EECE 210C Lecture, Fall 2008

8 Natural and Step Responses of RLC Circuits Assessment Problems 1 1 = = 500nF AP 8.1 [a] (RCP ~ LC’ therefore C = 500n 1 [b] a = 5000 = ORO therefore C=1,F 3) (106 $12 = —5000 + ys x 106 — ony") = (—5000 + 75000) rad/s [c] yaa = 20,000, therefore C = 125nF S12 = [-40 + (40)? — 20| 10°, $1=—5.36krad/s, sy = —74.64krad/s AP82 i, = —t+— | " [140-5000 4 26¢-20.0002) dr + 30 x 10-8 50 x 10-3/ 0 —14e—5000¢ t 26 e-20,000t t = 2 10-8 of —5000 lo’ —20,000 } + 30x 10 = 56 x 10-3(e~890 — 1) — 96 x 10-8(e-20.000" _ 1) + 30 x 1078 [56e—5000% _ 56 — 26e~20,000 + 26 + 30] mA = §6e75000 _ 2G e-20,000t m A, t>0 AP 8.3 From the given values of R, L, and C, s; = —10krad/s and sz = —40 krad/s. [a] v(0~) = v(0*) =0, therefore ig(0t) =0 8-1 8-2 CHAPTER 8. Natural and Step Responses of RLC Circuits [b] ic(0*) = —(iz(0*) + in(0")) = -(-4+0) =4A du(0t) aay dv(0*) 4 _ 847 J [ce] C a i,(0*) = 4, therefore a OF 4x 10°V/s [d] v = [Aye710 00 4. Ape 400004], t>0r + v(0*) = A, + Ad, =“ ) _ _10,000A; — 40,0004, Therefore A, + Ag = 0, —A, — 4A> = 40,000; A; = 40,000/3 V [e] Ay = —40,000/3 V [f] v = [40,000/3] [e720 — eA V, £0 1 AP 8.4 [a] aRG = 8000, therefore R= 62.52 10V « (Q+) = — = [b] tp(0") = 6259 160mA dv(0*) dt ig(O0t) = —(iz(0*) + ig(0*)) = —80 — 160 = —240mA =C dv(0t) — —240m dé Cc dv,(0*) dt Therefore 6000B2 — 8000B, = —240,000, By = (-80/3)V Therefore = —240kV/s [c] B, = v(0*) = 10V, = wg Bo _ aB, d [d] iL = —(ig + ic); ig = v/R; ig = OF v = e 810 cos 6000t — . sin 6000¢] V 1280 . Therefore ip = e~®[160 cos 6000¢ — sin 6000¢] mA ig = e780" 240 cos 6000t + = sin 6000¢] mA iz, = 10e78 [8 cos 60008 + = sin 6000¢] mA, t>0 1\2 1 ~~ 108 1 AP 8.5 [a] (sz) = TE= a therefore aRG 7 500, R=1002 [b] 0.5CV2 =12.5x 107%, — therefore Vo = 50V [c] 0.5217 =12.5x 103, Ip =250mA Problems 8-3 du(ot [d] Do = v(0*) = 50, uC ) = D, _ aD» 50 ty — 00 _ in(0*) = Fe = 500mA Therefore ic¢(0*) = —(500 + 250) = —750mA + -3 Therefore 2%") _ _750 x 19 = ~75,000V/s dt Cc Therefore D, — aD, = —75,000; a= —_ = 500, D, = —50,000 V/s fe] v = [50e75° — 50,000te °°] V in = z = [0.5e7°°% — 500te™"] A, t > OF . Vo 40 y+) — —— See AP 8.6 [a] ip(0*) = R = 500 0.08 A b] ic(0*) = I — ig(Ot) — 4, (0+) = —1 — 0.08 — 0.5 = —1.58A di,(0t) Ve 40 =~ = — =62. 5 dt L = 064 7 8254/8 1 1 . d] a= RO 7 1000; le 1,562,500; 51,2 = —1000 + 7750 rad/s e] ip = ip + Bie coswat + Bye sin wt, ij =I=-1A it (0*) = 0.5 = if + Bi, therefore BY, =1.5A diy,(0*) dt Therefore iz, (¢) = —1 + e70"[1.5 cos 750¢ + (25/12) sin 750¢]A, t2>0 Ldiz, dt AP 8.7 [a] i(0*+) = 0, since there is no source connected to L for t < 0. b ne(0*) = ve(0) = ( 15k = 62.5 = —aB, + waBs, therefore By = (25/12) A f] v(t) = = 40e7 10" [cos 750t — (154/3) sin 7504]V t > 0 15k+9k di(0*) di(0*) + c] 50+ 802(0T) + L a a 1 d] a = 8000; 55 = 100 x 10°; 8. = —8000 + j6000 rad/s e] i= is +e ™ |B) coswyt + By sin wat]; if =0, (Ot) =0 di(0*) dt Therefore Bl =1.67A; i1=1.67e%sin6000tA, t20 ) (80) = 50V = 100, = 10,000 A/s Therefore Bi = 0; = 10,000 = —aB! + waBy, 8-4 CHAPTER 8. Natural and Step Responses of RLC Circuits AP 8.8 u,(t) = vp + eo ™ [Bi coswat + By sin wat], vz = 100V dv,(0*) _ +) . ve(0*) = 50V; di 0; therefore 50 = 100+ By BL =-50V; 0 = —aBi+waB, a 8000 Therefore By = “Bi = (Fos) (—50) = —66.67V Therefore v,(#) = 100 — e~ 8 [50 cos 6000¢ + 66.67 sin 60004] V, Problems 1 10° 1 10° 2_ = ____=j0° “o= TG = Gas) 1° 81,2 = —12,500 + \/ (1.5625 — 1)108 = —12,500 + 7500 8, = —5000 rad/s S82 = —20,000 rad/s [b] overdamped [e] wa = we — 0? a? = w? — w2 = 10° — 36 x 10° = 0.64 x 108 a = 0.8 x 10* = 8000 1 _ goo, R= e250 2RC , “7 (16,000)(8) . [d] s; = —8000 + 76000 rad/s; 82 = —8000 — 76000 rad/s 1 fe] a = 10* = R=) = 62500 2RO 20 (104) P82 fa] —a+ \/a? —w? = —5000 —a — \/a? —w?2 = —20,000 —2a = —25,000 a: = 12,500 rad/s 1 10° 2RC 2R(0.05) 12,500 R=8002 2\/a? — w? = 15,000 A(a? — w?) = 225 x 10° *. Wo = 10,000 rad/s 1 2_ 198 — we = 10° = 7G; 1 {b] tz = wo) = —6.25e7 5008 4 95 e720,000 mA, = t > OF ott) = 1,25 e700 _ 29 ¢-20:000 mA > OF y= —(ig +ic) = be ig=C 50008 __ 5 e—20,0008 mn A | t>0r P83 = [a] a = 4000; wa = 3000 wg = \/w? — a? wi =wita? =9 x 10° + 16 x 10° = 25 x 10° 1 6 7g = 25x 10 1 U = Bs x 10(50 x 10-8) ~ OSH = 800mH 1 Ib] ¢ = oa 9 r= 2 = = 25000 2aC (8000)(50) [ce] Vo = v(0) = 125V Problems 8-5 8-6 CHAPTER 8. Natural and Step Responses of RLC Circuits [d] Io = ix (0) = —in(0) — ic(0) Vy 125 in(0) = 3 = 55 * 10-3 = 50mA . du ic(0) = C5, (0) a = 125{e~4[ 3000 sin 3000¢ — 6000 cos 30004] — 4000e~4% [cos 3000t — 2.sin 3000#] “(0) = 125{1(—6000) — 4000} = —125 x 104 of) = —125 x 104(50 x 10-°) = —6250 x 107° = —62.5mA I, = —50 + 62.5 = 12.5mA fe] “ = 125e740%[5000 sin 3000¢ — 10,000 cos 3000¢] = 625 x 10%e-4 [sin 3000¢ — 2 cos 30001] oF = 31,250 x 107-Fe4™ (sin 3000¢ — 2 cos 30004) ig(t) = 31.25e~4™ (sin 3000 — 2.cos 3000t) mA ip(t) = 50e~ 4 (cos 3000¢ — 2 sin 3000¢) mA in (t) = —in(t) — tc(t) = e100 (12.5 cos 3000t + 68.75 sin 3000¢)mA, t>0 CHECK: =z = {—4000e~ 4012.5 cos 3000t + 68.75 sin 30001] +e7 400 _37.5 x 10° sin 3000¢ +206.25 x 10% cos 3000¢] x 107% = 67 1000/1 56.25 cos 3000¢ — 312.5 sin 30004] Lee = e~ 400/125 cos 3000¢ — 250 sin 3000¢] = 125¢~40 (cos 3000¢ — 2 sin 30002] V i P84 [al (sha) | _ (4000) 2RC LC 1 1 oRO 4000 10° = ——_____- = 1] R= Tep00)(12.5) ~ LOK v(0) = D2 = 25V 25 i = — =2.5mA in(0) = = 25 ic(0) = -2.5 -5 = -7.5mA du -7.5 x 1073 5 D, = —6 x 10° + 4000(25) = —5 x 10° V/s [b] v = —5 x 10°te 40 4. 25e~ 400% “ = [20 x 10°t — 6 x 10°]Je40™ ig = cf = 12.5 x 1079/20 x 108t — 6 x 10°Je“40" = (25,000t — 7.5)e4" mA, +t>0 P85 [a] 2a = 200; a = 100rad/s 2\/a? —w? = 120; — w, = 80rad/s 1 1 C= dar ~ 300200) ~ 22K 6 L=4 10 _6.25H = WC (80)2(25) ig(0*) = Ay + Ap = 15mA dig diy, . din _ ata ta dic(0) __diz(0) __ dé (0) dt dt dt Problems 8-7 8-8 CHAPTER 8. Natural and Step Responses of RLC' Circuits di(0) 0 _ dt 7 6.257 A/S din(0) _1dv(0)_ 1ic(0) 15x 10% 3A/s dt R dt RC (200)(25x 10-6) — dic(0) _ i 3A/s 160A; + 40A_ = 3 4A, + Ag+ = 75 x 107%; <. A, =20mA; Ag = —5mA "tig = 200710 — 5e 4! mA, t>0 [b] By hypothesis v = Age! 4. Aye, t>0 v(0) = Ag + Ay = 0 dv(0) _ 15x 107% dt 25 x 10-6 = 600V/s v = —5e7 160 4 Ge AV, t>0 [ec] ig(t) = 500 = ~25e716 4 25e-4% mA, —- t > OF [d] i, = —tz — tc iz, = Se 16% _ 20e-4 mA, t>0 P86 [a] in(0) = aia = 45mA i,(0) = -30mA ic(0) = —i,(0) — éa(0) = 30 — 45 = —15mA 9 [b] a= 2. =? _ = 95,000 2RC (4000)(10) 9 1 _ (10°) (10°) °~ LC (250)(10) $1,2 = —25,000 + \/6.25 x 108 — 10°(4) = —25,000 + 15,000 8, = —10,000 rad/s; s2 = —40,000 rad/s w =4~x 108 v = Aje™ 10,000 4. 4, 40,0008 P 8.7 Problems v(0) = A; + A = 90 —15 3 “(0) _ ~10'A, — 4A, x 10* = 15 x 10™ =-15~x 10°V/s 10 x 10-° —A — 4Ag = —150 —3A_ = —60; Ag = 20; A; = 70 v = 70e7 10,000 4+ 20 ¢~ 400008 V7 t>0 d [ce] ic = ce = 10x 10-[—70 x 104e710.00% _ 89 x 104e— 40.000) —7e~10,000¢ _ 8e—40,000¢ mA i ip = 35e7 100008 4. 4 Qe 40,0008 mm A iy, = ig — tg = —28e7 1000 _ Qe40,000% mA t>0 1 10° a=— = ——~ = 2x 10* DRC ~ (000)(10) ~ 2 * 1° a? =4 x 108; v= uP Critical damping: v= Dyte~™ + Doe~** . 90 ir(0*) = 2500 = 36mA ic(0*) = —[i,(0*) +zp(0*)] = —[-30 + 36] = -6mA v(0) = D2 = 90 dv _ ie(0) _ -6 x 10-3 (0) = Di = Da = Or = Fo 109 =-6x 10° D, = aD» — 6 x 10° = (2 x 10*)(90) — 6 x 10° = 120 x 104 v = (120 x 104#+ 90)e™"' V,_ t D0 8-9 8-10 P 8.8 P 8.9 CHAPTER 8. Natural and Step Responses of RLC' Circuits 1 3 x 10° 2RC (25,000)(10) 12,000 1 8 To 74x10 $12 = —12,000 + 716,000 rad/s *, response is underdamped v(t) = Bye 1? cos 16,000¢ + Boe 1? sin 16,000¢ 90 y+) — = ; i +) — ss . v(0t) = 90V = B,; in(0T) (13,500/3) 21.6mA ic(O*) = [-i1(0*) + én(0t)] = —[-30 + 21.6] = 8.4mA dv(0t) 84x10 | al) = —12,000B, + 16,000B, = 840,000 or — 3B, +4Bo = 210; -. By =120V v(t) = 90e7 12000 cos 16,000¢ + 120e7 12 sin 16,000t V, t>0 a = 2000/2 = 1000 1 10° R= 56 = Bm) 727-789 v(0t) = —24V -24 . +) =— ip (0*) 78 864mA a = 2400-2 +4 21,60007180% + aut ) _ 2400 + 21,600 = 24,000V/s ic(Ot) = 18 x 10-6(24,000) = 432 mA ip (0*) = —[in(Ot) + ic(0*)] = —[-864 + 432] = 432mA