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Problem 16-13 Suppose a resistor, a coil, and a capacitor are connected in parallel Suppose the resistor has a conductance G = 010 S, and the susceptances are jBL = j0010 and jBC = j0020 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: 010 j0010 and 000 + j0020 Adding these gives a conductance component of 010 + 000 = 010 and a susceptance component of j0010 + j0020 = j0010 Therefore, the complex admittance is 010 + j0010 crystal reports qr code generator QR-Code Crystal Reports Native Barcode Generator - IDAutomation
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Generate QR Code in Crystal Report for .NET with control library. Problem 16-14 Suppose a resistor, a coil, and a capacitor are connected in parallel Suppose the resistor has a conductance G = 00010 S, and the susceptances are jBL = j00022 and jBC = j00022 What is the complex admittance of this combination Again, consider the resistor to be part of the coil Then the complex admittances are 00010 j00022 and 00000 + j00022 Adding these, the conductance component is 00010 + 00000 = 00010, and the susceptance component is j00022 + j00022 = j0 Thus, the admittance is 00010 + 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 100 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: 4: (Specific hardware and software used should be listed) crystal reports 8.5 qr code Print QR Code in Crystal Reports - Barcodesoft
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Window 10 Compatible Add native QR-Code 2D barcode generation to Crystal Reports without any special fonts. ISO/IEC 18004:2006 specification compliant. jBL = j[1/(628f L)] = j[1/(628 100 100)] = j000159 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 = 0000200 F Then: jBC = j(628f C ) = j(628 100 0000200) = j000126 Finally, consider the conductance, which is 1 100 = 00100 S, and the inductive susceptance as existing together in a single component That means that one of the parallel-connected admittances is 00100 j000159 The other is 00000 + j000126 Adding these gives 00100 j000033 if (XMLHttpRequestObjectreadyState == 4 && XMLHttpRequestObjectstatus == 200) { eval(XMLHttpRequestObjectresponseText); } } XMLHttpRequestObjectsend(null); } } function show() { var targetDiv = documentgetElementById("targetDiv"); targetDivinnerHTML = "Yep, it worked!"; } </script> </head> <body> <H1>Downloading JavaScript with Ajax</H1> <form> <input type = "button" value = "Get the JavaScript" onclick = "getData('javascriptphp')"> </form> <div id="targetDiv"> <p>The data will go here</p> </div> </body> </html> So when the user clicks the button, the application reads the text returned by javascriptphp and executes that text as JavaScript, calling the show function, which displays the message you see in Figure 4-5 This example demonstrates that you can download and execute JavaScript almost as easily as you can download text And that prepares you for working with Google Suggest, coming up next Problem 16-16 Suppose a resistor, a coil, and a capacitor are in parallel The resistance is 100 , the inductance is 100 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 = 1592 MHz Plug in the numbers as follows: Figure 4-5 jBL = j[1/(628f L)] = j[1/(628 1592 100)] = j00100 Next, calculate the capacitive susceptance Convert 1000 pF to microfarads to go with megahertz in the formula; thus C = 0001000 F Then: Try creating a new Ajax-enabled web page that includes a JavaScript function named adder, which adds two values and returns the sum: In the common gate circuit (Fig 23-12), the gate is placed at signal ground The input is applied to the source The illustration shows an N-channel JFET For other types of FETs, the same considerations apply as previously described for the common source circuit Enhancement-mode devices would require a resistor between the gate and the positive supply terminal (or the negative terminal if the MOSFET is P-channel) The dc bias for the common gate circuit is basically the same as that for the common source arrangement But the signal follows a different path The ac input signal enters through C1 Resistor R1 keeps the input from being shorted to ground Gate bias is provided by R1 and R2 Capacitor function adder(value1, value2) { return (item1 + item2) } crystal reports 2013 qr code Print QR Code from a Crystal Report - SAP Q&A
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