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vb.net barcode reader code Assumptions: Use rms values for all phasor quantities in the problem in Software
Assumptions: Use rms values for all phasor quantities in the problem QR Code JIS X 0510 Reader In None Using Barcode Control SDK for Software Control to generate, create, read, scan barcode image in Software applications. Make QR In None Using Barcode creator for Software Control to generate, create QR Code JIS X 0510 image in Software applications. Rline
QR Recognizer In None Using Barcode reader for Software Control to read, scan read, scan image in Software applications. Quick Response Code Maker In C#.NET Using Barcode creator for Visual Studio .NET Control to generate, create QRCode image in Visual Studio .NET applications. through the neutral line is zero, ie, Vn n = 0 The resulting perphase circuit is shown in Figure 751 Using phase a for the calculations, we look for the quantity Pa = I2 RL Painting QR Code JIS X 0510 In VS .NET Using Barcode creation for ASP.NET Control to generate, create Denso QR Bar Code image in ASP.NET applications. QR Code 2d Barcode Generation In Visual Studio .NET Using Barcode generator for .NET framework Control to generate, create Quick Response Code image in .NET framework applications. Analysis: Since the circuit is balanced, we can use perphase analysis, and the current
QR Code 2d Barcode Creator In VB.NET Using Barcode encoder for Visual Studio .NET Control to generate, create QR Code 2d barcode image in VS .NET applications. Draw Bar Code In None Using Barcode maker for Software Control to generate, create barcode image in Software applications. + ~ ~ VS _ Paint GTIN  128 In None Using Barcode generator for Software Control to generate, create UCC.EAN  128 image in Software applications. Generate ANSI/AIM Code 39 In None Using Barcode printer for Software Control to generate, create USS Code 39 image in Software applications. where I = 480 0 480 0 Va = = Zy + Rline 2 + j4 + 2 566 4 = 8485 A
Paint ECC200 In None Using Barcode generator for Software Control to generate, create Data Matrix ECC200 image in Software applications. Drawing GS1  13 In None Using Barcode creator for Software Control to generate, create EAN13 image in Software applications. Figure 751 One phase of the threephase circuit
USPS POSTNET Barcode Creator In None Using Barcode drawer for Software Control to generate, create Postnet image in Software applications. Data Matrix Maker In None Using Barcode encoder for Online Control to generate, create DataMatrix image in Online applications. and Pa = (8485)2 2 = 144 kW Since the circuit is balanced, the results for phases b and c are identical, and we have: PL = 3Pa = 432 kW Paint Barcode In None Using Barcode drawer for Online Control to generate, create barcode image in Online applications. Barcode Printer In None Using Barcode creation for Microsoft Word Control to generate, create barcode image in Word applications. Comments: Note that, since the circuit is balanced, there is zero voltage across neutrals
Recognizing EAN / UCC  13 In Visual C#.NET Using Barcode recognizer for Visual Studio .NET Control to read, scan read, scan image in VS .NET applications. Linear Maker In Visual Studio .NET Using Barcode generation for Visual Studio .NET Control to generate, create 1D Barcode image in VS .NET applications. This fact is shown explicitly in Figure 751, where n and n are connected to each other directly Perphase analysis for balanced circuits turn threephase power calculations into a very simple exercise Data Matrix ECC200 Scanner In None Using Barcode reader for Software Control to read, scan read, scan image in Software applications. Create Bar Code In .NET Using Barcode drawer for Reporting Service Control to generate, create barcode image in Reporting Service applications. Balanced Wye Loads In the previous section we performed some power computations for a purely resistive balanced wye load We shall now generalize those results for an arbitrary balanced complex load Consider again the circuit of Figure 746, where now the balanced load consists of the three complex impedances Za = Zb = Zc = Zy = Zy  (758) From the diagram of Figure 746, it can be veri ed that each impedance sees the corresponding phase voltage across itself; thus, since the currents Ia , Ib , and Ic have the same rms value, I , the phase angles of the currents will differ by 120 It is therefore possible to compute the power for each phase by considering the phase voltage (equal to the load voltage) for each impedance, and the associated line current Let us denote the complex power for each phase by S: S = V I so that S = P + jQ = V I cos + j V I sin (760) (759) where V and I denote, once again, the rms values of each phase voltage and line current Consequently, the total real power delivered to the balanced wye load is 3P , and the total reactive power is 3Q Thus, the total complex power, ST , is given by ST = PT + j QT = 3P + j 3Q = (3P )2 + (3Q)2 (761) Part I
Circuits
and the apparent power is ST  = 3 (V I )2 cos2 + (V I )2 sin2 = 3V I and the total real and reactive power may be expressed in terms of the apparent power: PT = ST  cos QT = ST  sin Balanced Delta Loads In addition to a wye connection, it is also possible to connect a balanced load in the delta con guration A wyeconnected generator and a deltaconnected load are shown in Figure 752 (762) Ia a
+ Van ~ _ n Ic
+ _ c Z Vab
_ +~
_ ~ Vcn +
Iab Ibc
+ b
Figure 752 Balanced wye generators with balanced delta load
It should be noted immediately that now the corresponding line voltage (not phase voltage) appears across each impedance For example, the voltage across Zc a is Vca Thus, the three load currents are given by the following expressions: 3V ( /6) Vab = Iab = Z Z  Vbc 3V ( /4) (763) Ibc = = Z Z  3V (5 /6) Vca Ica = = Z Z  To understand the relationship between deltaconnected and wyeconnected loads, it is reasonable to ask the question, For what value of Z would a deltaconnected load draw the same amount of current as a wyeconnected load with 7
AC Power
impedance Zy for a given source voltage This is equivalent to asking what value of Z would make the line currents the same in both circuits (compare Figure 748 with Figure 752) The line current drawn, say, in phase a by a wyeconnected load is (Ian )y = V Van = Z Zy  (764) while that drawn by the deltaconnected load is (Ia ) = Iab Ica = = = = Vab Vca Z Z 1 (Van Vbn Vcn + Van ) Z 1 (2Van Vbn Vcn ) Z 3Van 3V = Z Z  (765) One can readily verify that the two currents (Ia ) and (Ia )y will be equal if the magnitude of the deltaconnected impedance is 3 times larger than Zy : Z = 3Zy (766) This result also implies that a delta load will necessarily draw 3 times as much current (and therefore absorb 3 times as much power) as a wye load with the same branch impedance

