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VB .NET Dim x As Integer = 5 If x = 5 Then Console.WriteLine("it is true") Else Console.WriteLine("it is false") End If C# int x = 5; if (x==5) { Console.WriteLine("it is true"); } else { Console.WriteLine("it is false"); }

Analyze the thyristor-based variable voltage supply shown in Figure 1127 Determine: (1) the rms load voltage as a function of the ring angle and (2) the power supplied to the resistive load at zero ring angle and at ring angles equal to /2 and

2 0 Control circuitry iG(t) iAK(t) + vAK(t)

Of course, these estimates are just indicative and need to be adjusted to allow for other factors that impact productivity and are specific to your project and environment You should present any estimates as a range, or plus/minus 30 percent The obvious limitation of this equation-based technique for project estimation is that the equations cannot address the specific attributes of a particular project The equations model an average project derived from the ISBSG Repository It is very important to treat estimates obtained from the regression equations as ballpark figures only More accurate estimates will be obtained by using an estimation technique that considers the particular attributes of a planned project

Schematics, Diagrams, Circuits, and Given Data: VAK on = 0 V; RL = 240 pulsed gate current, iG (t), is timed as shown in Figure 1128 Assumptions: The thyristor acts as an ideal diode when on (VAK > 0)

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vS(t) 120 2

Analysis:

VB .NET Dim x As Integer = 5 If x = 5 Then Console.WriteLine("it is five") ElseIf x = 10 Then Console.WriteLine("it is 10") Else Console.WriteLine("it is neither") End If End If C# int x = 5; if (x==5) { Console.WriteLine("it is five"); } else if (x == 10) { Console.WriteLine("it is 10"); }

1 Load voltage calculation As explained in the preceding section, the load voltage will have the appearance shown in Figure 1129 The rms value of the load voltage as a function of the ring angle, , is therefore computed as follows: (120 2)2 2 VL ( ) = sin t d( t ) 2 (120 2) 1 (1 cos(2 t ))d( t ) = 2 1 (120 2) 1 + sin(2 ) = 2 2 2 Load power calculation We can now compute the load power for each of the three values of : PL = For = 0: V2 = RL V2 RL 120 2 2 240

= 30 W;

vL(t) 120 2

Summary

2 t

else { Console.WriteLine("it is neither"); }

11

PL = for = :

= 15 W

voltage This would not be the case if a resistive voltage divider were used to adjust the load voltage

The use of the regression equation estimation technique is useful to produce indicative or ballpark project estimates early in the life of a project

Diodes D1 and D2 form a full-wave recti er (see Figure 842); resistors R1 and R2 and thyristor T2 form a variable voltage divider Assume that thyristor T2 is not in the conducting state and that the anode voltage of D3 is such that D3 conducts Then T1 will be red near the beginning of the positive half-cycle of the AC source voltage, and its period of conduction will be long, providing a substantial current to the battery (resistors R4 and Rp are suf ciently large that most of the current owing through T1 will go to the battery) The potentiometer Rp is set so that when the battery voltage is low, the voltage VR is not suf cient to turn on the Zener diode, Z Thus, Z is effectively an open circuit, and T2 remains off (recall that we had initially assumed T2 to be off this con rms the correctness of the assumption) As the battery charges to a progressively higher value, Z

Select-Case Statement (VB .NET Only)

will eventually conduct; when Z conducts, a gate current is injected into T2 , which is then turned on When T2 conducts, the voltage across the R2 -T2 series connection becomes signi cantly lower, because T2 is now nearly a short circuit Resistors R1 and R2 are selected so that when T2 conducts, D3 becomes reverse-biased Once this condition occurs, T1 is turned off and charging stops You see that the circuit has built-in overcharging protection

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