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ch10_ssm - 10 Sinusoidal Steady State Power Calculations...

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10 Sinusoidal Steady State Power Calculations Assessment Problems AP 10.1 [a] V = 100/ 45 V , I = 20/15 A Therefore P = 1 2 (100)(20) cos[ 45 (15)] = 500 W , A B Q = 1000 sin 60 = 866 . 03 VAR , B A [b] V = 100/ 45 , I = 20/165 P = 1000 cos( 210 ) = 866 . 03 W , B A Q = 1000 sin( 210 ) = 500 VAR , A B [c] V = 100/ 45 , I = 20/ 105 P = 1000 cos(60 ) = 500 W , A B Q = 1000 sin(60 ) = 866 . 03 VAR , A B [d] V = 100/0 , I = 20/120 P = 1000 cos( 120 ) = 500 W , B A Q = 1000 sin( 120 ) = 866 . 03 VAR , B A 10–1

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10–2 CHAPTER 10. Sinusoidal Steady State Power Calculations AP 10.2 pf = cos( θ v θ i ) = cos[15 (75)] = cos( 60 ) = 0 . 5 leading rf = sin( θ v θ i ) = sin( 60 ) = 0 . 866 AP 10.3 From Ex. 9.4 I eff = I ρ 3 = 0 . 18 3 A P = I 2 eff R = 0 . 0324 3 (5000) = 54 W AP 10.4 [a] Z = (39 + j 26) ( j 52) = 48 j 20 = 52/ 22 . 62 Therefore I = 250/0 48 j 20 + 1 + j 4 = 4 . 85/18 . 08 A ( rms ) V L = Z I = (52/ 22 . 62 )(4 . 85/18 . 08 ) = 252 . 20/ 4 . 54 V ( rms ) I L = V L 39 + j 26 = 5 . 38/ 38 . 23 A ( rms ) [b] S L = V L I L = (252 . 20/ 4 . 54 )(5 . 38/+ 38 . 23 ) = 1357/33 . 69 = (1129 . 09 + j 752 . 73) VA P L = 1129 . 09 W ; Q L = 752 . 73 VAR [c] P = | I | 2 1 = (4 . 85) 2 · 1 = 23 . 52 W ; Q = | I | 2 4 = 94 . 09 VAR [d] S g ( delivering ) = 250 I = (1152 . 62 j 376 . 36) VA Therefore the source is delivering 1152 . 62 W and absorbing 376.36 magnetizing VAR. [e] Q cap = | V L | 2 52 = (252 . 20) 2 52 = 1223 . 18 VAR Therefore the capacitor is delivering 1223.18 magnetizing VAR. Check: 94 . 09 + 752 . 73 + 376 . 36 = 1223 . 18 VAR and 1129 . 09 + 23 . 52 = 1152 . 62 W
Problems 10–3 AP 10.5 Series circuit derivation: S = 250 I = (40 , 000 j 30 , 000) Therefore I = 160 j 120 = 200/ 36 . 87 A ( rms ) I = 200/36 . 87 A ( rms ) Z = V I = 250 200/36 . 87 = 1 . 25/ 36 . 87 = (1 j 0 . 75) Ω Therefore R = 1 Ω , X C = 0 . 75 Ω Parallel circuit derivation: P = (250) 2 R ; therefore R = (250) 2 40 , 000 = 1 . 5625 Ω Q = (250) 2 X C ; therefore X C = (250) 2 30 , 000 = 2 . 083 Ω AP 10.6 S 1 = 15 , 000(0 . 6) + j 15 , 000(0 . 8) = 9000 + j 12 , 000 VA S 2 = 6000(0 . 8) + j 6000(0 . 6) = 4800 j 3600 VA S T = S 1 + S 2 = 13 , 800 + j 8400 VA S T = 200 I ; therefore I = 69 + j 42 I = 69 j 42 A V s = 200 + j I = 200 + j 69 + 42 = 242 + j 69 = 251 . 64/15 . 91 V ( rms ) AP 10.7

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ch10_ssm - 10 Sinusoidal Steady State Power Calculations...

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