Solucionario Cap 10 Circuitos Dorf Sodoba 5 Edicion

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255Chapter 10 Sinusoidal Steady-State Analysis Exercises Ex. 1 0 . 3 -1 Ex. 10.3-2 Ex. 10.3-3 Ex. 10.4-1 Ex. 10.4-2 Ex. 10.5-1 KCL: i v R +CdvdtdvdtvRC = ICcos tsm+ Try v t = Acos t + Bsin t & plug into above D. E. Asin t + Bcos t +1RCAcos t +Bsin t = ICcos tequating sin t & cos t terms yields A RI1+ R C and B R C I1+ R Cfmm2 2 22m2 2 2 j L = j 3 1 = j 3 KVL : 10 + j3I + 2I = 0 I = 102+ j3 10 0 56.3 = 1013 i (t) = 1013cos(3t ) ,,,,1356 356 3..102 36 . e = 4.24e = 3 j3j45j45 122 2 1i 5cos 5t 12sin 5t = ( 5) +(12) cos 5t (180 tan ) 13 cos(5t 112.6 )5| ` + + . , v t = RI1+ R Ccos t +R C I1+ R Csin t v t = RI1+ R C cos t ( RC)fm2 2 22m2 2 2fm2 2 2 tan1( )2 2 14v 3cos 4t 4sin 4t = (3) (4) cos 4t tan 5 cos(4t 53 )3 + + (a) (b) T = 2 / 4 2 /v leads i by 30 ( 70) = 100 256Ex. 10.5-2 Ex. 10.6-1 Ex. 10.6-2 Ex. 10.6-3 ( )( )0.01 10 cos 1000.01 100 10107.071 4517.071cos 100 Adv v tdtjjv t+ + +V VV i = 8sin( t = 8cos( t 110 ) = 8Re{e e } I = 8e = 8j t j110j110 20110, ,) V = 80 + j75 = 109.7 43.2 = 109.7e v(t) = Re{109.7e e } = ( t +43.2j43.2j43.2 j t 109 7 . cos )jj8 32e8.54e = 328.54e = 3.75 ej90j111j (90-111) j21323 + (a) i = 4cos( t = Re{4e e } I = 4e = 4j t j80j80 8080,,)(b) i = 10cos( t + = Re{10e e } I = 10e = 10j t j20j20 2020 )(c) V = 10 = 10e v(t) = Re{10e e } = ( t 140j140j140 jwt 14010cos ) (a) (b) 257Ex. 10.6-4 Ex. 10.7-1 Ex. 10.7-2 Ex. 10.7-3 v = 1Cidt cos100tdt = sin100t = 50cos(100t 10 5 50 903) v cos100t = eKVL : i(t) +10 10di(t)dti(t)dt = vsj100tSt +

40 415 1033Re Assume i(t) = Ae where i is complex number to be determinedPlugging into D.E. yieldsAe jAe j2A)e e A = 41 j eso = tan i(t) = Re2 2e e e cos(100t +45 )j100tsj100t j100t j100t j100t j45j100t j45 j(100t -45 )+ + (Re4 2 211452 2 2 21 dv 6i = C 10 10 [100( 500)sin(500t+30 )]dt= 0.5sin(500t+30 ) = 0.5sin(500t+210 ) = 0.5cos(500t+120 ) From Figure Ex. 10.7 - 3 we get i(t) = I sin t ; I = I Av(t) = V cos t ; V = V Vi(t) = I sin t = I cos( tThe voltage leads the current by 90 , it is an inductor Z VI VI VIalso Z j L = L 90 L = VI or L = VI (H)m mm mm meqmmmmeqmmmm 9009009090)(a) v = Ri = 10(5cos100t) = 50cos100t(b) v = Ldidt sin100t sin100t = 5cos(100t +90 ) 0 01 5 100 5 . ( )(c) 258Ex. 10.8-1 ZR = 8 , ZC = 1 2.4 2.42.41512jjj j jj , ZL1 = j 5 (2) = j 10 , ZL2 = j 5 (4) = j 20 and VS = 5 -90 V. Ex. 10.8-2 ZR = 8 , ZC = 1 4 441312jjj j jj , ZL1 = j 3 (2) = j 6 , ZL2 = j 3 (4) = j 12 and IS = 4 15 V. Ex 10.9-1 ( )90 511105 3.98 10j jje ej +V ( )90 902205 5.6820 2.4j jje ej j V ( ) ( ) ( )51 901 2473.9 5.683.58j jje ee V V V Ex 10.9-2 ( ) ( )15 6818 64 19.28 6j jje ej +V ( ) ( )15 75212 44 2412 4j jj je ej j V ( ) ( ) ( )221 214.4 je + V V V 259 Ex. 10.10-1 Ex. 10.10-2 Ex. 10.10-3 KCL at V V4 jV Vj10j12 V j2 V j40aa a ba b:( ) ( ) + + + 214 4 20V V Vb a bKCL at V : .5 90 0 ( 2 j4)V (2 j6)V 10 j20a bbj10 2+j4( 20 j40) ( 4 j2)(10+j20) (2 j6) 200 j100Using Cramer's rule V 5 296.5a(4 j12) ( 4 j2) 80 j60( 2 j4) (2-j6)v (t)= 5 cos (100t+296.5 )= 5 cosa + + + + + + + (100t 63.5 )KVL I : j15I I I (10+ j15)I I (1)KVL I : j5I I I I j5)I j30 (2)1 1 1 21 22 2 2 11 2aa+ + + 10 2010 2010 30 9010 10( )( )(From Cramer' s ruleI = 20j30 10 j510+ j15j5 j200j100 Now V j15)I v t) cos( t +82 )V1L 1L ++ 101010 10200752 263 8115 90 2 263 81 24 2 8224 2. .( ( )( . . )( Writing mesh equations: + j50)I I j30 I j20)I j20I j50 j20I j10)I Solving these equations givesI j I j1.27, I j1.05ThenV I I ) = 14.3 VV V j50 = 36.6 83 V1 21 2 32 31 2 3a 1 2b a(((. . , . .(10 1010 1030 0087 0 09 132 0510 72 + + + + + + 260Ex 10.11-1 90 511105 3.98 10j jje ej +V 90 902205 5.6820 2.4j jje ej j V 51 901 2473.9 5.683.58j jtje ee V V V ( ) ( ) 8 10 2.4 208 10 2.4 20tj j jj j j + + +Z 4.9 + j 1.2 Ex 10.11-2 ( )90 511105 3.98 10j jje ej +V ( )90 902205 5.6820 2.4j jje ej j V ( ) ( ) ( )51 901 2473.9 5.683.58j jje ee V V V 261 ( ) ( )15 6818 64 19.28 6j jje ej +V ( ) ( )15 75212 44 2412 4j jj je ej j V ( ) ( ) ( )221 214.4 je + V V V Using superposition: v(t) = 3.58 cos ( 5t + 47 ) + 14.4 cos ( 3t - 22 ) Ex. 10.11-3 Ex. 10.12-1 Ex. 10.12-2 a) Turn off current source, use phasors with = 10 rad/sec Zeq1 1 111 j10 1010 j10 j)KVL : I j15I j)I I j10 i t) = 0.707cos(10t A + + + + 5 110 5 5 1 010100 707 4545((.( )ab) Turn off voltage source, = 0 rad/sec Current divider I = A2 10153 2So by superposition i(t) = 0.707cos(10t A 45 2 ) 23 361 11 10 1 1010 LC = 1000rad ( )( )secDiagram drawn with relative magnitudes arbitrarily chosen 262 Ex. 10.12-3 Ex. 10.14-1 Problems Section 10-3: Sinusoidal Sources P10.3-1 Two possible phasor diagrams for currents I = ILC CL25 15 202 2 Now if I I I I impossible) from case (2) I I I I LC L C CCL C L C + 6 20 1420 6 26(Z R X X jR )R +X and R = 1k , X C k Z j1) j kZ R kVV ZZ j12 j11 1 1 112121 1112 2os21 + (( )( )( 1 11000 1011 1 11 11212111216 (a) i(t) = 2 cos(6t +120 ) +4 sin(6t cos6t cos120 sin6t sin120 )+4 (sin6t cos60 cos6t sin60=2.46 cos6t +0.27 sin6t = 2.47 cos(6t 6026 26)( ). )(b) v(t) = 5 2 cos8t +10 sin(8t +45 )= 5 2 cos8t +10[sin8t cos45 +cos8t sin45 ]= 10 2 cos8t +5 2 sin8tv(t) = 250 cos(8t sin(8t + 2656 5 10 634 . ) . ) V 263P10.3-2 P10.3-3 P10.3-4 P10.3-5 22 2 f = = 6283 secT 31 10v(t) = V sin( t+ ) = 100 sin(6283t+ )m1v(0) = 10 = 100 sin = sin (0.1) 6So v(t) = 100 sin(6283t+6 )Vrad f = 2 2 Hzi(2 10 cos(2.4 +55 )but 2.4180 So i(2 10 cos(432 +55 ) = 300 cos(127 ) = mA +

1200600300 1200 2 10 55 3432300 18053 33) cos( ( ) )) .a) A = 10 T = 3.9ms ms = 3.3ms = 2T rad10 cos ( ) = 0.87 = 30vs(t) = 10cos (1900t +30 ) V 0 61900. sb) A = 10T = 12ms ms) = 5ms = 2T rad10 cos ( ) = 0.87 = 30 v(t) = 10cos (1260+30 ) V( . . 10 9 0 91260 s 264Section 10-4: Steady-State Response of an RL Circuit for a Sinusoidal Forcing Function P10.4-1 P10.4-2 P10.4-3 Section 10.5: Complex Exponential Forcing Function a) Complex Numbers P10.5-1 P10.5-2 LdidtR yields didti = cos 300tTry i A cos 300t + B sin 300tdidt A sin 300t +300B cos 300tyields A+120B = 0and 300B+120A = A = 0.46 B = 1.15so i(t) = cos 300t sin 300t = 1.24 cos (300t Ai sff+ +

v 120 4003003004000 46 115 68 . . )KCL : iv2Cdvdt dvdtv = 500 cos 1000tTry v A cos 1000t + B sin 1000tdvdtA cos 1000t +1000B cos 1000tyields A B=0and 1000B+500A=500 solving B = 0.4A = 0.2so v (t)=0.2 cos 1000t +0.4 sin 1000t =0.447 cos (1000t Vsff + + + +

0 50010001000 50063 )(j4) (.05) = j(0.2)j45 j4512e 12e3 j45~I( )= (210 )e i(t)=(2) cos (4t+45 ) mA6000+j(0.2) 6000 ( . ) ( . )(. ...5 36 9 10 531450 16 21010 16 25 26562 5 10 36 + j3)(6 j8) j53 2 4535 81.87 4 j3+ 5 81.87 [4 j3+ 36.87 ]55 2 8.13= 5 81.87 (4.48 j3.36) = 5 81.87 (5.6 36.87 ) = 28 +45 = 14 2 j14 2 ] ] + + ] ] + + + 265 P10.5-3 P10.5-4 P10.5-5 b) Response of a circuit P10.5-6 P10.5-7 Z j(25 10 j 4000 Zj(25 10j 0.004L6L6 )( ))( )160 1010 1066A CB j7)5e6e j* * j2.3j15 (. .30 65 6 31( ) ( ) .. .6 120 4 121121 213 + j3+2e j21.3so a= and b=j15a) Ae = j(3 b) = 4 +(3 b) e b b = 3+4 + tan (120 ) = A = 4 +(3 b) j120 2 2j tanb2 2 +

+

4120343934 3 393 8 0013412 2tan .( ( . )) .b) + + 4 8 154 8 1526 10 21 cos + j(b +8 sin ) = 3e j2.6 cos 2.58 = 72 b +8 sin (72 ) = b j120 .. cos.c) + 10102020 608 66j2a = Ae A cos 60 jA sin 60 A = cos 60 a = 2 j60;sin. Z Z j(10 j 10,000, Z j(10 j 10,000I( ) = VZ +Z +Z j 10000 j 10000 i(t) = 1 cos( t +90 ) mA, = 10 rad/ sec.R L7c7sR L C7 + 100 1 1010 1001 901000 001 90312, )( ))( ).. I( ) = VZ +Z j 0.004 i(t) = 0.1 cos( t + AR c 20 4