AC Steady State Analysis: Problems with Solution, Exercises of Electronic Circuits Design

Download AC Steady State Analysis: Problems with Solution and more Exercises Electronic Circuits Design in PDF only on Docsity! Chapter Eight: AC Steady-State Analysis  Chapter Eight: AC Steady-State Analysis 623 8.3 Given the following voltage and current i(t) = Ssin(377¢ ~ 20°) V v(t) = 10 cos(377¢ + 30°) V determine the phase relationship between i(7)} and v(r). SOLUTION: | lt) teaels 7 C4) ba Hye%| ee a en he 624 trwin, Basic Engineering Circuit Analysis, 8/E €.4 Determine the phase angles by which v,(7) leads i,(7) and 0, (1) leads i(7), where us) = 4sin(377t + 25°) V i(f) = 0.05 cos(3771 — 20°) A in(f) = ~O.L sin(377# + 45°) A SOLUTION: Telt= Dl Sin Cot ~ 15°) ye Gy, ~ Gg, = 160° | Veteeel tz i } Chapter Eight: AC Steady-State Analysis 625 &.5 Calculate the current im the resistor in Fig. P85 if the volt- age input is {a} v(t) = 10 cos(377r + 180°) 'V. (b) e2(7) = 128in(G77e + 45°) V. Give the answers in both the time and frequency Gomains. Figure P8.5 SOLUTION: AY 120g 1432 See Cart +40) A ts 5/180 A by titl)s & Sin (5776 448°) = bee lst dA Tag leu? A 628 irwin, Basic Engineering Circuit Analysis, 8/E &.© Find the frequency-domain impedance, 7. as shown in Fig. P88. © Figure P&.8 SOLUTION: Chapter Eight: AC Steady-State Analysis 629 &.9 Find the impedance, Z, shown in Fig. PS.9 at a frequency of 60 Hz. 10 mH 20 —_— i ——— me 10 F Figure P8.9 SOLUTION: iy 2 2 f= P99 ~~ By BE 2 Zt jee eV a Lt TE | Be ee wf OSA a pac Zw By 4% 630 Irwin, Basic Engineering Circuit Analysis, 8/E 8.70 Find the impedance, Z. shown in Fig. P8.9 at a frequency of 400 Hz. SOLUTION: Chapter Eight: AC Steady-State Analysis 633 &.73 Find the frequency-domain impedance, Z. shown in Fig. P8.13. FeSe bene © j2.0 ca -j20 Figure P8.13 SOLUTION: fie bez 2 oo. SHOT a G+ te, . g- FP 4 r. te €&, toy te a oe 634 irwin, Basic Engineering Circuit Analysis, 8/E 8.14 Find Z in the network in Fig. P8.14. @9 Figure P8.14 SOLUTION: 2,7 Bg+2e = jin 22,2 Red Da ryoP RD “er By @3> EsZ 2 ostyosn Bye yt 2g = Shon. 23 te 2 ‘ . ; 7 eee Bude 2 0.706 +] OBZ Ss Byte, Ghee Pr B= Z7ou 450. Fak | @= 23 Let’ 2. Lt Chapter Eight: AC Steady-State Analysis 635 &.15 Find Z in the network in Fig. P8.15. gy oO AAP ? res 12 fia S60 Py S40 & 4 Fy. Zao es [ O= ée Figure P8.15 SOLUTION: - Necliran:: 638 Irwin, Basic Engineering Circuit Analysis, 8/E 8.48 Draw the frequency-domain circuit and calculate v(r) for the circuit shown in Fig. P8.18 if is(t) = 20c08(377f4+ 120° AL SE iso) v(t) 10 $100 mH = =x 10 mF Figure P8.1& SOLUTION: . ~~ bOEY > Bile cae C8794 dae yt Sy Chapter Eight: AC Steady-State Analysis 639 i i t L 6.19 Draw the frequency-domain circuil and calculate v(r) forthe circuit shown in Fig. P8.19 if i,(2) = 2 cos (10007 + 120°) A. oe pi me + ve K : 2 isft) t) off L, 3 10 mH Le 3 2mH Figure P8.19 SOLUTION: g Be pang 4b 4 Tee 2 fleo® A Ve 640 Irwin, Basic Engineering Circuit Analysis, 8/E 8.20 Draw the frequency-domain network and calculate v,(1) in the circuit shown in Fig, P8.20 if v.(7) is A sin(S00r + 45°) V and is(r) is L cos(5007 + 45°) A. Also, use a phasor diagram to determine v,(1}. 333 pF 90 O te ~ hhh (1) + v(t) *) 20 mH Se U_lt) q is(t) Figure P8.26 SOLUTION: y_ ¥ m pea Ve 2 4Z9SP NV os 1 LS Ny Po 2 pe be oes ~br Bis Gols {lor Supeepess hin! Vo= Vg [sy Voy Ta. Ze ] Felt de J the Vo2 o Ls? V z oo Stow ( %00 cbt 9 vi Ny = Ve ~Ne = ~ 4 [ i) = 4 gh. (seed ASV | ~~ ee re i Chapter Eight: AC Steady-State Analysis 643 Lr 9% Se 89% Se In the circuit shown in Fig. P8.23, determine the value of the inductance such that the current is in phase with the source voltage. PS@ 4Q 12 cas (1000f + 75°) V G e OE, If iy 700 uF Figure P8.23 SOLUTION: . ~ r we DE Vs #E ore in phase dh A ry 4 . , ol 2 Les? vCE) a Set Ms. Regt jo = a Ze de Lod 5 oun ; ee Cr Gy tte no 4 | e- flO -) B= jor 644 Irwin, Basic Engineering Circuit Analysis, 8/E 224 The impedance of the box in Fig. P8.24is 5 + ja © at 1000 rad/s. What is the impedance at 1300 rad/s? Ca neeerecnme ier coset Cn Figure P&.24 SOLUTION: x= = Ras, a = chet i = = a joes @ loztooo vis => begs bre ak = Woo cls Zn Rag £7 (1300) (o-0a85 A Chapter Eight: AC Steady-State Analysis 645 8.25 The admittance of the box in Fig. P8.25 is 0.1 + 70.2 S at 500 rad/s.What is the impedance at 300 rad/s? a Yo ny Figure P8.25 SOLUTION: Y= O1 440-2 9 © Lb W& =0.2 @ sverls i 5 wo “ ° é Joly 648 irwin, Basic Engineering Circuit Analysis, 8/E £.28 The currents ip(t) and i¢(r) in the circuit shown in Fig. P8.28 can be drawn as phasors in a phasor diagram. Use the diagram to show that ig(4) + i¢(t) = ig(t). ic os aS is) vf) S19 =e 100 pF 10 cos (377f + 30°} A Figure P&.28 SOLUTION: i ' t i i Chapter Eight: AC Steady-State Analysis 649 & No iD ' Draw the frequency-domain network and calculate »,(1) in the circuit shown in Fig, P8.29 if io(2) is 300 sin(10*r ~ 45°) mA. Also, using a phasor diagram, show that i,(t) + (4) = it), S% fac | i5(2) gs 200 oe 3.33 BF is) volt) 6mH 3 100 She Figure P8.29 SOLUTION: Z= jbo Ze =-j3e z SPpt_ = 2orjyeon @_5 Rath = lo-j3en Eys Ty @e (Ore = 0.224 Lope? A Ty = Ed, C2, +h) = 0 eee Liga Vee U4, = aa [18 V Yolt\< )4-1 ce lieth + eu] 650 trwin, Basic Engineering Circuit Analysis, 8/E &.G0 Find the value of C in the circuit shown in Fig. P8.30 so that Z is purely resistive at the frequency of 60 Hz. 10 5 mH oO Sey a i—-— aC Cnr cr remnant riommenr ene Figure P8.30 SOLUTION: G2 er e+, = eg Mauser td foe ob Chapter Eight: AC Steady-State Analysis 653 €.23 In the circuit shown in Fig. P8.33. determine the value of the inductance such that v(f) is in phase with is(?). 20. ~ | ist) OD [—— 10 mF aL u(r) cos (377f + 30°) A +0 Figure P&,33 SOLUTION: 2 DT wll AQ 2, ap ag o-7o4 mt Regus i a 654 trwin, Basic Engineering Circuit Analysis, 8/E 8.84 Find the current E shown im Fig. P8.34. 10/0" A ) S10 = j10 I Figure P 8.34 SOLUTION: ~ po Vos | jo® jis ta To.7 fase a voter (It) | sc: 4| unereee tetra airtime Chapter Eight: AC Steady-State Analysis 655 6.35 Find the voltage V shown in Fig. P8.35. Figure P&.35 SOLUTION: Ve too Lat f LN PVs ey [eest y tad to 658 Irwin, Basic Engineering Circuit Analysis, 8/E &.38 Given the network in Fig. PS.38, determine the value of V, if Vy = 24 /0° Vv. 20 j2o é f A LOPS O & + : Vs é ia Vi 205%, Y, O Figure P8.38 SOLUTION: ~~ a @o=besd 2 a+janr foe Wi Fe 2 PTF Ly pate", Ey te fe gl Chapter Eight: AC Steady-State Analysis 659 8.39 Pind Vs in the network in Fig. P8.39. if V, = 4/0° V. 0 id, lank: he —-p i K 7 C, + @ S10 VV, 220 om 8 22 [* os Figure P8.39 SOLUTION: rm Za be $2 2 a T, Tips Tye HyZA Vs = Tete -V; = “8453 V [vs= 3.54 frisa.¢? vy 660 Irwin, Basic Engineering Circuit Analysis, 8/E +0 oO Figure P8.40 SOLUTION: Lor Thenensa’s Themrem ! NN penne ene ae em 2 4 e A “LT a Vous be ftsorv | L Dieserv Ee Vag, ~ iL cede eg tnee Ore Be phe . poe Le Vix “12282 f ks 2L0VG) 22 v, (Beez! Chapter Eight: AC Steady-State Analysis 842 ILL, = 4/0° A in the circuit in Fig 663 . P8.43, find £,. e109 Vite SP ke Figure P8.43 SOLUTION: res Khas aif'v Epps Te -2lete 2 Lora Vos Veet Vayse £0°V Bea Ve Aap joa Vetbin= Yegt4 => apes Vv Tee Ver/ bes WA Tt Tht Tey eye ieefe A Nos Ey Gh = b+ jley ~j tha 664 Irwin, Basic Engineering Circuit Analysis, 8/E 5.44 TfL, = 4/0° A in the network in Fig. P8.44, find 7,. 10 Figure P8.44 SOLUTION: y= OYE, = 4[9¢v ht SPT. a & LOA Wos iE, = by Le? Vv Na 2 Vie, apo lor y Epes Ve fost = jlwA Vas \ai8-Vea = Zia f3- Val, 2 2/24 Ty« Tar, -h, = aejiok Ve a(UTy = 4rjioa Ve ea Vy Ve 5 2 rjiov Des Ve A> Bryied Chapter Eight: AC Steady-State Analysis 665 8.48 in the network in Fig. P8.45, V, is known to be 4 /45° V, Find &. Xe © if O a, Yr, ~f1 2 + 12/0°V Z 192 ¢ Vo tt OQ Figure P8.4& SOLUTION: ~ Ths Yo. 4 LSTA Vas TC ee-jh = Weley fe, 7 Ty Tye 2 2as 83/7 A 668 lrwin, Basic Engineering Circuit Analysis, 8/E : &.4& Using nodal analysis, find 1, in the circuit in Fig. P8.48, ‘ Figure P8&.48 SOLUTION: Wale yp ee Ve eee lg ue BMV gE ae 3 . 3 —o a VarM Ye og Soe wy e¥e Cia) =4 Chapter Eight: AC Steady-State Analysis &4%9 Determine V, in the circuit in Fig. P8.49, 669 Figure P8.49 SOLUTION: y 1a 2 { K e a : c A jaa. 3 12,0°V | eS ~fet, 4+ B2(sej~s0 | Ted. te /33e7% A 670 Irwin, Basic Engineering Circuit Anaiysis, 8/E &.50 Using nodal analysis, find 1, in the circuit in Fig. P8.40. wa OV i2Aey 2/0" A ie @ 408% “Oss “4 ase) Figure P8,.50 SOLUTION: - ~ Chapier Eight: AC Steady-State Analysis 673 ' 6.53 Find ¥, in the network in Fig, P8.53 using nodal analysis, (PSV hy 1 Vv Bs : A if Ww 20 ito | 70 | j22 3 weeev (4) af A CL) o @)aerv 20S te Vo Figure P8.53 | t SOLUTION: Vel yoy Vio ved oo Vo = ¥ 62) Z-j! 2452 ! 2rju 674 Irwin, Basic Engineering Circuit Analysis, 8/E 6.54 Find L, in the circuit in Fig. P8.54 using nodal analysis. 10 aq Figure P8.54 SOLUTION: — @v, Wy, Cr jy) wVes epi (Ng, HNQ AEM, 2 1k Chapter Eight: AC Steady-State Analysis 675 &.55 Use the supernode technique to find I, in the circuit in Fig. P8.55, ay 120° V = K ey ] 193 20% © j20 ae 20 I, | 5.7 678 Irwin, Basic Engineering Circuit Analysis, 8/E &.58 Use nodal analysis to find V, in the circuit in Fig. P&.38. 6/orV¥ QT ° 1O5. la Vo 2/0 A ; Figure P&,58 SOLUTION: ° a VinVe = Ge vaya 12 Le VaNe = Ve Ve22Vo at Gu woe Vo a Ye 4¥a. gen? iptie Vitple7 [Wa pve 22 { aby oh 1 z 2 i SO \ ° cm Chapter Eight: AC Steady-State Analysis 679 8.89 The low-frequency equivalent circuit for a common- emitter transistor amplifier is shown im Fig. P8§.59. Vs. Compute the voltage gain V,/ Figure P8,39 SS SOLUTION: \eres/ | jOe term H ) 680 Irwin, Basic Engineering Circuit Analysis, 8/E £.60 Use nodal analysis to find V, in the circuit in Fig. P8.60. eS ¥ 124° V a ane CS jo ia y «FD vy +0 Figure P8,60 SOLUTION: VioV = cadet y VavYo Vom a7 %ol tay) Fo Yh 4 J @ sore: BY AV a Va 4 Vo oy where Uy > Ve, { ih i J eo v, #42 (241) No xo Py ~t © | vy ' | he le ' bap va ie = baa | Ve in 2434 L. ay Le ol