chap12

# chap12 - Chapter 12 Solution 1(a If Vab = 400 then Van =...

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Chapter 12, Solution 1. (a) If , then 400 ab = V = ° = 30 - 3 400 an V V 30 - 231 ° = bn V V 150 - 231 ° = cn V V 270 - 231 ° (b) For the acb sequence, ° ° = = 120 V 0 V p p bn an ab V V V ° = + = 30 - 3 V 2 3 j 2 1 1 V p p ab V i.e. in the acb sequence, lags by 30 ° . ab V an V Hence, if , then 400 ab = V = ° = 30 3 400 an V V 30 231 ° = bn V V 150 231 ° = cn V V 90 - 231 ° Chapter 12, Solution 2. Since phase c lags phase a by 120 ° , this is an acb sequence . = ° + ° = ) 120 (30 160 bn V V 150 160 ° Chapter 12, Solution 3. Since V leads by 120 ° , this is an bn cn V abc sequence . = ° + ° = ) 120 (130 208 an V V 250 208 °

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Chapter 12, Solution 4. = ° = 120 ca bc V V V 140 208 ° = ° = 120 bc ab V V V 260 208 ° = ° ° = ° = 30 3 260 208 30 3 ab an V V V 230 120 ° = ° = 120 - an bn V V V 110 120 ° Chapter 12, Solution 5. This is an abc phase sequence. ° = 30 3 an ab V V or = ° ° = ° = 30 3 0 420 30 3 ab an V V V 30 - 5 . 242 ° = ° = 120 - an bn V V V 150 - 5 . 242 ° = ° = 120 an cn V V V 90 5 . 242 ° Chapter 12, Solution 6. ° = + = 26.56 18 . 11 5 j 10 Y Z The line currents are = ° ° = = 26.56 18 . 11 0 220 Y an a Z V I A 26.56 - 68 . 19 ° = ° = 120 - a b I I A 146.56 - 68 . 19 ° = ° = 120 a c I I A 93.44 68 . 19 °
The line voltages are = ° = 30 3 200 ab V V 30 381 ° = bc V V 90 - 381 ° = ca V V 210 - 381 ° The load voltages are = = = an Y a AN V Z I V V 0 220 ° = = bn BN V V V 120 - 220 ° = = cn CN V V V 120 220 ° Chapter 12, Solution 7. This is a balanced Y-Y system. + 440 0 ° V Z Y = 6 j8 Using the per-phase circuit shown above, = ° = 8 j 6 0 440 a I A 53.13 44 ° = ° = 120 - a b I I A 66.87 - 44 ° = ° = 120 a c I I A 13 . 73 1 44 ° Chapter 12, Solution 8. , V 220 V L = + = 9 j 16 Y Z ° = + = = = 29.36 - 918 . 6 ) 9 j 16 ( 3 220 3 V V Y L Y p an Z Z I = L I A 918 . 6

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Chapter 12, Solution 9. = + ° = + = 15 j 20 0 120 Y L an a Z Z V I A 36.87 - 8 . 4 ° = ° = 120 - a b I I A 156.87 - 8 . 4 ° = ° = 120 a c I I A 83.13 8 . 4 ° As a balanced system, = n I A 0 Chapter 12, Solution 10. Since the neutral line is present, we can solve this problem on a per-phase basis. For phase a, ° = ° = + = 36.53 55 . 6 20 j 27 0 220 2 A an a Z V I For phase b, ° = ° = + = 120 - 10 22 120 - 220 2 B bn b Z V I For phase c, ° = + ° = + = 97.38 92 . 16 5 j 12 120 220 2 C cn c Z V I The current in the neutral line is ) -( c b a n I I I I + + = or c b a n - I I I I + + = ) 78 . 16 j 173 . 2 - ( ) 66 . 8 j 5 - ( ) 9 . 3 j 263 . 5 ( - n + + + + = I = = 02 . 12 j 91 . 1 n I A 81 - 17 . 12 °
Chapter 12, Solution 11. ° ° = ° = ° = 90 - 3 10 220 90 - 3 90 - 3 BC bc an V V V = an V V 100 127 ° = ° = 120 BC AB V V V 130 220 ° V 110 - 220 120 - BC AC ° = ° = V V If , then ° = 60 30 bB I ° = 180 30 aA I , ° = 60 - 30 cC I ° = ° ° = ° = 210 32 . 17 30 - 3 180 30 30 - 3 aA AB I I ° = 90 32 . 17 BC I , ° = 30 - 32 . 17 CA I = = CA AC - I I A 150 32 . 17 ° BC BC V Z I = = ° ° = = 90 32 . 17 0 220 BC BC I V Z ° 80 - 7 . 12 Chapter 12, Solution 12. Convert the delta-load to a wye-load and apply per-phase analysis.

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