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" " " " I1B B1 =12 I1C C1 =12 " " " " I2B B2 =12 I2C C2 =12
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and thus, by the principle of superposition, we have that the three line currents in Fig. 266 are equal to " " " " " IA I1A I2A A1 A2 =12 " " " " " IB I1B I2B B1 B2 =12 " " " " " IC I1C I2C C1 C2 =12 Now substitute, into the above three equations, the values of the positivesequence and negative-sequence voltages previously found. Doing this, the above three equations become " IA 102 18 j120 24 j240 =12 " IB 18 96 j120 30 j240 =12 " IC 24 30 j120 90 j240 =12 Recall now that the value of an indicated sum of a number of complex numbers can be found only if the numbers are expressed in rectangular form, because sum of complex numbers sum of real parts j sum of imaginary parts and thus, upon converting the above exponential forms into their equivalent rectangular forms,* then adding, you should nd that the above three equations become " " IA 6:750 j0:433 ; hence jIA j 6:764 amperes; answer: " " IB 3:750 j4:763 ; hence jIB j 6:062 amperes; answer: " " IC 3:000 j4:330 ; hence jIC j 5:268 amperes; answer: " " " " " " 212. In Fig. 266, A Vna 90=08, B Vnb 72=1208, C Vnc 54=2408, where " " " A V na is the reference vector. Then, using the same basic procedure as in section 10.8, we have (angles in degrees) " " " Vab B A 72=120 90 " " " Vbc C B 54=240 72=120 " " " Vca A C 90 54=240 In order to nd the values of the sums indicated above, the polar forms must rst be converted to rectangular forms; to do this, note that 1=120 cos 120 j sin 120 0:5 j0:8660 1=240 cos 240 j sin 240 0:5 j0:8660
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* Euler s formula, A 2 j A cos  j sin  .
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thus " Vab j126 j62:352j 140:58 volts; answer: " Vbc j 9:000 j109:12j 109:49 volts; answer: " Vca j 117 j46:76j 126:00 volts; answer: 213. The TOTAL power P is the sum of the powers produced in each of the 12-ohm loads. Hence, using the values of current found in problem 211, we have that P 12 6:764 2 6:062 2 5:268 2 1323 watts approx:; answer: 214. The given values of the generator voltages, and the values of the generator currents " " " IA , IB , and IC , were found in problem 211. Using these values in eq. (467) gives the following answers. PA r:p: 90 6:750 j0:433 607:5 watts PB r:p: 72 0:5 j0:8660 3:750 j4:763 72 1:875 4:125 432 watts PC r:p: 54 0:5 j0:8660 3:000 j4:330 54 1:500 3:750 283:5 watts and thus the total power output of the generator is PT PA PB PC 1323 W; as found in problem 213: 215. First, a comparison of the notation used in eqs. (460) and (463) with the notation " " " " " " used in Fig. 267 shows that A IA (thus A1 IA1 and A2 IA2 ), and also that " " " " B IB and C IC : thus, by eqs. (460) and (463), we have that " " " " "" IA1 1 IA a2 IB aIC 3 and " " "" " " IA2 1 IA aIB a2 IC 3 Now substitute these values into eqs. (468) through (470); upon doing this, and " " " noting that a3 1, a4 a, and also that (" a2 1, you should arrive at the a " following three simultaneous equations " " " " " " 2IA IB IC 3YA VN 3YA V " " " " " IA 2IB IC 3YB VN 3"YB V a" " " " " " IA IB 2IC 3YC VN 3"2 YC V a " " " " " Note that we have four unknowns (IA ; IB ; IC ; VN but only three equations; this " " " situation can be overcome by making the substitution IC IA IB , and upon doing this you should nd that the above three equations can be written in terms " " " of just three unknowns, IA , IB , VN , as follows: " " " " " 3IA 0IB 3YA VN 3YA V " " " " 0IA 3IB 3YB VN 3"YB V a" " " " " 3IA 3IB 3YC VN 3"2 YC V a "
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