R th is the Thevenin equivalent at the capacitor terminals.

Transcription

1 Chaper 7, Slun. Applyng KV Fg. 7.. d 0 C - Takng he derae f each erm, d 0 C d d d r C Inegrang, () ln I 0 - () I 0 e - C C () () r - I 0 e - () V 0 e C C Chaper 7, Slun. h C where h s he Theenn equalen a he capacr ermnals. Chaper 7, Slun. h Ω 60 0 ms (a) 0 //0 5kΩ, Th Th C 5x0 xx0 6 0 ms (b) 0 //(5 5) 8 0Ω, C 0x0. 6s Th Th Chaper 7, Slun 4. eq C eq CC where C eq, C C ( eq CC )(C C )

2 Chaper 7, Slun 5. () where (4) 4 4) (4)e -(-, C (0)(0.) () (0) 4e -(-4) 4e V Chaper 7, Slun 6. (0) 0 / 6 () e, C 40x0 xx0.5 () 4e V (4) 4V 5 Chaper 7, Slun 7. () (0) e -, h C where h s he Theenn ressance acrss he capacr. T deermne h, we nser a -V lage surce n place f he capacr as shwn belw. 8 Ω 0.5 V 0 Ω V 0., h 80 h C 0. () 0e V 6

3 Chaper 7, Slun 8. (a) C e d C d -4 C(0)(-4)e C 5-4 mf 4C 50 Ω (b) C 0.5 s 4 (c) - w C (0) CV0 (5 0 )(00) (d) - ( 0 w ) CV0 CV0 e e -80 e r e 80 0 ln () 86.6 ms 8 50 mj Chaper 7, Slun 9. () (0)e -, C eq eq Ω eq C (0.5)(8) () - 0e V

4 Chaper 7, Slun 0. 5 Ω 0 Ω 0 mf T 4 Ω (0)() e. f ( 0) A, hen (0) A (0) (0) (0) T 5 A A (0) 0(0) 4 T (0) V acrss he capacr ermnals. h - h C (0)(0 0 ) - -0 () (0)e 50e Ω d C C (0 0 d -0-5e A C - )(-500e By applyng he curren dsn prncple, 5 () (- C ) -0.6 C e -0 A 0 5 ) Chaper 7, Slun. Applyng KC he crcu, d 0 Dfferenang bh sdes, d 0 d A e - d d 0

5 If he nal curren s () (0) I0 A, - I0 e - - () d I 0 e - -I - I 0 e 0 e -, hen I 0 - Chaper 7, Slun. When < 0, he swch s clsed and he nducr acs lke a shr crcu dc. The 4 Ω ressr s shr-crcued s ha he resulng crcu s as shwn n Fg. (a). Ω V (0 - ) 4 Ω H (a) (b) (0 ) 4 A Snce he curren hrugh an nducr cann change abruply, (0) (0 ) (0 ) 4 A When > 0, he lage surce s cu ff and we hae he crcu n Fg. (b) Hence, () (0) e - 4e - A

6 Chaper 7, Slun. where h h s he Theenn ressance a he ermnals f he nducr. h Ω µ s Chaper 7, Slun 4 Cnerng he wye-subnewrk dela ges 6 Ω 80mH 0 Ω 0x0 0x50 50x / 0 85Ω, 4Ω, 70Ω //70 (0x70)/ Ω, 4//6(4x6)/ Ω Th 85x6.8 85//( ) 5.476Ω,.8 Th 80x ms

7 Chaper 7, Slun 5 (a) (b) 0 // 40 0Ω, 5/ 0 0. s Th 40 // Ω, (0x0 ) / ms Th 5 Th Th Chaper 7, Slun 6. (a) eq eq eq and ( ( eq ) ) ( ) (b) where eq and ( eq ( )( ( ) ) ) ( ) Chaper 7, Slun 7. (0) e - (), eq () e -6 d d -6 () 6e () -6 - e V ( 4)(-6)e -6

8 Chaper 7, Slun 8. If () 0, he crcu can be redrawn as shwn belw. 0.4 H eq () () 6 5 eq, (0) e - () e d - - () - (-) e.e - V d 5 Chaper 7, Slun 9. V 0 Ω / 40 Ω T fnd h we replace he nducr by a -V lage surce as shwn abe Bu and.e h 0 Ω 6 0. s 0 () h e -5 A 0

9 Chaper 7, Slun 0. (a) d d e (0)(-50)e H 50 5 Ω (b). 0 ms 50 (c). w (0) (0.)(0 ) 45 J (d). e p be he fracn - ( 0 I ) 0 p I0 e ()(0) p e e 0.96.e. p % Chaper 7, Slun. The crcu can be replaced by s Theenn equalen shwn belw. h V h H V 80 (60) 40 V h h Vh 40 I (0) ( ) w I () Ω h

10 Chaper 7, Slun. - () (0)e, eq eq Ω, () 0e -.5 A 5 Usng curren dsn, he curren hrugh he 0 hm ressr s (-) -e () e V Chaper 7, Slun. Snce he Ω ressr, / H nducr, and he () Ω ressr are n parallel, hey always hae he same lage. -.5 (0) -.5 The Theenn ressance h a he nducr s ermnals s 4 h ( ), h () (0)e -.5e, > 0 d (-4)(/) e d e -4 V, > 0-4 x 0.5e V, > 0 Chaper 7, Slun 4. (a) () - 5u() (b) () -0[ u() u( ) ] 0[ u( ) u( 5) ] - 0u() 0u( ) 0u( 5)

11 (c) x() ( ) [ u( ) u( ) ] [ u( ) u( ) ] ( 4 ) [ u( ) u( 4) ] ( ) u( ) ( ) u( ) ( ) u( ) ( 4) u( 4) r( ) r( ) r( ) r( 4) (d) y() u(-) 5[ u() u( ) ] u(-) 5u() 5u( ) Chaper 7, Slun 5. () [u() r( ) r( ) u( )] V Chaper 7, Slun 6. (a) () u( ) u() [ u( ) u() ] () u( ) u() u( ) (b) () (4 ) [ u( ) u( 4) ] () -( 4) u( ) ( 4) u( 4) () u( ) r( ) r( 4) (c) () [ u( ) u( 4) ] 4[ u( 4) u( 6) ] () u( ) u( 4) 4u( 6) (d) () - [ u( ) u( ) ] - u( ) u( ) 4 4 () (- ) u( ) ( ) u( ) 4 () - r( ) u( ) r( ) u( ) Chaper 7, Slun 7. () s skeched belw. () 0 4 -

12 Chaper 7, Slun 8. () s skeched belw. () Chaper 7, Slun 9 (a) x() (b) y() 7.8 0

13 (c) z( ) cs 4δ ( ) cs 4δ ( ) 0.656δ ( ), whch s skeched belw. z() δ ( ) Chaper 7, Slun 0. 0 δ 0 π ) δ( 0.5) d cs(π) 0. - (a) 4 ( ) d 4 4 (b) cs( 5 cs π - Chaper 7, Slun. 6-9 δ δ () e δ() cs π δ() d 5 e cs(π) (a) [ e ( ) ] d e e (b) [ ] ( ) Chaper 7, Slun. (a) u( λ ) dλ dλ λ 4 4 (b) r( ) d 0d ( ) d ( (c ) 6) δ ( ) d ( 6) 6 4

14 Chaper 7, Slun. () 0 () d (0) - 0 () - 0δ( ) d () u( ) A Chaper 7, Slun 4. (a) (b) (c) d [ u( ) u( ) ] δ( )u( ) d u( ) δ( ) δ( ) 0 δ( ) δ( ) d [ r( 6) u( ) ] u( 6)u( ) d r( 6) δ( ) u( 6) 0 δ( ) u( 6) d [ sn 4 u( ) ] 4cs 4 u( ) sn 4δ( ) d 4cs 4 u( ) sn 4xδ( ) 4cs 4 u( ) 0.566δ( ) Chaper 7, Slun 5. (a) () A e -5, (0) A () - e V (b) () A e, (0) A 5 () 5e V

15 Chaper 7, Slun 6. - (a) () A Be, > 0 A, (0) 0 B r B - - () e V, > 0 (b) () A Be, > 0 A -, (0) -6 - B r B - () - ( e ) V, > 0 Chaper 7, Slun 7. e h p, p 0. (a) 4s (b) ( ) 0 V / 4 0 h h Ae h 4 0 Ae 0.5 ( 0) 0 A A 8 0 8e 0.5 (c ) 0 8 e 0. 5 Chaper 7, Slun 8 e p h h h 0 h Ae u( ) e p ku( ), p 0, ku( ) u( ) k

16 p u( ) ( Ae ) u( ) If (0) 0, hen A / 0,.e. A-/. Thus ( e ) u( ) Chaper 7, Slun 9. (a) Befre 0, () (0 4 ) 4 V Afer 0, () ( ) [ (0) ( ) ] e - C (4)() 8, (0) 4, ( ) 0 () () 0 (8 0) e e V 8 (b) Befre 0,, where s due he -V surce and s due he -A surce. V T ge, ransfrm he curren surce as shwn n Fg. (a). -8 V Thus, 8 4 V Afer 0, he crcu becmes ha shwn n Fg. (b). F F 4 Ω 8 V V Ω Ω (a) (b)

17 [ (0) ( ) ] e - () ( ) ( ), (0) 4, C ()() 6 () () Chaper 7, Slun 40. (4 ) e e V (a) Befre 0, 6 V. Afer 0, [ ] () ( ) (0) ( ) e - ( ) 4, (0), C ()() 6 () (b) Befre 0, 4 ( 4)e () 4 8e V V. 6 Afer 0, () ( ) [ (0) ( ) ] e - Afer ransfrmng he curren surce, he crcu s shwn belw. Ω 0 4 Ω V 5 F Chaper 7, Slun 4. (0), ( ), C ()(5) 0 V 0 (0) 0, ( ) () 0 6 (6)(0) eq C (6 0)() 5 6 () [ (0) ( ) ] e - ( ) () 0 (0 0) e - 5 () -0. 0( e ) V

18 Chaper 7, Slun 4. () ( ) (0) ( ) e 4 (0) 0, ( ) () eq C eq, eq 4 4 () 4-4 () 8 8e - (a) [ ] () ( e ) V (b) Fr hs case, ( ) 0 s ha Chaper 7, Slun 4. - () (0) e 4 (0) () 8, C (4)() 4 - () 8e V Befre 0, he crcu has reached seady sae s ha he capacr acs lke an pen crcu. The crcu s equalen ha shwn n Fg. (a) afer ransfrmng he lage surce. 0.5, Hence, A Afer 0, he crcu s as shwn n Fg. (b). (), C C - C (0) e h T fnd h, we replace he capacr wh a -V lage surce as shwn n Fg. (c). 0.5 C V Ω (c)

19 0.5 C, h 60 Ω, C h C (0) 64 V () 64e C d C C -C - () d e Chaper 7, Slun 44. A e eq 6 Ω, C 4 () [ (0) ( ) ] e - ( ) Usng lage dsn, (0) (0) 0 V, ( ) () 4 V 6 6 Thus, - 4 e - 4 () 4 (0 4) e 4 6 d - 4 e () C ()(6) - e A d 4 Chaper 7, Slun 45. Fr < 0, s 5u() 0 (0) 0 Fr > 0, s 5, 4 ( ) (5) eq 7 4 0, eq C (0)( ) 5 () [ (0) ( ) ] e - ( ) () - 5.5( e ) V () d C d e - () e A

20 Chaper 7, Slun 46. C ( 6) x0.5 s, (0) 0, ( ) 6 6x5 0 Th s ( ) ( ) [ (0) ( )] e 0( e / / ) V Chaper 7, Slun 47. Fr < 0, u () 0, u ( ) 0, (0) 0 Fr 0 < <, C ( 8)(0.) (0) 0, ( ) (8)() 4 () () ) [ (0) ( ) ] e - ( e - ) ( 4 - Fr >, () 4( e ) ( ) ( ) 0 () () 0 ( e -(-) 0)e -(-) Thus, () 4 - ( e ) 0 4.8e V, -(-) V, 0 < < > Chaper 7, Slun 48. Fr < 0, u (-), (0) 0 V Fr > 0, u (-) 0, ( ) 0 h 0 0 0, h C (0)(0.) () () [ (0) ( ) ] e - ( ) - 0e V () () d C d - - e - (0.) 0e - A

21 Chaper 7, Slun 49. Fr 0 < <, (0) 0, ( ) ()(4) 8 eq 4 6 0, eq C (0)(0.5) 5 () ) [ (0) ( ) ] e - - ( e ) V ( 5 () 8-0. Fr >, () 8( e ). 45 ) () ( ) [ () ( ) ] e -( -() 5 ().45e, ( ) 0 V Thus, - 5 ( ) 8 e () -(.45e ) 5 V, V, 0 < < > Chaper 7, Slun 50. Fr he capacr lage, () ( ) (0) ( ) (0) 0 [ ] e - Fr < 0, we ransfrm he curren surce a lage surce as shwn n Fg. (a). kω kω 0 V kω (a) ( ) (0) 5 V ( ) kω h - C () 5 e, > h ( ) 0 4

22 We nw ban frm (). Cnsder Fg. (b). x x T kω 0 ma kω /4 mf kω Bu Thus, x 0 ma T d T C d T () () 7.5 T (b) ( e ) ma 0 (-5)(-4)e A 4 ( e ) ma -4 () e x ma -4 x () 7.5( e ) ma, > 0 Chaper 7, Slun 5. Cnsder he crcu belw. 0 V S Afer he swch s clsed, applyng KV ges d V S d d V S r - d d - d V Inegrang bh sdes, S

23 r - ln I 0 V S () V S - ln I0 VS VS - e I V 0 () S V V S S - I 0 e whch s he same as Eq. (7.60). Chaper 7, Slun 5. 0 (0) A, ( ) A 0 () ( ) (0) ( ) e - [ ] ( ) A Chaper 7, Slun 5. (a) Befre 0, 5 5 A Afer 0, () (0)e - 4, (0) 5 - () 5e A (b) Befre 0, he nducr acs as a shr crcu s ha he Ω and 4 Ω ressrs are shr-crcued. () 6 A Afer 0, we hae an crcu. () (0) e, () - 6e - A

24 Chaper 7, Slun 54. (a) Befre 0, s baned by curren dsn r 4 () ( 4 4 ) A Afer 0, () ( ) [ (0) ( ) ] e -, Ω eq.5 7 (0), () 6 7 eq 4 ( ) () () e () ( 6 e 7 )A (b) Befre 0, 0 () A Afer 0, eq eq (0) T fnd ( ), cnsder he crcu belw, a when he nducr becmes a shr crcu, 0 V 4 V H Ω 6 Ω Ω ( ) A () ( ) e -9 () -9 4 e A 4 9

25 Chaper 7, Slun 55. Fr < 0, cnsder he crcu shwn n Fg. (a). 0.5 H 0.5 H Ω 8 Ω 4 Ω Ω 4 V 0 V (a) (b) () 4 96 V 48 A Fr > 0, cnsder he crcu n Fg. (b). () ( ) [ (0) ( ) ] e - 0 (0) 48, ( ) A h 8 0 Ω, 0 () (48 )e 46 () () h -0 e e V 0 Chaper 7, Slun 56. eq Ω, () ( ) (0) ( ) e - [ ] (0) s fund by applyng ndal analyss he fllwng crcu.

26 5 Ω x 6 Ω A Ω 0 Ω 0.5 H 0 V 0 5 (0) 6 x x x 0 6 x x x A Snce 0 5 4, 4 ( ) (4) ().6 (.6)e.6 0.4e d () (0.4)(-0)e -0 d -0 () - 4e V -0 Chaper 7, Slun 57. A 0, he crcu has reached seady sae s ha he nducrs ac lke shr crcus. 6 Ω 0 V 5 Ω 0 Ω 0 0 0, () , (0).4 A, (0) 0.6 A

27 Fr > 0, he swch s clsed s ha he energes n and flw hrugh he clsed swch and becme dsspaed n he 5 Ω and 0 Ω ressrs. -.5 () (0) e, 5 () -.4e () e () A - (0), e A Chaper 7, Slun 58. Fr < 0, () 0 0 Fr > 0, (0) 0, ( ) 5 4 h 4 Ω, 4 () [ (0) ( ) ] e - ( ) -6 () 5( e ) A d () 5 d 4-6 () 5 5e V h ( e ) (-6)(5) e Chaper 7, Slun 59. e I be he curren hrugh he nducr. Fr < 0, s 0, (0) 0.5 Fr > 0, eq 4 6 6, ( ) () 4 () ( ) (0) ( ) () [ ] e - e -4 d () d () e 6-4 V (.5)(-4)(-e -4 ) eq

28 Chaper 7, Slun 60. e I be he nducr curren. Fr < 0, u () 0 (0) 0 8 Fr > 0, eq Ω, 4 ( ) 4 () () ( ) [ (0) ( ) ] e - ( e - ) 4 eq () () d d 6e (8)(-4) V e - Chaper 7, Slun 6. The curren surce s ransfrmed as shwn belw. 4 Ω 0u(-) 40u() 0.5 H, (0) 5, ( ) () ( ) (0) ( ) e - () () () e d d 0e -8 [ ] A (-5)(-8)e -8 V Chaper 7, Slun 6. eq 6 Fr 0 < <, u ( ) 0 s ha

29 (0) 0, () 6 ( e - ) ( ) () 6 ( ) 6 () 0.5 ( Fr >, ( e ) Thus, () Chaper 7, Slun e -(-) - ( e ) A () e 0.5)e -(-) -(-) A 0 < < > 0 Fr < 0, u (-), (0) 5 Fr > 0, u (-) 0, ( ) h Ω, 4 () () () () [ (0) ( ) ] e - ( ) e -8 A d (-8)()e -8 d Chaper 7, Slun e V h e be he nducr curren. Fr < 0, he nducr acs lke a shr crcu and he Ω ressr s shrcrcued s ha he equalen crcu s shwn n Fg. (a). 8 6 Ω 6 Ω 0 Ω Ω 0 Ω Ω Ω (a) (b)

30 0 (0).667 A 6 4 Fr > 0, h 6 4 Ω, 4 T fnd ( ), cnsder he crcu n Fg. (b) ( ) 6 () ( ) [ (0) ( ) ] e () e ( e ) A s he lage acrss he 4 H nducr and he Ω ressr 0 5 () d d ().667( e )V 6 e - h (4) (-)e e 6 - Chaper 7, Slun 65. Snce 0[ u() u( ) ] s, hs s he same as sayng ha a 0 V surce s urned n a 0 and a -0 V surce s urned n laer a. Ths s shwn n he fgure belw. s Fr 0 < <, (0) 0, ( ) 5 h 5 0 4, h 4 () [ (0) ( ) ] e - ( )

31 ( e ) A ( e ) () - () Fr >, ( ) 0 snce s 0 () ) ()e -( -() ().79e A Thus, - ( ) e ().79 e A 0 < < > -() A Chaper 7, Slun 66. Fllwng Pracce Prblem 7.4, - () V T e V T (0) -4, () -4e () -() 4e, > 0 - f C (0 0 )( 0 ) 6 () 4-50 () e 0.4e -50 ma, > Chaper 7, Slun 67. The p amp s a lage fllwer s ha as shwn belw. C

32 A nde, 0 A he nnnerng ermnal, d C 0 d d C d d d C - C () V e V T (0) 5 V, T () 5e - 00 V 6 - C ()(0 0 )( 0 ) 00 Chaper 7, Slun 68. Ths s a ery neresng prblem and has bh an mpran deal slun as well as an mpran praccal slun. e us lk a he deal slun frs. Jus befre he swch clses, he alue f he lage acrss he capacr s zer whch means ha he lage a bh ermnals npu f he p amp are each zer. As sn as he swch clses, he upu res g a lage such ha he npu he p amp bh g 4 ls. The deal p amp pus u whaeer curren s necessary reach hs cndn. An nfne (mpulse) curren s necessary f he lage acrss he capacr s g 8 ls n zer me (8 ls acrss he capacr wll resul n 4 ls appearng a he negae ermnal f he p amp). S wll be equal 8 ls fr all > 0. Wha happens n a real crcu? Essenally, he upu f he amplfer prn f he p amp ges whaeer s maxmum alue can be. Then hs maxmum lage appears acrss he upu ressance f he p amp and he capacr ha s n seres wh. Ths resuls n an expnenal rse n he capacr lage he seady-sae alue f 8 ls. C() V p amp max ( e -/(uc) ) ls, fr all alues f C less han 8 V, 8 V when s large enugh s ha he 8 V s reached. Chaper 7, Slun 69. e x be he capacr lage. Fr < 0, x (0) 0

33 Fr > 0, he 0 kω and 00 kω ressrs are n seres snce n curren eners he p amp ermnals. As, he capacr acs lke an pen crcu s ha x ( ) (4) h kω, C (0 0 )(5 0 ) 000 x () 48 x () - ) [ ] x (0) x ( ) e ( e ) x ( () x () ( e )V 0 h Chaper 7, Slun 70. e capacr lage. Fr < 0, he swch s pen and (0) 0. Fr > 0, he swch s clsed and he crcu becmes as shwn belw. S C () s 0 s d C () d where () s s Frm (), d s 0 d C - s d (0) C Snce s cnsan, - C s

34 -6 C (0 0 )(5 0 ) mv -00 mv 0. Frm (), s ( 0 ) mv Chaper 7, Slun 7. e lage acrss he capacr. e lage acrss he 8 kω ressr. Fr <, 0 s ha () 0. Fr >, we hae he crcu shwn belw. 0 kω 0 kω 0 kω 4 V 00 mf 8 kω Snce n curren eners he p amp, he npu crcu frms an C crcu. - C (0 0 )(00 0 ) 000 () () ) [ () ( ) ] e -( 000 ( e -( ) ) ) ( 4 As an nerer, -0k -( ) 000 ( e ) 0k -( ) ( e )A 8

35 Chaper 7, Slun 7. The p amp acs as an emer fllwer s ha he Theenn equalen crcu s shwn belw. C u() Hence, () ( ) - [ (0) ( ) ] e -6 (0) - V, ( ) V, C (0 0 )(0 0 ) 0. () (- - )e 5-0 e -0 d -6 C (0 0 )(-5)(-0) e d 0.5e -0 ma, > 0-0 Chaper 7, Slun 7. Cnsder he crcu belw. f C A nde, A nde, d C d ()

36 d C d () f Bu 0 and. Hence, () becmes d C d d C d C d r d C whch s smlar Eq. (7.4). Hence, T < 0 () - ( ) e > 0 T where T (0) and 4 () - C (0 0 )(0 0 ) 0. < e > 0 6 Frm (), d - f C (0 0 )(0 0 d -5-6e, 0 > -5-6e u() V -6 )(5e -5 ) Chaper 7, Slun 74. e capacr lage. f C

37 Fr < 0, (0) 0 Fr > 0, s 0 µ A. Cnsder he crcu belw. d s C () d () ( ) (0) ( ) e - () [ ] I s eden frm he crcu ha C ( 0 )(50 0 ) 6 0. C f s s A seady sae, he capacr acs lke an pen crcu s ha s passes hrugh. Hence, 6 ( ) (0 0 )(50 0 ) 0.5 V s Then, Bu ( e ) V -0 () 0.5 () 0 s - s f (4) f Cmbnng (), (), and (4), we ban - f d f C d - d -6 (0 0 )( 0 ) 5 d e ( 0 )(0.5) -0 0.e ( e )V -0 ( -0e )

38 Chaper 7, Slun 75. e lage a he nnnerng ermnal. e lage a he nerng ermnal. Fr > 0, 4 0 s s, 0 kω - () Als,.e. d C, 0 kω d, C µ F - s d C () d Ths s a sep respnse. () ( ) [ (0) ( ) ] e -, (0) -6 where C (0 0 )( 0 ) 50 A seady sae, he capacr acs lke an pen crcu s ha passes hrugh. Hence, as - s ( ) - -0.e. ( ) s (4) - 0 () - ( ) e -50 () - e -50 Bu r s 4 e s e V s ma 0k d r C - 0. ma d

39 Chaper 7, Slun 76. The schemac s shwn belw. Fr he pulse, we use IPW and ener he crrespndng alues as arbues as shwn. By selecng Analyss/Seup/Transen, we le Prn Sep 5 ms and Fnal Sep s snce he wdh f he npu pulse s s. Afer sang and smulang he crcu, we selec Trace/Add and dsplay V(C:). The pl f V() s shwn belw. Chaper 7, Slun 77. The schemac s shwn belw. We clck Marker and nser Mark Vlage Dfferenal a he ermnals f he capacr dsplay V afer smulan. The pl f V s shwn belw. Ne frm he pl ha V(0) V and V( ) -4 V whch are crrec.

40 Chaper 7, Slun 78. (a) When he swch s n psn (a), he schemac s shwn belw. We nser IPOBE dsplay. Afer smulan, we ban, (0) 7.74 A frm he dsplay f IPOBE.

41 (b) When he swch s n psn (b), he schemac s as shwn belw. Fr nducr I, we le IC By clckng Analyss/Seup/Transen, we le Prn Sep 5 ms and Fnal Sep s. Afer Smulan, we clck Trace/Add n he prbe menu and dsplay I() as shwn belw. Ne ha ( ) A, whch s crrec.

42

43 Chaper 7, Slun 79. When he swch s n psn, (0) / 4A. When he swch s n psn, 4 Th ( ) 0.5 A, Th ( 5) // 4 8 /, 80 / 5 ( ) ( ) [ (0) ( )] e e / / 80 A Chaper 7, Slun 80. (a) When he swch s n psn A, he 5-hm and 6-hm ressrs are shrcrcued s ha ( 0) (0) (0) 0 bu he curren hrugh he 4-H nducr s (0) 0/0 A. (b) When he swch s n psn B, Th // 6 Ω, / Th ( ) ( ) [ (0) ( )] e 0 e e / / 0.5 A (c) 0 ( ) A, ( ) ( ) A d ( ) ( ) 0 V d Chaper 7, Slun 8. The schemac s shwn belw. We use VPW fr he pulse and specfy he arbues as shwn. In he Analyss/Seup/Transen menu, we selec Prn Sep 5 ms and fnal Sep S. By nserng a curren marker a ne ermal f I, we aumacally ban he pl f afer smulan as shwn belw.

44 Chaper 7, Slun C -6 C Ω

45 Chaper 7, Slun ( ) 0, (0) 0, C 4x0 x5x0 50s ( ) ( ) [ (0) ( )] e Slng fr ges 50 ln s ( e / speed 4000m/67.6s 6.78m/s / 50 ) Chaper 7, Slun 84. e I be he fnal alue f he curren. Then ( ) I ( e / ), / 0.6 / 8 / I I ( e 50 ) ln ms. Chaper 7, Slun 85. (a) C -6 6 (4 0 )(6 0 ) 4 s Snce () ( ) [ (0) ( ) ] e - - [ ] () ( ) (0) ( ) e - [ ] ( ) ( ) (0) ( ) () () e Ddng () by (), ( ) ( ) e ( ) ( ) 0 ( ) ( ) ( ) ln ( ) ( ) ln 4ln () (b) Snce 0 <, he lgh flashes repeaedly eery C 4 s s

46 Chaper 7, Slun 86. () [ (0) ( ) ] e - ( ) ( ), (0) 0 () ( 8 0 ( e - ) - ( ) 0) e 0 8 e - ln() e Fr 00 kω, -6 C (00 0 )( 0 ) 0 0. ln() 0.97 s Fr MΩ, -6 C ( 0 )( 0 ) 0 ln().97 s 6 s 0. s Thus, 0.97 s < 0 <.97 s Chaper 7, Slun 87. e be he nducr curren. Fr < 0, 0 (0 ). A 00 Fr > 0, we hae an crcu 50 0., ( ) () ( ) (0) ( ) e - (). e -0 A 00 ms 0. s, - (0.).e [ ] 0.44 whch s he same as he curren hrugh he ressr. A

47 Chaper 7, Slun 88. (a) (b) - C (00 0 )(00 0 ) 60 µ s As a dfferenar, T > µ s 0.6 ms.e. Tmn 0.6 ms C 60 µ s As an negrar, T < 0. 6 µ s.e. Tmax 6 µ s Chaper 7, Slun 89. Snce < 0.T µ s < µ s -6 < 0-6 (00 0 )( 0 ) < 00 mh Chaper 7, Slun 90. We deermne he Theenn equalen crcu fr he capacr C s. s h, h s p s p h V h C s The Theenn equalen s an C crcu. Snce s h 0 0 Als, s 9 p 6 9 MΩ s p

48 where h Cs 5 µ s 6( ) h p s 0.6 MΩ 6 C s 6 h 5 pf Chaper 7, Slun 9. (0) 40 ma, ( ) 0 50 () ( ) (0) ( ) e - () 40e - ( ) 0 40 [ ] -0 0 e e ln (4) ln (4) ln (4) Ω Chaper 7, Slun 9. 0 d - 0 < < -9 C d -0 < < µ A 0 < < ms () - 8 ma ms < < ms 5 µ s whch s skeched belw. D () 0 µa 5 µs ms -8 ma (n scale)