Homework-8(1) P8.3-1, 3, 8, 10, 17, 21, 24, 28,29 P8.4-1, 2, 5

Transcription

1 Homework-8() P8.3-, 3, 8, 0, 7, 2, 24, 28,29 P8.4-, 2, 5 Secion 8.3: The Response of a Firs Order Circui o a Consan Inpu P 8.3- The circui shown in Figure P 8.3- is a seady sae before he swich closes a ime = 0. The inpu o he circui is he volage of he volage source, 2 V. The oupu of his circui is he volage across he capacior, v(). Deermine v() for > 0. Answer: v() = 6 2e.33 V for > 0 Soluion: Figure P 8.3- Here is he circui before = 0, when he swich is open and he circui is a seady sae. The open swich is modeled as an open circui. A capacior in a seady-sae dc circui acs like an open circui, so an open circui replaces he capacior. The volage across ha open circui is he iniial capacior volage, v (0). By volage division 6 v ( 0) = ( 2) = 4 V Nex, consider he circui afer he swich closes. The closed swich is modeled as a shor circui. We need o find he Thevenin equivalen of he par of he circui conneced o he capacior. Here s he circui used o calculae he open circui volage, V oc. 6 V oc = ( 2) = 6 V 6+ 6 Here is he circui ha is used o deermine R. A shor circui has replaced he closed swich. Independen sources are se o zero when calculaing R, so he volage source has been replaced by a shor circui. ( 6)( 6) R = = 3 Ω 6+ 6

2 Finally, oc ( oc ) Then R C / = = = 0.75 s.33 v = V + v 0 V e = 6 2 e V for > 0 P The circui shown in Figure P is a seady sae before he swich closes a ime = 0. Deermine he capacior volage, v(), for > 0. Answer: v() = 6 + 8e 6.67 V for > 0 Soluion: Before he swich closes: Figure P Afer he swich closes: 6 Therefore R = = 3 Ω so = 3( 0.05) = 0.5 s. 2 Finally, 6.67 v = v + ( v(0) v ) e = e V for > 0 oc oc

3 P The circui shown in Figure P is a seady sae before he swich opens a ime = 0. The inpu o he circui is he volage of he volage source, V s. This volage source is a dc volage source; ha is, V s is a consan. The oupu of his circui is he volage across he capacior, v o (). The oupu volage is given by v o () = 2 + 8e 0.5 V for > 0 Deermine he values of he inpu volage, V s, he capaciance, C, and he resisance, R. Figure P Soluion: Before he swich opens, he circui will be a seady sae. Because he only inpu o his circui is he consan volage of he volage source, all of he elemen currens and volages, including he capacior volage, will have consan values. Opening he swich disurbs he circui. Evenually he disurbance dies ou and he circui is again a seady sae. All he elemen currens and volages will again have consan values, bu probably differen consan values han hey had before he swich opened. Here is he circui before = 0, when he swich is closed and he circui is a seady sae. The closed swich is modeled as a shor circui. The combinaion of resisor and a shor circui conneced is equivalen o a shor circui. Consequenly, a shor circui replaces he swich and he resisor R. A capacior in a seady-sae dc circui acs like an open circui, so an open circui replaces he capacior. The volage across ha open circui is he capacior volage, v o (). Because he circui is a seady sae, he value of he capacior volage will be consan. This consan is he value of he capacior volage jus before he swich opens. In he absence of unbounded currens, he volage of a capacior mus be coninuous. The value of he capacior volage immediaely afer he swich opens is equal o he value immediaely before he swich opens. This value is called he iniial condiion of he capacior and has been labeled as v o (0). There is no curren in he horizonal resisor due o he open circui. Consequenly, v o (0) is equal o he volage across he verical resisor, which is equal o he volage source volage. Therefore v o ( 0) The value of v o (0) can also be obained by seing = 0 in he equaion for v o (). Doing so gives = V 0 e vo 0 = = 0 V Consequenly, V s = 0 V s

4 Nex, consider he circui afer he swich opens. Evenually (cerainly as ) he circui will again be a seady sae. Here is he circui a =, when he swich is open and he circui is a seady sae. The open swich is modeled as an open circui. A capacior in a seady-sae dc circui acs like an open circui, so an open circui replaces he capacior. The volage across ha open circui is he seady-sae capacior volage, v o ( ). There is no curren in he horizonal resisor and v o ( ) is equal o he volage across he verical resisor. Using volage division, 0 vo ( ) = ( 0) R + 0 The value of v o ( ) can also be obained by seing = in he equaion for v o (). Doing so gives vo = e = 2 V Consequenly, 0 2 = ( 0) 2 R+ 20 = 00 R= 40 Ω R + 0 Finally, he exponenial par of v o () is known o be of he form e where = R Cand R is he Thevenin resisance of he par of he circui conneced o he capacior. Here is he circui ha is used o deermine R. An open circui has replaced he open swich. Independen sources are se o zero when calculaing R, so he volage source has been replaced by a shor circui. so ( 40)( 0) R = 0 + = 8 Ω = R C = 8 C From he equaion for v o () 0.5 = = 2 s Consequenly, 2 = 8 C C = 0. = mf

5 P A securiy alarm for an office building door is modeled by he circui of Figure P The swich represens he door inerlock, and v is he alarm indicaor volage. Find v() for > 0 for he circui of Figure P The swich has been closed for a long ime a = 0. Figure P Soluion: Firs, use source ransformaions o obain he equivalen circui for < 0: for > 0: L So i ( 0 ) 2 A, 0, 3 9 2, 2 L = Isc = R = + = Ω = = = s R and i = 2 e > 0 L = = > 24 Finally v 9 il 8 e 0

6 Figure P P The circui shown in Figure P is a seady sae before he swich closes. The response of he circui is he volage v(). Find v() for > 0. Hin: Afer he swich closes, he inducor curren is i() = 0.2( e.8 ) A Answer: v() = 8 + e.8 V Soluion: Immediaely before = 0, i(0) = 0. Afer = 0, replace he circui conneced o he inducor by is Noron equivalen o calculae he inducor curren: I sc L 25 5 = 0.2 A, R = 45 Ω, = = = R 45 9 h.8 So i = 0.2 ( e ) A Now ha we have he inducor curren, we can calculae v(): d v () = 40 i () + 25 i () d = 8( e ) + 5(.8) e.8.8 = + e >.8 8 V for 0

7 P8.3-2 The circui in Figure P8.3-2 a seady sae before he swich closes. Deermine an equaion ha represens he capacior volage afer he swich closes. Figure P8.3-2 Soluion: For < 0, he swich is open and he capacior acs like an open circui because he circui is a seady sae. Consequenly, he curren in he 0 Ω resisor is 0A and so he volage across his v = 8 V. Immediaely before he swich opens we have resisor is 0 V. KVL gives v( 0 ) = 8 V. The capacior volage does no change insananeously so v v. 0 + = 0 = 8 V For > 0, he Thevenin equivalen of he par of he circui conneced o he capacior is characerized by 40 R = 0 40 = 8 Ω and, using volage division, v oc = ( 8) = 4.4 V A v oc 4.4 V = =, B v voc = 0 + = = 3.6 V and a = 5 = RC = = s LNAPTR, /3/07 P Consider he circui shown in Figure a and corresponding plo of he inducor curren shown in Figure b. Deermine he values of L, R and R 2. Answer: L = 4.8 H, R = 200 Ω and R 2 = 300 Ω. Hin: Use he plo o deermine values of D, E, F and a such ha he inducor curren can be represened as D for 0 i = a E + Fe for 0

8 Figure Soluion: From he plo D = i() for < 0 =20 ma = 0.2 A, and E + F = i(0+) = 20 ma = 0.2 A E = lim i = 200 ma = 0.2 A. The poin labeled on he plo indicaes ha i() = 60 ma when = ms = s. Consequenly ln a( ) = e a = = s Then 20 ma for e ma for 0 = 25 i

9 When < 0, he circui is a seady sae so he inducor acs like a shor circui. 24 R = = 200 Ω 0.2 As, he circui is again a seady sae so he inducor acs like a shor circui. 24 R R 2 = = 20 Ω = 200 R R = 300 Ω 2 2 Nex, he inducance can be deermined using he ime consan: = R R a = L 4.8 H τ = L = L = 25 = P Afer ime = 0, a given circui is represened by he circui diagram shown in FigureP a.) Suppose ha he inducor curren is 4 i = e ma for 0 Deermine he values of R and R 3. Figure P8.3-28

10 b.) Suppose insead ha R = 6 Ω, R 3= 20 Ω and he iniial condiion is i(0) = 0 ma. Deermine he inducor curren for 0. Soluion: The inducor curren is given by R a = = L. sc ( sc ) a i = i + i 0 i e for 0 where R 2.6 = sc = 36 R = 6 Ω and R + 4 a. Comparing his o he given equaion gives i R = R = Ω. Nex R ( R ) R3 R3 R = = + 4 = 0 = 40 Ω. 2 0 a = = 5 s τ 2 =. also 6 i sc = 36 = 28.8 ma. Then a i = i + i 0 i e = e = e. b. R = ( ) 20 = 0 Ω so sc 5 5 ( sc ) P Consider he circui shown in Figure P a.) Deermine he ime consan,, and he seady sae capacior volage, v( ), when he swich is open. b.) Deermine he ime consan,, and he seady sae capacior volage, v( ), when he swich is closed. Answers: a.) = 3 s and v( ) = 24 V; b.) = 2.25 s and v( ) = 2 V; Figure P Soluion: a.) When he swich is open we have

11 Afer replacing series and parallel resisors by equivalen resisors, he par of he circui conneced o he capacior is a Thevenin equivalen circui wih R = Ω. The ime consan is = RC = 33.33( 0.090) = 3 s. Since he inpu is consan, he capacior acs like an open circui when he circui is a seady sae. Consequenly, here is zero curren in he Ω resisor and KVL gives v( ) = 24 V. b.) When he swich is closed we have This circui can be redrawn as Now we find he Thevenin equivalen of he par of he circui conneced o he capacior: So R = 25 Ω and = RC = = 2.25 s Since he inpu is consan, he capacior acs like an open circui when he circui is a seady sae. Consequenly, here is zero curren in he 25 Ω resisor and KVL gives v( ) = 2 V.

12 Secion 8-4: Sequenial Swiching P 8.4- The circui shown in Figure P 8.4- is a seady sae before he swich closes a ime = 0. The swich remains closed for.5 s and hen opens. Deermine he capacior volage, v(), for > 0. Hin: Deermine v() when he swich is closed. Evaluae v() a ime =.5 s o ge v(.5). Use v(.5) as he iniial condiion o deermine v() afer he swich opens again e V for 0 < <.5 s Answer: v () = 2.5(.5) e V for.5 s < Figure P 8.4- Soluion: Replace he par of he circui conneced o he capacior by is Thevenin equivalen circui o ge: Before he swich closes a = 0 he circui is a seady sae so v(0) = 0 V. For 0 < <.5s, v oc = 5 V and R = 4 Ω so τ = = 0.2 s. Therefore 5 v = v + ( v(0) v ) e = e V for 0 < <.5 s oc e oc A =.5 s, v(.5) = = 5 V. For.5 s <, v oc = 0 V and R = 8 Ω so τ = = 0.4 s. Therefore (.5) 2.5(.5) v voc ( v(.5) voc ) e = + = 0 5 e V for.5 s < Finally e V for 0 < <.5 s v () = 2.5(.5) 0 5 e V for.5 s <

13 P The circui shown in Figure P is a seady sae before he swich closes a ime = 0. The swich remains closed for.5 s and hen opens. Deermine he inducor curren, i(), for > e A for 0 < <.5 s Answer: v () = (.5) e A for.5 s < Figure P Soluion: Replace he par of he circui conneced o he inducor by is Noron equivalen circui o ge: Before he swich closes a = 0 he circui is a seady sae so i(0) = 3 A. For 0 < <.5s, i sc = 2 2 A and R = 6 Ω so = = 2 s. Therefore i = i + ( i(0) i) e = 2 + e A for 0 < <.5 s sc sc 0.5(.5) A =.5 s, i(.5) 2 e 2 = + = 2.47 A. For.5 s <, i sc = 3 A and R = 8 Ω so = =.5 s. 8 Therefore (.5) 0.667(.5) i = isc + ( i(.5) isc ) e = e V for.5 s < Finally e A for 0 < <.5 s i () = 0.667(.5) e A for.5 s <

14 P The circui shown in Figure P is a seady sae before he swich opens a = 0. The swich remains open for 0.5 second and hen closes. Deermine v() for 0. Soluion: The circui is a seady sae before he swich closes. The capacior acs like an open circui. The iniial condiion is 40 v( 0 + ) = v( 0 ) = 24 = 2 V Figure P Afer he swich closes, replace he par of he circui conneced o he capacior by is Thevenin equivalen circui. Recognize ha R = 6.67 Ω and voc = 4 V The ime consan is = RC = ( 6.67 )( 0.05 ) = s = s The capacior volage is ( oc ) oc 3 3 v = v 0 + v e + v = 2 4 e + 4 = e V for s When he swich opens again a ime = 0.5 he capacior volage is 3( 0.5) v = v 0.5 = 4 + 8e = V Afer ime = 0.5 s, replace he par of he circui conneced o he capacior by is Thevenin equivalen circui.

15 Recognize ha R = 20 Ω and voc = 2 V The ime consan is = RC = 20( 0.05) = = s The capacior volage is ( ) ( oc ) 0.5 0( 0.5) v = v v e + v = e + 2 0( 0.5) = e V for 0.5 s so 2 V for 0 3 v = e V for s ( 0.5) e V for 0.5 s oc