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solution Chp8 Basic engineering circuit analysis, Exercises of Electronic Circuits Design

Basic engineering circuit analysis solution chapter 8

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Download solution Chp8 Basic engineering circuit analysis and more Exercises Electronic Circuits Design in PDF only on Docsity! ) A Download any solution manual for free All Departments , All Subjects www.myUET.net.tc CREATE YOUR OWN WEB PAGE VISIT www.myUET.net.tc Download any solution manual for free All Departments , All Subjects www.myUET.net.tc Irwin, Basic Engineering Circuit Analysis, 9/E 8.12 Find the impedance, Z, shown in Fig. P8.12 at a frequency of 60 Hz. 10 mH 20 > 10 uF Figure P8.12 SOLUTION: 10 mH 20 = 10 F 2.5 j(341)(}0m) = j3 772 2c] 5 265-25 2 30377) (10H) j 277 Dlr. - at, nN an oO NS S Z2 = [ 2 -j265.25 | + j 3.77 Chapter 8: AC Steady-State Analysis Problem 8.12 www.myUET.net.tc Request a solution Now And Get it within one week Absolutely free “We Feel Your Feelings..!” Irwin, Basic Engineering Circuit Analysis, 9/E 1 8.25 Find the frequency at which the circuit shown in Fig. P8.25 is purely resistive. Z—> $in 5 mH > 1 mF Figure P8.25 SOLUTION: = 1 mF = 447-2) xaklfZ Jce = JS5m(im) Fo ww = 447-21 IL 2m Fs Th2 Aa Chapter 8: AC Steady-State Analysis Problem 8.25 Irwin, Basic Engineering Circuit Analysis, 9/E T= 06 £53-26°A L(t) = 06 Cox (10004 53°26") A Ve = (0°6 £-53-26°) (-j 14-93) < " Ve = 8°96 /-143:26° V Ve = (0:6 453- 26°) (20) v Vp = Ite 5326°V CREATE YOUR OWN WEB Im 4 PAGE VISIT www.my UET.net.tc q0° mes > 7 —56-26 Re Ve 143.26 7 Vv Vs Problem 8.29 Chapter 8: AC Steady-State Analysis Irwin, Basic Engineering Circuit Analysis, 9/E 8.33 If v,(t) = 20 cos St volts, find p(t) in the network in Fig, P8.33. 30 05H re v(t) @) tHE ©1093 0.02F Vo(t) Figure P8.33 SOLUTION: Zi, = j5(05) = j 2-52 2.7 JCS = p52 Zz = a) Tj Jo JC5)(0-02) WMWA- Tana" 3 INS A | . toe 20 Lo"V jv 3 jon yar z= (-jlo lite) His = [ ett) Ilys lo- J 10 J = Pot Ztyse nl Chapter 8: AC Steady-State Analysis Problem 8.33 www.myUET.net.tc Request a solution Now And Get it within one week Absoluteley free “We Feel Your Feelings..!” Irwin, Basic Engineering Circuit Analysis, 9/E 8.38 Find é,(t) and i2(t) in the circuit in Fig. P8.38 10.cos (25f + 20°) V ® 50 0.01 F \W\——+ t in(t) i(t) 80 mH 40 f 200 mH sy 20 cos 25¢V Figure P8.38 SOLUTION: Zy= j 29 (8om) =j2-2 Zu= jQs)(vom) =j52 Ze= | = 7) 4 so 4 bet ond WW JN _ t, “JUD T, Ip £20°V @) jn Yn j5D *) 20 LV 50 AWW it _ |; = 102 20°V Ti y 34? 71, ©) j22 un f j52 T/=0A, Short across U2: T’ 52 104 20°v Chapter 8: AC Steady-State Analysis Problem 8.38 Irwin, Basic Engineering Circuit Analysis, 9/E a ° T's 102o = 156 [-1466A (j2 l|-]4) +5 tq! - 56 * ap" aa “) Ty’ = 3:12 1-16. 66°A Wt iy = Zz" “/** 3) 40 js © 20]o°v Problem 8.38 Chapter 8: AC Steady-State Analysis www.myUET.net.tc Request a solution Now And Get it within one week Absolutele free “We Feel Your Feelings..!” Download any solution manual for free All Departments , All Subjects www.myUET.net.tc Download any solution manual for free All Departments , All Subjects www.myUET.net.tc Irwin, Basic Engineering Circuit Analysis, 9/E 8.46 Find the frequency-domain voltage V,, as shown in Fig. P8.46. + -[§ 2 >R ions V5 0.525°A OD jen US Figure P8.46 SOLUTION: To ~ 05 L25°A zl 18-2 Mo y= j2u-J5 = jr (-V5) -j5 tyr Z= 322 2A? 2 iz 3-33 290° Coser . 3-332 90° +10 Tr = onsge 7658 A Voz Io (01582 96°58") Vo = 158 29658 V Chapter 8: AC Steady-State Analysis Problem 8.46 2 Irwin, Basic Engineering Circuit Analysis, 9/E Zr = 2}Iy) = 2 () 2-J) Zz O84 2763-4292 j2n a mn oO ’ ~64- t lo 73 2-63-43v an BVp Vo = 2. (10-73 2-63" 43°) 22+ OP4L-G3 43° . Vo = &L-90:)°V em Tne, $ 0 Problem 8.54 Chapter 8: AC Steady-State Analysis Irwin, Basic Engineering Circuit Analysis, 9/E 8.62 In the network in Fig. P8.62 1, = 4 {0° A, find I,. 193 -j1.0 2/2 A 19s Iy wfrv 310 Figure P8.62 SOLUTION: kcL at G) : I,= I,+ T, Vi-Vy = V2 = 1 JI -JI JI (W-9)) = V,-V3 7 Vit Va “jiy, +j1W+V, 7 Wy = 0 Chapter 8: AC Steady-State Analysis Problem 8.62 Irwin, Basic Engineering Circuit Analysis, 9/E KL at@: TT, +2l0° =I, Y-Y +2L0° = —Vy ! A- Vy JIC Vs-Wy ) = SW Yo -jlV2 + (-1452) Vy =9 Vu = 1( 420°) = 420°V My t+ C-14 §2) Wy = 12240° Vv) = 12£40° - Ylo® (-I4jU Vv, = 366 £ 45° Y Ui - Vy, = 2L0° Problem 8.62 Chapter 8: AC Steady-State Analysis Irwin, Basic Engineering Circuit Analysis, 9/E 3V,-V2 = 102 180° JIG, +(-INDV = 4290" VM = 822 L-179%3P Vv VW= O55 L-563°V To = 3-234-1773° { To = 3-22 21177:2°A Problem 8.66 Chapter 8: AC Steady-State Analysis Irwin, Basic Engineering Circuit Analysis, 9/E 8.68 Find V, in the network in Fig. P8.68 using nodal analysis. L yi WW t JW 20 ito 10 j2a 12°V i 2/0? A ) 2 ¢ 4fov ee V5 Figure P8.68 SOLUTION: J, 1 VW iv 20 -j10 12/0°V © 2/0? A KcLat O: I,+T, = 240°4+T, 220 - Vy + Y2ZO™ V, =2L04+ My 2-jI ) rts Vs 4+ M +YW = 122o° +4z20°-220° 2452 2) 2-)) D> ar Zba-uy® 65 LZA-7M® Chapter 8: AC Steady-State Analysis Problem 8.68 Irwin, Basic Engineering Circuit Analysis, 9/E VY, = 437 2212°V T= YW = YzZapee 2432 2452 T= 155 2 -23°8°A Vo = 20155 2-23-7%) Vo = 212-2328 VY Problem 8.68 Chapter 8: AC Steady-State Analysis 2 Irwin, Basic Engineering Circuit Analysis, 9/E -}!I, + jrT, = 620° -j2 7, +(2432)T, = 12245° T, = 1039 2115-26°A T= 325 L2157R5°A Vor 2(3 25 L157°5°) Vo = 65 2 157:5° MRAM S AGC CREATE YOUR OWN WEB PAGE VISIT www.my UET.net.tc Problem 8.77 Chapter 8: AC Steady-State Analysis Irwin, Basic Engineering Circuit Analysis, 9/E 8.82 Using loop analysis, determine V, in the network in Fig. P8.82. 4f°A ad 102 = 10 ‘ 12forv ia rn €>) TS + 310 2fea S20V, Figure P8.82 SOLUTION: KCL I= It 7, T= 34- Kee: p= T/T, I JI,7 J, kyet 220° + | (-T) SIC T)=9 4{3,-T,)-J! I, = 122190" Chapter 8: AC Steady-State Analysis Problem 8.82 2 Irwin, Basic Engineering Circuit Analysis, 9/E -T,t (FD T= [22180° I, = 420°A (I-j) T, = ladive +42o0° TL = 566 Z-!35° A kve: I(-T) - JI, +27 t (I) til =o {L-T,)- JIT, +2Ty =F T, 4 jl(T,-T, =o (-I-S) + (IT + Citi) Ty +2 Ty=O Oi) 420°) FC Ij) (5 66 £1359) t( It) +20,:0 (tj) Ty t 2Ty = 12°65 Liga KeL: Ty t2lo=T, yh- du = 2Lo° (I4j!) T, +27, = 12°69 218° 43° T,- Ty = 2L0° T, =5222-4-4° A T= 3:23 L713 A Problem 8.82 Chapter 8: AC Steady-State Analysis 2 Irwin, Basic Engineering Circuit Analysis, 9/E KeLat@: T= ith WM = Vi + Mo “JI | J — GR = TCW-W)-\” V, ti!" JIG =o Vo —2V, = YL1F° UV, #+(-1 +52) Va ~J2V3 = 4490° Vi + JIG, - JIM =O V2 — 203 = YL190° V, = 2:53 2 161 579V V, = 537 2-63-43°v V3 = Y 2-368 TY Voz Vv 6 Ver YWL- 3677 Problem 8.87 Chapter 8: AC Ste: Hee OS AMEKSZ- SSS Irwin, Basic Engineering Circuit Analysis, 9/E 8.88 Find V, in the circuit in Fig. P8.88 al, ji n= Figure P8.88 SOLUTION: kecat@: 2%, +1,t+ti=0 = Me -j\ aly y-Vy +tY¥=0 J! | -jl rv, = ji U,- V2) + Vv; st) QjuvVut jin, =o Chapter 8: AC Steady-State Analysis Problem 8.88 2 Irwin, Basic Engineering Circuit Analysis, 9/E Vi= 12Loev (2-5SDY, tJICi2Lo°) =o VY = 39 L£71-69V kc. ot @: JT, +25, +2L0°=9 V-V +2 [we +2Lo=0 2 =i ndor- Vy +2 PEbas") 4ole° =o 2 “J Y= 308 L8-96°V Vo = Ve Vz = 30728-96V Problem 8.88 Chapter 8: AC Steady-State Analysis Irwin, Basic Engineering Circuit Analysis, 9/E 8.101 Find V, in Fig. P8.101 using Thévenin’s theorem. na tom. 12fev(4) 320 ¢ 20g, =j10 16frv o @ 2/oa Figure P8.101 SOLUTION: Mt (=! - 1)vt Voc = 210° -jl J! 2 UV, = —16Lo°Vv Vo. Y= 12L0° Chapter 8: AC Steady-State Analysis Problem 8.101 Irwin, Basic Engineering Circuit Analysis, 9/E Voc = Vo.= UHZ1%0°V 12Z0° + 1621 80° Zryr 0-2 2m jin eG + Voc= YZ1 80% 702 &V% Vo = (Yzise) sire i Vp = 3:6 £15342 V Problem 8.101 Chapter 8: AC Steady-State Analysis Irwin, Basic Engineering Circuit Analysis, 9/E 8.107 Find V, in the network in Fig. P8.107 using Thévenin’s theorem. 1 a OV ) i a G 2Vx< | 19 iN=RV, $10 V3 Figure P8.107 SOLUTION: -jrxVy —jiV, +M=O0 «jl%, +Cl-jrayR=zo U,-Vy = 12L0° Here OS AMOK SSS Chapter 8: AC Steady-State Analysis Problem 8.107 Irwin, Basic Engineering Circuit Analysis, 9/E 4 8.108 Find the Thévenin’s equivalent for the network in Fig. P8.108 at terminals A-B. + ss CREATE YOUR OWN WEB 4/o A @ “fi aa5 V; PAGE 10 - a VISIT fa }> 2v; www.myUET.net.tc B | Figure P8.108 SOLUTION: (1451) G -sINe = 42-90 KCL ot @: TtT= 2VYy Chapter 8: AC Steady-State Analysis Problem 8.108 2 Irwin, Basic Engineering Circuit Analysis, 9/E Voc “WV + VV _ 2[Vo.-V J =jl Voe 7 “a JIC v,-) = ~pl Tod -j\NV, +(-1-JDM% + (1452) Yor =O kel ot @: T= YLo° Von ~V2 = 4LO I —H t+ Voc = YL-90° (451) Go JIN, = Y2-90° ~j1V, + (-1-JI) Me +( 42) oe = 0 —V, TW. =44°9e WY = &L 190° V Y= 794 ZNUGE OV Voe = 5°66 ZI35°V Problem 8.108 Chapter 8: AC Steady-State Analysis Irwin, Basic Engineering Circuit Analysis, 9/E Ty ie YLO0° ~ The Kve: ICI) -JIGI,) + 51CT,)=0 T-T, +il(420°-T) ti! =o -T, +(4i)) Th -s Tse = 42-90" KeL! 2Vxy +Ic= Th = Sl C420%~ Fe) -~j20 420° -Tye) + Tee = TZ Chapter 8: AC Steady-State Analysis Problem 8.108 2 Irwin, Basic Engineering Circuit Analysis, 9/E Cm > a. * 354 22-12 “12° i 120° Qty = 354 LPB TD Problem 8.114 Chapter 8: AC Steady-State Analysis Irwin, Basic Engineering Circuit Analysis, 9/E 8.115 Find the Thévenin equivalent for the network in Fig. P8.115 at terminals A-B. WW oO 10 I oA 10 = aly <F it af B oO Figure P8.115 SOLUTION: I, -)) C201209-V, J -31 CZ) = Y Chapter 8: AC Steady-State Analysis Problem 8.115 Irwin, Basic Engineering Circuit Analysis, 9/E VY, (1s) = JN(2) C1L0% J (120°) VY, = 212 2-45" V Zep = M= 22/-45° t 120° Zeg= 2112 L-45° 2 Zey = ps5 -jhd 2 Create your own web page Sign Up to www.myUVET.net.tc Problem 8.115 Chapter 8: AC Steady-State Analysis Irwin, Basic Engineering Circuit Analysis, 9/E 3 —V, + Wy = 1220? v, — JW +52 +(-142)Vy - flu, = 2 /- Gor Ve i 21ya7°v Vy = 35°38 Z- 137-39 V, = 174-727-137 3°v Vy, = 12-04 f-94-7°V Ve = 2554 <2-4q Rey = 35-38 Z-137-3° — 17-7 £-137-3° I To = 11:7 2-187-3°A CREATE YOUR OWN WEB PAGE VISIT www.my UET.net.tc Chapter 8: AC Steady-State Analysis Problem 8.123 Irwin, Basic Engineering Circuit Analysis, 9/E fia KCL ° I5ti= Ty ReL! 4, ta*= % Ket: T,= 7, the Ie = E,-L, KCL > I’t T re 5-G KCL L+h= 15 yy = Ts ~Ie Kkve: ICT) +1 CI) =2Nx Problem 8.123 Chapter 8: AC Steady-State Analysis Irwin, Basic Engineering Circuit Analysis, 9/E Vx = TCi) = Ty-Ts T,-T tty -Is =2 (Ty-5s) kvet = (-1) t IC T,) = 12 L0° {iy -Ts )+ T5-K = lelo [ae ee] vet 12") 441 ( Tg) 41(-t = 7h tit -( I5-I)=9o KCL + T,t 220° = Te Ta = 2x 7, =2( 15-4) I, -27, +2250 Nee OS aMEKS SES Chapter 8: AC Steady-State Analysis Problem 8.123 Irwin, Basic Engineering Circuit Analysis, 9/E Te Q ~ 2:0000; 6 0 12+ 0000 O -—2°0000i Yu —5-5000 +4: 5000; — 26-0000 — 240000) —13:0000 ~ |?’ 0000; =150000 — 42-Q000% 16-5 000 — 19*5000/ Lo= — {20660 — 12:0000/ Aer oe OS AMOK SESE Problem 8.123 Chapter 8: AC Steady-State Analysis Irwin, Basic Engineering Circuit Analysis, 9/E 8.126 The network in Fig. P8.126 operates at f = 60 Hz. Use PSPICE to find the voltage V, 19 3 20 > jin 2/°A 20 Pot * j2a +) 12/ev 31 ay, Figure P8.126 SOLUTION: F= Go Hz ) lo= 217 pack Ik ZU= jw l= ZB = Je =5:3mH Jw J(377) Z2co= sethares Swe | - _| sa, Jam) Chapter 8: AC Steady-State Analysis Problem 8.126 2 Irwin, Basic Engineering Circuit Analysis, 9/E C= 2:65 mF Re Sr c | 2-6 5mF R) \ R "OY _ wiv ET VI 2 QS! L 5’3MH PSPICE Schomotc Diagram ae: AC ANALYSIS FREQ vMC$N_0004) vPC$N_0004) 6.000E+01 4.3686+00 1.018E+01 WE NEED YOUR FEED BACK www.myUET net.tc Problem 8.126 Chapter 8: AC Steady-State Analysis
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