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In practice, loads are not always resistive, as will be seen in s 16 through 18, where circuit models for electromechanical actuators and electric motors are introduced A very common occurrence consists of a DC voltage supply providing current to a DC motor For the purpose of the present discussion, it will suf ce to state that a DC motor presents an inductive impedance to the voltage supply and requires a constant current from the supply to operate at a constant speed The circuit of Figure 1117 illustrates, as an example, a simple half-wave recti er connected to an RL load The circuit on top in Figure 1117, assuming an ideal diode, would present a serious problem during the negative half-cycle of the source voltage, since the requirement for continuity of current in the inductor (recall the discussion on continuity of inductor currents and capacitor voltages in 5) would be violated with D1 off Whenever the current ow through the inductor is interrupted (during the negative half-cycles of vAC ), the inductor attempts to build a yback voltage proportional to diL /dt Since the recti er does not provide any current during the negative half-cycle of the source voltage, the instantaneous inductor voltage could be very large and could lead to serious damage to either the motor or the recti er The circuit shown on the bottom in Figure 1117 contains a so-called freewheeling diode, D2 The role of D2 is to provide continuity of current when D1 is in the off state D2 is off during the positive half-cycle but turns on when D1 ceases to conduct, because of the yback voltage, LdiL /dt Rather than build up a large voltage, the inductor now has a path for current to ow, through D2 , when D1 is off Thus, the energy stored by the inductor during the positive half-cycle of vAC is utilized to preserve a continuous current through the inductor during the off period Figure 1118 depicts the load current for the circuit including the diode Note that D2 allows the energy-storage properties of the inductor to be utilized to smooth the pulselike supply current and to produce a nearly constant load current Analyzing the circuit on the bottom of Figure 1117, with vAC (t) = A sin( t) (115).

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How to Compress PDF Document in C# , VB.NET - E-iceblue
Spire.PDF allows us to compress PDF document in the following two ways: Compressing content; Compressing image. Compressing content. In Spire.PDF, we ...

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i have also tried to compress the .tiff first in jpeg and than send the stream to the pdf but the final size is even bigger and it consumes enormous ...

a b c d e 5e 4t cos 2 t sin( t + 2 ) 4e 2t sin(3t 50 ) e 3t (2 + cos 4t)

Using Table C-21 in Appendix C, the effort equations derived from the ISBSG s data for the different combinations of platform and development show the following ballpark effort estimates for a project of 500 function points:

a 5e 2t b 5e 2t cos(4t + 10 ) c 4 cos(2t 20 )

The String/string class is one that you will work with all of the time. This class is designed to hold an array of character data. Since character data is so common in a program, Microsoft has included a number of features in the String/string class that you should know about.

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C# tutorial: PDF compression - World Best Learning Center
In this C# tutorial you will learn to compress a new PDF file and existing PDF file in itextsharp.

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C# How to compress PDF images in C# for Visual Studio 2013
28 Feb 2018 ... There is no doubt that the PDF file which contains plenty of images will ... For the reasons above, two methods to compress images in PDF  ...

All the concepts and rules used in AC network analysis (see 4), such as impedance, admittance, KVL, KCL, and Th venin s and Norton s theorems, e carry over to the damped sinusoid case exactly In the complex frequency domain, the current I(s) and voltage V(s) are related by the expression V(s) = Z(s)I(s) (645)

6

where Z(s) is the familiar impedance, with s replacing j We may obtain Z(s) from Z(j ) by simply replacing j by s For a resistance, R, the impedance is ZR (s) = R For an inductance, L, the impedance is ZL (s) = sL For a capacitance, C, it is ZC (s) = 1 sC (648) (647) (646)

Impedances in series or parallel are combined in exactly the same way as in the AC steady-state case, since we only replace j by s

Platform and Language MF & 3GL MF & 4GL MR & 3GL MR & 4GL PC & 3GL PC & 4GL Multi & 3GL Multi & 4GL Estimated Effort for 500 FP Project 4,794 person hours 3,360 person hours 4,343 person hours 5,053 person hours 3,469 person hours 2,561 person hours 4,191 person hours 3,470 person hours

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7 ways to compress PDF files in C# , VB.NET | Syncfusion Blogs
25 Apr 2018 ... Syncfusion Essential PDF is a .NET PDF library that can be used to optimize or compress your PDF documents. Reducing the PDF file size can ...

You can read an individual character from a string using the same syntax we used on arrays: VB .NET Dim name As String = "bob" Dim letter As Char = name.Chars(0) Console.WriteLine(letter) ' Shows b C# string name = "bob"; char letter = (char)name[0]; Console.WriteLine(letter); // Shows b

Use complex impedance ideas to determine the response of a series RL circuit to a damped exponential voltage

Known Quantities: Source voltage, resistor, inductor values Find: The time-domain expression for the series current, iL (t)

Schematics, Diagrams, Circuits, and Given Data: vs (t) = 10e 2t cos(5t) V; R = 4 Assumptions: None Analysis: The input voltage phasor can be represented by the expression

V(s) = 10 0 V The impedance seen by the voltage source is Z(s) = R + sL = 4 + 2s Thus, the series current is: I(s) = 10 10 10 V(s) = = = = j 1 = 1 Z(s) 4 + 2s 4 + 2( 2 + j 5) j 10 2

method introduced in 4, with the complex frequency j (steady-state sinusoidal frequency) related by s (damped sinusoidal frequency)

Although you can read an individual character from a string, you cannot change it the same way: VB .NET Dim name As String = "bob" name(0) = "B" 'This will cause an error

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C# tutorial: PDF compression - World Best Learning Center
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