Table Of ContentChap 1 Problem Solutions
Telasingst + M0) V and $= U3 gayle — 20°) A. td Jor eet fat
ibe excemur value, (5) she Sas value and ie! the phagor expression i 3o-ar
canguier form sf volvage is the’ teferonge™ Is tae clctt iauuelive oF
: APs [ALA sn (te .
clusion: BGA Co Cotte “') ‘| L
wolreet
fe Masizmun vals
SEV hye LA
ee vas
= oy he
a
fe) Phasor expression in poles ead ractanguae ior
ie
Y
pide F -
The circus ie indyefive ae 2 awe V +30
aan
1.2 che oixculs of Prov, 1.1 consists of a purely resistive aad! a purely eeactive
Alement, Bn F end X {a} if che elements eve ir. series and (61 if thr clemensé
ae iu paralle
Selation:
{a} ments ers
?
BG
=
xe
2yte ~&-
Nae RA GIMP > eRe
—— ‘L.3 tn a single-phase circu Vy
te a reference node 0. Pd
wwe 2648
Teo a yenerst)
ith epee
10 polar form.
Solatsom:
reas 06 SF 25.88 — a4sn + 38H
= fiIOTE = IT ARZRN SY
2
1.4 A singleephase ac witage of 240 V 's apolied to 2 series circait chose impedance
1s DEBI? 2. Vind RX, P,Q end the power facto: of the cireate
Selution:
ou
1.5 (Fa capacitor is ccuanecced ia parallel wil the civeuit of Prob. £4 end if this
capacitor suppiies 1250 var, find the P and Q supphed by the 240-V sowee,
and finn Uhe revultane power factor
Selstion:
P= 380
@ = 4088 1250
pr = ow fant) — nat
16 A single-phase inductive load draws 10 MW at 0.6 power factor lagzing. Draw
the power rriang’e and determine the active power of a capasitar to be con-
sevt2d in parallel with the load to raise the power factor Lo 0.5.
Solan:
anfenrtos) = 1338
oa 0.85 To “
wens Tbe
@ _tgs—a2
> A718 Mer 7
iH
j
SLL =... __ a
a
1,7 A single-phase induction inntor s eperating at a very light juad éusing a large
par, of every day nud draws 10 4 “rom the scpply. A device is proposed to
“ineveage the effineney” of Lae maacor. Daring, 4 cemosistzation ue devine
placed i paralle) with te unloaded motor and che current érawn irom the
supply drops lo & A. When two o° she devices are placed ir. paraliel the currestt
Grops w 6 A. What simple device will cause this drop in eurrem.? Discuss the
advantages of :he devien, Is the efficiency of the motor increased by che device?
(Reva. shez an ingnecion motor draws lagging carrer},
Sohusiaa:
4 eapesitc will nun the drag in cuten th line because the lagi georaponent cfeurrent
dsr fy the motor willbe gary uflet by he Jeeding current drawn bythe cxpaztar.
hanged ifthe cexminel voltage remains
feumsten:, Se she moter eficeney rll emain tae some Ts ithe ba supplng he moar
‘ill be ls due co the Jneer ne eurret Hf the brs to the moter fror: Ube supply buss Inn,
the voltage drop he ire wil be reduced or ch may be desirable
The cursnt dram by she koter. Rowers, wih be
1.8 IP the iependancs between machines 1 and 2 of Example 2.1 is Z=0— j50
determine (a) whether each uwshine is generscing ot consuming power,
(3) whesher ewok machine is receiving or supplying postive reactor power and
the aragurt, and {c] the wlue of F and @ absorbed by the imperdance.
Solution:
f= 14664 580)
pa OER ais664 580)
Eu = 10010 — (208) = 1000~ 3268
z
(966 = y50)(10-
2) = coo — 268
Mace 3 genrates 1000 WW, ees 288 wap
‘Mactioe 2 ebeorte 1000 W, coeses 258 sar
(Capea: the ne sph (0m
1.9 Repos: Prodlom 18 if Z
er es ism
Exh = (s6s ~ 300) 2084 10)
Mactuce | gerarates 288 W, elivets 1400 var
aduoe 2 yanesares 265 W, sceoes £000 var
i
st in the lie absorbs 20.85} x
Buu scliues soy generar,
LL A ealtage soutge Bag = Bua" V and the rent through she sours is
yeu 8y Ju, = 10280° A, Find the values of P and @ and state whether the
sounse is delivesing or teveiving vac ea
Solution:
inant « 1,
1039 W defvered
400 var delivted
pase
(0 var absoril by saree sine Tou define positive curren: From ea and Ean defines
Pot oat hugher pocential chan ve wen ean Hs postive |
141 Solve Example 2.1 fF = 100 0° V and B= 120-80 V. Compare the results
‘vith Beample 1.1 and form some conclusions about the effect of variacion of
te magnitude of By in this circu,
p= WORMED OY 892900 a ae
7 zB
nani 978) = 1300 — ine
(103.994 360) (—12 = 30.78) = —42er ~ s700 - j21-4-508,
1200 = J
(Machine 1 absorbe (200 Wand 78 var
Mashine 2 delivers (200 WY ane 201 rar
0173.5 729 var absorbed by lave
a Exanpie 25 che howe ceceved 598 vac,
wf Srony each source. Ratemg [Ez eauesd sorte
ase in Q supplied to the Tin, bot the significant f
'sakatrosing [Bj cuused hat source ts supply not aly all he Q abeorbed by che ilne
sda 7 car elivered cathe [Ey soe
!DereS pw ranean som nz
Eraiuate the following expressions in polar form:
ie) jote?
Sclution:
20.88
Buin bes Lala
Sank =
$806) 05 = 90866 = 1-457.382 = 21mecact
054 yb.486 33 =
0.805 — 705-08.
1.588 1 386
50
LAS Three identica! impedances of £2, xe V-connecred 20 ‘alanced three.
phase line voltager of 208 V. Speskiy all the linc and phase voitages and the
currents ag phasors in po'ar forme with Ve, as reference for z phase sequence of
Ld Lara batanoed-thrge-phase ayatem the ¥-connected imapedances are 102308 9.
2 KE Wie = ABZANE VE speesty Jom im polar form.
6
1.16 The terninale
oir a vailaecer ox
sappiy are lageled 2, 6 and ¢. Betwees any
ares 115 V. A resistor of 10U 9 and a capacitor of 200 %
DPIY are conmected ‘x series ftom ¢ 10 6 wich the
resistor connected to @. Tho pict of connecuice of sh: elemeets ta each orice
is abeled x. Detersnine grepaically the volumeter reading Setween ¢ and x if
phase sequence is abs snd if phase wecuence ie ack
Selusion:
Sencenve abe
F
= Bats v a Re
r
lseaat = nev Pp
meter ceading +578 488 — STA ¥
Secuenon a
R= wey ;
HE = 75
meter radng = 996-525 — a1 ¥
1-16 Determine the curren drawn from a three-phase \MO-V line oy a three-phase
S-bp motor operating at Full oad. 90% efficiency and 80% power factor lagyion,
Find the values of P and Q draw, from the line
Solution:
1x6
Rrecrrresr]
P= Vix s40 + 2020x08 = 12,321
@ = V¥x M0 2030 x06
oa 098 A
Aras from tine
9,324 yar érasn om ine
7
LAT Ifthe suyeudanoe of each of the three lines connecting the toto: of Prob. 1.15
9 bus ig 0.3 ~ j1.D 0 find the Tine-to-tine voltage at the bus which supplies
440 V al the moter.
Sclatien:
Fa WS6(08—j06) = 1691—j22a8 A
‘Whou 2g refroncr i eltage ta neutral of the motor atthe veemieal where Ts caleioted,
gi, ¥. the supply bus veltage co neutral is
2404 OS+s1OUSAI— 1229, = A+ se
Linetoline vokage Wj = VBL 4 s1264] = 40 ¥
1.18 A balanced-A luod consisting of pure resistances of 15 0 per phase is in par-
lle! with » halanvesd-¥ load having pase impedauess of 8+ j6 0. Identical
impedances of 2+ 35 9 axe in earh of the threr lines connecting the combined
Jonds to a LLOV three-phese supp'y, Find the current dren from tie supply
and Tie voltage at the combired fads
Solution:
Covers Ao equities Y havang 28/3 =8 M/pase
R619) _ DEO aye eos eo
ease ~ Whoe "Ha 98 F
= dart jar = dans 2
‘Sueent dren a sep
2 = 2-38-aat 4 yar
103
by = Hees
5413573 = 788 /s088 2
2.08 frm supply
Letty Mi eual wolage atthe loud, netcbne voltage
%
Limetcline 4
2.2 Vo neutral
48
1.19 A three-phase load, draws 250 KW at a poster factor of 6.707 lagging from &
440-¥ ine, Is pataliel with this load in a three-phase capacitor hank which
dren 60 &VA. Find! che cotal current and resultant power factor,
Solution:
TLetlng §; and S; represent che load and capacitor tank, mapectivels,
Sy) = ome p50
1.20 A clree-passe motor draws 20 2VA at 0.707 power factor lagging from « 226
sos. Determine “ie kileyoltampene rating of eaparizocs to make the combines
power faccor 0.90 lagging. and deterraine the line cusreat before azid after the
Solatan:
From tee eure
Leute 28 =
14688
Whoa eapacor:
‘ich eapaciton:
1-21 A coal mining “drag line” machine in an open-pi: shine consumes 6,92 MVA
At O8 power factor lagg:ng when it digs coal, and it generaces (lelivers
the wlectric system’ G.10 MVA at 0.5 power faccar leading wheo the louied
shovel swings sway from the pic wall. At the end of the “dig" period, the
soange m0 supply cortent magnitude can cau tripping of a protective relay
which is constructed of sulid-scate circuitey. Therefore ic is desired ro minimive
she change in carent magnitude. Consider the placement of capacitors at
the machine terminals and find the amount of capasitive co-rection (in kvar)
to slimiate the change in steady-state current magnitude, ‘The machine is
energize from a 36.5 2, three-phase aupply. Start che solution by letting @ be
the tozal chree-phase aieyavars of the capacicors ceanected neross the machine
terminals, and write an expression for the magnitude of te line current drow
fy the muchine in terms of Q for both the digging and generacing operation
Assume Iaetovine voltage V1 is coustagt. ‘Than constant cuttent aymtude I means
lemstane where [S| = V3IY Uri (OF MA,
|S] = sees 208 — 301
In.zas— jn.ss2— 401
ase 0m — 1
(coos! + (ussee— py = 9.0003 +0008 —
gl 0 rE +6"
and oqvating [Sf forthe dig and eving pe
aagsy— 12089 4.9"
0987 = 0357
= 0.809 Meo: of 399 boar
122A generator (wiaich anay be represented by an era! in nusice with an induecive
tance) 's rated S00 MVA, 22 KV. Ite Y-conaected winengs have a reeetence
4 per urit, Find ube obnue value of the reactence of the wind mgs.
Solution:
1.23 The genezator of Prob. 1.22 is in a circuit for which che bases are
100 MVA, 20 XV, Starcing wich the pes-enit value given im Prob. 1
per-unit value of reactance
ecified as
Sn the
the yenerator windings on the specifiee hase
Souci: 2.60?
A
xe 196 (Se) (BY = p25 se wie
1.24 Draw the single-phase equivalent cizeu:t for the motor {an emf in series wits ire
uccive reeetasiot labeled Z,, andits connesticr: to she valiage supp'y described
1 Prot. 1G ané 2.17, Show on the dingeam the peruni: values of tne line
Snopedance and the witage a: the macar terminals an a bese of 20 kVA, 440 V.
‘Tyer using yer~anit values find she supply voltage in per unit and convert the
pernait value of the supply voltage to vol
10
Solution:
erst ase lesions
an 2m OH oer et
a= 92 2 00s perue
ve BB = a0 penne
noc
ee = = mms
Bucs = OO.
be revit
Vota execs
Yo = Lasts 6y(0031— jo1088)
10.777 x aorecata
100.604 + 70.0905 = 1862.97" per unit
| = 0086x140 = 70
1.25 Write the two noel admistanew equations, similar to Eqs, (2.57) and (1.88) for
the voluages at nodes @ aud @) of the ef-euit of Fig, 1.29. Then arrange the
odai adraittance equations for al four independent nodes of Fig, 1.23 into che
Yew lotsa of Eq. (1.81)
be@ iri Wave - moe
be HAM HIN THIN =
FRearraging equations for ous @ and bus @ yields
bus @ “WK GY = 0
bes AY, WAM HEM =
The Yi oor
© @
TfMrvew ¥ “
BS)” mame % 2
o % 4\T |B
a =A
