs erg | ; in CD at Vou avG al tg toes) po i wl th I (of = 0 ahyrk, Ly e)e12 2.dmde Ee 4 -Z Surg it | 4 Hs Z24¢-2.¢e mitt | | ee Chapter Seven: First- and Second-Order Transient Circuits 557 Use the step-by-step method to find v,(7) for ¢ the network in Fig. P7.43, PS¥ => O in ey an 60 4 . 7=0 2 2H Py = 20 Vo) 12V - | 2) Figure P7.43 SOLUTION: ay se 4g ee Cis 2 b= oO” bees Al £5 [a — Ae A rR, I + [ % Po alt ; = zw Ede (4) =S tector) 2s 4 ~~ ~~ Ry) 7 ‘ aod — $5 ne) een flee el 2A Uy Os 2 hg emt Vo (e032 0 2k R; = ths : Te aye fe 1 : [ ot Ri ah ~ “lg” ; S AJgs ~ we, v \ Reg? * Lo on oo — Lege Bahr = 2 558 irwin, Basic Engineering Circuit Analysis, 8/E i 7,44 Find i,(t) for ¢ > 0 in the network in Fig. P7.44 using the step-by-step method. 12kO Ae ey C 12mA = 6 KO Figure P7.44 FION: . > ~~ SOLUTION. yp ree Kale tao hese Saneag ten: ae . : “| Lg Cael n = dim dee i be J@ dA jd So, kp =O 1 Lot}=tmA pony = Ky ty, Tae -4ma | po - Chapter Seven: First- and Second-Order Transient Circuits 559 7.45 Use the step-by-step tectmique to find v,(1) fort > 0 in the circuit in Fig. P7.48. oC \° a | Figure P?_4§ SOLUTION: Vite Bre ~tfe hye bo Hip 8 re Ry = @i4by> BL BY Pek [7 . de tLe “Exo "Ry 2.414 Perky roe = "& r een , Lid Bye) oP LOD] tye Be te) 28 yoo Y Oh a ee. aby Zz 244 CO > as § te Vg « & ys [. rag Netord= 2.41 = TLL4V = iy eke