EEE471hw1sol

# EEE471hw1sol - EEE471/591 GSO 4 Edition Chapter 2 Solutions...

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EEE471/591 GSO 4 th Edition Chapter 2 Solutions 2.1 A 1 = 5/60 o A 2 = - 3 – j4 a) A 1 = 5/60 o = 2.5 + j4.33 b) A 2 = -3 – j4 = 5 /233.13 o c) A 3 = A 1 +A 2 = -.5 + j.33 = 0.6 /146.6 o d) A 4 = A 1 A 2 = 25/293.13 o = 9.82 – j23 e) A 5 = A 1 /A 2 * = 5/60 o /5 /-233.13 o = 1/293.13 o = .3928 – j .9196 2.3 Imax = 200A Vmax = 678.8 V Irms = 141.4 A Vrms = 480 V I = 141.4 /-5 o V = 480 /-105 o 2.10 a) p = vi = 678.8 cos (ωt – 105 o ) 200 cos (ωt – 5 o ) = 135,760 cos (ωt – 105 o ) cos (ωt – 5 o ) -I = (135,760)[ ½ cos (-100 o ) + ½ cos (2ωt – 110 o )] I = 67,880 [cos(-100 o ) +cos(2ωt – 110 o )] = -11,787 + 67,880 cos(2ωt – 110 o ) (a) V i = 200 cos(ωt- 5 o) v = 678.8 sin(ωt – 15 o ) v = 678.8 cos(ωt – 105 o ) 1

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2.10 b) S = V I * = 480 /-105 o 141.5 /5 o = 67872 /-100 o = -11,786 – j 66888 VA P = 11786 delivered (b) c) Q = 66,888 Vars delivered (c) d) PF = 0.174 lagging power factor load (see convention on Page 47) 2.12 V = 480 /45 o V I R = (480/10) /45 o = 48 /45 o A i R = (48) (2) 1/2 cos (ωt + 45 o ) p R = (678.8)(2) 1/2 (48) cos 2 (ωt + 45 o ) = (46078) [(1/2) cos (0) + ½ cos (2ωt + 90 o ) = 23039 [ 1 + cos (2ωt + 90 o ] (a) I C = ( 480 /45 o ) / (25 /-90 o ) i C = (19.2) (2) 1/2 cos (ωt + 135 o ) p C = [(480)(2) 1/2 (2) 1/2 (19.2] cos (ωt + 45 o ) cos (ωt + 135 o ) = (18432/2) [cos (-90 o ) + cos (2ωt + 180 o )] =9216 cos (2ωt + 180 o ) ( b) P R = (480)(48) cos (45 o – 45 o ) = 23,040 watts (c) Q C = (480)(19.2) sin (45 o – 135 o ) = -9216 vars (d) I = 48 /45 o + 19.2 /135 o = 51.7 /66.8 o PF = cos(45 o – 66.8 o ) = 0.928 Leading (e) 2 v(t) = 678.8 cos (ωt + 45
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## This note was uploaded on 02/21/2010 for the course EEE ??? taught by Professor Farmer during the Spring '10 term at ASU.

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EEE471hw1sol - EEE471/591 GSO 4 Edition Chapter 2 Solutions...

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