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java barcode reader library download Z line in Software
Z line Drawing PDF417 2d Barcode In None Using Barcode maker for Software Control to generate, create PDF 417 image in Software applications. PDF 417 Decoder In None Using Barcode decoder for Software Control to read, scan read, scan image in Software applications. 24 kV = (382 + j140 ) = 112 + j 411 14 kV
Drawing PDF417 In C#.NET Using Barcode encoder for VS .NET Control to generate, create PDF417 image in .NET applications. Encode PDF417 2d Barcode In VS .NET Using Barcode creator for ASP.NET Control to generate, create PDF417 2d barcode image in ASP.NET applications. The secondary current I S is given by
PDF417 2d Barcode Maker In VS .NET Using Barcode creator for .NET framework Control to generate, create PDF 417 image in .NET applications. PDF 417 Creation In Visual Basic .NET Using Barcode printer for VS .NET Control to generate, create PDF417 image in VS .NET applications. 90 kW = 4348 A (2300 V )(09 ) I S = 4348 258 A IS =
Code39 Creator In None Using Barcode printer for Software Control to generate, create Code 3 of 9 image in Software applications. Painting UCC  12 In None Using Barcode creator for Software Control to generate, create GS1128 image in Software applications. (a) The voltage at the power source of this system (referred to the secondary side) is
Barcode Printer In None Using Barcode drawer for Software Control to generate, create bar code image in Software applications. Make UPCA In None Using Barcode drawer for Software Control to generate, create UPC Symbol image in Software applications. Vsource = VS + I S Z line + I S Z EQ Vsource = 2300 0 V + ( 4348 258 A )(112 + j 411 ) + ( 4348 258 A )( 012 + j 05 ) Vsource = 2441 37 V Creating ANSI/AIM Code 128 In None Using Barcode creation for Software Control to generate, create Code 128 Code Set C image in Software applications. Paint UPC  13 In None Using Barcode generation for Software Control to generate, create GTIN  13 image in Software applications. Therefore, the voltage at the power source is
Painting RoyalMail4SCC In None Using Barcode encoder for Software Control to generate, create RoyalMail4SCC image in Software applications. Encoding Code 128C In ObjectiveC Using Barcode encoder for iPad Control to generate, create ANSI/AIM Code 128 image in iPad applications. Vsource = (2441 37 V ) EAN13 Creation In None Using Barcode printer for Online Control to generate, create EAN13 Supplement 5 image in Online applications. Encode GS1  13 In Visual Studio .NET Using Barcode maker for Reporting Service Control to generate, create GS1  13 image in Reporting Service applications. 14 kV = 1424 37 kV 24 kV
Barcode Reader In .NET Using Barcode Control SDK for ASP.NET Control to generate, create, read, scan barcode image in ASP.NET applications. Bar Code Decoder In Visual Basic .NET Using Barcode reader for VS .NET Control to read, scan read, scan image in .NET framework applications. (b) To find the voltage regulation of the transformer, we must find the voltage at the primary side of the transformer (referred to the secondary side) under full load conditions: Create Code128 In ObjectiveC Using Barcode creation for iPhone Control to generate, create Code 128 Code Set B image in iPhone applications. Code 3/9 Drawer In None Using Barcode generator for Online Control to generate, create Code 3/9 image in Online applications. VP = VS + I S Z EQ VP = 2300 0 V + (4348 258 A )(012 + j 05 ) = 2314 043 V
There is a voltage drop of 14 V under these load conditions Therefore the voltage regulation of the transformer is VR =
2314  2300 100% = 06% 2300 The power supplied to the load is POUT = 90 kW The power supplied by the source is
PIN = Vsource I S cos = (2441 V )(4348 A ) cos 295 = 9237 kW
Therefore, the efficiency of the power system is
35 POUT 90 kW 100% = 100% = 974% PIN 9237 kW
When travelers from the USA and Canada visit Europe, they encounter a different power distribution system Wall voltages in North America are 120 V rms at 60 Hz, while typical wall voltages in Europe are 220240 V at 50 Hz Many travelers carry small stepup / stepdown transformers so that they can use their appliances in the countries that they are visiting A typical transformer might be rated at 1kVA and 120/240 V It has 5001 turns of wire on the 120V side and 1000 turns of wire on the 240V side The magnetization curve for this transformer is shown in Figure P32, and can be found in file p32mag at this book s Web site Note that this turns ratio was backwards in the first printing of the text This error should be corrected in all subsequent printings 33 (a) Suppose that this transformer is connected to a 120V, 60 Hz power source with no load connected to the 240V side Sketch the magnetization current that would flow in the transformer (Use MATLAB to plot the current accurately, if it is available) What is the rms amplitude of the magnetization current What percentage of fullload current is the magnetization current (b) Now suppose that this transformer is connected to a 240V, 50 Hz power source with no load connected to the 120V side Sketch the magnetization current that would flow in the transformer (Use MATLAB to plot the current accurately, if it is available) What is the rms amplitude of the magnetization current What percentage of fullload current is the magnetization current (c) In which case is the magnetization current a higher percentage of fullload current Why SOLUTION (a) When this transformer is connected to a 120V 60 Hz source, the flux in the core will be given by the equation (t ) = VM cos t N P
(3104) The magnetization current required for any given flux level can be found from Figure P32, or alternately from the equivalent table in file p32mag The MATLAB program shown below calculates the flux level at each time, the corresponding magnetization current, and the rms value of the magnetization current % Mfile: prob3_5am % Mfile to calculate and plot the magnetization % current of a 120/240 transformer operating at % 120 volts and 60 Hz This program also % calculates the rms value of the mag current % Load the magnetization curve It is in two % columns, with the first column being mmf and % the second column being flux load p32mag; mmf_data = p32(:,1); flux_data = p32(:,2); % Initialize values S = 1000; Vrms = 120; VM = Vrms * sqrt(2); NP = 500; % % % % Apparent power (VA) Rms voltage (V) Max voltage (V) Primary turns
% Calculate angular velocity for 60 Hz freq = 60; % Freq (Hz) w = 2 * pi * freq; % Calculate flux versus time time = 0:1/3000:1/30; % 0 to 1/30 sec flux = VM/(w*NP) * cos(w * time); % Calculate the mmf corresponding to a given flux % using the MATLAB interpolation function mmf = interp1(flux_data,mmf_data,flux); % Calculate the magnetization current im = mmf / NP; % Calculate the rms value of the current irms = sqrt(sum(im^2)/length(im)); disp(['The rms current at 120 V and 60 Hz is ', num2str(irms)]); % Calculate the fullload current i_fl = S / Vrms; % Calculate the percentage of fullload current percnt = irms / i_fl * 100; disp(['The magnetization current is ' num2str(percnt) '% of fullload current']); % Plot the magnetization current figure(1) plot(time,im); title ('\bfMagnetization Current at 120 V and 60 Hz'); xlabel ('\bfTime (s)'); ylabel ('\bf\itI_{m} \rm(A)'); axis([0 004 05 05]); grid on;

