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qr code vb.net source Solutions to Problems in Visual Studio .NET
Solutions to Problems Decoding ANSI/AIM Code 128 In Visual Studio .NET Using Barcode Control SDK for .NET Control to generate, create, read, scan barcode image in Visual Studio .NET applications. Encoding USS Code 128 In VS .NET Using Barcode generator for Visual Studio .NET Control to generate, create Code 128C image in VS .NET applications. Thus, when the switch is closed: for RL 2 ohms; IL 8:33102=4:05715 2:05342 amperes; answer; for RL 3 ohms; IL 8:33102=5:05715 1:64738 amperes; answer; for RL 4 ohms; IL 8:33102=6:05715 1:37540 amperes; answer: Without the use of Thevenin s theorem it would be necessary to rework problem 53 for each new value of RL . 56. The rst step is to remove RL and nd the opencircuit voltage between terminals a and b. One way to do this is to write the two loopvoltage equations (see gure below), thus 20I1 12I2 100 12I1 24I2 0 from which I2 25=7 amperes, and hence the opencircuit voltage at a, b is equal to 9 25=7 32:143 volts. Code 128 Code Set B Decoder In Visual Studio .NET Using Barcode recognizer for Visual Studio .NET Control to read, scan read, scan image in .NET framework applications. Generating Barcode In .NET Using Barcode generator for .NET framework Control to generate, create bar code image in VS .NET applications. The next step is to nd the resistance looking back into terminals a, b (keeping RL disconnected, as before). The 8ohm and 12ohm resistances are in parallel when looking back from terminals a, b and hence combine together (product over sum, eq. (34), Chap. 2) to give 12 8 =20 4:8 ohms. The 4.8 ohms is now in series with the 3ohm resistance, giving us a total of 7.8 ohms in parallel with the 9ohm resistance. Therefore, looking into a, b we see 7.8 ohms in parallel with 9 ohms, which, using the product of the two, over the sum, is equal to 7:8 9 =16:8 4:179 ohms, approximately. Thus the equivalent Thevenin generator for Fig. 59 is therefore as shown in the following gure, answer. Reading Bar Code In .NET Using Barcode recognizer for .NET Control to read, scan read, scan image in VS .NET applications. Code 128C Printer In C#.NET Using Barcode creator for .NET Control to generate, create Code 128A image in Visual Studio .NET applications. Solutions to Problems
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Scanning UPC Code In .NET Using Barcode recognizer for VS .NET Control to read, scan read, scan image in .NET framework applications. Paint Bar Code In Java Using Barcode creation for BIRT Control to generate, create bar code image in BIRT reports applications. We must next nd the value ofthe conductance component Gg ofthe Norton equivalentgenerator. To do this, we now remove the shortcircuit from the terminals a, b in the pre ceding gure; Gg now equals the conductance seen looking to the left into the now opencircuited terminals a, b. From inspection of Fig. 59 (disregard the 9ohm branch for the moment), note that we look into one branch consisting of 3 ohms in series with the parallel combination of 8 ohms and 12 ohms, that is, into 3 8 12 =20 7:8 ohms. Since this resistance is in parallel with the 9ohm resistance, the total resistance seen looking into the opencircuited terminals a, b is Rg 9 7:8 =16:8 4:178571 ohms Recognizing Data Matrix 2d Barcode In Java Using Barcode recognizer for Java Control to read, scan read, scan image in Java applications. Print ANSI/AIM Code 39 In None Using Barcode creation for Font Control to generate, create Code 39 Full ASCII image in Font applications. Solutions to Problems
Printing Barcode In Java Using Barcode generator for Java Control to generate, create barcode image in Java applications. GTIN  13 Creator In None Using Barcode maker for Excel Control to generate, create EAN13 image in Excel applications. thus, Gg 1=Rg 0:239316 mhos; approx: and hence the answer is that the Norton equivalent generator for Fig. 59 is The Thevenin equivalent of Fig. 59 appears in the solution to problem 56, and the Norton equivalent appears above. Setting RL 10 ohms in the Thevenin equivalent generator and applying Ohm s law, and setting GL 1=10 0:1 mho in eq. (67), we have that Thevenin case: IL 32:143=14:179 2:267 amperes; approx: Norton case: IL 0:7692308=0:339316 2:267 amperes; approx: there being some di erence beyond the third decimal place because the same degree of accuracy was not used in calculating in problem 56 as in problem 58. 59. Let a, b be the output terminals of any given network; to convert the network into the Thevenin equivalent generator we must take the following two steps. (a) (b) Find the opencircuit voltage at the terminals a, b; this is the generated voltage, Vg , of the equivalent generator (Fig. 57). Find the resistance looking into the opencircuited terminals a, b; this is the internal resistance, Rg , of the equivalent generator (Fig. 57). Here we use the basic Ohm s law formula V RI 1=G I, thus V (b) Isc 6:155 56:468 volts Vg Gg 0:109 Now apply the two steps to Fig. 60 (keeping the switch open), as follows. (a) We must now replace all generators with their internal resistances. In this case we must remember that a constantcurrent generator has INFINITELY GREAT internal resistance. Thus, as far as resistance is concerned, the constantcurrent generator has no shunting e ect on Gg , and thus 1=Gg 1=0:109 9:174 ohms Rg hence the Thevenin equivalent generator is

