Chaper 8 he Complee Response of R and RC Ciruis Exerises Ex 8.3-1 Before he swih loses: Afer he swih loses: 2 = = 8 Ω so = 8 0.05 = 0.4 s. 0.25 herefore R ( ) Finally, 2.5 ( ) = o + ( (0) o ) = 2 + V for > 0 v v v v e e 157
Ex 8.3-2 Before he swih loses: Afer he swih loses: herefore Finally, 2 6 R = = 8Ω so = = 0.75 s. 0.25 8 1 1 1.33 i ( ) = is + ( i(0) is ) e = + e A for > 0 4 12 Ex. 8.3-3 A seady-sae before = 0: i = 10 10+40 + 12 16 40 10 = 01. A 158
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 40 2 2 2 Finally: i() = (0.1 0.3)e + 0.3= 0.3 0.2e A Ex. 8.3-4 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
Ex. 8.3-5 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 = 25 45 =5 9 i () = 0.2 (1 e 1.8 ) A d 1.8 1.8 1.8 Finally: v() = 40 i()+25 i() = 8(1 e ) + 5(1.8)e = 8+e V d Ex. 8.3-6 < 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 =.1 640 = 1 6400 160
So 6400 i () = 406.25e + 93. 75 ma Finally d 6400 6400 6400 v() = 400 i() + 0.1 i() = 400 (.40625e +. 09375) + 0. 1( 6400) (0. 40625 e ) = 37. 5 97. 5e V d Ex. 8.4-1 3 6 = 210 110 = 2 10 ( )( ) ( ) 2 3 v () = 5+ 1.5 5 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. 8.4-2 i (0) = 1mA, I s = 10mA 500 i 9e ma R h = = 500 *K 16 500 = 10 Ω, v () = 300 i = 3 27. e V R 16 ( )K 500 We require ha v R = 1.5V a = 10ms = 0.01 s ha is 15. = 327. e 5 500 ( 001. ) e 1.5 3 = = 0.555 27. 5 = ln (0.555) =. 588 5 = 0.588 = 8.5 H 161
Ex. 8.6-1 0 < < 1 /(1)(.1) /RC v() = v( ) + Ae where v( ) = (1A)(1 Ω )= 1V v() = 1+Ae = 1+Ae 10 + 10 Now v(0 ) = v(0 ) = 0 = 1+ A A = 1 v() =1e V > 1 =.5s v() = v( )e = v(.5)e 1 1 10(.5) Now v(.5) = 1 e =.993V.5 (1)(.1) 10(.5) 10(.5) v() =.993e V Ex. 8 6-2 < 0 no soures v(0 ) = v(0 ) = 0 0 < < 1 + /RC v() = v( ) + Ae = v( ) + Ae 5 7 2 10 (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
Ex. 8.6 3 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. 8.7-1 v s = 10sin 20 V dv dv KV a: 10sin20+ 10.01 + v = 0 + 10v = 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 = 20 1 2 f 1 2 Complee response: v() = v () + v () + n 10 v() = Ae 40 os 20 + 20 sin 20 Now v(0 ) = v(0 ) = 0 = A 40 A = 40 10 v() = 40e 40 os 20 + 20 sin 20 V f n 10 Ex. 8.7-2 5 i s = 10e 5 KC a op node: 10e + i + v / 10 = 0 Now v =.1 di di 5 + 100i = 1000e d d Naural response: s + 100 = 0 s = 100 i () = Ae 5 f 5 5 5 Be Fored response: ry i () = Be & plug ino D.E. 5 B = 10.53 + 100 Be = 1000e 100 Complee response: i() = Ae + 10.53e + Now i(0 ) = i (0 ) = 0 = A + 10.53 A = 1053. 5 100 i() = 10.53 (e e ) A 5 n 100 163
Ex. 8.7-3 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 16 1 6 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
P8.3-2 We know ha 16 1 6 16 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 = 0.5 10 1333 = 0.67ms Finally v = 4e.67 +4 V 16 6 4 9 where has unis of ms. P 8.3-3 Before he swih loses: 165
Afer he swih loses: 6 = = 3 Ω so = 3 0.05 = 0.15 s 2 herefore R ( ) Finally, 6.67 ( ) = o + ( (0) o ) = 6 + 18 V for > 0 v v v v e e. P 8.3-4 Before he swih loses: Afer he swih loses: 166
herefore Finally, 6 6 R = = 3Ω so = = 2 s. 2 3 10 0.5 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, = 20 10 4 10 = 0.08 s. 12.5 C ( ) = o + ( (0) o) = 10 10 V for > 0 v v v v e e Finally, v = v = e > 12.5 0 ( ) C ( ) 10 10 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 3 20 10 3 4000 Nex, s s ( ). i ( ) = i + ( i(0) i) e = 0.5 10 1 e A for > 0 d vo = i = e > d Finally, () () 4000 5 10 V for 0 167
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 P8.3-10 < 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
P8.3-11 a = 0 (seady-sae) + 4 9 4 9 4 9 v 0 = 0 i 0 = 6A = i 0 for > 0 16 16 (R ) 20 i = i 0 e = 6e A P8.3-12 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 = 6.74 10. V Now deermine C from disharging ondiion ( o ) 0.5 v v e C Rbulb 6.74 10 3 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 <100 10 A R >60 kω R hen for he harging k.. 993 = 1e 5/ RC 496. = 5 RC R = and see ha R ~ 100k Ω > 60kΩ. 5 4.96 (10 ) = 100.8 k Ω 5 1 6 169
P8.3-13 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 12 24 () 24 and i = 2e > 0 () () 24 Finally v = 9 i = 18 e >0 h P 8.3-14 Before he swih opens, v () v ( ) C = 0 V 0 = 0 V. Afer he swih opens: C 10 = = 20 kω so = 20 10 4 10 = 0.08 s 3 0.5 10 3 6 herefore R ( )( ) Nex, 12.5 C ( ) = o + ( (0) o) = 10 10 V for > 0 v v v v e e. Finally, v = v = + e > 12.5 0 ( ) C 10 10 V for 0 170
Seion 8-4: Sequenial Swihing P 8.4-1 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) = 5 + 5 V for 0 < < 1.5 s v v v v e e A =1.5 s, herefore Finally ( ) 0.5 1.5 = + e. For 1.5s <, v o = 10 V and R = 8 Ω so 8 0.5 4 s v(1.5) 5 5 = 7.36 V 1.5 0.25( 1.5) = o + o = < = =. v ( ) v ( v(1.5) v ) e 10 2.34 e V for 1.5 s + e < < v () = 0.25( 1.5) 10 2.34 e V for 1.5 s < 0.5 5 5 V for 0 1.5 s P 8.4-2 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 6 0.5 ( ) = 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
1.5 0.667( 1.5) = s + s = < i ( ) i ( i(1.5) i) e 3 0.64 e V for 1.5 s Finally 1.5 0.667( 1.5) = s + s = < i ( ) i ( i(1.5) i) e 3 0.64 e V for 1.5 s + e < < i () = 0.667( 1.5) 3 0.64 e A for 1.5 s < 0.5 2 A for 0 1.5 s P8.4-3 = 0 (seady-sae) KV : 52+18i+ (12 8 )i=0 i(0 ) = 104 39 A 6 + i = i =2A = i (0 ) 6+ 2 0 < < 51ms i () = i (o) e R = 6 12 + 2 = 6Ω i () = 2e 6 A (R ) 6 6 i() = i() = 2 3 e A 6+ 12 > 51ms i () = i (51ms) e ( R )(.051) i (51ms) = 2e = 1.473 6(.051) i () = 1.473 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 100 400 + a = 0, q + q = 300µ C 100 400 Now using q = CV (100µ F) (V) + (400µ F) (V) = 300µ C V = 0.6 V 172
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 P8.4-5 20 40 V o = V = (40 ) 5i 1= 20 5 = = 15 V 20+20 40 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%: 500 500 500 ) A 500 for e =.01 or 500 = 4.605 = 9.2ms 3 = R = 15 10 = 2ms 75. P8.4-6 5 =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 5 100 kω 10 1 / 173
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 1 1+4+2+1 7 = 7 A 8 (0 ) = v (0 ) = 2i = 7 / 4 V + 2 0 < <.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 07. 07.. 7 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 + 7 10 v = 0 >.35 ( 07. ) (. 35) v (. 35) = 7 4e = 1.37V now = v = v x KC: v + 5i + v 4 2 + i = 0 (1) also : i = 0.2 dv d (2) (.35)5 8.625(.35) From (1) & (2) dv d + 5 8 v = 0 v ( ) = v (.35) e = 1.37 e 174
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 2 1 1 2 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 1 2 2 h When R = 4k Ω and R = 1k Ω, hen A < 5 is required o guaranee sabiliy. 1 2 ( ) 175
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 1 2 2 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 1 2 2 RR 1 2 = R +R +R B 1 2 2 h 1 2 when R = 6k Ω and R = 3k Ω, B > 3is required o guaranee sabiliy. 1 2 2 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
Seion 8-6: he Uni Sep Response P8.6-1 10u() 4 = % & ' 10( 0) 4=4 < 0 10(1) 4= 6 > 0 < 0: > 0: P8.6-2 6u( ) + 4u() = 6(1)+4(0)=6 < 0 % &' 6(0)+4(1)=4 > 0 < 0: > 0: 177