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204. One way is to make use of eq. (440), as follows. First, by eq. (435) VL 1:732 330 571:56 volts Next, p " IL Ip Vp =jZ j 330= 306 18:865 amperes p " cos  R=jZ j 15= 306 0:8575 PT 1:732VL IL cos  16014:03 watts; answer; or; 16:014 kilowatts; answer: " 205. First consider the phase voltage Vna ; note that the phase current here is the same as " 0 as shown in the gure below. the line current Iaa
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Since the system is given to be completely balanced, the voltage at junction point n 0 is the same as the voltage at point n (a wire connected from n to n 0 would show zero current). Hence, by Ohm s law, " Vna Vna " 0:0098 5 j3 Vna Iaa 0 " 15 j9 Z " thus  arctan 3=5 30:968, showing that line current Iaa 0 LAGS phase vol"na by approximately 318. Hence, as inspection of Fig. 253 shows, line current tage V " " Iaa 0 lags line voltage VAB by approximately 618. Likewise, the same procedure will " " " " show that Ibb 0 and Icc 0 , respectively, will lag line voltages VBC and VCA by 618. 206. Remember that V 0 and I 0 in eq. (445) are maximum (peak) values. Then note that eq. (445) can be written as p 3 0 0 V0 I0 V I cos  3 p p cos  3VI cos  2 2 2
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where V and I are now rms values (rms values are used in AVERAGE POWER calculations). Next, inspection of Fig. 257 will show that V VL , and also that p " I IL = 3, where V and I are the rms values for each of the three impedances Z ; thus p IL p 3VL p cos  3VL IL cos  3 showing that in a balanced three-phase system the instantaneous power p is the SAME as the total average power PT . 207. To express the SUM of any two sets we rst label the sets in A, B, and C notation. We then go in the ccw sense, in both sets, rst nding the sum of the two A vectors,
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then the sum of the two B vectors, then the sum of the two C vectors, the SUM of the two sets then being expressed in the form of eq. (449). In the particular case here, it s given that both sets are to be labeled in the sequence ABC ; hence, after applying eq. (446), eq. (449) becomes*
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0 0 0 " " S1 S2 A1 A1 A1 A1  j120 A1 A1  j240 0 00 or, letting A1 A1 A1 , the above becomes 00 00 00 " " S1 S2 A1 A1  j120 A1  j240
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which is exactly the basic form of eq. (446), showing that the SUM OF TWO POSITIVE-SEQUENCE SETS is a BALANCED set of three vectors. Thus an unbalanced set of three vectors (Fig. 261) cannot be expressed as the sum of two positive-sequence sets (or the sum of two negative-sequence sets). 208. The same general discussion, given at the start of the solution to problem 207 above, applies here also. Then, referring to Figs. 262 and 263, we have, in this case BY FIG: 262: " " B1 A1  j120 " " C1 A1  j240 BY FIG: 263: " " B2 A2  j240 " " C2 A2  j120 A
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and, upon substituting these values into eq. (449), you can show that " " " " " " " " S1 S2 A1 A2 A1 A2 j 120  j120 A1 A2  j120  j240 " " " " However, since (A1 A2 does not, in general, equal either A1 A2  j120 or " "1 A2  j120 , it follows that eq. (A) is not of the general form of eq. (446) and (A thus must represent an unbalanced set of three vectors. 209. We wish to express the given unbalanced set as the sum of positive, negative, and zero sequence sets. To do this we ll make use of eqs. (460) through (466), where, in this case (angles in degrees), it s given that " A 15=0 15 j0 " B 9=100 9 j100 " C 24=215 24 j215
Let us begin by substituting the above values into eq. (460), which gives us " A1 5 3 j340 8 j335 Now, in order to nd the value of the indicated sum of the above complex numbers, they must rst be put into the rectangular form (a jb), which is done by applying Euler s formula  j cos  j sin . Upon doing this, you should nd that " A1 15:070 j4:407 " showing that the vector A1 lies in the fourth quadrant, thus (continuing to use degrees) " A1 15:701= 16:301 or, in exponential form, " A1 15:701 j16:301
* In the solution here we ll omit the overscore (vector) notation on the As because of the presence of so many prime marks that will be used with them.