Chapter 8 The Complete Response of RL and RC Circuits

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1 Chaper 8 he Complee Response of R and RC Ciruis Exerises Ex Before he swih loses: Afer he swih loses: 2 = = 8 Ω so = = 0.4 s herefore R ( ) Finally, 2.5 ( ) = o + ( (0) o ) = 2 + V for > 0 v v v v e e 157

2 Ex Before he swih loses: Afer he swih loses: herefore Finally, 2 6 R = = 8Ω so = = 0.75 s i ( ) = is + ( i(0) is ) e = + e A for > Ex A seady-sae before = 0: i = = 01. A 158

3 Afer = 0, he Noron equivalen of he irui onneed o he induor is found o be 20 1 = = Ω = = so I s 0.3 A, R h 40, = R h Finally: i() = ( )e + 0.3= e A Ex A seady-sae for < 0 Afer = 0, replae he irui onneed o he apaior by is hèvenin equivalen so V = 12V, R = 200 Ω, = R C = (200)(20 10 ) = 4 ms o h h 4 Finally: v() = (12 12)e + 12 = 12 V 6 159

4 Ex Before = 0, i() = 0 so I o = 0 Afer = 0, replae he irui onneed o he induor by is Noron equivalen So I = 0.2 A, R = 45 Ω, = s h R h = =5 9 i () = 0.2 (1 e 1.8 ) A d Finally: v() = 40 i()+25 i() = 8(1 e ) + 5(1.8)e = 8+e V d Ex < 0: ( ) i 0 = 0.5 A > 0: Replae he irui onneed o he induor by is Noron equivalen o ge Coninued I s = 93.75mA, R = 640Ω, = h R h = =

5 So 6400 i () = e ma Finally d v() = 400 i() i() = 400 (.40625e ) ( 6400) ( e ) = e V d Ex = = 2 10 ( )( ) ( ) 2 3 v () = e where is in ms v (1) = 5 3.5e 2 = 2.88V 1 So v () will be equal o v a =1 ms if v = 2.88 V Ex i (0) = 1mA, I s = 10mA 500 i 9e ma R h = = 500 *K = 10 Ω, v () = 300 i = e V R 16 ( )K 500 We require ha v R = 1.5V a = 10ms = 0.01 s ha is 15. = 327. e ( 001. ) e = = = ln (0.555) = = = 8.5 H 161

6 Ex < < 1 /(1)(.1) /RC v() = v( ) + Ae where v( ) = (1A)(1 Ω )= 1V v() = 1+Ae = 1+Ae Now v(0 ) = v(0 ) = 0 = 1+ A A = 1 v() =1e V > 1 =.5s v() = v( )e = v(.5)e (.5) Now v(.5) = 1 e =.993V.5 (1)(.1) 10(.5) 10(.5) v() =.993e V Ex < 0 no soures v(0 ) = v(0 ) = 0 0 < < 1 + /RC v() = v( ) + Ae = v( ) + Ae (10 ) where for = (seady-sae) apaior beomes an open v( ) = 10V 50 v() = 10 + Ae 50 Now v(0) = 0 = 10+A A = 10 v() = 10(1e ) V >, =.1s 1 1 v() = v(.1)e 50(.1) where v(.1) = 10(1e 50(.1) 50(.1) v() = 9.93e V ) = 9.93 V 162

7 Ex for < 0 i = 0 0 < <.2 >.2 + ( ) K *K KC: 5 v/2 + i = 0 also: v = 0.2 di d 10 i() = 5+ Ae i(0) = 0 = 5+A A = 5 10 so have i () = 5 (1e ) A di + 10i = 50 d 10(.2) i (.2) = 4.32 A i() = 4.32e A Ex v s = 10sin 20 V dv dv KV a: 10sin v = v = 100 sin 20 d d Naural response: s + 10 = 0 s = 10 v () = Ae Fored response: ry v () = B os 20 + B sin 20 plugging v () ino he differenial equaion and equaing like erms f yields: B = 40 & B = f 1 2 Complee response: v() = v () + v () + n 10 v() = Ae 40 os sin 20 Now v(0 ) = v(0 ) = 0 = A 40 A = v() = 40e 40 os sin 20 V f n 10 Ex i s = 10e 5 KC a op node: 10e + i + v / 10 = 0 Now v =.1 di di i = 1000e d d Naural response: s = 0 s = 100 i () = Ae 5 f Be Fored response: ry i () = Be & plug ino D.E. 5 B = Be = 1000e 100 Complee response: i() = Ae e + Now i(0 ) = i (0 ) = 0 = A A = i() = (e e ) A 5 n

8 Ex A urren i = v / 1 flows in he induor wih he swih losed. When he swih opens, i s anno hange insananeously. hus, he energy sored in he induor dissipaed in he spark. Add a resisor (say 1 k Ω aross he swih erminals.) Problems Seion 8.3: he Response of a Firs Order Cirui o a Consan Inpu P8.3-1 We know ha i = I I e + I o SC o SC where I o= i16 o and = R. In his problem o = 0 and I o = i(0)=3ma. H he Noron equivalen of is So R = 1333 Ω and I = 5mA. h 5 = 5 H so = = = 3.75 ms R 1333 () h s 3.75 Finally i = 2e + 5 ma > 0 where has unis of ms. 164

9 P8.3-2 We know ha v = V V e + V 0 o 0 where v = v and = R C, 0 C 0 h o In his problem, = 0 and v = v (0) = 8V. 0 0 he hèvenin equivalen of is so R h = 1333 and V o = 4 V. Nex, C = 0.5µ F so = = 0.67ms Finally v = 4e V where has unis of ms. P Before he swih loses: 165

10 Afer he swih loses: 6 = = 3 Ω so = = 0.15 s 2 herefore R ( ) Finally, 6.67 ( ) = o + ( (0) o ) = V for > 0 v v v v e e. P Before he swih loses: Afer he swih loses: 166

11 herefore Finally, 6 6 R = = 3Ω so = = 2 s i ( ) = is + ( i(0) is ) e = 2 + e A for > 0 3 P8.3-5 Before he swih opens, v () v ( ) o = 0 V 0 = 0 V. Afer he swih opens he par of he irui onneed o he apaior an be replaed by i's hevenin equivalen irui o ge: o 3 6 herefore ( )( ) Nex, = = 0.08 s C ( ) = o + ( (0) o) = V for > 0 v v v v e e Finally, v = v = e > ( ) C ( ) V for 0 P8.3-6 Before he swih opens, v () v ( ) o = 0 V 0 = 0 V. Afer he swih opens he par of he irui onneed o he apaior an be replaed by i's Noron equivalen irui o ge: o herefore 5 = = 0.25 ms Nex, s s ( ). i ( ) = i + ( i(0) i) e = e A for > 0 d vo = i = e > d Finally, () () V for 0 167

12 P8.3-7 < 0 Sine he inpu o his irui is onsan, he apaior will a like an open irui when he irui is a seady-sae: > 0 P8.3-8 Sine he inpu o his irui is onsan, he induor will a like a shor irui when he irui is a seady-sae: < 0 > 0 i = 2ma i = 4mA P8.3-9 Sine he inpu o his irui is onsan, he induor will a like a shor irui when he irui is a seady-sae: < 0 > 0 i = 2mA i = 2mA P < 0 Sine he inpu o his irui is onsan, he apaior will a like an open irui when he irui is a seady-sae: > 0 168

13 P a = 0 (seady-sae) v 0 = 0 i 0 = 6A = i 0 for > (R ) 20 i = i 0 e = 6e A P V 5 Assume he apaior is harged a = 5, i.e., = 1 e =. 993 V V 5 3 Similarly, assume he apaior is disharged when = e = V Now deermine C from disharging ondiion ( o ) 0.5 v v e C Rbulb e CRbulb C = 10 5 = = F = 10 µ F () ( ) o max max Now deermine a ondiion for R from harging irui a he insan 6V 6 v = 0 < A R >60 kω R hen for he harging k = 1e 5/ RC 496. = 5 RC R = and see ha R ~ 100k Ω > 60kΩ (10 ) = k Ω

14 P Firs, use soure ransformaions o obain he equivalen irui for < 0: for > 0: i = 2A 1 1 So I 2 0= 2A, I s = 0, R h = 3 Ω+9 Ω =12 Ω, = = = R () 24 and i = 2e > 0 () () 24 Finally v = 9 i = 18 e >0 h P Before he swih opens, v () v ( ) C = 0 V 0 = 0 V. Afer he swih opens: C 10 = = 20 kω so = = 0.08 s herefore R ( )( ) Nex, 12.5 C ( ) = o + ( (0) o) = V for > 0 v v v v e e. Finally, v = v = + e > ( ) C V for 0 170

15 Seion 8-4: Sequenial Swihing P Replae he par of he irui onneed o he apaior by is hevenin equivalen irui o ge: Before he swih loses a = 0 he irui is a seady sae so v(0) = 10 V. For 0 < < 1.5s, v o = 5 V and R = 4 Ω so = 4 0.5= 2 s. herefore 0.5 ( ) = o + ( (0) o) = V for 0 < < 1.5 s v v v v e e A =1.5 s, herefore Finally ( ) = + e. For 1.5s <, v o = 10 V and R = 8 Ω so s v(1.5) 5 5 = 7.36 V ( 1.5) = o + o = < = =. v ( ) v ( v(1.5) v ) e e V for 1.5 s + e < < v () = 0.25( 1.5) e V for 1.5 s < V for s P Replae he par of he irui onneed o he induor by is Noron equivalen irui o ge: Before he swih loses a = 0 he irui is a seady sae so i(0) = 3 A. For 0 < < 1.5s, i s = 2 A and R = 6 Ω so 12 = = 2 s. herefore ( ) = s + ( (0) s ) = 2 + A for 0 < < 1.5 s i i i i e e A =1.5 s, 0.5( 1.5) i(1.5) 2 e 12 = + = 2.47 A. For 1.5 s <, i s = 3 A and R = 8 Ω so = = 1.5 s. herefore 8 171

16 ( 1.5) = s + s = < i ( ) i ( i(1.5) i) e e V for 1.5 s Finally ( 1.5) = s + s = < i ( ) i ( i(1.5) i) e e V for 1.5 s + e < < i () = 0.667( 1.5) e A for 1.5 s < A for s P8.4-3 = 0 (seady-sae) KV : 52+18i+ (12 8 )i=0 i(0 ) = A 6 + i = i =2A = i (0 ) < < 51ms i () = i (o) e R = = 6Ω i () = 2e 6 A (R ) 6 6 i() = i() = 2 3 e A > 51ms i () = i (51ms) e ( R )(.051) i (51ms) = 2e = (.051) i () = e A 14(.051) P8.4-4 = 0 Assume V 1 = volage aross 10µF apaior = 3V 0 < < 10mS Wih R negligibly small, we may assume a sai seady-sae siuaion is obained in he irui nearly + insananeously ( = 0 ). hus wih boh apaiors in parallel, he ommon volage is obained by onsidering harge onservaion. a = 0, q 100µ F = CV = (100µ F) (3V) = 300 µ C q 400µ F = CV = (400µ F) (0) = 0 q = q + q = 300µ C o a = 0, q + q = 300µ C Now using q = CV (100µ F) (V) + (400µ F) (V) = 300µ C V = 0.6 V 172

17 10ms < < ls Combine 100µF & 400µF in parallel o obain wih V(10ms) = 0.6V 2(.01) (.01) RC (.01) (10 3 ) (5x10 4 ) v() = V(10ms) e = 0.6e v() = 0.6 e V P V o = V = (40 ) 5i 1= 20 5 = = 15 V for R, kill soure wih V = 0 Noe i = 1 2 A R = 1 v0 1 = 1(10 Ω ) 5( 1 2 A ) = 7.5 Ω R = 7.5Ω eq Fored response i = 2A naural: i = Be = Be oal : i = Be + 2 now i(0) = 0 B = 2 i() = 2(1e ime o 99%: ) A 500 for e =.01 or 500 = = 9.2ms 3 = R = = 2ms 75. P =RC=10 10 =.1s v (0) =5 V 6 ( ) K * K = Now 5/2 =v ( )=5e 1 1/ 1/ v () 5e e =.5 =.0693s i( ) = v( 1 ) 52 / 1 = = 25µ A kω 10 1 / 173

18 P8.4-7 = 0 1seady-sae6 + v(0 ) = v(0 ) = (2A)(5 Ω) = 10V 0 < < 100ms v() = v(0)e RC (5) (.01) 20 v() = 10e = 10e V > 100ms v()=v(100ms) e v(100ms)=10e (1) (4) (.01) 20(.1) 25(.1) = 1.35 V v() = 1.35e V P8.4-8 = 0 (seady-sae) i = 2 v = 7 A 8 (0 ) = v (0 ) = 2i = 7 / 4 V < <.35 Closing he righmos swih shors ou 1Ω in parallel wih 7A soure and isolaes 7A soure leaving vx KC a a : + 5i + i + v 4 2 = 0 (1) KC a b : v x 5 + v + vx i = 0 (2) 4 1 also: i =.2 dv d (3) Plugging () 3 ino (1) & (2) & hen eliminaing v yields dv d So v () = v (0) e = 7 4 e, i =.2 dv =. 245e d.7 So from (1) we have v () = 24i + 2v = 2.38 e x v = 0 >.35 ( 07. ) (. 35) v (. 35) = 7 4e = 1.37V now = v = v x KC: v + 5i + v i = 0 (1) also : i = 0.2 dv d (2) (.35) (.35) From (1) & (2) dv d v = 0 v ( ) = v (.35) e = 1.37 e 174

19 Seion 8-5: Sabiliy of Firs Order Ciruis P8.5-1 his irui will be sable if he hèvenin equivalen resisane of he irui onneed o he induor is posiive. R V h = I ( )K R1 i() = R +R I 1 2 R h V = Ri() 2 Ri() *K = (R R)R R +R hen R > 0 requires R > R. In his ase R = 400Ω so 400 > R is required o guaranee sabiliy. h 2 2 P8.5-2 he hèvenin equivalen resisane of he irui onneed o he induor is alulaed as R V h = I v() = RI R R +R AR V = I( R+R 1 2) ARI* = he irui will be sable when R > 0, ha is, R h R 1+R2 > 0 > A R 2 h h When R = 4k Ω and R = 1k Ω, hen A < 5 is required o guaranee sabiliy. 1 2 ( ) 175

20 P8.5-3 he hèvenin equivalen resisane of he irui onneed o he induor is alulaed as R V h = I V = R i() = R ( i() + Bi() + I ) i() = V R 1 2 R2 R +R +R B I = R 1 h R2 R +R +R B I he irui is sable when R > 0, ha is R +R R 1+R 2+R2B > 0 B > R RR 1 2 = R +R +R B h 1 2 when R = 6k Ω and R = 3k Ω, B > 3is required o guaranee sabiliy P8.5-4 he hèvenin equivalen resisane of he irui onneed o he induor is alulaed as R = h V I v() = RR 1 2 R +R I 1 2 R V = v () Av() ( )K *K he irui will be sable when R h = RR 1 2 ( 1 A) R +R 1 2 h > 0, ha is, when A<1. 176

21 Seion 8-6: he Uni Sep Response P u() 4 = % & ' 10( 0) 4=4 < 0 10(1) 4= 6 > 0 < 0: > 0: P u( ) + 4u() = 6(1)+4(0)=6 < 0 % &' 6(0)+4(1)=4 > 0 < 0: > 0: 177

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