barcode in vb.net 2010 When conductance in Software

Encoder Code-39 in Software When conductance

16-6 When conductance
Code 39 Recognizer In None
Using Barcode Control SDK for Software Control to generate, create, read, scan barcode image in Software applications.
Drawing Code 39 Extended In None
Using Barcode creation for Software Control to generate, create Code 3/9 image in Software applications.
is present along with susceptance, admittance vectors point at angles; they are neither vertical nor horizontal.
Read Code 3 Of 9 In None
Using Barcode scanner for Software Control to read, scan read, scan image in Software applications.
ANSI/AIM Code 39 Maker In Visual C#
Using Barcode printer for Visual Studio .NET Control to generate, create ANSI/AIM Code 39 image in .NET applications.
Parallel GLC Circuits 253
USS Code 39 Drawer In .NET
Using Barcode generator for ASP.NET Control to generate, create Code 3/9 image in ASP.NET applications.
Making Code 3 Of 9 In Visual Studio .NET
Using Barcode drawer for .NET framework Control to generate, create Code 39 Full ASCII image in .NET applications.
You ve seen how vectors add in the RX plane. In the GB plane, the principle is the same. The net admittance vector is the sum of the component admittance vectors.
Making Code 39 Full ASCII In VB.NET
Using Barcode encoder for Visual Studio .NET Control to generate, create Code 3/9 image in Visual Studio .NET applications.
Making EAN 128 In None
Using Barcode generator for Software Control to generate, create EAN / UCC - 14 image in Software applications.
Formula for Complex Admittances in Parallel Given two admittances, Y1 = G1 + jB1 and Y2 = G2 + jB2, the net admittance Y of these in parallel is their vector sum, as follows:
Encoding Code 128 In None
Using Barcode generator for Software Control to generate, create Code-128 image in Software applications.
DataMatrix Generation In None
Using Barcode generator for Software Control to generate, create DataMatrix image in Software applications.
Y = (G1 + jB1 ) + (G2 + jB 2 ) = (G1 + G2 ) + j(B1 + B 2 ) The conductance and susceptance components add separately. Just remember that if a susceptance is inductive, then it is negative imaginary in this formula.
Barcode Encoder In None
Using Barcode generation for Software Control to generate, create barcode image in Software applications.
Bar Code Encoder In None
Using Barcode creator for Software Control to generate, create barcode image in Software applications.
Parallel GLC Circuits
USPS POSTal Numeric Encoding Technique Barcode Printer In None
Using Barcode creation for Software Control to generate, create USPS POSTal Numeric Encoding Technique Barcode image in Software applications.
Printing Code 128B In C#
Using Barcode generator for Visual Studio .NET Control to generate, create USS Code 128 image in Visual Studio .NET applications.
When a coil, capacitor, and resistor are connected in parallel (Fig. 16-7), the resistance should be thought of as a conductance, whose value in siemens (symbolized S) is equal to the reciprocal of the value in ohms. Think of the conductance as all belonging to the inductor. Then you have two vectors to add, when finding the admittance of a parallel GLC (conductance-inductance-capacitance) circuit: Y = (G + jBL ) + (0 + jBC) = G + j(BL + BC) Again, remember that BL is never positive! So, although the formulas here have addition symbols in them, you re adding a negative number when you add in an inductive susceptance.
Decoding UPC-A Supplement 5 In Visual C#.NET
Using Barcode reader for .NET Control to read, scan read, scan image in .NET applications.
Paint Code 3 Of 9 In Java
Using Barcode creation for Java Control to generate, create Code-39 image in Java applications.
Problem 16-13 Suppose a resistor, a coil, and a capacitor are connected in parallel. Suppose the resistor has a conductance G = 0.10 S, and the susceptances are jBL = j0.010 and jBC = j0.020. What is the complex admittance of this combination Consider the resistor to be part of the coil. Then there are two complex admittances in parallel: 0.10 j0.010 and 0.00 + j0.020. Adding these gives a conductance component of 0.10 + 0.00 = 0.10 and a susceptance component of j0.010 + j0.020 = j0.010. Therefore, the complex admittance is 0.10 + j0.010.
Decode Code 128 Code Set B In None
Using Barcode scanner for Software Control to read, scan read, scan image in Software applications.
Barcode Printer In .NET
Using Barcode generation for Reporting Service Control to generate, create bar code image in Reporting Service applications.
16-7 A parallel
Read European Article Number 13 In Java
Using Barcode decoder for Java Control to read, scan read, scan image in Java applications.
Code 3/9 Drawer In Visual Studio .NET
Using Barcode generator for Reporting Service Control to generate, create Code39 image in Reporting Service applications.
conductanceinductance-capacitance (GLC) circuit.
254 RLC and GLC Circuit Analysis
Problem 16-14 Suppose a resistor, a coil, and a capacitor are connected in parallel. Suppose the resistor has a conductance G = 0.0010 S, and the susceptances are jBL = j0.0022 and jBC = j0.0022. What is the complex admittance of this combination Again, consider the resistor to be part of the coil. Then the complex admittances are 0.0010 j0.0022 and 0.0000 + j0.0022. Adding these, the conductance component is 0.0010 + 0.0000 = 0.0010, and the susceptance component is j0.0022 + j0.0022 = j0. Thus, the admittance is 0.0010 + j0. This is a purely conductive admittance. Problem 16-15 Suppose a resistor, a coil, and a capacitor are connected in parallel. The resistor has a value of 100 , the capacitance is 200 pF, and the inductance is 100 H. The frequency is 1.00 MHz. What is the net complex admittance First, you need to calculate the inductive susceptance. Recall the formula, and plug in the numbers as follows:
jBL = j[1/(6.28f L)] = j[1/(6.28 1.00 100)] = j0.00159 Megahertz and microhenrys go together in the formula. Next, you must calculate the capacitive susceptance. Convert 200 pF to microfarads to go with megahertz in the formula; thus C = 0.000200 F. Then: jBC = j(6.28f C ) = j(6.28 1.00 0.000200) = j0.00126 Finally, consider the conductance, which is 1 100 = 0.0100 S, and the inductive susceptance as existing together in a single component. That means that one of the parallel-connected admittances is 0.0100 j0.00159. The other is 0.0000 + j0.00126. Adding these gives 0.0100 j0.00033.
Problem 16-16 Suppose a resistor, a coil, and a capacitor are in parallel. The resistance is 10.0 , the inductance is 10.0 H, and the capacitance is 1000 pF. The frequency is 1592 kHz. What is the complex admittance of this circuit at this frequency First, calculate the inductive susceptance. Convert the frequency to megahertz; 1592 kHz = 1.592 MHz. Plug in the numbers as follows:
jBL = j[1/(6.28f L)] = j[1/(6.28 1.592 10.0)] = j0.0100 Next, calculate the capacitive susceptance. Convert 1000 pF to microfarads to go with megahertz in the formula; thus C = 0.001000 F. Then:
Copyright © OnBarcode.com . All rights reserved.