ch10_ism - 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 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 10–1
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10–2 CHAPTER 10. Sinusoidal Steady State Power Calculations AP 10.3 From Ex. 9.4 I ef = I ρ 3 = 0 . 18 3 A P = I 2 ef R = ± 0 . 0324 3 ² (5000) = 54 W AP 10.4 [a] Z = (39 + j 26) k ( 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 . . 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 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 =1Ω ,X C = 0 . 75 Ω
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Problems 10–3 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 [a] The phasor domain equivalent circuit and the Thévenin equivalent are shown below: Phasor domain equivalent circuit: Thévenin equivalent: V Th =3 j 800 20 j 40 =48 j 24=53 . 67/ 26 . 57 V Z Th =4+ j 18 + j 800 20 j 40 =20+ j 10=22 . 36/26 . 57 For maximum power transfer, Z L = (20 j 10) Ω
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10–4 CHAPTER 10. Sinusoidal Steady State Power Calculations [b] I = 53 . 67/ 26 . 57 40 =1 . 34/ 26 . 57 A Therefore P = 1 . 34 2 ! 2 20=18 W [c] R L = | Z Th | =22 . 36 Ω [d] I = 53 . 67/ 26 . 57 42 . 36 + j 10 . 23/ 39 . 85 A Therefore P = 1 . 23 2 ! 2 (22 . 36)=17 W AP 10.8 Mesh current equations: 660 = (34 + j 50) I 1 + j 100( I 1 I 2 )+ j 40 I 1 + j 40( I 1 I 2 ) 0= j 100( I 2 I 1 ) j 40 I 1 + 100 I 2 Solving, I 1 =3 . 536/ 45 A , I 2 . 5/0 A ; · .. P = 1 2 (3 . 5) 2 (100) = 612 . 50 W AP 10.9 [a] 248 = j 400 I 1 j 500 I 2 + 375( I 1 I 2 ) 0 = 375( I 2 I 1 j 1000 I 2 j 500 I 1 + 400 I 2 Solving, I 1 =0 . 80 j 0 . 62 A ; I 2 . 4 j 0 . 3=0 . 5/ 36 . 87 A · = 1 2 (0 . 25)(400) = 50 W
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Problems 10–5 [b] I 1 I 2 =0 . 4 j 0 . 32 A P 375 = 1 2 | I 1 I 2 | 2 (375) = 49 . 20 W [c] P g = 1 2 (248)(0 .
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This note was uploaded on 03/26/2008 for the course ELEN 214 taught by Professor Zou during the Spring '08 term at Texas A&M.

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ch10_ism - 10 Sinusoidal Steady State Power Calculations...

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