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In a COMPLETELY BALANCED three-phase system the three line voltages have equal magnitudes, displaced from each other by 1208, and the three load impedances have equal " values of Z ohms. If these conditions do not exist, the three-phase system is said to be UNBALANCED. Thus, any one of the following three conditions of unbalance may be encountered. 1. 2. 3. The line voltages are unequal, either in magnitudes, or phase angles, or both. The three load impedances are not all equal. A combination of (1) and (2) is present.

The updateButtons( ) method updates the state of all the buttons on the button panel based on the state of the selected download. The updateButtons( ) method is shown here:

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To solve such problems by the method of symmetrical components, the basic procedure is to rst resolve the unbalanced system into the sum of three balanced systems, then separately nd the solution to each of the balanced systems. The nal answer is then the SUM OF THE SEPARATE SOLUTIONS, in accordance with the principle of superposition (as in problem 50, Chap. 4, for example). (This assumes that, for practical purposes, the circuit is a linear system; see footnote in section 8.6.) The following problems, in which Y-connected generators will feed Y-connected loads, will serve to illustrate the basic procedures. Such problems can be worked either in terms of line voltage or phase voltage, but in the problems here we ve elected to specify phase voltages instead of line voltages (the relationships between the two were developed in section 10.8 and are summarized in Fig. 253, which shows the relation between phase and line voltages in a balanced Y-connected generator). Problem 210 " In Fig. 265, a zero-sequence voltage of V0 volts is applied, through a three-wire line, to a Y-connected load as shown. Explain, from inspection of the gure, why no zero-sequence current can ow in any such three-wire Y-connected load.

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.

/* Update each button's state based on the currently selected download's status. */ private void updateButtons() {

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.

if (selectedDownload != null) { int status = selectedDownload.getStatus(); switch (status) { case Download.DOWNLOADING: pauseButton.setEnabled(true); resumeButton.setEnabled(false); cancelButton.setEnabled(true); clearButton.setEnabled(false); break; case Download.PAUSED: pauseButton.setEnabled(false); resumeButton.setEnabled(true); cancelButton.setEnabled(true); clearButton.setEnabled(false); break; case Download.ERROR: pauseButton.setEnabled(false); resumeButton.setEnabled(true); cancelButton.setEnabled(false); clearButton.setEnabled(true); break; default: // COMPLETE or CANCELLED pauseButton.setEnabled(false); resumeButton.setEnabled(false); cancelButton.setEnabled(false); clearButton.setEnabled(true); } } else { // No download is selected in table. pauseButton.setEnabled(false); resumeButton.setEnabled(false); cancelButton.setEnabled(false); clearButton.setEnabled(false); } }

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 *

" " 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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