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

Download Steady State Power Analysis: Problems with Solution and more Exercises Electronic Circuits Design in PDF only on Docsity! Chapter Nine: Steady-State Power Analysis  Chapter Nine: Steady-State Power Analysis 741 &.3 The voltage and current at the input of a circuit are given by the expressions : v(t) = 170 cos(wt + 30°) V i(t) = Scos(wr + 45°) A Determine the average power absorbed by the circuit. SOLUTION: 742 Irwin, Basic Engineering Circuit Analysis, 8/E &.4 The voltage and current at the input of a network are given by the expressions v(t) = 6cosat V i@) = 4sinar A Determine the average power absorbed by the network. SOLUTION: « P= Vn a a a => GO; 2-46" Chapter Nine: Steady-State Power Analysis 743 9.8 Find the average power absorbed by the resistor in the cir- cuit shown in Fig, P9.5 if v(r) = 10 cos(377 + 60°) V and v)(r} = 20 cos(377t + 120°) V. 0) Figure P§.§ SOLUTION: Chapter Nine: Steady-State Power Analysis 745 8.8. Determine the average power supplied by each source in the network shown in Fig. P9.8. ©8 AM + “Oo 40 I + 20a nas ¥, 4L0° A 220 9 | = | © Figure P9.8 Z * SOLUTION: i+ B= 2% 2 7 4144. 169 J7nse <Q pr NEA 8 Baes ets 2s 0.834 hy Oba Zn. = Reg +4 Meg Onda He reas tae pack of B cen sumes aiseage Priater P+ Ine. P2 GW (ost) [rs dow | E ey 2 746 twin, Basic Engineering Circuit Analysis, 8/E 9.9 Find the average power absorbed by. the network shown in Fig, P99. Te c. t e jek 4/60° A CG) yo =20 4 =e -/4 0 Figure P9.9 SOLUTION: amie Vee -heoe | o—- |. 2a Lista Lots J Va ZR, = Ske iis? Vv Po Vo Dew Chapter Nine: Steady-State Power Analysis 747 8.10. Given the network in Fig. P9.10, find the power supplied and the average power absorbed by each element. PSY + T, oY FS 40 Re S20 4/30° ah) Vro po as -j10 Figure P9.10 SOLUTION: Let 2, 2443S gnd t2 = 2 fle ¢ gps Aze . 12S -9 D2o Brey Ve= 4h 30° Ze = Dor L2id? v £029) te i, Snore t Ps) a | eee (aut + 6B.13°) + ca 8.8 “| Pet = F474 Coe C2mt SiS) + fdin w | Te VW /a, = 14 Lee ak Fae Ure Ble Lake? a 2 am Peo = Ew pg, = 4.0 Ww Pea= Cen ks to Ww Fa z 750 Irwin, Basic Engineering Circuit Analysis, 8/E §.73 Calculate the average power absorbed by the 1-0 resistor in the network shown in Fig, P9.13. fea fio pee ie KR A A, ery (+) A ,320 A gsi9 v, gy x Be i Figure P9.13 SOLUTION: bles TU C2rjpey- 28, Fo 74M 4 Bap) =o ‘gaeyn ~e (ha tif : Tose. -45° A } yi wt le PD eg Fe EYE ETS Loe SE ALES US _ : aaa Tat Tem a) | Pas 22EW Z Neen ctteneeen ere Chapter Nine: Steady-State Power Analysis £.74 Given the network in Fig. P9.) 14, find the average power supplied to the circuit. 12/0° A G) Ey 4/02 V e Vs Figure P9.14 | —f2n > 751 SOLUTION: tet €e ba ya than Samrce. trims dec mation, zx, ¢ Psepplted -b eri = Pp, + Pag = liebe r _ t 2 BE nVs nom #42440) 8 f-S34°8 Gow 752 Irwin, Basic Engineering Circuit Analysis, 8/E 8.15 Given vs(t} = 100 cos 100/ volts, find the average power supplied by the source and the current i,(r) in the network in Fig. P9.15. 100 in(t) %s am Ry 50 mH vs(t) ©) 15QS ae 1 mF Figure P9,15 SOLUTION: UT a Le, F. 2.7 Ss Ble fion Ry + roo OV ae Vy SO Ze bet 2, 2 Pp Se 2 xcgte x 4-1 J Wake Fe5h Var tooo 2x = Ux ef [~48.B y Gt Bi 4b Tyee /a, = sit LaLeea Tz)? 564 won lio th ede) A | bef Zoe Bete Be = 147 fone? O. Ty Vs / By ~ 679] 1.49? 4 Pus Vos om Diag com (8 ~ 7.44") Ree sae 2 i Chapter Nine: Steady-State Power Analysis | 755 $.18 Find the average power supplied and/or absorbed by each element in Fig, P9.18. By g.. A Af 12 20 Vs + Us ~, . A oem vara) GQ yeha ti “ian ¢ 2/0°V x, : ty t, Figure P9.18 SOLUTION: T= idea jit, 2 leq ys -2 18° f a Tf 1 y3e* Tz poe (30° A yh g talel oe 2 ten poe fuer A fe 5h 2 LS Zio? Tye 0.892 fprg? A con. (bO4-Ta} x O,660 WwW Sup plecel Tyg t 7 Ver Tem cee (0~ 174°) = OIMW Supe dat z PP ota Spy biel. FP tise tbe sorbed 756 trwin, Basic Engineering Circuit Analysis, 8/E 9.78 Determine the average power absorbed by the 4-© in the network shown in Fig. P9.19. PS¥ vy \ Ve. AA ’ 1 @ ~j4n d J () rez 5 j20 20° A 40 i, Figure P§.18 = SOLUTION: Vy tebe Vig Wee oo Ss (enjes of Vas z . fe “Cp i VaoYs ) 240° + Yee a AV, a Vel isis 8 Chapier Nine: Steady-State Power Analysis 8.20 Given the network in Fig. P9,20, find the total average power supplied and the average power absorbed in the 4-0, resistor. Figure P9.20 “fin 757 wesc SOLUTION: Vr V2. - Moe & le? 2 wyh Ve7%s 4 Var¥8 4 Ve no “Zz je & -V Var Mig Verve Ve ce my Vi CLE) -Ve my Ve eae ~2OM bby (3-j2ltj4 Vez o -Yo4Ve + 2Vz70 Ve 1G bTaFy BL. y 760 Irwin, Basic Engineering Circuit Analysis, 8/E 9.22 Determine the average power absorbed by a 2-2. resistor connected at the output terminals of the network shown in Fig. P9.23. I, 20 fo 2 LOO, tT, ~j20 Qe x, oe == 2. 12/0°V Figure P9.23 SOLUTION: a2 Ty (@yj2)tet + jw @ nies Ee (dpe) jem 2% jae te dey » @2 ry (eee) + eyo | ie 9.4aw | eee el Chapter Nine: Steady-State Power Analysis 761 $.24 Determine the average power absorbed by the 2-kQ output resistor in Fig. P9.24, us(t) = 2 cos wf V Figure P9.24 SOLUTION: Vo “IL - Re -2, We CAV 94 Cee (wt) = 4 tes, (wt teot)v Us Uf) ey Vo 2 4& 1Be" V Pee Vo, Tey = 4mwW / ZR, | 762 lrwin, Basic Engineering Circuit Analysis, 8/E 9.25 Determine the average power absorbed by the 4-kO resistor in Fig. P9.25. Us(f) = 2 cos wi ¥ e Figure P&.25 SOLUTION: Vez 2derv ve2 Ve (14 Bie )- &Lerv 2 rman Ye = Yom Pe, 2 4S mw 2G, Chapter Nine: Steady-State Power Analysis 765 9.28 Determine the impedance Z, for maximum average power transfer and the value of the maximum average power absorbed by the load in the network shown in Fig. P9.28, joo SLE CC) ar Vy o 4/30° A ~j20 Zr 45 Figure P9.28 SOLUTION: Therein 6. ~ ~ et a byu QtB) Fe eo Bag hee sp th ey ee 4 Vee . - pee mrt ang Sry 7 2 y2 SL. Superposition! Vers Ts & te Ys Be a ¢lof-imiey Ry +e rte Ts Vee erate. 766 Irwin, Basic Engineering Circuit Analysis, 8/E 9.29 Determine the impedance Z, for maximum average power transfer and the value of the maximum average power absorbed by the load in the network shown in Fig. P9.29. 6forV po “ =o Aor, ie S10 Zy S10 - Figure P¢.29 SOLUTION: Thivenin Sg. rye le poy jee ys ba are sagen T “4 ’ de “Nee Lippe Vine ia Vg ten tm redraw 8 pt Be C 4 [20] Chapter Nine: Steady-State Power Analysis 767 $.30 Repeat Problem 9.29 for the network in Fig, P9.30. 6/0°V 6 20 fia $A" seo |Z, S10 , | Figure P9.30 SOLUTION: “Thewenin ¢. Zz “~ =a Vi 2 Vs a Va7 Vs wih es aa] fh] Veet MeVa2b4o lipv? v Lederer for 2TH yen Tsvec 2 2 Ob / hy A tre tS. os : 1 Ay “ta it Ra Zep, oo Zz me 2G), pe Pleo sow ers basi ee Bm OF 03 VS> Bate a z ——— | Zu eni*: 0.44 jo3 a \ 770 Irwin, Basic Engineering Circuit Analysis, 8/E 9.33 Repeat Problem 9.32 for the network.in Fig. P9.33 os R, AQ : Ae 10 gz aR a vecev (+t), = (]) 4a Zp, 19 fo fh nal ~ - a Zia Figure P9.33 SOLUTION: boo ~ ¥g 2 di. Vy t+¥ec 4 Pedy ' : or a = ; “Moe a iode 74) Gu) (ais #) +h ney Yee ae ry (Ts Veo = jt v ry ge I | bet 2, ° &r&,= leg] ae Srme tee 8p 2 12 4-52, 2 447 [-tee A Chapter Nine: Steady-State Power Analysis — : 771 €.34 Repeat Problem 9.32 for the network in Fig. PO.34. © Figure P9.34 SOLUTION: » eee ds , Ti, 22 iva we om JRE rd, Cheyne) = 6 Lo" @ 4 1@y x ‘ TZ, * DT, os ~G-i¥ “ f-~ 43.8 7 Ey 3 a 'S oS a ob 43-8 A Vous Ti @, +, ge £¢.-32Vv J r Orie Re (BDL 2.8 - fot L AN h Bre. . Be oe _ L Zi> ane By = 28 tats! te eee ee ees | 2 Te Ven 2 0.7e ] 560° A © a 3 I Sat te. Poe key | Ps Laz | 772 Irwin, Basic Engineering Circuit Analysis, 8/E 9.35 Repeat Problem 9.32 for the network in Fig. P9.35, 20 . “ee + te Z; + 4/0 A @) 2 FE -p0 e 6/02 V . “e ‘ Ly Ve > Figure P9.35 SOLUTION: , Toy oe Vi -Vee Vib ie [~ = ry a” e ) BR y rR Use Suter pos his te Cina yy x Aes io > 3 n Eo mc ® Vs Vi2 {tae + et fit karte By ny Vez -o.d *jOaV wd Nowe “G44 jaBV md Be Bre Be Chak / [Ritts tae] yaub-—4y ———t 4  Chapter Nine: Steady-State Power Analysis 775 9.38 Find the impedance @, for maximum average power transfer and the value of the maximum average power transferred to Z,, for the circuit shown in Fig. P9.38. 12/0°V ae a he a, 19 me << at By lg it Figure P9.38 SOLUTION: tzis*y anne WON ay a ~~ , 4 Loop LT . ree ° “oe ; in set Berea Beye Vee/teee a forn |e sbnt oe 4; \ lenl— ery Meee B 4 so | ee bar 2 94 +5 te san rennet ame ttl Po 2 mb P ro 2 $57 Re [ Re asl 776 Irwin, Basic Engineering Circuit Analysis, 8/E 9.39 Repeat Problem 9.38 for the network in Fig. P9.39, Z, t oe LAS no 10 NZ ove Be Ao fq, @ 420° A A, + |e, Ve S40 + Figure P9.39 SOLUTION: Le. \ Bae Pe Ue Ny a Ty (Rpt! de €-j4v ~ ha ae ~ y a. y . Ve. ye Ey hye 410 Reéty we & © s hy Sr? ne . , . \ . et enh Vora Vy + 2¥x 8 12 35 ' Zeale® Lede o Vat Fy (o,-0y 2x = Tz Ce-siy- 5, Ciegd yiekds Tye = 3.07 fatto” A Chapter Nine: Steady-State Power Analysis 777 8.40 Find the value of #; in the circuit in Fig. P9.40 for maximum average power transfer. jaa 2/5? A Ch). BOS Zi, As I, al, bes Figure P9.40 1 SOLUTION: 780 Irwin, Basic Engineering Circuit Analysis, 8/E 9.43 Calculate the rms value of the waveform shown in Fig. PO.43. v(t) &) a i | ad I ' i : 1 pop oo i fo [o 2 4 6 8 10 12 14 ifs) Figure P9.43 SOLUTION: 2 1 Onabean Ts: Bs Vlts & Rebeg i 4 gat cy t a bog 2 & os Ma arm OU fy aw ca cent | tens VAP eithak = [efow LAY CRS FA C2\ 34 Chapter Nine: Steady-State Power Analysis 781 8.44 Calculate the rms value of the waveform in Fig. P9.44, v(t) (Vv) . 6 U \ ra : i : : we {c 2 3 4 7 te) Figure P9.44 SOLUTION: ; - Be octez 9 2 § 4? e<bee Tz 48 wots é ze t <3 t Ws fay pet a% 6 act ey (oo nee <§ 2 37 '8 . A Vems: |b) de? | > » see] > b| 2 fk 24 + Bi mee , lf) i4 j yo ey  trwin, Basic Engineering Circuit Analysis, 8/E 782 §.45 Calculate the rms value of the waveform in Fig. P9.45, u(t) (A) 4 B prmcrescg a E 1 \ oN __ - 0 2 3 4 6 7 is) Figure P9.45 SOLUTION: : . ( a aacteze | ” , 4 i Teds reeves | ba2t 9 reter bob UA 2d sy -zete et? { © patad ( o } g 2 / . say * did ae)® 4 far)? - 2a eae |): ra ie] sl: Voie > x =e J] § ¢ e 7 4 12 Teng = 241 8 + 30- bo + 283% \ (41 \3 Chapter Nine: Steady-State Power Analysis 785 $48 Calculate the rms value of the waveform shown in Fig. P9.48. 68 vy) 4 ~ / f lo + 2 8 4 5 6 t(s) Figure P9.48 SOLUTION: wilt) consists 4] Z feleaticel triangles: 1s Fevangé, © 0, Cs At and VEled= 16k* ‘ ; f uu, Ub ak zt ee? | re ig (sens for 2nd triangle} } - fa i i i ! 786 trwin, Basic Engineering Circuit Analysis, 8/E 9.49 The current waveform in Fig. P9.49 is flowing through a 5-Q resistor. Find the average power absorbed by the resistor, PSV i(t) (A) onamnscres te " "ey Figure P9.49 SOLUTION: Tras TLE) cerpaa gts f a triangle ane a Accding he, . a Trvanylt Ulta Zt ond Fb) = eE™ 2 .% a Py LED okt s 433|" 1a. 67 } ; ; a) 3 > ~ | +2 , Kestangte © Tpbts -# gud 1d + Ie » | 4 ( leet = jet | 2 32 Ly ly - 7% OK vy Lems > ya Uhde > } j [ouresz] | peers | J Chapter Nine: Steady-State Power Analysis 787 9.80 Calculate the rms value of the waveform shown in Fig. P9.50. ey i(t) (A) i i 0 1 12 3 4; 5 a = ~~ f ( is) f i / f F 2 b eo Figure P9.50 SOLUTION: T= 38 CLES com sists oad a brands “| lgecancA duradion. god. Be rectangle l eomd. chiaedhan . 2 Cems -fft celle ¥ Trrangle ' Tiths 24-2 9 duet fitis 4th stad (orange | +é4l oa 4 ftaak = at) - 4a] J, hi 1 Ly oe \ 2, 1b ant tla ft atte I a “17 “4, - . 7 Tams + } if ign f | Bens 0 7424 | 790 Irwin, Basic Engineering Circuit Analysis, 8/E 2.59 A plant draws 250 A rms from a 240-V rms line fo supply a load with 26 kW. What is the power factor of the load? SOLUTION: pet os a f) ay Fe Mil pf Zu) Me , po Ue en ~ P fe Po LP ps 0.833 Chapter Nine: Steady-State Power Analysis 731 &.44 The power company supplies 80 kW to an industrial load. The load draws 220 A rms from the transmission line. If the load voltage is 440 V rms and the load power factor is 0.8 lagging. find the losses in the transmission line. & ® SOLUTIGN: ASSu rms Oy =” } - * . } [2c] Mu Yo Ath 20° MV rceg eon td Tihs gto (-8 Ams Pl MA) pfs 27440 B= dee! (oe 3, 4° ee Pore, 2 KB “P= Soooa ~77 to | Cie ® 25ho w | 792 Irwin, Basic Engineering Circuit Analysis, 8/E 9.55 An industrial load that consumes 40 kW is supplied by the power company through a transmission line with 0.1 © resistance, with 44 kW. If the voltage at the load is 240 V rms, find the power factor at the load. SOLUTION: ~ Cc hee Pos 4xot Ww INCE Be V ring . at, + Ve P fel vo f= Mlisl pf Py = thew ro eee , chp Poe fe Be te = (ord [ele Zoo Acmg [pe eae] ivr, eee rn Chapter Nine: Steady-State Power Analysis 795 §.58 Determine the real power, the reactive power, the complex power, and the power factor for a load having the following characteristics. (a) I = 2/40° Arms, V = 450/70° V rms. (b) T = 1.5/-20° Arms, Z = 5000/15° O (c) V = 200 /+35° V rms, Z = 1500 /-15° O SOLUTION: a) Pe loti vi em (Oy-Or) = Z) (460) ewe (70-40) Qs | vlit{ svn (év-8z) = Q= 450 VAgs S= V2" = (450 lh S= 400 J 30° va | bY Vettes 7500 LoS" Vins Pe Ps 68) 5m) caalis) fo to.Fe 10, 9k | pbs cna! (20%) ae | pt 2 AS 736) Sin (15) = @= 2412 VAGS! Se yr* »— | S= l2zso / is? [Pp O Es Ve = 0.133 / 80° Se vVU¥ > [S= 26.7 £-15° VA «(Sle (®y-Oz) = 26.7 con (is) [P= 25.80 ®= \s sin (ev-Ge\ > > LQerea Vat obsess. (- 2°} 796 Irwin, Basic Engineering Circuit Analysis, 8/E 9.89 An industrial load operates at 30 kW, 0.8 pf lagging. The load voltage is 240 /0° V rms. The real and reactive power losses in the transmission-line feeder are 1.8 kW and 2.4 kvar, respectively. Find the impedance of the trasmission line and the input voltage to the line. PSY SOLUTION: te [Mle Ro, Beet. 2 (Se Acms ez _ Wipk z4ace-B) Vs $ a ES ve ALP —) = 8, > cea (pt) = 346.9% L iT, | = Sap Tr= |sé 1236.9? Aras Chapter Nine: Steady-State Power Analysis 797 9.60 A transmission line with impedance 0.1 + j0.2 2 is used to deliver power to a load. The load is capacitive and the load voltage is. 240 /0° V rms at 60 Hz. lf the load requires 15 kW and the real power loss in the line is 660 W, determine the input voltage to the line. SOLUTION: . AR ns mR & t+ Vs @ a [7] Vu T ~ “T 800 Irwin, Basic Engineering Circuit Analysis, 8/E 9.83 Given the network in Fig. P9.63, compute the input source voltage and the input power factor. oe § poet ws er AS ee. p03 20 YS jorge 10050 yaa = , Bo kW | + 2oOkW Ve Vs © 0.86 pt | ¥, opi | - 220/0° Vins lagging | - lagging l ! - Figure P$.63 SOLUTION: ~ loa g. (Tpit Fe = Zoe dt Arma } Vol (pts) 2ze 6-8} vt eb) ee Tye 4186-9 "hee Or = Oy, - tet ( pf.) =~ 369° . Lead: (E,{= fe. 8, = B- tet C ph) 7 IV OF L T,= 3 [+ 247° piety SA; te : \ fe Vis fo-at t joos Yn + vie 224 f f00° Very J ms Tye Dyrtze gat 3h’? Ars Pts = tae ( Bys - 8) 2 bee Cut - i-3ne)) [ pbs = 0.78 Laing | Chapter Nine: Steady-State Power Analysis 801 §.64 Use Kirchhoff’s laws to conipute the source voltage of the network shown in Fig. P9.64. q AAS ee oo¢Q /0.250 Tz, . vx, _ 24 kW VL 36 kW Vs C ) 0.85 pt | 220/0°Vims | 6.78 pt lagging lagging Figure P9,64 SOLUTION: > 3 itels Cee FRME 2 zo deme Lead Z * IVi( Upfz) B20 (0-78) ( Tye Zhe [~38.7" Avr i J - aN - l Ons Oy.- eo (Cptajs ~ 387% f tds Pe eke vzg Arms ) Leada. 4 WiCek) 226 Co.e) pe Pe kasi Bes Px, > By, ~ ext" (pF, ) = -T.8° Af tye Ott, « 337 36.1? Aree prey Vs = fo.04 + paeesy Tet Vy [Ver 248 L 4d? Vern | S i 802 Irwin, Basic Engineering Circuit Analysis, 8/E %.85 Given the network in Fig. P9.65, determine the input voltage Vy. PSY aAA eye, oS 010 jot QO rs Te + ty : 30 kVA 40 kW Me Vs © 0.9 pt 0.795 pf| 240/0° V rms lagging lagging 4 i 5 Figure P8.65 SOLUTION: Cltie Pu. = 300 159 Aves } Loads: DNC) 24060.4) pe 139 | 28:3" Arig - Oy» By, ~ tea" (8.3) =~ 25.8% ( tf fee. 2 Zio Arms y Leodd 2: } [MA Cph) yee (32.5 Arr i . * \ Bp: Oye co Co.ms)s - 37.38 J Toe Crt, 247 32.7% heme Ver (ob rjpond Dy a Vy Be ant a Aenea > 9 Yom 288 e088 Vrms | oO QO ao Chapter Nine: Steady-State Power Analysis §.68 Given the circuit in Fig. P9.68, find the complex power supplied by the source and the source power factor. If f = 60 Hz, find »,(x). — » 619 f020 $x, +z, Yrs . : 30 kW 20 kVA 10 kW Ve © V5 0.8 0.9 0.8 480/0° V rms leading lagging lagging Figure P9.68 SOLUTION: P {Tgl = 3, fosdo t,o Arms : Load3+ \MLCpds) 4 Ba(e.8) Tye 24-0 1368 “Arms Opy> Ov~ coe! (pts) = ~ 368 IDy{s Zoowo, . 4b.7 Arog Load 27 480 (0-4) ~ Dh. 40.3 Lease? Anny, Or, 2 o- eer! (04) = -25.6° IT l= Boeee , Te.) Acs ; Load 480 (0-8) Tye 284 36:20 bring Brg2 Ove oe (ph) = 36.9' Dye dyptttfy= 125 /5.25° Aras = &iejo2Te +, = 47) (B06? S.= Ve Do = O14 2? kVA \s. - CA Lz leva [efi om pts = Ce2 ( Bys- Bre) = 0.999 Sina Oy~8, <2, Lonely ee us (= YS Ve] oa Conf e 2 069V : antes 806 Irwin, Basic Engineering Circuit Analysis, 8/E @.69 In the circuit in Fig. P9.69, the complex power supplied by source V, is 2000 /-30° VA. Tf V, = 200 /10° V rms, find V3. Figure P9.69 SOLUTION: S\ 2 EM =? ove [8% Zee fie * T¥e lo [cite De tof do* Ares, vV,< 2 fs tpt, B® Vue -Gtfiey T ee | Ver ass [nig 48 ven Chapter Nine: Steady-State Power Analysis 807 9.79 For the network in Fig. P9.70, the complex power absorbed by the source on the right is OQ + j1582.5 VA, Find the value of & and the unknown element and its value if f = 60 Hz. df the element is a capacitor, give its capacitance; if the element is an inductor. give its inductance.) R CG) 120H10°Vims 7) (4) 15020: Vrms Vv ¢ rt Ul 3] 32 Figure P9.70 SOLUTION: 222 jiseZ.e VA « Vs2 0¥ iso fg re Is = io, ole" & Tes jou Le" Aras Vei Ven » 147 “4 %.00 2, cat eo . [| k= | dys “JR ee jy > 377 ls ———--- | Oe Fad, F | 810 irwin, Basic Engineering Circuit Analysis, 8/E 8.73 An industrial load is supplied through a transmission line that has a line impedance of 0.1 + 70.2 Q. The 60-Hz line voltage at the load is 480 /0° V rms. The load consumes 124 kW at-0.75 pf lagging. What value of capacitance when placed in parallel with the load will change the power factor to 0.9 lagging? oS SOLUTION: iSual= fag taewe® 2 663 EVA Goce cor ph) 44° Sods 165.3 /4n¥ leA Gor = Tn (Sota) tet ke vac, \Sumols 1avxto® 379 uta Crees oa (ode 258" 049 Sue? (B28 (258° EVA Ganges HS = Go. kee Bee Oner- Quid % ~wcly Cosi 199) x0 us 37D Fes Wyle 468 Vea Vee S63 pF | | eA Been ne cnet Chapter Nine: Steady-State Power Analysis BIT 9.74 The 60-Hz line voltage for a 60-kW, 0.76-pf lagging industrial load is 240 /0° V rms. Find the value of capacitance that when placed in parallel] with the load will raise the power factor to 0.9 lagging. SOLUTION: Soal= fe. beeen 2 7h eva ) Bele= 78-9 [4a.0° GVA pHi Pe . ON Gigs Sh 5 eume Beis = bea “Cp Bid) = tea" C 0. mje YOS? | bo. &VA 1 Snears bib L2S8" bua Oneass Cee lode 29.87% J Yee Qeeng ~ Wort @ WOW = 247- obT ot tue 300 ris Wil = 24%0V ras 812 Irwin, Basic Engineering Circuit Analysis, 8/E 2.73 An industrial load consumes 44 kW at 0.82 pf lagging from a 240 /0°-V-1ms 60-Hz line. A bank of capacitors totaling 600 pF is available. If these capacitors are placed in parallel with the load, what is the new power factor of the total load? SOLUTION: ISeuals Fo 44000 =< SP DEVA Solas $3.9 L78U Leva Qua? Dm ESett § phic 82 Bria = eve (phy) = 249° Qald= 30.7 kVA De a2 wO We = 370 (ooona®’y ( 240)" =-j 4, @ ke Wye, Q. = One Veta 9 Quy t 7 kv ae, Breuer tant ( e) = 24 aging Gina Oro” nos P, 4 W's PL sete 2 tye Byte | Phy * 0.726 Hag ing Chapter Nine: Steady-State Power Analysis 815 &.78 A 5.1-kW household range is designed to rate On 240-V rms sinusoidal voltage, as shown i . POLT8a, However. the electrician haz mislakenly connected the range to 120 V ims. as shown in Fig. P9.78b, What is the effect of this error? A 7 i ye0foe Vie (+ 120/402 ¥ rms e b B : {a) a A 120/0°V rms. ® Be N (b) Figure Pg.?8 SOLUTION: Q3 tea yaad, [Trangel = Feangs . Stee 2 21.25 Arms : Wab{ [2422 | Os connected = 47.0 Phas: The cam rank cheawn boy dy vange Is doubled + Rak ! tho reduetin te unl tage bey add» i  816 lovin, Basic Engineering Circuit Analysis, 8/E ! §.7& A man and his son are flying a kite. The kite becomes entangled in a 7200-V rras power line close to a power pole. The man crawls up the pole to remove the kite: While trying to remove the kite. the man accidentally touches the 7200-V rms line. Assuming the power pole is well erounded, what is the potential current through the man’s body? ©* SOLUTION: BE rms I oped ¢ af \ Lontact Bsc] Rares Currant Flows RA Nprmmmnnennn NR emi From onechandy + 0 . ~ . - 4 hroush bate i F200 lol Oe Laan | arms, dso thar i Rosle PO A Mf hand. d dau bse, Rare ds ole Bast Case: skin ts hry Ese PSK IL | Laem tle a ZIFMmA lasers Cane: silclucts auk f Psa e (40 > Fare = loon. \Timan b = 7208 = ihe Ares BLigd) 2b IaS) 24a mA diew Ral i4¢ A bth Bhp a £. 3 SS = ‘ ee, Chapter Nine: Steady-State Power Analysis 817 8.80 A number of 120-V rms household fixtures are to be used to provide lighting for a large room. The total lighting load 1s 8 kW. The National Electric Code requires that no circuit breaker be larger than 20 A rms with a 25% safety margin. Determine the number of identical branch circuits needed for this requirement. SOLUTION: Leora |? Pos £2. be. 7 Aras ivi 12.0 geenuey|* Oo. 18 C20} = 15 Ams (25°% safle ) moepin) We ef branch. « [Brena | = HY Fe venue @ \ te 5 keane! | ! Lot 8 beancen "| i 820 irwin, Basic Engineering Circuit Analysis, 8/E &.83 A 5-kW load operates at 60 Hz, 240 V rms and has a power factor of 0.866 lagging, We wish to create a power factor of at least.0.975 lagging using a single capacitor. Can this requirement be met using a single capacitor from Table 6.1? SOLUTION: ~ (Sela) = fe = Sse 5 SWAP Sok: S171 30° eva pteid Bale P \ Log 28h kVA ; wg O oud = Sg cor Cptyy)= Be J ‘ a os ~ Is abs S882 2 5.13 eva 1 Sime Sid fizge leva BATS 5 bd tab le Way Lorene = tos Lov) = i2.8° } Pates = Le ie Vad Bee - weil = Our “Mois © 74 kVA Vul= 24eV SIRF ange site | Chapter Nine: Sieady-Staie Power Analysis 821 $.86 Use an RC combination to design a circuit that will reduce a 120-¥V rms line voltage to a voltage between 75 and 80 V rms while dissipating less than 30 W. SOLUTION: & 5 cn. \ : yooh] This cicant should do Vg) £ 7 : WVal2 72S Vems : {Ne \Ws le 20 Vems Ve wy, fitS\A, oy Chee t+ fo RY fls0 Lfu* @ 280 = Substhk OD fade C2) yf elds We)? Lwedte Nel *LwlPR 230 Ge bi trans eluset2, LS (ear | ¢ cig Cel do gx SSivL 822 Irwin, Basic Engineering Circuit Analysis, 8/E SFE-? An industrial load consumes 120 kW at 0.707 pf , lagging and is connected to a 480 /0° V rms 60-Hz ‘line. Determine the value of the capacitor that, when connected in parallel with the load, will raise the power fuctor to 0.95 lagging, © sging. SOLUTION: Suid t= on = iZome, 2 7a kVA Solas ito LESS kita P Pot ae? , Qa, = lteo eVAe, PN ag i Sous coat Opty) = 45° J ° 1Sasuy t > feone®/ 6.4¢ a 126k VA \ S nears ize LBZ? EVA Pra = Coe! ofe)= 142° Z Q eas ® 34.3 kvAe Qe ne wl We = Oren” God = - PO.? kv ae we 327 ols [Vibe 4 Vem MeTIIT TS 4249 td Chapter Nine: Steady-State Power Analysis 825 SFE-4 An rms-reading voltmeter is connected to the output of the op-amp shown in Pig. 9PFE-4. Determine the meter reading. 36 kQ 12 kO i DAY A No i Vs i 1.414 60s af V ©) rms-reading volimeter Figure §PFE-4 SOLUTION: ~ Va = - Be vy Vee taid Loy se }fe® Vers a Vo.e 7 BVs =~ 3/o° Vem [mete ands av | 826 irwin, Basic Engineering Circuit Analysis, 8/E SFE-8 Determine the average power delivered to the resistor in Fig. 9PFE-Sa if the current waveform is shown in Fig. 9PFE-5b. © H(i) (A) 2 Yi) $40 a ed EE a =e s {a} (b) Figure 9PFE-&§ SOLUTION: a ¢ a, Pa Deny ® tT 2s Dears PA eet y { ty ostet, Test gd tMatt Pl tte Clie & foe Jeter blele-2 preted [ete 4 '@ tof t Fems 3 5 Lfaeayt = £474 Pe Ute le le 8.67 Ws]