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CHAPTER 10 Magnetic Coupling. Transformers
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Problem 211 An unbalanced Y-connected three-phase generator is connected, by means of a three-wire line, to a balanced Y-connected resistive load of 12 ohms per phase. It is given that the generator phase voltages are, in volts, equal to " " " A 90=08 B 72=1208 C 54=2408 " " " " in the same ccw sense A to B to C as in Fig. 261, with A the reference voltage. This is shown in schematic diagram form in Fig. 266.
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Fig. 266. The given three-phase system.
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Using the method of symmetrical components, show that the magnitudes of the line currents, in amperes, are equal to " jIA j 6:764 answer " " jIB j 6:062 answer jIC j 5:268 answer
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Problem 212 " " " Let us, in problem 211 (Fig. 266), denote the line voltages by Vab , Vbc , and Vca . Show " " "ab j 140:58 jVbc j 109:49 jVca j 126:00: that, in volts, jV Problem 213 One formula for calculating average power is P RI 2 , where I is magnitude of rms current. Using this formula, nd the total power P produced by the generator in problem 211.
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CHAPTER 10 Magnetic Coupling. Transformers
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Problem 214 It can be shown that the true average sinusoidal power output P of a generator is " equal to the REAL PART (r.p.) of the PRODUCT of the generator voltage V and ", that is the CONJUGATE of the generator current I "" P r:p: V I 467 *
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" " where I ( double overscore ) denotes the conjugate of I . Verify that the use of eq. (467) gives the same answer as found in problem 213. The preceding problems dealt with Fig. 266, which is a case in which an UNBALANCED GENERATOR feeds a BALANCED LOAD. Now let us consider the opposite case, in which a BALANCED GENERATOR feeds an UNBALANCED " LOAD, as illustrated in Fig. 267, where voltage V is the reference phase voltage with "N is the voltage at the load junction point, also reference to the junction point O. Also, V with respect to the point O.
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Fig. 267
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" Keep in mind that VN denotes the voltage drop from the junction point in the load to the junction point at O. Therefore, since the generator voltage is equal to the sum of the voltage drops in any closed path, we have the following three voltage equations in Fig. 267: " " " V Z A IA V N " " " " aV ZB IB VN " " " " a2 V ZC IC VN Here we re dealing with the general case of unequal load impedances; thus the three line (and phase) currents will, in general, be unequal, and hence can be resolved into the sum of positive and negative sequences; thus " " " IA IA1 IA2 " " " IB IB1 IB2 " " " IC IC1 IC2
* See note 29 in Appendix.
CHAPTER 10 Magnetic Coupling. Transformers
which, upon making use of eqs. (461), (462), (464), and (465), can also be written as " " " IA IA1 IA2 " "" " " IB aIA1 a2 IA2 " " " "" IC a2 IA1 aIA2 " " " and thus, upon substituting these values of IA , IB , and IC into the rst set of equations following Fig. 267, you should nd that " " " " " ZA IA1 ZA IA2 VN V " "" " " " " " aZB IA1 a2 ZB IA2 VN aV " "" " " " " " a2 ZC IA1 aZC IA2 VN a2 V 468 469 470
Problem 215 Here you are asked to complete the foregoing discussion concerning Fig. 267 as follows. Making use of eqs. (460) and (463), and the fact that, by Kirchho s current " " " law, IA IB IC 0, show that the values of the line currents in Fig. 267 are given by the equations " " " " 1:732 YA YC  j30 YA YB  j30 V " IA " " "A YB YC Y " " " " 1:732 YB YC  j90 YA YB  j30 V " IB " " " YA YB YC " " " " 1:732 YA YC  j30 YB YC  j90 V " IC " " " YA YB YC where, in terms of the admittances " " YA 1=ZA " " YB 1=ZB " " YC 1=ZC 471 472 473
in which the reciprocal ohms are called mhos. Problem 216 " " " In Fig. 267 let V 125 volts, ZA 3 j4 ohms, ZB 8 ohms, and ZC 5 ohms. Using symmetrical components, verify that the magnitudes of the line currents, in amperes, are equal to " jIA j 25:99 " jIB j 22:85 " jIC j 17:37
Problem 217 In three-phase work we often deal in terms of line voltage instead of phase voltage. Thus, in Fig. 267, suppose that the reference voltage is taken to be the " line voltage, VAB VAB =08 VAB , this being the voltage from the center wire to the top wire in the diagram. What changes would be required to express eqs. (471) through (473) in terms of line voltage instead of phase voltage
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