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Else MessageBox.Show("There was an error") End If C# // // // // // // if { Call the Save() method If objReport.Save() = True Then MessageBox.Show("Bug reported!") Else MessageBox.Show("There was an error!") End If (objReport.Save() == true) MessageBox.Show("Bug recorded"); txtDescription.Text = ""; } else { MessageBox.Show("There was an error"); } 13. Add code that calls the FormatData() method just before you call the Save() method: VB .NET 'Call the FormatData() method objReport.FormatData() C# //Call the FormatData() method objReport.FormatData(); 14. Run the program and verify that it still saves the bug report data when the BugReport.csv file is not in Read Only mode and gives an error when it is. 15. Once you have verified that that program is working as it was before you moved the code, go back and remove all of the code you commented out. In this exercise, you saw how to create a new class that included fields, properties, a constructor, and a method. You also saw how to create an object from your class, set the properties, and call the methods. While you have not fundamentally changed the way the program behaves, you have made it easier to reuse the code in the ReportToFile class in another project. For example, you could create a web application and use the ReportToFile class in that project without modifications. You will learn more about web applications in 9, but if you would like to see an example, you will find the completed version of the files in the 6 DemoCode\Exercise6-4WebSite folder (see Appendix A for download instructions).

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Figure 1711 This gure is a cross section in which the wire connections between crosses and dots have been cut away In Figure 1711, the dashed line indicates the axis of the stator ux according to the right-hand rule, indicating that the slotted stator in effect behaves like a pole pair The north and south poles indicated in the gure are a consequence of the fact that the ux exits the bottom part of the structure (thus, the north pole indicated in the gure) and enters the top half of the structure (thus, the south pole) In particular, if you consider that the windings are arranged so that the current entering the right-hand side of the stator (to the right of the dashed line) ows through the back end of the stator and then ows outward from the left-hand side of the stator slots (left of the dashed line), you can visualize the windings in the slots as behaving in a manner similar to the coils of Figure 1710, where the ux axis of Figure 1711 corresponds to the ux axis of each of the coils of Figure 1710 The actual circuit that permits current ow is completed by the front and back ends of the stator, where the wires are connected according to the pattern a-a , b-b , c-c , as depicted in the gure Another important consideration that facilitates understanding the operation of electric machines pertains to the use of AC currents It should be apparent by now that if the current owing into the slotted stator is alternating, the direction of the ux will also alternate, so that in effect the two poles will reverse polarity every time the current reverses direction, that is, every half-cycle of the sinusoidal current Further since the magnetic ux is approximately proportional to the current in the coil as the amplitude of the current oscillates in a sinusoidal fashion, so will the ux density in the structure Thus, the magnetic eld developed in the stator changes both spatially and in time This property is typical of AC machines, where a rotating magnetic eld is established by energizing the coil with an alternating current As we shall see in the next section, the principles underlying the operation of DC and AC machines are quite different: in a direct-current machine, there is no rotating eld, but a mechanical switching arrangement (the commutator) makes it possible for the rotor and stator magnetic elds to always align at right angles to each other.

Mesh 1: KVL requires that vS v1 v2 = 0, where v1 = i1R1, v2 = (i1 i2)R1 R1 + v1 vS + _ i1 + v2 R2 i2 R4 R3

1, 2

3

Mesh 2: KVL requires that v2 + v3 + v4 = 0 where v2 = (i2 i1)R2, v3 = i2R3, v4 = i2R4 R1 R3 + v3 v2 i2 R4 + + v4

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