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Unformatted text preview: P5.10 VU)=10005nﬁ5003T)V
p(r=) v (ﬂ/R—20005in2(5007rf)=1000[1—sin(10002rf)] w a”; =(Vm]2/R=1000W
111+) (v)
1000
9 ’i if (has)
—[000
am 6"” “‘0
£000
2 1:
ffhs) * — 1T .2 — 1 2 4 —
P5.13 rm _ fﬁ; (ﬂa’r— /E[£25dr+£4dr]_3.808A
p513 Vm — _/_}v2(r)=a«r \/j[58xp( r)]2dr_ Jj[25exp(— 2am =,J[— 12 5exp( 2r)]:j)_ — 12 .—5[1 exp(— —2)] =3.288 v P5.24* v1(1‘) = 100 cos(c0 1‘]
v2 (1°) 2 100 sin(.:0 r) = 100 cos{w r — 90°)
1.11 =10040° = 100
1/2 =100z — 90° = 1100
v = v; + 1.12 = 100 — 1100 =141.4z —45° 3 1», (1°) 2 141.4 (105(14):° —45°)
l 0
° ‘4 V2 lags V1 by 90°
V; lags V'l by 45”
V; leads V; by 45” [DO P5.31 V1(f)=10005(mf — 30°) v I”, = 45x rm = 14.14 A v; =104—30° v L =14.144—70° A
1; (r) = 14.14 003(0)?” — 70°) .4 P5.35* vi (1) = 10 005(2000111‘)
c0 = 20001: Z1 2 J'ml = 01200:: = 200nz90°
vi =1040°
It = v,i /ZL = [1/2001 — 90° 1; (r) = (1/20n)cos[2000111‘ — 90°): [1/201c)sin(20001t1‘) I; (1‘) lags v1 (1‘) by 90” 23+)
1103
0.5 1' (as) P5.36 (a) V 2100130" I = 2430” Z 2 ¥ 2 5010“ = 50 + J'O Because Zis pure real, The elemenT is a resisTance of 50 £1. (b) NoTice ThaT The currenT is a sine raTher' Than a cosine. V 2100430” I = 3/_’ — 60” Z : ¥ 2 33.33190" 2 J'33.33 Because Z is pure imaginary and posiTive, The elemenT is an inducTance. (0:400 L=——83.33 mH (c) NoTice ThaT The volTage is a sine r'aTher Than a cosine. V=IOOZ—60° I=2Z30° Z=¥=5OZ—90°=—J50 Because Z is pure imaginary and negaTive, The elemenT is a capaciTance.
I : 200 C = — = 100 F
‘” ‘29 *“
P5_37* v; (1") = 10 cos(2000:rf)
a) = 2000?: Z: = 1 = 415.92 =15.924 —90° 52 (of Va 21040”
IC = vC/z = 0.6283490“
W) = 0.6283 cos(2000:rr + 90° ) = —0.6283 sin(2000:n‘) HT) leads v5 (1') by 90° P5.40 (a) z =¥ 2540“ =5 Because Zis pur'e real, The elemenT is a r'esis’rance of 5 Q. (b) 2 =¥= .4490" = 14 Because Z is pure imaginary and positive, The element is an inductance.
0021000 L=E=4mH
(0
(c) 22%2201—90 =—J'20 Because Z is pure imaginary and negaTive, The elemen'r is a capacitance.
(0:1000 C=i=50yF
\Z'iw ...
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 Spring '08
 DABBAGH

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