Solution Manual for Power Systems
Analysis, 2/E 2nd Edition Arthur R. Bergen, Vijay Vittal
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PART I. SOLUTIONS TO PROBLEM SETS
2 / v =zxi2o co(t+3o) =s V= v20/3o° v i = zxio (t-3°) = I= to [-30° A
@) pte= VI/r/[e9+ (2ot +4¥+4 )] = (oo u2oo ct w
S= vz" =1200 [@o° = P+ => P= 60ow, = 103 A+
(6) Z= Y/T = 12 l60° = %+310 3 R+3 2R=6,X=0 39
2.2(e n (2.3) c 4=4 P,= 7707 =//Ir/0 +t)
0-d pk.:= -23= I)r](a9 -t)7he.,oVl = too, ee
4·t ITl= io a4 0= 45 Dd g> s°= Z= i0ls°
7 07+3707 =R+3K 2=7 07, x= 7 07
0) 5.v1"= Zy>(07+35207) io P= 707, > 707
() op&easy · - if)= [r] e A
?6 = bite)= L di z. -2L[I/at a.t =-u[z/~2at
h» = u//= 7o = Q.Th> Q.As..e!
= Jr '-5
2. B ·9pl 0a4 Yg-208 /@o, Ya.= 7o[@_,
°
--g(7) • Fs . " , va =r20/30 V,-to/-30°, T.= 2//os° =>
Ya = t2o -/so Ua9 p- eh·=
F,=r2/is" z,= '2ls:
2• s-vr = v(xv=v = Y{+Y" = 35to+o1 =01+3$
28(040. 4o,» n od0 0-% e +-J, /kc»d.k,
T = 0972s /@.3si"°, T= 09r23 /go.6s° z.=092/so°
5 p el< as ) 6 = 7/9/t.1=0.o1 Lo° Ta
Wo--, T= oo/ [-0, T.= 00/ /3o°• 2.10
