Homework7 - P10.5—2 40052 V Apply KCL at node a fl 025iv...

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Unformatted text preview: P10.5—2 40052? V Apply KCL at node a: fl+025iv+g=0 1 dr Apply KVL to the right lllESll: 4i+4ii—v=0 => v=4i+4ii {it of: After some algebra: d2 d , i+5—i+5i=49052r dr‘ dr Now use 5': Im Re{ej(2r+5]} and 4 I208 2 f = 4Re{eflf} to write d—me Re{e~“33"33'}]+ 5 i [1," Re{e“‘”’+‘9]}]+5 [Em Re{e*"13’+‘“}] = 4Re{efl’} dr dt Re{:r_2 _Im gills—6] :|+ 5 % ‘Im eJ-II2:+5]:|+ 5 _Im ejIIEI—S] 1}: Rte-{4 (ad-2r} Re {—4 #31“ efl' + 5(12 eJfiIm 333')+ 5 955;,” em} = Re{4 5”} —4 gig!“ + 5 (fl 2591m]+5 e591“ = 4 . 4 4 4 Ire-“"— m — = = =0.3934—34° —4+5(j2}+5 1+j10 10.05454 5(r]=0.398cos[2r—35°) A 2D kfl I VS = 25— 90° ‘5 + _ . -J -J . .- 2. =11: z. =—=—=— 160009 VS “3} 1 C wC (500)(0.125><10_fi} J —'15000 ' 100004—90o 24—90" V119}: J— (24—900) = # = 1354—1411311 20000—315000 256 121—39” tllfil'EfOI'E 0(1) =1.25cos(500r—141°) 1; 1110.54 0' dr (0.01)(;100]1-'+1-'=10 _ 10 ‘E v = 1.011 :05 (100 1—450] ‘0’ 0.01 1l+1==l0c05100r V =?.0?1 4—45‘3 P10.fi—1 P1fl.fi.3 Plfl.fi—4 2.5 Z ‘1353 A PlflJ—fi Replace series and parallel capacitors by an . . . . 4 H EQILLTBIEIIT capacflor and serves mduc’rors by an equivalent inductor: l—’ 100 Q 5 mF Then 2 1 . 100 . 2'00 _ 200 2 J'm(5><10_3) IDO[_J?] ‘J— 1+}— Z=jw4+ 1 =jw4+ mm- =jm4+ w x (f: 100+ _3 lDO+[—_}"—] l—j— 1+}; jm(5 x10 ) c:- I co (9 11-3 1 — '7' 7' z = jru4+1OUm—m=jm4+100 4 j“? = 400,, +j[4m— “(303] 1+ 4 4+8)" 4+m‘ 4+m“ m2 P1fl.3—14 chrcscu’r the circuit iii The flcqucncy dciuain using iiupcdauccs and p113 scu‘s. Lct _ _ ism—gm) Zl _3 fl jlflXED-Clfl 413—350 and . 1 . . I1fl[25) zw =_J—3+_;m||25=—;2c+—=12.5z—33.7 fl ' 1D[5><1{T] 25+;2c Z1 and 222 are cc-uucctcd in parallel. Clu‘i‘cnt division gis‘cs I1 = —>< flfllifli‘: = UDELMSITD A .3‘;1+2",1 11 3-12) Mr]=D.Ul4ccs{10f+32.?°) A Plfl.8—1fi Repreaen’r the circuit in The irequency dmuain thing impedances and p113 501‘s. Let 1 2 =25 '20 2 —=25 15:29.2.13109 1 Jr“ ) +j(20)[fl.flfi2] w z? = 2fl+é= 2a— I10 = Hafiz—25.5“ fl - Hmfiams) z} = 40+ j(2fl)2 = 4D+ Iii-U = 56.5Tfi45° Q alldlflt ZF =32 II 2.3 =13.36£—3° =13.67—jl.67’ (2 Then T '3' Z I=£x—1=U.118£6.1°A ZI+ZP 13112+Z3 3'0 i(r)=0.113cos(zflr+6.10) A PIES—EU (a) Uiing voltage divisinn tn-‘ice 4'1] IUD = K24——:<24=—12 v 4n+3n in+1nn 1’0) (b) Represent the circuit in the fi‘eqnency deninin naing phnaci‘a and impedance-3,. Where 21:21?!) 1 Z = 1D 4+ —||l'[:|' =12.2+ '7D.2=71.3fl£3fl.2°§1 1 J“ ) [mun-am) ] "I I3 =j(lfl)3+;+25=25+j5fl= 55.9fli’63.4°fl j [2D][D.DD5) 1 2“ = j [2n){nnn4) +15 =15—j12.5 =19.53£ —39.3° fl Using voltage dii'iaicn twice 1E1a Z ' x24£45°——4 x24£450= 24.8186)" V 21+ 3';a 23 +334 17: '30 fit] = 24.8ccs [20m an”) 1? P111342 (a) The node equations are 24 — Va 4'3 24 — vb 25 01' l 1 1 4G 2'1] 1 5 1 _EE 5011:ng Llaing MATLAB give-s 1? = 3.713 V 3. 2D 1 1 l 25 2D 5D vb and vb 212.69 "1-" 24 40 24 25 {13] Use 1111354215 and impedanees to repreaent the circuit in the frequency domain as Z 1 1where EL . 25+;[21214 = 25+;211 = 2232222.? 12 K 2.1: 41111 +;{20]5 = 3.56+j33.fi = 22.53232? 11 ‘ “2111111201141. 23 = 211 12 21 = 15+;{21212 = 15+;411 = 42.222524“ , 1 2 = ' 21:1 3+—= '51::5112901112 i H ] j{20]{0.005} “r The node equations are 24145°—T, '1; Wig—Yb —= + 24221524;b 1'34}, Th —+ =— 2; 23 IE 1 1 1 1 24245“ +—+— —— 1" I: 23 24 23 = 21 1 1 1 1 , _ 2424? —— —+—+— ‘11, 23 2] 23 25 2; Selving 115ng hL'fl'Lu-EB gives 1', = 222144.12“ 1'], = 3.45.5421“ 50 v: [I] =T.39cus[2{}1'+4—’1-°] ‘J v, [I] =3.45e05{2[|1‘+45_1] V Pius-1:5 Represent the circuit in the flequenejt domain using phasors and impedanees. Apply KCL at the top node of R and L to get [fiflz—?5°j—‘t'+35.{1fl-E}°—T_ 1' 1-411 40 RIIJ'coL sea—15° 355111? [-“1 1 1 1" _ = _ _ __ ._ T 411590“ + 411 k_,t=1o+4t1U2: Jase, Using the given equation for 1111') we get 21.255—163.E°=T= Lafiilfilj fi {1.1123 1—' __ -_ 1* “+111 J 21:11: then 1—1; =w—oossti—j] = U.fl4—jfl.flll?fi R 20L 21231—1533“ finally 1 1 R=—=2'fl 3.11:1 L=—=4.ssH o. 3 2D[fl.flll?fij Pitts—1s Represent the circuit in the fiequenejr domain using phasors and impedanees. Lahel the node voltages. The node equations are 24115“ — Tu Tu Tu — Tb — + 25 j 413 H} 24111213, "Lg—1; 1'], + :— 1,115.25 111 45 01' 1 _1 1 1 T 24.51? ——_,r—+— —— a 25 411 111 1:1 _ 25 1 . 1 1 1 ‘ 24151:?“ —— J—,+—_+— Th — 10 +5.1: 4: 111 ease—911° {1140— 111.015 41.111 f1; __‘11.95r1.g15° 41.111 fl.1112+jfl'.lfifl'_Th ‘133411111151’ Solving gives T1 = 24.5?[32.fi° V and T1, = 25.595252“ RF Then T, — Tb = D.334?1134_9° A 50 :{r}: fl.334?cos{101+134.9°}£ P1 [LID—9‘ H i! .I'I. M 5:! - z._= [13]“): 145—53.? r1 —_,i'3+4 T-P‘f-fi O 1”“ 444—11320 I I: =1] +J'4 =1.44+J'2.IJS ass/155.3“ r: z =3_515—3?_9°fl Vii! —fti.-'1 .|" I I '? ii! ,3' LE 3 “’1— ‘ 3 J =2??—J2.16§l 4' ' _ - t. ’ D =[2.ssz—?B.4°‘i tile—3” Lisa—9M .r’ I I= 2.35z—T3A” I: jiklII‘T—jllllhflj “15245—244”; PIIJJIIJ—ll Use superposition in the time deflmjn. Let flit} be the part of fit) due to- vilfi} and hit} be the part efffij due to vim. To determine f]{:‘)__ set viii} = I]. Represent the resulting eireuit in the fiequenejr' domain to get EI 9 Hum“ v 3: 1.=.'1_”iere 2.: = 2fl+ fit}: 32.461Tfi“ fl I:=1G+[J4‘U‘II15:I=23.15+j4.93=23.57£12fifl l 13 = zs+_— ; [rename] = ZD—jlfl = 22.355— 26.6“ 0 Next, using Dhm's law and current division gives = scum” x 13 = Ejfiflfl‘flc“) 1 114:2”13] zl+zg sizz+zzzg+1113 I ={}.132.x_’—1T.IS° A 50- f{f]=fl.132cos{2flI—1T.fi°] A To dflermint ism, sat Valli!) = CI. Represent the rasultins circuit in Hill? fi'equenq' domain to gist where 2.4 = 2D+j4fl = 443211534" 5'! 2.5 =113|+{J'2{}||15] =19.fi+fl.2 = 25.33.5202“ 5'! l 15 =2D+T=2flqlfl= 23285—45” n JiisHssss}. cht, using lElizlnl's law and current division gives L=—13‘5‘15D x 1* =—zl[13£_15ai =ssiins~=s ' 15+{z4uz,] L+L zlzi+zizj+z;13 50 f:{f:i={}.3TTcos[lflf+13°j A. Using sup-exposilion, .iIII] = i][i:i+f1[r]|= fl.132cns{2flr —l?.fi°] + fl.3i‘i‘cos[lfli‘ +13”) A. P'lfl‘JZ—ElI Represent the circuit in the frequency domain as Applyr KC'L at the top nnde nf the impedance nf the enpaeitnr to get Via—4T: +% => %1'5=[1+;[5x1n5)c)v leUC Applyr KCL at the inverting nnde nf the op amp tn get 1—4+‘”=n : Eff—i4? in R in SCI R Va —2><1{J4 r5 =1+j(5><1{]5]C Converting the input and nutth sinnmids tn phat:an give-3. YE, _ 3513? 1-1 45:1“ .1 =2fi135° SCI _ 31 R 35133“ =A — #51303—155‘1[[35105]C] 1+:‘(5X1fi5]c — 1+[(5XIDS]CT Equating angle-3 givaa 1 43° 133° =18UD—1011'1((5><1{]5]C) 5 C =L.) = .1510“5 = 2 1:]: 5510* Next. equating 11mg1i1‘11dea givea R R 2=—3><MJ4 =2”)4 5 R=104=10 151 1+[35105)(2><104) ~45 13115—5 00: 05:5 0.1 11:]: 00(0.2—:] 0.15:5 0.2 _0 0.2 5:503 1 0.1 1 02 3 9131 +0.1 0.2 2 K; = EDD (00:) dr+ID_1[9{J(D.2—r]] :1:]=7[Jfl rldr+Im [0.2—:) 13:] 2 = £[fl+fl]=lg 1: .3 3 3 V =Jfi= 4.241? I'fl‘JE KVL: [lfl+j'2fl][l ago—“:21: :a {row-sop] +jsr1 =ssb° KCL: 11+11=sao° SolViJig these equations using Cramer's rule: ID '21} '2 = +3 J =lEJ+J13 l l ' '2 '_ - '3 r. =1 3“? = 3' J?- = Assess? A =—o.ss—;o.5A ' d s 1 113+le rl= 5—1.: s+o.ssr+ 1.5: sssqs = 45.41544?“ A Now we are read}r to calculate the powers. First. the powers delivered: sm. =%{sgo'][—r; ] = 2.5[o.411{18{l—4.4?]] =—rs.o+;1.ss VA sm.=%[s—j2r.]{szo'] =[s—;2{s.3s+;.51:|s = iso—jsss VA sm. + sm. = so J31: VA 51ml dolls-moi Next, the powers absorbed: It] . l 2 2 5m: Eio|rl =Et.ss] =2.oVA 51m = gm 3 = “:41: VA s_j._., =%{—j2} I1 3 = —; [+5.41]I =—;41.1 VA sm = 2.0—;er VA absorbs-cl To our numerical accuracy, the total complex power deliVered is equal to the total complex power absorbed. P 1 1 .6—6 (Using 11115 V0 Ines} 120 |szng|=g=1+12 Q HOLD Vlms “Range 21%.! = 14.12445°=10+j10[2 RR 1 1 ange ‘f— R =—= =144 Q 1m” P 100 120,32 Vrms 240 2 Rm =£ ) :40 Q g“ 11.000 {a} Ina- = ufl‘éfi =3.SZ—4S°Anns, It”? = 1mg” =111+.8_‘Lill31'=“ Anus ’ 10+;m and 24010“ I = =5EJ£EJ° A "3‘5"" 4.3 From KCL: Il =InfiE+Iw =Sfi—jfi= 55.31451“ A L = 4m 4m = 50.33.5130” A ; IN = —I._ —11 = 1921131“ A {‘33 Pnfi3=|lmfi€2 Rflfisflzzjw and gm=|lmixfls=mjum Pm=1mw and gm=0 PM = ?22+ma+12_000 =123-2 m1 ' > 5 =12 3-22 1’22 = 12.34432“ INA gm=m+fl+fl=m2mfl j 2’ ’ H The [menu pawm factorispf=ms[3_2°j= 11998, {C} TUE! 1P[J,:"[_J U:- o “Mr”; 1 l‘lHnI'mgH _.I '| L'- 5;: K Range 2L] 1;: '1 .11 fl 1mm Meshequatinns: 3mm —20 —1{}—j1{}__[_.._ T120100 —2<:+ 154 444 -1B =-12{Lfifl° —1C|—_;'1{l —144 153mm i_I,: i o Solve to get: IA = 64.3—j15? =6432—1J” Arm IEl = m.3-;n.19= Mid—QTAJIIE I: = frfl+jfl= fiflfi’fl“ Arms The vultage across the lamp is |11m| = Rm |IB —IC| = 144|1.2M—s.6°|=133.2 1.! P11.fi—12 Represent the circuit 111 the frequency 11011121111 :15 .man [hm-n" {a} * 2(14.12+;3.42j 1 = — =13M44°4 “ [ 24:2? ] PF 2 3.42 14.12 IE: W =1.3?flfi°A 241:2? [=1A+IB ={1.3?£44°_}+[1.3Télfi°_}={fl.9315+jfl.954_}+[l.319+ffl.3??} = 2215+ “11.231: 2.662131)” 14 =£=9fllfiz45° 2.552.231? s =%{24g25°}[2.652g3e°}* = 219.545“ = 2259+;‘2259 v11 2[s.42+ 214.121“ 1 = —« =132g15°14 {b} "‘ [ 2422? 13:24‘5? =132z44°21 15+J9 1=1_.L+1B 4.4424311” .41 24515“ =— = 9131115545“ E! 11515215343” 5 = 22.59+j22.59 V251 24|IA| {c} P=14_12 W: cosfifi—HA} ,3 0.35?=cos{?5—afi}l :> .50 = r5°—31°=44° IFS—EA :: 0 | Then 2(1412] —=1.3]II 24ccs[31°] lhl= 5-0 rA =13?ng A A150 _ 2455“ _ 9+ :15 I=IA+IB = 2.552130“ A [B =131£1fi°£ {I111 24.55“ IA = .. _ 13+}? _ 34:?5“ 9+J'15 I=I_4_ +IB = 115152.530” A =1.3?z_’44° I =1.3T.§1Ifr° B 341’???“ = — = 9.0115545“ 5."! 115152630” 5 = 22.59 +j23.59 VA Pll.8—4 ZT=400+j800 Q and zL=2000 || —j1000=400—j800 9 Since ZL = the average power delivered to the load is maximum and camiot be increased by adjusting the value of the capacitance. The voltage across the 2000 Q resistor is VR :5 ZL Z +Z r L P: E Lasmw JE 2000 = 2.5715 2 5.59615“ V So is the average power delivered to the 2000 Q resistor. ...
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This note was uploaded on 01/24/2010 for the course ECE 2025 taught by Professor Juang during the Spring '08 term at Georgia Tech.

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Homework7 - P10.5—2 40052 V Apply KCL at node a fl 025iv...

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