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qr code vb.net source Solutions to Problems in VS .NET
Solutions to Problems Code 128 Code Set C Decoder In .NET Framework Using Barcode Control SDK for Visual Studio .NET Control to generate, create, read, scan barcode image in Visual Studio .NET applications. Code 128 Code Set C Drawer In Visual Studio .NET Using Barcode generator for .NET Control to generate, create Code 128A image in Visual Studio .NET applications. 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 Reading Code 128 In Visual Studio .NET Using Barcode scanner for Visual Studio .NET Control to read, scan read, scan image in VS .NET applications. Bar Code Printer In Visual Studio .NET Using Barcode creator for Visual Studio .NET Control to generate, create barcode image in Visual Studio .NET applications. Then
Bar Code Decoder In VS .NET Using Barcode scanner for .NET framework Control to read, scan read, scan image in .NET applications. Print Code 128B In C# Using Barcode maker for Visual Studio .NET Control to generate, create ANSI/AIM Code 128 image in VS .NET applications. Hence
Generate Code 128 In .NET Using Barcode maker for ASP.NET Control to generate, create Code128 image in ASP.NET applications. Printing Code 128B In VB.NET Using Barcode creation for .NET framework Control to generate, create Code 128 Code Set C image in VS .NET applications. 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 Draw GS1 128 In .NET Framework Using Barcode creation for VS .NET Control to generate, create UCC  12 image in .NET framework applications. GS1 DataBar Expanded Creator In Visual Studio .NET Using Barcode printer for Visual Studio .NET Control to generate, create GS1 DataBar Expanded image in VS .NET applications. 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 threephase 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, Make Matrix 2D Barcode In .NET Framework Using Barcode maker for .NET Control to generate, create Matrix 2D Barcode image in Visual Studio .NET applications. ANSI/AIM ITF 25 Encoder In Visual Studio .NET Using Barcode encoder for .NET Control to generate, create ITF image in Visual Studio .NET applications. Solutions to Problems
Paint Bar Code In Java Using Barcode encoder for BIRT reports Control to generate, create barcode image in BIRT applications. Barcode Drawer In None Using Barcode printer for Font Control to generate, create bar code image in Font applications. 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* Code 3/9 Generation In VB.NET Using Barcode maker for VS .NET Control to generate, create USS Code 39 image in .NET applications. Code 3 Of 9 Recognizer In VS .NET Using Barcode reader for VS .NET Control to read, scan read, scan image in VS .NET applications. 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 Code 128 Encoder In None Using Barcode drawer for Excel Control to generate, create ANSI/AIM Code 128 image in Office Excel applications. Data Matrix Creation In VB.NET Using Barcode generation for .NET framework Control to generate, create Data Matrix ECC200 image in VS .NET applications. which is exactly the basic form of eq. (446), showing that the SUM OF TWO POSITIVESEQUENCE 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 positivesequence sets (or the sum of two negativesequence 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 Generate Barcode In Visual Studio .NET Using Barcode encoder for ASP.NET Control to generate, create bar code image in ASP.NET applications. GS1 DataBar Limited Drawer In Java Using Barcode encoder for Java Control to generate, create GS1 DataBar Truncated image in Java applications. 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.

