AC Circuits Problems

8/2/2019 AC Circuits Problems

1/17

tha Slnu$ldrl St..dy,st.te Ra.pon5. 2r,Eil Problem3 10.7. For the circuit of Fig. 10-15, sketch, as a fonction of ot (a) lV,ll(b) anc v.; (c) ll.l.Anrj lv,U3)J = 25.7 V; ans V,(-jto): -26.(t;1r"05) : 7.91 A10.t, Replace borh to-O resisrors in rhe circuir of Fig. 10-15 wirh l00,pFcapacitors and let I. be the desired response. Sketch the magnitude andphase angl of the response as functions of a, and determint all criticalfrequencies of the response.

-4rr. I.080) : I.031:2E A; 0, 170.7, 1100 rad/softo

Hg. lO.t5. See Oriltftobs. 1G7 snd 10-8.

I.I rof,l I\ roa{ ,H lv,oo 1o: v

identifirst thatsay thatat w

said to haresponseon the (l)or

asbycomple

ProblemaFlg. lct6: Se Prob. 1.

, Use nodal analysis on rhe circi[ shown in Fis. 10-16 to find the phasor voltaee2Q -i4{t

lo

(a) Find Yr in Fis. 10-17. (r) To what idetrticat values should the capacitiveimpedances b chanSed so ihat Vr is 180'out ofphase with rhe source vottaee?

1oo 1d v

roo tq: v

ar rhe ori

appare

Flg. lGlT! Ses Prob. 2.

Flg, to.t8! See prob.3.

of sil is si- erlols r i.(r) = loj cos idr A in rhe circuir shown in Fig. 10-18, fiod u:(0.that 500 rl

iro () -ilo I _iro f}

^ 0.5.,11ko


2/17

ztl E !tr..?lng Cllcult ln.ty.lrFlg. tGl9: See Probs.4and 5.

I6

10 cos (10.r + 30o) A

Find udr) in the circuit of Fis- 10-19 by usins nodal analysis.Write three mesh equations and solve them.to determine tr(0 for the circuitshow! in Fis. 10'19.ffo = 100 mdls for the source in Fig. 10-20, find I,.

(a) Draw a tree for the circuit shown h Fig. 10-21 so lhat i. is a link cunent,assisn a complete set of liok currents. and find i(r). (,) Construct another tree ihwhich ur is a tree-branch voltage, assign a complete sel of tre-branch voltages,and find u,(,).

125 sin lott v8 The op-amp shown in Fig. lG22 has an infinite input impedance, zero outputimpedance. and a iarse bu. finit {posilive, real) eain, A = - V"/Y . (a) Construct

F.. to.23: See P

HgL tG2+ Se6 Pr

Ea. lG25r Sse Pr

H9. lC2& Se6 P

Flg. lG2O! See Prob. 6.

Flg. tG2l: SsB Prob. 7.

Fla. 10.22: See Prob. 8.

v, I_t=- l,nff'*

0.4 mti

4OO,rF


3/17

th.9lnu.oldalStordr.Si.teBe.pon.e 279a basic differe.tiaror by lettiltgZr = Rr. frndY.,lY,, and then show that V,,/V" +-jaqRra; A+ -- bJ Let Zr represenr qand R/in parallel,6nd Y./v,, and then/ 'ho$ rhar v, v, - /darfl, rl i-C,Rr,a.A-.a.A t 'e superpo\ilron ro find l. dnd l.' r in lhe circuit of l-ig. l0 71.

H!. lO23: Se Prob. 9.

f9. lG2.lr Se Prob. 10.

tig. lG25! Ss Prob. 11.

Flg. lG2G! S6s Prob. 12

t0 Find the Thevenin equivalent oflhe niwork lhownt0-24h _12o i8o

-i0.006 v1

U ter @ - I rad. in l-i8. 1G25. and frnd rhe tlevenin equi!alent a. \een arlerminals a-r. Drsw the equivalent circuir as avolrage source V,e in series with aresistance R, and either an inductance I; or a capacitance C,r.

(a) Fig. 10-24d; (b) Fig.

t2 lG26 at: (a) o =n Fic.ind ih Thdvenin equivalent of thl0 rad/s; (r) @ = 20 rad/s.

s(,f

o.25 tc1,.

20fl


4/17

Enet!...tra CNroldr tr.ty.t

Il!. lG2?; Se Prob. 13.

fEr tc2e See Prob. 14.

;ttL lG2$ Seo Prob. 15.

Fltl lG3O! See Prob. 16.

13 Find V, in $e circuit of Fis. lG27 by usins: (a) superposirioD; (r) repeated source

jsa 20 tcvt

I+) i$aA -j2oa v: +-i5o (n14 Use supr?osition to help fiod ir(r) in the circuit of Fis.(a) lm cos l(l,tr Vi (r) l0o sin t0a, Vi (.) lm cos 500Or V.

0.ol F1,"

ira10128 if u"(r)

5kQ217J\ ,,II (d0.Ol lrF

15 Lr o = 50 rad/s for lhe source of Fis. tG2s. ra) Con\rruct a phasor diagramshowing the three curents. Us a scale of I A/in. (b) Construct another phasordiisram showins the three voltags, using a scale of Mid_

I,=2diA

i2A

10s},o mH

Er 'Oi3

li$ i(\3,

Hs. rF3

Flg.lG34r

16 Io Fig. 1G30, let I, be a current source with an amplirude of 3 A at 1000 rad/s.Using this current as the refrence phasor, find the four indicated voltages andshow them on an accu.ate phasor diagmm. Identify rhe equality, Vr + V1 =

v.i20 o tv,+) t-2sfi*vz

17 For the chuit shown in Fi8. tGlt. ter Vr = tmzq v. lvrl _ 80 V. and lV, _50 v. (a) Use a phasor diagram m Aetermine V, ;; V;. rire.e a,e r*o po"siUte

o-o2 H


5/17

lh. l|rusoidai aioadv.Slaie AolPon.. 2Alanswers: aive both. (b) If Zr is af, induclive iinPedance and Z, is a capacitiveimpedance, 6nd Yr and V:.

z\v,lz,

l8 In the circuit of Fis. 10"12, it is known that lV! = 50 v, V,l = 125 v' and Vr =200l]!l V. Use sraphical methods (straishtedse, ruler, protractor' comPass,and all that tun stuff) lo find the angle ofvr. Although there mav seem to be twoanswers to this problem, on is ri8ht, and the other is wrong.R, x, L,

2,,

19 A 4-O resistor is in parallel wilh a 125-pF capaci.or. The parallel combimtion isin series with a z-mH inductor and a siousoidal voltage source V,. Caiculate a fewpoints, and then plot a curve ofthe magnitud ofthe ratio ofthe capacitor voltageto the source voltase. Cover the frequency inlerval. 0 < o < 5 krad/s.

Flg,lo.3t: See Prob. 17.

Fig.lO'32! See Prob. 18.

Fig, 19.33r See Prob. 20.

Ftg. lG34: See Prob. 21.2, Plot a curve of %",/Vh vs. o,l0-34. -5000 < o < 50m rad/s. for the circuit of Fis.

1000 md/s-

and %l =

ro0lEl v

20 For the circuil shown in Fis. 10-13: (a) locate all the crilical frequenciesthe .esponse lV./I, i (r) calculate the value of the response at one point b-:tween each pair of critical frequenciesi (.) sketch the respoose over the range,i-10. o< l0rad/r.

l2t /rF


6/17

s4 Eislno.rlns Cl?cult Araly.i.stant, this would not result in a lower bill; lhis cure therefore doesinterest the coosumer. A purely reactive load must be added to the syand il i\ clear thar il musr be added in parallel. \rnce lhe \upply vollage loinductor motor must not change. The circuit of Fig. I l - l I is thus applicablewe interpret Sr as the induction motor's complex power and 52 as theplex power drawn by the corrective device. Lel us assume a voltage2301E V rms.The complex power supplied to the induction motor must have a realof50 kW and an angle ofco\ rr0.8,. Hence.

solsqs l(o.E)

P.oblenllg. I l.l i

sr- = 50 + j17.5 kvAln order to achieve a PF of 0.95, the total complex power must become

lnS 5-r, /cos 't0.95t = 50 - i16.41 kVAThus, the complex power drawn by the corrective load is

s, = -j21.07 kvAThe necessary load impedance Zr may be found ir several simple steps.curnt drawn by Z, is

0.8

t|! I l.t{

S,_Y j 21 tJ70Ii: : j91.6 Ar,:j91.6 Aa:!=ffi- izst a

230

and, therefore,

Drlll Problem

Flg.ll.l2! See DrillProb.11-9.

lfthe operating frequency is 60 Hz, lhis load can be provided by a Icapacitor. The ioad may also be simulated by a synchronous capacilype of rotating machine. Thrs is usudlly lhe more economical proceduresmall capacitive reactances. Whatever device is slected, its initialmaintenance, and depreciation must be covered by the reduction inelectric bill.t1-9. For thq circuit of Fig. li-l2.findrhecomplexpo*erabsorbedby(d) dependent source; (b) inductor; (c) independent source; (d) resistor: t,fl-rCcapacitor. Ans: 42 j10.5;j25.5; -3 + jtst 45t -j30


7/17

AYer.E

source rn rhe Lircuil of ti8. Il-ll.ib,E\aludrethdtpoqer,ll.001 .. Ir------o-- |I asin2orA I I"-l If,m.-r="*roocos2orv(l ,ont ,.",,4 |L____l____Y IFor the circuu sho*n rn trg. ll-14: ra, nnir,ne po\. ': 'x'orUea Uy rrre Irnducrora\afuncuonol:imeilb).kelchlhepoq r,'rime.0 r' I2 ms: (. ' and caiculale lhe value oi lhe potre' a. rn'. Iffi I{0.,n }roon },oon $r,,,,^ Itltt {A vohage source. l0 co\ i000t v, a 0.4-! I re$.tor. a l-mH intuctor. ana a f .Z5- |mF capacno. are alijn:eries. Find Ihe power beins ab.orbed by each of rhe four Ielemenl'al/ - lm\. 1A l0O-V barlery. an open swrrch, a l0.O resiror. and an uncharged l-mF capaci- 'lror are in


8/17

eoa lneln .rftrg ClrcuB lr.lr.l./7 Find the average power absorbed by each of the five leEents in the circuit of Fig.tt-11.F.lg. ll.l7! Se Prob. 7.

fb.ll'18! See Prob. 8.

tla. tl.t9r S Prob. 11.

Flg. li.2O! S6s Prob. 12.

6r, 10f,l

-i5 s)

E A square wave of voltage is 100 V for 0


9/17

circuit ofFiB.

to'a cune.t ivatue oIthe resisrori (d)

capaciror

sourceon aload,sribject ao the(.) RzxL - 0.for rhe foltow-

E, tl.2l: See Prob. 13.

Hg.'l'1.22: See Frob. 14.

lla.ll.23: Sse Prob. 15.

Flt. tt.2+ So Prob. 16.

100 tl o.4IIc F

' Av.r.r. Pow.' rnd BM.!u.. 30,12 In Fig. 11-20, find the value of Zr to which a maximum average power can bedelivered, and tne value ofP2,,.."".l.f Selecl values ofi and C in Fig. 11-21 so that a maximum average power isdelivered to R, and thn calculate the value of that power.

250 rd v

I :l For the circuit of Fig. I 1-12, make a sketch of ihe averase power delivered to Rz\'. Ihe .dlue ol Rr, 0 - R. : 40 O.tor, 50Y X,o.1H

l5 Wilh refrence to the circuit of Fis. 11'23: (a) what load impedance should beconnected belween r aod ) to absorb a maximum averase power? (r) What is

10 tol Y =) v,r (:r

16 For the periodic voltase waveform shown in Fis. ll-24,6ndr (a) the averasevalue ol lhe rohage: (6, rhe erecrire value of rhe ,lohagel (, I rhe average powerIhe rollae kould deliver to a 6-O re\i\ror.

17 A pe.iodic curreni waveform is 3e ,/aAfor0


10/17

18 A periodic current i,(r) is I A for0


11/17

Avor.g6 Pos...rd hMS Value. aOOIr."r/. It.27: See Prob. 23.

ia. t t.28! See Pro5.24

tlg, tt.29! Ss6 prob. 29.

o

i) lloo1ti

24 ia ) Find the power factor of th. loa.l ro the right ot r,n in thc circuit shown in Fig.ll-2E. (r) What value of capacilance should be connected across a-, !o achieve alasging PF of0.95?a 2OA

5 io,,,b

r20/d Y rms60 Hz

6a Hz

,0 l01A

l0 mH

loo Q

(c)for 2

in Fig.

(.) .hthewhat0.92

25 A source, u,(r) = 100los 100, V, is in series with 8 O, l00orH, ar( 2.5 mF. Fi.dl(4) the PF at which the sou.ce is operatin8i (r) lhe val,Je of capacitarce thatwould have to be connected directly across the source so rhat it ope.ares with apower factor of 0.9 leadins.26 A cerrain rwo-terminal network has a.ross ir a sleady-state sinusoidal vokagewhosefreguncy is fO Hz and whos effeclive voliage is 120 V. The inpedanc;fthe network is :l + j5 O. (a) Ar whar power faclor is ir operating: (b) Whataverase po{,er is it absorbing? (.) Whar vahrc of capacibn.e should be ptaced inparallel \yirh it so thar the powe. factor becomes 0.9 leadin!?27 The foilowinsrhree eieclricai deviccs are all cornectd in paralet:j tVAat0 707PF leading, 7 kVA at 0.8 Pf'hssing. and a series aI nerwork wirh .It = 2 {, andl, = lf).61 H. This composire load is supplied with powr from an ida; 100,Vrms. 50-Hz, sinosoidal vohag source. Findi (a) th pawer faoor at which rhesource is operatingi 1a') rh effectiv. vai.re of lbe source cuffenti (.) the capaci-tance rhat musi be placed in paratlel wirh the composire load so that rhe powerfacior wili be 0.95 laggins.28 A lwo-terminal network has a vollag. ,r(, = 4.os 00r - 30") V. across irsterminals, and a current. ir(, = 2 cos (10, + 20") A, enlering the terminal at whichrhepositive voltase .eference is ]ocated. Find: (a) yr.",": (r) Ir,h.; (c) the sveragpower absorbed by the nelwork; (d) the power facror at which the networklsoprating; (e) rhe apparent polver drawn by the neHorkl (,a) rbe complex powerdmwn by lhe network.2, (a ) -Find the complex power absorbd by each of ahe five circuii elements in Fig.ll-29. (r) Al what PF b the source operaring?


12/17

3lO Engltro..ins Clrcu AnalY'la30 A source of 500/0' V rs rn series wirh an impedance of -r l0 O rd a load Z' r

'r'" '"""" " 'l*^*L'i., ";;;;;;;;; ,, + Jr 5 kvA' rd) what comprei-'youe' r' aAi'e'"n ro Zr ' ond rDt whdr '! lhc rodo rrj.tyto t o'"'nt t"nJJ"'-'r''" " i" t't r l l0 and fi nd rhe Pl- of: ra' rhe neruorl ro'n" n'nt ""'toi'i'iJ"*'* io'J t'"ine *ppl'"a ur rt" 'ource. (') sdecttbe value" * *^ ' *"ti "''ii'l "'"'" "'r' toit'" '''srt or ii'" u'r o rc!slo' is 0'es raggins'

Fla. tl.3or See Prob 31 t,.ro I lac+ i4r}32 A 230 \ ,m*ource hd tu" l.'::"'1:'.'-':.':.'] ',1,111i:i; ffi;";;ii;::laasinE oower tacror of0 a and the other draws:iiiil'1",i"i;;";; '.""urce 'unenr cmpurude rb' AI sh'r PF i' rhe 'our=

60 Hz

12.1htroductionI'|l


13/17

3|!t fnatro.dn! Clrcutt AntlYtl'flr. 12.23 See Probs 8ro 11.

ri.l?2+ See Prob' 12

i*

l1=. Se

8 Tle baranced hreephaie,h e+q,,e.ir-c:' fl,Iliri;ifff l: Ii)'"Id; :!008q v rms wirh po'irt!: chl-':'q5T'-^;:;;,ii"li*l..rla*".,,4, v,.0.8i;s8ins.lr n" 0.6o find: (a).rherorarpo*j]"ii",:- q;00A v,.. *,bq Refer (o the batanced Y Y crrcu'r or r-r8. !L)-o'*i',oii r.va ",rn.:-n - l.: ar+'.h2se \equence. Lel the sour(e supply r. r$;' :iY.|,#ix i'ii: i:i r ff;: #*tiiii"'J,^r;r'r:;*umed lr I,r = 20Eq A rms ano rne IL#^ilo,ii'i)' ^" i,a.;_* comprer power 'uPpried bv rhe \ource:r'r vi'r (dt,r i'n"ii*. ':-z:. t", z, ."r,:::ll .1*1'i',1: Tffil::r:i:.lkfl#"'l fffwirh 0.25 H tf v.^ = 2101q:V rm5 attr' trz'lrr the loral power dehvered lo lne loao: r') Ihe rotal power lort in lhe wIrE.".rlson..;,al rf'" pn "t "hich lhe \ource operale\-...- ,- Li" tr-rr A\suces,srance:,d, ,rc r, 'u is lhown in Li8. 12-24. A\suGr2 An unbaranced rhre+Pha'e for-wrte $'l:1,.;;;;;;.i,,i,bez^{. r0-po"rrive pha\e 'equence w'llv- - l^*g^'^' .,-, 'frii, i#'= i;li o, -d Zcrv = 12 + j0 o Find r^r'

I r A Y-connccred toad hs * T'*11': :lI" #,'$i:::i.l'J;.1'": l;;trais lhe load power faclor to 0 9J lagerng


14/17

placed in parallelwith each phase impedance if/: 60 Hz? (r) Ler the line voltasear lhe load ha\e an ampjitude of500 V rms. Ho\^ many kVAR tcapacrri\e, arerequiredat the load to oblain PF = 0.95 laCgi.cl14 The balanced lhree-wire, three-phase circuit of Fig. 12-25 has R- = 0 and V,, =288.7L1q V rms. Each phase of the load absorbs 2.4 kw ar 0.8 PF leadins.Assume ar. spquence. Find,. (a) 1,bi @) Zpt (c) r".i (d) the total load power.

Polyph.. Cl?cultr a2a

10f,}

-i10 o

with3.2 O.

: 0.94(d)parallel

I

;la. t2.25: See Probs. 14b 16.

12.26! See Probs. 17't9.

15 The balancd three,phase load shown in Fig. 12-25 requires a rotat power of 6 kWat PF = 0.83 laggitrC. IfR" : 0.8 O afld Vcj = 100l2M V rms with (+) phasesequence, find: (a) V.6 i (b) the totalcomplex powerbeirs supplied by the source.16 Let R" = 4 r) in th balanced three-phase circuit ofFis. 12-25. The load draws atotal complex power, S - 2400 +j2100 VA, and an additioDal 300 W is dissiparedin the lines. Determine the rms amplitude ofr (a) Id i (b) Iali (c) V.,.l7 fhe \ource in Fi8. l2-2o i\ baianced and ha\ po\ilr!e phase sequence \rirhJ - t'0Hz. Find I,,r, Ir,, l c. and the power furnished by each phase of the sourc.b

8r.65 4E v

lE A balanced, th.ee-phase, Y,connected volrage source has V,,,, = ,,10/60" V rmswilh I ' phd\e \equenLe. and rl \upplie\ a bdlanceo. .\*onnecled. lh-ree-pha,eload. Th total power drawn by the Ioad is t2 kW a. 0.75 lassins power facror.Find: (4) the line current I,,; (r) the Ioad phase current Ic,; (c) the liDe voltage19 Add I O ofresistance in each line of Fig. 12-26 and again work hob. 17.

vbr


15/17

a.3wrr to orld-Iumtr.tld probtsn 633l3 (a)

(b) iQ) --Y,cosor + Ri +tr]'.

cos (or +- dt RC',]\tan ) _)v?,- +lj@

35'79IIt3

l5t71921232527

Cn l.. 9 la) s.60/-s 2!'; (b) 1.585r!-f; (.) 5.95/r06.9.; (d) 1.s00 -, J.&31(e) 7 .02 + j3.84162.8e/(:or+47.r1 V(a) 1.25 cos{1000r ll4") A; (6) t.25 cos 00001 64.) A;(c) I.25nrm'*i)A. (d) o.729er0orl rrr) Ala) 4.81 A; lbJ -0.980 A; (r) 10.42/15.62., 10.42 cos 000, + t5.62.) A(a) -(4.0 Y: (b) -4.02 A; (c) 257 W3.54 - j2.83 A50 cos (5000r - 135') V,69.5 cos (5000r + 40.7.) V,2l cos (5000r + 38.21 V(a) 1/45 H; (r) 0.896 H(a) 183.5 - j55.0 O; (r) 0.717 H; (c) 20.3 and 196.7 radls(a) 2.78 mH; (r) 0.750 mH(d) 50 krad/s; (b) 45.9 and 54.5 krad/s(a) 436 radls, (b, 816 radls; (c) 500 and 2000 rad/s!a) a0 llrad^; (r) 40 and 10 krad/s; (.) 98.0 krad/s; (d) 120 krad/s(a) 90 O, 20 mH: (r) 50 O. 100 mH

Ch.ptar 10 14.61/ts7.6. y0.285 cos (lY/ + 4.09") VI35

7

9t113l5

0.808 cos (loar - 46.0") A(a)^Irsg; 464 *1. ,., central 5 e: l/ lo righl in 5 mH: r,(r, _E.9E co^s f1000/ - i58.2"r A: tb, lree: c,..400sF. righl 5e:.,Lwilh! ar letl: l,rr) - 44.9 cos I10001 isg 2") \2/-146.6'and 2 cos (500r - 146.61 A0.894/-63..{'V in series with 0.4 O and I.2j F(a) and (r) Y, = -7.5 + j0 VSee sketches belowPt o.r 5


16/17

PEngl .rHii Clrcut ln lt.l.

Fro.l9 lEl1

17 (a) Vr = 8013C(29 69') v. v, = JO/-U(-52 41") v or vr : 80l'1[ v'v? = 5062 v. (b) v, = 80DC V' vu: 501--1I v19 See sketch below

1000 2000:l See sketch belo\t

?10.21

PIG23

l%l

-4000 -2000 0


17/17

lr!r.r. ao od+iMb...d pEtt.h.25 See sketches below

Pto.z3

Chapter l l I3579II

13l511192123252729

3l

(a) 23.1 - 21.1 cos 40r 42 sin 40r W; (r) -70.6 W- 103.9 W, t47.8 W. -2r9.6 W, 175.7 W(a) 0; lb) 20; i) 22.5; (/) 600-271.2 W,88.9 W,0, I82.3 W.0(l1) 0.690 w; (r) 4.5 w(a) 20 w; (b) 30 w; (c) 40 w; {d) 50 w500 o, 8 pF, 78.i Wb) t.96s - j0.262 Ai (b) 6.25W(a) 0.M74 At (bJ 2.14 Atu) 21.91: (b\ 2t.21i G) 22.98(a) 81.6 Y ,lb) 44a W ./rar 2.8c k!A. 2-7a kt/dtbt 0.949 lag; 161 i4.9E or41.6 mH(ar 0.8 lagl (b) q8't tF(d) 0.906 lag; (b) 96.3 A rmsi k) 322 r,F(a) -530.8 + j.16.2 VA, 184.6 +j0 vA,0 +j230.8 VA,346.20 - j276.9 VA; (D) 0.996 lead(a) 0.6 laci (r) 0.636 las; (c) 320 rrF

+ j0 vA,

Chapier 12 13579

5, 6, and -7 A(a) 34.014..C A rms; (b) 14.14/-73.j. A tms(d) 23.5 A rms; (r) 48.4 A rms; (.) 187.7 V rms(a) 159.2 t Ft (b) 3.46kVArd) 4540lQ.Iq V rmr: rb) t7.9J/ -14.04, A rms: (,(1) 98.7 percent 246 + j62.2 A;11 (:l ?:81+41" Arj,ns; (r) 1713 w; (c) 60.3 w; (d) 0.e06 tagt3 (a) t3.0s pFt (b) 1.230 kVAR15 ta) 151.5/ 112.3" V rm\l (r, I0.18 . j4.0J kVAl1 l? 8l4l4g. 5.t82q". and ls.8J/-7?.5'A rms: 1496.42J, and r4e6 w19 16.401120]{, 5.90DA, and 16.40/-80.4. A rns; 1261.482. and i26i w

AC Circuits Problems

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