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Fig. 254
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In Fig. 254, A 0 , B 0 , and C 0 denote the three wires of the outgoing transmission line. Actually, in diagrams such as the above, in which operation is at 60 Hz, it s understood that it will be necessary to use iron-core transformers. Hence, in practical drawings the iron-core symbol is omitted, and the above Y transformer would be drawn as shown in Fig. 255.
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Fig. 255
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In Figs. 254 and 255, note that the full generator output (the line voltage) is applied to each of the three -connected primary coils. This voltage, after being stepped up by each individual transformer, then becomes the phase voltage on the Y-connected secondary side. Problem 202 Given that there is no external load on the -connected generator of Fig. 247, explain why there is no current ow around the closed-loop circuit formed by the three generators.
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CHAPTER 10 Magnetic Coupling. Transformers
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Problem 203 (a) In Fig. 248, nd the phase voltage if the line voltage is 3300 volts. (b) Suppose, in Fig. 254, that the transmission-line voltage is required to be 66,000 volts. If each of the three transformers has a turns ratio of 1-to-12 (primary turns to secondary turns), what value of generator phase voltage is required
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Current and Power in Balanced Three-Phase Loads
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Here we take up the case in which a balanced Y-connected generator feeds a BALANCED " three-phase load of Z ohms per phase, taking, rst, the case of a balanced Y-connected load. Let us begin by redrawing Fig. 248, now adding some additional notation, as shown in Fig. 256.
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Fig. 256
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It s apparent, by inspection, that the above can be a totally symmetrical, balanced system only if the generator phase voltages are equal to the corresponding voltage drops in the load; that is, only if " " Vna Vn 0 a 0 " " Vnb Vn 0 b 0 " " Vnc Vn 0 c 0
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" " " Also, in the gure, the three line currents are denoted by Icc 0 , Iaa 0 , and Ibb 0 as shown. Also note, from direct inspection of the gure, that the three line currents are actually equal to the phase currents (this is true only for a balanced Y-connected load). Since we re dealing with a balanced system, it follows that the line currents (and also the phase currents in this case) all have equal magnitudes; that is " " " jIcc 0 j jIaa 0 j jIbb 0 j IL Ip 437
Next, the POWER, P, produced in the above balanced Y-connected load can be found as follows. First, as before, let Vp be the equal magnitudes of the three phase voltages. Then, since Vp and Ip denote the magnitudes of the rms voltages and currents in each of the
CHAPTER 10 Magnetic Coupling. Transformers
three load impedances, it follows (from section 8.5) that the POWER Pp produced in each of the three impedances is equal to Pp Vp Ip cos  438
and thus the TOTAL POWER PT produced in all three impedances in Fig. 256 is equal to PT 3Vp Ip cos  439
where cos  is the same power factor of each of the three equal impedances. Or since, by eq. 435, p Vp VL = 3 and also since, Ip IL eq. (439) can also be written as PT p 3VL IL cos  440
which gives the total power produced in the balanced Y-connected system of Fig. 256 in terms of line voltage and line current. Problem 204 " In Fig. 256, suppose the generator phase voltage is 330 volts and Z 15 j9 ohms. Find the total power output of the generator. (Answer: 16,014.03 watts) Problem 205 In problem 204, show that the line currents lag the line voltages by approximately 618. Next, suppose the load in Fig. 256 were delta-connected instead of Y-connected. In such " a case the situation at the load-end of the line would be as shown in Fig. 257, where VAB , "BC , and VCA denote the three line voltages (as in Fig. 248). Also, let us denote the three V " " " line currents by IA , IB , and IC , as shown.
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