CHAPTER 5. Exercises. the coefficient of t so we have ω = 200π

Transcription

1 HPTER 5 Exerises E5. (a) We are given v ( ) 5 s(π 3 ). The angular frequeny is he effiien f s we have ω π radian/s. Then f ω / π Hz T / f ms m / 5 / 6. Furhermre, v() aains a psiive peak when he argumen f he sine funin is zer. Thus keeping in mind ha ω has unis f radians, he psiive peak urs when π ωmax 3 max.8333 ms 8 (b) P avg / R 5 W () pl f v() is shwn in Figure 5.4 in he bk. E5. We use he rignmeri ideniy sin( z ) s( z 9 ). Thus sin(3π 6 ) s(3π 3 ) E5.3 ω πf 377 radian/s T / f 6.67 ms m 55.6 The perid rrespnds 36 herefre 5 ms rrespnds a phase angle f ( 5 /6.67) Thus he vlage is v ( ) 55.6 s(377 8 ) E5.4 (a) 9 j 4.4 s( ω ) sin( ω ) 4.4 s( ω 45 ) 45 (b) j 5.5 j j s( ω 3 ) 5 sin( ω 3 ).8 s( ω 3.44 ) () 5 6 j 7.5 j j sin( ω 9 ) 5 s( ω 6 ) 3.4 s( ω 5.8 ) 4

2 E5.5 The phasrs are 3 3 and 3 45 v lags v by 6 (r we uld say v leads v by 6 ) v leads v 3 by 5 (r we uld say v 3 lags v by 5 ) v leads v 3 by 75 (r we uld say v 3 lags v by 75 ) E5.6 (a) E5.7 (a) jω j / / j 5 9 (b) The phasr diagram is shwn in Figure 5.a in he bk. / jω j / /( j 5) 9 (b) The phasr diagram is shwn in Figure 5.b in he bk. E5.8 (a) R R 5 5 R R R / R /(5) (b) The phasr diagram is shwn in Figure 5. in he bk. E5.9 (a) The ransfrmed newrk is: s 9 5 j m 43

3 i ( ) 8.8 s(5 35 ) m R R (b) The phasr diagram is shwn in Figure 5.7b in he bk. () i() lags v s () by 45. E5. The ransfrmed newrk is: jω / /( j 5) /( j) /( j) R /() Ω /( j 5) E5. The ransfrmed newrk is: We wrie K equains fr eah f he meshes: j ( ) j j ( ) Simplifying, we have j ) ( ( j) and. ( ) 45 ) and i ( ) s( Slving we find i.44 s( ). Thus we have 44

4 E5. (a) Fr a pwer far f %, we have s( θ ), whih implies ha he urren and vlage are in phase and θ. Thus, Q P an( θ ). ls m P /[ s θ ( )] 5 /[5 s( )]. 4.4 and Thus we have (b) Fr a pwer far f % lagging, we have s( θ )., whih implies ha he urren lags he vlage by θ s (.) Thus, Q P an( θ ) 4.49 kr. ls, we have P /[ s( θ )] 5.. Thus we have 7.7 and m () The urren raings wuld need be five imes higher fr he lad f par (b) han fr ha f par (a). Wiring ss wuld be lwer fr he lad f par (a). E5.3 The firs lad is a µf apair fr whih we have /( jω ) The send lad absrbs an apparen pwer f kwih a pwer far f 8% lagging frm whih we have θ s (.8) Nie ha we sele a psiive angle fr Ω θ 9 sin( θ P s( θ ) Q ) 3.77 kr / 3.77 beause he lad has a lagging pwer far. Thus we have P s( θ ) 8. kw and Q sin( θ ) 6 kr. θ Nw fr he sure we have: P P P 8 kw Q Q Q.3 s s kr s Ps Qs far Ps /( s 8.35 k s s pwer ) % 96.33% / 8.35 E5.4 Firs, we zer he sure and mbine impedanes in series and parallel deermine he Thévenin impedane. 45

5 5 j5 5 j5 5 j 5 / / j j Then we analyze he irui deermine he pen-irui vlage j n / E5.5 (a) Fr a mplex lad, maximum pwer is ransferred fr * j5 R jx. The Thévenin equivalen wih he lad aahed is: The urren is given by j5 j5 The lad pwer is P R (.3536 / ) 6.5 W 46

6 (b) Fr a purely resisive lad, maximum pwer is ransferred fr R 5 3. Ω. The Thévenin equivalen wih he lad aahed is: The urren is given by j5 The lad pwer is P R 3.(.3456/ ) 6.57 W E5.6 The line--neural vlage is / N phase angle was speified in he prblem saemen, s we will assume ha he phase f an is zer. Then we have an bn n The irui fr he a phase is shwn belw. (We an nsider a neural nnein exis in a balaned Y-Y nnein even if ne is n physially presen.) The a-phase line urren is an a j The urrens fr phases b and are he same exep fr phase. b Y P s( θ ) 3 s(37. ) 3.88 kw 47

7 Y Q sin( θ ) 3 sin(37. ).44 kr E5.7 The a-phase line--neural vlage is an / The phase impedane f he equivalen Y is Y / 3 5 / Ω. Thus he line urren is an a Y Similarly, and b Finally, he pwer is P 3( a / ) Ry 3. kw Prblems P5. (a) nreasing he peak ampliude m?. Srehes he sinusidal urve verially. (b) nreasing he frequeny f? hriznally. 4. mpresses he sinusidal urve () Dereasing θ? 5. Translaes he sinusidal urve he righ. (d) Dereasing he angular frequeny ω? urve hriznally. 3. Srehes he sinusidal (e) nreasing he perid? hriznally. 3. Srehes he sinusidal urve P5. The unis f angular frequeny ω are radians per send. The unis f frequeny f are herz, whih are equivalen inverse sends. The relainship beween hem is ω πf. 48

8 P5.3* v ( ) sin( π 3 ) s( π 6 ) ω π rad/s f 5 Hz phase angle θ 6 T f ms 7.7 P m ( ) R W π 3 radians The firs psiive peak urs fr π π 3 peak peak.3333 ms P5.4 v ( ) 5 sin( 5π ) 5 s( 5π 3 ) ω 5π rad/s f 5 Hz phase angle θ 3 T f 4 ms P m ( ) R 5 W π 6 radians Psiive peak urs fr 5π π ms The firs psiive peak afer is a T ms peak 49

9 P5.5* Sinusidal vlages an be expressed in he frm v ( ) s( ω θ). The peak vlage is m 8.8. The frequeny is 4 f / T khz and he angular frequeny is ω π f π radians/s. The phase rrespnding a ime inerval f µs is 4 θ ( / T ) Thus we have v ( ) 8.8 s(π 7 ). m P5.6 The angular frequeny is ω πf 5π radians/s. We have T / f 8 ms. The sinusid reahes a psiive peak ne quarer f a yle afer rssing zer wih a psiive slpe. Thus, he firs psiive peak urs a 3 ms. The phase rrespnding a ime inerval f 3 ms is θ ( / T ) Thus, we have v ( ) 5 s(5π 35 ). P5.7 i ( ) s( 657π 3 p () Ri () 9 s ( π ) 95[ s( 4π )] P ) R avg ( ) 3( ) 95 W W 5

10 P5.8 v p ( ) 53 sin( 596π ) ( ) v ( ) R 48 sin ( 596π ) 64[ sin( 9π )] ( ) R 64 W P avg W P f T Hz m ω πf 68.3 rad/s i ( ) 7.7 s(π 8 ) θ 36 max 8 T P5. sequene f MT insruins amplish he desired pl fr par (a) is Wx *pi; Wy *pi; Thea 9*pi/8; :.:; x s(wx*); y s(wy* Thea); pl(x,y) y hanging he parameers, we an bain he pls fr pars b,, and d. The resuling pls are 5

11 (a) (b) () (d) T i P5.* v ( ) d d T P5.* T.5 v T ( ) d [ 5 sin(π )] d [.5.5 sin(4π )].5 d.5.5 s(4π ) 4π T P5.3* i ( ) d d d T 4 5

12 T T P5.4 v ( ) d [ s(π ) sin(π )] [ s (π ) s(π ) sin(π ) sin (π )].5 d d i P5.5 v ( )d T T [6 7 s(π )] d [36 4 s(π ) 89 s (π )] d.. 36d 36d [ ].. 4 s(π ) d 4 s(π ) d.. 89 s (π )] d 44.5d s(π )] d T T P5.6 v ( ) d [ 3exp( )] d [ 9 exp( )] d [ 4.5 exp( )] 4.5[ exp( )].973 P5.7 ( ) (.5 ) T T v d d

13 P5.8 The values f all peridi wavef are n equal heir peak values divided by he square r f w. Hwever, hey are fr sinusids, whih are impran speial ases. P5.9 T add sinusids we:. nver eah f he sinusids be added in a phasr.. dd he phasrs and nver he sum plar frm using mplex arihimei. 3. nver he resuling phasr a sinusid. ll f he sinusids be added mus have he same frequeny. P5. Tw mehds deermine he phase relainship beween w sinusids are:. Examine he phasr diagram and nsider i rae unerlkwise. f phasr pins in a given direin befre phasr by an angle θ, we say ha leads by he angle θ r ha lags by he angle θ.. Examine pls f he sinusidal wavef versus ime. f reahes a pin (suh as a psiive peak r a zer rssing wih psiive slpe) by an inerval befre reahes he rrespnding pin, we say ha leads by he angle θ ( / T ) 36 r ha lags by he angle θ. P5.* v ( ) s( ω ) v v s ( ) sin( ω ) s( ω 9 ) s 9 ( ) 4.4 s( ω 45 ) j j lags by 9 lags by 45 s leads s by 45 54

14 P5.* ω πf 4π ( ) s( 4π 3 ) v v ( ) v 3 5 s(4π 5 ( ) s( 4π 9 ) v( ) lagsv( ) by v( ) lagsv3( ) by 6 v ( ) leads v ( ) by 6 3 ) P5.3* We are given he expressin 5 s( ω 75 ) 3s( ω 75 ) 4 sin( ω ) nvering phasrs we bain j (.7765 j.8978) j j Thus, we have 5 s( ω s( ω 8.9 ) 3s( ω 75 ) 4 sin( ω ) ) P5.4 The magniudes f he phasrs fr he w vlages are 6 and. The phase angles are n knwn. f he ph ase angles are he same, he phasr sum wuld have is maximum magniude whih is 7. On he her hand, if he phase angles differ by 8, he phasr sum wuld have is minimum magniude whih is 5. Thus, he maximum value f he sum is 7 and he minimum is 5. P5.5 v ( v v s ) s( ω 45 ) ( ) 5 sin( ω 6 ) 5 s( ω 3 ) j j 75 s.6 j ( ).6 s( ω.3 ) 55

15 lags by 75 lags by 46.3 s s leads by 8.77 P5.6 3 T. 5 s m f T Hz ω πf 4π θ 36 max 45 T v ( ) 3s(4π 45 ) rad/s P5.7 v i ( ) s( ω 3 ) m ( ) 7.7 s( ω 5 ) P5.8 We are given he expressin 5 sin( ω ) 5 s( ω 3 ) 5 s( ω 5 nvering phasrs we bain j j j.5 Thus, we have 5 sin( ω ) 5 s( ω 3 ) 5 s( ω 5 ) ) 56

16 P5.9 sequene f MT mmands generae he desired pl is: :.:; v s(9*pi*) s(*pi*); pl(v,) The resuling pl is Nie ha he firs erm s(9π) has a frequeny f 9.5 Hz while he send erm s(π) has a frequeny f.5 Hz., he raing vers are in phase and add nsruively..5, he ver fr he send erm has raed ne half urn mre han he ver fr he firs erm and he vers anel., he firs ver has raed 9.5 urns and he send ver has raed.5 urns s hey bh pin in he same direin and add nsruively. P5.3 T / f / ms i leads i by he angle θ ( / T ) 36 (4 / 4.49) Therefre, he angle fr i is θ θ θ whih is equivalen P5.3 Fr an induane, we have jω. Fr a apaiane, we have. jω 57

17 P5.3 Fr a pure resisane, urren and vlage are in phase. Fr a pure induane, urren lags vlage by 9. Fr a pure apaiane, urren leads vlage by 9. P5.33* v ( ) s( π ) i jω j π π 9 ω π ( π) 9 ( ) ( π) s( π 9 ) ( π) sin( π ) i ( ) lagsv ( ) by 9 P5.34* v i i ( ) s( π ) j j ω ω π ( ).683 s( π 9 ).683 sin( π ) ( ) leads v ( ) by 9 Ω 58

18 P5.35 (a) Nie ha he vlage is a sine raher han a sine j eause is pure imaginary and negaive, he elemen is a apaiane. ω 5 µ F ω (b) j eause is pure real, he elemen is a resisane f 5 Ω. () Nie ha he urren is a sine raher han a sine j eause is pure imaginary and psiive, he elemen is an induane. ω 4.5 H ω P5.36 Malab m-file ha prdues he desired pls is: f::; *pi*f*.;./(*pi*f*e-6); R5 pl(f,) axis([ ]) hld pl(f,) pl(f,r) The resuling pl is: 59

19 P5.37 (a) 9 j eause is pure imaginary and negaive, he elemen is a apaiane. ω 5 µ F ω (b) 4 9 j 4 eause is pure imaginary and psiive, he elemen is an induane. ω 5 8 mh ω () eause is pure real, he elemen is a resisane f Ω (a) Frm he pl, we see ha T 4 ms, s we have f / T 5 Hz and ω 5π. ls, we see ha he urren lags he vlage by ms r 9, s we have an induane. Finally, ω m / 5 Ω, frm whih we m find ha 3.8 mh. (b) Frm he pl, we see ha T 8 ms, s we have f / T 5 Hz and ω 5π. ls, we see ha he urren leads he vlage by ms r 6

20 9, s we have a apaiane. Finally, / ω m / 5 Ω, frm whih we find ha.593 µf. m P5.39 The seps in analysis f seady-sae a iruis are:. Replae sures wih heir phasrs.. Replae induanes and apaianes wih heir mplex impedanes. 3. Use series/parallel, nde vlages, r mesh urrens slve fr he quaniies f ineres. ll f he sures mus have he same frequeny. P5.4* jω R j ω ω 565 : j j j 4.6 Ω ω 7 : 54 j 37.8 Ω ω 77 : 54 j36.55 Ω P5.4* s R jω R j R jω lags s by 45 m 6

21 P5.4 s R jω j 5 R m R jω lags s by 6.56 P5.43 jω jω ω 59 : j34.3 Ω ω 758 : j 4. Ω ω 55 : j 69.4 Ω P5.44 / R / R jω jω ω 57 : 3.58 j3.6 Ω eause he imaginary par is negaive, he impedane is apaiive. ω 66 :.669 j 4.63 Ω eause he imaginary par is negaive, he impedane is apaiive. ω 648 :.69 j.339 Ω eause he imaginary par is negaive, he impedane is apaiive. P5.45 s R j ω R j m R ( j ω ) leads s by 45 6

22 P5.46* s R j ω R j m R ( j ω ) leads s by P5.47 s.5 s R R leads s j jω by P5.48* s R s m ( j) R m ( jω ) m lags s by

23 P5.49* s m s R jω jω j.5 j.5 R R m jω 5 9 m ( jω ) 5 9 m The peak value f i ( ) is five imes larger han he sure urren! This is pssible beause urren in he apaiane balanes he urren in he induane (i.e., ). P5.5 s jω R j ω j 5 j 5 s R. j j The peak value f v ( ) is five imes larger han he sure vlage! This is pssible beause he impedane f he apair anels he impedane f he induane. P j j ( 86.6 j 5) j R jω j 5 R j5 j 5

24 lags by 56.6 lags by 9 P5.5 al jω R jω j. j. j 5 j 5 5 j R al R 9 R R ( ) ( ) j j j P5.53* Firs we wrie he K equain: Then we enlse ndes and in a lsed surfae frm a supernde and wrie a K equain: 65

25 j 5 j 5 find The sluin hese equains is: P The K equain is. 4 j5 j6 Slving, we 4 P5.55 The K equain is 4. Slving, we find 8 j P5.56 Wriing K equains a ndes and we bain 5 j5 3 j 5 j5 Slving hese equains, we bain P5.57* Wriing K equains arund he meshes, we bain 4 j5( ) 3 83 j 6 j5( ) 5 Slving, we bain: P5.58 The urren hrugh he urren sure is 4 Wriing K arund he perimeer f he irui, we have ( j 7) 4 8 Slving, we bain:.64.3 and

26 P5.59 Wriing K equains arund he meshes, we bain j ( ) j ) 5( ) ( 3 j 53 5( 3 ) Slving, we bain: P5.6 Malab prgram ha prdues he desired pls is: w::; e-3; 5e-6; magaabs(w*-./(w*)); magbabs(./((./(i*w*))i*w*)); pl(w,maga) hld n pl(w,magb) axis([ ]) hld ff The resul is: 67

27 P5.6 Malab prgram ha prdues he desired pls is: w::5; e-3; R5; magaabs(ri*w*); magbabs(./((./(i*w*))/r)); pl(w,maga) hld n pl(w,magb) hld ff The resul is: P5.6 The unis fr real pwer are was (W). Fr reaive pwer, he unis are vl-amperes reaive (R). Fr apparen pwer, he unis are vlamperes (). P5.63 Pwer far is he sine f he pwer angle. is fen expressed as a perenage. PF s( θ ) % 68

28 P5.64 (a) Fr a pure resisane, he pwer is psiive and he reaive pwer is zer. (b) Fr a pure induane, he pwer is zer and he reaive pwer is psiive. () Fr a pure apaiane, he pwer is zer, and he reaive pwer is negaive. P5.65 lad wih a leading pwer far is apaiive and has negaive reaive pwer. lad wih a lagging pwer far is induive and has psiive reaive pwer. P5.66 See Figure 5.3 in he bk. P5.67 Real pwer represens a ne flw, ver ime, f energy frm he sure he lad. This energy mus be supplied he sysem frm hydr, fssil-fuel, r nulear sures. Reaive pwer represens energy ha flws bak and frh frm he sure he lad. side frm lsses in he ransmissin sysem (lines and ransfrmers), n ne energy mus be supplied he sysem reae he reaive pwer. Reaive pwer is impran mainly beause f he inreased sysem lsses assiaed wih i. P5.68 Usually, pwer far rrein refers adding apaianes in parallel wih an induive lad redue he reaive pwer flwing frm he pwer plan hrugh he disribuin sysem he lad. T minimize pwer sysem lsses, we need % pwer far fr he mbined lad. Ulimaely, an enmi analysis is needed balane he ss f pwer far rrein agains he ss f lsses and addiinal disribuin apaiy needed beause f reaive pwer. P5.69* This is a apaiive lad beause he reaane is negaive. R (5).5 kw P Q X (5) ( 5).5 kr Q θ an an (.5) 6.57 P pwer far s( θ) 89.44% apparen pwer Q P 5.6 K 69

29 P5.7 This is an induive lad beause he reaane is psiive. 7 j P R (3.9) kw Q X (3.9) (38) 38.7 kr θ 54.6 pwer far s( θ ) 57.9% apparen pwer K P5.7 θ i θ v θ pwer far s( θ) 86.6% leading P s( θ).99 kw Q sin( θ) 7.5 kr apparen pwer 5 5 K P5.7 9 k θ i pwer far sθ % 78.8% lagging This is an induive lad. θ v θ 5 ( 3 ) P s( θ) 48 kw Q sin( θ) 8 kr pparen Pwer 8 K P5.73 (a) Fr a resisane in series wih an induane, real pwer is psiive and reaive pwer is psiive. (b) Fr a resisane in series wih a apaiane, real pwer is psiive and reaive pwer is negaive. 38 P5.74 f he induive reaane is greaer han he apaiive reaane, he al impedane is induive, real pwer is zer, and reaive pwer is psiive. f he induive reaane equals he apaiive reaane, he al impedane is zer, and he urren is infinie. Real pwer and reaive 7

30 pwer are indeerminae. This ase is rarely, if ever, f praial imprane. f he induive reaane is less han he apaiive reaane, he al impedane is apaiive, real pwer is zer, and reaive pwer is negaive. P5.75 f he induive reaane is greaer han he apaiive reaane, he al impedane is apaiive, real pwer is zer, and reaive pwer is negaive. f he induive reaane equals he apaiive reaane, he al impedane is infinie, and he urren is zer. Real pwer and reaive pwer are zer. f he induive reaane is less han he apaiive reaane, he al impedane is induive, real pwer is zer, and reaive pwer is psiive. P j Delivered by Sure : P 68 s( ) kw Q 68 sin( ).97 kr bsrbed by Sure : P 86 s( ) 3.7 kw Q 86 sin( ).77 kr bsrbed by resisr: P R kw R bsrbed by indur: Q X.8 kr P5.77 j ( 6 3) Delivered by Sure : 6.(3) s( ).93 P Q kw 6.3(3) sin( ).53 kr bsrbed by Sure : 8(3) s( 6 6) P kw

31 Q 8(3) sin( 6 6).9 kr bsrbed by resisr: P R.74 kw R bsrbed by indur: Q X.74 kr P5.78 pparen pwer P 5 s( θ ) θ 53.3 (We knw ha θ is psiive 5 beause he impedane is induive.) j5. 49 R 5.49 R.6 Ω. 4 H π 6 jω P P Q.63 pparen pwer 585 j66.5 sθ.85 kw sinθ 5.9 kr Pwer far s k ( ) % lagging P5.8* P Q Pwer far s j 3 sθ.85 kw sinθ kr pparen pwer 3.39 k ( ) % leading 7

32 P5.8* ad : P kw θ s (.9) Q P an θ kr ad : θ Sure: s Q P 5 k (.8) sin( θ ) s( θ ) kw 9 kr P P P kw s Q Q Q 3.84 kr s pparen pwer ( P ) ( ) Ps Pwer far pparen pwer s Q s 6 k % lagging P5.8 ad : P 5 kw θ s (.9) Q P an θ kr ad : P kw θ s (.8) ( θ is negaive fr a leading pwer far. ) Q P an θ 7.5 kr Sure: P P P 5 kw s Q Q Q 5.79 kr s pparen pwer ( P ) ( Q ) s Ps Pwer far pparen pwer s 5.84 k % leading 73

33 P5.83 P5.84 pparen pwer ( ) P R Q Q Q 4.38 ( ) ( ) ( ) X R X 447 ( 9.3) W 3633 W 37.5 R pparen pwer P Q 365 P Pwer far 99.48% leading pparen pwer R jω j ω 55 j8 j9.3 P R Q Q Q X ( ) X ( ) 93 Pwer far s % ( ) leading P5.85* (a) sθ.5 θ 75.5 P 4 P s s ( θ) ( θ) k(.5) 75.5 kw 4 (b) Q Q lad al Q X.58 ω 7 µf sin θ kr Q Q lad 3 ( ) X 74

34 The apair mus be raed fr a leas kr. Wih he apair in plae, we have: P kw () The line urren is smaller by a far f 4 wih he apair in plae, reduing R lsses in he line by a far f 6. P5.86 The a seady sae Thévenin equivalen irui fr a w-erminal irui nsiss f a phasr vlage sure in series wih a mplex impedane. The a seady sae Nrn equivalen irui fr a w-erminal irui nsiss f a phasr urren sure n in parallel wih a mplex impedane. is he pen-irui vlage f he riginal irui. n is he shr irui urren f he riginal newrk. The impedane an be fund by zering he independen sures and finding he impedane lking in he erminals f he riginal newrk. ls, we have. P5.87 T aain maximum pwer, he lad mus equal (a ) he mplex njugae f he Thévenin impedane if he lad an have any mplex value; (b) he magniude f he Thévenin impedane if he lad mus be a pure resisane. P5.88 The lad vlage is given by he vlage divider priniple. The lad vlage is larger han he Thévenin vlage in magniude if he magniude f is less han he magniude f whih an happen when he imaginary pars f and have ppsie signs. This des n happen in purely resisive iruis. 75

35 P5.89* (a) ering he urren sure, we have: Thus, he Thévenin impedane is j Under pen irui ndiins, here is zer vlage arss he induane, he urren flws hrugh he resisane, and he Thévenin vlage is n Thus, he Thévenin and Nrn equivalen iruis are: Ω (b) Fr maximum pwer ransfer, he lad impedane is j 5 P lad lad lad R lad lad lad j 5 j 5 ( ) ( / ) 5 W () n he ase fr whih he lad mus be pure resisane, he lad fr maximum pwer ransfer is.8 P lad lad lad R lad lad ( ) j 5.8 lad 47. W 76

36 P5.9 ering sures, we have: Thus, he Thévenin impedane is j 4 j 5 Wriing a urren equain fr he nde a he upper end f he urren sure under pen irui ndiins, we have 45 5 j 5 n Thus, he Thévenin and Nrn equivalen iruis are: Fr he maximum pwer ransfer, he lad impedane is j 4 lad lad P lad R lad lad j 4 j 4 ( ) 44.6 W lad n he ase fr whih he lad mus be pure resisane, he lad fr maximum pwer ransfer is 4.47 lad P lad lad R lad lad j ( ) 5. W lad 77

37 P5.9 he lwer lef-hand nde under pen-irui ndiins, K yields. 5 frm whih we have. Then, he vlages arss he 5-Ω x x resisane and he j5-ω induane are zer, and K yields ba 3 3 x Wih shr irui ndiins, we have 3 3 x j n ba s The Thévenin impedane is given by Finally, he equivalen iruis are: x x P5.9 Under pen-irui ndiins, we have ab (4 j3) Wih he sure zered, we lk bak in he erminals and see j3 j 3 4 4Ω Nex, he Nrn urren is n

38 P5.93 Fr maximum pwer ransfer, he impedane f he lad shuld be he mplex njugae f he Thévenin impedane: lad j 8 R Seing real pars equal: Ω R lad j Seing imaginary pars equal: 8 ω lad ( ) lad ( 8ω ) 38.6µ F ( ω ) lad lad P5.94* Fr maximum pwer ransfer, he impedane f he lad shuld be he mplex njugae f he Thévenin impedane: Y Y lad lad lad j 8 R lad lad.4 j.46 jω lad.4 j.46 Seing real pars equal: R.4 R lad lad 4.93 Ω Seing imaginary pars equal: ω.46 lad lad 44.7 µ F P5.95 We are given ( ) 8 s( 3 66 ) v an 79

39 (a) y inspein, ω πf 3 f 48.6 Hz (b) () v v v v bn n bn n ( ) 8s( 3 54 ) ( ) 8s( 3 74 ) ( ) 8 s( 3 74 ) ( ) 8 s( 3 54 ) P5.96 We are given: v s ω 6 v ( ) bn s ω 6 v ( ) s( ω ) an ( ) ( ) ( ) n The phasr diagram is: Fr unerlkwise rain, he sequene f phasrs is ab. Thus, his is a negaive sequene sure. Frm he phasr diagram, we an deermine ha v v v ab b a ab b a an bn n bn n an ( ) 3 s( ω 9 ) ( ) 3 s( ω 3 ) ( ) 3 s( ω 5 ) 3 5 8

40 P5.97* 3 Y R 3 P 3 s Y Y 9.36 kw ( θ) s( ) P5.98* Y R jω 5 j j Y Ω P5.99 Tal pwer flw in a balaned sysem is nsan wih ime. Fr a single phase sysem he pwer flw pulsaes. Redued vibrain in generars and mrs is a penial advanage fr he hree-phase sysem. n addiin, less wire is needed fr he same pwer flw in a balaned hreephase sysem. P5. This is a psiive sequene sure. The phasr diagram is shwn in Figure 5.4 in he bk. Thus, we have 43 an 3 The impedane f an equivalen wye-nneed lad is Y j Ω 3 The equivalen irui fr he a-phase f an equivalen wye-wye irui is: 8

41 Thus, he line urren is an a. j. P n lad P line an (. j.) ( ) 9 ( ). 9 W a a Y 8.58 kw P5.* This is a psiive sequene sure. The phasr diagram 5.4 in he bk. Thus, we have: is shwn in Figure an 43 3 The impedane f a equivalen wye-nneed lad is 3 Y.667 j.6667 Ω The equivalen irui fr he a-phase f an equivalen wye-wye irui is: 8

42 Thus, he line urren is: an a. j. P lad P n line (. j.) an ( ) kw ( ) kw a a Y 83

43 P Y an n a Y b Y bn an ab n bn 84

44 44 P5.3 Y an bn n ab b a a b an 5 j (.3) P W Q 3 Y Y 43 rsm R s ( θ) s( 66.5 ) ( θ) sin( 66.5 ) P5.4 The line--line vlage is The impedane f eah arm f he dela is / /( j 5) sin The equivalen wye has impedanes f Y Then wrking wih ne phase f he wye-wye, we have 4 line 6. Y pwer far s(63.44 ) % 44.7% P 3 (4)(6.)(.447) 5.83 kw 85

45 P5.5 s suggesed in he hin given in he bk, he impedanes f he iruis beween erminals a and b wih pen mus be idenial. Equaing he impedanes, we bain: () b a ) /( / Similarly fr he her pairs f erminals, we bain () a ) /( / b ) /( / (3) Then adding he respeive sides f Equains and, subaing he rrespnding sides f Equain 3, and dividing bhs sides f he resul by, we have: Similarly we bain: a and b 86

46 P5.6 s suggesed in he hin, nsider he iruis shwn belw. The admianes f he iruis beween erminals mus be idenial. Firs, we will slve fr he admianes f he dela in e f he impedanes f he wye. Then we will inver he resuls bain relainships beween he impedanes. () a b b a b b a Y Y / / Similarly wrking wih he her erminials, we bain () (3) Then adding he respeive sides f Equains and 3, subaing he rrespnding sides f Equain, and dividing bhs sides f he resul by, we have: a b b a b a Y Y a b b a a Y Y nvering bh sides f his resul yields: a b b a a Y Similarly, we bain: and a a b b a b a b b a a b b a 87