Additional Analysis Techniques: Problems with Solution, Exercises of Electronic Circuits Design

Download Additional Analysis Techniques: Problems with Solution and more Exercises Electronic Circuits Design in PDF only on Docsity! Chapter Five: Additional Analysis Techniques  Chapter Five: Additional Analysis Techniques 321 2.3 Find J, in the network in Fig. P5.3 using linearity and the assumption that J, = |mA, —PS% 2mA Gd) t, Figure P5.3 SOLUTION: sf 2 =~ imA ty ’ Vo > Heoe T, = ay Sas Tye Mo, thm A {24407 . Vi* Boe0 ey +¥Vy 2 vy Vs r My ge Ama 3 OR But, Te achalle epgunels 2mh, Se g 322 Irvin, Basic Engineering Circuit Analysis, 8/E 5.4 Find J, in the network in Fig. P5.4 using Imearity and the assumption that 4, = I mA. 2kO x, 2kO Ap Ah I open Chapter Five: Additional Analysis Techniques 323 3.0 In the network in Fig. P5.5, find J, using superposition. ©% 6kO 6 kO oe Ay rev) 6kaS Ch.) ema As 6kO Pr ° Figure P5.5 SOLUTION: Legs - p te eg, = boo +{ fovos £2,000) Bp tell 326 Irwin, Basic Engineering Circuit Analysis, 8/E 5.8 Use superposition to find V, in the circuit in Fig. P5.8. 6V Figure P5.8 SOLUTION: Aw mofo fxs 180m /4oce he Ry S27 ' f of g OPS | | L a ~ 1.2gV Chapter Five: Additional Analysis Techniques 327 =.8 Find /, in the circuit in Fig. P5.9 using superposition. HS 6kO % Figure P5.9 SOLUTION: ige Ey, Cb ae Aye tm A a . / ~ = = : De pyetede egies x - oe byte ~ [ae F ve, i + Guage disisite , J oxi | bor = es eel we S bewn thon f Le = tko 4 L = > fo heng” fy } Teg2 7 am A { bea AY bee { i joe bee ene te Ney t+ Top = 328 Irwin, Basic Engineering Circuit Analysis, 8/E 5.10 Use superposition to find f, in the circuit in Fig, P5.10, 12V Figura P5.170 SOLUTION: } Pk | | | ! | ene nen ees arimenameebint nema, | Tos = [ ee } 2 Bell Chapter Five: Additional Analysis Techniques 331 5.72 Given the network in Fig. P5.13, use superposition to find V,. oO 1KOS 12V S2Kn Vy 1kO 2ko _. * fl chk O BV i) S1kQ 2mA Figure P5.13 SOLUTION: 2V Vo oMay © Vani bs 332 Irwin, Basic Engineering Circuit Analysis, 8/E me A 5.74 Use superposition to find V, in the circuit in Fig. P5.14. ema) 1koO $y Oo + 6V +) Vo 4 -O Figure P5.14 SOLUTION: g cae Bye tee, an e By = 24 Bee bHes - - VYorb| & [2 zav Le Bend, 2 ey =& Abs Sen5b Moo | * foo SL. Chapter Five: Additional Analysis Techniques 333 5.15 Find V, in the circuit in Fig. P5.15 using superposition. 2kn= ae ° 2 Kas Vo o Figure P5.15 SOLUTION: a - - ry AE i 336 Irwin, Basic Engineering Circuit Analysis, 8/E 5.78 Use superposition to find /, in the circuit in Fig. P5.18. Lh | 4kO Figure P5.13 SOLUTION: Bakes Lee Ys ska Pye Re a2hen + ne RS Roo yt Wee Gh y | A tye Li bs Cp = 15, @ . “Le a9 ry \- : _. 3 Nae -12 Fg ZC Rgtt2) > -2.aAV Ee Poet Roe B My Bea | ied 1 te . 23 By. we ah ees Po> Rs AlPyrPs = Zien. | fl $2 ; Fp Ips amc Fe A Retbp\- 2.4 ma Top Tp Be) ee pa Ry rey ey Ree 278, ey = 1k Rez Re tle =3.5kR Te = 2xto”? Re act Pe) * Lo7md Gis % Gy 4, Ga Ves To Tairerds [Sero2ut A] Chapter Five: Additional Analysis Techniques 337 5.19 Find £, in the circuit in Fig. P5.19 using superposition. 2V S4kO av@) Figure P&5.19 SOLUTION: Pp Ris Ppa he = tlk by 2 AL Bye (RAMA Es 4ASkA Daye exe be “ty Es) Tie o.seand bye be (2yth) = par Vas Z Re W(t ree 2 O.4sv ge XN ; ea ~VeQ /S Obert) = -O fem A  a BH Ber tay © dha es ri pie) \ v—| w ay = 340 Invin, Basic Engineering Circuit Analysis, 8/E 5.22 Use superposition to find /, in the network in Fig. P5.22. Psy 24V ©) S&S 2kO 2mA 2mA 3kO S4kO “ey Figure P5.22 SOLUTION: ~ ys Bee ype GWh Bye ohn Pye Rhye Says tae 3 Gs Mis Gat Gas 4a) = -6.¢m% Gee tae Gy Git i+ ig the) = o.dmd tae Dols Bae bon a kth 4 £ Aur ‘ a Ven vd Ba SC la rG)e bay  Chapter Five: Additional Analysis Techniques 341 5.23 The loop equations for a two-loop network are NR + Rp = \ : LRy + ER» = V; What is the relationship among V,, V2, and Ry for 1, = 0. SOLUTION: 344 Irwin, Basic Engineering Circuit Analysis, 8/E = 5.26 Use the results of Problem 5.25 to determine the value of Vz such that V, is zero in the network in Fig. P5.26. Figure P5.26 Explain why the values of A, and Ry have no impact on your analysis, SOLUTION:  The turret threat Lets freak bg the bom Curnedt Saute, Thus, the vatucad by haa he tmpad ia sun vebeh a nabeyts. ‘ Use, ts hen. Vo20 hee. turrénck Hrrough Ry is teen Ot 4aecl Lose 4 the “padne, “4d Le, Chapter Five: Additional Analysis Techniques 345 3.27 (a) Given the network in Fig. P5.27, find the value of R, such that V, = 0 V. (b) Then find the Thévenin and Norton equivalent circuits at A-B as seen by R, using the results of (a). ; Ry = 30 A + t 1 av) La(]) Ry=1098V Nx 2 0 Ro B Figure P5,27 SOLUTION: , DD Scgseposibo: f d+ tay CE 2 Uy B Nos Ne __d ove ih U/l tea tbs) YoutVer eVoso = \4Pa | Rtey By by rEg thy z Uyreores fr sy conn AR pen : DY . ee Voc 7 W4- ER ~ 4h © ov yay (2) fay ¥. 14 We 2° OR oe . . 4 ie ee Prt = 24, © 270 fn ce a : pore cy 2% on G) Sezer — Rerim & 348 Irwin, Basic Engineering Circuit Analysis, 3/E 5.80 Find /, in the circuit in Fig. P5.30 using Thévenin’s theorem. 12V e) 4kO 4 A 6kOS 26k 6 mA ’ Ly Figure P5.30 SOLUTION: ®&, =f, 7 4ki Foye but Py © GE A kG fhe Veg ¥ Gwe TPS Ugg = -2d iy hvu ny" FL. es s \y a ets Dima 2 CE ge hy bys Let : ' Chapter Five: Additional Analysis Techniques 349 3.41 Find V, in the network in Fig. P5.31 using Thévenin’s theorem. PSY 2maA G) $y" AW ~O 2kO | 2kO + wv 4ka = 6kQS V _ <, < o | | 5 Figure P5.37 SOLUTION: Bp Cs 2hn. Ry = 4h pr een £5) = Gooet, - Lo00L The ZhA G Zh Th Lys 1btma + want | a Moe Ge Re CT, Hl 4 Ke L-T) oe 2d Ves f0.07V Vans Bat (ie) = 3. 33k0 a ——— a \ it Noe Vocka (Lb tH) | Verb. aa XQ) 8 ve i 350 Irwin, Basic Engineering Circuit Analysis, 8/E 5.32 Pind V, in the circuit in Fig. P5.32 using Thévenin’s theorem. +0 PAS a x jo) Orv 2kO v—_—_—/ 12 kD Cf) ama S 2kO ry oO! Figure P5.32 SOLUTION: Eye Rye Zhan Bye Het Bye eh eT Cryrh\ = BD, Tz 2 tnA E,=> 3. 33mA Ray + Ry T= Von = 2). 938 Prts @, HPV aks o 3 23ke ey eWay - oneness Ag by ee we Pye eae Yl Ry Ne 2 et. r an bey Gay en Chapter Five: Additional Analysis Techniques 353 5.28 Find V, in the circuit in Fig, P5.35 using Thévenin’s theorem, ema(f) » 2 kQ Dy mA 2kO< ni Ay * fh 3 > i) x= iy oo Figure P5.35 SOLUTION: Eypeeme Le gmt dye -tma D4 by fort) Mero Ty tg + yl Dyas Vou * Zov 8.38 Find V, in the theorem. Irwin, Basic Engineering Circuit Analysis, 8/E network in Fig. P5.36 using Thévenin’s Figure P5.36 SOLUTION: or 4a) imn Dye @mA Te - IMA by 2 - ima Rat, +8, OF3-E)- Rr tpeweoze Voe = BV ‘Rw Cyt edy = Sku fou . Lt a= Vor Bap Vac) 59 Ne : Eayy The i tooo — Le OSV 4 fi Chapter Five: Additional Analysis Techniques 355 oa? Find V, in the network in Fig. P5.37 using Thévenin’s theorem. O Figure P5.37 SOLUTION: A dy = 4 i Weed. 7 fhe Ey ty + Eph) e, 4 os bmi ( tas Vea~ Bs hy 4 Bz OT, -h2) ba Yoo = Bi2V fan (etek, = Pe. yell 358 Irwin, Basic Engineering Circuit Analysis, 8/E 5.44) Find V, in the network in Fig. P5.40 using Thévenin’s theorem. 2 On 1kO 1kQ ? fy DAK O + 1kOS 1 ma(f) 1kKOS OV, O Figure P5.40 SOLUTION: Bu f= tka Ee 2mA oT, --i m4 Dye - Ima ZEA CR ind x 2. “t 2 oe eo ~ ra Ad Put, +Be( Ty gw Foe Von BY x) bey (2 -D Ye Mecee Chapter Five: Additional Analysis Techniques 359 3.47 Solve Problem 5.12 using Thévenin’s theorem. SOLUTION: &, =G =€o- hte a= Gy zen apn : z Vand ay bp, PoE 4 Ty Cb, ety lly) oO : t _ . We alt TGs) - Ee, -Ty ty =° $b +t n@ LD ana Vee Ty = - 2m a IDEM y-HV Ry — Be tee | tgs BY po td Ty (Gstyta) - Fe, - Fs <0 OF Dee A Fe 7% wi Pat, tele Grig) EG © ey ts + oe Ox . Be -Tx dt Lys bn L Tea 3h at = ° , Fri = Ye fp, m= 1 1FhO yt &y (4-G) 48s Gy -t 0 mo Ai £ a Ww R, AE Ves Von m= Ves Corey | Ee vey peer) f 360 Irwin, Basic Engineering Circuit Analysis, 8/E : 5.42 Solve Problem 5.13 using Thévenin’s theorem. SOLUTION: B= = Ge tea hye le EZ + [ Aiev Voey Voc = 12V Me Me Vocd = ov he. by 6@ i pve : ee Vows + Ze Fy = AV H)2mA 2 Voe= [bv Chapter Five: Additional Analysis Techniques 363 Given the linear circuit in Fig. P5.45, it is known that when a 2-kO load is connected to the terminals A-B, the load current is 10 mA, If a 10-kO load is connected to the terminals, the load current is 6 mA. Find the current in a 20-kO, load. OA OB : Figure P5.45 SOLUTION: Erk ' pt vas : AS ete tye Mee CD Bs Bry i 364 Iwin, Basic Engineering Circuit Analysis, 8/E 5.46 If an 8-kO load is connected to the terminals of the network in Fig. P5.46, Viz = 16 V. Ifa 2-kQ load is connected to the terminals, Vi, = 8 V. Find ab ifa 20-kQO, load is connected to the terminals. SOLUTION: fee’ = t Vee fZoml, yy Beas f Loon +h Brent Bony 20> + Oy fan \ $a Hee / Chapter Five: Additional Analysis Techniques 365 5.7 Find J, in the network in Fig. P5.47 using Norton’s theorem, 2 kD 4kQ ‘ AYA yy = 6 Figure P5.47 CG 24 SOLUTION: Res Zhe Paw tln by e¥ere fy fe a ee i Arh 1d Hse Dye Fy Tee k ama Tye BE baad WOE 2 Or é, z ba Tex FmA & fe eA Pl ap a le camer - —— po eng | Very = Va Oy © 4 La Fin | , ae cone eg / ee 4. A Ft Tas brn | R287 mA + fy o & 3 rn ay BEP muy 24 Ey tes a a ee Irwin, Basic Engineering Circuit Analysis, 8/E 3.50 Find J, in the network in Fig. P5.50 using Norton’s theorem, &% vA o 2kQ Ty po fe 6 kA 3kO 12V *) “T 4kQs Figure P5.50 SOLUTION: @, > 3ka Eat i%e Coat a - 9, By ea oY ° ay ; Fy, Chapter Five: Additional Analysis Techniques 369 5.51 Use Norton’s theorem to find V, in the network in Fig. P5.51. Figure P&.51 SOLUTION: Bt Ry HZ ae Rye te g qr ) | Tyee PA/RY = GmA 370 irwin, Basic Engineering Circuit Analysis, 8/E 5.52 Find V, in the network in Fig. PS.52 using Norton’s 0 é = theorem. 4kO Ay) rev(t Figure P5.52 SOLUTION: Re 3k Bg kes kn My =h~ inke, @, > the Raa Sle Ree 4en Rae Ba Ve tha Pgs he He = tier Ty. = Ty te= T, (@2+@a)- Py Ty T2-L3- oma Pes h Zima 2¥se Fal Pa +R4) - Reh - Paty =0 t2= Dk, +P, Ry Eng = [ee 7tde PR] F (Mee) © B.s7LA ~ ple Vo = Ese ( By 7/29) Noe eau | a) Pas eal Ve ee