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barcode in vb.net 2005 Formula for complex admittances in parallel in Software
Formula for complex admittances in parallel QR Code Reader In None Using Barcode Control SDK for Software Control to generate, create, read, scan barcode image in Software applications. Generate QRCode In None Using Barcode creation for Software Control to generate, create Quick Response Code image in Software applications. Given two admittances, Y1 G1 jB1 and Y2 in parallel is their vector sum, given by Y (G1 G2) G2 jB2, the net admittance Y of these B2) Scan QR Code In None Using Barcode decoder for Software Control to read, scan read, scan image in Software applications. QR Code JIS X 0510 Encoder In C#.NET Using Barcode generator for .NET framework Control to generate, create QRCode image in .NET framework applications. j(B1
Quick Response Code Creation In Visual Studio .NET Using Barcode encoder for ASP.NET Control to generate, create Quick Response Code image in ASP.NET applications. Quick Response Code Creation In .NET Using Barcode creator for .NET Control to generate, create QR image in Visual Studio .NET applications. 292 RLC circuit analysis
Denso QR Bar Code Printer In VB.NET Using Barcode generator for .NET Control to generate, create QR image in .NET applications. GS1128 Creator In None Using Barcode encoder for Software Control to generate, create UCC128 image in Software applications. 166 When conductance is present along with susceptance, admittance vectors point northeast or southeast. UPC Code Encoder In None Using Barcode maker for Software Control to generate, create UCC  12 image in Software applications. Making Code 3 Of 9 In None Using Barcode generator for Software Control to generate, create Code 39 Full ASCII image in Software applications. The susceptances B1 and B2 might both be inductive; they might both be capacitive; or one might be inductive and the other capacitive. GTIN  13 Maker In None Using Barcode generator for Software Control to generate, create EAN13 image in Software applications. Bar Code Creation In None Using Barcode maker for Software Control to generate, create barcode image in Software applications. Parallel GLC circuits
USPS POSTal Numeric Encoding Technique Barcode Encoder In None Using Barcode creation for Software Control to generate, create USPS POSTNET Barcode image in Software applications. Generating Code 39 Full ASCII In Java Using Barcode drawer for Java Control to generate, create USS Code 39 image in Java applications. When a coil, capacitor, and resistor are connected in parallel (Fig. 167), the resistor should be thought of as a conductor, whose value in siemens is equal to the reciprocal of the value in ohms. Think of the conductance as all belonging to the inductor. (Thinking of it all as belonging to the capacitor will also work.) Then you have two vectors to add, when finding the admittance of a parallel GLC (conductanceinductancecapacitance) circuit: Y (G jBL) (0 G j(BL BC) jBC) UPC  13 Maker In None Using Barcode printer for Office Word Control to generate, create EAN13 Supplement 5 image in Word applications. Data Matrix 2d Barcode Decoder In C# Using Barcode scanner for .NET framework Control to read, scan read, scan image in VS .NET applications. 167 UPCA Supplement 5 Creation In ObjectiveC Using Barcode maker for iPad Control to generate, create GTIN  12 image in iPad applications. Encode Code 39 Extended In ObjectiveC Using Barcode printer for iPad Control to generate, create ANSI/AIM Code 39 image in iPad applications. A parallel GLC circuit.
EAN 13 Maker In Java Using Barcode creator for Java Control to generate, create EAN 13 image in Java applications. Print Matrix 2D Barcode In VS .NET Using Barcode drawer for ASP.NET Control to generate, create Matrix 2D Barcode image in ASP.NET applications. Problem 1613 A resistor, coil, and capacitor are connected in parallel with G 0.1 siemens, jBL j0.010, and jBC j0.020. What is the net admittance vector Consider the resistor to be part of the coil, obtaining two complex vectors, 0.1 j0.010 and 0 j0.020. Adding these gives the conductance component of 0.1 0 Parallel GLC circuits 293 0.1, and the susceptance component of j0.010 mittance vector is 0.1 j0.010. j0.020 j0.010. Therefore, the ad Problem 1614 A resistor, coil, and capacitor are connected in parallel with G 0.0010 siemens, jBL j0.0022 and jBC j0.0022. What is the net admittance vector Again, consider the resistor to be part of the coil. Then the complex vectors are 0.0010 j0.0022 and 0 j0.0022. Adding these, the conductance component is 0.0010 0 0.0010, and the susceptance component is j0.0022 j0.0022 j0. Thus, the admittance vector is 0.0010 j0. This is a purely conductive admittance. There is no susceptance. Problem 1615 A resistor, coil, and 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 and capacitive susceptances. Recall jBL j(1/(6.28fL), and plug in the values, getting jBL j(l/(6.28 1.00 100) j0.00159 Megahertz and microhenrys go together in the formula. As for jBC, recall the formula jBC j(6.28fC). Convert 200 pF to microfarads to go with megahertz in the formula; C 0.000200 F. Then jBC j(6.28 1.00 0.000200) j0.00126 Now, you can consider the conductance, which is 1/100 0.0100 siemens, and the inductive susceptance to go together. So one of the vectors is 0.0100 j0.00159. The other is 0 j0.00126. Adding these gives 0.0100 j0.00033. Problem 1616 A resistor, coil, and 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 j(1/(6.28fL)). Convert the frequency to megahertz; 1592 First, calculate jBL kHz 1.592 MHz. Then jBL j/(l/(6.28 1.592 10.0)) j0.0100 Then calculate jBC j(6.28fC). Convert picofarads to microfarads, and use megahertz for the frequency. Therefore jBC j(6.28 1.592 0.001000) j0.0100 Let the conductance and inductive susceptance go together as one vector, 0.100 j0.0100. (Remember that conductance is the reciprocal of resistance; here G 294 RLC circuit analysis 1/R 1/10.0 0.100.) Let the capacitance alone be the other vector, 0 j0.0100. Then the sum is 0.100 j0.0100 j0.0100 0. 100 j0. This is a pure conductance of 0.100 siemens. Converting from admittance to impedance
The GB plane is, as you have seen, similar in appearance to the RX plane, although mathematically the two are worlds apart. Once you ve found a complex admittance for a parallel RLC circuit, how do you transform this back to a complex impedance Generally, it is the impedance, not the admittance, that technicians and engineers work with. The transformation from complex admittance, or a vector G jB, to a complex impedance, or a vector R jX, requires the use of the following formulas: R X G/(G2 B2) B/(G2 B 2) If you know the complex admittance, first find the resistance and reactance components individually. Then assemble them into the impedance vector, R jX. Problem 1617 The admittance vector for a certain parallel circuit is 0.010 j0.0050. What is the impedance vector In this case, G 0.010 and B 0.0050. Find G2 B2 first, because you ll need to ( 0.0050)2 0.000100 0.000025 use it twice as a denominator; it is 0.0102 0.000125. Then R X G/0.000125 0.010/0.000125 80 B/0.000125 0.0050/0.000125 40 jX 80 j40.

