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barcode printing using vb.net Kirchhoff s first law in Software
Kirchhoff s first law Scanning QR In None Using Barcode Control SDK for Software Control to generate, create, read, scan barcode image in Software applications. Draw QR Code ISO/IEC18004 In None Using Barcode encoder for Software Control to generate, create QR Code image in Software applications. The physicist Gustav Robert Kirchhoff (18241887) was a researcher and experimentalist in electricity back in the time before radio, before electric lighting, and before much was understood about how currents flow. QR Code JIS X 0510 Recognizer In None Using Barcode reader for Software Control to read, scan read, scan image in Software applications. QR Code JIS X 0510 Creator In Visual C# Using Barcode creation for VS .NET Control to generate, create QR Code image in VS .NET applications. 90 Directcurrent circuit analysis Kirchhoff reasoned that current must work something like water in a network of pipes, and that the current going into any point has to be the same as the current going out. This is true for any point in a circuit, no matter how many branches lead into or out of the point. Two examples are shown in Fig. 55. Create QR Code In Visual Studio .NET Using Barcode generation for ASP.NET Control to generate, create QR Code ISO/IEC18004 image in ASP.NET applications. Denso QR Bar Code Printer In .NET Using Barcode maker for VS .NET Control to generate, create Denso QR Bar Code image in VS .NET applications. In a network of water pipes that does not leak, and into which no water is added along the way, the total number of cubic feet going in has to be the same as the total volume going out. Water can t form from nothing, nor can it disappear, inside a closed system of pipes. Charge carriers, thought Kirchhoff, must act the same way in an electric circuit. This is Kirchhoff s First Law. An alternative name might be the law of conservation of current. Generate QR Code 2d Barcode In VB.NET Using Barcode creator for .NET framework Control to generate, create QR image in VS .NET applications. Code 3/9 Creation In None Using Barcode printer for Software Control to generate, create Code 3/9 image in Software applications. Problem 511 GTIN  12 Generator In None Using Barcode maker for Software Control to generate, create UPCA Supplement 2 image in Software applications. ECC200 Creator In None Using Barcode printer for Software Control to generate, create Data Matrix ECC200 image in Software applications. Refer to Fig. 55A. Suppose all three resistors have values of 100 , and that I1 2.0 A while I2 1.0 A. What is the battery voltage First, find the current I drawn from the battery. It must be 3.0 A; I I1 I2 2.0 1.0 3.0 A. Next, find the resistance of the whole combination. The two 100 resistors in series give a value of 200 , and this is in parallel with 100 . You can do the EAN / UCC  13 Creator In None Using Barcode generation for Software Control to generate, create UCC  12 image in Software applications. Print Code 128 In None Using Barcode drawer for Software Control to generate, create Code 128C image in Software applications. TeamFly
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Code 39 Recognizer In Java Using Barcode reader for Java Control to read, scan read, scan image in Java applications. Create Barcode In Java Using Barcode printer for BIRT reports Control to generate, create bar code image in Eclipse BIRT applications. 55 Kirchhoffs First Law. At A, the current into either X or Y is the same as the current out of that point: I = Il + I2. At B, the current into Z equals the current out of Z: Il + I2 = I3 + I4+ I5. Also see quiz questions 13 and 14. UPCA Supplement 5 Generation In None Using Barcode creation for Font Control to generate, create Universal Product Code version A image in Font applications. EAN / UCC  14 Generation In Java Using Barcode encoder for Java Control to generate, create UCC.EAN  128 image in Java applications. Kirchhoff s second law 91 calculations and find that the total resistance, R, across the battery, E, is 66.67 . Then E IR 66.67 3.0 200 volts. (Some battery.) Data Matrix 2d Barcode Generation In Java Using Barcode encoder for BIRT reports Control to generate, create Data Matrix image in Eclipse BIRT applications. Drawing Barcode In .NET Framework Using Barcode printer for Reporting Service Control to generate, create bar code image in Reporting Service applications. Problem 512 In Fig. 55B, suppose each of the two resistors below point Z has a value of 100 , and all three resistors above Z are 10.0 . The current through each 100 resistor is 500 mA. What is the current through any of the 10.0 resistors, assuming it is equally distributed What is the voltage, then, across any of the 10. 0 resistors The total current into Z is 500 mA + 500 mA 1.00 A. This must be divided three ways equally among the 10 resistors. Therefore, the current through any one of them is 1.00/3 A 0.333 A 333 mA. The voltage across any one of the 10.0 resistors is found by Ohm s Law: E IR 0.333 10.0 3.33 V. Kirchhoff s second law
The sum of all the voltages, as you go around a circuit from some fixed point and return there from the opposite direction, and taking polarity into account, is always zero. This might sound strange. Surely there is voltage in your electric hair dryer, or radio, or computer. Yes, there is, between different points. But no point can have an EMF with respect to itself. This is so simple that it s almost laughable. A point in a circuit is always shorted out to itself. What Kirchhoff really was saying, when he wrote his second law, is a more general version of the second and third points previously mentioned. He reasoned that voltage cannot appear out of nowhere, nor can it vanish. All the potential differences must balance out in any circuit, no matter how complicated and no matter how many branches there are. This is Kirchhoff s Second Law. An alternative name might be the law of conservation of voltage. Consider the rule you ve already learned about series circuits: The voltages across all the individual resistors add up to the supply voltage. Yes, they do, but the polarities of the EMFs across the resistors are opposite to that of the battery. This is shown in Fig. 56. It s a subtle thing. But it becomes clear when a series circuit is drawn with all the components, including the battery or other EMF source, in line with each other, as in Fig. 56. Problem 513 Refer to the diagram of Fig. 56. Suppose the four resistors have values of 50, 60, 70 and 80 , and that the current through them is 500 mA. What is the supply voltage, E Find the voltages E1, E2, E3, and E4 across each of the resistors. This is done via Ohm s Law. In the case of E1, say with the 50 resistor, calculate E1 0. 500 50 25 V. In the same way, you can calculate E2 30 V, E3 35 V, and E4 40 V. The supply voltage is the sum El E2 E3 E4 25 30 35 40 V 130 V. Kirchhoff s Second Law tells us that the polarities of the voltages across the resistors are in the opposite direction from that of the supply in the above example.

